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
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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:

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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.

90
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:

95
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?

99
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;

100
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:
Your schedule was unrealistic or you have fallen behind schedule
due to unforeseen incidents;
There is an imposed end date by the customer;
There is a market demand to complete the project earlier;
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.

104
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?

110
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.

115
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

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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?

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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

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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

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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.

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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.

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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:
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

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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:
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:
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:
variance of the activity?
A. 1.568
B. 1.833
C. 3.361
D. 7.564

130
Correct Answer: C
Solution:
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
T
.
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:

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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:

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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.
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:

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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:
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:
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:

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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:
.
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

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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

156
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:
.

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

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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

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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

PMP Certificate Math Practice - Sample - Time management

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PMP Certificate Math Practice - Sample - Time management