Homework: Chapter 3 Project Management home / shud Score: 0 of 3 pts 10 of 10 (9

Question

Homework: Chapter 3 Project Management home / shud Score: 0 of 3 pts 10 of 10 (9 complete) Prin Problem 3.29 What is the minimum cost of crashing the following project that Roger Solano manages at Slippery Rock University by 4 days? Normal Crash Time Normal Total Cost Costwith Crashing Immediate Predecessorfs) $800 Activity Time (days) (days) $1,100 $600 $650 $1,500 $1,650 $750 $1,200 By how many days should each activity be crashed to reduce the project completion time by 4 days? Fil in the table below. (Enter your n Each Activity Should be Reduced BY (days) ?! s. | 1.1 : /i ? :/? | '. (M.) More Enter your answer in the edit fields and then click Check Answer Clear All MacBook Air 80 5

Explanation / Answer

Path C-E = 6+5 =11 days (critical)

Path A-D = 4+6 = 10 days

Path B = 7 days

First, crash the critical path activity

Among the critical activities, C is having least cost slope i.e $50 so first crash it by one day and the maximum crash time of C is reached

Next, there are two critical paths i.e. A-D and C-E both having 10 days

C cannot be crashed anymore as maximum limit reached

Next minimum cost slope is E i.e $150 but if we crash it by 3 days again that will leave the critical path as A-D with 10 days co no benefit by crashing

!st critical path- A-D = 10 days

2nd critical path - C-E = 10 days

Find the least cost values in two critical paths

1st path A = $300

2nd Path E = $150 (As C is crashed to the limit)

So crash them by 1 day each

So now Critical path1 becomes A-D = 9 days

Critical path C-E = 9 days

Cost = $50(C)+ $150(E)+$300(A) = $500

Now we have two critical paths having 9 days each

A is crashed to the limit but D can be crashed by 2 more days and in critical path 2 E can be crashed by 2 more days (As E can be crashed by 3 days)

So we can crash both D and E by 2 days each to reach our target to 7 days

cost = 2*375(D) + 2*150(E) = 1050

So total cost = 500+1050 = $1550

Crashing details

A = 1 day

B= 0 day

C = 1 day

D = 2 days

E= 3 days

Please comment if any doubt

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Activity Predecessor Normal time Crash time Normal cost Crash Cost Cost Slope A 4 3 800 1100 300 B 7 5 200 600 200 C 6 5 600 650 50 D A 6 4 750 1500 375 E C 5 2 1200 1650 150

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