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Exam 17: Project Management

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Question 61

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Concerning PERT and CPM:

A) These are two different names for essentially the same technique.
B) CPM tends to focus more on the critical path; PERT focuses on the parallel paths.
C) PERT gives a clearer indication of activity slack, while CPM gives a better picture of path slack.
D) Because of its recent discovery, much less is known about CPM than PERT.
E) CPM is useful for project manager selection; PERT is useful for project team selection.

F) All of the above
G) A) and B)

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Question 62

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If the expected length of a path is more than 2.5 standard deviations less than a desired project completion time, the path will not be a factor in assessing the probability of finishing the project on time.

A) True
B) False

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Question 63

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The earliest finish time for an activity is equal to the latest finish time minus the activity time.

Which of the following are limitations of PERT? (I) Time estimates may include a fudge factor. (II) Important activities may be overlooked. (III) It is an after-the-fact analysis.

Suppose a project is subject to the following risk events (each with a given probability of occurring and a cost associated with that occurrence) : $\begin{array} { l l l } \text { Event } & \text { Probability } & \text { Cost } ( \$ 000 \mathrm {~s} ) \\\text { I } & .2 & 53 \\\text { II } & .6 & 54 \\\text { III } & .1 & 12 \\\text { IV } & .3 & 21\end{array}$ Given these, Event _____ should probably get the most attention and Event _____ should probably get the least.

The standard deviation of the critical path is equal to the sum of the standard deviations of all of the activities on the critical path.

Consider the following work breakdown structure: $\begin{array}{r}\text { Duration Estimates (Days) }\quad\quad\quad\\\begin{array} { l c c c c } \text { Activity } & \text { Precedes } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { Start } & \text { U, V } & - & - & - \\\text { U } & \text { W } & 35 & 50 & 65 \\\text { V } & \text { W, X } & 28 & 40 & 52 \\\text { W } & \text { Z } & 26 & 35 & 44 \\\text { X } & \text { Y } & 28 & 40 & 52 \\\text { Y } & \text { Z } & 26 & 29 & 38 \\\text { Z } & \text { End } & 36 & 60 & 84\end{array}\end{array}$ What is the estimated expected (mean) time for activity Y?

Consider the following work breakdown structure: $\begin{array}{l}\begin{array} { c c c c c } &&&\text { Time Estimates (days) }\\\text { Activity } & \text { Precedes } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { Start } & \text { A, B, C } & - & - & - \\\text { A } & \text { D } & 38 & 50 & 62 \\\text { B } & \text { E } & 90 & 99 & 108 \\\text { C } & \text { End } & 85 & 100 & 115 \\\text { D } & \text { F } & 19 & 25 & 31 \\\text { E } & \text { End } & 91 & 100 & 115 \\\text { F } & \text { End } & 62 & 65 & 68\end{array}\end{array}$ What is the probability that this project will be completed within 205 days?

The basis for use of the beta distribution in PERT is:

Project risk events tend to be unlikely __________ a project, but that's also when they tend to be the __________.

Activity D has an optimistic time of three days, a pessimistic time of nine days, and a most likely time of four days. Determine its expected time and variance.

Slack time is equal to LS-ES or LF-EF.

The project computing algorithm requires both a forward and a backward pass.

The task of identifying risks should involve everyone associated with the project.

Consider the following work breakdown structure: $\begin{array}{r} \text { Duration Estimate (Days) }\quad \quad \quad \quad \\\begin{array}{lccccc} & \text { Start } & \text { Finish } \\\text { Activity } & \text { Node } & \text { Node } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { A } & 1 & 2 & 9 & 15 & 21 \\\text { B } & 1 & 3 & 13 & 21 & 23 \\\text { C } & 2 & 4 & 6 & 9 & 18 \\\text { D } & 2 & 5 & 23 & 35 & 47 \\\text { E } & 3 & 5 & 19 & 26 & 27 \\\text { F } & 4 & 6 & 23 & 29 & 41 \\\text { G } & 5 & 6 & 14 & 20 & 26 \\\text { H } & 6 & 7 & 2 & 5 & 8\end{array}\end{array}$ Within what amount of time (in weeks) is there a 90 percent probability that the critical path for this project will be completed? 95 percent? 99 percent?

Consider the following work breakdown structure: $\begin{array}{cccccc}&\text { Start }&\text { Finish }&&\text { Duration (Days) } \\\text { Activity } & \text { Node } & \text { Node } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { A } & 1 & 2 & 41 & 50 & 59 \\\text { B } & 1 & 3 & 54 & 60 & 66 \\\text { C } & 2 & 4 & 58 & 70 & 82 \\\text { D } & 3 & 4 & 32 & 41 & 44\end{array}$ What is the estimated standard deviation in the time for activity D?

Consider the following work breakdown structure: $\begin{array}{l}\begin{array} { c c c c c } &&&\text { Time Estimates (days) }\\\text { Activity } & \text { Precedes } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { Start } & \text { A, B, C } & - & - & - \\\text { A } & \text { D } & 38 & 50 & 62 \\\text { B } & \text { E } & 90 & 99 & 108 \\\text { C } & \text { End } & 85 & 100 & 115 \\\text { D } & \text { F } & 19 & 25 & 31 \\\text { E } & \text { End } & 91 & 100 & 115 \\\text { F } & \text { End } & 62 & 65 & 68\end{array}\end{array}$ What is the estimated slack time for activity A?

The path in a network with the average length of time to completion is called the critical path.

A project is represented by the following diagram: The critical path for the network shown is:

Consider the following work breakdown structure: $\begin{array}{r} \text { Duration Estimate (Days) }\quad \quad \quad \quad \\\begin{array}{lccccc} & \text { Start } & \text { Finish } \\\text { Activity } & \text { Node } & \text { Node } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \\\hline \text { A } & 1 & 2 & 9 & 15 & 21 \\\text { B } & 1 & 3 & 13 & 21 & 23 \\\text { C } & 2 & 4 & 6 & 9 & 18 \\\text { D } & 2 & 5 & 23 & 35 & 47 \\\text { E } & 3 & 5 & 19 & 26 & 27 \\\text { F } & 4 & 6 & 23 & 29 & 41 \\\text { G } & 5 & 6 & 14 & 20 & 26 \\\text { H } & 6 & 7 & 2 & 5 & 8\end{array}\end{array}$ What is the estimated standard deviation (in weeks) for the critical path completion time?