Costs for a project are $12,000 per week for as long as the project lasts. The project manager has supplied the cost and time information shown. Use the information to:
a. Determine an optimum crashing plan.
b. Graph the total costs for the plan.

Activity Crashing
Potential
(weeks) Cost per Week to
Crash

a 3 $11,000

b 3 3,000 first week

$ 4,000 others

c 2 6,000

d 1 1,000

e 3 6,000

f 1 2,000

Question

Costs for a project are $12,000 per week for as long as the project lasts. The project manager has supplied the cost and time information shown. Use the information to:
a. Determine an optimum crashing plan.
b. Graph the total costs for the plan.

Activity	Crashing Potential (weeks)	Cost per Week to Crash
a	3	$11,000
b	3	3,000 first week
		$ 4,000 others
c	2	6,000
d	1	1,000
e	3	6,000
f	1	2,000

Accepted Answer

a. (1) Compute path lengths and identify the critical path:

Path	Duration (weeks)
a–b	24 (critical path)
c–d	19
e–f	23

(2) Rank critical activities according to crash costs:

Activity	Cost per Week to Crash
b	$ 3,000
a	11,000

Activity b should be shortened one week because it has the lower crashing cost. This would reduce indirect costs by $12,000 at a cost of $3,000, for a net savings of $9,000. At this point, paths a–b and e–f would both have a length of 23 weeks, so both would be critical.
(3) Rank activities by crashing costs on the two critical paths:

Path	Activity	Cost per Week to Crash
a–b	b	$ 4,000
	a	11,000
e–f	f	2,000
	e	6,000

Choose one activity (the least costly) on each path to crash: b on a–b, and f on e–f, for a total cost of $4,000 + $2,000 = $6,000 and a net savings of $12,000 − $6,000 = $6,000.
(4) Check to see which path(s) might be critical: a–b and e–f would be 22 weeks in length, and c–d would still be 19 weeks.

(5) Rank activities on the critical paths:

Path	Activity	Cost per Week to Crash
a–b	b	$ 4,000
	a	11,000
e–f	e	6,000
	f	(no further crashing possible)

Crash b on path a–b and e on e–f for a cost of $4,000 + $6,000 = $10,000, for a net savings of $12,000 − $10,000 = $2,000.
(6) At this point, no further improvement is possible: paths a–b and e–f would be 21 weeks in length, and one activity from each path would have to be shortened. This would mean activity a at $11,000 and e at $6,000, for a total of $17,000, which exceeds the $12,000 potential savings in costs.
b. The following table summarizes the results, showing the length of the project after crashing n weeks:

Path	n=0	1	2	3
a–b	24	23	22	21
c-d	19	19	19	19
e–f	23	23	22	21
Activity crashed		b	b,f	b,e
Crashing costs ($000)		3	6	10

A summary of costs for the preceding schedule would look like this:

Project Length	Cumulative Weeks Shortened	Cumulative Crashing Costs ($000)	Indirect Costs ($000)	Total Costs ($000)
24	0	0	24(12) = 288	288
23	1	3	23(12) = 276	279
22	2	3 + 6 = 9	22(12) = 264	273
21	3	9 + 10 = 19	21(12) = 252	271
20	4	19 + 17 = 36	20(12) = 240	276

The graph of total costs is as follows:

Question 17.p.5: Costs for a project are $12,000 per week for as long as the ...

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