pmp certificate math practice - sample - time management

Time Management Chapter

This is a sample of my book

PMP Certificate Math Practice

That includes

Earned Value Measurement problems, Schedule

compression, Critical Path, Schedule Crashing and

Fast tracking Problems, Contracts Problems, Risk

Problems, and More.

This book aims to guide you through the different types of math-

based questions that you can expect to find on the exam, including

a deep collection of earned value management, decision tree,

critical path and float problem examples. With the practice

questions provided within, you can gain those very insights that are

required to maintain momentum as you progress through the exam

on your way to achieving the certification.

if you are interested in buying the book please visit

https://gumroad.com/l/oguo

Or at amazon.com

http://www.amazon.com/PMP-Certification-Math-

Practice-Compression-

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Chapter Two

Time Management

This Chapter Covers

2.1 Three Point Estimating

2.1.1 Triangular Distribution

2.1.2 Beta Distribution

2.2 Critical Path Method

2.3 Schedule Compression

2.3.1 Crashing

2.3.2 Fast Tracking

PMP Certification Math Practice

84

Time Management Chapter may be difficult for those who rarely had to

deal with scheduling and duration estimating of tasks. It is really important

for a project manager to understand how to schedule, the importance of

creating a network diagram, the compression of schedule. The exam will

test your knowledge of these topics. You will find a plenty of practice

problems in this chapter to help you understand and answers the time

management questions for the exam.

2.1 Three Point Estimating

During project planning, we need to estimate time duration and cost for the

project activities. We can use a single-point estimation to estimate activity

duration, where the estimator submits one estimate per activity: Activity A

will take 5 days to complete. Things do not always go as we plan and we

need to incorporate uncertainties and risks in our estimation. So, we use a

Three-point estimation technique to come up with an approximate range for

the activity duration. With this the project manager can better understand

the potential variations of the activity and overall project estimates.

Program Evaluation and Review Technique (PERT) is one of the concepts

that use three-point estimating technique. Here we use three estimates to

define the approximate range of estimated duration for an activity. The

estimator gives three estimates most likely (tM), Optimistic (tO) and

Pessimistic (tP) estimates for an activity.

CHAPTER TWO Time Management

85

2.1.1 Triangular Distribution

The expected duration of the activity can be calculated using a simple

average of the three estimates which gives triangular distribution of values

within the range. The formula to calculate expected duration using

triangular distribution:

tE = (tO + tM + tP)/3 [or simply O+M+P/3 ].

Example: Duration of an activity has the following estimates: Optimistic – 5

days; Most Likely – 8 days; Pessimistic – 10 days. The time estimate for

that activity based on the triangular distribution is:

A. 7 days.

B. 8 days.

C. 6 days.

D. 9 days.

Correct Answer: A

Solution:

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Triangular distribution problems

1. For an activity, we think that in the best case we need 3 days to

finish a task, most likely this is going to be 5, but in the worst

scenario, in case where we need to perform much more work

because not all the details have been provided, we believe that it is

going to take 10 days. What is the estimated duration of the activity

using triangular distribution?

A. 5.5

B. 6

C. 5

D. 4

2. The following four tasks represent the Critical Path of a project. The

estimates of each of these tasks are shown below. What is the length

of the Critical Path triangular distribution is used?

Task: Optimistic, Most likely, Pessimistic

A. 17 22 33

B. 12 25 32

C. 18 29 34

D. 8 22 26

A. 80

B. 86

C. 90

D. 96


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2.1.2 Beta Distribution

With a Beta distribution technique stronger consideration is given for the

most likely estimate. This technique is derived from the traditional PERT

technique which calculates the weighted average for the expected duration

of the activity. The formula to calculate expected duration using Beta

distribution (PERT technique) is:

tE = (tO + 4xtM + tP)/6 [or simply O+4xM+P/6 ].

Standard deviation for an activity using Beta distribution= tP – tO/6 [or

simply P-O/6].

Variation for an activity using Beta distribution = [(tP – tO)/6]2 [or

square of Std Dev].

Example: You are managing a software development project and your team

is estimating activity durations for each of the tasks identified. A task has

the most optimistic estimate as 7 days, the most pessimistic as 15 days and

the most likely as 13 days. What are the three-point estimate, Standard

deviation and variation for this task?

A. PERT = 11.67; SD = 8; Variation = 16.

B. PERT = 12.5; SD = 8; Variation = 16.

C. PERT = 11.67; SD = 1.33; Variation = 1.77.

D. PERT = 12.5; SD = 1.33; Variation = 1.77.

Correct Answer: D

Solution:


88

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PERT, Standard deviation, and variance problems

1. An activity in your project has the following estimates. Optimistic:

15 days; Most probable: 21 days; Pessimistic: 26 days. What is the

expected duration of the activity using three point estimates?

A. 28.16

B. 20.83

C. 21.27

D. 27

2. You have completed estimates for all your project tasks and have

arrived at the total project duration. The optimistic estimate for the

project is 52 weeks and pessimistic estimate is 64 weeks. What is

the standard deviation of the project?

A. 1.5

B. 2.0

C. 2.5

D. 5.0


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variance of the activity?

A. 1.568

B. 1.833

C. 3.361

D. 7.564

4. You are managing a software development project and have come

up with the PERT estimates (in days) for the Critical Path activities

as shown below. What is the standard deviation of the allover path?

Tasks Optimistic Most likely Pessimistic

Requirement collection 10 12 16

Design 22 28 34

Code 35 45 59

Testing 20 22 26

A. App 6.8 days

B. App 5.2 days

C. App 4.7 days

D. You cannot derive the path standard deviation from the

information given.


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5. You are working on estimated activity durations for your project

tasks. You have applied a three-point estimation on a Critical Path

which contains two activities, assuming ±3sigma Confidence

interval. The duration uncertainty (pessimistic – optimistic) for the

two activities are 12 weeks and 18 weeks. What is the duration

uncertainty for the entire path?

A. 22 weeks

B. 18 weeks

C. 26 weeks

D. It is not possible to calculate this from the information given.

6. Sara is conducting a stress test for an application. She has to connect

to 3 servers and 5 applications to complete the test activities.

Optimistically she can complete the test in 4 days. However when

the server and applications have high traffic then she could take 12

days. She is most likely to take 8 days to complete the testing

activities. What is the expected duration and the standard deviation

for the test if Sara wishes to use weighted average method to

compute the same?

A. 8.3, 1.5

B. 8.0, 1.33

C. 9.1, 1.0

D. 9.7, 2.2


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7. One of the tasks in your project has an optimistic estimate of 25

days and pessimistic estimate of 40 days. What is the most likely

estimate for the task?

A. 20 days

B. 32.5 days

C. 11 days

D. Unknown, the most likely estimate is a separate estimate.

8. Your project has an optimistic duration estimation of 12 weeks,

pessimistic duration estimation of 24 weeks and most likely duration

estimation of 18 weeks. Your company has a quality requirement of

3 sigma. What is the duration estimation within which your project

should complete?

A. 10 weeks to 18 weeks.

B. 10 weeks to 20 weeks.

C. 12 weeks to 24 weeks.

D. 18 weeks to 24 weeks.

9. A task has the following information: pessimistic estimation: 10

hours; optimistic estimation: 5 hours; Most likely estimation: 8

hours. Calculate the expected duration of the task using three-point

Beta distribution estimation technique.

A. 7.83 hours.

B. 8.5 hours.

C. 9.32 hours.

D. 7.38 hours.


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10. A task in your project has pessimistic estimate of 12 days, optimistic

estimate of 8 days and most likely estimate of 10 days. What is the

mean of the expected duration using PERT technique?

A. 12 days.

B. 9.5 days.

C. 11 days.

D. 10 days.

2.2 Critical Path Method

The Critical Path method is a method used to calculate the minimum project

duration and the amount of flexibility that the tasks have on the logical

networking path with respect to scheduling. The Critical Path is the

sequence of activities that represent the longest path through the project

which will determine the shortest possible duration for the project.

Total float is the amount of time that an activity can be delayed from its

earliest start date without delaying the project’s finish date. Free float is the

amount of time an activity can be delayed from its earliest start date without

delaying the start date of any successor. The activities on the Critical Path

have zero float.

Float is calculated using the formula; Float = Late start – Early start or Late

finish – Early finish.


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Example1: You are the project manager and you have the following

dependencies for the activities in your project.

a. Activity 1 with a duration of 4 weeks can start immediately.

b. Activity 2 with a duration of 5 weeks can start after the completion of

Activity 1.

c. Activity 3 with a duration of 3 weeks can start after the completion of

Activity 1.

d. Activity 4 with a duration of 7 weeks can start after the completion of

Activity 2.

e. Activity 5 with a duration of 3 weeks can start after Activity 3 and

Activity 4 are complete.

What is the duration of the project?

Solution:

Let us develop the project network diagram.

We will use the following convention to draw the network diagram:

Activity�Name

Float

ES EF

LS LF

Duration


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We will use forward pass to determine early start and early finish for the

activities. Here we start with Activity 1, ES is 0 and EF is 4 as the duration

is 4 weeks. Activity 2 will start after activity 1, so ES for activity 2 is 4 and

EF is 9. Similarly we can determine ES, EF for Activity 3 and 4. For

activity 5, as it has two predecessors, ES will be the later EF of the

predecessors which is 16.

We will use backward pass to determine late start and late finish for the

activities. We start with Activity 5. LF for activity 5 will be 19 and late start

16. For activity 3 and 4 LF will be 16. Similarly we can determine LS and

LF for all the activities. At Activity 1, as there is convergence, LF will be

the early LF of the successors which is 4. Now compute the float for all the

activities as LS-ES or LF-EF.

So we get the network diagram for the project. Here there are 2 paths:


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1. Start-Activity 1-Activity 2-Activity 4-Activity 5 –End, which has a

duration of 19 weeks.

2. Start-Activity 1-Activity 3-Activity 5-End, which has a duration of

10 weeks.

Since the first path is the longest path; that is the Critical Path. The duration

of the project will be the duration of Critical Path which is 19 weeks. Note

that the float of the activities on the Critical Path is zero.

Example2: In Example 1, if for some reason Activity 3 takes 7 weeks to

complete, what is the effect on the project?

Solution: It will have no effect on the project. This activity is on a non-

Critical Path with a float of 9 weeks. So taking 4 weeks more for the

activity will not have any effect on the duration of the project.

Example3: In Example 1, a new activity, Activity 6 is added, which has to

be performed after Activity 3 and should complete before Activity 5 can

start and the duration of this activity is 10 weeks. What is the effect of this

on the project?

Solution: Let us alter the network diagram to reflect this.


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We can see here that the Critical Path has changed. Now path 2, which is

Start-Activity 1-Activity 3- Activity 6-Activity 5-End, has become the

Critical Path with a duration 20 weeks.

The project duration has increased by 1 week (not 10 weeks which is the

duration of the activity added), as the new activity was added on the non-

Critical Path with a float of 9 weeks.

Example 4: In Example 3, if the duration of Activity 6 is 9 weeks, what is

the effect of this on the project?

Solution: We can see that the duration of the project does not change it will

remain 19 weeks. But now both path 1 and path 2 have become Critical

Paths.

We can have more than 1 Critical Path in a project, but this increases the

risk as there will be more activities with float 0, which will give little

flexibility for the project manager in terms of schedule.


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Critical Path and Float problems

1. You are the project manager and you have the following dependencies

for the activities in your project.

a. Activity 1 with duration of 4 weeks can start immediately;

b. Activity 2 with duration of 5 weeks can start after completion of

Activity 1;

c. Activity 3 with duration 3 weeks can start after completion of

Activity 1;

d. Activity 4 with duration of 7 weeks can start after completion of

Activity 2;

e. Activity 5 with duration 3 weeks can start after Activity 3 and

Activity 4 are complete.

1. A Stakeholder has added an Activity 6 with a duration of 10 weeks

which can start after Activity 3 and needs to be completed before

Activity 5. What is the effect of this new activity on the project?

A. The project schedule will increase by 10 weeks.

B. The project will be delayed by 6 weeks.

C. The project will be delayed by 1 weeks.

D. There is no effect on project schedule.

2. Activity A has an Early Start of 10, Early Finish of 16, Late start of

19 and Late finish of 25. Which of the following statements is true

for Activity A?


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A. Activity A has a free float of 9 days.

B. Activity A has a total float of 9 days.

C. Activity A has a total float of 4 days.

D. Activity A is on the Critical Path.

3. The Following is the network diagram of your project that you have

just developed. What is the minimum duration of the project?

FS means Finish to Start;

FF means Finish to Finish;

A. 36

B. 39

C. 44

D. 51

4. You are managing a construction project and have come up with the

project schedule. The details on the activities of the project are given

below. Your stakeholder approaches and asks you if the project

duration can be shortened by 3 weeks. Which of the following

activities can you try and shorten to reduce the project duration by 3

weeks?


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Activity Preceding Activity Duration

Start None 0

A Start 2

B Start 3

C A 5

D B 8

E C,D 2

F D 3

G E,F 5

End G 0

A. Activity B

B. Activity D

C. Activity G

D. Activity C

5. You have a project with the following activities:

a. Activity A takes 30 days and can start after the project starts;

b. Activity B takes 30 days and can start after the project starts;

c. Activity C takes 40 days and can start after A and B are

complete;

d. Activity D takes 15 days and can start after C is complete;

e. Activity E takes 20 days and can start after C is complete;

f. Activity F takes 10 days and can start after D is complete;


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g. Activity G takes 35 days and can start after E and F complete.

Which of the following is true if activity D actually takes 5 days to

complete?

A. The Critical Path is decreased by 5 days;

B. The Critical Path is decreased by 10 days;

C. The Critical Path is increased by 5 days;

D. The Critical Path changes to Start, B, C, D, F, G, and End.

6. A task in your project has an early start of day 5 and late start of day

7, early finish of day 10 and late finish of day 12. Which of the

following statements is true?

A. The task is on the Critical Path.

B. The task is on the Critical Path and has a float of 5 days.

C. The task is not in Critical Path and has a float of 2 days.

7. A project involves four tasks as given below: Task 1 can start

immediately and has an estimated duration of 2 weeks. Task 2 can

start after Task 1 is completed and has an estimated duration of 5

weeks. Task 3 can start after Task 2 is completed and has an

estimated duration of 6 weeks. Task 4 can start after Task 1 is

completed and must be completed when Task 3 is completed. The

estimate for Task 4 is 10 weeks. What is the shortest amount of time

required to complete a project?

A. 26

B. 31

C. 13


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D. 18

8. A portion of a project network diagram is given below. What is the

late finish for activity 3?

A. 18

B. 11

C. 9

D. 16

9. You are managing a webpage design project and have just

completed the project scheduling. The Critical Path of the project

has a duration of 26 days with a standard deviation of 4 days. The


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customer wants the project completed in 30 days. What is the

maximum project float?

A. 0 day

B. 2 days

C. 4 days

D. 8 days

10. You have a project consisting of three tasks, Task A, having a

duration of 3 months, Task B having a duration of 5 months and

Task C having a duration of 2 months. Task A and Task B can be

performed concurrently and Task C will have to be performed after

Task A and Task B are complete. The project begins in January and

the customer has imposed the end of the year as the project

completion date. What is the total duration of the Critical Path?

A. 7 months

B. 10 months

C. 12 months

D. 15 months


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2.3 Schedule Compression

Delivering a project on-time is always challenging for the project managers.

There are many reasons for which you might need to complete the project

earlier than expected. Schedule compression techniques are used to shorten

the schedule duration without reducing the project scope. There could be

many reasons for shortening the project duration:

x Your schedule was unrealistic or you have fallen behind schedule

due to unforeseen incidents;

x There is an imposed end date by the customer;

x There is a market demand to complete the project earlier;

x You see an opportunity to get another project if you are able

complete the project early.

There are two schedule compression techniques:

1. Crashing;

2. Fast tracking.

2.3.1 Crashing

Crashing is a technique where you add additional resources to the project to

compress the schedule. It is the technique where you can shorten the

schedule duration for the least incremental cost by adding resources.


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Examples of crashing include bringing in more resources to the project,

approving overtime, paying additional costs to expedite certain activities.

Some of the points to be remembered while crashing are:

1. You need to crash activities on Critical Path to get the desired result;

2. You need to watch for the dependencies from other paths before

deciding on the activities to crash;

3. There will always be additional cost involved in crashing;

4. Additional resources in turn translate into increased communication

channels resulting in coordination challenges;

5. There is always risk associated with project crashing.

Near critical paths: A Near critical path is a path with small amount of

float. The duration of the near critical path will be nearer to the critical path.

You need to be watchful of these paths as there is possibility that these will

turn into critical paths during schedule compression.

Steps to solve problems on crashing:

1. Determine all the paths of the project with the help of network diagram

with their duration. Identify the Critical Path (the path with the highest

duration);

2. Identify the activities on the Critical Path that can be crashed

considering the dependencies that the activities have with activities on

non-critical path;


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3. Determine the crash cost/month (or week or day) i.e. slope for the

activities identified;

4. Starting from the activity with the lowest slope, identify the activities

that can be crashed to get the desired schedule compression;

5. Check for the duration of the near-critical paths to ensure that the

duration is less than the duration of the Critical Path after crashing the

identified activities;

6. If the duration of any near-critical path is more than the duration of

crashed Critical Path duration, then activities on these paths have to be

crashed using the same procedure;

7. Additional costs (crash costs) of the project for shortening the duration

can be found by adding crash costs for all the activities identified for

crashing;

8. Total cost of the project after crashing can be found by adding the crash

costs of the activities identified for crashing and normal cost of the

remaining activities.

Example1:

1. The Network diagram of a project is shown as below. The customer

has asked you to complete the project 5 weeks earlier. None of these

activities can be performed in parallel. What is the best option that

you have?


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A. Add additional resource to Activity E, to complete it in 15

weeks;

B. Add additional resource to Activity B, to complete it in 25

weeks;

C. Add additional resource to Activity D, to complete it in 10

weeks;

D. Tell the customer that it is not possible to complete the project

earlier.

Correct Answer: C

Solution: The project has four paths:

Start-A-C-E-G-End = 125.

Start-A-C-D-F-G-End = 130 (Critical Path).

Start-B-C-E-G-End = 125.

Start-B-C-D-F-G-End = 130 (Critical Path).

Reducing the duration of Activity E by 5 weeks will not reduce the

project duration as it is on a non-critical path. Reducing Activity B

by weeks will not help as Activity C has a dependency with A and

B. Reducing duration of Activity D by weeks will reduce the


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Critical Path duration by 5 weeks and in turn the project duration by

5 weeks. So C is the correct option.

Example2:

The projects task details are given in the table below. There is a market

demand which mandates you to complete the project in 12 days. What is the

best possible option to achieve this?

A. Crash A by 2 days, B by 1 day, D by 1 day;

B. Crash A by 2 days, C by 2 days;

C. Crash A by 2 days, B by 1 day, D by 1 day, C by 1 day and E by

1 day;

D. Crash A by 2 days, B by 1 day, D by 1 day, C by 2 days.

Correct Answer: C

Solution:


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The project has 2 paths ABD – 16 days and ACE – 14 days. ABD is

the critical path. Crash A,B D by 2,1 and 1 days respectively. Now

ABD is 10 days. But ACE is 12 days, we need to reduce that by 2

days. If we reduce C by 2 days cost= 400. If we reduce C by 1 day

and E by 1 day, cost is 300 So choose this option. So the correct

answer is Crash A by 2 days, B by 1 day, D by 1 day, C by 1 day

and E by 1.

2.3.2 Fast Tracking

Fast tracking is a schedule compression technique in which activities or

phases which are normally done in sequence are done in parallel. You can

apply fast tracking by re-scheduling activities to be worked on

simultaneously. This works only when activities can be overlapped to

shorten the duration. You can also overlap activities partially, for e.g.: you

can start an activity after the predecessor activity is 60% complete.

Some of the points to be remembered while crashing are:

1. You need to fast track activities on Critical Path to get the desired

result;

2. You need to watch for the dependencies from other paths before

deciding on the activities to fast track;

3. There will be no additional cost involved in fast tracking;

4. This method should only be used when activities can actually be

overlapped;

5. There is always a risk of rework in fast tracking if the parallel

activities have dependencies;


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6. You need to be watchful of near critical paths as there is possibility

that these will turn into Critical Paths, during schedule compression.

Steps to solve problems on fast tracking:

1. Determine all the paths of the project using a network diagram with

their duration. Identify the Critical Path (the path with the highest

duration);

2. Identify the activities on the Critical Path that can be fast tracked

considering the dependencies that the activities have with activities

on non-critical path;

3. Determine the activities to fast track based on the duration that need

to be shortened for the project;

4. Check for the duration of the near-critical paths to ensure that the

duration is less than the duration of the Critical Path after fast

tracking the identified activities;

5. If the duration of any near-critical path is more than the duration of

compressed Critical Path duration, then activities on these paths

have to be fast tracked to using the same procedure.

Example 1:

The Network diagram of a project is shown as below. The customer has

asked you to complete the project 5 weeks earlier. You do not have

additional costs to spend on the project. What is the best option that you

have?


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A. Perform Activities D and F in parallel to reduce 5 weeks;

B. Perform Activities E and G in parallel to reduce 5 weeks;

C. Add additional resource to Activity D, to complete it in 10

weeks;


earlier.

Correct Answer: A

Solution: You cannot add additional resources as it will incur

additional cost. So option C is incorrect. Activities E and G are not

on the Critical Path, hence fast tracking them will not reduce the

project duration. So Option A is correct.

Example 2:

The network diagram and duration of the tasks are given below. If you fast-

track activities C, E and F, what will be the duration saved for the project?


111

A. 15 days

B. 10 days

C. 12 days

D. 5 days

Correct Answer: D

Solution: If you complete tasks C, E and F in parallel, it will take 13

days instead of 28 days to complete. But tasks B and D will take 23

days to complete D and F both should be completed before G can

starts we can save 28-23 = 5 days.

Schedule crashing problems




you have?


112

A. Add additional resources to Activity E, to complete it in 15

weeks;

B. Add additional resources to Activity B, to complete it in 25

weeks;

C. Add additional resources to Activity D, to complete it in 10

weeks;


earlier.

2. You have completed the Development project schedule process for

your project and you see that the project has a float of -2 months.

The activities mentioned below are all on the Critical Path. Which

activities presented below would you crash to save 2 months on the

project?


113

A. Tasks 3 and 6.

B. Tasks 1 and 6.

C. Task 1.

D. Task 6.

3. You have come up with your project schedule. During discussion of

the same with the customer, he has asked you options to complete it

4 weeks earlier and what additional costs you will incur to do that.

The crash details for the activities are given below. What will be the

additional cost required to complete the project 3 months earlier?

Activities Original

duration

(weeks)

Crash

duration

(weeks)

Original cost

($)

Crash cost

($)

B 8 5 $10,000 $13,000


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C 10 8 $15,000 $18,000

E 12 11 $8,000 $10,000

G 9 7 $12,000 $15,000

I 5 4 $5,000 $6,000

A. $5,000

B. $6,000

C. $3,000

D. $4,000

4. You have 4 tasks in your project and 2 paths; Tasks 1, 2 and 4 are in

one path and Tasks 1 and 3 are on the other path. Assuming cost is

important, in which sequence should crashing of activities be

planned?

Tasks Normal time

(months)

Crash time

(months)

Normal

cost

Crash cost

1 4 2 1,000 4,000

2 5 3 3,000 4,000

3 5 2 2,000 2,500

4 3 2 5,000 7,000

A. Tasks 2, 1, 4.

B. Tasks 1, 2, 4.

C. Tasks 3, 2, 1, 4.

D. Tasks 4, 2, 1.


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5. The following data shows the project tasks, crash times/costs. The

network diagram for the project is given below. Calculate the cost of

the project until you can no longer crash the project any further.

Tasks Normal

time

(weeks)

Normal

cost

Crash time

(weeks)

Crash

cost

Slope

A 6 $1000 5 $1100 $100

B 12 $1500 8 $2300 $200

C 15 $2000 12 $2450 $150

D 10 $3000 8 $3600 $300

E 6 $400 5 $600 $200

F 12 $8000 10 $9000 $500

G 7 $3500 7 $3500 $0

A. $22,550

B. $16,600

C. $19,400


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D. $21,100

6. You are given the following data about the project tasks, network,

and crash times/costs. What is the total duration of the project after

crashing all the activities?

Tasks Normal

time

(Weeks)

Normal

cost

Crash time

(Weeks)

Crash

cost

Slope

A 6 $1000 6 $1000 $0

B 12 $1500 10 $2000 $250

C 15 $2000 11 $2400 $100

D 10 $3000 8 $3600 $300

E 6 $4000 5 $6000 $2000

F 8 $8000 6 $9000 $500

G 5 $3500 4 $3900 $400


117

A. 42 weeks

B. 33 weeks

C. 30 weeks

D. 37 weeks

7. The projects task details are given in the table below. There is a

market demand which mandates you to complete the project in 12

days. What is the best possible option to achieve this?

A. Crash A by 2 days, B by 1 day, D by 1 day.

B. Crash A by 2 days, C by 2 days.


1 day.


8. The network diagram for a project and the details for crashing are

given below:

What will be the additional costs required to crash the project by 4

weeks?


118

Task Normal

cost

Crash cost/

week (slope)

Maximum

crash time

A $100 $20 2

B $200 $25 1

C $400 $10 3

D $350 $30 3

E $500 $25 2

F $600 $50 1

G $450 $15 1

A. $120

B. $70

C. $85

D. $60


119

9. The below shows the details of the Critical Path tasks for a project.

The project has a total float time of 3 months. What will be the cost

of the project after crashing to reduce the schedule by 3 months?

Activity Original

duration

(mths)

Crash

duration

( mths)

Original

cost

Crash

Cost

Cost per

month

A 14 12 $10,000 $14,000 $2,000

B 9 8 $17,000 $27,000 $10,000

C 3 2 $25,000 $26,000 1,0000

D 7 5 $14,000 $20,000 $3,000

A. $66,000

B. $69,000

C. $74,000

D. $71,000


given below. You have additional $1000 that you can spend to

complete the project earlier. What will be the new project duration?


120

Tasks Normal

time

Normal

cost

Crash

time

Crash

cost

Slope

A 4 $1000 4 $1000 $0

B 5 $1500 3 $2000 $250

C 6 $2000 2 $2400 $100

D 10 $3000 8 $3600 $300

E 8 $4000 7 $6000 $2000

F 5 $8000 3 $9000 $500

G 8 $3500 7 $3900 $400

A. 27

B. 25

C. 24

D. 23


121

Schedule fast Tracking problems


has asked you to complete the project 5 weeks earlier. You do not

have additional costs to spend on the project. What is the best option

that you have?

A. Perform Activities D and F in parallel to reduce by 5 weeks.

B. Perform Activities E and G in parallel to reduce by 5 weeks.


weeks.


earlier.

2. You are the manager of a software improvement project. In the

middle of the schedule timeline, you discover that you are behind

schedule. Your senior management told you that it is crucial to get

project done on time, and if not done on time, it will be a big loses

for your company. You start quick efforts to compress your

schedule. After analyzing the schedule, you find out that you have a


122

lot of discretionary dependencies in your schedule. You decided that

the best thing to do is:

A. Use the same relationship and Crash the schedule.

B. Use the same relationship and apply Fast Track the schedule.

C. Remove the old relationship between activities Crash the

schedule.

D. Remove the old relationship between activities and Fast Track

the schedule.

3. A project has the following activities planned to be done in

sequence.

Design – 2 weeks.

Coding – 4 weeks.

Testing – 3 weeks.

You decide to design and coding in parallel to save time. How many

weeks of effort can you save by doing this?

A. 4 weeks.

B. 2 weeks.

C. 3 weeks.

D. 5 weeks.

4. You are managing a software development project and you have

come up with the following duration estimation for the activities.

Design – 14 days.


123

Coding – 32 days.

Testing – 20 days.

To save some time in the project, you decide to start coding once

design is 50% complete and start testing once coding is 75%

complete. What will be the new project duration?

A. 40 days.

B. 48 days.

C. 51 days.

D. 58 days.

5. The network diagram for a project is shown below. Currently the

project takes 20 weeks to complete. The customer has asked you to

come up with options to reduce the project’s duration by 2 weeks.

What will be your answer?

A. Fast-track activities F and H.

B. Fast-track activities E and G.

C. Fast-track activities D and E.

D. Fast-track activities F and H and also E and G.


124

6. The network diagram of a project is given below. You need to

complete the project 5 weeks earlier. What is the best possible

option to accomplish this?

A. Perform activities A and B in parallel.

B. Perform activities A and C in parallel.

C. Perform activities D and E in parallel.

D. It is not possible to reduce the project schedule by 5 weeks.

7. You have come up with the network diagram for your project which

takes 16 days to complete. You can benefit by completing this

project 3 days earlier and will save some money. What is the best

possible option that you have?


125

A. Fast-track activities B and E.

B. Fast-track activities E and H.

C. Fast-track activities H and J.

D. None of these.

8. You have four tasks A, B, C and D on the Critical Path with a

duration of 5, 7, 6 and 9 weeks respectively. Tasks A-B and C-D

have mandatory dependency and B-C has discretionary dependency.

We need to compress the schedule of the project by 1 week. What

can be done to achieve this?

A. Perform A and B in parallel.

B. Perform B and C in parallel.

C. Perform C and D in parallel.

D. None of these.

9. The network schedule of a project is shown below. If we fast-track

all the tasks what will be the duration of the project?


126

A. 15 days.

B. 38 days.

C. 25 days.

D. 20 days.

10. The network diagram and duration of the tasks are given below. If

you fast-track activities C, E and F, what will be the duration saved

for the project?

A. 15 days.

B. 10 days.

C. 12 days.

D. 5 days.


127

ANSWERS:

Triangular distribution solved problems

1. For an activity, we think that in the best case we need 3 days to

finish a task, most likely this is going to be 5, but in the worst

scenario, in case where we need to perform much more work

because not all the details have been provided, we believe that it is

going to take 10 days. What is the estimated duration of the activity

using triangular distribution?

A. 5.5

B. 6

C. 5

D. 4

Correct Answer: B

Solution:

7KUHH�SRLQW�HVWLPDWLRQ�XVLQJ�WULDQJXODU�GLVWULEXWLRQ�

�RSWLPLVWLF� � �PRVW�OLNHO\� � �SHVVLPLVWLF��

��

2. The following four tasks represent the Critical Path of a project. The

estimates of each of these tasks are shown below. What is the length

of the Critical Path triangular distribution is used?

Task: Optimistic, Most likely, Pessimistic


128

A. 17 22 33

B. 12 25 32

C. 18 29 34

D. 8 22 26

A. 80

B. 86

C. 90

D. 96

Correct Answer: D

Solution:

7ULDQJXODU�HVWLPDWLRQ�IRU�WKH�IRXU�DFWLYLWLHV�DUH�� DQG�� 6R�WKH�OHQJWK�RI�WKH�&ULWLFDO�3DWK ��

PERT, Standard deviation, and variance solved problems



expected duration of the activity using three point estimates?

A. 28.16

B. 20.83

C. 21.27

D. 27


129

Correct Answer: B

Solution:

7KUHH�SRLQW�HVWLPDWH�RU�3(57�HVWLPDWH� �� W3 � � � W0 � W2��

2. You have completed estimates for all your project tasks and have

arrived at the total project duration. The optimistic estimate for the

project is 52 weeks and pessimistic estimate is 64 weeks. What is

the standard deviation of the project?

A. 1.5

B. 2.0

C. 2.5

D. 5.0

Correct Answer: B

Solution:

6WDQGDUG�GHYLDWLRQ�LV� �W3 W2� ޔ� � �� ޔ� � ��



variance of the activity?

A. 1.568

B. 1.833

C. 3.361

D. 7.564


130

Correct Answer: C

Solution:

$FWLYLW\�YDULDQFH �W3 W2 ޔ�� ኼ �� ޔ�� ኼ ��

4. You are managing a software development project and have come

up with the PERT estimates (in days) for the Critical Path activities

as shown below. What is the standard deviation of the allover path?

Tasks Optimistic Most likely Pessimistic

Requirement collection 10 12 16

Design 22 28 34

Code 35 45 59

Testing 20 22 26

A. App 6.8 days

B. App 5.2 days

C. App 4.7 days

D. You cannot derive the path standard deviation from the

information given.

Correct Answer: C

Solution:


131

TDVN�9DULDQFH� � ᗗዚዅዙዀ ᗘኼ�3URMHFW�6WDQGDUG�GHYLDWLRQ

6XP�RI�WKH�WDVN�YDULDQFHV�൘൲ൠ൬ൣൠ൰ൣ�ൣ൴൧ൠ൲൧൭൬�൭�൲߶�൭൴൰ൠ൪൪�൮ൠ൲߶ �

ᔄᓋ�� ᓏኼ� ᓋ�� ᓏ

ኼ��ᓋ�� ᓏ

ኼ��ᓋ�� ᓏ

ኼ

�ኼ ��ኼ ��ኼ ��ኼ ��ൣൠ൷൱.

5. You are working on estimated activity durations for your project

tasks. You have applied a three-point estimation on a Critical Path

which contains two activities, assuming ±3sigma Confidence

interval. The duration uncertainty (pessimistic – optimistic) for the

two activities are 12 weeks and 18 weeks. What is the duration

uncertainty for the entire path?

A. 22 weeks

B. 18 weeks

C. 26 weeks

D. It is not possible to calculate this from the information given.

Correct Answer: A

Solution:

െൢ൲൧൴൧൲൷��൳൬ൢ൰൲ൠ൧൬൲൷ ൕ ൔ �� ൘൲ൠ൬ൣൠ൰ൣ�ൣ൴൧ൠ൲൧൭൬ ኻኼ

ዀ �� ൛ൠ൰൧ൠ൬ൢ �ኼ �


132

െൢ൲൧൴൧൲൷��൳൬ൢ൰൲ൠ൧൬൲൷ ൕ ൔ

�� ൘൲ൠ൬ൣൠ൰ൣ�ൣ൴൧ൠ൲൧൭൬ �� ൛ൠ൰൧ൠ൬ൢ �ኼ �

൙൭൲ൠ൪�൮ൠ൲߶�൴ൠ൰൧ൠ൬ൢ �� ൲൭൲ൠ൪�൮ൠ൲߶�൱൲ൣ�ൣ൴൧ൠ൲൧൭൬ �� ൱൧൦൫ൠ �� q��൱൧൦൫ൠ�൵൧൪൪�൦൧൴�ൠ�൳൬ൢ൰൲൧ൠ൬൲൷�൰ൠ൬൦�൭�� ࣵ ��൵൩൱�

6. Sara is conducting a stress test for an application. She has to connect

to 3 servers and 5 applications to complete the test activities.

Optimistically she can complete the test in 4 days. However when

the server and applications have high traffic then she could take 12

days. She is most likely to take 8 days to complete the testing

activities. What is the expected duration and the standard deviation

for the test if Sara wishes to use weighted average method to

compute the same?

A. 8.3, 1.5

B. 8.0, 1.33

C. 9.1, 1.0

D. 9.7, 2.2

Correct Answer: B

Solution: 3(57�H[SHFWHG�GXUDWLRQ�

� � �� 6'� ��


133

7. One of the tasks in your project has an optimistic estimate of 25

days and pessimistic estimate of 40 days. What is the most likely

estimate for the task?

A. 20 days

B. 32.5 days

C. 11 days

D. Unknown, the most likely estimate is a separate estimate.

Correct answer: D

Solution: The most likely estimate is one of the three point

estimates which is separate from pessimistic and optimistic

estimates.

8. Your project has an optimistic duration estimation of 12 weeks,

pessimistic duration estimation of 24 weeks and most likely duration

estimation of 18 weeks. Your company has a quality requirement of

3 sigma. What is the duration estimation within which your project

should complete?

A. 10 weeks to 18 weeks.

B. 10 weeks to 20 weeks.

C. 12 weeks to 24 weeks.

D. 18 weeks to 24 weeks.

Correct Answer: C

Solution:

3(57�HVWLPDWLRQ� ��


134

�� 9DULDQFH� ��

6R�IRU��VLJPD�WKH�UDQJH�VKRXOG�EH�� DQG�� DQG��ZHHNV

9. A task has the following information: pessimistic estimation: 10

hours; optimistic estimation: 5 hours; Most likely estimation: 8

hours. Calculate the expected duration of the task using three-point

Beta distribution estimation technique.

A. 7.83 hours.

B. 8.5 hours.

C. 9.32 hours.

D. 7.38 hours.

Correct Answer: A

Solution:

൙߶൰�൮൭൧൬൲�ൡ൲ൠ�ൣ൧൱൲൰൧ൡ൳൲൧൭൬�൧൱�ൕൊൗ൙�

�൱൲൧൫ൠ൲൧൭൬ ��

10. A task in your project has pessimistic estimate of 12 days, optimistic

estimate of 8 days and most likely estimate of 10 days. What is the

mean of the expected duration using PERT technique?

A. 12 days.

B. 9.5 days.

C. 11 days.

D. 10 days.


135

Correct Answer: D

Solution:

0HDQ�XVLQJ�3(57�

� RSWLPLVWLF� � �� PRVW�OLNHO\� � �SHVVLPLVWLF�

��

Critical Path and Float solved problems

1. A Stakeholder has added an Activity 6 with a duration of 10 weeks

which can start after Activity 3 and needs to be completed before

Activity 5. What is the effect of this new activity on the project?

A. The project schedule will increase by 10 weeks.

B. The project will be delayed by 6 weeks.

C. The project will be delayed by 1 weeks.

D. There is no effect on project schedule.

Correct Answer: C

Solution:

The initial network diagram is as shown below:


136

The network diagram after activity 6 is added is as shown:

We can see here that the Critical Path has changed. Now path 2,

which is Start-Activity 1-Activity 3- Activity 6-Activity 5-End, has

become the Critical Path with a duration of 20 weeks.

The project duration has increased by 1 week (not 10 weeks which

is the duration of the activity added), as the new activity was added

on the non-critical path with a float of 9 weeks.


137

2. Activity A has an Early Start of 10, Early Finish of 16, Late start of

19 and Late finish of 25. Which of the following statements is true

for Activity A?

A. Activity A has a free float of 9 days.

B. Activity A has a total float of 9 days.

C. Activity A has a total float of 4 days.

D. Activity A is on the Critical Path.

Correct Answer: B

Solution:

7RWDO�IORDW� �/DWH�VWDUW� �(DUO\�VWDUW� �� .

3. The Following is the network diagram of your project that you have

just developed. What is the minimum duration of the project?

FS means Finish to Start;

FF means Finish to Finish;

A. 36

B. 39

C. 44

D. 51


138

Correct Answer: A

Solution: Pay attention to the relationship between activities.

Total duration= (15+10+3 (this is due to FS+3)) + (12- 4 (this is due

to FF-4)) + (0) (as Activity D (12-4=8 days) will be completed

before activity C (7days) finishes) =36.

Note: Dependencies between activities are assumed to be Finish to

Start unless told otherwise. However in this question the

dependencies are finish to start and finish to finish.

4. You are managing a construction project and have come up with the

project schedule. The details on the activities of the project are given

below. Your stakeholder approaches and asks you if the project

duration can be shortened by 3 weeks. Which of the following

activities can you try and shorten to reduce the project duration by 3

weeks?

Activity Preceding Activity Duration

Start None 0

A Start 2

B Start 3

C A 5

D B 8

E C,D 2

F D 3


139

G E,F 5

End G 0

A. Activity B

B. Activity D

C. Activity G

D. Activity C

Correct Answer: B

Solution:

The network diagram for the project will be:

We have 3 paths Start-A-C-E-G-End = 14; Start-B-D-E-G-End =

18; Start-B-D-F-G-End = 19;

So; Start-B-D-F-G-End is the Critical Path. We need to shorten the

activity on the Critical Path to shorten the project duration. We

cannot shorten B and F as they are of 3 weeks only. So D is the

better option.


140

5. You have a project with the following activities:

a. Activity A takes 30 days and can start after the project starts;

b. Activity B takes 30 days and can start after the project starts;

c. Activity C takes 40 days and can start after A and B are

complete;

d. Activity D takes 15 days and can start after C is complete;

e. Activity E takes 20 days and can start after C is complete;

f. Activity F takes 10 days and can start after D is complete;

g. Activity G takes 35 days and can start after E and F complete.

Which of the following is true if activity D actually takes 5 days to

complete?

A. The Critical Path is decreased by 5 days;

B. The Critical Path is decreased by 10 days;

C. The Critical Path is increased by 5 days;

D. The Critical Path changes to Start, B, C, D, F, G, and End.

Correct Answer: A

Solution: The original network diagram for the project is as shown

below:


141

The project has four paths:





Since Activity D is reduced by 10 days, Critical Path now changes

to Start-A-C-E-G-End and Start-B-C-E-G-End = 125 which has a

duration of 125 days. Hence the Critical Path and the project

duration is reduced by 5 days.

6. A task in your project has an early start of day 5 and late start of day

7, early finish of day 10 and late finish of day 12. Which of the

following statements is true?

A. The task is on the Critical Path.

B. The task is on the Critical Path and has a float of 5 days.

C. The task is not in Critical Path and has a float of 2 days.

Correct Answer: C

Solution:

The Activity has a float of LS-ES=2 days, and therefore it is not on

the Critical Path.

7. A project involves four tasks as given below: Task 1 can start

immediately and has an estimated duration of 2 weeks. Task 2 can

start after Task 1 is completed and has an estimated duration of 5

weeks. Task 3 can start after Task 2 is completed and has an


142

estimated duration of 6 weeks. Task 4 can start after Task 1 is

completed and must be completed when Task 3 is completed. The

estimate for Task 4 is 10 weeks. What is the shortest amount of time

required to complete a project?

A. 26

B. 31

C. 13

D. 18

Correct Answer: D

Solution:

There are two paths here; Start-Task 1-Task 2-Task 3-End which

has duration of 13 weeks; Start-Task 1-Task 4-Task3-End which

has duration of 18 weeks so this is the Critical Path and the shortest

time required to complete the project is 18 weeks.

8. A portion of a project network diagram is given below. What is the

late finish for activity 3?


143

A. 18

B. 11

C. 9

D. 16

Correct Answer: C

Solution:

Late start of the successive activities are 10 and 17, one with the

lowest is 10. So late finish of Activity 3 is 10 -1 = 9.

9. You are managing a webpage design project and have just

completed the project scheduling. The Critical Path of the project

has a duration of 26 days with a standard deviation of 4 days. The


144

customer wants the project completed in 30 days. What is the

maximum project float?

A. 0 day

B. 2 days

C. 4 days

D. 8 days

Correct Answer: D

Solution:

Total project float= Project float 4 days (because customer wants

the project in 30 days and Critical Path is 26 days) + standard

deviation 4 days = 8 days.

10. You have a project consisting of three tasks, Task A, having a

duration of 3 months, Task B having a duration of 5 months and

Task C having a duration of 2 months. Task A and Task B can be

performed concurrently and Task C will have to be performed after

Task A and Task B are complete. The project begins in January and

the customer has imposed the end of the year as the project

completion date. What is the total duration of the Critical Path?

A. 7 months

B. 10 months

C. 12 months

D. 15 months

Correct Answer: C


145

Solution:

There is an imposed end date for the project which is 12 months

from the project beginning. Hence the total duration of Critical Path

will be 12 months.

Schedule crashing solved problems




you have?

A. Add additional resources to Activity E, to complete it in 15

weeks;

B. Add additional resources to Activity B, to complete it in 25

weeks;


weeks;


146


earlier.

Correct Answer: C

Solution: The project has four paths:





Reducing duration of Activity E by 5 weeks will not reduce the

project duration as it is on non-critical path. Reducing Activity B by

5 weeks will not help as Activity C has a dependency with A and B.

Reducing duration of Activity D by weeks will reduce the Critical

Path duration by 5 weeks and in turn the project duration by 5

weeks. So C is the correct option.

2. You have completed the Development project schedule process for

your project and you see that the project has a float of -2 months.

The activities mentioned below are all on the Critical Path. Which

activities presented below would you crash to save 2 months on the

project?


147

A. Tasks 3 and 6.

B. Tasks 1 and 6.

C. Task 1.

D. Task 6.

Correct Answer: B

Solution:

Additional cost to crash 3 and 6=2,500; additional cost to crash 1 by

1 month and 6 by 1 month = 1,500; additional cost to crash

1=2,000;Task 6 cannot be crashed for 2 weeks. So Option B is

better.

3. You have come up with your project schedule. During discussion of

the same with the customer, he has asked you options to complete it

4 weeks earlier and what additional costs you will incur to do that.


148

The crash details for the activities are given below. What will be the

additional cost required to complete the project 3 months earlier?

Activities Original

duration

(weeks)

Crash

duration

(weeks)

Original cost

($)

Crash cost

($)

B 8 5 $10,000 $13,000

C 10 8 $15,000 $18,000

E 12 11 $8,000 $10,000

G 9 7 $12,000 $15,000

I 5 4 $5,000 $6,000

A. $5,000

B. $6,000

C. $3,000

D. $4,000

Correct Answer: D

Solution:

There are 3 options to reduce project duration by 4 weeks; Tasks B

and E which will cost $5,000; Tasks B and I which will cost $4,000;

Tasks C and G which will cost $6,000. Crashing tasks B and I will

be most economical with an additional cost of $4,000. So option D

is correct.

4. You have 4 tasks in your project and 2 paths; Tasks 1, 2 and 4 are in

one path and Tasks 1 and 3 are on the other path. Assuming cost is


149

important, in which sequence should crashing of activities be

planned?

Tasks Normal time

(months)

Crash time

(months)

Normal

cost

Crash cost

1 4 2 1,000 4,000

2 5 3 3,000 4,000

3 5 2 2,000 2,500

4 3 2 5,000 7,000

A. Tasks 2, 1, 4.

B. Tasks 1, 2, 4.

C. Tasks 3, 2, 1, 4.

D. Tasks 4, 2, 1.

Correct Answer: A

Solution:

Here the critical path is 1-2-4 as it has the maximum duration. So

option C is incorrect.

Let us find per month crash cost for each task.

Task1=3000/2=1500; task2=1000/2=500; task 4 =3000/1=3000; so

the sequence will be Tasks 2-1-4.

5. The following data shows the project tasks, crash times/costs. The

network diagram for the project is given below. Calculate the cost of

the project until you can no longer crash the project any further.


150

Tasks Normal

time

(weeks)

Normal

cost

Crash time

(weeks)

Crash

cost

Slope

A 6 $1000 5 $1100 $100

B 12 $1500 8 $2300 $200

C 15 $2000 12 $2450 $150

D 10 $3000 8 $3600 $300

E 6 $400 5 $600 $200

F 12 $8000 10 $9000 $500

G 7 $3500 7 $3500 $0

A. $22,550

B. $16,600

C. $19,400

D. $21,100

Correct Answer: D


151

Solution: There are two paths here: ABDG with a duration of 35

and ACEFG with a duration of 46. Total crash time for the activities

on Critical Path = 7. So if we crash ACEFG by 7 weeks, the

duration of the project will become 39. To get the total cost for the

project, we need to add crash costs for the activities on the Critical

Path and normal costs for the activities on non-critical path �� .

6. You are given the following data about the project tasks, network,

and crash times/costs. What is the total duration of the project after

crashing all the activities?

Tasks Normal

time

(Weeks)

Normal

cost

Crash time

(Weeks)

Crash

cost

Slope

A 6 $1000 6 $1000 $0

B 12 $1500 10 $2000 $250

C 15 $2000 11 $2400 $100

D 10 $3000 8 $3600 $300

E 6 $4000 5 $6000 $2000

F 8 $8000 6 $9000 $500

G 5 $3500 4 $3900 $400


152

A. 42 weeks

B. 33 weeks

C. 30 weeks

D. 37 weeks

Correct Answer: B

Solution:

There are six paths here and the Critical Path is BCDG with a

duration of 42 weeks. The duration that we can crash on Critical

Path = 9 weeks.

So duration of the project after crashing all the activities = 42 – 9 =

33.

7. The projects task details are given in the table below. There is a

market demand which mandates you to complete the project in 12

days. What is the best possible option to achieve this?


153

A. Crash A by 2 days, B by 1 day, D by 1 day.

B. Crash A by 2 days, C by 2 days.


1 day.


Correct Answer: C

Solution:

The project has 2 paths ABD – 16 days and ACE – 14 days. ABD is

the critical path. Crash A, B, and D by 2,1 and 1 days respectively.

Now ABD is 10 days. But ACE is 12 days, we need to reduce that

by 2 days. If we reduce C by 2 days cost = 400.If we reduce C by 1

day and E by 1 day, cost is 300 so choose this option.

So the correct answer is Crash A by 2 days, B by 1 day, D by 1 day,

C by 1 day and E by 1.


given below:


154

What will be the additional costs required to crash the project by 4

weeks?

Task Normal

cost

Crash cost/

week (slope)

Maximum

crash time

A $100 $20 2

B $200 $25 1

C $400 $10 3

D $350 $30 3

E $500 $25 2

F $600 $50 1

G $450 $15 1

A. $120

B. $70

C. $85

D. $60


155

Correct answer: C

Solution:

The Critical Path here is ADFG. As per the crashing details, the

most economical is G, which reduces the duration by 1 week at a

cost of $15. Next is A, and we can crash this by 2 weeks at an

additional cost of $40. Then we can crash D by 1 week for an

additional cost of $30. So the additional cost required to crash the

project by 4 weeks is $85.

9. The below shows the details of the Critical Path tasks for a project.

The project has a total float time of 3 months. What will be the cost

of the project after crashing to reduce the schedule by 3 months?

Activity Original

duration

(mths)

Crash

duration

( mths)

Original

cost

Crash

Cost

Cost per

month

A 14 12 $10,000 $14,000 $2,000

B 9 8 $17,000 $27,000 $10,000

C 3 2 $25,000 $26,000 1,0000

D 7 5 $14,000 $20,000 $3,000

A. $66,000

B. $69,000

C. $74,000

D. $71,000

Correct Answer: D


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Solution:

Original cost of the project 66,000. A and C are the activities with

least crash costs to save 3 months. Costs to crash A by 2 months =

4000, costs to crash C by 1 month = 1000.So cost of the project

after crashing = 66000+4000+1000 = 71000.


given below. You have additional $1000 that you can spend to

complete the project earlier. What will be the new project duration?

Tasks Normal

time

Normal

cost

Crash

time

Crash

cost

Slope

A 4 $1000 4 $1000 $0

B 5 $1500 3 $2000 $250

C 6 $2000 2 $2400 $100


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D 10 $3000 8 $3600 $300

E 8 $4000 7 $6000 $2000

F 5 $8000 3 $9000 $500

G 8 $3500 7 $3900 $400

A. 27

B. 25

C. 24

D. 23

Correct Answer: C

Solution:

The Critical Path here is ADFG with a duration of 27 weeks. As per

the crashing details ,the most economical is D, which reduces the

duration by 2 weeks at a cost of $600. Next is G and we can crash

this by 1 week at an additional cost of $400. So for $1000,we can

reduce the duration by 3 weeks so the new project duration is 24

weeks.

Schedule fast Tracking solved problems


has asked you to complete the project 5 weeks earlier. You do not

have additional costs to spend on the project. What is the best option

that you have?


158

A. Perform Activities D and F in parallel to reduce by 5 weeks.

B. Perform Activities E and G in parallel to reduce by 5 weeks.


weeks.


earlier.

Correct Answer: A

Solution: You cannot add additional resources as it will incur

additional costs. So option C is incorrect. Activities E and G are not

on the Critical Path, hence fast tracking them will not reduce the

project’s duration. So Option A is correct.

2. You are the manager of a software improvement project. In the

middle of the schedule timeline, you discover that you are behind

schedule. Your senior management told you that it is crucial to get

project done on time, and if not done on time, it will be a big loses

for your company. You start quick efforts to compress your

schedule. After analyzing the schedule, you find out that you have a


159

lot of discretionary dependencies in your schedule. You decided that

the best thing to do is:

A. Use the same relationship and Crash the schedule.

B. Use the same relationship and apply Fast Track the schedule.

C. Remove the old relationship between activities Crash the

schedule.

D. Remove the old relationship between activities and Fast Track

the schedule.

Correct Answer: D

Solution: Since there will not be any increase in the resources of

the project then you need to use fast tracking. You can use fast

tracking by adjusting the relationships between activities.

3. A project has the following activities planned to be done in

sequence.

Design – 2 weeks.

Coding – 4 weeks.

Testing – 3 weeks.

You decide to design and coding in parallel to save time. How many

weeks of effort can you save by doing this?

A. 4 weeks.

B. 2 weeks.

C. 3 weeks.


160

D. 5 weeks.

Correct Answer: B

Solution: By doing activities in parallel the time taken will be the

duration of the longest activity. So we can save 2 weeks by this.

4. You are managing a software development project and you have

come up with the following duration estimation for the activities.

Design – 14 days.

Coding – 32 days.

Testing – 20 days.

To save some time in the project, you decide to start coding once

design is 50% complete and start testing once coding is 75%

complete. What will be the new project duration?

A. 40 days.

B. 48 days.

C. 51 days.

D. 58 days.

Correct Answer: C

Solution:

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161

5. The network diagram for a project is shown below. Currently the

project takes 20 weeks to complete. The customer has asked you to

come up with options to reduce the project’s duration by 2 weeks.

What will be your answer?

A. Fast-track activities F and H.

B. Fast-track activities E and G.

C. Fast-track activities D and E.

D. Fast-track activities F and H and also E and G.

Correct Answer: D

Solution: There are 5 paths in the project: D-E-G = 17; B-C-F-H =

20; B-C-E-G = 20; A-C-E-G = 18; A-C-F-H = 18. Both B-C-F-H

and B-C-E-G are Critical Paths and both of them need to be reduced

by 2 weeks. So we need to fast track both F-H and E-G. So option D

is correct.


162

6. The network diagram of a project is given below. You need to

complete the project 5 weeks earlier. What is the best possible

option to accomplish this?

A. Perform activities A and B in parallel.

B. Perform activities A and C in parallel.

C. Perform activities D and E in parallel.

D. It is not possible to reduce the project schedule by 5 weeks.

Correct Answer: D

Solution: The Critical Path here is A-B. Even if A and B are done in

parallel it will take min 8 weeks of time and we can reduce the

schedule only by 4 weeks. So it is not possible to reduce the

schedule by 5 weeks.

7. You have come up with the network diagram for your project which

takes 16 days to complete. You can benefit by completing this


163

project 3 days earlier and will save some money. What is the best

possible option that you have?

A. Fast-track activities B and E.

B. Fast-track activities E and H.

C. Fast-track activities H and J.

D. None of these.

Correct Answer: C

Solution: All the possible options are on the Critical Path. By fast-

tracking B and E, we can only save 2 days. Activity H can be

started only after 4 days because of A and D. So, if we fast-track E

and H again we can save only 2 days. We can fast-track H and J and

can save 3 days. J can be started in parallel with H as all dependent

activities will complete by 7 days. So option C is correct.

8. You have four tasks A, B, C and D on the Critical Path with a

duration of 5, 7, 6 and 9 weeks respectively. Tasks A-B and C-D


164

have mandatory dependency and B-C has discretionary dependency.

We need to compress the schedule of the project by 1 week. What

can be done to achieve this?

A. Perform A and B in parallel.

B. Perform B and C in parallel.

C. Perform C and D in parallel.

D. None of these.

Correct Answer: B

Solution: We can fast-track activities only with discretionary

dependency and not with mandatory dependency. So option B is

correct.

9. The network schedule of a project is shown below. If we fast-track

all the tasks what will be the duration of the project?

A. 15 days.

B. 38 days.

C. 25 days.

D. 20 days.

Correct Answer: A


165

Solution: If we fast-track all the tasks, the duration taken will be the

task with the highest duration which is 15 days.

10. The network diagram and duration of the tasks are given below. If

you fast-track activities C, E and F, what will be the duration saved

for the project?

A. 15 days.

B. 10 days.

C. 12 days.

D. 5 days.

Correct Answer: D

Solution:� If you complete tasks C, E and F in parallel, it will take 13 days

instead of 28 days to complete. But tasks B and D will take 23 days

to complete. D and F both should be completed before G can start.

So we can save 28-23 = 5 days

Time Management Chapter

This is a sample of my book

PMP Certificate Math Practice

That includes

Earned Value Measurement problems, Schedule

compression, Critical Path, Schedule Crashing and

Fast tracking Problems, Contracts Problems, Risk

Problems, and More.

This book aims to guide you through the different types of math-

based questions that you can expect to find on the exam, including

a deep collection of earned value management, decision tree,

critical path and float problem examples. With the practice

questions provided within, you can gain those very insights that are

required to maintain momentum as you progress through the exam

on your way to achieving the certification.

if you are interested in buying the book please visit


Or at amazon.com

http://www.amazon.com/PMP-Certification-Math-

Practice-Compression-

ebook/dp/B010ERKRLK/ref=sr_1_3?ie=UTF8&qid=143

9387991&sr=8-3&keywords=pmp+math






pmp certificate math practice - sample - time management

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