# Solutions to Week 4 Homework (Crashing and Ev)

Topics: Critical path method, Project management, Crash Bandicoot Pages: 5 (817 words) Published: April 6, 2013
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 (from 2 days to 1 day) on ABE, the project cost would increase to \$1500 + \$50 =\$1550 and crashing Activity C by 1 day would cost an additional \$250. Making total project cost = \$1550 + \$250 = \$1800.

Therefore, the activities that should be crashed to meet a project deadline of 10 days at minimum cost = Crash B (1 day) and then crash E (1 day) at a cost of \$300.

c) Find the new budget (or cost of the project) [2 points]

Sum of normal costs:\$1450
Crash Costs:
B by 1 day: \$ 50
E by 1 day:\$ 250
_________ Total new Cost of the Project: \$1750

PROBLEM 2

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

Activity A: Crash Cost = (\$800 - \$300) / (5-3)
= \$500 / 2
= \$250 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) / (5-2)
= \$300 / 3
= \$100 per day

b) Which activities should be crashed to meet a project deadline of 13 days at minimum cost? [3 points]

The diagram shows two paths: AD and BCD. Based on normal duration estimates, AD will take 10 days to complete and BCD will take 14 days. Therefore the earliest the project can be completed (on critical path BCD) is 14 days.

The total project cost based on normal time is:

\$300 + \$250 + \$400 + \$200 = \$1150

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

If Activity B is crashed by 1 day (from 3 days to 2 days), the total project duration would become 13 days and project cost would amount to \$1200 (i.e. \$1150 + \$50).

Activity B should be crashed to meet a project...