Question

# Activity A Duration 6 Predecessor - Crash weeks - Crashing cost per week - B 4...

 Activity A Duration 6 Predecessor - Crash weeks - Crashing cost per week - B 4 A 2 \$480 C 5 A 1 \$380 D 6 A 3 \$450 E 5 B 2 \$430 F 4 B,C 3 \$350 G 5 C,D 2 \$410 H 6 D 2 \$570 I 5 E,F 2 \$540 J 6 E,F,G 2 \$530 K 3 G,H 1 \$550 L 4 I,J 1 \$560 M 5 I,J,K - \$620 N 6 H,K,J 1 \$650 O 4 M,N - - The time for each activity tocomplete, the possible weeks by which that time can be reduced and the cost per week to decrease the activity time. The cost to complete the project in normal time is \$10,000 (crashing) 1. Draw the project network for this project?what are the slacks in the steps? 2. what is the minimum duration of the project? 3. what are the critical paths of this project? 4. How much will be the most economical way to complete the project in 27 weeks?

1. Project Network and ES, EF, LS and LF are shown in the snapshot below:

Numbers notation on each node/activity:

Top Left - ES (Early Start)

Top Right: EF (Early Finish)

Bottom Left: LS (Late Start)

Bottom Right: LF (Late Finish)

Center: Activity Names

Bottom Middle: Activity times (in weeks)

Slack = (LF - EF) or (LS - ES)

All the above values can be tabulated and calculations can be done for Slack as shown below:

 Activity Duration Predecessor Crash weeks Crashing cost per week ES EF LS LF Slack Start 0 - - \$0 0 0 0 0 0 A 6 - - \$0 0 6 0 6 0 A B 4 A 2 \$480 6 10 8 12 2 C 5 A 1 \$380 6 11 7 12 1 D 6 A 3 \$450 6 12 6 12 0 D E 5 B 2 \$430 10 15 12 17 2 F 4 B,C 3 \$350 11 15 13 17 2 G 5 C,D 2 \$420 12 17 12 17 0 G H 6 D 2 \$570 12 18 14 20 2 I 5 E,F 2 \$540 15 20 19 24 4 J 6 E,F,G 2 \$530 17 23 17 23 0 J K 3 G,H 1 \$550 18 21 20 23 2 L 4 I,J 1 \$560 23 27 25 29 2 M 5 I,J,K - \$620 23 28 24 29 1 N 6 H,K,J 1 \$650 23 29 23 29 0 N O 4 M,N - \$0 29 33 29 33 0 O Finish 0 O - \$0 33 33 33 33 0

2. Minimum duration of the project = 33 weeks

3. Critical Paths:

All those activities which have ZERO SLACK lie on the CRITICAL PATH. [That's the rule and definition of critical path, since activities on critical path HAVE TO BE started and finished on their scheduled times]

So, Critical Path activities are: A, D, G, J, N, O (as also highlighted in the snapshot below:

So, Critical Path is: A - D - G - J - N - O

Critical path time = 6 + 6 + 5 + 6 + 6 + 4 =

There can be only 1 critical path with those activities.

4. Crashing the non-critical activities will not have much impact. So, we should crash the critical activites and ensure that the non-critical activities don't become critical activities. Also, we crash starting with the least expensive activity to the most expensive.

Out of the critical activities, least expensive is G with \$420 as crashing cost per week. So, we crash G by maximum weeks, i.e., by 2 weeks.

Now, Project Duration become = 33 - 2 = 31 weeks

Next candidate for crashing is D with \$450 as crashing cost per week, So, we crash D by maximum weeks, i.e., by 3 weeks.

Now, Project Duration become = 31 - 3 = 28 weeks

Next candidate for crashing is J with \$530 as crashing cost per week, So, we crash J by 1 week only since we need to complete the project in target 27 weeks which gets achieved by crashing J just by 1 week.

Now, project duration = 28 - 1 = 27 weeks.

So, New cost = Normal Cost + Crashing Cost

=> New Cost = \$10,000 + \$420*2 + \$450*3 + \$530*1 = \$12,720

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