Refer to the following table: Activity Predecessor Normal Time Normal Cost Crash
ID: 381470 • Letter: R
Question
Refer to the following table:
Activity
Predecessor
Normal Time
Normal Cost
Crash Time
Crash Slope
A
-
4
30
3
10
B
-
7
60
5
70
C
A
1
80
1
N/A
D
B
4
40
2
20
E
B
5
110
2
50
F
C
5
90
2
200
G
D
2
60
1
30
H
E
2
70
1
40
I
F, G, H
2
140
2
0
Note: Crash time indicates the time it could be crashed to (e.g. A could be crashed to 3 units but not less than that)
While trying to crash project duration, an intermediate step had the following activity durations:
Activity
Intermediate time
A
4
B
5
C
1
D
3
E
4
F
5
G
2
H
1
I
2
Based on the above information, find the activity(activities) that will be selected next after this intermediate step. In addition, find the crash time (minimum project duration) and corresponding cost. (hint: continue the crash time calculation from the intermediate step provided above but note that you have final crash times provided in the first table so cannot crash few activities any further. In addition, the final cost can be calculated based on all the activities that were reduced in terms of their durations).
Activity
Predecessor
Normal Time
Normal Cost
Crash Time
Crash Slope
A
-
4
30
3
10
B
-
7
60
5
70
C
A
1
80
1
N/A
D
B
4
40
2
20
E
B
5
110
2
50
F
C
5
90
2
200
G
D
2
60
1
30
H
E
2
70
1
40
I
F, G, H
2
140
2
0
Explanation / Answer
With the normal activity duration, BEHI is the critical path having duration of 16 days. After crashing as in table 2, we have 3 paths each having duration of 12 days.To make it leaner, we need to crash one day one by one from each paths
We can crash activities A, D and E by one day each. The project duration now becomes 11 days. No further crashing is possible.
Activities crashed
A-1, B-2, D-2,E-2, H-1
Cost after crashing if A - 40
Cost after crashing of D-40
Cost after crashing of B- 200
Cost affer crashing of E- 86.6x2 = 173.2
Cost after crashing H -110
Total cost = crashed costs of A, D,B, E, H + normal costs of CFGI
= 563.2+370 = 933.2
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