Question

Here is a list of activity times for a project as well as crashing costs for its activities. Determine which activities should be crashed and the total cost of crashing if the goal is to shorten the project by three weeks as cheaply as possible. There are three paths (top, middle and bottom). Each path contains three sequential activities which must be performed in the order they are listed (e.g., A before B, B before C). Note that Activity C is contained on both the top and middle paths. (Omit the "$" sign in your response.)

Path	Activity	Duration (weeks)	First Crash		Second Crash
Top	A	5	$8		$10
	B	6	7		9
	C	3	14		15
Middle	D	3	9		11
	E	7	8		9
	C	3	14		15
Bottom	F	5	10		15
	G	5	11		13
	H	5	12		14

	Activity	Cost
First crash	G  B  F	$
Second crash	B-G  D-E  A-C	$
Third crash	B-E  F-G  C-H	$

Answer 1

Duration of path A-B-C = 5+6+3 = 14

Duration of path D-E-C = 3+7+3 = 13

Duration of path F-G-H = 5+5+5 = 15

F-G-H is currently the critical path.

The least of crashing any activity is of activity F. So, to reduce the project by 1 week,

First Crash = Activity F, Cost = 10

Now, there are two critical paths = A-B-C and F-G-H. Duration = 14 weeks each

So we need to crash both of these paths by 1 week each to reduce the project duration by 1 week.

The least cost activity in A-B-C is B and in F-G-H is G (note that F is already crashed, and second crash cost is high (15). the next least crash cost activity is G)

Second crash = Activity B-G , Cost = 7 + 11 = 18

Now all three paths are critical paths having duration of 13 weeks each. So we need to crash each of these three paths by 1 week in order to reduce the project duration by 1 week.

In the first two paths (i.e. A-B-C and D-E-C) , activity C is common, so that can be crashed at least cost.

In path F-G-H, the next least cost activity is H. so that should be crashed.

Third Crash = Activity C-H , Cost = 14 + 12 = 26

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