Question

What is the minimum cost of crashing the following project that Roger Solano manages at Slippery Rock University by 4 days?

NORMAL CRASH TIME ACTIVITY (DAYS) (DAYS) TIME NORMAL CRASH COST IMMEDIATE COST PREDECESSOR(S) 900 $1,000 B 300 400 C 4 3 500

Please show work and calculations.

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Answer #1

Critical path = Longest path of the project = C-E = 12 days

Other Paths = AD = 11 days

To reduce project time by 4 days , we need to crash C by 1 day, E by 3 days and A by 1 day and D by 2 days so that total project duration is 8 days.

Cost of crashing = Crashing cost of A, C, D and E = (1000-900) + (600-500) + (1200-900) + (1600-1000) =100+100+300+600 = $1100

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