IE5541-HW7

# Time direct cost indirect cost total cost 50 850 9 f

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Time Direct Cost Indirect Cost Total Cost Week 21 980 Week 22 800 50 850 9 F 5x C 4x A 9 F 5x C 4x

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HW -7 IE5411 Sunny Ahmed Week 23 620 60 680 Week 24 470 120 590 Week 25 380 180 560 Week 26 330 240 570 Week 27 300 300 600 The curve looks as follows: From the plot we can come to a conclusion that the optimum time cost schedule should finish the project in 25 weeks with a total cost of \$560. It can be seen that if we decrease the project from 27 weeks to 24 weeks we will still be lower than the 600 cost we had for 27 weeks. This basically means if we decrease project by 1 week we save \$30, by 2 weeks we save \$40 and by 3 weeks we save \$10 respectively. Now if the customer wants to pay us \$10 per week for shortening the project we can reduce it by 3 weeks which will mean our total cost will be 590-3x10 = \$530. So we will be reducing cost by \$70 if we reduce the project by 3 weeks and customer decides to pay us \$10 per week of reduction. We can agree to the customer to reduce the project by 3 weeks. If we reduce 2 weeks we save \$60 and if we reduce it by 4 weeks we pay \$40 more than what we initially had at \$600. Thus we can reduce project by 3 weeks if customer agrees to pay \$10 a week. International Capital Part B Act Normal Cost Time t E Max. Crash Time Crash Cost(slope) A 3000 7 3 500 10
HW -7 IE5411 Sunny Ahmed B 5000 4 2 1000 C 6000 5 0 0 D 20000 20 3 3000 E 10000 11 2 1000 F 7000 7 1 1000 G 20000 12 2 3000 H 8000 6 1 2000 I 5000 9 1 2000 J 7000 5 1 1000 K 12000 30 6 1000 103,000 The AON network diagram for the project looks as follows. 11

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HW -7 IE5411 Sunny Ahmed A 7 B 4 C 5 D 20 E 11 G 12 H 6 F 7 I 9 J 5 K 30 The critical path is A-C-D-H-J-K which is 73 days. The other paths are as follows: A-C-E-F-H-J-K = 7+5+11+7+6+5+30 = 71 A-C-G-H-J-K = 7+5+12+6+5+30 = 65 A-C-G-I-K = 7+5+12+9+30 = 63 B-C-D-H-J-K = 4+5+20+6+5+30 = 70 B-C-E-F-H-J-K = 4+5+11+7+6+5+30 = 68 B-C-G-H-J-K = 4+5+12+6+5+30 = 62 B-C-G-I-K = 4+5+12+9+30 = 60 Now upper management at International Capital set an average of 70 days for a project to complete with a possibility of 95%. In order to meet that Brown needs to reduce his project from 73 days to 70 days. Now there are many ways he can do that but not necessarily each way will be the most cost effective. He needs to reduce the Critical Path duration by 3 days. First way he can do it by crashing Activity A. Let us start by crashing activity A by 3 days. 12
HW -7 IE5411 Sunny Ahmed A 4X B 4 C 5X D 20 E 11 G 12 H 6 F 7 I 9 J 5 K 30 Cost increase = \$103,000 + 3(\$500) = \$104,500. Duration = 70 days. If he does this all the paths becomes as follows: A-C-D-H-J-K = 70 days A-C-E-F-H-J-K = 68 days A-C-G-H-J-K = 62 days A-C-G-I-K = 60 days B-C-D-H-J-K = 70 days B-C-E-F-H-J-K = 68 days B-C-G-H-J-K = 62 days B-C-G-I-K = 60 days In this case A-C-D-H-J-K and B-C-D-H-J-K both become critical. Next what he can do is crash activity K by 3 days. 13

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HW -7 IE5411 Sunny Ahmed A 7 B 4 C 5X D 20 E 11 G 12 H 6 F 7 I 9 J 5 K 27 Cost increase = \$103,000 + 3(\$1000) = \$106,000. Duration = 70 days. If he does this all the paths becomes as follows: A-C-D-H-J-K = 70 days A-C-E-F-H-J-K = 68 days A-C-G-H-J-K = 62 days A-C-G-I-K = 60 days B-C-D-H-J-K = 67 days B-C-E-F-H-J-K = 65 days B-C-G-H-J-K = 59 days B-C-G-I-K = 57 days Critical Path remains to be A-C-D-H-J-K.
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