Below are the data for a Time-Cost CPM Scheduling model analysis.The time is in days and the costs include both direct and indirect costs. $\begin{array}{|l|l|l|l|l|l|}
\hline & \text { Immediate } & \text { Normal } & \text { Crash } & \text { Normal } & \text { Crash } \
\hline \text { Activity } & \text { Predecessor } & \text { Time } & \text { Time } & \text { Cost } & \text { Cost } \
\hline \text { A } & \text { None } & 3 & 2 & \$ 200 & \$ 400 \
\hline \text { B } & \text { A } & 4 & 3 & \$ 300 & \$ 600 \
\hline \text { C } & \text { A } & 1 & 1 & \$ 200 & \$ 200 \
\hline \text { D } & \text { B and C } & 3 & 2 & \$ 500 & \$ 550 \
\hline \text { E } & \text { D } & 2 & 1 & \$ 500 & \$ 900 \
\hline
\end{array}$
What are the total time of this project and total normal cost?
A)Total time is 13 days,total cost is $1,200
B)Total time is 12 days,total cost is $1,700
C)Total time is 11 days,total cost is $1,600
D)Total time is 10 days,total cost is $1,750
E)Total time is 9 days,total cost is $1,700

Question

Quizplus · Accepted Answer

To obtain the total time, we need to run CPM analysis on the given activity network. The results of CPM analysis are as follows:

\begin{array}{|l|l|l|l|}
\hline 	ext { Activity } & 	ext { ES } & 	ext { EF } & 	ext { Duration } \
\hline A & 0 & 3 & 3 \
\hline B & 3 & 7 & 4 \
\hline C & 3 & 4 & 1 \
\hline D & 7 & 10 & 3 \
\hline E & 10 & 12 & 2 \
\hline 	ext { Total } & - & - & 12 \
\hline
\end{array}

ES denotes earliest start time, EF denotes earliest finish time, and duration is the time required to complete an activity.

The critical path is A-B-D-E, which has a duration of 12 days. Hence the total time of this project is 12 days.

To obtain the total normal cost, we need to sum up the normal costs of all activities on the critical path. Hence the total normal cost is $300 + $500 + $500 + $900 = $1,700.

Now we can investigate the crash options to select the best choice(s). By inspecting the data, we can see that the crash cost for activity A is higher than the crash cost for activity B, and it will not reduce the duration of the critical path. Similarly, the crash cost for activity C is already at its minimum, and crashing it will not reduce the duration of the critical path. On the other hand, crashing activities B, D, and E will reduce the duration of the critical path. However, since we are only interested in the minimum cost to complete the project within 10 days, we need to find the cheapest combination of activities to crash such that the critical path duration is reduced to 10 days.

One possibility is to crash activity B by 1 day and activity D by 1 day. This would reduce the critical path duration from 12 days to 10 days, but it would increase the total cost by $300 + $50 = $350. Another possibility is to crash activity E by 1 day, which would also reduce the critical path duration from 12 days to 10 days, but it would increase the total cost by $400. These are the only two combinations of activities that can reduce the critical path duration to 10 days. However, the combination involving the crashing of activities B and D is cheaper than the combination involving the crashing of activity E. Hence the best choice is to crash activities B and D by 1 day each.

Therefore, the best choice(s) is:
Answer: B
 Crash activities B and D by 1 day each to reduce the duration of the critical path from 12 days to 10 days, and increase the total cost by $350.

Below Are the Data for a Time-Cost CPM Scheduling Model