rs preparation course

PREPARATION COURSE

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asManagement Consulting

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Prepared By.·Reda A. Sabry CPD,PMP,MBA,RMPM.Sc. of Construction Management

Certified Projects DirectorOCTOBER 201 1

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- Chapter 1: Project Life Cycle & Frame Work.

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- Chaoter8:

Index

Cash Flow (In & Out).

Progress & Follow Up.

PERT & Precedence Diagram. (& CPM Exercises)

Cost & Resource Management.

Critical Path Method (AOA & AON Diagrams).

Project Planning.

Communication & Documentation Management

Procurement Management.

Quality Management.

Project Risk Management & SWOT Analysis.

Human Resource & Organization Breakdown Structure (OBS).

Scope Management & Work Breakdown Structure (WBS).

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Index

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Text Box

Feasibility Study

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lPMA Preparation Course

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,,~---------------Chapter (1): Framework

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Project Life Cycle & Framework

Table of Content:

1. Introduction

I-I What is a Project?

1-2 What is Project Management?

1-3 Relationship to other Managemen: Disciplines

1-4 Related Endeavors

2. The Project Management Context

2-1 Project Phases and the Project Life Cycle

2-2 Project Stakeholders

2-3 Organizational Influences

2-4 Key General Management Skills

2-5 Socioeconomic Influences

3. Project Management Processes

3-1 Project Processes

3-2 Process Groups

3-3 Process Interactions

3-4 Customizing Process Interactions

Triple Constraint:

Balanced Project

TimeScope

Quality

Cost

Chapter (1): Framework IPMA Preparation Course


What is a Project

A temporarv endeavor undertaken to create a unique product or serviceTemporary

• Has a definite Beginning and End, not an on-going effort.• Ceases when objectives have been attained. Team disbanded upon completion.

- Unique:• The product or service is different in some way from other products or services.

Progressively Elaborated:• Progressively means "proceeding in steps; continuing steadily by increments".• Elaborated means "worked out with care and detail; developed thoroughly".

Projects Versus Operations

Similarities between Projects and Operations• Both perfom1ed by people (even 1 person).• Both are constrained by resources.• Both are planned, executed and controlled.• Both are done for a purpose and have interrelated activities

Differences between Projects and Operations• Operations are on going and repetitive.• Projects are temporary endeavor undertaken to create unique products or services that are

progressively elaborated.

What is Project Management?

The application of knowledge, skills, tools, and techniques to project activities in order to meet or exceedstakeholder needs and expectations.

• Project Management• Program Management• Stakeholder Management

ApplicationArea Knowledge

and Practice

GeneralManagement

Knowledgeand Practice

Relationship of Project Management to Other Management DisciplinesThe Project

ManagementBody of Knowledge

Ttiis figure is a conceptual vie'\\! of t!lese reiationsllips.Ttle overlaps. shown are not proportional. L,-------------------------------,



Related Management Disciplines

- General Management: Planning, organizing, staffing, executing, and controlling the operations ofan ongoing enterprise

- Application Areas: Common elements that are significant in some Projects but not present in otherprojects

• Technical• Management• Industry groups

Related Endeavors

Program: Consists of a group of projects supporting broad and general goals mar-aged In acoordinated way

- Sub-projects: Components of a project that are often contracted out

Life Cycle

- There are different Life Cycles• Project Life Cycle• Process Groups Life Cycle.• Product Life Cycle.

Project Life Cycle

- Project Life Cycle. A collection of generally sequential project phases whose name and numberare detennined by the control needs of the organization or organizations involved in the project.

- Project Phase. A collection of logically related project activities, usually culminating in thecompletion of a major deliverable

Cost andStaffing

Level

InitialPl'lase

Statt

Chapter (1): Framework

Intermediate PI-lases(one or more)

FinalPhase

Fi nish

IPMA Preparation Course


Characteristics of Project Life Cycle• Cost and Staffing level is lower at the start, higher towards the end.• Probability of successful completion is lower at the start, higher toward the end.• Risk and uncertainty is higher at the start, lower towards the end.• The ability of stakeholder to influence the project product is higher at the start, lower

towards the end.Key Concepts

Project Phase: "A collection of logically related project activities usually culminating in thecompletion of a major deliverable".

- Project Life Cycle: "Collectively the project phases are known as the project Life Cycle." Itserves to define the Beginning and the End of a project and it may include several Product LifeCycles.

- Product Life Cycle: The natural grouping of ideas, decisions, and ac:ions into Product Phases,from product conception to operations to Product Phase-out. It serves t'J define the Beginning andthe End of a tangible and verifiable deliverable.

Project Management Life Cycle

- This Life Cycle describes what you need to do to manage the project.o Initiatingo Planning• Executingo Controllingo Closing

Stakeholders

- A stakeholder is someone whose interests may be positively or negatively impacted by the project.They may also include those who may exert influence over the project but would not otherwise beconsidered stakeholders. Examples of Stakeholders are:

o Project Manager: The individual responsible for managing the project.o Customer: The individual or organizational that will use the product of the projecto Perfonning organization, government agencies.o Sponsor: The individual/group who provides the financial resources for the project.o Team: Those who will be completing work tasks on the project.o Internal/External: Stakeholders can come from inside or outside the organization.o End users - The individuals or organizations that will use the product of the project when

it is completed.o Society, citizens - Often members of society can be stakeholders - e.g., instances when a

new road is being build through a community or a new emergency phone number (such asa 911 number) is being set up.

o Others - Owners, fenders, sellers.Sponsor Versus Customer

Similarities between a Sponsor and a CustomerThe customer, although not the same as the Sponsor, Both fulfill the same role

• Both fonnally accept the product of the project.• Both may provide major key dates and milestones.o Both risk threshold should be taken into account.• Both don't sign the project charter, that is the role of senior management I____________________________________________'L~

Chapter (1): Framewark IPMA Preparation Caurse


Forms of Organization

There are different types of organizational structure, based upon project manager's level ofauthority.

- Organizational structure can be characterized as scanning a spectrum from functional toprojectized as follows:

• Functional Organization• Weak Matrix Organization• Balanced Matrix Organization• Strong Matrix Organization• Projectized Organization

Functional Organization

This is the most common form of organization. The organization is grouped by areas of specializationwithin different functional areas (e.g., accounting, marketing and manufacturing).

Chief ProjectExecutive Coordll1ati011

_------------------------l--_/ ,I Functional Functional Functional \I Manager Manager Manager J~ I, --------- --------- -------~

(Blacl, boxes represent staff engaged in project acth/ities.)

Advantages of Functional Organizations- Easier Management of specialists- Team reporting to I supervisor

Similar resources are centralized and grouped by specialty.- Clear career path in areas of specialty- Job security is guaranteed.

Chapter (1): Framework JPMA Preparation Course


Disadvantages of Functional Organizations- More focus on the specialty than over the concern for the project success.- No career path in PM.

PM has no authority.

Projectized Organization

All organization is run by projects. The Project Manager has total control of projects. Personnel areassigned and report to a Project Manager.

ProjectCoordination

,..-- --~---I

ProjectManager

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ChiefExecutive

ProjectManager

ProjectManager

\. ,/....... _-------

(Black boxes represent staff engaged in project activities.)

Advantages & Disadvantages of Projectized Organizations

- Advantages:• Project Manager has the power and full authority over the project's resources.• Communications are easier and faster.

- Disadvantages:• No career path in areas of specialty.• No job security.

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Matrix Organization

- This form is an attempt to maximize the strengths and weaknesses of both the functional andproject forms.

- In a strong matrix, power rests with the Project Manager. In a weak matrix, power rests with thefunctional manager. In a balanced matrix, the power is shared between the functional manager andthe Project Manager.

Organization Structure versus Project Characteristics

~roje\;t Manager's Little or None Limited Low to Moderate High toAuthority Moderate to High AlmostTotal

Percent of Performing Virtually 0-25% 15-60'-" 50-95% 85-10Jk,Organization's NonePersonnel AssignedFull Time to Proje\;t Work

Proje\;t Manager's Role Part-time Part-time FUll-time Fuil-time FUll-time

Common Tttles for Project Proje::t Project Project ProjectProje\;t Manager's Role Coordinatorl Coordinatorl Mana~! Managerl Manager!

Project Leader Project Lead er project Officer Pro~am Manager Program Manager

Proje\;t Management Part-time Part-time Part-time Full-time Full-timeAdministrative Staff

'Neak MatI ix 0, ganizationChapter (1): Framework IPMA Preparation Course

ffiS Management Conswlting Elcuse

ChiefExecutive

FunctionalManager

FunctionalManager

FunctionalManager

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{Blacli b*xes re1:res*nt stafr engageci in proje c.t actii,ities. j

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Balanced Matrix

In this fonn of organization, the project expeditor acts primarily as a staff assistant and communicationscoordinator. The expeditor cannot personally make or enforce decisions.

ChiefExecutive

FunctionalManager

FunctionalManager

FunctionalManager

(Black boxes represent staff engaged In project activities.)

~ --------- --------- --------,/ ,I I, /'--------------------------£---

ProjectCoordination

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This form of organization is similar to the Balanced Matrix Organization, except that the coordinator hassome power to make decisions, some authority, and reports to a higher-level manager

ChiefExecutive

Manager ofProject Managers

Project Manager

Project Manager

FunctionalManager

FunctionalManager

(Black boxes represent staff engaged in project activities.)

, -------- -------- -------- ----------,r. . . _ \, I--------------------------------~-----

ProjectCoordination

Advantages & Disadvantages of Matrix Organizations

- Advantages:• Both Project Manager and functional managers share the responsibility of the project's

resources.Disadvantages:

• There is more than one boss for the project team.• Communications are complex as the team members should make dual reporting.

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Project Office: There is a range of uses for what constitutes a project office.A project office may operate on a continuum from providing support functions to projectmanagers in the form of training, software, templates, etc. to actually being responsible for theresults of the project Samples:

• Providing the policies, methodologies and templates for managing projects within theorganization.

• Providing support and guidance to others in the organization on how to manage projects,training others in project management or project management software and assisting withspecific project management tools.

• Providing Project Managers for different projects and being responsible for the results ofthe projects. All projects (or projects of a certain size, type or influence) are managed bythis office.

SOCIOECONOMIC INFLUENCES

- Standards and Regulations:

• A STANDARD is a " document approved by a recognized body, that provides, forcommon and repeated use, rules, guidelines, or characteristics for products, processes orservices with which compliance is not mandatory. " There are numerous standards in usecovering everything from thermal stability of hydraulic fluids to the size of computerdiskettes" .

• A REGULATION is a "document which lays down product, process or servicecharacteristics, including the applicable administrative provisions, with which complianceis mandatory." Building codes are an example of regulations.

- Internationalization

As more and more organizations engage in work which spans national boundaries, moreand more projects span national boundaries as well. In addition to the traditional concernsof scope, cost, time, and quality, the project management team must also consider theeffect of time zone difference, national and regional holidays, travel requirements for faceto-face meetings, the logistics of teleconferencing, and often volatile political differences.

- Cultural Influences

• Culture is the "totality of socially transmitted behavior patterns, arts, beliefs, institutions,and all other products of human work and thought".

• Every project must operate within a context of one or more cultural norms. This area ofinfluence includes political, economic, demographic, educational, ethical, religious, andother areas of practice, belief, and attitudes that affect the way people and organizationsinteract.

Project Management Processes

l'rojeel l'1 ocessesChapter (1): Framework IPMA Preparation Course


- Process Groups

Process Interactions f

InitiatingProcesses

ControllingProcesses

PlanningProcesses

ClosingProcesses

ExecutingProcesses

Overlapping the Project Management Processes

ExeclltingProcesses

Levelof

Activity InitiatingProcesses

PhaseStart

Cantrall ing Processes

Time

ClosingProcesses

PhaseFinish

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i-------------------------------,,Chapter (1): Framework lPMA Preparation Course


Project Management Knowledge Areas

Project Integration Management

Project Scope Management

Project Time Management

Project Cost Management

- Project Quality Management

Human Resource Management

Communication Management

Project Risk Management

Project Procurement Management

PROJEctMANAGEMENT

4.1 !'roject F1an D"",lcpment4.2 !'roject F1an Execution4.3 Integral@d Cllan"",, Control

5.1 Initiation5.2 SoDP~ F1anning5.3 SX)P~ Dennilion5.4 SX>pe Verincation5.5 SX>pe Change Control

6.1 Acfrvirl D=!finition6.2 Act""~ SequEncing6.3 ActM~ Duralion Estimating6.4 Schedule DevelcpmEl1l6.5 Schedule Control

7.1 ReSOUf1:8 Planning7.2 Cost E.-<timating7.3 Cost Budgeting7.4 Cost Control

10.1 Communicalions Planning10.2 Infomlatiol1 Distribution10.3 pg~onnance ","porting10.4 Mministrati1/fl Closure

Chapter (1): Framework

8.1 Quality F1anning8.2 Quality Assuran",8.3 Qual",; Control

11.1 Risl< Mana~ment Planning11.2 RlsI~ Identifi""tion11.3 Qualitati,'e Risk Analysis11.4 Quantitative Risk ~naIJ'Sis

11.5 Risl~ Re'Ponse F1anning11.6 Risl< r',(onitoring and Control

9.1 Olganirational F1anning9.2 Staff Acquisiticn9.3 Team Development

12. Project PfllCU'ementMan~nt

12.1 Procurement Planning12.2 Solicitatton F1anning12.3 Solicitation12,4 Source S9lecbon12.5 Contract Admin istJation12.6 Contract Closeout


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Chapter (2): Project Scope Mnagement IPMA Preparation Course

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Project Scope Management

Scope Management:

Initiation Planning Controlling

1. 2. 4.

Initiation Scope Scope

Planning Verification

3. 5.Scope ScopeDefinition Change

Control

• Processes nsed to identify all the work required and only the work required to successfullycompletiug the project:

);> Initiation);> Scope Planning);> Scope Definition);> Scope Verification);> Scope Change Control

• Constantly checking to make sure you are completing all the work.• Saying no to additional work not included in the project or not part of the project charter.• Preventing extra work or gold plating.

You should give the customer what they asked for, no more and no less. Giving any extras is a waste of timeand adds no benefit to the project. Make sure you understand this approach and why it is a good idea.

Scope:

• Product scope - The features and functions that are to be included in a product or service• Project scope - The work that must be done in order to deliver a product with the specified features and

functions• Successful completion of product scope is measured against the requirements; proiect scope is measured

against the plan

Chapter (2): Praject Scape Management IPMA Preparation Course

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IHot Topics:

I, Work breakdown structure (WBS)2, Initiation3. Scope verification4. Scope management plan5. Project charter6. Scope definition and decomposition7. Project selection methods8. Constraints9. Scope statement10. Definition of scope managementII. WBS dictionary12. Delphi technique13. Management by objective, (MBO)

Initiation:

• Formal authorization that the project exists• Recognition the project should continue into the next phase• The major steps most likely required in the project initiation process are as follows:

- Identifying Project Alternatives- Selecting the Project- Identifying the Project Sponsor- Appointing the Project Manager

Inputs:

• Product description - Documents characteristics ofthe product or service that the project wasundertaken to create and its relationship to the business need or other stimulus that gave rise to theproject

• Strategic plan - Describes the organization's mission, vision, and goals for the future, which theproject supports

• Project selection criteria - Defined in terms of the product and covers the full range of managementconcerns

• Historical information - Results of previous project decisions and performance

Tools & Techniques:

• Project Selection MethodsThere should be a formal process for selecting projects in all companies in order to make the best use oflimited corporate resources. Without such formal methods, projects are often selected not for their value,but for less quantitative reasons such as personal relationships.

-----------------------------------[Chapter (2): Project Scope Management IPMA Preparation Course


1. Benefits measurement methods (comparative approach)• Murder board - a panel of people who try to shoot down a new project idea• Peer review• Scoring models• Economic models• Benefits compared to cost

2. Constrained optimization methods (mathematical approach)• Linear programming• Integer programming• Dynamic programming• Multi-objectives programming• Expert judgment - Experts with specialized knowledge or training assess the inputs to this process

Outputs:

• Project charter• Project manager selected• Constraints - Factors that limit the project management team's options regarding scope, staffing,

and schedule• Assumptions - Factors that, for planning purposes, will be considered to be true, real, or certain

Project Charter:

• Formally recognizes the existence of a project• Refers to the business need the project is addressing• Describes the product to be delivered• Gives the project manager the authority to apply resources to the project• A project charter provides, at minimum, the following benefits:

-Gives the project manager authority. On the exam, this is the most commonly described benefit or use ofthe project charter. In most project situations, the project team does not report to the project manager in thecorporate structure. This leads to issues of "how to gain cooperation and performance."-Formally recognizes (authorizes) the existence of the project, or establishes the project. This means that theproject does not exist without a project charter.-Provides the goals and objectives of the project. Most project managers are not provided with the basicinfonnation (what is in the project charter) to complete the project. The information provided in the projectcharter is considered vital to the project. The information provided in the project charter is considered vitalto the success of the project. Without the project charter information, it is like being told to get into a car anddrive without being told where to go.

• The project charter is also:-Created by a manager who is external and higher in the corporate hierarchy, and not by the project manageror the team.-Created during initiation.-Broad enough so it does not NEED to change as the project changes.-An output of scope initiation.

Chapter (2): Project Scape Management IPMA Preparation Course

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Sample Project Charter:

Project Title: InforITlation Technology (IT) Upgrade ProjectProject Start Date: March 4, 2002 Projected Finish Date:DeceITlber 4, 2002Project Manager: KjITl Nguyen, 691-2784, [email protected] Objectives: Upgrade hardvvare and softvvare for all eITlployees(approxiITlately 2,000) "Within 9 months based on nevv corporate standards.See attached sheet describing the nevv standards. Upgrades ITlay affectservers and ITlidrange cOITlputers as vvell as netvvork hardvvare andsoftvvare. Budgeted $1,000,000 for hardvvare and softvvare costs and$500,000 for labor costs.Approach:• Update the IT inventory database to deter=ine upgrade needs• Develop detailed cost estimate for project and report to CIa• Issue a request for quotes to obtain hardvvare and softvvare

Use internal staff as ITluch as possible to do the planning, analysis, andinstallation

Scope Planning:

Process of progressively elaborating and documenting the project work that produces the product of theproject

Inputs:

• Product description - Contains the characteristics of the product or service in which the project willresult

• Project charter - Formally recognizes the existence of a project• Constraints - Factors that limit the team's options such as resources, budget, schedule and scope.

Constraints are identified and managed and input to many aspects of project management.Management identifies some constraints, and the stakeholders and the team identify others.

• Assumptions - Factors that, for planning purposes, will be considered to be true, real, or certain

Tools & Techniques:

• Product analysis - Techniques to develop a better understanding of the product (e.g., systemsengineering, value engineering, function analysis, quality function deployment)

• Benefit/cost analysis - Estimating the tangible and intangible costs (or outlays) and the benefits (orreturns) of various project alternatives

• Alternative identification - Techniques used to generate different approaches to the project (e.g.,brainstorming and lateral thinking)

• Expert judgment

--------------------------------iLChapter (2): Project Scope Management IPMA Preparation Caurse

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Outputs:

• Scope statement - Written statement of project. It contains:>- Project objectives>- Project justification>- Project deliverables

• Provides a basis for a common understanding of the scope among the stakeholders and to determine ifthe project phase has been completed. Loosely the same as a scope of work or statement of work, thescope statement is an output of scope planning.

• Supporting detail- Supporting documentation containing identified requirements, plans, assumptions,and constraints

Scope Management Plan:

Guidelines for how scope is to be managed and how scope changes are to be integrated into the project

• It includes:• An assessment of the stability of the project scope• A clear description of how scope changes will be identified and classified

Delphi Technique:

-A method most commonly used to obtain expert opinions on technical issues, scope of work needed,estimates or risks.-To use the Delphi technique, a request for infonmation is sent to experts, the responses they return arecomplied, and the results are sent back to them for further review. The Delphi technique has three rules:

1. Do not get the experts in the same room.2. Keep the experts identifies anonymous.3. Try to build conseusus.

Scope Definition:

• Subdividing major project deliverables into manageable components, in order to improve theaccuracy of cost, time, and resources estimates

• Provides a baseline and assigns responsibility• A scope baseline is the original plan, plus or minus approved changes

Inputs:

• Scope statement• Constraints• Assumptions• Other planning outputs - Outputs of the processes in other knowledge areas should be reviewed for

possible impact on project scope definition• Historical infonmation - About previous projects

Chapter (2): Project Scope Management IPMA Preparation Course

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Tools & Techniques:

• WBS templates - Use a WBS from a previous project or a standard template to develop a WBS forthis project

• Decomposition - Subdividing major project deliverables into smaller, more manageable componentsuntil deliverables are defined in sufficient detail for supporting future project activities-Identify major elements of the project-For each element, decide if adequate cost and duration estimates can be developed at this level ofdetail-Identify constituent elements (e.g., tangible, verifiable results)-Verify the correctness of the decomposition - is it necessary and is it sufficient for completion ofthe item decomposed

Scope Definition & Decomposition:

• Some students confuse these terms with the WBS. The guide defines scope definition and decompositionas "subdividing the major project deliverables into smaller, more manageable components. " The bestway to handle these terms is to think of scope definition and decomposition as what you are doing, andthe WBS as the tool to do it.

WBS Definition:

Deliverable oriented grouping of project elements that organizes and defines the total scope of the project

Hardware

-Code of accounts - Uniquely identifies each element of the WBS·Work packages - A deliverable at the lowest level of the WBS·WBS dictionary - Created with team members' assistance, a WBS Dictionary is designed to control whatwork is done and when, to prevent scope creep and to increase understanding of the effort for each task.Sometimes entries are also called task descriptions. They help the project putting boundaries on what isincluded in the task (or work package).

Chapter (2): Project Scope Management IPMA Preparation Course


• The creation of a WBS has only a few "rules":

-It is created with the help of the team.-The first level is completed before the project is broken down further.-Each level of the WBS is a smaller segment of the level above.-The entire project is included in each of the highest levels. However, eventually some levels will

be broken down further than others.-Work toward the project deliverables.-Work not in the WBS is not part of the project.-A WBS can become a template for future projects.

-Break down the project into work packages or activities that:

• Can ~e realistically and confidently.• Can,lOt be logically subdivided further.• Can be completed quickly.• Have a meaningful conclusion and deliverable.• Can be completed without interruption (without the need for more information).• The top level of the WBS is project title. The first level is most commonly the same as the project

life cycle (for a software project-requirement analysis, design, coding, testing, conversion andoperation). The second and later levels break the project into smaller pieces. The lowest level thatthe project manager will manage the project to call the work package or activity. This work maybe broken down again by the person or persons completing the work.

InformlltionSystem

1.

[]Project Systems Hardware Software Facilities Training

Management Engineering Acquisition Development Modifications Development1.1 1.2 1.3 1.4 1.5 1.6

Project Product CPU Operating Facility Training

I- Planning Design Acquisition System Plans Plans1.1.1 1.2.1 1.3.1 1.4.1 1.5.1 1.6.1

Project Systems Auxiliary Database Facility TrainingI- Contro! Integration Equipment 1.4.2 I- Modification "- Courses

1.1.2 1.2.2 1.3.2 1.5.2 1.6.2

Project Test & Printer Application Facility.... Data - Evaluation Acquisition - Development - Installation

1.1.3 1.2.3 1.3.3 1.4.3 1.5.3


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r-------------------------------1Outputs:

• Work breakdown structure• Scope Element Updates

Scope Verification:

• "The process of formalizing acceptance of the project scope by the stakeholders." A description ofScope verification might include any of the following phrases:

• Reviewing work products and results to ensure that all completed correctly and satisfactorily.• Conducting inspections, reviews and audits.• Determining whether results confonn to requirements.• Determining if work products are completed correctly.• Documenting completion of deliverables.• Gaining formal sign-off.

Inputs:

• Work results - Information concerning which deliverables have been fully or partially completed• Product documentation - Documents describing the project's products (e.g., plans, specifications,

technical documentation, drawings)

• WBS• Scope statement• Project plan

Tools & Techniques:

-Inspection - Includes activities, such as measuring, examining, and testing, undertaken to determinewhether results conform to requirements-Also referred to as reviews, product reviews, audits, and walk-throughs

Outputs:

Formal acceptance - Documentation of the client or sponsor's acceptance of the product

Scope Change Control:

• A process for controlling changes to project scope by:-Influencing factors, which create scope changes, in order to ensure the changes are beneficial-Determining that a scope change has occurred-Managing the changes when they occurThe system includes the paperwork, tracking systems, and approval levels necessary for authorizing changes

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Chapter (2): Praject Scape Management IPMA Preparation Course


Change Control Process:

• Steps for changing official project documents• Change Control Board (CCB) - Responsible for approving or rejecting change requests• Procedures for handling those changes that may be approved without prior review• The scope change control process is a subsidiary process of the overall project control process

Inputs:

• WBS• Performance reports - Provide infonnation on scope perfonnance, such as: which interim

products have been completed and which have not• Change requests• Scope management plan

Tools & Techniques:

• Scope change control system - Defines procedures for changing project scope• Performance measurement - Techniques for assessing the magnitude of any variation and

determining the cause of the variance and whether the cause requires corrective action• Additional planning - Scope changes may require changes to the WBS or analysis of alternative

approaches

Outputs:

• Scope changes Any modification to the agreed upon project scope, as defined by the approvedWBS

• Corrective action - Any action taken to bring future project perfOlmance back in-line with theproject plan

• Lessons learned - Documentation of the reasons behind the corrective action and other types ofinformation; to be included in the historical database for this project and for other projects of theperforming organization

• Adjusted baseline - the corresponding baseline document may be revised and reissued to reflectthe approved change and form the new baseline for future changes

Summary:

• The importance of adequate planning and control of the scope cannot be enough emphasized.Revising other project management process shows that the outputs of the scope managementprocess, especially at least on or both of the scope statement and the WBS, are key inputs to otherprocess,

• For example, both are inputs to activity definition process, whish is the leading process towarddeveloping the project schedule and the WBS is an input for resource planning and costestimating process, which are leading processes towards development of the project budget,

• The Scope statement is very important in producing the quality management plan and theprocurement management plan as well as for developing the statements of works (SOW) which isthe base for tendering and contracting activities.


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Chapter (3): Human Resource Management IPMA Preparation Course

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Human Resource Management

Processes required making the most effective use ofthe people involved with the project

Organizational Planning

Planning

StaffAcquisition

lETean1 Development

xe\C1Ul~mg

Organizational Planning:

Planning, identifying, documenting, and assigning project roles, responsibilities, and reporting relationships

• Individuals and groups may be part of the organization performing the project or may be external toit.

• Internal groups are often associated with a specific functional department, such as engineering,marketing, or accounting.

Inputs:• Project interfaces

Organizational interfaces - Formal and informal reporting relationships among differentorganizational unitsTechnical interfaces - Formal and informal reporting relationships among technicaldisciplinesInterpersonal interfaces - Formal and informal reporting relationships among differentindividuals working on the project

• Staffing requirements: - Definition of the kinds of competencies required and in what time frame andfrom which kinds of individuals.

• Constraints: - Factors that limit project team's options.

Organizational structure of the performing organization.• Strong matrix• Weak matrix

Collective bargaining agreements with unions or other employee groups.

Preferences of the project management team.

Skills and capabilities ofthe staff.


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Tools & Techniques:

• Templates: - oRole and responsibility definitions or reporting relationships from a similar project.

• HR Practices: - Variety of policies, guidelines, and procedures used as organizational- planning aid.

• Organizational Theory: - Body ofliterature describing how an organization can, and should, be

Structured.

• Stakeholder Analysis: - Ensures their needs are met.

Organizational Theory:

• Organizational Theory describes how a company can be organized to complete its work.

• PM] talks about five types of organizational structure.

• Each type is described in terms of the project manager's level of the authority.

• Questions on the exam related to organizational theory included:

I. Who has the power in each type of organization - the project manager of the functional

manager?

2. Advantages of each type of organization

3. Disadvantages of each type oforganization.

Functional Organization:

• This is the most common from of organization.

• The organization is grouped by areas of specialization within different functional areas specialization

within different functional areas (e.g., accounting, marketing, and Manufacturing). A functional

organizational chart might look like the figure in the next figure.

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I--------------------------------------------!uChapter (3): Human Resource Management lPMA Preparation Course

RJS Management Consulting House

Projectized Organization:

All organization is by projects. The project manager has total control of projects/ Personnel are assigned andreport to a project manager.

Matrix Organization:

• This form is an attempt to maximize the strengths and weakness of both functional and project forms.

• The team members report to two bosses: the project manager and the functional manager.

• This forms may look like.

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• In a strong matrix, power rests with the project manager. In a balanced matrix, the power is shared

between the functional manager and the project manager.

• A Tight matrix has nothing to do with a matrix organization. It simply refers to locating the offices for

the project team in the same room.


Management Consulting Houser---------------------------------------------1l

Organizational Structure influences on Projects:

Project Managers'sLittle or None Limited

Low to Moderate to High to AlmostAuthority Moderate High Total

Percent of PerformingOrganization'sPerscnnel Assigned Virtually None 0-25% 15-60% 50-95% 85-100%Full Time to ProjectWork

Project Manager'sPart-Time Part-Time Full-Time Full-Time Full-Time

Role

ProjectProject Project Project

Common Titles forCoordinator!

Coordinator/ Manager/ Manager/ Project Manager/Project Manger's Role

Project LeaderProject Project Project Project OfficerLeader Officer Officer

Project ManagementPart-Time Part-Time Part-Time Full-Time Full-Time

Administrative Staff

Outputs:

Roles and Responsibility Assignments: Roles (Who does what) and responsibilities (Who decides what)assigned to the appropriate stakeholder.

• Staffing Management Plan: - Describes when and how human resources will be brought onto, and

taken off the project team.

• Organization Chart: - Graphic display of project reporting relationships. OBS indicates which

organizational unit is responsible for which work items.

• Supporting Detail: -

Organizational impact: What alternatives are precluded by organizing in this manner?

Job descriptions: Written outline of characteristics involved in perfonning a given job.

Training needs: To develop skills needed to perfonn job.

I----------------------------------..;cChapter (3): Human Resource Management IPMA Preparation Course

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Resource Histogram:

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Reward Systems:

• Extrinsic (External)SalaryWork ConditionsStatus

• Intrinsic (Internal)AchievementResponsibilityAdvancement

• Rewards given timely and in public


Management Consulting Houser-------------------------------,

Staff Acquisition:

Getting the needed human resources assigned and working on the project.

Inputs

• Staffing Management Plan: -

• Staffing-pool description: - Characteristics of the potentially available staff

Previous available staff

Personal experience

Personal interests

Personal characteristics

Availability

Competencies and proficiency

• Requirement practices: - Policies, guidelines, and procedures governing staff assignments.

Tools & Techniques:

• Negotiations: -With FMs or other project management teams; to ensure receiving appropriate resources

within required time frames.

• Pre-assignment:- Variety of policies, guidelines, and procedures used as organizational- planning aid.

• Procurement- Body of literature describing how an organization can, and should be structured.

Outputs

• Project staff assigned: - Appropriate people have been assigned to work on the project full-time, part

time, or variably.

• Project team directory: - lists ofall project team members and other key stakeholders.

Team Development:Enhancing the ability of stakeholders to contribute as individuals as well as enhancing the ability of the teamto function as a team.

Inputs

• Project staff

• Project plan

• Staffing management plan

• Performance reports

• External feedback: Team measure itself against the expectations of those outside the project.

("

Ii-------------------------------!,..,



Tools & Techniques• Team-building activities: - Include management and individual actions taken to improve team

performance.

• General Management Skills

• Reward and recognition systems: - Formal management actions that promote or reinforce desired

behavior.

• Collaction: - Placing all or most of the team members in the same physical location.

• Training: - Includes all activities designed to enhance the skills, knowledge, and capabilities to the

project team.

Outputs• Performance improvements

Improvements in individual skills.

Improvements in team behaviors

Improvements in either individual skills or team capabilities

• Inputs to performance appraisals

Management Styles:

Autocratic

• StrengthMature, well defined projects

Quick decisions required

• Weakness

Limits staff buy-in leading to low moral.

Possible arbitrary decisions.

Laissez-Fair

• Strength

Innovative projects

High morale of self-motivated staff

• Weakness

Confusion about objectives of the project

Inability to make decisions


RS Management Consulting Hou~e________________________________1

Maslow's Hierarchy ofNeeds:

5. SelfActual ization

4. Esteem

3. Sodal

2. Safety

1. Physiological

t Self-Actualization - Need to grow and use abilities to the fullest

and most creative extent

Esteem - Need for respect, prestige, recognition, sense of

competence

Social- Need for love, affection, and sense of belonging

Safety - Need for security, protection, and stability

Physiological- Need for biological maintenance (food, water etc.)

("--------------------------------------------c..


RS, Management Consulting House

McGregor's Theory:• TheoryX

Traditional view of management; top-down

Managers: Control the people

Workers: Viewed as inherently self-centered, lazy

• Theory Y

Workers: Viewed as willing and eager to accept responsibility

Managers: Create environment that aids workers in achieving goals.

• Theory Z

Introduced in 1981 and is based on the Japanese approach to motivating workers, emphasizing trust,

quality, collective decision making and cultural values.

Sources of Authority and Control: (POWER)Formal:

• A legitimate form of power.

• Based on a person's formal position in the company.

Reward:

• Refers to positive consequences or outcomes that a person can offer.

Coercive (Penalty):

• Refers to negative consequences that a person can inflict on others. (Firing, docking,

reprimand, etc.)

Referent:• Refers to earned power when people admire a person and want to follow that person

as a role model. Also called charisma.

Expert:- Refers to earned power that a person acquires based on his/her technical knowledge, skill,

or expertise on some topic or issue.

Conflict Management:The PM must carefully select the method of managing conflict appropriate for his/her organization so that an

atmosphere conductive to constructive results is developed.

Five methods of managing conflict:

Problem Solving (Confrontation)

Compromising

Smoothing

Withdrawal

Forcing


iRS Management Consulting Houser"---------------------------------------------t

Managing People on the Project:

Select techniques that are appropriate for personal and organizational relationships that are temporaryand newThe nature and number of project stakeholders will often change as the project moves from phase tophase. Choose techniques that are appropriate to the current needs ofthe projectTeam must be sufficiently aware of HR administrative requirements to ensure compliance

Inputs of(Organizational Planning):

Project interfaces:

Organizational interfaces - Formal and informal reporting relationships among differentorganizational units.Technical interfaces - FOlmal and informal reporting relationships among technical disciplines.Interpersonal interfaces - Formal and informal reporting relationships among differentindividuals working on the project.

Project Coordinator:

The coordinator has some power to make decisions, some authority, and reports to higher-levelmanager. This form may look like this

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Hertzberg's Motivation Theory:

.. HygienePay; working conditions; boss's attitude

Poor hygiene factors negatively impact motivation

Good hygiene factors increase motivation

.. MotivatorsPositive motivation leads to achievement and self-actualization

Workers have a sense of personal growth and responsibility

-------------------------------1[;Chapter (3): Human Resource Management IPMA Preparation Course

RSS Management Consulting House

Chapter (4): Communication Management IPMA Preparation Course


Communications Management

Processes required to ensure timely and appropriate development, collection,dissemination, storage, and, ultimately, disposition ofproject information.

Communications Planning

Information Distribution

Performance Reporting

Administrative Closure

Planning Executing Controlling Closing


!IRS Management Consulting HOUS~

l

Communications Planning:

Determining the information and communications needs of the stakeholders: who needs what information,when they will need it, and how it will be given to them

Inputs:

• Communications requirements - The sum ofthe information requirements of the project stakeholders• Combining - "Type and Format" with: "Analysis of the Value"• Project Organization• Responsibility relationships• Disciplines, departments, specialties• Logistics (number of, where)• How many individuals and locations• External needs - Media

Requirement

• Communication technology - Used to transfer information back and forth among project elements

~-I~

Availability ofTechnology

Immediacy of the need

• Constraints

Limit Option:• Procured Resources - Contracts• Technologies - copiers, fax, telephones, E-mail• Organizational preferences

• Assumptions

Involve some risk:• Resources are non contract• All team members speak English• Have access to required technologies• Understand organizational preferences

Chapter (4): Communication Management

Expected Project Staffing

~R~{Length of the Project


R;;, Management Consulting House

Tools & Techniques:

• Stakeholder analysis - A method for developing a systematic and logical view of the information needsof the stakeholders and of the sources for meeting those needs

Outputs:

Communication management plan - provides:

• Collection and filing structnre - Methods used to gather, update, and store various types ofinformation

• Distribution structure - Specifies to whom information will flow and what method will be usedto distribute various types of infonnation.

• Descriptiou of informatiou to be distributed - Includes format, content, level of detail, andconventions and definitions to be used

• Production schedules - Show each type of communication• Methods for accessing information• Method of updating and refining the communication management plan as the project

progresses

Information Distribution:

Making needed information available to project stakeholders in a timely manner

• Includes implementing the communications management plan, as well as responding to unexpectedrequests for information

The Impact of the Number of People on Communications Channels:

0"'%'" .

0 0

o0 0

4 people, 6 communications channels, etc.

number of communications channels = n(n-1)2

..0 0

2 people, 1 communications channel

o0 0

o

3 people, 3 communications channels


Input:

• Work results• What has been done• What has not been done• level of quality• What you've paid• What you're going to pay

• Communication management plan• Project plan

Tools & Techniques:

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Management Consulting Houser

• Communications skills - Skills for exchanging information• Written, oral, listening, and speaking• Internal and external communication• Formal reports, briefings and informal memos, ad hoc conversations• Vertically, up the organization, and horizontally, with peers

• Information-retrieval systems - Manual filing systems, electronic-text databases, projectmanagement software, and systems which allow access to such technical documentation asengineering drawings

• Information-distribution systems - Methods such as project meetings, hard-copy documentdistribution, shared access to networked electronic databases, fax, electronic mail, voice mail, andvideo conferencing

Output:

• Project Records - Organized storage and maintenance of correspondence, memos, reports, anddocuments describing the project

• Project reports - Fonnal project reports on project status and/or issues• Project presentations - Provide infonnation formal or informally to any or all of the project

stakeholders

Performance Reporting:

Collecting and disseminating pelformance information to provide stakeholders with information about howresources are being used to achieve project objectives

• This includes status reporting (describing where the project now stand) , progress reporting(describing what the project team has accomplished) , and forecasting reporting (predicting futureproject status and progress)

• Provides information on scope, schedule, cost, and quality, and possibly on risk and procurement

Input:

• Project plan - Contains the various baselines used to assess project performance• Work results - Accurate infonnation on project status, such as information about fully, or partially,

completed tasks and costs incurred or committed• Other project records - Any infonnation pertaining to the project context

LJChapter (4): Communication Management IPMA Preparation Course

ftS, Management Consulting House

Tools & Techniques:

• Performance reviews - Meetings held to assess project status or progress• Variance analysis - Comparing actual project results to planned or expected results

Trend analysis Examining project results over time to determine ifperformance is improving ordeteriorating

• Earned-value analysis - Integrating scope, cost, and schedule measures to assess projectperformance

Planned

Actual

Earned Value

Planned value =

Actual costs =

Eamed value

BCWS

ACWP

BCWPBCWP

Task A, which I was supposed to complete today, is scheduled to cost $1000.I am only 85% done on this task. So, I have done $850 worth of work, which is my earned value

• Information-distribution tools and techniques

• Performance reports are distributed using the tools and techniques described inInformation Distribution.

• Communications skills• Information retrieval system• Information distribution methods

Output:

• Performance report - Organizes and summarizes the information gathered and presents the results ofany analysis. Reports should provide the kinds of information and the level of detail required by variousstakeholders and documented in the communications management plan

• Change requests - Requests for changes to some aspect of the project

Performance report

80,000,000.00 .-70,000,000.00./Q) 60,000,000.00::s ./'iii 50,000,000.00 ./> 40,000,000.00

.,~> ""30,000,000.00 ....'

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Time >O";'IX.Xo<XActual Cost

Chapter (4): Communication Management lPMA Preparation Course

RS Management Consulting House_________________________________________---;1

l

Administrative Closure:

Documenting project results in order to formalize the acceptance ofthe product by the sponsor, client, orcustomer.

• It includes collecting project reports, ensuring they reflect final specifications and analysis of projectsuccess and effectiveness, and archiving such information for future use

• Each phase ofthe project should be properly closed to ensure that important and useful informationis not lost

Input:

• Performance-measurement documentation - All documentation produced to record and analyzeproject performance, including the planning documents which established the framework forperformance measurement

• Product documentatiou - Documents produced to describe the product• Other project records

Tools & Techniques

• Performance-reporting tools and techniques• Project reports• Project presentations

Output:

• Project archives - Complete set of indexed project records• Project closure - Confirmation that the project has met all customer requirements for the product of

the project• Lessons learned


" RSS Management Consulting House

1-------------------------------------

Chapter (5): Quality Management IPMA Preparation Course

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Quality Management

Processes necessary to ensure that the project will satisfy customerrequirements

Quality PlanningQuality Assurance

Quality Control

Planning Executing Controlling

l. 2. 3.Quality Quality QualityPlanning Assurance Control

• Addresses both the management of the project and of the product of the project

• Improves project management as well as the quality of the product

• Delivery of both high quality and high grade of product

• Quality is planned in, not inspected in

• Investments in product quality improvement must often be borne by the performing

organization


[as Management Consulting House________________________________________1

Philosophies:

• Deming

• Japanese

• Juran

• Crosby

• Theory X

• Theory Y

Leadership, long-term company position; continuous improvement;participatory; zero defects

Similar to Deming

Decrease cost of quality

Decrease cost of quality; authoritarian; zero defects

Workers lazy, motivated by money

Workers good and trustworthy

ISO 9000:

• 1809000

lS09001 - Design/Develop/Produce/lnstall & Service

IS09002 - Produce & Install

IS09003 -Inspection & Testing

• Cost

Prevention costs: Training, surveys, implementation of quality system

Failure costs: Rework, rejects, scrap

External Costs: Warranty, recall, handling complaints

Note - 85 % of costs are responsibility of management

Project Management &Quality Management:

Customer satisfaction - Understanding, managing, and influencing needs so that

customer expectations are met or exceeded

Prevention over inspection - The cost of avoiding mistakes is much less than the cost of

correcting them

Management responsibility - Success requires participation of all members of the team,

but it is the responsibility of management to provide resources needed

Processes within phases - Plan/Do/Check!Act cycle



Process Improvement:

ProcessImprovement

Check

Quality Planning:

Identifying the quality standards that are relevant to the project and determininghow to meet them

Quality is planned in - NOT inspected in

Inputs:

• Quality policy - "Overall intentions and direction of an organization with regard to quality as

formally expressed by top management" International Organization of Standards

• The quality policy of the organization can be adopted for use by the project

• Stakeholders should be fully aware of it

• Scope statement - Documents the major product deliverables and objectives that serve to define

important stakeholder requirements

• Product description - Contains details of technical issues and of other concerns that may affect

quality planning

Standards and regnlations - Specific standards or regulations that may affect the project

Other process outputs - Processes in other knowledge areas may produce outputs that should be

considered part of quality planning


Management Consulting Housef--------------------------------]

Tools & Techniques:

Benefit/cost analysis - Cost and benefit tradeoffs of meeting quality requirements

Benchmarking - Comparing actual or planned project practices to other projects, in order to

generate ideas for improvement and to provide a standard against which to measure performance

Design of experiments - Analytical technique that helps identify which variables have the most

influence on the overall outcome and helps detennine an optimal solution from a relatively limited

number of cases

• Flow charting

Cause-and-effect diagram - Illustrates how causes and subcauses relate and create

potential problems or effects

Process flow chart - Shows how various elements interrelate

Cause and Effect Diagram:

Factor Factor Factor Factor

\ \ \ EffectPerformance

Factor Factor' Factor Factor

Examplepeople

\policy

organization

methods

information

\equipment

materials

\environment

EfficientMail

Service



Process Flow Chart:

Verify

Outputs:

Quality management plan - Describes how the PM team will implement quality policy.Addresses quality control, quality assurance, and quality improvement for the project.

• Operational definitions (Metrics) - Describes what something is and how it is measured by thequality control process

• Checklist - Used to verify that a set of required steps has been performed• Inputs to other processes - Quality planning may identify a need for further activity in another

area

Quality Assurance:

Planned and systematic activities implemented within the quality system in order toprovide confidence that the project will satisfY the relevant quality standards

• What- Managerial audit function

• How- Prepare and implement an organization QA program- Tailor a practical program to meet requirements

• Why- Quality improvement- Increase effectiveness and efficiency

Inputs

• Quality management plan• Results of quality-control measurements - Records of quality-control testing and measurement,

presented in a format useful for comparison and analysis• Operational definitions


Management Consulting HouserI

Tools & Techniques:

Quality-planning tools and techniquesBenefit/cost analysisBenclunarkingFlowchartingDesign of experiments

Quality audits - Structured review of QM activities in order to identify the lessons learned thatcan improve the performance ofthis project and of other projects within the organization

Outputs

• Quality improvement

Taking action to increase the effectiveness and efficiency of the project and to provideadded benefits to the project stakeholders

Preparing change requests or taking corrective action in accord with procedures for overallchange control

Quality Control:

Monitoring results to determine if they comply with relevant quality standards andidentifYing ways to eliminate problem causes

• Project results are deliverables and management results, such as cost and schedule performance

Inputs

• Work Results• QMP• Operational definition• Checklists

Tools & Techniques

• Flowcharting

• Inspection - Measuring, examining, and testing to determine whether results conform with

requirements

• Statistical sampling - Choosing for inspection a part of a population of interest

• Acceptance sampling - Evaluating a portion of a lot to detennine acceptance or rejection of the

entire lot

Pluses:

Minuses:

Good if" inspection costs high, destructive testing required, inspection procedures

tedious

Not good iflots small



Control Charts

Graphic displays ofthe results, over time, ofa process; used to assess whether the process is "in control"

Rule of Seven - Run of seven points toward upper or lower control limit may indicate thatprocess is out of control

• Assignable causes - Variations may be caused by differences in machines, workers overtime, etc.Identify the root cause.

Pareto Diagram

Histogram, ordered byfrequency ofoccurrence, that shows how many results were generated by type orcc.tegory ofidentified cause

Pareto's Law - A relatively small number ofcauses will typically produce a large majority oftheproblems or defects (80/20 rule)

Pareto Diagram

4 10

oEDcBA

o

3 Cumulativ7

e

2 5Number of Percentage ofDefectives Frequency

Defectives1 by Cause 2

Defective

Trend Analysis

Using mathematical techniques to forecast future outcomes based on historical results

Technical performance - How many errors or defects have been identified; how many remainuncorrectedCost and schedule performance - How many activities per period were completed with significantvariances


Management Consulting Houser------------------------------------11I..

Statistical Quality Control

Prevention - Keeping errors out of theprocessInspection - Keeping the errors out of thehands ofthe customer

Special causes - Unusual eventsRandom causes - Normal process variation

Using Standard Deviation

• Mean (f.l) - Average

Attribute sampling - The result conforms or itdoes notVariables sampling - The result is rated on acontinuous scale that measures the degree ofconformity

Tolerances - The result is acceptable if it fallswith the range specified by the toleranceControl limits - The process is in control if theresult falls within the control limits

• Variance - Sum of squared differences between mean and each value, (x-f.l)2, divided by numberof samples less one or (9-1=8)

• Standard deviation - Square root of the variance

Upper control limit (UCL) - Three standard deviations

Statistics

2 standard deviations (+/- 1) = 68.26 %4 standard deviations (+/- 2) = 95.46 %6 standard deviations (+/- 3) =99.73 %

-3 -2 -I o +1 +2 +3



l~ --.J

Example of Standard Deviation

Item12345678

9

Total

Weight (x) (x-/-l)24.9 0.027785.0 0.004445.1 0.001115.2 0.017785.3 0.054445.5 0.187784.7 0.134444.8 0.07111

D 0.00111

45.6 0.50000

Example of Standard Deviation

Mean (/1) = 45.6 / 9 = 5.066; (rounded to 5.07)

Variance - Sum of squared differences between mean and each value, (x- /1)2, divided by numberof samples less one or (9-1=8); thus 0.5 / 8 = 0.0625

Standard deviation - Square root of the variance; thus 0.0625 = 0.25

• Upper control limit (UCL) = Three standard deviations; thus 3(0.25) + /1 (or 5.07) = 5.82.• The lower control limit (LCL) equals 4.32 or 5.07 - 3(0.25).

Outputs

• Quality improvement

• Acceptance decisions - Items inspected will either be accepted or rejected. Rejected items mayneed rework

• Rework - Action taken to bring a defective or non-confonning item into compliance withrequirements or specifications

• Completed checklists - Become part ofproject's records

• Process adjustments - Immediate corrective or preventive action, in accordance with changecontrol procedures


r RSS Management Consulting House

Chapter (5): Risk Management IPMA Preparation Course

R<S Management Consulting House

Project Risk ManagementIn the past We treated risk as a "let's live with it."

Today Risk Management is part of doing business. It forces us to focus on the futurewhere uncertainty continues to increase.

Project Risk Management:

Is the systematic process of Identifying, Analyzing, and Responding toproject risk.

Maximizing the probability and consequences of positive events

&Minimizing the probability and consequences ofadverse events

To Project Objectives


Major Processes:

Planning

[Management Consulting House

r--

Controlling

Integrating Risk with other project management functions:

Integration

Scope

Life

~ationsEnviron.Ex. ~

ycleariables

Commnnication

Quality RegnirementsStandards

AvailabilityProductivity

HR

Time

stObj ctives

Cost

I Procurement I

Chapter (6): Risk Management /PMA Preparation Course


Risk:An uncertain event or condition that, if it occurs, has a positive or negative effect on aproject objective.

The severity of any risk can be defined in terms of two quantities:

• Likelihood:The extent to which the risk effects are likely to occur.

• Impact (Consequence):The effect that a risk will have on the project if it occurs.

Risk Management Planning:

The process ofdeciding how to approach and plan the risk managementactivities for a project.

Risk Management Plan

Describes how risk identification, qualitative and quantitative analysis, response planning,monitoring, and control will be structured and performed during the project life cycle.

Methodology

Roles andResponsibilities

Reporting Formats

Chapter (6): Risk Management

Budgeting

Scoring andinterpretation

Timing

Tracking


Management Consulting Houser'--------------------------------1

Risk Identification:

Involves determining which risks might affect the project and documenting theircharacteristics

Risk Categories:

Risks that may affect the project for better or worse can be identified and organized intocategories.

Macro ]'---r----

Environment

Market

Project

TargetInformation

Design

Managerial

Team

Control

Political

Financial

Social

GatheringProcurement

Technical

- Job Site

Funding

Contractual

Act of God

Techniques:

•

•


Delphi Technique

•



Information-Gathering Techniques:

Interviewing (Nominal Group Technique)

SWOT Analysis:

External Factors

Opportunities (0)

Threats (T)

Qualitative Risk Analysis:

SOGrowth

STStability

WOCooperate

WTMitigation

The process of assessing the impact and likelihood of identified risksPrioritizes risks according to their potential effect on project objectives


Probability-Impact Matrix:

rRS Management Consulting HOUS~

Likely

Probable

Occasional

Remote

Improbable

Minor Major Severe Catastrophic


Impact

lPMA Preparation Course


Quantitative Risk Analysis:

Process aims to analyze numerically the probability of each risk and its consequence onproject objectives.

Expected Monetary Value (EMV):

Status Payoff Probability %

Good Market· Good Quality 80,000 15

Good Market· Poor Quality 50,000 45

Poor Market· Good Quality 20,000 25

Poor Market· Poor Quality -20,000 15

Expected Payoff for good market = 80,000*0.15 + 50,000*0.45 = LE 34,500

Expected Payoff for good quality = 80,000*0.15 + 20,000*0.25 = LE 17,000

EMV = 80,000*0.15 + 50,000*0.45 + 20,000*0.25 + (-20,000)*0.15 = LE 36,500


Management Consulting House,.--~

1

80,000

30,000

80,000

30,000Bad QualityP:30%

Good QualityP:70%

Bad QualityP:50%

55,000

65,000

SubcontractLE 15,000

Purchase MachineLE 17,500

47,500

Risk Response Planning:The process of developing options and determining actions to enhance opportunities andreduce threats to the project's objectives.

Decision Tree Analysis:

Risk response planning must be :

• Appropriate to the severity ofthe risk• Cost effective in meeting the challenge• Timely to be successful• Realistic within the project context• Agreed upon by all parties involved• Owned by a responsive person

Risk Response Strategies:

Avoidance(Prevention)

Mitigation(Corrective action)


Transference (Shift Responsibility)

Risk Transference:


Acceptance(Accept consequences)

Risk Monitoring and Control:

The process of keeping track of the identified risks, monitoring residual risks and identifying new risks,ensuring the execution of risk plans, and evaluating their effectiveness in reducing risk

Workaround Plans:Unplanned responses to emerging risks that were previously unidentified or accepted.

Chapter (6): Risk Management /PMA Preparation Course

rt

R5 Management Consulting House

Chapter (7): Procurement Management IPMA Preparation Course


Procurement ManagementProcesses required to acquire goods and services from outside the organization

Planning Executing Closing

3.1. ~ Solicitation 6.

Procurement ~Contract

Planning Closeout

--

L4. Source

2. Selection

Solicitation4-

f--Planning

5. ContractAdministration

Buyer-seller Relationship:

• Buyer - The customer• Seller - Contractor; vendor; supplier external to the performing organization• The tenns and conditions of the contract become key inputs to many of the seller's processes

Procurement Planning:

Determining what project needs can best be met by procuringproducts or services outsidethe organization

• Includes consideration of potential subcontracts

Inputs:

• Scope statement - Describes current project needs and strategies• Product description - Defines the end product ofthe project; provides important information about

any technical issues or concerns• Procurement resources - Resources and expertise needed to support project procurement activities• Market conditions - The products and services available in the marketplace• Other planning outputs-Cost and schedule estimates

• Quality management plans• Cash-flow projections• WBS• Identified risks• Planned staffing

Chapter (7): Pracurement Management IPMA Preparation Course

• Constraints - Factors that limit the buyer's options, such as availability of funds• Assumptions - Factors that, for planning purposes, will be considered to be true, real, or certain

Tools & Techniques:

• Expert judgment - Groups or individuals with specialized knowledge or training• Make-or-buy analysis - Detennining if the product can be produced cost-effectively by the

performing organization

Break-Even Analysis:

At Break-Even Point:

Revenue =SPxQ

SP = Selling PriceFC = Fixed CostVC = Variable CostTC = total CostUC = Unit CostQ = Quantity

Quantity

Fe

TC

VC

ProfitArea

Revenue = TC

SPXQBE = FC+ UCXQBE

FCQBE =

SP- DC

VC=UCxQ

Revenue = SP x Q

TC=FC+VC

LossesArea

Incomeor Expenses

i-------------------------------uChapter (7): Procurement Management IPMA Preparation Course


Contract-Type Selection:Risk Allocation (or Construction Contractors:

Contract Price

Quality, Workmanship, Site, Subcontractor

RisksIII

Employer's Ri~ks

II

III

Contractors RisIIIIIII

All RiskTransferred

ToM.e

Fixed Time&

Fixed Cost

1-------l--------1-------=-\.------ -- - ---------

Finarice + Force Ma~eureI I

F.M F

• Cost Plus Contract:This is a contract in which the contractor is reimbursed his actual costs incurred, plus specifiedpercentages on the actual costs in respect of his overhead charges and gross profit, without anytotal contract price being quoted.

• Bill O/Quantity (BOQ Contract):This a common form of contract in which the design has been completed by the employer and canspecified in the tender.

• Target Cost Contract:This is an elaboration of the "Cost Plus" concept, which aims to overcome the main defects of thelatter by introducing incentives to the contractor to be as efficient as possible by having a ceilingprice.

• Turnkey Contract:It is a package contract in which all the separate design, disciplines, civil, mechanical, & electricalfor the setting up of a major entity are placed in the hands of a main contractor, who may subcontractor specialist works to appropriate sub-contractors.

• BOT&BOOT:- Built, Own, Operate & Transfer.- All Risk belongs to Main contractor.- Time is fixed & Cost is fixed.


RSS Management Consulting Hous~

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I

• Fixed price - A fixed total price for a well-defined product. May include incentives for meeting orexceeding selected project objectives.

• Cost reimbursable contacts - Payment to the seller of actual costs

• Direct costs - Costs incurred in implementing the project• Indirect costs (Overhead) - Costs allocated to the project by the perfonning organization, as a

cost of doing business. May include incentives for meeting or exceeding selected projectobjectives.-

• Time and Material contract - Seller is paid a preset amount per unit of service, and the total valueofthe contract is a function of the quantities needed to complete the work

Types of Contract:Contract Definition:

A biding agreement that obligates the supplier to provide certain product and the client to pay forit.A Contract is legal relation ship-subject to remedy at court.

1- Classified by method of evaluation contract price:

1.1 Fixed-price contract1.2 Price-adjustment contract1.3 cost-plus Contract1.4 target-cost contract1.5 bills of quantity contract1.6 Schedule of rates contract

2- Classified other than by method of evaluating coutract price:

2.1 Competitive contract2.2 Negotiated contract2.3 Package contract (package deal)2.4 Turnkey contract2.5 Continuation contract2.6 Serial contract2.7 Running contract2.8 Service contract

The Conditions of contract are the "Rules" by which the contract is run.They set out the rights and obligations of the parties and agree the action that will be taken by theparties if various eventualities arise during the course of contract.

Outputs:

• Procurement management plan - Describes how the remaining procurement processes will bemanaged

• Statement of work• Describes the procurement items in sufficient detail for prospective sellers to determine if

they are capable ofproviding the item• Should include a description of any collateral services required for the procured item

J--------------------------------------------LChapter (7): Procurement Management IPMA Preparation Course

rI


Solicitation Planning:Documenting the product requirements and identifying potential sources

Inputs:

• Procurement management plan• Statements of work• Other planning outputs - Any modifications to other planning processes, as a result of procurement

planning.

Tools & Techniques:

• Standard forms - Include standard contracts, standard descriptions ofprocurement items, andstandardized versions of all, or part, of the needed bid documents

• Expe, t judgmentOutput:>:

• Procurement documents - Used to solicit proposals from potential buyers• Invitation to bid (IFB)• Request for proposal (RFP)• Request for quotation (RFQ)• Invitation for negotiation• Contractor initial response

• Procurement documents should include:• The relevant statement of work• Description of the desired fonn of the response• Required contractual provisions

• Contract Documents:1- A Statement of the scope of the Contract2- Data (e.g. site conditions. Climate) affecting the execution of the Works3- A Technical Specification4- Bills of quantities (especially in civil engineering and building construction

Contracts); Priced Schedules of Plant (in plant supply contracts)5- Programmed for the construction - completion of the Works6- The conditions of Contract7- Site Regulations8- The form of Tender: the offer9- Unconditional acceptance of the tender10- Guarantees or Bonds11- A formal agreement12- Additions or variations ofthe above made during negotiations or subsequently

By agreement between the parties.

• Evaluation criteria - Used to rate or score proposals (objective or subjective)-Understanding oftheneed

••••

Overall or life-cycle costTechnical capabilityManagement approachFinancial capacity

Chapter (7): Procurement Management lPMA Preparation COllrse

_______________________R_~_,_'_M_a_n_a_B_e_m_e_n_t_C_o_n_S_u_lt_;_n_B_H_O_U---;S~

• Statement of work updates - Modifications to one or more statements of work may be identifiedduring solicitation planning

Solicitation:

Obtaining responses (e.g., bids, quotes, offers) from prospective sellers on how projectneeds can be met

• Effort usually expended by the prospective seller.

Inputs:

• Procurement documents.• Qualified-seller lists - Lists or files with information on the relevant experience and other

characteristics of prospective sellers.

Tools & Techniques:

• Bidders' conference:• Meetings with prospective sellers prior to preparation of a proposal-Ensure all

Prospective sellers have a clear, common understanding of the procurement• Responses to questions may be incorporated into the procurement documents as

amendments

• Advertising - Placing advertisements in general-circulation publications, such as newspapers, or inspecialty publications, such as journals

Outputs:

• Proposals - Seller-prepared documents that describe the seller's ability and willingness to provide therequested product

Source Selection:The receipt ofbids or proposals, and the application ofthe evaluation criteria in order to

select a providerInputs:

• Proposals• Evaluation criteria - May include samples of the suppliers previously produced products/services

for the purpose of providing a way to evaluate their capabilities and quality of products• Organizational policies - Formal or informal policies that can affect the evaluation of proposals

Tools & Techniques:

• Contract negotiation - Involves clarification and mutual agreement on the structure andrequirements of the contract, prior to the signing of the contract

- Final contract language should reflect all agreements reached- Include responsibilities and authorities, applicable terms and laws,

technical and business management approaches, contract financing, and price

• Weighting system - Select a single source, who will be asked to sign a standard contract- Rank and order all proposals to establish a negotiating sequence- Select a short list of qualified sellers, based on a preliminary proposals and,

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Point Rating System:

• Technical proposal 33 1/3 %

Technical 50%Company 25%Previous experience 25%

• Presentation 33 1/3 %

Team 60%Format 40%

• Price 33 1/3 %

Day rates 85%Expenses 15%

• Overall 100%

Outputs:

- A mutually binding agreement that obligates the seller to provide the specified product,and obligates the buyer to pay for it-A contract is a legal relationship subject to remedy in the courts-It may be called a contract, an agreement, a subcontract, a purchase order, or amemorandum of understanding-The review-and-approval process should ensure that the contract language describes aproduct or service that will satisfy the need identified



---------------------------------------------1'

A Form of Agreement:

• To emphasis the formality and importance of the Contract .to summaries the highlights of a complexcontract.

• To stress that the contract is with the employer in cases in which the enquiry, tenders. Negotiations.• To schedule the items forming the contract documentation.• To tidy-up a contract situation when there have been expectative negotiations following the tender.

• The Agreement besides listing the contract documents normally sets out the main feature ofthecontract. E.g. the agreed contract price and its terms of payment, completion dates. Guarantees andbonds. Agreed liquidated damages and suchlike.

Definitions:

• Request for Proposal (RFP)- Request for a price and a proposed method of meeting requirements- Often used when both price and technical evaluation determine winners

• Request for Quote (RFQ)- Request for materials or services- Price-driven

• Warranty- A promise about aproduct made by either a manufacturer or seller.- Part of a contract, the truth of which is necessary to the validity of the contract-Implied Warranty - a guarantee imposed by law in a sale- Express Warranty - an assertion or promise concerning goods or services

Bonds and Guarantees:

I. A deed is a general term for a contract documents under seal2. A Bond is a deed in which one party promises to perform something under stated circumstances

in a specified way for the benefit of second party.If the undertaking is to pay money it is called a "'common money bond"; ifit is to produce some act or event it is called a

"bond upon condition",

A bond only needs written acceptance by the party benefiting to become binding. It is dischargedand becomes void as soon as the undeltaking has been performed as specified.

3. A guarantee is a promise by a third party "c" to be answerable to a party "A" in a manner and toan extent specified, for the payment of debt or for a default or miscarriage in the performance of aduty owed by a party "B" to the party "A" in the event that party "B" fails to perform hisengagement.A guarantee does not have to be a bond (i.e..a formal documents under seal) but in engineeringcontracts it is often so required in order to emphasis the validity of the actual words written.A guarantee must however (at any rates in most of the cases met with in engineering projects) beeither in writing or "evidenced in writing" I.e. recorded as to its major features in a memorandumof some sort.

--------------------------------------------L,Chapter (7): Pracurement Management IPMA Preparation Course

DJ~ Management Consulting House:I~

Contract Administration:

Ensuring the seller's performance meets contractual requirements.

• The legal nature ofthe contractual relationship makes it imperative that the project team be acutelyaware of the legal implications of actions they take during administration of the contract

• Payment terms should be defined within the contract and should link progress made withcompensation paid

Applying the appropriate PMprocesses to the contractual relationships

• Project plan execntion - To authorize the contractor's work at the appropriate time• Performance reporting - To monitor contractor's cost, schedule, and technical performance• Quality control- To inspect and verify the adequacy of the contractor's product• Change control- To ensure that changes are properly approved and that the appropriate stakeholders

are made aware of such changesInputs:• Contract• Work results - Status of deliverables, to what extent quality standards are being met, and costs

incurred or committed are collected as part of the project plan execution• Change requests - Modifications to the terms of the contract or to the description of the product or

the service to be provided. Contested changes are called claims, disputes, or appeals.• Seller invoices - Submitted to the buyer to request payment for work perfonned. Invoicing

requirements are usuall y specified in the contract.

Tools & Techniques:

• Contract change control system - Defines the process by which the contract may be modified. Itincludes:

-Paperwork- Tracking systems- Dispute-resolution procedures- Approval procedures- Performance reporting- Payment system

• Performance reporting - Provides management with information about how effectively the seller isachieving the contractual objectives

• Payment system - Handled by the accounts payable system of the performing organization. Thesystem must include the appropriate reviews and approvals by the project management team.

Outputs:• Correspondence - Contract terms and conditions require written documentation of buyer-seller

communications, such as warnings of unsatisfactory performance, contract changes, and clarifications• Contract changes - Approved and unapproved changes are fed back through the appropriate project

planning and project procurement processes, and the project plan, or other relevant documentation, isupdated as appropriate

• Payment requests


_______________________RS_,_'_M_a_n_a_g_e_m_en_t_C_o_n_S_U_lt_i_ng_H_O_U_S_~i·····I

Contract Close Out:

Completion and settlement ofthe contract, including resolution ofany open items.

• Involves product verification and administrative closeout

Inputs:

• Contract documentation - Includes, but not limited to:ContractSupporting schedulesRequested and approved contract changesSeller-developed technical documentationSeller performance reportsFinancial documents (e.g., invoices and payment records)Results of any contract-related inspections

Tools & Techniques:

• Procul'ement audits -Structured reviews of the procurement process, from procurement planningthrough contract administration, conducted to identify successes and failures Oessons learned)

Outputs:

• Contract file - Set of indexed records, prepared for inclusion with the final project records• Formal acceptance and closure - Person or organization responsible for contract administration

provides the seller with formal written notice that the contract has been completed. Normally specifiedin the contract.

--------------------------------------'LChapter (7): Procurement Management IPMA Preparation Course


Different Kinds ofagreements

> The Main Characteristic of the Joint Venture.

1- Participation for one Project.2- It has not a considerable legal personality; it is just an Association ofPersons.3- They usually deal depending on written agreement for a ce11ain Project.4- They share project assets & the project got his own finance.5- Bank account by name of Joint Venture.6- They participate in the right of mutual supervision on the project.7- Distribute the losses & the profit according to the ratio agreed upon.8- They responsible jointly or severally in front of the third party.9- The J.V Project is established on the fiduciaries and accordingly to the legal

impact which display in (Confidence / Secrecy / ....etc ) .10- The J.V Project has a legal representative in front of the client & the Joint Ventures parties.11- The J.V.Project got his Employees.

> The Characteristic ofthe Consortium:

1- A written contract between the main contractors to form the consortium2- In Confrontation of the client: Every contractor signed by his name.3- Every contractor has his own part of work (Civil, Erection, ....etc )4- Appointing a common leader for the consortium (not a manager) his missions are co-ordination &

speaking by the name of consortium's members and to deal with the correspondences.5- Each member is responsible jointly & severally in front of the client.6- It has not a considerable legal personality or name.7- There is no common bank account for the group or for the consortium's name.

> Definition of the Government contract

1- Dealing with Government, Organizations, or Institutions.2- To include exceptional conditions.3- Concerning public Project.

Following Law 9 / year 1983 which replaced by law 89 / year 1998 for Tenders procedures.


roes ofContracts

Contract Definition:

A biding agreement that obligates the supplier to provide certain product and the client to pay for itA Contract is legal relationship subject to remedy at court.

1- Classified by method of evaluation contract price:

1.1 Fixed-price contract1.2 Price-adjustment contract1.3 cost-plus Contract1.4 target-cost contract1.5 bills of quantity contract1.6 Schedule of rates contract

2- Classified other than by method of evaluating contract price:

2.1 Competitive contract2.2 Negotiated contract2.3 Package contract (package deal)2.4 Turnkey contract2.5 Continuation contract2.6 Serial contract2.7 Running contract2.8 Service contract

The Conditions ofcontract are the "Rules" by which the contract is run.They set out the rights and obligations of the parties and agree the action that will be taken by theparties ifvarious eventualities arise during the course ofcontract.

-----------------------------------jLChapter (7): Procurement Management IPMA Preparation Course

as Management Consulting House

Different Kinds ofagreements

);. The Main Characteristic of the Joint Venture.

10- Participation for one Project.11- It has not a considerable legal personality; it is just an Association of Persons.12- They usually deal depending on written agreement for a certain Project.13- They share project assets & the project got its own finance.14- Bank account by name of Joint Venture.15- They participate in the right of mutual supervision on the project.16- Distribute the losses & the profit according to the ratio agreed upon.17- They responsible jointly or severally in front of the third party.18- The J.V Project is established on the fiduciaries and accordingly to the legal

impact which display in (Confidence / Secrecy / ....etc ) .10- The J.V Project has a legal representative in front of the client & the Joint Ventures paIiies.11- The J.V.Project got his Employees.

);> The Characteristic of the Consortium:

5- A written contract between the main contractors to fonn the consortium6- In Confrontation of the client: Every contractor signed by his name.7- Every contractor has his own part of work ( Civil, Erection, ....etc )8- Appointing a common leader for the consortium (not a manager) his missions are co-ordination &

speaking by the name of consortium's members and to deal with the correspondences.5- Each member is responsible jointly & severally in front of the client.6- It has not a considerable legal personality or name.7- There is no common bank account for the group or for the consortium's name.

);> Definition of the Government contract

4- Dealing with Government, Organizations, or Institutions.5- To include exceptional conditions.6- Concerning public Project.

Following Law 9/ year 1983 which replaced by law 89/ year 1998 for Tenders procedures.


RSS Management Consulting HOUS~~

------------------------------------------11

FIDIC

FIDIC refers to (Federation Internationale Des Ingenieurs Conseils).

The Editions Before rear 1999:

1. The Red Book (Conditions ofContract For works ofCivil Engineering Construction): 72 Clauses

• First Edition at 29 August 1957.• Second Edition at 1969.• Third Edition at 1977: (for Engineer interests).• Forth Edition at 1987: (gives the Contractors his rights).• Forth Edition Reprint at 1988, 1992, and 1996.

The Addendum of Forth Edition at 1996 has changed Clause 67 to be (Dispute AdjudicationBoard), This Clause enables the Employer to chose - with the Contractor- Dispute Adjunction Boardconsists ofone Adjudicator or three according to kind & volume ofworks).

The red book consists of 2 parts, first one contains general conditions and the second part isguides for writing & preparing Particular conditions.

2. The yellow book (Conditiol/s of Contract for Electrical and Mechanical Works - IncludingErection on Site): 51 Clauses

• First Edition at 1963.• Second Edition at 1980.• Third Edition at 1987: reprint at 1988 & 1997

At Third Edition reprinted at 1997 contains Clause 50 (Dispute Adjudication Board)

The yellow book consists of2 Parts.

3. The Orange Book (Conditiol/ ofContract For Design & Build (Turnkey}):

• One Edition at 1995.

4. The White Book (Client / COl/sultant Model Services Agreement): 44 Clauses

• First Edition at 1990.• Second Edition at 1991.• Third Edition at 1998:

Important modifications in Third Edition:

1. The Client's right to end the contract (clause No. 40).2. Dispute Adjudication Board (Clause No. 43).

-------------------------------1'L~;



September 1999 Editions (The Last):

1. The New Red Book (Conditions ofContract for Construction Works):

• First Edition at September 1999.

In this book, the Employer or his Representative prepares the design documents.Clauses were reduced to be 20 clauses only.The English was simplified.

2. The New Yellow Book (Conditions ofContract for Plant (!nd Design - Build):


In this book, the contractor prepares the design documents by his followers according to theEmployer's requirements.

3. The Silver Book (Colldition ofContract ofEPC / Turnkey Projects):


This Book is suitable for Turnkey Projects such as power plan, Factories, infrastructures and

development Projects.

The contractor has the total responsibility for design and execution for project with little

involvement from the employer or the project company of BOOT.

By usual arrangements the contractor does all Design Engineering, Procurement &

Construction (EPe) then hands over the project to be ready for operation to the Project

Company or The employer.

4. The Green Book (Short Form ofContract):15 Clauses


In This Book is recommended to the relatively small capital value (Not more than 2Million $) or short duration Project (around one year).

The Contractor does all works from design provided by the Employer or his Representatives.


30 Units [RSS Management Consulting Houser

~--------------------...:-_--------------_!i

The Editions until year 1999:

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----------------------------------IL.Chapter (7): Procurement Management IPMA Preparation Course


Chapter (8): Planning Management IPMA Preparation Course


Project Planning Management

Project Time Management:

Processes required ensuring timely completion ofthe project

>- Activity Definition>- Activity Sequencing>- Activity Duration Estimating>- Schedule Development>- Schedule Control

Estimating

>- Project Time management involves a bottoms-up approach. Estimate the duration of eachactivity and combine that with the network diagram to build up the schedule to determineoverall project duration.

>- Top-Down estimating is not a recommended method for establishing overall project duration.Noted/qualified exception: When there is a limited amount of detailed information about theproject.

Activity Duration Estimating:

• Expert judgment guided by historical information used whenever possible• Anything less is inherently uncertain and risky

Quantitativelv Based DurationsExample (Pipeline Construction)

200 m ofpipe can be welded per day30000 m of pipe are required30000 m / 200 m per day = 150 days

Quantitatively Based DurationsExample (IT Planning)

I program debugged and unit tested per week20 programs to be corrected20 programs / 1 program per week = 20 weeks

2 Days 12 Days 5 Days

Network Diagram Analysis:

There are many ways to do this.This is just one approach.Find one you like and learn it.

A --> B -.. C

\Start Finish

6 Days 7 Days 3 Days /D --> E -.. F

Chapter (8): Ptanning Management IPMA Preparation Course

Project Time Management:

Planning

RSS Management Consulting House r

I

Controlling

2. "-Activity

/ Sequencingl. 4. 5.Activity Schedule ~ ScheduleDefinition

\ /Development Control

3.ActivityDurationEstimating

Activity Definition:

• Identify the activities that must be perfonned to produce the project deliverables• Define the activities that must be perfonned to meet the project objectives

Inputs:

• WBS - Primary input to activity definition• Scope statement - Project justification and project objectives• Historical information - The activities that were actually required on previous, similar

projects• Constraints - Factors that will limit the PM team's options• Assumptions - Factors that, for planning purposes, will be considered to be true, real, or

certain• Expert judgment

Tools & Techniques;• Decomposition - Subdividing project work packages into smaller, more manageable

components• Template - An activity list from a previous project or an activity list for a WBS element from

the current project

Outputs:• Activity list - A list ofall the activities that will be performed on the project and a description

of each• Supporting detail- Documentation that contributes to the process, including all identified

assumptions and constraints• WBS updates - Refinements to the existing WBS



Activity Sequencing:IdentifYing and documenting interactivity logical relationships

Inputs:

• Activity list• Product description - Product characteristics affect activity sequencing• Mandatory dependencies (Hard logic) - Determined by the qualities of work to be done• Discretionary dependencies (Soft logic) - Defined by the project management team• External dependencies Relationships between project activities and non-project activities• Milestones - Events need to be part of the activity sequencing to assure that the requirements

for meeting the milestone (s) are met.

Tools & Techniques:

• Precedence diagramming method (PDM)• Nodes represent activities and arrows show dependencies

• Arrow diagramming method (ADM) - Uses arrows to represent activities and connectingnodes to show dependencies

A\ Start

~/ EFinish

• ADM usesfinish-to-start dependencies only and uses dummy activities to show logicalrelationships·

• Network templates - Can include an entire project or just a portion of it (i.e., subnets andfragnets)

Outputs:

• Project network diagram - Schematic display of project's activities and dependencies• Activityiist updates - Dividing or redefining activities so that the relationships are correctly

diagrammed


R$ Management Consulting House r

Activity Duration Estimating:

Process oftaking information on project scope and resources and then developingdurations for input to schedules

• Elapsed time (Delay) - Work periods between the finish of one activity and the start of anotheractivity.

Inputs:

• Activity list• Copstraints• Assumptions• Resource requirements - Duration estimates are influenced by resource effort and assignments• Resource capabilities - Duration estimates are influenced by the capability of the people and the

material resources assigned to them• Historical Information-Project files:

o Commercial duration estimating databaseso Project team knowledge

• Identified risks - Determine the extent to which the effect of risks is included in the baselineduration estimate for each activity

Tools & Techniques:

• Expert judgment - Used with historical information• Analogous estimates (Top-down estimating) - Uses duration of a previous, similar activity as

the basis for the of estimate of a future activity• Quantitatively based durations - Quantities to be performed for each specific work category

defined by the engineering/design effort, multiplied by the productivity unit rate• Reserve time (contingency) - An additional time frame, called time reserve, contingency, or

buffer, that can be added to the activity duration or elsewhere in the schedule as recognition ofschedule risk

Outputs:

• Duration estimates - Quantitative assessments of the likely number of work periods required tocomplete an activity

• Basis of estimates - Documentation of the assumptions used for developing the estimates• Activity list updates

Schedule Development:

Determining the start andfinish dates ofproject activities

• If start and finish dates are not realistic, the project is unlikely to be finished on schedule

f--------------------------------------------i:c~'"



Inputs:

• Project network diagram• Activity Duration estimates• Resource requirements• Resource pool - Description of the available resources and the times they are available to

work on the project• Calendars - Identify periods when work is allowed• Project calendars: Affect all resources• Resource calendars: Affect specific resources or categories of resources• Constraints

- Imposed dates- Key events or milestones

• Assumptions• Leads and lags - Dependencies that require lead or k!g values to accurately define the

relationship• Risk management plan• Activity attributes - Includes who has responsibility for perfonning the work, geographic

area or building (location for the work), and activity type (summary or detail level)

Tools & Techniques:

Mathematical analysis -:

Calculating theoretical early and late start and finish dates for all activities

• Critical Path Method (CPM) - Calculates a single, deterministic early and late start and finishdate for each activity, to be used to determine which activities must be completed on time toavoid impacting the finish date of the project

• Program Evaluation and Review Technique (PERT) - Uses sequential network logic and aweighted-average duration estimate to calculate duration. Uses he probability of an estimate'saccuracy.

• Resource-leveling heuristics (Resource-based method) - Changing the schedule toaccommodate resources

• PM Software - Used to assist schedule development and to display schedule-developmentoutputs

• Coding structure - Allows for sorting/and/or extractions based on difference attributes assignedto activities

• Duration compression - Looks for ways to shorten the schedule without changing the projectscope

• Crashing - Analyzing cost and schedule tradeoffs to determine how to obtain the greatestamount of compression for the least incremental cost

• Fast tracking - Doing activities in parallel that would normally be done in sequence• Simulation- Calculating multiple project durations with different sets of activity assumptions


RSS Management Consulting House ,.

IOutputs:

• Project schedule - Includes planned start and expected finish dates for each activity• Supporting detail- Documentation of all identified assumptions and constraints• Schedule management plan - Defines how changes to the schedule will be managed• Resource requirement updates - Updates based on the results of resource leveling and

On updates to activity lists.

Gantt Charts:

• Bar chart - Displays activity start and end dates, as well as expected durations• Milestone chart - Displays scheduled start or completion of major deliverables

Typical Gantt chart:

Typical Milestone Chart:

CurrentDate

Event Jan Feb Mar~ .. Apr May Jun JuF Aug

Subcontracts Signed DVSpecifications Finalized

Design Reviewed 6Subsystem Tested 6First Unit Delivered 6Production Plan Completed 6

There are many other acceptable wa}'s to display project infonnation on a milestone chait.

I Planned 6 Actual T I

Figure 6-7. lvIileslone Chart

Schedule Control:• Influencing factors that cause schedule changes• Determining that the schedule has changed and managing the changes

Inputs:

• Project schedule - The approved project schedule is called the schedule baseline; provides thebasis for measuring and reporting schedule performance

• Performance reports - Provide information on schedule performance, such as which planneddates have been met and which have not

• Change requests - Occur in many forms• Schedule management plan

LChapter (8): Planning Management IPMA Preparation Course


Tools & Techniques:

• Schedule change control system - Defines the procedure for changing the project schedule• Progress measurement (Earned Value) - Assessment of schedule variations to determine

whether they require corrective actions• Additional planning - Prospective changes may require new or revised activity duration

estimates, modified activity sequences, or analysis of alternative schedules• PM software - Useful tool for tracking planned dates against actual dates and for forecasting the

effects of schedule changes• Variance Analysis - comparing target dates with actual start and finish dates to detect deviations

and implement corrective solutions in case of delays

Outputs:

• Schedule updates - Modification to the schedule infonnation used to manage the project• Corrective action - Anything done to bring expected future schedule performance back in-line

with the project plan• Lessons learned - Documentation of the causes ofvariances, of the reasons the corrective actions

were chosen, and of other infonnation learned through controlling the schedule

Basic Terminology:

• Schedule revisions - Changes to the scheduled start and finish dates in the approved (or baseline)schedule


RSS Management Consulting House i

Planning Processes

Core Processes

From theInitiating

Pro~s.~

(Figure 3-4}

=>=>FIJ)mth€'

ControllingPrclt:Bss.:-s

(FigUl;? 3-7)

Facilitating Processes

1l!9.-1',..!!J!!I!!.,.! I!!, ,·,·9-""- 9.2

0t9nizoticocl r ~ff kq..si:itk-nFbnning

Relationships among the Planning Processes

To thE'&~::uting

ProC$$S8S

(F~",

=>

WBS - Work Breakdown StructureSS - Scope Statement



Chapter (9): Project Planning IPMA Preparation Course


Project Planning

Activity on Arrow (AOA) or Arrow Diagram Method (ADM):

The direction of the arrow represents a precedence relationship.In the network segment

iA

• j

Event i must occur before activity A can commence. Similarly, eventj cannot occur until activity A has beencompleted.The precedence relationship is transitive among nodes. If i precedes j and j precedes k, then i precedes k:

Example:

1-2

1-3

2-3

2-4

3-5

4-5

Example:

----;.- j

1-2

1-2

1-3&2-3

2-4

k

An activity Network

B B G)2

C

E BG)

G C&B An activity Network

Chapter (9): Praject Planning IPMA Preparation Course

[Management Consulting HouseVJ~______________________"'l.f\"'~= ___1r

I,Arrow Network Techniques

Dummy Activity (Dashed Arrow):

Case 1:A<C,B & B<D & C<D

(2 3) = activity C A C D

CD '~,8

(2 3) = activity B

BXThe Right:

DA C

0----1,~ ':0)--_38~ ~0---

B E

E is Dummy Activity (Duration = 0)

Case 2:A<C & B<E & C<D,E

D

Ex

Because B is not <

A C D

)----0\\\\\

}--....:::E_-Or"

--------------------------------------il

LChapter (9): Project Planning IPMA Preparation Course


Case 3:A<D,EB<E,FC<F

D -0A

E XC F A is not < F

D ,0A

-----':0 E ,0

C F

Case 4:A<C & B<D & C<D

A C D2 4 0 Dummy x is unnecessary

B , ,

3X

Activity A = (1 - 2 )Activity B = (1 - 3 )Activity C = (2 - 3 )Activity D = (3 - 4 )


RSS Management Consulting House---- =="- -11"'"

lCase 5:

A<CB < C, D, EC<E

AJ_----=c=--- , 91--....;E=--~,0

,I ,

I "x: ./I z",,,,"I "I "I "

D

B is a redundant restriction on E

c'0 E ,0

AII

x:IIII

B

D

No need for Dummy Z

r-------------------------------L;,Chapter (9): Praject Planning IPMA Preparation Course


Project Planning byProgram Evaluation and Review Technique (PERT)

&Critical Path Method (CPM)

1. Program Evaluation and Review Technique (PERT):

PERT was developed by the U.S. Navy during the late 1950 to accelerate the development of the Polaris

Fleet Ballistic Missile. The development of this weapon involved the coordination of the work of thousands

of private contractors and other government agencies.

2. Critical Path Method (CPM):

The critical path method closely resembles PERT in many aspects but was developed independently by E. 1.

du Pont de Nemours Company. As a matter of fact, the two techniques, PERT and CPM were developed

almost simultaneously. The major difference between them is that CPM does not incorporate uncertainties

in job times Instead, it assumes that activity times are proportional to the amount of resources allocated to

them, and that by changing the level of resources, the activity times and the project completion time can be

varied. Thus, CPM assumes prior experience with similar projects from which relationships between

resources and job times are available. CPM then evaluates the trade-off between project costs and project

completion time.

CPM is mostly used in construction projects where one has prior experience in handling similar projects.

Although PERT and CPM were independently developed and come from different origins, the methods are

quite similar. In current practice, the words PERT and CPM are sometimes used interchangeably the

definitions and procedures that follow apply to both PERT and CPM.

Chapter (9): Project Planning lPMA Preparation Course

______________________R"""'ts::::,""''-_M_u_n_u_9_e_m_e_n_t_c_o_n_S_u_lt_;n_9_H_O_U_S-i~'

tTotal Float:

The Total Float TF of activity is the maximum time that activity can be delayed without causing delay in thefinal project completion. It is computed by:

Free Float:

Free Float is a measure of the maximum time activity may be delayed without affecting the start of the,successor activities. Free Float differs from total float in that it measures the time available without delayingsucceeding jobs. It is determined by: t •

An activity's free float can, of course, never be larger than its total float.

0 25

B 0

FS=5 I 0 250 0 10

A 0

0 10 10

FS= 10 0 25

C 0

5 25

r"-------------------------------..,LChapter (9): Project Planning IPMA Preparation Course

rt

Problem:


A Procure parts for subassembly 1 5 None

B Procure parts for subassembly 2 3 None

C Procure parts for subassembly 3 10 None

0 Build subassembly 1 7 A

E Build subassembly 2 10 B

F Build subassembly 4 5 D&E

G Build subassembly 3 9 B&C

H Final assembIy 4 F&G

1 Final inspection and test 2 H

Solution:

7 )_-=2:":("",,n,--,~G)

Activity on Arrow Network

A 5 0 5 2 7 2

B " 0 3 1 4 1~

C 10 0 10 0 10 0 *0 7 5 12 7 14 2

X 0 3 3 10 10 7

E 10 " 13 4 14 I~

F 5 13 18 14 19 1

G 9 10 19 10 19 0 *H 4 19 23 19 23 0 *1 2 23 25 23 25 0 *


Management Consulting Housef

Problem for Activity on Arrow:

Q Lead time 10

R Line available 30

A Measure and sketch Q 2

B Develop material list A 1

C Procure pipe B 30

D Procure valves B 45

E Prefabricate sections C 5

F Deactivate line R&B 1

G Erect scaffold B 2

H Remove old pipe and valves F&G 6

I Place new pipe H&E 6

J Weld pipe 1 2

K Place valves D&F&G 1

L Fit up pipe and valves K&J 1

M Pressure test L 1

N Insulate K&J 4

0 Remove scaffold L&N 1

P Cleanup M&O 1

Solution:

Q ..:.;.R -r

o

Project Network

-----------------------------------------1 ,

Chapter (9): Praject Planning IPMA Preparation Course


Activity - On - Node (AON):

The primary difference in this method is that activities are represented by nodes, with the arrows merelyrepresenting precedence relationships. In other words, time is consumed at the nodes and not the arrows there isno need at any time to create dummy activities. Any precedence relationship can be expressed correctly withoutthem.

The construction of the network is facilitated by adding two fictitious activities ("Start" & "Finish").

Problem for Activity - on - Node:

Q Lead time 10

R Line available 30

A Measure and sketch Q 2

B Develop material list A I

C Procure pipe B 30

D Procure valves B 45

E Prefabricate sections C 5

F Deactivate line R&B I

G Erect scaffoId B 2

H Remove old pipe and valves F&G 6

I Place new pipe H&E 6

J Weld pipe I 2

K Place valves D&F&G I

L Fit up pipe and valves K&J I

M Pressure test L I

N Insulate K&J 4

0 Remove scaffold L&N I

P Cleanup M&O I


(0)

Chapter (9): Project Planning


Activity - On - Node Representation


""~"~'-''''1

[

RS,S Management Consulting House

Chapter (10): Pert, Precedence,CPM & Crashing IPMA Preparation Course


Program Evaluation & Review technique (PERT):

Higher

Longer

Pessimistick

Beta Distribution

Possible Durations

Most Likely(Used in OriginaI~PM Calculations)

PERT Weighted Average =

~ Optimistic + 4 X Mos; Likely + Pessimistic )

Shorter

Lower

ProbabilityOf

Occurrence

Mean or Average Duration (!!l:

(a + 4 m + b)/1=

6

[(b - a)Z I

(6) 2

(f = (b - a)/6


Illustrative Example:

A (1-2) 2 5 8

B (2-4) A 6 9 12

C (2-3) A 6 7 8

D (4- 5) B,C I 4 7

E (2- 5) A 8 8 8

F (5 -6) D,E 5 13 17

G (3 - 6) C 3 12 21

H (6-7) F,G 3 6 9

I (7 - 8) H 5 8 II

5A

8

Project Network

Standard Deviation (er):

--------------------------------------LChapter (10): Pert, Precedence,CPM & Crashing IPMA Preparation Course


Solution:

A 5 1 1

B 9 I I

C 7 0.333 0.11

D 4 1 1

E 8 0 0

F 13 2 4

G 12 3 9

H 6 I I

1 8 I 1

A 5 0 5 0 5 0*

B 9 5 14 5 14 0*

C 7 5 12 7 14 2

D 4 14 18 14 18 0*

E 8 5 13 11 19 6

F 13 18 31 18 31 0*

G 12 12 24 19 31 7H 6 31 37 31 37 0*1 8 37 45 37 45 0"'

Critical Path = A, B, D, F, H, 1

Total Duration = 45 days

The Sum ofC.P Variances = Project Variances = 1 + 1 + 1 + 4 + 1 + 1 = 9

The Standard deviation for The Project = 3


Management Consulting HouseR,~__________________-'W -lf

Probability of Completing the Project:

In our example, T is normally distributed with mean 45 and standard deviation 3. For any normaldistribution, the probability that the random variable lies within one standard deviation from the mean is0.68. Hence, there is a 68 % chance that the project duration will be between 42 and 48 days. Similarly,there is a 99.7% chance that T will lie within three standard deviations (between 36 and 54).

1I··3a 11-2a II-a 11+ 3a

(T - ~)

Za -/Iz)

Za 0.00 0.01 0.02 ...................................

0.0 0.5000 0.4860 0.4920 ...................................

0.1 0.4602 0.4562 0.4522

0.2 0.4207 0.4168 0.4129

:

1 0.1587 0.1562 0, I539 ...................................

3 ,00135 .0 3 988 .03687 ...................................


Ifn

• Tabulalion or Ihe ,'alul:1 of a nnU11. for (he slandardilClf normal curve:.

, Itl u PC::> :,l .... f -,- ~~ll J J:.' 1.\277

"" ~HC3 under Ihe SI:lnd:udill:d normal cur\'c from t "" t. 10 : ... Cf.".

~y;~:::l

c§S~....~:::l

~....~.~

~

~

,

:, .0 .1 .2 .J .4 .l .6 I .7 .8 .9

------,- -"-

3 .oom .O}9R8 .0'687 .0'48) .01))1 .O'l)) .OJ159 .0 1108 .o~n) .04 481

4 ,0'317 .OJ:07 .0'lll .0 '85-1 .O'5W .0 13-10 .0'211 ,0'130 .0'193 .f1'n9

51.0":87 .O"l1n .((996 .01 579 .O'Jl3 .0'190 .0'107 .0'599 .O'~) ~ .0"826 ,0'981 .0"'.530 .O·::!52 .O'IH .O lf77i OIO~O'1 (PO'06 .Olol~ .O"!~3 .O"26C. -1'-

·.Ol5~1 mC:lns .oo5-11סס0.

z, .00 .01 .02 .Ol .IJ.I .05 .06 .07 .08 .09

- --------------------1.5 .0668 .0655 .0&43 .06l0 .0618 .0606 .0594 .0582 .om .om1.6 .0548 .OSJi .Ol26 .0516 .Ol05 .().l95 .0<85 .0<75 .().l6S .0<551.7 .0+16 .().ll6 .0427 .0<18 .().l09 .().j01 .Ol92 .om .om .Ol671.8 .om .om .0344 .0336 .0329 .0321 .0314 .0)07 .0lO/ .02941.9 .0287 .0281 .O:!1~ .0268 .0262 .0256 .0250 .0244 .O~)9 ,0133

2.0 .0228 .0222 ,0217 .0212 .0207 .0:02 .0197 .0/92 .om .0lBl2.1 .0119 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .01U

2.J .0139 ,0136 .om .0129 .011l fl121 .0119 .0116 .01 J.1 .0110

203 ,0107 .Ol().j .0102 .OO<J!XJ .0096< .OM]? .00914 .00889 .00166 .OOS~2

104 .00820 .00798 .00776 .00755 .oom .007/4 .00695 .00676 .00657 .006l9

1.5 .00621 .006Q.l .00587 .00570 .00554 .00539 .00513 .00508 .(l(~9~ .~80

2.11 .~66 .~5l .00·l-l0 .(X}..I27 .0041~ .(x}'I0~ .(0)91 .00379 .OOl68 .OOll7

1.7 .OOJ4~ .00Jl6 .00326 .oom .00lOl .00298 .00289 .00280 .oo~n ,O02~

2,8 .(Xl2Sf, .00248 .00240 .t1:~J~ .00226 .OO~19 ,00212 .00.105 .00199 .0019J2.9 .00187 .0018/ .00175 .00169 .OOl&-! ,00159 .OOI.5~ .001<9 .OOloW .001l9

hblll A·1 (COM. I

o :0

Norm.1 dis~ribution [1]-

./"

-, .00 .01 .02 ·.Ol .04 .05 .06 .07 .08 .09- --------------------0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4121 .4681 .46<10.1 .4602 .4562 ,·U22 .448l .444l .4404 .4l64 .4l25 .4286 .42470.2 .4107 04168 041)9 ,4090 .4Ol2 AOIl .J974 .J9l6 .J897 .l819O.l .Jill .l783 .JW .J707 .3669 .J631 .Jl94 .Jll7 .Jl20 .J4830.4 .3l-l6 .l409 .JJ72 .JJJ6 .llOO .l264 .m8 .ll92 .JJl6 .)121

u.S .J08l .lOlO .lOll .2981 .2946 .2912 .2877 .284l .2810 .21760,6 .274l .2709 .2616 .264l .2611 .2578 .2546 .2l14 .248l .24510.7 .2420 .~J89 .2l58 .2327 ;22961 .2266 .22l6 .2206 .2117 .21480.8 .2119 .2090 .2061 '.103l '~200l .1977 .1949 .1922 .1894 .18670.9 .1841 .1814 .1788 .1762 .17J6 .1711 .1685 .1660 .16l5 .1611

1.0 .ll87 .1S62 .l5l9 .llll .1492 .1469 .1446 .124l .1401 .ll791.1 .1ll7 .1Jl5 .lll< .1~92 .1271 .1251 .12l0 .1210 .1190 .11701.2 .I III .III I .I I12 .109l .107l .1036 .IOl8 .1010 .100l .098lI.J .0968 .09ll .0934 .0918 .0901 .0885 .0869 .085J .om .08231.4 .0808 .om .0778 .0764 .0749 .0)l5 .0721 .0708 .0694 .0681

r.bl. A.1

9{j~;:;,~~,::.

",.lti~,lti

"~""[:;'

IQ

;;~

~~g'~

"i;l'1\

Calculation ofProject Standard Deviation

1. Calculate Variances for each Critical Path activity (cr2)

36

2. Project Variances =:E rl for Critical Activities.

3. Project Standard deviation = (J = ~ 2:E cr C.P.A

----------------------------------------',LChapter (lO): Pert, Precedence,CPM & Crashing IPMA Preparation Course


For the Previous Example:

);> What is the Probability that the Project will be completed within 50 days?

);> What is the Completion time having a probability of 80 %?

Za = (T - ~) / () = (50 - 45) / 3 = 5/3 = 1.666

From the Chart: area under the curve = .0485

1 - a = 1- 0.0485 =0.9515

Probability for Completing within 50 days = 95 %

80% ---+ 1 - a = 0.8 ---+ a = 0.2

---+ Za =0.84 = (T-~) / (J = (T - 45)/3

---+ T = 2.52 + 45 = 47.52 days '" 48 days.


______________________R--'-t5_;_'_M_a_n_a_y_e_m_e_n_t_c_o_n_S_U_l_t_in_g_H_O_U_S-:~..l

Example:

I • .• •.I ~~.. IlLY I 0

1-2 6

1-3 8

2-3 2

2-4 6

3-4 5

3-5 5

4-6 4

5-6 2.5

2

Mod T ;L'nh, • • "

7 14

10 12

3 4

7 85.5 9

7 9

6 8

3 3.5

The table above gives the optimistic, most likely and pessimistic duration for each activity in theactivity- on- arrow diagram.

Required:

(a) Compute the critical path.(b) What is the probability that the project will be completed within 25 days?(c) What is the completion time having probability of73%?


RS,S Management Consulting House

Solution:

fII

A 6 7 14 8 1.78 *

B 8 10 12 10

C 2 0 4 0 0.11 *.J .J

D 6 7 8 7

E 5 5.5 9 6 0.44 *

F 5 7 9 7

G 4 6 8 6 0.44 *H 2.5 3 3.5 3

• The Critical Path:C.P =

• Project Duration = 8 + 3 +6 + 6 = 23 days

A,C,E,G

Project Standard Deviation = ( V ) 0.5 = 1.667

For T = 25 Za = (25 - 23) / (J = 2/ 1.667 = 1.2

FromtheChartZ a = 1.2 --> a = 0.1151 (areaunderthecurve)

Probability to finish = remaining area = (l - a ) x 100 = 88.5%

• Probability to finish 73% means 1 - a = 0.73

a = 0.27 from the chart

Z a 0.61 = (T - 23) /1.667 ---+ T = 24 days

Chapter lID): Pert, Precedence,CPM & Crashing IPMA Preparation Course

RS Management Consulting House------------------------I[

Precedence Diagram Method (PDM):

This is a Method of Constructing a Project network diagram that uses boxes to represent the activities andconnects them with arrows that show the dependencies.

ES I TF I EF

Activity Title

LS IDuration I LF

1) Finish to Start:

• Activity A must finish before Activity B can start, the initiation of the work of the successor dependsupon the completion of the work ofthe predecessor.

• A LeadsBbyx• The Gap between B & A = x

A

B

FS= x

2) Start to Start:• Activity A must start before Activity B can start, the initiation of the work ofthe successor depends

upon the initiation of the work of the predecessor• There are overlap between A & B = Y• SS = x = (Duration of A) - Y

A



3) Finish to Finish:• Activity A must finish before Activity B can finish, the completion of the work of the successor

depends upon the completion of the work of the predecessor.

A

B

4) Start to Finish:

• Activity A must start before Activity B can finish, the completion of the successor is dependent uponthe initiation of the predecessor.

A

BI

SF


R$ Management Consulting HOUS~

------------------------------------------11(

Example:

Figure-2 shows the activity-an-node network and the precedence relationships for a project. Find thecritical path by calculating ES, EF, LS, and LF for each activity.

FS=OD12

SS=2 FS=5A

FF=3 B E10 8 5

SS, SF=2,7

SS=8F

C 43

SS 6


Solution:

o

o

SS, SF=2,7


..FS=O .. .. .. 1---.--,..--.\....

5 0 13 ...-1--'-_-'--(.,;' .. "..-

FS=5

SS=6

• The Critical Path:A,B,D

• Project Duration = 25 days


Example:

Complete the forward and backward path calculations on the precedence diagram given below.Assume that the project statts at T=O, and that we have scheduled project duration of 30 days.Also, assume that no splitting of activates is allowed.

Il ;

D FTest & Debug SFIS Document

program program6 12

FSO

A BObserve SS3 Write ;,;omputer

system cases Program8 FF4 12

SS 0 CCollect system E

cases FS 0 Run Program

4 6

Required:

(a) Give detailed forward and backward path calculations.

(b) Show ES, EF, LS, LF and the critical path on the diagram.

----------------------------------LChapter (10): Pert, Precedence,CPM & Crashing IPMA Preparation Course


Solution:

1) Forward and Backward Calculations:

A 8 0 8 0 8 0

B 12 3 15 3 15 0

C 4 3 7 20 24 11

D 6 15 21 15 21 0

E 6 21 27 24 30 3

F 12 18 30 18 30 0

2) The Critical Path on The Diagram:

15 21 18 30

D FTest & Debug SF15 Document

program program6 12

0 8 3 15 15 18 30

A BObserve SS3 Write computer

system cases ------- Program8 FF4 12

0 8 3 15 3 7 27

SSO CCollect system E

cases FS 0 Run Program

4 6

20 24 24 30


RSS Management Consulting Hous~_

-------------------------"-----------------11

Pro;ect Crashing:

It is often true that the perfonnance of some or all project activities can be accelerated by the allocation ofmore resources at the expense of the higher activity direct cost. When this is so, there are many differentcombinations of activity durations that will yield some desired schedule duration. However, eachcombination may yield a different value of total project cost. Time/cost trade-off procedures are directedat detennining the least-cost schedule for any given project duration.

For example, consider the simple eight-activity project shown in Table. Each activity can be performed atdifferent durations ranging from an upper "normal" value, at some associated "nonnal" cost, down to alower, "crash" value, with an associated higher cost. Note that if time/cost trade-off values for eachactivity are assumed to be linear, the cost of intermediate activity durations between the normal and crashdurations is easily determined from the single cost "slope" value for each activity [c..g., the cost ofperforming activity (0,2) in 7 days instead of 8 equals $400 + $80, or $480].

1\.T, •nucmaln....u ...]

~. ., I Cost I Time (days) Cost Slopeume ~uay~}

(0, I) 4 $ 210 3 $280 $ 70

(0,2) 8 400 6 560 80

(1,2) 6 500 4 600 50

(1,4) 9 540 7 600 30

(2,3) 4 500 I 1100 200

(2,4) 5 150 4 240 90

(3, 5) 3 ISO 3 ISO -'

(4,5) 7 600 6 750 ISO

$ 3050Table: Illustrative Network With Activity Time/Cost Data.

$ 4280

a This activity cannot be expedited

Solutions:

1'Z

223

,IS, , ,IS " 22

7', , ,22',

144 19

9 IS13

//

1://

/

// 5/

10// 5

64

Chapter (10): Pert, Precedence,CPM & Crashing IPMA Preparatian Caurse


l.,.tp<;fII A "

~

~1" ..1r'ILJ' •~.."ll

ES EF LS LF(0, 1) 4 0 4 0 4 0 *(0,2) 8 0 8 2 10 2

(1,2) 6 4 10 4 10 0 *(1,4) 9 4 13 6 15 2

(2,3) 4 10 14 15 19 5

(2,4) 5 10 15 10 15 0 *(3,5) 3 14 17 19 22 5

(4,5) 7 15 22 15 22 0 *Total Duration = 22

Critical Path = (0 - 1), (1- 2), (2 - 4) & (4 - 5)

Project Duration = 4 + 6 + 5 + 7 = 22 days

Project Paths = C. P +

2nd Path (0 -1), (l - 4) & (4 - 5)

3'd Path (0 - 2), (2 - 3) & (3 - 5)

4th Path (0 - 2), (2 - 4) & (4 - 5)

=4+9+7

=8+4+3

=8+5+7

= 20 days

= 15 days

= 20 days

sth path (0-1),(l-2),(2-3)&(3-5) =4 + 6 + 4 + 3 =17days

Min Project duration = C.P duration after Crashing = 17 days

For C. P Extra Cost = 70 + 100 + 90 + 150 = 410 $

2nd Path after Crashing C. P = 3 + 9 + 6 = 18 days

4th Path after Crashing C. P = 8 + 4 + 6 = 18 days

So we need another 1 days to be Crashed from those Paths, actually from Activity (l - 4) & (0 - 2)

Total Crashed Cost = 410 + 30 + 80 = 520 $

Project Total Cost to reduce the overall duration to be 17 days = 3050 + 520 = 3570 $


RSS Management Consulting House----------------------=..:::-.:~------------_ll

Example:

The table given below shows the duration's and direct costs, for each activity in the network under bothnonnal and crash conditions.

Required:

>- Compute the early start, early finish, late start, late finish and the float for each activity.>- Establish the minimum direct cost for the project if it's required to reduce the overall duration by two

weeks.

I . ~~ . . . .:... shA

'w....Q, L.rasn'J n",.~ J. mal UlreC[ cost 0 ... Total :cost

F' (L. E.) (L. E.)A 1~2 4 500 3 750B 2~3 4 100 2 300C 2~4 2 200 2 200D 2~5 5 600 4 760E 3-6 6 700 5 830F 4-8 4 200 3 300G 5-7 7 140 5 200H 6-9 4 200 2 300I 7-8 2 80 2 80J 8-9 I 100 I 100K 9-10 7 600 6 670

2

IPMA Preparation CourseChapter (10): Pert, Precedence,CPM & Crashing

f

---------------------------------------l' .L"


Solution:

A Earlit:st Latest'UJI ES EF LS LF

A 0 4 0 4 0 * 4 0 500 750.)

B 4 8 5 9 1 4 2 100 300

C 4 6 12 14 8 2 2 200 200

D 4 9 4 9 0 * 5 4 600 760

E 8 14 9 15 1 6 5 700 830

F 6;

10 14 18 8 4 3 200 300

G 9 16 9 16 0 * 7 5 140 200

H 14 18 15 19 1 4 2 200 300

I 16 18 16 18 0 * 2 2 80 80

J 18 19 18 19 0 * 1 1 100 100

K 19 26 19 26 0 * 7 6 600 670

Project Paths

c.p = A,D,G,l,J,K= 4+5+7+2+1+7 = 26 Weeks

2'nd Path [A,C,F,J,K] = 4+2+4+1+7= 18 Weeks

3'rd Path [A,B,E,H,K] = 4+4+6+4+7= 25 Weeks

To reduce the overall duration by 2 weeks it will become 24 weeks

so, we need 2 weeks from C.P and I week from 3'rd path.

the lowest cost for crashing C.P is activity G [I week=30LE], then k [1 week = 70LE]

So, we take K (as it is included in the two paths}+ 1 week from G

So, the extra crashing cost = 70 + 30 = 100 LE

So, total direct cost for 24 weeks duration

= [500 + 100 + 200 + 600 + 700 + 200 + 140 + 200 + 80 + 100 + 600 ] + 100 = 3520 LE


I

1Management Consulting House

!"".,

Example:

Data have been established for the following set of activities:

.. ,

A "taays)

..JI r •

. , " " ip~..~.. 'UJ'- . Crash A

NUllU.ll I 'r"~h 1 'UIllHll .

A 7 4 $ 95 $ 100 None

B 6 3 90 97 None

C 5 4 86 104 None

D 7 7 92 98 A&C

E 6 5 87 93 A,B&C

F 7 5 112 120 B

G 8 5 101 113 F

H 9 6 97 109 K,E&D

I 12 10 95 100 A&C

J 10 7 100 110 I

K 9 8 105 114 F

(a) Draw the CPM node methods.(b) Find the Critical Path using the activity - on - node representation.(c) Find the Critical Path using the activity - on - arrow representation.(d) Crash the network to minimum - time duration along a minimum - cost schedule. Plot the results.


Solution:


Because F > B only 8

H9 x

7A

,,,,,,, B,,

6',, ,,------

F

------7

G8

H

9


Jq>5~ Management Consulting House__________________n.....:::...~=__ _jr'

Project Paths:

C, I, J 5 + 12 + 10 = 27

A, I, J = 7 + 12 + 10 = 29

C, D, H = 5 + 7 + 9 = 21

A, D, H = 7 + 7 + 9 = 23

C, E, H = 5 + 6 + 9 = 20

A,E,H=7+6+9=22

B, E, H = 6 + 6 + 9 = 21

B, F, G = 6 + 7 + 8 = 21

B, F, K, H = 6 + 7 + 9 + 9 = 31

Activity Duration ES EF LS LF TF

A 7 0 7 2 9 2

B 6 0 6 0 6 0

C 5 0 5 4 9 4

D 7 7 14 15 22 8

E 6 7 13 16 22 9

1 12 7 19 9 21 2

F 7 6 13 6 13 0

G 8 13 21 23 31 10

H 9 22 31 22 31 0

J 10 19 29 21 31 2

C.P. = B,F,K,H = 31 days

Project Duration after max. Crashing = 22 days

Crashing the Critical Path Activity only first to get the min. Cost

Project Cost = 1060 $ after Crashing C.P.

Extra Cost = 7 + 8 + 12 + 9 = 36 $ for c.p

Chapter (10): Pert, Precedence,CPM & Crashing /PMA Preparation Course

Project Paths before Crashing:

2nd Path = C, I, J = 27 days

3rd Path = A 1 J = 29 days, ,

4th Path = A, D, H = 23 days

5th Path = A, E, H = 22 days .y

6th Path = B, E, H = 21 days .y

7'h Path =B, F, G=21 days .y

8th Path = C, D, H = 21 days .y

9th Path = C E H = 20 days .y, ,

As "A" is the lowest Crashing Cost,


after Crashing the c.p:

27 need 5 days

29 need 7 days

23 need I day

We Crash 3rd Path (A, I, J) by taking 3 days from A, 2 days from 1 plus only 2 days from J to reach 22days.

But in this case still 2nd Path will be 23 days as we deduct 2 days from 1 & 2 days from J 27 - 2 - 2 = 23day

So we have to Crash another day from J as the Lowest case.

So the Extra Cost = 10 $ from J + 5 $ from 1 + (5 x 2)/3 $ from A = 20 $

Total Cost after Crashing to reach 22 days = 1060 + 36 + 20 = 1116 $

Chapter (10): Pert, Precedence,CPM & Crashing /PMA Preparation Course

EXERCISES

Ex. (1);

A building contractor is trying to plan construction activities for a custom home. The sequence ofactivities, precedence relationships, and time durations are given in the table, Construct the appropriateCPMJPERT diagram for the sequence of activities using both (a) activity-on-arc and (b) activity-on-nodemethods.

0

T ... 0 .'OoL I Time" -- "J' • ~ , .. .'• • '" IlJaYS1

A Start - 0

B Excavate and pour footings A 4

C Erect wooden frame, including rough roof B 2

D Pour concrete foundation C 4

E Lay brickwork D 6

F Install basement drains and plumbing C I

G Pour basement floor F 2

H Install rough plumbing F 3

I Install rough wiring D 2

J Install heating and ventilating D&G 4

K Fasten plaster board and plaster (including drying) I&J &H 10

L Lay finish flooring K 3

M Install kitchen fixtures L I

N Install finish plumbing L 2

0 Finish carpentry L 3

P Finish roofing and flashing E 2

Q Fasten gutters and downspouts P I

R Lay storm drains for rain water C 1

S Sand and varnish flooring O&T 2

T Paint M&N 3

U Finish electrical work T I

V Finish grading Q&R 2

W Pour walks and complete landscaping V 5

x Finish S&U&W 0


R'SS Management Consulting House

Ex. (2):

Consider a project to promote a new product. The activity durations to complete the project are given inthe table. Find the minimum total time to complete the project.

Time•......J ,«nVll

I n ..."c.

A Lead-lime planning 3

B Develop training plan 6 A

C Select trainees 4 A

D Draft brochure 3 A

E Conduct training course I B,C,D

F Deliver sample products 4 A

G Print brochure 5 D

H Prepare advertising 5 A

M Release advertising I H

N Distribute brochure 2 G

Ex. (3):

A project consists of the activities shown in the table. The durations represent the optimistic (a), mostlikely (m). and pessimistic (b) times (days) for each activity. Find the critical path for this PERTNetwork.

,A

,. -J-;

".

(1,2) 5 8 10

(1,3) 18 20 22

(1,4) 26 33 40

(2,5) 16 18 20

(2,6) 15 20 25

(3,6) 6 9 12

(4,7) 7 10 12

(5,7) 5 7 8

(6, 7) 3 4 5


______________________RS-"'_,=--'_M_a_n_a_g_em_e_n_t_C_o_n_S_u_l_t_in_g_H_O_U_S-l~ ..

Ex. (4):

The possible activity durations of a certain project are given in the table. These figures representoptimistic (a), most likely (m), and pessimistic (b) times respectively. The scheduled completion time is17.5 days. Find the probability' of completing the project in the scheduled time.

"A

lJuranon . -,._--- --,

ba

(1,2) 6 8 10

(1,3) 4 6 7

(1,4) 4 8 12

(2,5) 5 6 8

(3,5) 7 8 9

(4,6) 7 10 14

(5,6) 3 4 5

Ex. (5):

Consider the activity network shown. Assume that the latest allowable project completion date is set at 49days. (Ts(L) #- Ts(E).] Find the critical path for this network.

23

35

210

114

4

2

o

(-----------------------------------------LChapter (10): Pert, Precedence,CPM & Crashing IPMA Preparation Course

[peoplJe1 -1

Ex. (6):


Consider a project to promote a new product. The probabilistic time durations to complete the activitiesare given in the table.

..~ ",-'

"; A. .vuy VI Most Likely ,ioti..~~'''''J •

a m bNone 0 Lead-time planning 2 3 5

0 1 Develop training plan 2 6 10

1 2 Select trainees 3 4 5

0 3 Draft brochure 1 3 4

9,10,5,8 4 Conduct training course I 1 1

13 5 Deliver sample products 3 4 4

3 6 Print brochure 4 5 6

3 7 Prepare advertising 2 5 7

7 8 Release advertising 1 1 1

6 9 Distribute brochure 2 2 3

6,2 10 Train sales force 3 5 6

4 11 Review market survey 2 4 5

0 12 Develop prototype product 5 7 8

12 13 Manufacture sample products 2 3 4

(a) Compute the mean and variance for each activity.(b) Compute the critical path using the activity-an-node representation.(c) Compute the critical path using the activity-an-arc representation.(d) What is the probability that the network will be realized in less than 30 days? 40 days? 50 days?(e) Detennine the number of project days for which the probability of exceeding this duration is only

100.



Solution of Exercise (1):

6 12017 19 3 pI

R 12 V 0

261 1 p7 21 2~~9~ 211 Spow 8

101 8 116 161 8 ~8 1818 ~9 2915 P4 1\

E~ f~'1-4 Q~,."",,", 1/ f~

,,'

o 10 14 41 016 6 I0 110 181 6 p4 . 2~ 2 6 2611 ~1B 0 ,. (' 0

DO~o 14 14 41 2 16 61 4 110 "" . 10 2 112 J2]1133 1\

2711 ~" I 2

~iI I U I,,12 2 114 3~ 1134 ,,

2~ I P9,,,,, ,6 II 7 7 I 19 1\ 101 0 14 1\ 141 0 124 24 0 j27 271 0 129 29f 0132

~F 0 C.:. II-- J 0 K 0 L 0 L.,l. N 01--. T 0 Xo7 I I 8 8 2 po 101 4 14 14110 124 24 3 127 2712129 29f 3~2 304,

/,

\,,

71 3 10 \ 2112 30 \ 321034

\ H 3[.1 o 2 SOl;

11 13 If 291 3 j32 321 2 34

~o

• A 01o 0

Critical Path = A, B, C, D, J, K, L, N, T & SProject Duration = 0 + 4 + 2 + 4 + 4 + 10 + 3 + 2 + 3 + 2 = 34 days


~"=.,"<'\.


o. _._ ,..6 10 1110130 6 11

I D 0--- __________s-

G 0 --------- -- N 001313['\.'

6 I 5 11 I 2 133 6 11

Solution of Exercise l1.l:.

3 3 9

" B 0

6 6 12

5 73 10c: 2

4E 3

1212 1 13

~F 64 13

4 8 181 41 9

H 0

7 lECritical Pa th = A, D, G & NProject Duration = 3 + 3 + 5 + 2 = 13 days

Chapter (10): Pert, Precedence,CPM & Crashing IPMA Preparatian Course


A 1(1,2)1 5 1 8 1 10 1 8--

B (1,3 ) 18 20 22 20C (1,4 ) 26 33 40 33

D (2,5) 16 18 20 18

E I (2,6) 15 20 25 20

F I (3,6) 6 9 12 9

G 1(4,7) 7 10 12 10

H 1(5,7) 5 7 8 7 8 D 26>

M 1(6,7) 3 4 5 4 8 18 18 36A 8

36~18 19 E0/ 8 20 18

7 33_c 410

~900 B 2(~0)20 F 29 1\1 33, 10 20 9

-~ 639 4 43, , 30 . 30 39,

" 0" 43,

.... , (1,- 33 G _--, ,0 , 33, , _--<n,

33, , _------ 10, , , -- 33

3~'·- -~



A I '~ri~n.<"1~" "'J ~ ~

(1-2) A 6 8 10 8* 16/36 = 0.44

(I -3) B 4 6 7 6 9/36 = .25

(1-4) C 4 8 12 8° 64/36= 1.77

(2-5) D 5 6 8 6* 9/36~O.25

(3-5) E 7 8 9 8 8/36 = 0.22

(4-6) F 7 10 14 10° 49/36 = 1.36

(5-6) G 3 4 5 4* 4/36 =0.11


6

18F _~810

S

S

s

..,

E

6

6 ..,c•.-

8 D 14__ - __ n n -3::S 6 14

S --"I. (. ", .. I'..... -,,-o c~·-c'.....0__ {l-

---0~---O------6------6-

'" 0'" Co "'" 8

8 "'"~h

1Project Variance =:E the Variances for C. P. Activities

= 0.44+0.25+1.77+0.25+0.22+1.36+ 0.11 = 4.4

Critical Path = A & 0 & G

Critical Path = B & E & G

Critical Path = C & F

Project (j = 2.1

Z. = (17.5 - 18) / 2.1 = - 0.5 /2.1 = - 0.238

a. = 0.3669 = area under the curve not (I - a.)

Probaility for 17.5 = 36.7%



Solution of Exercise ill.30

3 \ 15 2 17 J 645 47

22 3~rlfln X ",,29 183 / kOJ .-< N ~I~ 47"" 2

<:>""or, 29 12 15 10 ~lfl

2 2 ""'- 31o / 47 25 4922 0 0 10 N

.... N "'"8 39 21 24

10 10 2 10 1 5 24 4 28---------------18 31 45 49

I ..... I "o I ........ I ,,-)0004, )-ol. ...... I .. '00. 0 ..... WI N .-

I ..... Nl .s;.. 41 "25" 25 ' '-'~ I I 8.-

4 ' "/,,'::00 27 '-lOO 10,:/'49, ' ,I 2 I ,"

, ' 31', , 37 12 ' "4 N, N 21." ("I, ("I ,\Ow .... ' '"".... , /"

29" ",," 29 17 39 "Y 394 21 1 22 7 '

38 39

C.P = (1-2), (2-4), (4-5), (5-7) & (7-8)C.P = 8 min float




Critical PathProject Variance (a')

0, 1, 2, 10, 4, 11 = 23 days0.25 + 1.77 + 0.111 + 0.25 + 0.25 2.631 & (J 1.622

6 12 tl -- 1l I5 13

6 0 9 5

g I ~ 13 1612 18~

312 6

3 0 6 I6 11 II 16 12

5 I3 & 7 0 8 6

12 I 5 17 17 I 1 1& r-....:::---<

Io 10 3

3 I0 9 9 10 13

o 0\ I °-2 0 ____

01 33 3169 91 4 13



A"uv"J' D. ,I IeI ~~~n'lq ~,IA I .eviatio~

I -"I

0 None 2 3 5 3 0.5 0.25

1 0 2 6 10 6 1.33 1.77

2 1 3 4 5 4 0.333 0.111

3 0 1 3 4 3 0.5 0.25

4 9,10,5,8 I I I I 0 0

5 13 3 4 4 4 0.1667 0.0278

6 3 4 5 6 5 0.33 O. II

7 3 2 5 7 5 0.833 0.694

8 7 I I I 1 0 0

9 6 2 2 3 2 0.1667 .0278

10 6, 2 3 5 6 5 0.5 0.25

I I 4 2 4 5 4 0.5 0.25

12 0 5 7 8 7 0.5 0.25

13 12 2 3 4 3 0.33 O. III

Chapter (10): Pert, Precedence,CPM & Crashing

Project Standard Deviation = 1.62

Note:

.T - fl > + 3 (J ~ Probability = 100%

.T - J.l < - 3 (J ~ Probability = 0%

For 30 days:

Zu = (30 - 23) / 1.62 = 4.32 > 3

ProbabilitY' 100%

Also for 40 & 50 days ProbabiFity100%

e) Probability'" 90% -t I - a = 0.9

a = 0.1 -t Z u = 1.285

1.285 = (T - 23) / 1.62

T = 25 days

IPMA Preparatian Course


Chapter (11): Course & Resource Management IPMA Preparation Course


Cost & Resources Management

Type of Cost:

• Fixed Cost

• Indirect Cost

• Dry Cost

SeIling Price:

•

•

• Variable Cost

Controllable Cost

Time Related cost

• Direct Cost

• Uncontrollable cost

• Manufacturing Cost

Mar -up or Mar ·u

Dry ost

Selling Price

SeIling price = Dry Cost + Mark-up

Dry Cost

Mark-up

Material cost + Man power+ Machinery cost

= Site O.H + Office O.H M+ Contingency M+ Cost ofInvestment (InterestM M

cost) + Profit

; M : Management Decision



Project Cost Management:

Planning Controlling

Processes required to ensure the project is completed within tile approved budget

Resource Planning

Cost Estimating

Cost Budgeting

Cost ControlResource Planning:

Determining physical resources needed (i.e., material, equipment, and people) and what quantities of eachshould be used and when they would be needed to pelform project activities

Inputs:

• Work breakdown strncture (WBS) - Identifies the project elements that require resources• Historical information - Used to identify the types of resources that were required for similar work

on previous projects• Scope statement - Contains proj ect justification and the project objectives, which need to be

considered• Resource pool description - Description of resources available, if necessary, to work on a project

f------------------------------------------1't, ..u'

Chapter (11): Caurse & Resaurce Management IPMA Preparation Course


• Organizational policies - Of the performing organization, regarding staffing and the rental andpurchase of supplies and equipment

• Activity duration estimates

Tools & Techniques:

• Expert jndgment - Expertise, provided by any group or individual, used to assess the inputs to thisprocess

}> Other units within organization}> Consultants}> Professional and technical associations}> Industry groups

• Alternatives identification• PM software - Used to organize resource pools, define resource availabilities and rates, and defines

resource calendars

Outputs:

• Resource requirements - Description of the types (e.g., skill levels) and numbers of resources required byeach element of the WBS

Cost Estimating:

Process ofdeveloping an approximation (or estimate) for the cost ofthe resources necessary to completethe project activities.

• Difference between cost estimating and pricing:}> Cost estimating: Assessing how much it will cost the organization to provide the

product or service}> Pricing: Assessing how much the organization will charge for the product or service

• Cost estimating also includes identifying and considering cost alternatives

Inputs:

• WBS - Used to organize the cost estimates and to ensure that the cost of all identified work has beenestimated

• Resource requirements• Resource rates - Unit rates for each resource• Activity duration estimates - Affects cost estimates ifproject budget includes an allowance for the cost of

financing (i.e., interest)• Chart of accounts - Coding structure used by the organization to report financial information. Cost

estimates must be assigned to the correct accounting category.• Historical information - Information on the cost of resources

}> Project files - Records of previous project results that are detailed enough to aid indeveloping cost estimates

}> Commercial cost-estimating databases - Historical information availablecommercially

}> Project team knowledge

Chapter (11): Course &Resource Management IPMA Preparation Course

RS Management Consulting Hous~

• Estimating pnblications - Commercially available data on cost estimating• Risks - Since risks can have a significant impact on cost, the effect of risk on the cost estimates for each

activity must be considered

Tools & Techniques:

Analogous estimating (Expert judgment) - Used to estimate total project costs if there is a limited amount ofdetailed infonnationParametric modeling - Using project characteristics (or parameters) in a mathematical model to predict costs(e.g., price per square foot)Bottom-up estimating - Estimating the cost of individual work items and then rolling up the costs to arrive ata project totalComputerized tools - PM software and spreadsheets

Outputs:

• Cost estimates - Quantitative assessments of the cost of resources (e.g., units of currency or staff hours).• Types of estimates:

>- Order of magnitude (-25% / +75%)>- Budgetestimate (-10%/+25%)>- Definitive estimate (-5% / +10%)

• Supporting detail>- Description of estimated scope of work>- Documentation of the basis for the estimate>- Documentation of any assumptions made>- Range of possible results

• Cost management plan>- Describes how cost variances will be managed>- Part of the overall project plan

------------------------------------------LChapter (11): Course & Resource Management IPMA Preparation Course

RS1S Management Consulting House

Cost Budgeting:Allocating the overall cost estimate to individual activities or work packages, in order to establish a costbaseline for measuringproject pe/jormance

Inputs:

• Cost estimates• WBS - Identifies the project elements to which the costs will be allocated• Project schedule - Used to assign costs to project elements for the time period when costs will be incurred• Risk management plan - Often includes cost contingency, which can be determined on the basis of the

expected accuracy of the estimate

Tools & Techniques:

• Cost budgeting tools and techniques

Example for Resource Loading (Cost Budgeting Tools):

Bar chart: 140 r120-

100

Data DateS-curve:

ACWP

80

~ 60d

14 16 18 20 22 2410 12

Week

40

20~0"""""1

2 4 6: 810 12 14 16 18 20 22 24

Week86

k

4

14 k

2o

4

Histogram:

j"i;j10

6k5

o o 2 4 6 8 10 14

Week22

\V ••k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Cost 0 6 14 8 8 8 8 10 2 2 2 2 2 6 4 4 4 4 7 3 3 3 3

Total cost 0 6 12 26 34 42 50 58 68 70 72 74 76 78 84 88 92 96 100 107 110 113 116

Chapter (11): Course & Resource Management /PMA Preparation Course

Resources or Cost Leveling (smoothing):

R:S Management Consulting Hous~

RsS Management Consulting House

Outputs:

• Cost baseline - Time phased budget that will be used to measure and monitor the cost performance of theproject,-------------------------,

140

120

100~e 80

~ 600III

40

20

0Jan Feb Mar Apr May Jun

Cost Control:

Reporting Period

• Influencing the factors that create changes to the cost baseline to ensure that changes are agreed upon• Determining that the cost baseline has changed• Managing the actual changes when and as they occur

• Includes:~ Monitoring cost performance to detect variances from the plan>- Ensuring that all appropriate changes are recorded~ Preventing incorrect, inappropriate, or unauthorized changes~ Infonning the appropriate stakeholders of authorized changes~ Analyzing positive and negative variances and how they affect the other control processes

Inputs:

• Cost baseline• Performance reports - Infonnation on cost performance• Change requests - Oral or written requests to change the budget• Cost management plan

Tools & Techniques:

• Cost change control system - Procedures used for making changes to the cost baseline• Paperwork• Tracking system• Approval levels

• Performance measurement - Assessment of the magnitude of any variation, the cause of the variance, andthe corrective action needed

• Additional planning - Changes may require new or revised cost estimates or an analysis of alternativeapproaches

Chapter (11): Course & Resource Management lPMA Preparation Course

RSS Management Consulting Hou~:

• Computerized tools - Used for tracking the value ofplanned costs compared to actual costs and forpredicting the effects of cost changes r

• Earned Value Management - Continuously measuring project performance by relating the planned value,1the earned value for work accomplished, and the actual costs incurred

Performance MeasurementCost

A

...Time

EAC

i i

Projected SchedlIlei Slippage !

(At Completion)! • • '

CV

,I

-----------------------------~]-----I ... _ - I

I ~~ II ... I

I ... '" I UCI'" II ........ BAC I

i "..,..,..,.;- ~------_r:--------; -------I II II II II II II II II II II II II II I

ETC

ScheduledSlippage

(To dates)

Key:

,,,,,,Data'Date

• BCWS (PY): Budgeted Cost for Work Scuduled• BCWP (EY) : Budgeted Cost for Work Performed• ACWP (AC) : Actual Cost of Work Performed• SV: Schedule Variance• CV: Cost Variance• ETC: Estimate to Complete• BAC: Budget at Completion• EAC: Estimate at Completion (LRE- Last Revised Estimate)• VAC: Variance at Completion

Data Date• Slippage = ----- - Data Date

SPI• Estimate Duration At Completion (EDAC) = Project Duration I SPI• CV=EV -AC• SV=EV -PV• CPI = EV/AC• SPI = EV/PV

r"-------------------------------iL,;Chapter (11): Course & Resource Management IPMA Preparation Course


% of Work scheduled:

• Project Duration = 12 Months.• Data Date or (time now) = 6 Months.• Where we are now???• Are we 50% work scheduled!!!

100%

90%

80%

100%

BCWS

30%

15%

BCWS

Time Now Time Now

BCW~S B_AC

PV

Time Now

PV0/0 of Work Scheduled =

BAC


RSS Management Consulting Hous~I

% of Budget Spent:

t

AC 1-----7f'

PV I<-------,,.e-.,,..r

__ -EACACWe---....

BAC

Time Now

AC0/0 of Budget Spent =

BAC

% of Work Accomplished:

BAC

__ -EACACWP .... --............

AC 1-----::>1"

PV I------"L.-..::".......

EV I--~",,::;---:::::....

Time Now

EV0/0 of Work Accomplished =

BAC r'------------------------------------\LChapter (11): Course &Resource Management lPMA Preparation Course


Case (1):EAC = AC + ETC

Case (2):

EAC AC + BAC - EVCase (3):

EAC = AC + (BAC - EV) / CPI= AC + (BAC / CPI) - (EV / CPI)= AC + BAC/CPI- AC= BAC/ CPI

Outputs:

• Revised cost estimates• Budget updates• Corrective action• Estimate at completion - Forecast of most likely total project costs based on project performance and risk

quantification• Project closeout - Processes and procedures for closing or ending a project• Lessons learned - Documented causes of variances

Earned Value (EV):

A method ofmeasuringproject pelformance by comparing the amount ofwork planned with that actuallyaccomplished, in order to determine ifcost and schedule pelformance are as planned

EVTerms:

n"t" I'" It term Term~

Scheduled Work Planned Value (PV)Budgeted Cost Of

Work Scheduled (BCWS)

Work Accomplished Earned Value (EV)Budgeted Cost of

Work Performed (BCWP)Actual Cost of Work

Actual Cost (AC)Actual Cost of

Accomplished Work Performed (ACWP)

Authorized Work Budget at Completion Same

Forecasted Work Estimate at Completion Same

Work Variance Scheduled Variance Same

Cost Variance Cost Variance Same

Completion Variance Variance at Completion Same


Establishing Control Accounts:

• Control account - Level of the WBS at which a budget is assigned and a control account manager (CAM)is given responsibility for delivering the item

• Create control accounts at the level that senior management wants to track cost and schedule• Use the responsibility assignment matrix (RAM) as a tool

Responsibility Assignment Matrix:

• RAM - The integration ofthe WBS and the OBS of a project• Proper creation ofthe RAM results from locating the intersection of the OBS unit that is assigned

responsibility for the work and the specific WBS element that defines the work to be performed

Control Account Example:

CA#l:

• WBS Element: 1.1 Frame

• OBS Element: Frame Shop

• CAM: Mr. 1. M. Smart

• Budget: $ 125,000

• Schedule: Jan - Jun 00

Control Account Managers:

• Receive work authorization• Develop work and planning packages• Assign a budget and schedule for each work and planning package• Build a baseline by summing the planned value (PV) for each reporting period

Control Account Elements:

Work PackagesDetailed, short-spantasks or material items;Required to accomplishthe CA objectives;Typically for the near term

Task 1

Planning PackagesFuture work that has not been plannedin detail as work packages; Alwaysscheduled to occur in the future

Work Packages

Planning Packages

Task 2

Task 3

Task 4

Task 5

A

A

A

A

III ,i



Cost Coding:

Standard Cost code

Recognize & categorize /Physical & geographical

Delete nnnecessary items

Modify or expand important partsFeatures in project

/\

\ ;/XXXXX - XX - XXXXX - X

--...._-"""'--j --...._-v"'--./ --...._-v-"'--j --...._-v...--...jProject number

Project

88NB04

Area Facility

Area - Facility

11

Work type code

Work Type

03320

Distribution code

0= totalI = Labor2 = Material3 = Equipment4 = Subcontractor

Distribution

code

2

, ., I , ., I • I

88 = Jop Start 1988

N = Negotiated

B = Building

II th Floor { ConcretelightweightAggregate}

Material Cost

04 = 4 th Building project during this year


IRS Management Consulting Hou~e

--------------------------'-----------------11I.

Construction Codes for Work Type:

01 General Requirements

01346 Computer Software

03 Concrete

03210 Reinforcing Steel

05 Metal

05090 Metal Fastenings

07 Element Protection

07310 Shingles

09 Finishes

09400 Terrazzo

11 Equipment

11460 Unit Kitchens

13 Special Construction

131 75 Ice Rinks

15 Mechanical

15110 Valves

02 Site Construction

02070 Geosynthetics

04 Masonry

04400 Stone

06 Woods & Plastics

06060 Wood Materials

08 Doors & Windows

08210 Wood Doors

10 Specialties

10350 Flagpoles

12 Fur nishings

12100 Art

14 Conveying Systems

14200 Elevators

16 Electrical

16270 Transformers

-------------------------------------1b·,-i

Chapter (11): Caurse & Resaurce Management lPMA Preparation Course

r

l

Construction Codes for Work Type:

02700 Bases, Ballasts, Pavements,and Appurtenances (23)

02730 Aggregate Surfacing (I)

02740 Flexible Pavement (26)

02750 Rigid Pavement (2)

02760 Paving Specialties (16)

02762 Pavement Marking (16)

02772 Curbs and Gutters (2)

02775 Sidewalks (I)

02780 Unit Pavers (18)

02785 Flexible Pavement Coating and Micro-Surfacing (2)

02790 Athletic and Recreational Surfaces (47)

02795 Porous Pavement (I)

02800 Site Improvements and Amenities (70)

02815 Fountains, Pools and Water Displays (59)

02820 Fences, Gates and Hardware (150)

02830 Retaining Walls (33)

02840 Walk, Road and Parking Appurtenances (84)-- "

02850 Prefabricated Bridges (14)

02862 Highwav Noise Barriers (4)-- ,02870 Site, Street & Mall Furnishings (83)

02872 Street/Mall Clocks, Bells & Carillons (I 6)

02873 Outdoor Sculpture & OrnamentallVork (5)

02874 Bicycle Racks & Lockers (16)

02875 Site and Street Shelters (38)

02882 Recreational Facility & Playground Equipment (52)

02884 Amusement Park Equipment (0)

02890 Traffic Signs and Signals (37)

02900 Planting (8)

02912 Soil Preparation & Materials (6)

02920 Lawn and Grasses (0)

02940 Simulated Plants & Rocks (0)

02945 Planting Accessories (22)

02950 Site Restoration and Rehabilitation (I)

02955 Restoration of Underground Piping (I)

02975 Flexible and Bituminous Rehabilitation Pavement (0)

02980 Rigid Pavement Rehabilitation (0)

Chapter (11): Caurse & Resaurce Management


02700 Bases, Ballasts, Pavements,and Appurtenances (23)

02730 Aggregate Surfacing (I)

02740 Flexible Pavement (26)

02750 Rigid Pavement (2)

02760 Paving Specialties (16)

02762 Pavement Marking (16)

02772 Curbs and Gutters (2)

02775 Sidewalks (I'

02780 Unit Pavers (18)

02785 Flexible Pavement Coating and Micro-Surfacing (2)

02790 Athletic and Recreational Surfaces (47)

02795 Porous Pavement (I)

02800 Site Improvements and Amenities (70)

02815 Fountains, Pools and Water Displays (59)

02820 Fences, Gates and Hardware (150)-- '

02830 Retaining Walls (33)

02840 Walk, Road and Parking Appurtenances (84)

02850 Prefabricated Bridges (14)

02862 Highwav Noise Barriers (4)-- '

02870 Site, Street & Mall Furnishings (83)

02872 Street/Mall Clocks, Bells & Carillons (16)

02873 Outdoor Sculpture & Ornamental Work (5)

02874 Bicycle Racks & Lockers (16)

02875 Site and Street Shelters (38)

02882 Recreational Facilitv & Playground Equipment (52)-- '

02884 Amusement Park Equipment (0)

02890 Traffic Signs and Signals (37)

02900 Planting (8)

02912 Soil Preparation & Materials (6)

02920 Lawn and Grasses (0)

02940 Simulated Plants & Rocks (0)

02945 Planting Accessories (22)

02950 Site Restoration and Rehabilitation (I)

02955 Restoration of Underground Piping (I)

02975 Flexible and Bituminous Rehabilitation Pavement (0)

02980 Rigid Pavement Rehabilitation (0)



Schedule Variance:

• Comparing the EV, the amount originally budgeted for the work that has been completed or is in process,the PV, the amount budgeted for the work that was planned to have been accomplished

• SV=EV -PV• A negative result means less work has been performed than was planned

SV Example:

• PV = $42,000• EV = $38,000• AC = $48,000

• SV = EV - PV $38,000 - $42,000 = - $4,000

• SV% = SV / PV = - $4000 / $42,000 = - 0.095 = - 9.5%

Cost Variance:

• Comparing the amount originally budgeted for the work completed or in process, the EV, to the actualof that work, the A C

• CV=EV-AC• A negative CV means more dollars were spent to accomplish the work than was planned

Cost Variance Example:

• PV = $42,000• EV = $38,000• AC = $48,000

• CV = EV - AC = $38,000 - $48,000 = - $10,000

• CV%. = CV / EV = - $10,000 / $38,00 = - 26%

Performance Indices:

Definition: Cost- and schedule-performance efficiency calculations; expressed in $

• Cost Performance Index (CPI)CPI=EV/AC

• Schedule Performance Index (SPI)SPI=EVIPV

CPI Example:

• PV = $42,000• EV = $38,000• AC = $48,000• CPI = EV / AC = $38,000/ $48,000 = 0.79• $0.79 worth of work was actually done for each $1.00 spent

------------------------------------------L.Chapter (11): Course & Resource Management IPMA Preparation Course


SPI Example:

• PV = $42,000• EV = $38,000• AC = $48,000• SPI = EV / PV = $38,0001 $42,000 = 0.90• $0.90 worth of work has been done for each $1.00 worth of work that was planned to be done

Estimate at Completion (EAC):

Definition: The management's assessment ofthe cost ofthe project at completion• After variance analysis, the estimated cost at completion is determined

EAC Example:

One methodology:• EAC=BAC/CPI• BAC = $80,000• CPI = 0.79• EAC =$80,00010.79= $101,265

Variance at Completion:

• BAC = $80,000• EAC = $101,265• VAC=BAC-EAC=$80,000-$101,265 =-$21,265• Based on past perfonnance, project will exceed planned budget by $21,265

Earned Value Examples:

PV$1$2$1$1$1$1$3$2$2

EV$1$2$1$2$2$2$2$1$1

AC$1 On schedule On cost$1 On schedule Under cost$2 On schedule Over cost$2 Ahead of schedule On cost$3 Ahead of schedule Over cost$1 Ahead of schedule Under cost$1 Behind schedule Under cost$3 Behind schedule Over cost$1 Behind schedule On cost

Chapter (11): Caurse & Resaurce Management lPMA Preparation Course

RS,S Management Consulting House-------------------------'---------------------'1

Example:

PV EV AC EAC BAC100 125 75 600 560125 100 100 850 80075 75 75 700 560100 75 100 570 600

Sample Cost Problem:

• BAC

• EV

• PV

• AC

= $ 40k

= $ 20k

= $28k

= $ 26k

Calculate% of Work Scheduled% of Budget Spent% of Work AccomplishedCost VarianceSchedule Variance

Sample Solution:

Calculate:

• % of Work Scheduled• % of Budget Spent• % of Work Accomplished

• Cost Variance• Schedule Variance

Chapter (11): Course & Resource Management

PV /BAC = $ 28K/ $ 40K= 70 %AC / BAC = $ 26K / $ 40K = 65 %EV /BAC = $ 20K/$ 40K= 50 %

EV -AC = $ 20K-$ 26K= - $ 6KEV -PV= $ 20K-$ 28K= - $ 8K



Exercise (1):

Consider the simplified activity-on-arrow description of a project network given in the table.

A •• ~u:u..v..-J (oays)

A (1,2) 8

B (2,3) 10

C (2,4) 2

D (3,4) 16

E (3,5) 4

F (4,5) 8

G (3,6) 7

H (4,6) 12

1 (5,7) 3J (6,7) 8K (7,8) 2

(a) Draw the CPM network for this task structure.(b) Determine the project duration.(c) Assume that each task requires exactly one person daily to perfonn the associated activity. Draw a bar

chart and a resource-loading diagram that profiles the resource usage through time (assume that all jobsstart at the earliest possible time).

(d) Use the resource-leveling procedures to minimize the, total manpowerusage per day over the entire project duration.

o A 81 ------- 2 """'~~~

o 8 8

Solution:

E22

51

6 ,46" J46"" 54

8 "54'

3

54 K 567 ------- 8

54 2 56

Project Duration = 56 days & Critical Path = A, B, D, H, J & K



Solution:

Bar Chart for (Early & late)

I/Dav

A f--------J

B

c8 10

H

18 2223 27

25 27

34

..~42 51

134 46

II~42 45 54

J

o 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56

DaysResources Resource Histogram

5

4

3

2

1 +-_-::""_...J

3

o D~

o 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56

2

Resource Leveling

:i:11+-----------

ofu~

o 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56

Resources



Exercise (2):

12

S

)------..-1,@---------+\71

3 12 8

0<f------< C )-------I,@26 3

Consider the following CPM network (activity-on-node) with the activity durations (weeks) above the node andthe crew sizes per week required to perform the activity below the node.

S 9

B )------,0---- _8 3

3

E

7

-------{r2 6

(a) Find the critical path for this project network.(b) Compute the total float & free float for all activities.(c) Draw a bar chart and a resource-loading diagram showing the resource requirements profile for this

project over the total time period (assume all jobs at earliest possible start).(d) Use resource-leveling techniques to "smooth out" the manpower requirements.

Solution:3 116 8 1168 17

B F 160

191 5 24 241 9 33

3 I 0 IS lsi 0 23 331 0 45

Y C 0 ------ ----~ G 0 K 0

3 112 lsi 8,

3311215 23, 45, ,

010, ,

3, ,,, , ,

A, , ,

07 111

,231 0

,013 3 \11 12

,33

,3 7 , ,,, ,....

D H II J 0,

0

14 1 4 18 18 1 5 23 23110 --'JJ

3 110 6 6 110 13

E I0 10

13 I 3 16 16 1 7 23

Critical Path = A, C, G, J & K & Project Duration = 45 Weeks


R$ Management Consulting Hous1e------------------------------------------1,

33

334/W

24

24

23

19

3/W15

15

Bar Chart for Early & Late

6/W

7 7/W 12 23

I «"»».&':~6 6/W 13 23

I~23,---_...::..:..'------,

3

3 8/W 8

3 1/W 7 18I~32/W6 16

c=~m8 3/W 17

J33 4/W 45

Kl".,-.,"-.,.-,-,,,.,,-,,,-.,,,-.,",-,,,-.,-.,,,-.,,,-.,-.-,,,-.,,,-.,,,-r'-lrT'!'~,~,~,~,~,~,~,~,~,~,~,2,

2/W

A

B

C

D

'" E.~-:E F-"-< G

H

I

o I 234567 8 9101112131415161718192021222324252627282930313233343536373839404142434445

Resource MW:0gram

30 l272421181512

96 -

~ t-"-:-',

27

o 3 6 9 12 15 18 21 24 27

Resource Leveling30 33 36 39 42 45

3027

24

21

18

15

12

963 2

oo 3 6

19

9 12 15 18 21 24 27 30 33 36 39 42 45



Exercise (3):

Given the data for a small project.

~ ,,10 On Cost Rate-J (Dav~) . I T. Ii' !ngV

A 5 - 200B 6 A 300C 4 A 400D 7 C 250E 9 C 250F 5 D,B 300G 3 E,F 700

a) Plot the S- Curve of direct costs derived from the Early Schedule.b) Plot the S- Curve of direct costs derived from the Late Schedule.

Solution:

11 16 0

F

5 10 6 16 ,,,,,,,9

,,,D

,,

9 7 16

12 9 21

24

24

Critical Path = A, C, D, F & G & Project Duration = 24 days


R$ Management Consulting Hous~

------------------------------------------1\


====74o 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

F

E

Ai- -'

G

Days

0 5 9 10 11 12 16 18 21 24ES 0 1000 3800 4600 5400 5900 7900 9000 9900 12000LS 0 1000 2600 2850 3400 3950 7150 8250 9900 12000

9900

8250900

150

5900

12 16 18 21 24

________________________________1;h,)



Chapter (12): Update & Follow Up IPMA Preparation Course


Update & Follow up Exercises

Exercise (1):

Data for a small project is tabulated below:

Ar, <11> ..+, Total Cost

- on I II.. li'..illay) (L.E)

A 2 ------ 200 400B 10 A 300 3000C 7 A 400 2800D IS B 400 6000E 5 B 600 3000F 6 C 500 3000G 3 E,F 800 2400H 6 F 700 4200K 3 D,G,H 200 600

Required:The project scheduler has decided that the target schedule will be an early start schedule except foractivities D, E & G which will be scheduled at their late times.Plot the S-curve of costs for this project derived from the target schedule.The project has been worked on for ten days. During this time, L.E. 6000 has been charged against thecosts. Using the target S-curve developed. What is your opinion as to the status of the project at this time?Calculate the estimation at completion (EAC)Calculate the estimation of duration variance at completion (EDAC)

(Assume BCWP = 5500 LE)Solution:

21 6 9 9 16 15 lsi 6 21

C 0 F 0 H 6

8 17 15 151 6 21 21 16 27

o 10 2 21 0 12 121 0 27 271 0 30

A 0 B 0 D 0 K 0-- - - - - - _.".. - --------------------

01 2 2 2 110 12 121

15 27 271 3 30

121 7 17 171 7 20,E 0 G 7

191 5 24 241 3 27


R~ Management Consulting HOUSj


l200 / DAI I

BJC

D'".~-:E Eti< F

G

H

300/D

400/

800/DII

200/DK LI_----'

o 1 2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18192021222324252627282930

Date

Date 2 19 12 15 19 21 24 27 30 FCC;

400 I 5300 10400 24800f :.:-

Cumulative Cost 7700 14800 18200 21200 254q,--K::3

30000

25000

20000

-'"0 15000U

10000

ACWP(AC)

5000

0 00i i

S-curve

iii i ) i

I--------------------------------------:, .

o 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Date

Chapter (12); Update & Follow Up /PMA Preparation Course

r}

l

ft,!; Management Consulting House

After 10 days:

BCWS = PV = 5300 + 800 = 6100 LE

AC = ACWP = 6000 LE & EV = BCWP = 5500 LE

CV = EV - AC = 5500 - 6000 = - 500

CPI = EV / AC = 5500 / 6000 =0.916

SV = EV - PV = 5500 - 6100 = - 600

SPI = EV / PV = 5500 / 6100 = 0.9016

• The Project is over budget by 500 LE & Late with amount = 600 LE

• Estimated Cost At Completion (ECAC) = 25400 / 0.916 = BAC / CPI = 27730 LE

• Estimated Duration At Completion (EDAC) = Project Duration / SPI = 30 / 0.916 = 33 days

Chapter (12): Update & Follaw Up IPMA Preparation Course

Management Consulting HouseI

Exercise (2):The time- scaled diagram shown represents a portion of a construction project. The estimated cost of eachactivity is indicated on the diagram. The project has been worked on for 10 months. During this time, thefollowing activities have been completed, with actual costs as shown:

Activity A B C DDate completed (month) 3 10 4 10

Actual cost (L.E.) 20,000 75,000 25,000 30,000

a. Plot the s-curve of costs for this project.b. On the same chart, plot the budgeted cost of work perfonned (BCWP) and the actual cost of work

performed (ACWP) curves.c. What is your opinion as to the status of the project at this time?d. Predict the level of expenses and the final duration expected at the end of the project.

• Duration's in months.

8 A 3

(L.E. 15,000)

9 l---=-F-,-(4:..c0:..:.,0..:..00'-'-)_-l 13 \-...I..L.--l

(10,000)

3

3 D (20,000)

7

7E (25,000)

12 ---- 13

------------------------------------------iLChapter (12): Update & Fallaw Up IPMA Preparation Course


Solution:

8 A 3

(L.E. 15,000)

5K

B (60,000)

10K9

F (40,000)1----------{13

10K

5K

3

3D (20,000)

5K7

E (25,000)1---------1 12 - - -- - 13

5K

Actual Schedule after 10 Months

o 1 2 3 4 5

Month,

6 7 8 9 10

Activity A B C D E F GBudget Cost / Month 5000 10000 15000 5000 5000 10000 5000Actual Cost / Month 6666.7 10714.39 25000 4285.7Budget CWP Slope 5000 8571.4 30000 2857.1

Chapter 112}: Update & Follow Up IPMA Preparation Course

RSS Management Consulting Housr--------------------=---------------11-

Month 3 4 5 7 9 10 12 13 15 JBCWS 15000 - 75000 105000 135000 150000 180000 190000 2000001BCWP 15000 56428.5 - - - 125000 L(

ACWP 20000 60000 - - - 150000 J

S - Curve

250000

200000

150000

100000

50000

o 1 2 3 4 5

BCWS

6 7 8 9 10 11 12 13 14 15

Months

00

At Duration = 10 Months:

PV = 135000 + 5000 + 1000 = 150000 & AC = 150000

CV = BCWP - ACWP = 125000 - 150000 = - 25000

CPl = 125000/150000 = 0.833

SV = BCWP - BCWS = - 25000 & SPI = EV / PV = 0.833

The Project Cost overrun by 25000 LE & Late by 25000 LE

Estimated Duration At Completion (EDAC) = 15/0.833 = 18 Months

Estimated Cost At Completion (ECAC) = 200000 / 0.833 = 240096 LE

& EV = 125000

i----------------------------------,'.Chapter (12): Update & Fallow Up IPMA Preparation Course


Exercise (3):

o

o

5

5 9

5

8

12

D

10 5 15

You are given the above network for a small construction project. The progress report after 10days shows the following data:

• Activities Band C are finished on time.• Activity F has 3 days left.• Activities g and E are on schedule.• Activity D: its material will not be delivered before 3 days from now.

Required:

a. Calculate the original project duration.b. Update the network and determine the project duration, the new critical path, and the new total

and free floats.


30 Units

RSS Management Consulting HOUSf

l

Solution:

a. The original project duration = 17 days, Critical Path = A, B, F, H

o 5

5 9

B

13

13

9

0 20

, , , , , , , , , , , , ,8 12 , , 20 0 22, ,

8 H 0

3 20 2 22

b. The new project duration = 22 days,

The new Critical Path = A, D, H

G 7

15 5 20

!------------------------------------------iLChapter (12): Update & Follow Up IPMA Preparation Course


GhapGev 43

Chapter (73): Cash Flow IPMA Preparation Course


Project Cash Flow

1. Cash Flow Projection:

The projection of income and expense during the life of a projeci can be developed from several time-scheduling aids used by the contractor. The sophistication of the method adopted usually depends on thecomplexity of the project. One of the most commonly used and simplest aidJ is the so-called S-curve. Inmany contracts (e.g., public contracts such as those used by the Army Corps of Engineers), the ownerrequires the contractor to provide an S-curve of his estimated progress and costs u"io5 the life of theproject. The contractor develops this by constructing a simple bar chart of the project, assigning costs tothe bars, and smoothly connecting the projected amounts of expenditures over time.

Consider the following highly simplified project (Figure l) in which four major acrivities are scheduledacross a four-month time span. Bars representing the activities are positioned along a time scaleindicating start and finish times. The direct costs associated with each activity are shown above each bar.It is assumed that the monthly cost of indirect charges (i.e., site office costs, telephone, heat, light, andsupervisory salaries, which cannot be charged directly to an activity) is $5000. Aisuming for simplicitythat the direct costs are evenly distributed across the duration of the activity, the monthly direct costs canbe readily calculated and are shown below the time line. The direct charges in the second month, forexample, derive from activities A, B, and C, all of which have a portion in the period. The direct charge issimply calculated based on the portion of the activity scheduled in the second month, as:

Activity A: ll2 x 50,000 = $25,000Activity B: I 12 x 40,000 = 20,000Activity C: I /3 x 60,000 = 20,000

$ 65,000

II

3"50,000

s30,000D

Mqnth{V dirfct costslrdnthlv indirect eosrs.

Tsral monrhly corrtfurirulatjve monthly cosrs

s25,000 s65,000 s75,000 315,000$:!.,qgg g5r0oo g5.ooo 95;g@930i06-o $7o;000

. SEO,QoO $20,000

330,000 st00to00 s180,000 s?00.000

Chapter (73): Cash Flow I PMA P re pd ration Cou rse


s20d

315f

3?mr(

$ 50x

2Time in months

lhe letrer "k" is usud to indicate thousands of dollars.

The figure shows the total monthly and cumulative monthly expenditures across the life of the project.The S-curve is nothing more than a graphical presentation of the cumulative expenditures over time. Acurve is plotted below the time-scaled bars through the points of cumulative expenditure. The name S-curve comes from the fact that the curve of cumulative expenditures has the appearance of a "lazy S" Thisgeneral shape characteristic results are.

23Tlme {monthrl

Figure (2) Expenses and income profiles (Without Down payment).

J'r0csJ.E 2m

': 150!Ermo,2s50E6o

Expente

Chopter (73): Cosh Flow IPMA Preparotion Course


The income profile has a stair-step appearance, since the progress payments are transferred in discrete

amounts based on the above equation. The crosshatched area between the income and expense profiles

indicates the need on the part of the contractor to finance part of the construction until such time as he is

reimbursed by the o*nei. This difference between income and expense makes it necessary for the

contractor to obtain temporary financing. Usually a bank extends a line of credit against which the

contractor can draw to -buy

materials, make payments, and pay other expenses while waiting forreimbursement. This is similar to the procedure used by major credit card companies in which they allow

credit card holders to charge "rp"nr",

and carry un outttrnding b?lance for payment Interest is charged

by the bank (or Credit Card Company) on the amount of the outstanding balance or overdraft' It is, ofcourse, good policy to try to minimize the amount of the overdraft and, therefore, the interest payments.

The amount of the overdraft is influenced by a number of factors including the amount of markup or

profit the contractor has in his bid, the amount of retainage withheld by the owner, and the delay between

billing and payment by the owner.

Similar examples of this type of inventory financing can be found in many cyclic commercial

undertakings. Automobile dealers, for instance, typically borrow money to finance the purchase ofinventorieJof the new car models and then repay the lender as cars are sold. This is often achieved by a

floor plan whereby a major distributor guarantees a specific- purpose loan or overdraft with the dealer's

brokeis. Clothing stores buy large inventories of spring or fall fashions with borrowed money and repay

the lender as sales are made.

Interest on this type of financing is usually quoted in relationship to the prime rate. The prime rate is the

interest rate charged preferred customers who are rated as very reliable and who represent an extremely

small risk of default (i.e., General Motors, Exxon, etc.). The amount of interest is quoted in the number ofpoints (i.e., the number of percentage points) above the prime rate. The higher-risk customers must pay

more points than more reliable borrowers. Construction contractors are norrnally considered high-risk

borrowers; if they default the loan is secured only by some materials inventories and partially

completed construction. In the event that a manufacturer of household appliances defaults, the

inventory of appliances is available to cover part of the loss to the lender' Additionally, since

construction contractors have a historically high rate of bankruptcy they are more liable to be charged

additional interest rates in most of their financial borrowings.

Some contractors offset the overdraft: borrowing requirement by requesting front or mobilization money

from the owner. This shifts the position of the income profile so that no overdraft occurs (Figure 3). Since

the owner is normally considered less of a risk than the contractor, he can borrow short-term money at a

lower interest rate. ff the orn"r agrees to this approach, he essentially takes on the lnterim-financing

requirement normally carried by the contractor. This can occur on cost-reimbursable contracts where

Chopter (13): Cash Flow IPMA Preparation Course


lncoma profile shi{ted toleft of exo€ns curve

Ng overdraft

Expenrectrrvg

$200 K

$.t.50

3rm

K

$so KfMotilization IpryL

Figure 3 Influence of front or down payment on expense and income profiles

The owner has great confidence in the contractor's ability to complete the project. In such cases, itrepresents an overall cost savings to the owner, since otherwise he will ultimately be back-billed for the

contractor's higher financing rate if the contractor must carry the overdraft.

2. Overdraft requirements

In order to know how much credit must be made available at the bank, the contractor needs to know what

the maximum overdraft will be during the life of the project. With the information given regarding the

four-activity project, the overdraft profile can be calculated and plotted. For purposes of illustration, the

interest rate applied to the overdraft will be assumed to be one percent per month. That is, the contractor

may pay the bank lVo per month for he amount of the overdraft at the end of the month. More, commonly,

Oaiiy interest factors may be employed for the purpose of calculating this interest service charge. Month-

end balances might otherwise be manipulated by profitable short-term borrowings at the end of the

month. The calculations required to define the overdraft profile are summarized in Table 1. The table

indicates that the payment by the owner occurs at the end of a month based on the billing at the end of the

previous month. It ir usso-"d that the interest is calculated on the overdraft and added to obtain the

amount financed. This amount is then reduced by the amount received from the owner for previous

billings, To illustrate: The overdraft at the bank at the end of the second month is $100,300. The interest

on tttii amount is $1003 and is added to the over-draft to obtain the total amount financed ($101,303). To

obtain the overdraft at the end of the third month, the progress payment of $28,350 is applied to reduce

the overdraft at the beginning of the third month to $72,953. The overdraft at the end of the period is,

then, $ 72,953 plus the costs for the period. Therefore, the overdraft is $72,953 plus $ 80,000 or

$152,953. The information in the table is plotted in Figure 4. The overdraft profile appears as a saw-tooth

curve plotted below the base line. This profile shows that the maximum requirement is $154,483.

Therefore, for this project the contractor must have a line of credit that will provide at least $155,000 at

the bank plus a margin for safety, say $175,000 overall to cover expenses.

Requirements for other projects are added to the overdraft for this project to get a total overdraft or cash

commitment profile. The timing of all projects presently under construction by the contractor leads to

overlapping olerdraft profiles that must be considered to find the maximum overdraft envelope for a

given period of time.

Chopter (73): Cash FIow I PMA Pre pa ration Cou rse


Table L Overdraft Calculation:

" simple illustration only most lenders would calculate interest charges more precisely on the amount / time involved employing daily.

$ 25,0005,000

2

$ 6s,0005,000

Month34

$ 75,000 $ 15,0005,000 5,000

80,000 20,0004,000 1,000

84,000 21,00000

28,350 66,150 84,000

189,000

200,000

210,000

108,333

1,093

$109,416

Direct costIndirect cost

SubtotalMarkup 57o

Total billedRetainage withheld 107o

Payment received

Total cost to date

Total amount billed todate

Total paid to date

Overdraft end of month

Interest on overdraftbalance u

Total amount financed

100,000

105,000

28,350

100,300

1,003

101,303

30,0001,500

70,0003,500

31,5003,150

73,5007,350

30,000

31,500

30,000

300

30,300

31,500

200,000

t210,000

210,000

25,416

254

$25,670

94,500 178,500

r52,953

1,530

I54,483

(-) s830

0

Chapter (73): Cash Flow IPMA Preparotion Course


s2tox

stoox

s3t.5K

$(tr,3m s25.670Ovcrdraf t

profih

$10t,3Gt r(xt.416

3152,953 3154,4S1

Figure 4 PIot of maximum overdraft.

Bids submitted that may be accepted must also be considered in the projection of total overdraftrequirement. The plot of total overdraft requirements for a set of projects is shown in Figure 4.

Cash flow management involves all of the techniques described in this chapter-and very much more. It isfairly true to say, for example, that you cannot budget the other fellow's payments! That is, cash inflowsare affected by a significant degree of uncertainty. A cash flow management model of a relatively simplekind involves making provision for a set of at least so variables and requires a computer program tosecure sufficiently timely and usable output cash management decision making and control.

Chapter (73): Cash FIow I PMA P re pa ration Co urse

RS Management Consulting Hause

Exercise:

The programme for the construction of a small workshop building is displayedprecedence diagram, shown in Fig. 1. The value of the work contained in eachcalculated from the rates contained in the bill of quantities and is listed in Table below.

in theactivity

form of ahas been

Overlap between finish

L. Precedence diagram for construction of workshop.The contractor undertaking this project would like you to prepare graphs of cumulative cost and moniesreceived to date against time for activities starting as early as possible, in order that he may have a clearerpicture of the financial implications of this contract From these graphs calculate the net profit if interest ischarged at \Vo p.a. on outstanding monies. The gross profit margin is 107o of contract value and retentionis 57o up to a maximum limit of f,3000. Measurement is made monthly with a payment delay of onemonth. Half of the retention is paid on practical completion and the remaining half six months later. Tosimplify the calculation you may assume that all costs must be met at the instant they are incuned.What is the maximum amount of cash the contractor needs to execute this contract and when does herequire this amount?

Rate of work throughout any activity isuniform

of X & start of Y

E Fix cladding

Chopter (13): Cash Flow IPMA Preporotion Course


Solution:

Calculation time of each activitv:

Precedence diagram of small workshop with activity times calculated.

Bar Chart of activities and monthly value calculation:

Value 4 500 r0000 20000

ActivityMonths

1 2 3 4 5 6

A

B

C

D

E

F

4 500 4 500

4 000 4 000

6000

-4 000

20 000

4 000

r2 000

15 000

2000

-

Chapter (73): Cosh Flow

8 500 4 000 33 000



Cash Flow calculations:

Months t2

Monthly value 4500

of Work8500 4000 33000 10000 20000

Monthly Cost 4050 7650 3600 29700 9000, 18000

CumulativeCost

I 1700 15300 45000 54000 72000

MonthlyPayment

4275 807s 3800 31350 9500 21500 1500

Overdraft endof the month

4050 11727.14 11130.7 32830.3 38250.3 25156.6 15825.15 -ss6e -706e

Interest on0verdraftBalance

27.14 78.57 74.5 22Q 256.3 168.ss 106.03

TotalFinancialreouired

4077.r4 11805.7 11205.3 33050.3 38506.6 25325.r5 15931.2

Monthly Interest = 8Vo I 12 month = (0.0067) Monthly.Net Profit = 7069 $.

Max Cash needed = working Capital = Max. Overdraft required = 38506.6 $ at the end of Month 5

and before receiving Fourth Payment.

20000

10000

0

-10000

-20000

-30000

.40000

.50000

89101112

-15E25.15-1S931.2

Chopter (73): Cosh Flow IPMA Preparation Course

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Chapter (14): Feasibility Study IPMA Preparation Course


Feasibility Study

The project development cycle (Fig 1) comprises the Pre-investment / Investment / and theOperational Phases.

Each of these three major phases is divisible into stages, some of which constituteimportant industrial activities. The major objective of this Manual is to give a betterunderstanding of the problems encountered in carrying out the various tasks under the preinvestment phase of industrial projects.

Several parallel activities take place within this phase and even overlap into the succeedinginvestment phase. Thus, once the early stages of pre-investment studies have produceddependable indications of a viable project, investment thrust to the final evaluation stageand the investment phase.

Before dealing with the pre-investment phase, the various stages of the investment andoperational phases are considered briefly together with promotional activities as these havea bearing on the nature and scope ofpre-investment studies.No single from a small-scale unit producing a particular product or component to a largemulti-product complex.

Pre-Investment Phase:

The pre-investment phase comprises several stages: identification of investmentopportunities (opportunity studies); preliminary project selection and definition (prefeasibility studies); project formulation (feasibility studies); the final evaluation andinvestment decision.Support or functional studies are a part of the project fonnulation stage. These are usuallydone separately, one common reason since the agency carrying out the feasibility studymay not have the qualified manpower or expertise to conduct studies in the areasconcerned.

These stages assist a potential investor in the decision-making a process and provide thebase for project decision and implementation.

To differentiate between an opportunity, a pre-feasibility and feasibility study is not an easytask in view of the frequent, inaccurate use of these tenus

Chapter (14): Feasibility Study /PMA Preparation Course

9.g~;:;~

[5'

~e'~

~~i3g."G't::;;:'"

Pre-investment phase Investment phase Operational phllse

Identification Preliminarv Proje':l Evaluation Ncooliation Project Construction SUrt·upof investment uleclion shoe formulation and decision and design HaDe stageopportunities (Pro·feasibility nage (Techno- stagc coni/acting stageIProjecl ideasl Jludyl economic IEvaluation stagefeilsibility teporl)studyl

Investment promotion activities

J II

Implementation planning and follow-up

II

Capital investment expenditures

II

Project Development Cycle

~'-=''"=1

~~:::s

c§t'e3§....~:::stil-=N:'~.

~

~


Preliminary Project Selection & Definition(Pre-Feasibility Studies)

Technical Studies and Cost Forecasting

1- Analysis ofpast and present demand1.1- Quantitative information1.2- Qualitative information1.3- Sources of information1.4- Assessment and preliminary processing of data

2- Estimating Future Demand (Different Methods)2.1- Projection of the trend2.2- Utilization of technical coefficients2.3- International comparisons2.4- Possibilities for export of to impOlt substitution2.5- Econometric methods

2.5.1- Relationship between demand and price2.5.2- Relationship between demand and income2.5.3- Relationship between demand (overall or per head) for a given good and price and

income (overall or per head) of consumers2.6- Forecasting without statistical data

2.6.1- Analytical Forecasts by sector of use2.6.2- Consideration of development targets2.6.3- Consideration of political factors2.6.4- The final evaluation of market size

2.7- Final Assessment of the Market

3- The problem of uncertainties in Demand Estimates and how to take them into account3.1- Uncertainty in estimates of present and past demand3.2- Uncertainty due to the method used for forecasting3.3- Uncertainty concerning the trend of future demand3.4- Allowing for uncertainty in the assessment of demand

4- Conclusion: the Choice of a Production Target



Project Formulation Stage (Economic Feasibility Study)Financial & Economic Evaluation

Project preparation should be geared towards the requirements of financial and economic evaluation. Onceal I the elements of a feasibility study are prepared, the next step is to compute the total investment costs.In many cases it has to be assumed that project financing is available at the feasibility stage; the financialimplications are then calculated and included in the total production costs. Financial evaluation shouldpreferably rely on discounting methods and incorporate sensitivity analysis. Projects should also beevaluated from the aspect of their direct and indirect effects on the national economy.

Total Investment Costs*Calculate the total investment costs by summarizing all investment componer,ts.

Project FinancingPrepare cash-flow table for financial planning, then Estimate annual financial costs.

Total Production CostsCalculate total production costs by summarizing all cost items plus unit costs.

Financial Evaluation

Compute commercial profitability criteria

• Simple Methods ofFinancial Evaluation:

We call it simple methods since they do not consider the entire life of the project, but only briefperiods. In addition, the annual data used are taken at the actual value and not at the discounted value.

I- Break-even Analysis2- Pay-Back Period

• Discounting Methods:

I- Net Present Value (NPV)2- Internal Rate of Return (IRR)



Simple Methods

1. Break-Even Analysis:

Quantity

SP = Selling PriceFC =Fixed CostVC =Variable CostTC =Total CostUC =Unit CostQ = Quantity

FC

TC

VC

Break-even analysis detennines the break-even point (BEP)-the point at which sales revenues equalproduction costs. The break-even point can also be defined in tenns of physical units produced, or of thelevel of capacity utilization at which sales revenues and production, costs match each other.Prior to calculating the break-even point, the following conditions should be observed:Production costs are a function of the volume of production or of sales (e.g. in the utilization ofequipment);The volume ofproduction equals the volume of sales;Fixed operating costs are the same for every volume of production;Variable unit costs vary in proportion to the volume of production, and consequently total production costsalso change in proportion to the volume of production;The unit sales prices for a product or product mix are the same for all levels of output (sales) over time.The sales value is therefore a linear function of the unit sales prices and the quantity sold;Data from a normal year of operation should be taken;The level of unit sales prices, variable and fixed operating costs remain constant;A single product is manufactured or, if several similar ones are produced, the mix should be convertibleinto a single product;The product-mix should remain the same over time.These conditions will not always exist in practice and the results of break-even analysis may, in tum, beinfluenced negatively. Therefore, break-even analysis should only be considered as a tool supplementaryto other project evaluation methods.

Profit RevenueArea

IncomeExpenses

VC=UCxQ

Revenue = SP x Q

TC=FC+VC



--------------------------'----------- ...,'I

At Break-Even Point:

Revenue~TC

SPXQBE = FC+ UCXQBE

FCQBE =

SP- DC

A high break-even point is inconvenient since it renders a finn vulnerable to changes in the level ofproduction (sales); the higher the fixed costs, the higher the break-even point;The larger the difference between unit sales price and variable operating costs, the lower the break-evenpoint. In this case the fixed costs are absorbed much faster by the difference between unit sales price andvariable unit costs.

Example:Assume FC = 3280 $

UC=3.25 $SP = 6.25 $And Full Production Capacity = 2000 unit

Using the data of the example, the break-even point (BEP) would be reached at a production of

3,280,0001,093,333 units

6.25-3.25

Expressed in terms of sales revenue, becomes:FC

Revenue at BEP = SP x QSE = SP ( )SP- UC

6.25 x3,280,000

6.25 - 3.25= $ 6,833,331

Break-even analysis lends itself easily to sensitivity analysis, particularly with the following modifiedequation, which is used to calculate the rate of capacity utilization at the break-even point:

FCBEP at Capacity Utilization =

( SP x Max Q- UC x Max Q)


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For the example the break-even point would be reached at a capacity utilization of

3,280BEP at Capacity Utilization = ----------

6.25 x 2,000 - 3.25 x 2,000= 55%

In this way, break-even analysis can be useful in determining the impact of changes in unit prices, and infixed and variable production costs, on the break-even point of a project.The above approach has the advantage of enabling the project planner to calculate several break-evenpoints, taking into account alternative investment proposals resulting from different installed capacities oralternative technological processes. Changes in the installed capacity cause variations in the fixed costs.Changes in the technological process also have an impact on the variable costs since, for example, atechnologically more advanced (and more costly) process normally leads to lower variable unit costs,particdlarly as far as labor costs are concerned.


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2. Pay-Back Period Method:

This is a well tried, widely used and simple method of evaluation, which computes the time taken by aproject to pay back from its cash flow, the amount of capital invested. The project that pays back soonest isthe preferred investment. Projects that cannot repay within a certain time. Say three years are rejected, e.g.two projects, A and B. are under con consideration:

Project A Project B

Life 5 years 9 years

Cost £ 160,000 £ 150,000

Net Cash flow + £ 40,000 + £ 30,000

160/40 150/30Pay back period

= 4 years = 5 years

Therefore, Project A is preferred to Project B since the capital expenditure is returned in a shorter time.This method suffers from two impOl1ant disadvantages in that it treats cash flow in successive years withequal significance, and thus rails to take account of the rime value of money, and it ignores all cash flowafter the point at which pay back is reached. This crude and simplistic method cannot consider workingcapital in the computation. In the example above, management must decide whether it prefers A with atotal cash flow of £ 200 000 over five years and the flexibility to reinvest at that point, or B. which earns £270 000 over nine years.The pay back method fails to focus attention on this aspect of the problem since it does not consider thereturn over the full life of the rival projects. Pay hack is still widely used in industry, often as a 'coarsescreen to reject weak projects from the range or opportunities available to a firm, and derive a short list formore detailed analysis. In a situation where technological change can dramatically reduce the economiclife of projects, or where cash shortages in a business promote the need for cash flow in early years, thepay back method can still be useful method of capital project evaluation, since if favors projects that covertheir cost as quickly as possible.By the same token it may also reject sound long-tenn projects. Because of the failure to consider theoverall return on the capital invested.

Again the pay-back period is defined as the period required to the original investment outlay through theprofits earned; by the project; "profit" is defined as net profit after tax, adding financial cost anddepreciation.

Il


When calculating the pay-back period, the computation usually starts with the construction period duringwhich the initial investments will be made. The calculation of the pay-back period for the example isshown below.

Calculation of pay-back period:

Total investment costs

Annual net profit plus interest plus depreciation

Year 1 (construction period)

Year 2 (construction period)Year 3Year 4Year 5Year 6Year 7

10300 $

Amountpaid back

(= "profit")

:'E

:'E

8702030233035003500

Balance at End ofyear

10300

10300943074005070I 570

The calculation indicates that the original investment costs will be recovered after a little less than 6.5 years,including the construction period. The same result can also be obtained using the cumulative net cash-flow:

The initial investment outlay of $10.3 million will be repaid shortly before 6.5 year

There are two ways of calculating the pay-back period: one is a modified version of the one described above, exceptthat it does not include the construction period The pay-back period for the example would thus be after 6.5 - 2 =4.5 years.

In the second method, the value ofland ($0.3 million) and of working capital ($2.0 million) are deductedfrom the total investment costs on the assumption. These values can be fully regained at the end of theproject. Thus only $8.0 million investment outlay must be recovered; this consists mainly of fixed assetssuch plant and equipment, as well as buildings and civil works. In this case the pay back period would be5.2 years. If the construction time is disregarded, the pay-back period becomes 3.2 years.Total investment cost = 8000 $Annual net profit plus interest plus depreciation:

A single project proposal may be accepted if the pay-back period is smaller than or equal to an acceptable timeperiod; this period is usually derived from past experience with similar projects.

The major merit of the pay-back period as a project selection criterion is its easy calculation. It is particularly usefulfor risk analysis, which is relevant in politically unstable countries and in branches of industry that face rapidtechnological obsolescence. The main shortcomings of this method are that it does not consider what will happenonce the project has paid for itself and that it overemphasizes quick financial returns. Furthermore, this method doesnot measure the profitability of project proposal but is mainly concerned with its liquidity. In summary, this methodis not a reliable criterion for project selection, but can be a useful supplementary tool in some cases.


Discounting methods

1. Net present value:The net present value (NPV) of a project is defined as the value obtained by discounting, separately foreach year, the difference of all cash outflows and inflows accruing throughout the life of a project at afixed, pre-determined interest rate. This difference is discounted to the point at which the implementationof the project is supposed to start. The NPVs obtained for the years of the life of the project are added toobtain the project NPV as follows:

NPV=NCFl +(NCF2 x a2)+(NCF3 x a3)+...+(NCF; x a;)+...+(NCFn +an)

Where NCFj is the net cash flow of a project in years 1,2,3,... , I, ..., n, and aj is the discount factor in years2, 3, ..., I, ..., n, appropriate to the discount rate applied. Discount factors are obtained from presentvalue tables.The discount rate (or cut-off race) should be equal either to the actual rate of interest on long-term loans inthe capital market or to the interest rate paid by the borrower. Since capital markets frequently do not exist,the discount rate should reflect the opportunity cost of capital: the possible return on the same amount ofcapital invested elsewhere. Expressed differently, this should be a minimum rate of return below which anentrepreneur considers that it does not pay for him to invest.The discounting period should be equal to the life of the project. For instance, the useful life of equipmentis generally between 10 and IS years. Factory buildings of solid material will usually last 30 or 40 years,vehicles 4 or 5 years etc.The practical solution is to take the life of the most essential part of the fixed assets. Obviously, in afactory this is the basic equipment. The value of fixed assets that last longer, assets like buildings, forexample, must be given at their salvage value at the end of the discounting period. This is also true of thevalues ofland and working capital. Which remain practically constant during the life of the project.The replacement of assets with a shorter life during the discounting period must be considered. In mostcases, the discounting period includes the construction period (say, two years) plus some 10 years ofproject life.If the NPV is positive. The profitability of the investment is above the cut-off discount rate. If it is zero,the profitability is equal to the cut-off rate. A project with a positive or zero NPV can thus be consideredacceptable. If the NPV is negative, the profitability is below the cut-off rate, and the project should bedropped.Using the data of the example, the NPV of the total investment outlay (schedule 10-13) and the NPV of theequity capital (schedule 10-14) can be determinated. The relevant schedules are given at the end of thischapter. It should be noted that depreciation is not taken into account since it does not involve any cashmovement. However, repayments of credits are considered, since they are cash outflows.Schedules 10-13 and 10-14 show that a total working capital of 52.0 million will be recovered by the endof the project and that the entire bank overdraft of 51.5 million will be repaid (schedule 10-14). If theoverdraft was not repaid, the terminal value would only be 50.5 million (which is covered by the equitycapital), but in this case interest payment would have to be taken into account throughout the wholediscounting period.The calculation of the NPV for the total investment costs (schedule 10-13) is identical to the case wherethe project is undertaken without any outside financing. However, the calculation of the NPV for theequity capital (schedule 10-14) corresponds to the case where outside financing (loans) is involved. In bothcases, a supporting table should be prepared in addition to the cash-flow tables to calculate the

-------------------------------L"Chapter (14): Feasibility Study lPMA Preparation Course


corporate tax. The "Net Income Statement", bearing in mind that in the case of project financing withoutoutside funds the production costs will not contain any financial costs.The profitability rates of the total investment outlay as well as of the equity capital are above 10 per cent,since both present values are positive. As mentioned earlier, a project can be accepted if the NPV is greaterthan or equal to zero.If one of several project alternatives has to be chosen, the project with the largest NPV should be selected.This needs some refinement, since the NPV is only an indicator of the positive net cash flows or of the netbenefits of a project. In cases where there are two or more alternatives, it is advisable to know how muchinvestment will be required to generate these positive NPVs. The ratio of the NPV and the present value ofthe investment (PVI) required is called the net present value ratio (NPVR), and yields a discounted rate ofreturn; this should be used for comparing alternative projects. The fonnula is as follows:

NPVNPVR =

PVI

If the construction period does not exceed one year, the value of investment will not have to be discounted.Comparing the two alternative ways of financing the project in the example, the following NPVRs areobtained:

Scheduled 10 - I3(End ofline D)

Scheduled 10 - 14

NPV

1473

1026

PYI

2871 + 3 780 + 928 + 154+43 + 94 + 327 = 8 197

2871 + I 890 + 327 = 5 088

NPVR

0.1 79

0.201

Thus. It is more profitable for the entrepreneur's equity capital to finance the project from outside fundsthan to rely exclusively on own funds. Given alternative projects, the one with the highest NPVR shouldbe chosen. When only one project is being considered a positive choice should only be made if the NPVRis greater than or equal to zero. When comparing alternatives, care should be taken to use the samediscounting period and rate of discount for all projects.In summary, the NPV has great advantages as a discriminatory method compared with thePay-back period or the annual rate ofreturn, since it takes account of the entire life of the project and ofthe timing of the cash flows. The NPV can also be considered as a calculated investment rate which theprofit rate of the project should at least reach. The shortcomings of the NPV are the difficulty in selectingthe appropriate discount rate and that the NPV does not show the exact profitability rate of the project. Forthis reason the NPV is not always understood by businessmen used to think in terms of a rate of return oncapital. For these reasons, it is advised to use the internal rate of return.


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l2. Internal rate of return:

The internal rate of return (IRR) is the discount rate at which the present value of cash inflows is equal tothe present value of cash outflows; by another way, it is the rate at which the present value of the receiptsfrom the project is equal to the present value of the investment, and the net present value is zero. Theprocedure used to calculate the IRR is the same as the one used to calculate the NPV. The same kind oftable can be used and, instead of discounting cash flows at a predetermined cut-off rate, several discountrates may have to be tried until the rate is found at which the NPV is zero. This rate is the IRR, and itrepresents the exact profitability of the project.

The calculation procedure begins with the preparation of a cash-flow table. An estimated discount rate isthen used to discount the net cash flow to the present value. If the NPV is positive, a higher discount rate isapplied. If the NPV is negative at this higher rate, the IRR must be between these two rates. However, ifthe higher discount rate still gives a positive NPV, the discount rate must be increased until the NPVbecomes negative.If the positive and negative NPVs are close to zero, a precise (the closer to zero, the more precise) and lesstime-consuming way to arrive at the IRR uses the following linear interpolation formula:

PV + NV

Where i, the lRR, PV is the NPV (positive) at the low discount rate of i j ,and NV is the NPV (negative) atthe high discount rate of hThe numerical values of both PV and NV used in the above fonnula arepositive. It should be noted that i j and h should not differ by more than one or two per cent. The abovefonnula will not yield realistic results if the difference is too large, since the discount rate and the NPV arenot related linearly.In the project without outside financing of the example (schedule 10-13), the NPV = 771,000 at a 15 percent discount rate. In order to find the IRR, several discount rates greater than I 5 per cent should be trieduntil the NPV is equal to zero. The NPVs at discount rates of 17 per cent and 18 per cent are shown below.

Net cash flow DiscountNPV

DiscountNPV

Year Scheduled factor at($ Thousands)

factor at($ Thousands)

($ Thousands) 17% 18%

I - 3 300 0.854 - 2 818 0.847 = 2 7952 - 5 000 0.730 - 3 650 0.718 = 3 5903 535 0.624 - 334 0.609 3264 I 755 0.533 935 0.516 9065 2 240 0.456 I 021 0.437 9796

,270 0.389 I 272 0.370 I 210.)

7 3 500 0.333 I 165 0.314 I 0998 I 140 0.284 324 0.266 3039 2 140 0.243 520 0.225 48210 2 140 0.208 445 0.191 40911 2 140 0.177 379 0.162 34712 5 640 0.151 851 0.137 773

110 - 203

----------------------------------'LCChapter (14): Feasibility Study IPMA Preparation Course


The above table shows that, discounted at 17 per cent, the net cash flow is still positive, but it becomesnegative at 18 per cent. Consequently, the IRR must lie between 17 per cent and 18 per cent. For practicalpurposes this would be sufficiently close to be able to calculate the exact IRR using the formula given andgraphical interpolation.Thus,

If = 17 +110(18-17)

110 + 203= 17.35%

In the graphical method the positive and the negative NPVs are plotted on the ordinate, and the discountrates are plotted on the abscissa. This is shown below.

202 312 )0 202 = 31218-(l7+X) 18 - 17 (l-X) 1

202 = 312 - 312 X )0 312X=312 -202 )0 X = 0.352

IRR = 17.35 %

NPV($ thousand)~II..,I 110

• 1001

0 17.35 lRR! \ I \ )

17 .2 .4 .6 .8 18 %

- 100

- 200 J

~

Chapter (14): Feasibility Study

-202


The line connecting the negative and the positive NPV cuts the abscissa (NPV = 0) at a discount rate equalto the IRR. In the example, this is above 17.3 per cent.

The IRR indicates the actual profit rate of the total investment outlay and, if required, of the equity capital,The IRR of the total investment outlay can also be used to determine the conditions ofloan financing sinceit indicates the maximum interest rate that could be paid without creating any losses for the projectproposal. In order not to endanger the liquidity of the project, it would be necessary, however, to adjust theloan repayment schedule to the cash inflows.The investment proposal may be accepted if the IRR is greater than the cut-off rate, which is the lowestacceptable investment rate for the invested capital. If several alternatives are being compared, the projectwith the highest IRR should be selected ifIRR is greater than the cut-off rate.

-------------------------------l....Chapter (14): Feasibility Study IPMA Preparation Course

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yinimum Attractive Rate ofReturn (MARRI:-

Actually after calculating the IRR , we shall compare it with the company MARR to take "YES or NO"decision for the project.

This means, if:

IRR > MARR

IRR<MARR )0

"YES"

"NO"

MARR is the minimum attractive rate of return, there is a number of things to consider in this regard:

• ~'irst, most organizations have more investment opportunities than available funds so the projectswhich are approved have large positive NPV values,

• Secondly, cost and benefit estimates are always subject to uncertainty, When the NPV is zero, thereis no room to offset the effect if actual costs are slightly higher or actual benefits are slightly lowerthan estimated, In general, an organization would be satisfied if on average, across the totality of allprojects implemented, it earned MARK

The reality is that some projects will ultimately provide a return equal to or greater that that originallypromised; whereas others will do considerably, more poorly than promised and may well have an actualreturn well below MARR,

Therefore, in the overall picture, a project, which is at the margin NVP (MARR) = 0, may not beparticularly appealing,

Note that the decision rule is for the evaluation of independent projects where you can estimate therevenues as well as the costs associated with a project.


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r

Time Value ofMonev:

Every one is aware that the amount of money held in a saving account will increase with time if interestpayments are allowedTo remain on deposit compound in the account. The value of .a sum of money left on deposit after anyperiod oftime may be calculated using the Following Equation:

F = P (1 + i)"

Where, F = value at end of it periods (future value)P = present valueI = interest rate per periodn = number of periods

The expression (1 + i)" is often called the single -payment compound interest factor.

The Previous Equation can be solved to find the present value (present worth) of some future amount,resulting in the following Equation:

FP

(1 + i) n

Tile expression 1/ (1 + i) n is called the single-payment present worth factor. Expressions have also beendeveloped that yield the value of a series of equal periodic payments at the end of any number of periods(uniform series compound amount factor), the present worth of such a series (uniform series presets wOlthfactor) the periodic payment required to accumulate a desired amount at some future date (sinking fundfactor), and the annual cost to recover an investment including the payment of interest over a given periodof time (capital recovery factor). Other expressions have been developed to find the present and futureworth of gradient (nonuniform) series ofpayments.These equations form the basis of a type of economic analysis commonly called engineering economy Themethods or engineering economy are widely used to analyze the economic feasibility of proposed projects,to compare alternative investments, and to determine the rate of return on an investment. However,because of their complexity and the difficulty of accounting for the effects of inflation and taxes, thesetechniques have not been widely used within the construction industry. A present worth analysis, however,is very helpful when corn-paring the cost of different altematives.

In addition, the following equation is for determining the present value of Annual amount:-

A (Hi) n - 1p=

i (Hi) n

Which lead to : -

P [ i (Hi) n]A=

i (l+i) n - 1

-------------------------------L.Chapter (14): Feasibility Study IPMA Preparation Course


FP =

(1+ i) n

F = P (1 + i) n

(1+ i) n - 1P = A

i (1+ i) n

i (1 + i) n

A = P(1+ i) n - 1

(1 + i) n - 1F = A

A = F----(1+ i) n - 1

Same as above mentioned

(P/F) ;=,n=

(A/F) ;=,n=

(F / A) ;=,n=

(A / P) ;=, n=

To find P given F

To find A given F

To find F given A

To find P given A

To find A given P

So, we should draw the cash flow diagram and then transfer the figures to the present to get the NPV inorder to economic evaluation of the project:-

Salvage Value

Net Annual Cash Flow

Example of Cash Flow Diagram

Investment Cost


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iExample No#: 1

A small construction firm entered into a 10-year contract with a quarry for gravel supplywith an agreement to buy quantities equal to $250,000, and the payment terms agreed are

The company will pay 20% immediately as down payment, and the rest of the amount to bepaid on equal yearly installments, where the first installment will due by the end of the thirdyear from the contract's date.

After paying the first four installments, the firm requested that it would be allowed to payoff the rest of the contract with a lump sum by the end of the seventh year. What shouldthis lump sum be if the yearly interest rate is 13%7

SOLUTION (J)

50,000

250,000

A

1 2 6 7 8 9 lOrL

I I I I I I I

[J

t

R

= 255,380.0 LE

To Calculate the Annual Payment (A), we need first to calculate the remaining amount after 2years (R)

R = 200,000 (1 + 0.13)2

A = 255,380 [ 0.13 (1+0.13/] = 53,217.8 LE

(1+0.13)8 - I

-------------------------------:LChapter (14): Feasibility Study IPMA Preparation Course


SOLUTION (1)

7

6

F

F = Final Remaining Payment

8 9 10

F = Yearly Payment + The Remaining 3 Payments ( P / A) i=13, n=3

F = 53,277.8 + 53,217.8 (1+0.13)3 - 1

0.130+0.13)3

Chapter (14): Feasibility Study

= 178,873.146LE


Example No#:2

A small construction firm entered into a 10-year contract with a quarry for gravel supply with anagreement to pay $50,000 immediately plus $10,000 per year beginning with the end of the sixth year andending with the end of the contract period. Because of substantial profits at the end of the fourth year, thefirm requested that it be allowed to pay off the rest ofthe contract with a lump sum. Both the firm and thequarry agreed to an interest rate of 8% compounded annually. What should this lump sum be?

SOLUTION (2)

50,000

1 2 3 4

z

5 6 7

10K

8 9 10

First, we get the Present Value of the Annual Installment at year (5); then, we translated tothe end of year (4).

z = [ ( P / A) i~8, n=5 (10,000) ] X (p / F) i~ 8, n=l

z= [ 3.9927 x (10,000) ] x 0.9259

= 36968.4 LE

!----------------------------------------LCChapter (14): Feasibility Study IPMA Preparation Course


; = SO/C

II PlF PIA AIF AlP FlP FIA

0.9259259 0.9259259 1.0000000 1.0800000 1.0~00()() 1.0000002 0.8573388 I.n32650 0..+807692 0.5607692 I. I66-1()() 2.0800003 0.7938322 2.5770970 0.3080335 0.3880335 1.159712 3.246-1004 0.7350199 3.3111270 0.2~ 19~08 0.3019208 1.3604~9 -1.5061115 0.6805831 3.9927100 0.1704565 0.2504565 I.-Ioy 3?8 5.86660 I

6 0.6301696 -1.6118800 0.1363154 0.116315-' 1.586874 7.3359197 0.5834904 5.1063700 0.1120714 . 0.192072-1 1.713824 8.9228038 0.5402689 5.7466390 0.0940148 0.1740148 1.850930 10.6366309 0.5002490 6.2468880 0.0800797 0.1600797 1.999005 12.487560

10 0.4631935 6.7100810 0.0690295 0.1490'295 2.158925 14.-\86560

11 0.4288829 7.1389640 0.0600763 0.1-100763 2.331639 16.645490I? 0.3971138 7.5360780 0.0526950 0.1326950 2.518170 18.97713013 0.3676979 7.9037760 0.0465218 0.1265218 2.71962-1 21.49530014 0.3404610 8.2442370 0.0412969 0.1212969 '2.937194 24.11491015 0.3152417 8.5594790 0.0368295 0.1168295 3. J 71169 '27.151110

16 0.1918905 8.8513690 0.0329769 0.1129769 3.425943 30.32420017 0.2702690 9.1216380 0.0296294 . 0.1096294 3.700018 33.75013018 0.2502490 9.3718870 0.0267021 0.1067021 3.996020 37.45024019 0.2317121 9.6035990 0.0241276 0.1041276 4.315701' 41.44616020 0.2145482 9.8181470 0.0218522 0.1018522 4.660957 45.761%0

21 0.1986557 10.0168000 0.0198323 0.0998323 5.033834 50.42292022 0.1839405 10.2007400 0.0180321 0.0980321 5.436540 55.45676023 0.1703153 10.3710600 0.0164222 0.0964222 5.871464 60.89330024 0.1576993 10.5287600 0.0149780 0.0949780 6.341181 66.76476025 0.1460179 10.6747800 0.0136788 0.0936788 6.848475 73.105940

26 0.1352018 10.8099800 0.0125071 0.0925071 7.396353 79.954-12027 0.1251868 10.9351600 0.0114481 0.0914481 7.988061 87.35077028 0.1159137 11.0510800 0.0104889 0.0904889 8.627106 95.33883029 0.1073275 11.1584100 0.0096185 0.0896185 9.317275 103.96590030 0.0993773 11.2577800 0.0088274 0.0888274 10.062660 113.283200

35 0.0676345 1J .6545700 0.0058033 0.0858033 1-I.7853...1Q 172.31680040 0.0460309 11.9246100 0.0038602 0.0838602 11.724520 259.05650045 0.0313279 12.1084000 0.0025873 0.08'25873 31.920450 386.50560050 0.0213212 12.2334800 0.0017429 0.0817429 46.901610 573.770200

60 0.0098759 12.3765500 7.979E-40 0.0807979 101.257100 1.253.2 J 300()70 0.0045744 12.4428200 3.676E-40 0.0803676 218.606-100 2.720.08000080 0.0021188 12,4735100 1.699E-40 0.0801699 471.954800 5.886.93500090 9.814E-40 12.4877300 7.859E-50 0.0800786 1.0 I8.':1 15()O[) 12.7239·10000

100 -I.546E-40 12.4943200 3.638E-50 0.080036-1 2.199.761000 2: ,48-1.520000


RS Management Consulting HouserI

i=9%

1/ fir PI.-\ ,.1/ F ..lIP FlP FI.·\

n.9174312 O.\) 174., 12 I.OO()OOOO I.OL)O()OOO I.()900()() I.O()OO()O, O.X416XtJlJ 1.7591110 O.-+7X6X\)0 O.56X46X9 I.IXXIOO 2.0900()()-,

O. 77~ I~35 2.5312950 0.305054X O.395054x 1.29:i02\) 3.27X 100-'4 O. 70X42:i 2 3.23 tJ7200 0.21::;M~7 0.30X6687 1.411582 4.5731'29

5 n.t,4993 \ 4 3.8X\)65 10 O.1670':l25 0.2570':l25 1.53X624 5.9847\1

6 0.5\)62673 4.-+85\) 190 0.1329198 0.2229198 1.677 Ion 7.523333

7 0.54703-!2 5.0329530 O. In~6905 O.19~6905 1.828039 9.:200-+3.5

8 0.5018663 5.534~ 190 0.0906744 0.1806744 1.992563 11.028470

9 0.-+604278 5.9952470 O.076198~ 0.1667988 2.171893 l3.0·2 1040

\0 0.-+224 IO~ 6.-+ 176580 0.065820 I O.155~20 I 2.36736-J. 15.192930

II 0.3875329 6.8051910 0.0569467 0.1469467 2.580426 17.560290

12 0.3555347 7.1607250 0.0496507 0.1396507 2.812665 20.140720

13 0.3261786 7.-+869040 0.0435666 0.1335666 3.065805 22.953380

14- 0.2992465 7.7~61500 0.038-+332 0.1284332 3.341727 26.019190

15 0.2745380 8.0606880 0.0340589 0.1240589 3.642482 29.360920

16 0.2518698 8.3125580 0.0302999 0.1202999 3.970306 33.003400

17 0.2310732 8.5436310 0.0270462 0.1170462 4.327633 36.973700

18 0.2119937 8.7556250 0.0242123 0.1142123 4.717120 41.301340

19 0.1944897 8.9501150 0.0217304 0.1117304 5.141661 46.018460

20 0.1784309 9.1285460 0.0195465 0.1095465 5.604411 51.160120

21 0.1636981 9.2922440 0.0176166 0.1076166 6.108808 56.764530y, 0.1501817 9.4424250 0.0159050 0.1059050 6.658600 62.873340~o 0.1377814 9.5802070 0.0143819 0.1043819 7.257874 69.531940-~

24 0.1264049 9.7066120 0.0130226 0.1030226 7.911083 76.789810

25 0.1159678 9.8225800 0.0118063 0.1018063 8.623081 84.700900

26 0.1063925 9.9289720 0.0107154 0.1007154 9.399158 93.323980

27 0.0976078 10.0265800 0.0097]49 0.0997349 1O.245080 102.723100

28 0.0895484 10.1l61300 0.0088520 0.0988520 11.167140 112.968200

29 0.0821545 10.1982800 0.0080557 0.0980557 12.172180 124.135400

30 0.0753711 10.2736500 0.0073364 0.0973364 13.267680 136.307500

35 0.048986l 10.5668200 0.0046358 0.0946358 20.-+ 13970 215.710S00

40 0.0318376 10.7573600 0.0029596 0.0929596 31 A09420 337.882400

45 0.0206922 10.8812000 0.0019017 0.0919017 48.327290 525.858700

50 0.01344S5 1O.9616800 0.0012269 0.0122690 74.357520 815.083600

60 0.0056808 11.0479900 5.142E-40 0.0905142 176.031300 1.944.792000

70 0.0023996 11.0844500 2.165EAO 0.0902165 416.730100 4.619.223000

80 0.0010136 11.0998500 9.132E-50 0.0900913 986.551700 10.950.570000

90 4.282E-40 11.1063500 3.855E-50 0.0900386 2,335.527000 25.939.180000

100 I.S09E-40 1l.l091000 1.62SE-50 0.0900163 5.529.041000 61,422.670000 c·-

L,oChapter (14): Feasibility Study IPMA Preparation Course


1= 10%

n Plr PIA Air AlP FIP FIA

0.9090909 0.9090909 1.0000000 1.l000000 1.100000 1.000000'2 0.8264463 1.735537 0.4761905 0.5761905 UI0000 2.1000003 0.7513148 2..:186852 0.302 1148 0.402J 148 1.331000 3.310000

.4 0.6830135 3.169865 0.2154708 0.3154708 1.464100 4.6410005 0.6209213 3.790787 0.1637975 0.2637975 1.610510 6.105100

6 Oj644739 4.355261 0.1296074 0.2296074 1.771561 7.7156107 Ojl31581 4.868419 0.1054055 (1.2054055 1.948717 9,4871718 0,4665074 5.334926 0.0874440 0.1874440 2.143589 11.4358909 0.4240976 5.759024 0.0736405 0.1736405 2.357948 13.579480

10 0385::433 6.144567 0.0627454 0.1627454 2.593742 15.937420

II 03504939 6.495061 0.0539631 0.1539631 2.853117 18.53117012 03186308 6.813692 0.0467633 0.J467633 3.138428 21.38428013 0.2896644 7.103356 0.0407785 0.1407785 3.452271 24.52271014 0.2633313 7.366687 0.0357462 0.1357462 3.797498 27.97498015 0.2393920 7.606080 0.0314738 0.13J4738 4.177248 31.772480

16 0.2176291 7.823709 0.0278166 0.1278166 4.594973 35.94973017 0.1978447 8.021553 0.0246641 0.124664 I 5.054470 40.54470018 0.1798588 8.201412 0.0219302 0.1219302 5.559917 45.59917019 0.1635080 8.364920 0.0195469 0.1l95469 6.115909 51.15909020 0.1486436 8.~ 13564 0.0174596 0.1l74596 6.727500 57.275000

21 Q1351306 8.648694 0.0156244 0.1156244 7.400250 64.00250022 QI228460 8.771540 0.0140051 0.1140051 8.140275 71.40275023 0.11l6782 8.883218 0.0125718 0.1125718 8.954302 79.54302024 QlOI5256 8.984744 0.0112998 0.1l12998 9.849733 88.49733025 0.0922960 9.077040 0.0101681 0.1101681 10.834710 98.347060

26 0.0839055 9.160945 0.0091590 0.1091590 11.91818 109.18180027 0.0762777 9.237223 0.0082576 0.1082576 13.10999 121.09990028 0.0693433 9.306567 . 0.0074510 0.1074510 14.42099 134.20990029 0.0630394 9.369606 0.0067281 0.1067281 15.86309 148.63090030 0.0573086 .9.426914 0.0060792 0.1060792 17.44940 164.494000

35 0.0355841 9.644159 0.0036897 0.1036897 28.10244 271.02440040 0.0220949 9.77905 I 0.0022594 0.1022594 45.25926 442.59260045 0.0137192 9.862808 0.0013910 0.1013910 n89048 718.904800SO 0.0085186 9.914814 8.592E-40 0.1008592 117.39090 1.163.909000

60 0.0032843 9.967157 3.295E-40 0.1003295 304.4816 3.034.81670 0.0012662 9.987338 1.268E-40 0.1001268 789.7470 7.887.47080 4.882E·40 9.9951180 4.884E-50 0.1000488 2.048.400 20.474.0090 1.882E·40 9.998118 1.883E-50 0.1000188 5.3 13.023 53.120.23

100 7.257E·50 9.999274 7.257E-60 0.1000073 13.780.61 137.796.1

Chapter (14): Feasibility StUdy IPMA Preparation Course

RS

1= 11%

" Plr PIA ..IfF AfP FlP Ff.·\

o.l)OO9009 0.9009009 1.0000000 1.1100000 I. 11()()OO 1.0000002 0.811622-1 1.7125230 0.-+739337 0.5:)39337 1.232100 2. I I00003 0.731191-1 2.-1-137150 0.2992131 0.-109213 I 1.367631 3.3-12100-I 0.6587310 3. 102-1-160 0.2 I 2326-1 0.3223'16<j. 1.51 ~070 -1.7097315 0.593-1513 3.6958970 0.1605703 0.2705703 1.685058 6.22780 I

6 0.53-16-10::\ -1.23053::\0 0.1263766 0.2363766 I.no-l 1S 7.9128607 0.481658-1 -1.7121960 0.1022153 0.2122153 2.076160 9.78327-18 0.-1339265 S.I-I61230 0.08-+3211 0.194321 I 2.304538 11.859-1309 0.3909}4:3 5.5370480 0.0706017 0.1806017 2.55::\037 I-'Ll63970

10 O.35218-1S 5.8891320 0.059801-1 0.1698014 2.S3942I 16.722010

II 0.3172833 6.2065150 0.0511210 0.1611210 3.151757 19.561-13012 0.2858-108 6,-1923560 0.0440273 0.1540273 3.498-"51 22.71319013 0.25751-13 6.7498700 0.0381510 0.1481510 3.883280 26.21164014 0.2319948 6.9818650 0.0332282 0.1432282 4.310441 30.0949:2015 0.2090043 7.1908700 0.0290652 0.1390652 4.784589 34.405360

16 0.1882922 7.3791620 0.0255167 0.1355167 5.310894 39.18995017 0.1696326 7.5487940 0.0224715 0.1324715 5.895093 44.5008-1018 0.1528222 7.7016170 0.0198429 0.1298429 6.543553 50.395'94019 0.1376776 7.8392940 0.0175625 0.1275625 7.263344 56.93949020 0.1240339 7.9633280 0.0155756 0.q55756 8.062311 64.202830

21 0.1117423 8.0750700 0.0138379 0.1238379 8.949166 72.26514022 0.1006687 8.1757390 0.0123131 0.1223131 9.933574 81.214310')0 0.0906925 8.2664320 0.0109712 0.1209712 11.026270 91.147880~~

24 0.0817050 8.3481370 0.0097872 0.1197872 !2.239160 102.17420025 0.07360~ I 8.4217450 0.0087402 0.1l87402 13.585460 114.413300

26 0.0663136 8.4880580 0.0078126 0.lI78126 15.079860 127.99880027 0.0597420 8.5478000 0.0069892 0.1169892 16.738650 143.07860028 0.0538216 8.6016220 0.0062571 0.lI62571 18.579900 159.81730029 0.0484879 8.6501100 0.0056055 0.1156055 20.623690 178.39720030 0.0436828 8.6937930 0.0050246 0.1150246 22.892300 199.020900

35 0.0259236 8.8552400 0.0029275 0.1129275 38.574850 341.58960040 0.0153844 8.9510510 0.0017187 0.1117187 65.000870 581.82610045 0.0091299 9.0079100 0.0010135 0.1110135 109.530200 986.63860050 0.0054182 9.0416530 5.992E-40 0.1l05992 184.564800 1.668.771000

60 0.0019082 9.0735620 2.103E-40 0.1102103 524.0572 4.755.06670 6.720E·40 9.0848000 7.397E-50 0.1100740 1,488.0 I9 13,518.36SO 2.367E-40 9.0887570 2.604E-50 0.1100260 4.225.113 38,401.0290 8.336E-50 9.0901510 9. 170E-60 0.1100092 11.996.87 109.053.4

lOO 2.936E-50 9.0906420 3.229E-60 0./100032 34.064.17 309.665.2 rL,j


M Management Consulting House

1= 120/0

/I PIF PIA AIF AlP FIP FI.-\

O.S9~~571 (1.8',)~~571 1.0000000 I.I~OOOOO I. I20000 1.000000

" 0.7971',)39 1.6900510 0.4716981 0.591698 I I.~5-l-l00 2.120000-,0.7117802 2.'1018310 0.2963490 0.'11634',)0 IAO-l',)2S 3.37-1400.'

4 0.6355181 3.0373490 0.201J234-+ O.329:!3-l-4 1.573519 4.7793~S

5 0.567426',) 3.60-17760 0.1574097 0.~77-l097 1.7623-L2 6.35~S47

6 0.5066311 4.111-1070 0.1232257 0.2432257 I.l)/::;g23 8.1151897 0,4523492 4.5637570 0.0991177 0.2191177 2.210681 10.0890108 0..)038832 4.9676400 0.0813028 0.20l30~8 2.475963 12.2996909 0.3606100 5.328~500 0.0676789 0.1876789 2.773079 1-1.775660

10 0.3219732 5.6502230 0.0569842 O. 1769842 3.1 058...n~ 17.5-187,)0

II 0.2874761 5.9376990 0.0484154 0.1684154 3.478550 20.65458012 0.2566751 6.19437-10 0.0414368 0.1614368 3.895976 24.133130l3 0.2291742 6.4235480 0.0356772 0.1556772 4.363493 28.02911014 0.~046198 6.6281680 0.0308712 O. 150871~ 4.887112 32.39260015 0.1826963 6.8108640 0.0268242 0.1468242 5.473566 37.279710

16 0.1631217 6.9739860 0.0233900 0.1433900 6.130394 4~.753280

17 0.1456443 7.1196300 0.0204567 0.1404567 6.866041 48.88367018 0.1300396 7.2496700 0.0179373 0.1379373 7.689966 55.74971019 0.1161068 7.3657770 0.0157630 0.1357630 8.612762 63.-:13968020 0.1036668 7.'1694440 0.0138788 0.1338788 9.646293 72.052440

21 0.0925596 7.5620030 0.012:2401 . 0.1322401 10.803850 81.69874022 0.08264~5 7.6446460 0.0108105 0.1308105 12.100310 92.50~580

~" 0.0737880 7.7184340 0.0095600 0.1295600 13.552350 104.602900--'24 0.0658821 7.7843160 0.0084634 0.1284634 15.178630 118.1552007- 0.0588233 7.8431390 0.0075000 0.1275000 17.000060 133.333900_J

26 0.0525208 7.8956600 0.0066519 0.1~66519 19.040070 150.33390027 0.0468936 7.9425540 0.0059041 0.1259041 ~ 1.3~4880 169.37400028 0.0418693 7.9844230 0.0052439 0.1252439 ~3.883870 190.69890029 0.0373833 8.0218060 0.0046602 0.1246602 26.7-19930 214.58280030 0.0333779 8.0551840 0.0041437 0.1241437 ~9.9599~O ~41.33270()

35 0.0 I89395 8.1755040 0.0023166 0.1223166 52.7996~0 431.66350040 0.0107468 8.2437770 0.0013036 0.1213036 93.050970 767.U91-10U-IS 0.0060980 8.28~5160 7.363 E-40 0.1207363 163.987600 1.358.23000050 0.003460~ 8.30-14980 4.167E-40 0.1204167 289002200 2.-l00.0 ISOOO

60 0.00111-11 8.3~-I0490 1.33SE-40 0.1 ~O 1338 897.5%') 7..1r71.6-H70 3.587E--I0 8.3303440 4.306E-50 0.1 ~00431 2.787.800

.... .., '"t') -. -. ...,:,,_1. ___' __"'_'

W I. 1551:·-10 S.:;Y~371 0 1J86E-50 0./200139 8.658.4};3 72.145.6990 3.71\/E·50 8.3330~30 4:46~E-60 0.1 ~00045 26.8\/1.93 224.{ll) I. ,

!OO I. 1\/7 E·50 S.':;3~2340 1.-I37E-60 0.120001-1 83.522.27 6%.010.6

Chapter (14): Feasibility Study lPMA Preparation Course

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I" 0£89t6'OS060£<:,'118£>£lP1'OS[S;[ZIO'O01:,LPWL£1:8L980'O01:0lv6tCOL0<:tL61'OItttl171"1'0tr[11"10'00696L[6'901'90860'061OtI,ZL'19891:"W'6600Z9P1'O60OZ910'O0,066£8'9£1:18011'0SI0906£n,8L0986'L,,80nnOtS09SIO'00£606<:L'96L11:,1:1'0LIO£L!L9'9t91:£L90'LZ9Zt'1,1'0Z9aleO'O0,L8£09'9Z96tlt'I'O91

09tLlt'OtOLZHZ'9SltLt,I'OSI,.LtWO06L£<:9t'980686,1'0,I 01L<:8S't£"Lt£n,L998,l'O,L99SWOOS8t<:0£'999L9081'O..I00LtS6'6Z110S68't£0>££91'0£0£:£££0'00;:1811:1'9,t91to;:'O£10810,9',;:t<::S-t:£Ct1986S91'OI9868WOOLt9LI6',6,OLO£;:'OZlOZ£rln<:198,:[8'[,ItS,L1'0'1,.8'WOo1!'69~9',LL6909;:'O11

O,L61t'81L9'1"6£'[968<:t81'O96~::-tmOO[t"9Zt',f}{~~t6Z'O01o[L,It'"ZtOrOOT68981'61'06:::tj~t90'OO~,t)!ll',~rS~Z~CO6O'Ji:L,C;:1ttt~f'9'ZL98£80;:'OLq~i"~LO'OOOLL,6L't66,19LCO80'.1C)tllt'(J1,09;:,C<:80119;:;:'0801196()'O00I9<:;:t'tlJll90,ZniL9()'::'~Z~~'~c~6J,(rz"f~llJ,"'O;:\.';;Ill:1'(100~;;!.6(,f,,!fOi-:t'Ot)

I..'.Z(}~t·~)~~'I~tS'1,tJ£tS;:'O~t1£t~I'()()It"LLI~.~.6()~LlTS:-'();;':'()L()t-~·1-t!.to£'J'I;:tc>19££,()Ztf1Il)O;:'O01LttU,;:!.i-:1~'\J'HI

~,

(l(!()l)()t't-!'()l'Ztt'lO;:;:~[;:t'O0:;:;;£(,;:'0m'"Il)~-';:ZO~Ol(NO~

(}(I(J(I,'I'Z00(,t)!.Z'!9~'~'t6i)n)C)l~ti>\WO(IZIII,'-.It)'I-,'itJ,'SCO-c()lJ(J(IiIll'I()OOll,'I'!OOO(JOl'J'IOOll()OOO'1X~~()ts~-OS~~61~,;\'(l

\/:1df.:ldl'c'.:III'\-u:lidI(

·1%fl=!j

;.,J

asnoHBumnsuo]1uaUlaBvuvwtS1J


i = 140/(1

11 PIF PIA AIF AlP FlP FI,4

O.S77 1L).1() 0.8771930 1.0000000 1.1400000 1.1400()O I.001)0000 0.7694(,75 l.n46661 0.467~897 0.607!897 1.2LJ9600 2.140000:; 0.n749715 2.]216~2 0.2907.315 0.4307315 lAg 15-J"'; 3.-1396004 0.)920S03 2.913712 O.~O3~O4~ 0.3432048 1.688960 4.9211-W5 OJ 193687 3.-+33081 0.1512835 0.2912835 1.925415 6.610104

6 0.-1555865 3.RB8668 0.1171575 0.2571575 2.194973 8.5355197 0.3996373 4.288305 0.09319~4 0.2331924 2.502269 10.730-1908 0.3505591 4.638864 0.0755700 O.~ 155700 2.S5~5S6 13.2327609 0.3075079 4.9.+6372 0.0621684 0.2021684 3.251949 16.085350

10 0.2697438 5.216116 0.0517135 0.1917135 3.707221 19.337300

II 0.236617-1 5.-152733 0.Q433943 0.1833943 4.226232 23.0'+-1520I~ 0.2075591 5.660~92 0.0366693 0.1766693 4.817905 27.27075013 0.1820694 5.842362 0.0311637 0.1711637 5.-192'+ II 32.0(;865014 0.1597100 6.0m072 0.0266091 0.1666091 6.261349 37.581060J5 0.1400965 6.142168 0.0228090 0.1628090 7.137938 43.8-+2410

16 0.12~S917 . 6.265060 0.0196154 0.1596154 8.1372-19 50.98035017 0.1077997 6.372859 0.0169154 0.1569154 9.27646-1 . 59.117600IS 0.0945611 6.467420 0.0146212 0.1546212 10.575170 68.394070

.19 0.0829484 6.550369 0.0126632 0.1526632 12.055690 78.96923020 0.0727617 6.623131 0.0109860 0.1509860 13.743490 9l.024930

21 0.0638261 6.686957 0.0095449 0.1495449 15.667580 104.76840022 0.0559878 6.742944 0.0083032 0.1483032 17.861040 120.43600023 0.0491121 6.792056 0.0072308 0.1472308 20.361580 138.29700024 0.0430808 6.835137 0.0063028 0.1463028 23.212210 158.65860025 0.0377902 6.872927 0.0054984 0.1454984 26.461920 181.870800

26 0.0331493 6.906077 0.0048000 0.1448000 30.166580 208.33270027. 0.0290783 6.935155 0.0041929 0.1441929 34.389910 238.-19930028 0.0255073 6.960662 0.0036645 0.1436645 39.204490 272.88920029 0.0223748 6.983037 0.0032042 0.1432042 44.693120 312.09370030 0.0196270 7.002664 0.0028028 0.1428028 50.950160 356.786800

35 0.0101937 7.0700-15 0.0014418 0.1414418 98.100180 693.5727()O40 0.0052943 7.105041 7.-15 IE--40 0.1407451 188.883500 1.342.0250DO45 0.00~7497 7.123217 3.S60E--40 0.1403860 363.6791 DO 2.590.56500050 0.0014281 7.132656 2.002E-40 0.1402002 700.233000 -1.99·<.521000

60 3.852£-40 7./-10106 5.395E-50 0.1400540 2.595.91 'J ID.S}5.l:;70 1.039E-4() 7.1-12115 1.455E-50 0.1400145 9.623.645 1>8.733.18SO 2.803E·50 7.1-t26:;7 3.9"4E·60 0.1400039 -'5.676.LJ8 ~5~.S2S.";

90 7.56/ E·60 7.142803 1.059E·60 0.1400011 132.262.5 t);';-.4.T24.S

100 2.039E·60 7.14:2h.i3 2.~55E· 70 0.1400003 -190.326.2 3.S0=:.32~.O


rI


i = 200/0

1/ PIF PIA AIF AlP FIP FlA

1 0.8333333 0.8333333 1.0000000 1.2000000 1.200000 1.0000002 0.694++44 1.5277780 0.4545455 0.6545455 1.-140000 2.2000003 0.5787037 2.1064SI0 0.2747253 0.4747253 1.728000 3.6400004 0,482253 I 2.5887350 0.1862891 0.3862891 2.073600 5.3680005 0.4018776 2.9906120 0.1343797 0.3343797 2.488320 7.441600

6 0.3348980 3.3255100 0,1007057 0.3007057 2.985984 9.9299!07 0.2790816 3.6045920 0.0774239 0.2774239 3.583181 /2.9159008 0.2325680 H371600 0.0606094 0.2606094 4.299817 16.4990809 0.1938067 4.0309670 0.0480795 0.2480795 5.159780 20.798900

10 0.1615056 4.1924720 0.0385228 0.2385228 6./91736 25.958680

II 0.1345880 4.3270600 0,0311038 0.2311038 7.430084 32.15042012 0.1121567 4.4392170 0.0252650 0.2252650 8.916100 39.58050013 0.0934639 4.5326810 0,0206200 0.2206200 10.699320 48.49660014 0.0778866 4.6105670 0.0168931 0.2168931 12.839180 59.19592015 0.0649055 4.6754730 0,0138821 0.2138821 15.407020 72.035110

16 0.0540879 4.7295610 0.0114361 0.2114361 18.488430 87.44213017 0.0450732 4.7746340 0.0094401 0.2094401 22.186110 105.93060018 0.0375610 4.8121950 0.0078054 0.2078054 26.623330 /28.16700019 0.0313009 4.8434960 0.0064625 0.2064625 31.948000 154.74000020 0.0260841 4.8695800 0.0053565 0.2053565 38.337600 186.688000

21 0.0217367 4.8913160 0.0044439 0.2044439 ' 46.005120 225.02560022 0.0181139 4.9094300 0.0036896 0.2036896 55.206140 271.03070023 0.0150949 4.9245250 0.0030653 0.2030653 66.247370 326.23690024 0.0125791 4.9371040 0.0025479 0.2025479 79.496850 392.484200?- 0.0104826 Vi475870 0.0021187 0.2021187 95.396220 471.981100-)

26 0.0087355 4.9563230 0.0017625 0.2017625 114.475500 567.37730027 0.0072796 4.9636020 0.0014666 0.2014666 137.370600 681.85280028 0.0060663 4.9696680 0.0012207 0.2012207 164.844700 819.22330029 0.0050553 4.9747240 0.0010162 0.2010162 197.813600 984.06800030 0.0042127 4.9789360 8,461 E-40 0.2008461 137.376300 1./81.882000

35 0.0016930 .:\.9915350 3.392E-40 0.2003392 590.668200 2.948.34100040 6.804E-.:\0 .:\.9965980 J.362E-40 0.2001362 1,.!69.772 7,343.85800045 2.734E-40 4.9986330 5A70E-50 0.2000547 3.657.262 18.281.31000050 1.099E-40 4.9994510 2.198E-50 0.2000220 9,100.438 45.497.190000

60 1.775E·50 5.9999110 3.529E-60 0.2000035 56.347.51 28 I.732.670 2.866E-60 4.9999860 5.732E-70 0.2000006 348.889.0 J .744.440.080 4.629E-70 4.9999980 9.258E-80 0.2000001 2.160.228.0 10.801.137.090 7A76E-SO 5.0000000 1,495E-80 0.2000000 13.375.565 66.877.822

100 1.207E-SO 5.0000000 2.145E-90 0.2000000 82.817.975 cl.1409E8

Chapter (14): Feasibifity Study lPMA Preparation Course

R$ Management Consulting House!j -!

i = 25%

11 ?iF PIA ..IfF AfP FlP FlA

O.SOOOOOll 0.::100000 1.0000000 1.2500000 I.25000ll 1.00000ll, O.6400ll()O 1..\40000 0.-144.\44..\ 0.694..\4..\.\ 1.562500 2.250000-J 0.5 120()OO 1.952000 0.262295 1 0.5122951 1.953125 3.~ I250ll-I 0,,\Ol)60()() 2.361600 O. 1734.\ I7 0.'123.\417 2..\41406 5.7656255 1>.3276800 2.689280 0.1218467 0.3718467 3.051753 S.207031

6 0.2621..l.\O 2.951.\24 0.OB8S 195 0.3388195 3.814697 11.2581907 0.20<)7152 3.161139 0.0663417 0.3163417 4.768372 15.073.\908 O.167T722 3.328911 0.0503985 0.3003985 5.960464 19.8418609 0.1342177 3..\63129 0.0387562 0.2887562 7.4505S I 25.802320

10 0.1073742 3.570503 0.0300726 0.2300726 9.313226 33.252900

II 0.0358993 3656403 0,0234929 0.1734929 1\.64153 42.56613012 0.0687195 3.725122 0.0184476 0,2684476 1-1-.55 192 5.\.20766013 0.054<)756 3.780098 0.0145434 0.2645434 18.18989 6ll.7595S014 0.0439805 3.824078 0.01 \5009 0.2615009 22.73737 86.94947015 0.Q35184:.1- 3.859263 0.0091169 0.2591l69 23.'1217 \ \09.686800

16 0.0281475 3.887410 0.0072407 0.2572407 35.52714 138.108500\7 0.0225180 3.909928 0.0057592 0.2557592 44.40892 173.63570018 0.0180144 3.927942 0.0045862 0.2545862 55.51115 213.04460019 0.0144115 3.942354 0.0036556 0.2536556 69.38894 273.55530020 0.0115292 3.953883 0.0029159 0.2529159 86.73617 342.94.\700

:n 0.0092234 3.963107 0.0023273 0.2523273 108.42020 429.68090022 0.0073787 3.970485 0.0018584 0.2513584 135.52530 538.10 11 0023 0.0059030 3.976388 0.0014345 0.2514845 169.40660 673.62640024 0.0047224 3.981111 0.0011862 0.2511862 211.75820 843.03290025 0.0037779 3.984888 9.481 E-40 0.2509481 264.69780 1.054.791000

26 0.0030223 3.937911 7.579E-40 0.2507579 330.87220 [,319.48900027 0.0024179 3.990329 6.059E-40 0.2506059 413.59030 1.650.36100028 0.0019343 3.992263 4.345E-40 0.2504845 516.98790 2.063.95200029 0.0015474 3.993310 3.875E-40 0.2503875 646..23490 2.530.93900030 0.0012379 3.995043 3.099E-40 0.2503099 807.79360 3.227.174000

35 4.056E-40 3.998377 L015E-40 0.2501015 2.465.19000 9.856.7610040 1.329E-40 3.999468 3.324E-50 0.2500332 7.523.16400 30.088.66000045 4.356E-50 3.999826 1.089E-50 0.2500109 22.958.37000 91.831.50000050 IA27E-50 3.999943 3.568E-60 0.2500036 70,064.92000 280.255.700000

60 1.532E-60 3.999994 3.831 E-70 0.2500004 652.530.'1 2.610.118.070 I.M6E·70 3.999999 4.114E-30 0.2500000 6.077 .163.0 24.308.64980 l.767E-80 4.000000 4.417E-90 0.2500000 56.597.994 2.2639E890 1.897E·90 4.000000 4.74E-100 0.2500000 5.2711E8 2.IOS4E9

lOO 2.04E·100 -1.000000 5.09E- II 0 0.2500000 4.9091 E9 1.964EI0IfL~,'



i = 30%

n PIF PIA AIF AlP FiP FIA

0.7692308 0.7692308 1.0000000 1.3000000 1.300000 . 1.0000002 0.5917160 1.3609470 0.4347826 0.7347826 1.690000 2.3000003 0.4551661 1.8161130 0.2506266 0.5506266 2.197000 3.9900004 0.3501278 2.1662410 0.1616292 0.4616292 2.856100 6.1870005 0.269329 [ 2.4355700 0.1105815 0.4105815 3.712930 9.043100

6 0.207 [762 2.6427460 0.0783943 0.3783943 4.826809 12.7560307 0.1593663 2.1'.021120 0.0568736 0.3568736 6.274852 17.5828408 0.1225895 2.)247020 0.0419152 0.3419152 8.157307 23.8576909 0.0942996 3.0190010 0.0312354 0.3312354 10.604500 32.015000

10 0.0725382 3.0915390 0.0234634 0.3234634 13.785850 42.619500

11 0.0557986 3.1473380 0.0177288 0.3177288 17.921600 56.40535012 0.0429220 3.1902600 0.0134541 0.3134541 23.298090 74.32695013 0.0330169 3.2232770 0.0102433 0.3102433 30.287510 97.62504014 0.0253976 3.2486750 0.0078178 0.3078178 39.373760 /27.91250015 0.0195366 3.2682110 0.0059778 0.3059778 51.185890 167.286300

16 0.0150282 3.2832390 0.0045772 0.3045772 66.541660 218.47220017 0.0115601 3.2948000 0.0035086 0.3035086 86.504160 285.01390018 0.0088924 3.3036920 0.0026917 0.3026917 112.455400 371.51800019 0.0068403 3.3105320 0.0020662 0.3020662 146.192000 483.97340020 0.00526 [8 3.3157940 0.0015869 0.3015869 190.049600 630.165500

2[ 0.0040475 3.3198420 0.0012192 0.3012192 247.064500 820.21510022 0.0031135 3.3229550 9.370E-40 0.3009370 321.183900 1.067.28000023 0.0023950 3.3253500 7.202E-40 0.3007202 417.539100 1.388.46400024 0.0018423 3.3271920 5.537E-40 0.3005537 542.800800 1.806.00300025 0.0014172 3.3286090 4.257E-40 0.3004257 705.641000 2.348.803000

26 0.0010901 3.3297000 3.274E-40 0.3003274 917.333300 3.054.44400027 8.386E-40 3.3305380 2.5[8E-40 0.3002518 1,192.533000 3.971.77800028 6.450E-40 3.3311830 I.936E-40 0.3001936 1.550.293000 5,164.31100029 4.962E-40 3.3316790 1.489E-40 0.3001489 2.015.381000 6,714.60400030 3.817E-40 3.33206 [0 1.145E-40 0.3001145 2.619.996000 8.729.985000

35 1.028E-40 . 3.33299 [0 3.084E-50 0.3000308 9.727 .~(;O 32,422.87000040 2.769E-50 3.3332410 8.306E-60 0.3000083 36.11 ~ ..';;) 120.393.90000045 7.457E-60 3.3333080 2.237E-60 0.3000022 /34.106.0 447,019.40000050 2.00SE-60 3.3333270 6.025E-70 0.3000006 497,929.2 1,659.761.000000

60 1.457E-70 3.3333330 4.370E-80 0.3000000 6,S64.377 .0 22.881.25470 1.057E-80 3.3333330 3.170E-90 0.3000000 94.631.268 3. J544ES80 7.67E-lOO 3.3333330 2.30E-IOO 0.3000000 1.3046E9 4.34S6E990 5.56E- [10 3.3333330 1.67E-I10 0.3000000 1.798EIO 5.995ElO

100 4.03E·120 3.3333330 l.21E-120 0.3000000 2.479E 11 8.264Ell



i = 35%

1/ PIF P/..\ .·I/F AlP FlP FlA

O.7~Oi~()7 0.1--101--107 1.000000 1.3500000 \ .350000 I.OOOO()I),

O.5~B(')t)6;-\ 1.2~<)-+3~O OA255319 0.7755319 \X,2500 .2.3:'OO(){J-3 O. --IO()--I--I 2 I 1.695~~nO n.23966-+5 0.5~966-15 2.--IbOJ75 ~.11251)1)

--I n.30 I06~2 I. <) <)6 <)--1X0 0.15076-+2 0.50076-+2 3.321500 6.o3~:)75

5 O.2230lJ5 2.21<)%10 O.I0045R.3 0,4504583 -+.-+X--IO}3 l).95--10~ 1

6 0.1651952 2.3851570 n.0692597 0,4 19'2597 6.053--1--15 l·qJS--IIO1 O. (~:2.J66X 2.5075230 0.0487999 0.3%7999 8.172151 20. ~c) IS6n~ 0.0906421 2.5981650 0.0348870 0.3848870 11.03240 28.66·+0109 0.0671423 2.6653080 0.0251912 0.3751912 14.89375 _,9.6964 10

10 0.0497350 2.7150430 0.0183183 0.3683183 20.10656 54.590160

II O.036840S 2.7518840 0.0133875 0.3633875 27.14385 74.69672012 O.O~7:2894 2.7791730 0.0098193 0.3598193 36.64420 10 l. 840600(J 0.0202144 2.7993870 0.0072210 0.35n210 49,46967 138,48~8()014 0.0149736 2.8143610 0.0053204 0.3553204 66.78405 187.95.140015 0.0110916 2.8254530 0.0039256 0.3539256 90.15847 254.738500

16 0.0082160 2.8336690 0.0028994- 0.3528994 121.7139 3-1·U9700017 0.0060859 2.8397550 0.0021431 0.3521431 164.3138 -166.61090018 0.0045081 2.8-+42630 0.0015850 0.3515850 221.8236 630.9'2470019 0.0033393 2.8476020 0.0011727 0.3511727 299.4619 852.74830020 0.0024736 2.8500760 8.679E-40 0.3508679 404.2736 1.152.210000

21 0.0018323 2.85\9080 6.425E-40 0.3506425 545.7693 1.556.48400022 0.0013572 2.8532650 4.757E-40 0.3504757 736.7886 2,102.25300023 0.00t0054 2.8542700 3.522E-40 0.3503522 . 994.6646 2.839.04200024 7.447E-40 2.8550150 2.608E-40 0.3502608 1.342.7970 3.833.70600025 5.516E-40 2.8555670 L932E-40 0.350 \932 1.812.7760 5.176.504000

26 4.086£--+0 2.8559750 lA31E-40 0.350143\ 2.;W7.2480 6.989.28000027 3.027E--i0 2.8562780 1.060£-40 0.3501060 3.303.7850 9.436.52800028 2.242E--i0 2.8565020 7.849E-50 0.3500785 4.460.1090 12,740.31000029 1.66 t E-40 2.8566680 5.814E-50 0.3500581 6.02 L 1480 17.200.420000030 1.230E-40 2.8567910 4.306E-50 0.3500431 8,118.5490 13.111.5700000

35 2.744E-50 2.8570640 9.603E-60 0.3500096 36.448.69 j 04.136.300000040 6.119£-60 2.8571250 2.142E-60 0.3500011 163,437.10 466.960.400000045 1.365E-60 2.3571390 4.776E-70 0.3500005 732,857.6 2.093.876.000000050 3.043£-70 2.8571420 l.065E-70 0.3500001 3.286.158.0 9.389.020.0000000

60 1.513E-30 1.8571430 5.297E-90 0.3500000 66.073.317 1.3878£370 7.53£-100 1.8571430 2.63E-IOO 0.3500000 1.3285E9 3.7957E980 3.74E·IIO 2.8571430 UI E- II 0 0.3500000 2.671 £10 7.631E 1090 1.86E-110 1.8571430 6,52E-130 0.3500000 5.371EII l.535EI2

100- 9.26E· 1-i0 2.8571430 3.24£-140 0.3500000 1.080EI3 3.085£ 13

-----------------------LChapter (14): Feasibility Study lPMA Preparation Course


i = -10%

1/ PlF PIA AIF AlP FlP FlA

0.71-12857 0.71-12857 1.0000000 104000000 1AOOO()O 1.0000002 0.51020-11 1.2244900 OAI66667 O.S 166667 1.960000 2.-l000003 0.J6-1-1315 1.5889210 0.2293578 0.6293578 2.7-14000 -I.3bOOOO4 0.2603082 1.8492290 0.1407658 0.5407658 3.841600 7.10-10005 0.1859344 2.0351640 0.0913609 004913609 5.378:240 10.945600

6 0.1328103 2.1679740 0.0612601 0.461 :2601 7.529536 16.3238407 0.09486-15 2.2628390 0.04192:!8 0.4419228 10.541350 23.8533808 0.0677604 2.3305990 0.0290742 0.4290742 -Ll57890 34.3947309 0.048-1003 2.3789990 0.0::03448 0.4203448 20.661050 49.152620

10 0.0345716 2.4135710 0.0143238 0.4143238 28.925470 69.813660

II 0.0246940 2.4382650 0.0101277 0.'1101277 40.49565 98.73913012 0.0176286 2.4559040 0.0071821 0.4071821 56.69391 139.234800I' 0.0125990 2.4685030 0.0051039 0.4051039 79.37143 195.928700.'14 0.0089993 2.4775020 0.0036324 0.4036324 111.12010 275.300200IS 0.0064281 2.4839300 0.0015879 0.4025879 155.56810 386.420200

16 0.0045915 2.48852 j 0 0.001845 I 0.4018451 217.7953 541.98830017 0.0032796 2.4918010 0.0013162 0.4013162 304.9135 759.78370018 0.0023426 2.4941440 9.392E-40 0.4009392 426.8789 1.064.69700019 0.0i116733 2.4958170 6.704E-40 0.4006704 597.6304 1.491.57600020 0.0011952 2.4970120 4.787E-40 0.4004787 836.6826 2.089.206000

21 8.537E-40 2.4978660 3.418E-40 0.4003418 1.171.356 2.925.88900022 6.098E-40 2.4984760 2.441£-40 0.4002441 1.639.898 4.097.24500023 4.356E-40 2.4989110 1.743E-40 0.4001743 2.295.857 5.737.14200024 3.111£-40 2.4992220 1.245E-40 0.4001245 3,214.200 8,032.99900025 2.222£-40 2.4994440 8.891£-50 0.4000889 4.499.880 11.247.200000

26 1.587E-40 2.4996030 6.350E-50 0.4000635 6.299.831 15.747.08000027 U34E-40 2.4997170 4.536£-50 0.4000454 8.819.764 22.046.91000028 8.099£-50 2.4997980 3.240E-50 0.4000324 12.347.670 30.866.67000029 5.785E-50 2.4998550 2.314E-50 0.4000231 17.286.740 43,214.34000030 4.132£·50 2.4998970 1.653E-50 0.4000165 24.20 1.-l30 60.510.080000

35 7.683E-60 2.4999810 3.073£-60 OAOOO031 130,161.100 325.400.30000040 1.428E-60 2.4999960 5.714E-70 0.4000006 700.037.700 1.750.092.00000045 2.656E-70 2.4999990 1.062E-70 0.4000001 3,764.971.000 9.-l12A24.0000(JO50 4.939E-80 2.5000000 1.975E-80 0.4000000 20.248.916.000 50.622.288.000000

60 1.707E·liO 2.5000000 6.83E-100 0.4000000 5.8571E8 1.-l643E970 5.90E·II0 2.5000000 2.36E-110 OAOOOOOO 1.694E I0 -+.235E 10SO 2NE-120 2.5000000 8.16E-130 0.4000000 -I.90IEII 1.225E 1290 7.05E-140 2.5000000 2.82E-140 0.4000000 1.418E13 3.544E 13

100 2.44E-1 SO 2.5000000 9.76E·160 0.4000000 4.100E14 1.025EI5


IManagement Consulting House

r

i = ..5 '70

" PIF 1'/. \ .I/F .. \/1' FIP Fl."

O.6K9(,552 O.h;)L)6S52 1.1l0000O 1...500000 1.-150000 1.0ooilOO0 O.-I7:'\h2"3 I. I6:'\ 27<}{) O,.)OX 1633 O.X5X lo~;3 2.102:'\00 2.-I5(01)()-; tU2XO 1h7 1AlJ 32960 0.21%5'1:5 O.66W>:i95 3.0-l~625 ..,l..55~:'OO

-I {L~2()21~4 1.71'):'\1:'\0 O.1315:,\cl5 IJ.5X 15595 ... -120:'\06 ;.bilI125:'\ 0.1 :,\(,Ill 27 1.S7S5270 1l.0X" I83-1- n.s,:;:; I ~.1.f 6."0')73-1 12.021 I">}O

6 11.10759S0 1.9~J 1220 O.OS42553 1).50-l255} l) .29-'\1 1-1 I SA} 1.3/0~ lW7420}4 2.0:'\73260 0.0360679 0.4~606711 13.-I7G-I70 27.725'+:-)0S 0.05117-13 2.IOX5000 0.0242707 0.-17..2707 1<) .5-1l18SIl -11.201950') O.0}:'\2931l 2.1-137930 0.016-1629 O.-lfG-I629 28.334270 60.7-12820

10 Il.02-13-100 2.1681330 0.0112262 0.-1612262 -11. OX..Hi90 StJ.077090

II 0.16nf>2 2.1849200 0.0076<1:27 0.-1576827 59.572S00 130.16\SOO12 0.115767 2.1964960 0.0052705 0.-1552705 86.380560 180.73..160013 0.0079839 2.20-14800 0.0036217 0.'1536217 125.25\800 276.115\001-1 0.0055061 2.2099860 0.002-1915 0.-1524915 181.615.100 -101.367000\5 0.0037973 2.2 I37S-l0 0.0017153 0.'1517153 263.341900 582.982100

16 0.00261S9 2.2164030 0.0011816 OA51l816 381.845800 8-16.32..00017 0.00\8061 2.2 \82090 8.142E ..40 0.-1508142 553.676400 1.228.17000018 0.0012456 2.2194540 5.612E-40 0.45056\2 802.830800 1,781.S"600019 8.590E...:\0 2.2203130 3.869E..40 0.4503869 1.164.105000 2.584.67700020 5.924E..40 2.2209060 2.668E..40 0.4502668 1.687.952000 3.748.782000

21 4.086E..40 2.2213140 1.839E..40 .0.4501839 2.447.530000 5.436.734000~~ 2.818E....0 2.2215960 1.26SE.....O 0.4501268 3,548.919000 7.88...26-1000~" 1.943E....0 2.2217900 8.746E ..50 0.4500875 5.145.932000 11,433.1 SOOOO~~

24 1.340E....0 2.2219240 6.032E..50 0.4500603 7.461.602000 16.579.11000025 9.243E.. 50 2.2220170 4.160E..50 0.4500416 10.819.320000 24.0..\0.720000

26 6.374E ..50 2.2220810 2.869E..50 0.4500287 15,688.0200 34.860.04000027 4.396E .. 50 2.2221250 1.978E.. 50 0.-1500\98 22.747.6300 50.548.06000028 3.032E.. 50 2.2221550 U64E..50 0.4500136 32.984.0600 73.295.68000029 2.091 E.. 50 2.2221760 9.409E..60 0.4500094 47.826.8800 106.279.70000030 1.442E·50 2.2211900 6,489E..60 0.4500065 69.348.9800 154,106.600000

35 2.250E ..60 2.2222170 1.012E..60 0.4500010 444.508.5 987.79-1.50000040 3.510E..70 2.2~22210 1.579E-70 0.4500002 2.S49.18\.0 6.33 L512.000000-15 S.476E..SO 2.2222:220 2,464E..80 0.4500000 18.262.495 40.583.319.00000050 8.543E-90 2.2222220 3.844E..90 0.4500000 1.1706E8 2.6013E8000000

60 2.08E-100 :2.22222'20 9.36E .. 110 0.-1500000 4.8093E9 1.069E I000000070 5.06E-120 2.2222220 2.28E.. 120 0.-1500000 1.976E II 4.391 E II 00000080 1.23E-130 2.:2222220 5.54E.. 140 0.4500000 8.IISEI2 1.804E 1300000090 3.00E-150 2.2222'220 U5E- \50 0.4500000 3.335E 14 7.412E 1-1000000

100 7.30E.. 170 2.2222220 3.28E.. 170 0.4500000 l.370EI6 3.045E 16000000

--------------------------'L,Chapter (14): Feasibility Study IPMA Preparation Course


Economic Exercises

Exercise No.# (1)

A construction finn is studying buying new machinery to replace the old one. Three mutually exclusivealternatives were provided. If the opportunity cost of Capital ( i) ~ I2%, and the useful life of themachinery is 5 years, which equipment should be selected? Use the Present Worth Method.

Equipment Initial Cost Annual Costs Annual Income Salvaee ValueA 35,000 8,000 23,000 3,500B 40,000 7,000 23,000 4,000C 100,000 2,000 34,000 10,000

3,500

Solution No# (11: - i = 12%

o

r

15K

11 21 31 41 51ALTA

35K

NPV (A) = - 35,000 + 15,000 (PIA) i~l2, 0=5 + 3,500 (P/F) i~12, n=5 1

4,000= - 35,000 + 15,000 (3.6047) + 3,500 (0.5674)

= 21,056.4 $o

r40K

16K

11 2t 31 41 51ALTB

NPV (B) = - 40,000 + 16,000 (PIA) i~l2, 0=5 + 4,000 (PIF) i~l2, n~5

= - 40,000 + 16,000 (3.6047) + 4,000 (0.5674)

= 19,944.8 $ o

r

32K 10,000

It 21 31 41 sfALTC

lOOKNPV(C) = - 100,000 + 32,000 (PIA) i~12, 0=5 + 10,000 (P/F) i~l2,n~5

= - 100,000 + 32,000 (3.6047) + 10,000 (0.5674)

= 21,024.4 $

From Above Alternative (B) is Acceptable


Exercise No.# (2)

An engineering firm is considering buying special Computer Aided Design (CAD) equipment which willcost 250,000 LE has an estimated salvage value of 28,000 LE in 10 years.Maintenance costs are estimated at 4,000 LE per year. If the firm can generate 55,000 LE annually inincome from this equipment, what will be the net present value of the firm of such investment if its MARRis 9%?

Solution No# (2): -

51K 28K

o

250K

1 2 3

i=9%

10

NPV = - 250,000 + 51,000 (PIA) ;=9, n=IO + 28,000 (PIF) i=9, n=IO

= - 250,000 +51,000 (6.41) + 2,000 (0.422)

= 89,128.05 LE

f"-------------------------------L.



Exercise No.# (3)

An investment of $1 0,000 can be made in a project that will produce a uniform annual revenue of $5,310for 5 years and then have a salvage value of $2,000. Annual disbursements will be $3,000 each year foroperation and maintenance costs. The Company is willing to accept any project that will earn 10% ormore, before income taxes, on all invested capital. Show whether this is a desirable investment or not usingthe annual worth method.

Solution No# (3): -

2.310 2.000

i = 10%

0r-__.J,;1~ 2;;....J__.....;;3_-L. 4-L.__,.;;5...L

10.000

NPV = - 100,000 +2,310 (PIA) i=10,0=S + 2,000 (PIF) i=10,0=5

- - 100,000 +2,310 (3.7907) + 2,000 (0.6209)

= - 1.683 $

So as the result ofNPV is negative, It means the project is NOT acceptable for the company.


Management Consulting House"

Exercise No.# (4)

In building the landing strip at small municipal airport, which is not used by commercial planes, onemethod of construction will cost $500,000 and have an estimated life, with proper maintenance of 40years. The annual cost of such maintenance is estimated of $10,000 per year. An alternative type ofconstruction is estimated to be $ 275,000 . but at the end of every 10 years another.

Solution No# (4.): -

I

I

Alt# 11 2 3 40

10K

SOOK

10 20 30 40Alt#2

r I I I27SK 27SK 27SK 27SK

Assume 1=10%

NPV for Alt 1 = 500,000 + 100,000 (PIA) i~lO. n=10

= 500,000 + 100,000(9.779)

= 1,477,900 $

NPV for All 2 = 275,000 [ 1 + (P/F) i~IO, n~IO + (P/F) ;=10, n~20 + (PIF) i~IO, n~30 ]

= 275,000 [1 + 0.3855 + 0.1486 + 0.0573 ]

= 437,635 $

-----.;,)... Alternative (2 ) should be used as its cost is less than the other one

------------------------- -_~I



Exercise No.# (5)

The Jack n' Jill Toy Company currently purchases the metal parts which are required in the manufacture ofcertain toys. The purchase price of metal parts needed for one toy is $ 0.5. There has been a proposal thatthe company make these parts themselves from the scrap material they got. Two machines will be requiredfor the operation:Machine A will cost $ 18,000 and have a life of 6 years and a $2,000 salvage value. Machine B will cost $12,000 and have a life of 4 years and $ 5,00 Salvage value.Machine A will require an overhaul after 3 years costing $3,000. The annual fixes operating costs ofmachines A and B are expected to be $6,000 and $ 5,000 respectively.A total of four laborers will be required for the two machines at a cost of $2.5 per hour per worker. In anormal 8-hour day the machines can produce parts sufficient of manufacture 1,000toys. Use MARR of12% per year. How many toys must be manufactured each year in order to justify the purchase of themachines?

Solution No# (5): -

• Annual Cost for Manuf.We will take 12 year as project's life

= 6,000 + 5,000 + (4 x 2.5 x 8) for each 1,000 toys= 11,000 + 80X

C 2K 2K

lRK

12K.-

lRKAssume X -

Annual Cost A =

0.5K 0.5K0.5K.

1 2 3 4 5 7 8 9 10 11 12

1 1 1 1 1 1 1C 3K C C 3K C

12K 12K

if

Numbers of toys / 1,000

30,000 (AlP) ;=12, n=12 + C + I 1,500 (P/F) ;=12, n=4 (AlP) ;=12, n=12

+ 16,000 (PIF) ;=12, n=6 (AJP) ;=12, 0~12 + I 1,500 (P/F) ;=12, n=8 (AlP) ;=12, n=12

- 2,500 (AIF) i=l2, 0=12 + 3,000 [(PIF) ;=12, n=3 (AlP) i=l2, 0=12 + (P/F) ;=12, n=9 (AlP) ;=12, n=12]

Annual CostA = 30,000(0.1614) +C + 11,500(0.6355)(0.1614) + 16,000(0.5066)(0.1614)

+ I 1,500 (0.4038)(0.1 61 4) - 2,500 (0.04 I4) + 3,000 [ (0.71 17)(0. I614) + (0.361 )(0.1 61 4 I= C + 8,495

AnnualCostA= 11,000 +8,495 +80X = 19,495 + 80X

Purchasing Cost = 0.5 + 1,000 X = 500 X

500X-80X 19,495 X=46.417Number of Annual Toys to be manufactured = 46.417 x 1,000 = 46.417 Units.


RC' Management Consulting House~ f--------------------'-------------:l

Exercise No.# (6)

A company offer a few number of houses. With selling price equal to 250,000LE, the buyer showed pay30% in advance, and the rest to be in two unequal payments with interest of 11 % yearly.First payment is 30% from in the house price and to be paid after seven years. Second payment to be paidafter ten years. Calculate the value of the second payment?

SOLUTION (6)

250,OOOLE

o

75,OOOLE

'L= second payment

We transfer all the amount to Future Value after 10 years:-

(i) = 11%

7

75,OOOLE

10

z

'L= 250,000 ( F / P) ;= 11, n=10 - 75,000 (F / P) ;= 11, n=IO - 75,000 ( F / P) ;= II, n=3

'L= 250,000 (2,84) - 75,000 (2.84+1.3676)

= 394,430LE



Exercise No.# (7)

A plant manager is trying to decide whether to buy a new unit of equipment now or wait and purchase asimilar one 3 years from now.

The piece of equipment at the present time would cost LE 75,000, but three years from now it is expectedto cost LE 120,000.

1fthe interest rate the company uses is 20% per years, should the plant manager buy now or buy it 3 yearslater?

SOLUTION (7)

(i) =20%75,OOOLE

1

NPV for Case 1 = 75,000 LE

NPV for Case 2 = 120,000 LE ( P / F) ;= 20,0=3

= 120,000 (0.5787)

= 69.444.44 LE

Case 2 < Case 1

------.1 Then, He should buy it after 3 years.

Chapter (14); Feasibility Study

23

120,OOOLE


IRS Management Consulting Housf

Exercise No.# (8)

A dealer sells a refrigerator at six annual installments of$1250 each plus a down payment of$500.Calculate the interest rate if the cash price was $5000.

SOLUTION (8)

5,000$

12 3

I

4

I

5

I

6

1,250$

5,000 = 500 + 1,250 ( P / A) i~?, n~6

500$ ( P / A) i=?, 0=6 = 4,500 / 1,250 = 3.6

From the Table (i) = ???? %

- 8.71 + 4.355 X

4.62

z

3.6

..-. -- ....

4.355

10

............................

.......~ ..............

8

= 32.878.71

0.265

- 4.355

X

4.62 X

4.62

2+X

X

As 4.355

32.87

3.6

z

z = 3.6 (30.56) = 25.26

4.355

X

(i) = 10 + X - Z = 15.3%

-------------------------------------:LChapter (14): Feasibility Study IPMA Preparation Course


Exercise No.# (9)

A company is considering constructing a plant to manufacture a proposed new product. The land costs$300,000 the building costs $600,000, the equipments costs $250,000 and $100,000 working capital isrequired.It is expected that the product will result in sales of $750,000 per year for 10 years, at that time the landcan be sold for $400,000, the building for $400,000. the equipment for disbursements for labor, materials,and all other expenses are estimated total $475,000. If the company requires a minimum return of25% onprojects of a comparable risk, detennine if it should invest in the new product line. Use the internal rate ofreturn method.

SOLUTION (9)

Salvage Value= 800,OOOLE

MARR= 25%

Assume (i) = 8%

275,OOOLE

1,150,000

NPV = -1,150,000 + 800,000( P / F) ;=8, n=IO + 275,000 (P / A) i=8.n=IO

NPV = -1,150,000 + 800,000(0.4632) + 275,000 (6.71)

= 1,065,810$

Assumed (i) = 12% :-

NPV = - 1,150,000 + 800,000 (0.322) + 275,000 (5.65)

= 6,61,350$


NPV

.. 1,065,810


661,350

8

1,065,810 = 661,350

X+4 X

X(l,065,810) = 661,350 (X+4)

12

...................................

x

NPVX= 2,645,400

1,065,810 - 661,350

=654,

(i) = 12 +6.54 = 18.5%

IRR<MARR

"N0" for New Production Line

f

----------------------------------lChapter (14): Feasibility Study IPMA Preparation Course