Question

A-1. Determine the critical path.

multiple choice 1
• A-C-E-G
• A-B-E-G
• A-B-D-G
• A-C-F-G
a-2. Determine the early completion time in weeks for the project.
For the data shown, reduce the project completion time by three weeks. Assume a linear cost per week shortened.
ACTIVITY NORMAL
TIME NORMAL
COST CRASH TIME CRASH COST
A 5 $ 7,000 3 $ 13,000 B 10 12,000 7 18,000 C 8 5,000 7 7,000 D 6 4,000 5 5,000 E 7 3,000 6 6,000 F 4 6,000 3 7,000 G 4 7,000 3 9,000 b-1. Which activities in order of reduction would be shortened?
multiple choice 2
• G-D-A
• D-G-A
• A-G-D
• D-B-C
b-2. Find the resulting cost to crash.

Answer 1

The critical path for the project is A-B-E-G and the early completion time is 26 weeks. The activities to shorten in order of reduction would be G-D-A, and the resulting cost to crash is $7,000.

Step-by-step explanation:

a-1. Determine the critical path.

The critical path is the sequence of activities that determines the shortest time to complete a project. It is the longest path through the project network, meaning that if any activity on this path is delayed, the entire project will be delayed.

In this case, the critical path is A-B-E-G, as it has a total duration of 26 weeks, which is the longest duration among the given options.

a-2. Determine the early completion time in weeks for the project.

The early completion time is the shortest possible time to complete the project without considering any delays. It is found by adding the duration of each activity on the critical path.

In this case, the early completion time is 26 weeks, which is the total duration of the critical path activities A, B, E, and G.

b-1. Which activities in order of reduction would be shortened?

To determine the activities to shorten, we need to find the activities that are on the critical path and have the lowest crash cost per week shortened.

In this case, the activities in order of reduction would be G-D-A, as they are all on the critical path and have the lowest crash cost per week shortened (G has a crash cost of $2,000 per week, D has a crash cost of $1,000 per week, and A has a crash cost of $2,000 per week).

b-2. Find the resulting cost to crash.

To find the resulting cost to crash, we need to multiply the crash cost per week shortened by the number of weeks to crash for each activity.

In this case, the resulting cost to crash for activities G, D, and A would be:

G: $2,000 (crash cost per week) * 1 (weeks to crash) = $2,000

D: $1,000 (crash cost per week) * 1 (weeks to crash) = $1,000

A: $2,000 (crash cost per week) * 2 (weeks to crash) = $4,000

Therefore, the resulting cost to crash is $2,000 + $1,000 + $4,000 = $7,000.