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Determine optimum cost of optimum duration for the following data of a project .
written 5.7 years ago by gravatar for teamques10 teamques10 ★ 69k •   modified 4.6 years ago

Indirect Cost of project = Rs. 1000/day.

Activity 1-2 1-3 2-3 2-4 3-4 4-5
National Duration(days) 6 8 4 5 5 7
Crash Duration (days) 3 5 1 3 3 4
Normal Cost (Rs). 7000 4000 6000 8000 5000 8000
Crash Cost ( Rs) 14500 8500 9000 15000 11000 15000

construction management
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written 5.7 years ago by gravatar for teamques10 teamques10 ★ 69k
Activity Normal Duration $(N_T)$ Crash Duration $(C_T)$ Normal Cost $(N_C)$ Crash Cost $(C_C)$ Cost slope $\frac{C_c-N_c}{N_T-C_T}$
1-2 6 3 7000 14500 2500
1-3 8 5 4000 8500 1500
2-3 4 1 6000 9000 1000
2-4 5 3 8000 15000 3500
3-4 5 3 5000 11000 3000
4-5 7 4 8000 15000 2333.33
38000

1st Iteration

enter image description here

Critical path

$1)1,-2-3-4-5 = 6+4+5+7 = 22$

$2) 1-3-4-5 = 8+5+7 = 20 $

$3) 1-2-4-5 = 6+5+7 = 18$

Workout For Path $1-2-4-5$

$\therefore$ Project Duration = 22 Days

Total Project Cost =$38,000 + 22 \times 1000 = 60,000 Rs$

Critical Path Critical Activity Crash limit $(N_T-C_T)$ Cost slope
1-2-3-4-5 1-2 3 2500
2-3 3 1000
3-4 2 3000
4-5 3 2333.33

2nd Iteration

Crash Activity by one only

enter image description here

$1-2-3-4-5 = 6+3+5+7 = 21 \ days$

$1-3-4-5 = 8+5+7 = 20 \ day$

Project duration = 21 day

Cost = Previos Cost + Cost Slope - Indirect Cost

= $60000 + 1 \times 1000 - 1000 \times 1$

$Cost= 6000 \ Rs$

Critical Path Critical Activity Crash limit Cost slope
1-2-3-4-5 1-2 3 2500
2-3 2 1000
3-4 2 3000
4-5 3 2333.33

3rd Iteration

Crash activity 2-3 one day only

enter image description here

$1-2-3-4-5 = 6+2+5+7 = 20$

$1-3-4-5 = 8+5+7=20$

Project Duration = 20 Day

$Cost = 60,000 + (1 \times 1000) - (1 \times 1000) = 60,000 \ Rs$

Critical Path Critical Activity Crash limit Cost slope
1-2-3-4-5 1-2 3 2500
2-3 2 1000
3-4 2 3000
4-5 3 2333.33
1-3-4-5 1-3 3 1500
3-4 2 3000
4-5 3 2333.33

4th Iteration

Crash Activity 1--3 and 2-3 by one only

enter image description here

$1-2-4-5 = 6+5+7 = 18$

$1-3-4-5 = 7+5+7 = 19$

$1-2-3-4-5 = 6 + 1+5+7 = 19$

Project duration = 19 day

$Cost = 60,000 + (1 \times 1500 + 1 \times 1000) - (1 \times 1000) = 61,500 \ Rs$

Since the total cost of this Iteration Four is more than the previous Iteration.STOP The procedure and treat the solution of previous Iteration '3' Cost the best solution for optimization.The final Crash project duration is 20 Days and Optimum cost is 60,000 Rs,.
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