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Solutions to Week 4 Homework (Crashing and Ev)

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Solutions to Week 4 Homework (Crashing and Ev)
PROBLEM 1
Use the network diagram below and the additional information provided to answer the corresponding questions. [15 points]

a) The crash cost per day per activity [10 points]

Activity A: Crash Cost = ($800 - $300) / (7-3) = $500 / 4 = $125 per day

Activity B: Crash Cost = ($350 - $250) / (3-1) = $100 / 2 = $50 per day

Activity C: Crash Cost = ($900 - $400) / (6-4) = $500 / 2 = $250 per day

Activity D: Crash Cost = ($500 - $200) / (3-2) = $300 / 1 = $300 per day

Activity E: Crash Cost = ($550 - $300) / (2-1) = $250 / 1 = $250 per day

b) Which activities should be crashed to meet a project deadline of 10 days at minimum cost? [3 points]

The diagram shows two paths: ABE and CDE. Based on normal duration estimates, ABE will take 12 days to complete and CDE will take 11 days. Therefore the earliest the project can be completed (on critical path ABE) is 12 days.

The total project cost based on normal time is: $300 + $250 + $400 + $200 + $300 = $1450

Activity on the critical path that can be accelerated at the lowest cost per day is Activity B @ $50.

If Activity B is crashed 1 day (from 3 days to 2 days), the total project duration = 11 days and project cost = $1500 (i.e. $1450 + $50)

There are now two critical paths ABE and CDE; both with 11 days. The activity with lowest cost per day on the CDE path is Activity C or E; both $250. Since Activity E is however on both paths, we should crash it by the one day available bringing the project completion to 10 days. This increases the total project cost to:

$1500 + $250 = $1750

This beats the alternative of crashing Activity C on the CDE path and Activity B on the ABE path. By crashing Activity B for 1 day

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