class_crash-problem_ch9

# class_crash-problem_ch9 - 3.5.10.1 Path Duration Crash 01...

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Unformatted text preview: 3.5.10.1 Path Duration Crash 01 Craah 02 Crash 03 Craah lb ,A-C-F muons-671:1. 63 * 601* 36.1 31 . A—D-li 16+16+10-60 36 36 L 36 35 , 1.6-0-6 16+12+6+1o-u. 60 60 ‘1 36 i 37 . l-G-II 20+6+10-36 36 36 36+ 37 ...~..—.- "- Hin(t1:> Coat/Heck Initial Crash 01 Crash '2 Crash '3 Crash 16 Activity Duration Reduction Duratio Duration Duration Duration Duration A 10 2.000 14 W10? 10 1o 10 6 12 9.000 20 26 20 20 20 c 12 5.000 16 16 6 15 . 13 . 12 0 11 4.000 16 16 16 16 16 6 9 3.000 12 12 12 . 10 10 r 6 7,000 15 15 15 15 15 c 5 10,000 6 6 6 6 6 n 5 6.000 10 10 10 10 t 9 Total Craah Coot: I \$8.000 323.000 \$39,000 \$50,000 * Show: activity craahcd Refer to the network and the Table above: Which activity to crash? It should be on critical path (CP) — A-C-F = 47 weeks. It should have lowest reduction cost (cost to crash per week). Therefore, crash A to 10 weeks (by 4 weeks) making A-C-F = 43 weeks. Note that other paths containing A are also reduced by 4 weeks. Cost to reduce A to 10 weeks @ \$ 2000/week = (14-10) x 2,000 = \$ 8,000. Now project duration is 43 weeks at extra cost of \$ 8,000. . A-C-F is still critical. A cannot be reduced further (10 weeks is minimum duration). 10. Therefore, reduce C or F till other path(s) become critical. 11. Thus, reduce C to 15 weeks (by 3 weeks) @ \$5,000/week = \$ 15,000. 12. Now project duration is 40 weeks at extra cost of \$ 23,000. 13. Now there are two critical paths and further reduction should reduce both paths. 14. Alternatives are C + E = \$8000 C + G = \$ 15,000 C + H = \$ 11,000 P + E = \$ 10,000 F + G = \$ 17,000 f. F + H = \$13,000. 15. Crash lowest cost alternative —- C + E till other paths become critical. 16. Therefore crash C and E each by 2 weeks @ \$8000/week = \$ 16,000. 17. Now project duration is 38 weeks at extra cost of \$ 39,000. 18. Continue as above as required. §°°°>‘.°‘.U‘:>E”!°t‘ 9999‘s» ...
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