chapter 3 project management. projects are typically characterized as: –one-time, large scale...

Chapter 3

Project Management

Project Management

Projects are typically characterized as:– one-time, large scale operations– consuming large amount of resources– requiring a long time to complete– a complex set of many activities

3 Important Project Management Functions:– Planning – determine what needs to be done– Scheduling – decide when to do activities– Controlling – see that it’s done right

PERT/CPM project management technique(Program Evaluation & Review Technique)/(Critical Path Method)

• Inputs– list of activities– precedence relationships– activity durations

• Outputs– project duration– critical activities– slack for each activity

1 2Excavate

& pour footings

Pour foundation

Install drains

Project Network for House Construction

3

6

7

4

8

9

5

10

11

12

16

1813

1715

14

Install roughelectrical & plumbing

Pourbasement

floorInstall

cooling &heating

Erectframe & roof

Laybrickwork

Laystormdrains

Installdrywall

Layflooring

Installfinished

plumbing

Installkitchen

equipmentPaint

Finishroof

Installroof

drainage

Finishgrading

Finishfloors

Pourwalks;

Landscape

Finishelectrical

work

Finishcarpeting

CPMA project has the following activities and precedence

relationships:

Immediate Immediate Predecessor Predecessor

Activity Activities Activity Activitiesa -- f c,eb a g bc a h b,dd a i b,de b j f,g,h

Construct a CPM network for the project using:1.) Activity on arrow2.) Activity on node

Activity on Arrow(Initial Network)

Activity on Arrow(Final Network)

a

b

c

d

e

f

g

hi

j

Activity on Node

Critical Path

path any route along the network from start to finish

Critical Path path with the longest total duration

This is the shortest time the project can be completed.

Critical Activity an activity on the critical path

*If a critical activity is delayed, the entire project will be delayed. Close attention must be given to critical activities to prevent project delay. There may be more than one critical path.

To find critical path: (brute force approach)

1. identify all possible paths from start to finish

2. sum up durations for each path

3. largest total indicates critical path

1

2 6

4 7

53

b = 2

d = 4

g = 9

h = 9

f = 8c = 5

a = 6

k = 6

j = 7

i = 4

e = 3

Slack Times

Earliest Start (ES) – the earliest time an activity can startES = largest EF of all immediate predecessors

Earliest Finish (EF) – the earliest time an activity can finishEF = ES + activity duration

Latest Finish (LF) – the latest time an activity can finishwithout delaying the project

LF = smallest LS of all immediate followers

Latest Start (LS) – the latest time an activity can start without delaying the project

LS = LF – activity duration

Slack Times

Slack how much an activity can be delayed

without delaying the entire project

Slack = LF – EF or Slack = LS – ES

Slack

EF LF

ES LS

c = 10

g = 12

f = 1

7b =15

a = 10 e = 15

i = 7

d = 20

h = 9

b = 4

d = 5

h = 5

i = 3

c = 5

a = 5g = 4

j = 6

e = 5

f = 6

Input Table for Microsoft Project(Example 10.1, page 387)

Gantt Chart for Microsoft Project(Example 10.1, page 387)

Project Network for Microsoft Project(Example 10.1, page 387)

Activity Crashing(Time-Cost Tradeoffs)

An activity can be performed in less time than normal, but it costs more.

Problem: If project needs to be completed earlier than normal, which activity durations should be decreased so as to minimize additional costs?

Guidelines:• Only crash critical activities• Crash activities one day at a time• Crash critical activity with lowest crashing cost per day

first• Multiple critical paths must all be crashed by one day

Activity Crashing Example

Crash project as much as possible.

Activity DurationCrashedDuration

ActivityCost

CrashedCost

CrashingCost/day

a 3 2 40 45

b 4 3 50 54

c 8 5 50 68

d 5 4 30 33

c = 8 d = 5

b = 4a = 3

Minimum duration = 9 days; Total additional cost = $30

Program Evaluation & Review Technique(PERT)

3 duration time estimates

– optimistic (to), most likely (tm), pessimistic (tp)

Activity duration:

mean te = (to + 4tm + tp) / 6

variance Vt = [(tp – to) / 6]2

Path duration:

mean of path duration = T = Σ te

variance of path duration = σ2 = Σ Vt

X = T ± Zσpath

Z is number of standard deviations that X is from the mean.

Example: If the mean duration of the critical path is 55 days and the variance of this path is 16, what is the longest the project should take using a 95% confidence level?

probabilityof being late

.05

actualproject

duration

T55

X

Zσcp

PERT Example

If the expected duration of a project is 40 days and the variance of the critical path is 9 days, what is the probability that the project will complete in less than 45 days?

in more than 35 days?

in less than 35 days?

in between 35 and 45 days?

probabilityof being late

actualproject

duration

T40

45

Zσcp

PERT Example

The expected duration of a project is 200 days, and the standard deviation of the critical path is 10 days. Predict a completion time that you are 90% sure you can meet.