Crashing

In: Business and Management

Submitted By suhaas87
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Pages 4
Table 2.8 contains data for the installation of new equipment in a manufacturing process at your customer’s plant. Your company is responsible for the project. Indirect costs are $10,000 per week, and a penalty cost of $10,000 per week will be incurred for every week the projects is delayed beyond week 9.
1. What is the shortest time duration for the project regardless of cost?
2. What is the minimum total cost associated with completing the project in 12 weeks?
3. What is total time of the minimum-cost schedule?
TABLE 2.8 DATA FOR THE EQUIPMENT INSTALLATION PROJECT
Activity Normal Time (days) Normal Cost ($) Crash Time (days) Crash Cost ($) Immediate Predecessor($)
A - 3 2 7,000 10,000
B - 1 1 3,000 3,000
C A 4 2 12,000 40,000
D B 2 1 12,000 28,000
E C 1 1 8,000 8,000
F D,E 4 2 5,000 15,000
G E 2 1 9,000 18,000

Based on the data in the Table, we can make the following network diagram:

We see from the diagram that there are three paths for completion:
Path 1: A – C – E – G
Path 2: A – C – E – F
Path 3: B – D – F
We now calculate the time taken to complete each path to determine the critical path.
Time taken for completion of Path 1 (A – C – E – G): 3 + 4 + 1 + 2 = 10 days
Time taken for completion of Path 2 (A – C – E – F): 3 + 4 + 1 + 4 = 12 days
Time taken for completion of Path 3 (B – D – F): 1 + 2 + 4 = 7 days
From this we can see that Path 2 takes the longest time and hence is the Critical Path. We will start crashing the process on the critical path. To decide which activity to crash first, we look at the crashing cost per day and the activity with the least crashing cost per day will be crashed first.
We see that from the activities on the critical path, Activity A has the least crashing cost per day. Hence we crash activity A. Now, the new completion times are:
Time taken for completion of Path 1 (A – C – E – G): 2 + 4 + 1 + 2 =…...

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