Refer to the following table: Note: Crash time indicates the time it could be cr
ID: 381599 • Letter: R
Question
Refer to the following table:
Note: Crash time indicates the time it could be crashed to (e.g. A could be crashed to 3 units but not less than that)
While trying to crash project duration, an intermediate step had the following activity durations:
Based on the above information, find the activity(activities) that will be selected next after this intermediate step. In addition, find the crash time (minimum project duration) and corresponding cost. (hint: continue the crash time calculation from the intermediate step provided above but note that you have final crash times provided in the first table so cannot crash few activities any further. In addition, the final cost can be calculated based on all the activities that were reduced in terms of their durations).
Activity Predecessor Normal Time Normal Cost Crash Time Crash Slope A - 4 30 3 10 B - 7 60 5 70 C A 1 80 1 N/A D B 4 40 2 20 E B 5 110 2 50 F C 5 90 2 200 G D 2 60 1 30 H E 2 70 1 40 I F, G, H 2 140 2 0Explanation / Answer
Following activities have been reduced so far and by the time duration as written in parenthesis:
B(2), D(1), E(1), H(1)
Now there are three critical path in the project. These paths and their resp. duration written in parenthesis are:
A-C-F-I (12), B-D-G-I (12), B-E-H-I (12)
Activities that can be crashed now and their remaining crash times are: A (1), D (1), E (2), F (3), G (1)
We need to crash an activity on all three of these paths to reduce project duration further. The least cost to do so is by crashing activities: A, D, E
Cost of crashing these activites by 1 time unit = 10+20+50 = 80
Next we need to crash activities E, F, G. Cost = 50+200+30 = 280
Now critical path B-E-H-I is fully crashed. So project duration cannot be reduced further.
Now project duration is 10 time units. All three paths are still critical.
Total cost of project = normal cost (680) + crash cost (10+70*2+20*2+50*3+200*1+30*1+40*1)
= 1290
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