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Determine the minimum cost and optimum duration for the project.

Determine the minimum cost and optimum duration for the project. The data for each activity of the network is given in the following table.

Indirect cost = Rs. 800/- per day

Activity Normal Duration (days) Crash Duration (days) Normal Cost (Rs.) Crash Cost (Rs.)
1-2 7 4 10000 15100
1-3 9 6 7000 9400
2-3 5 2 7000 10000
2-4 6 4 12000 16000
3-4 6 4 9000 12000
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Activity Name Activity Normal Duration (days) Crash Duration (days) Normal Cost (Rs.) Crash Cost (Rs.) Cost Slope
A 10-20 2 2 1000 1000 0
B 10-30 7 3 500 900 100
C 20-30 6 3 300 420 40
D 20-40 5 4 200 250 50
E 30-40 0 0 0 0 0
F 30-50 9 4 600 900 60
G 40-60 11 6 600 1000 80
H 50-60 6 3 700 910 70

$\text{Cost Slope}=\frac{\text{Crash Cost-Normal Cost}}{\text{Normal Time-Crash Time}}$

Stage 1 : - All Normal

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Critical path = 10-20-30-50-60 i.e. A-C-F-H

Project duration = 23 days.

Stage 2 : - All Crash

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Critical path :-

1) 10-20-30-50-60 (A-C-F-H)

2) 10-20-40-60 (A-D-G)

Project duration = 12 days

Stage 3 :- Stage 1 + Crash C by 1 day

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Critical paths :- 10-20-30-50-60 (A-C-F-H)

10-30-50-60 (B-F-H)

Stage 4 :- Stage 3 + Crash F by 4 days

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All critical

Duration = 18 days

Stage 5 : - Stage 4 + Crash F by 1 day & Crash G by 1 day

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All critical

Project duration = 17 days

Stage Description Duration Direct Cost Indirect Cost Addition Crash Cost Total
1 All Normal 23 3900 $80 \times 23 = 1840$ - 5740
2 All Crash 12 5830 $80 \times 12 = 960$ - 6340
3 Stage 1 + Crash C by 1 day 22 3900 $80 \times 22 = 1760$ $1 \times 40=40$ 5700
4 Stage 3 + Crash F by 4 days 18 3900 $80 \times 18 = 1440$ $40+4 \times 60=280$ 5620
5 Stage 4 + Crash,F by 1 day & Crash G by 1 day 17 3900 $80 \times 17 = 1360$ $280+1\times 60+1 \times 80=420$ 5680

Optimum cost = Rs. 5620

Optimum duration = 18 days

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