Optimal Project Crashing Using the following information, develop the optimal time–cost solution. Project costs are $1,000 per day.

Question

Optimal Project Crashing
Using the following information, develop the optimal time–cost solution. Project costs are $1,000 per day.

Activity	Normal Time	Crash Time	Cost per Day to Crash
a	6	6	-
b	10	8	$500
c	5	4	300
d	4	1	700
e	9	7	600
f	2	1	800

Accepted Answer

a. Determine which activities are on the critical path, its length, and the length of the other path:

Path	Length
a–b–f	18
c–d–e–f	20 (critical path)

b. Rank the critical path activities in order of lowest crashing cost, and determine the number of days each can be crashed. Note: Available days = Normal time – Crash time

Activity	Cost per Day to Crash	Available Days
c	$300	1
e	600	2
d	700	3
f	800	1

c. Begin shortening the project, one day at a time, and check after each reduction to see which path is critical. (After a certain point, another path may equal the length of the shortened critical path.) Thus:
(1) Shorten activity c one day at a cost of $300. The length of the critical path now becomes 19 days.
(2) Activity c cannot be shortened any more. Shorten activity e one day at a cost of $600. The length of path c–d–e–f now becomes 18 days, which is the same as the length of path a–b–f.
(3) The paths are now both critical; further improvements will necessitate shortening both paths.
The remaining activities for crashing and their costs are:

Path	Activity	Crash cost per day
a–b–f	a	No reduction possible
	b	$500
	f	800
c–d–e–f	c	No further reduction possible
	d	$700
	e	600
	f	800

At first glance, it would seem that crashing f would not be advantageous, because it has the highest crashing cost. However, f is on both paths, so shortening f by one day would shorten both paths (and hence, the project) by one day for a cost of $800. The option of shortening the least expensive activity on each path would cost $500 for b and $600 for e, or $1,100. Thus, shorten f by one day. The project duration is now 17 days.
(4) At this point, no additional improvement is feasible. The cost to crash b is $500, and the cost to crash e is $600, for a total of $1,100. This would exceed the indirect project costs of $1,000 per day.
(5) The crashing sequence is summarized as follows:

	LENGTH AFTER CRASHING n DAYS
Path	n=0	1	2	3
a–b–f	18	18	18	17
c–d–e–f	20	19	18	17
Activity crashed		c	e	f
Cost		$300	$600	$800

Question 17.7: Optimal Project Crashing Using the following information, de...

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