If the critical path is {A, C, D, F} and we need to cut 3 weeks, which task(s) s
ID: 381931 • Letter: I
Question
If the critical path is {A, C, D, F} and we need to cut 3 weeks, which task(s) should we expedite ("crash"). Assume we can do partial expediting if necessary. If you expedite it at all, though, mark it on the list.
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
1200
Task Tiime (regular) Fastest Time ("Crashed")Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A 6 41200
B 4 4 0 C 7 5 800 D 8 6 900 E 2 2 0 F 9 6 1500 G 3 1 600Explanation / Answer
ask
Tiime (regular)
Fastest Time ("Crashed")
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A
6
4
1200
B
4
4
0
C
7
5
800
D
8
6
900
E
2
2
0
F
9
6
1500
G
3
1
600
The critical path is given as :- {A, C, D, F}
Total completion time = 6+7+8+9 =30
Among the critical activities, the least crash cost is associated with activity C.
So for the first week, Activity C will be crashed having the crash cost of 800
So the activity C’s completion time = 7-1=6, cost = 800
The completion time of critical path will be 30-1 = 29
So the above table will be as below:-
ask
Tiime (regular)
Fastest Time ("Crashed")
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A
6
4
1200
B
4
4
0
C
6
5
800
D
8
6
900
E
2
2
0
F
9
6
1500
G
3
1
600
It can further be lowered by 1 week due to least crash cost:-
Thus for second week
, Activity C will be crashed having the crash cost of 800
So the activity C’s completion time = 6-1=5, cost = 800
The completion time of critical path will be 29-1 = 28
So the above table will be as below:-
ask
Tiime (regular)
Fastest Time ("Crashed")
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A
6
4
1200
B
4
4
0
C
5
5
800
D
8
6
900
E
2
2
0
F
9
6
1500
G
3
1
600
Now we cannot crash activity C any more as the activity C’s crash time has exhausted.
Now we will look at the lest crashing cost of the remaining activities which is associated with the activity D.
Thus for third week
, Activity D will be crashed having the crash cost of 900
So the activity D’s completion time = 8-1=7, cost = 900
The completion time of critical path will be 28-1 = 27
So the above table will be as below:-
ask
Tiime (regular)
Fastest Time ("Crashed")
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A
6
4
1200
B
4
4
0
C
5
5
800
D
7
6
900
E
2
2
0
F
9
6
1500
G
3
1
600
Thus if the crashing of three week is required, we will crash the activities C by two weeks and D by one week.
ask
Tiime (regular)
Fastest Time ("Crashed")
Cost to Crash ($)
[difference from regular cost to crashed cost total; not per week]
A
6
4
1200
B
4
4
0
C
7
5
800
D
8
6
900
E
2
2
0
F
9
6
1500
G
3
1
600
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