Project Time-Cost Trade-Off 7.1 Introduction In the previous chapters ...

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800007000060000Cost (LE)50000400003000020000100000100 110 120 130 140 150Example 7.3Project duration (days)The durations and direct costs for each activity in the network of a small construction contractunder both normal and crash conditions are given in the following table. Establish the least costfor expediting the contract. Determine the optimum duration of the contract assuming theindirect cost is LE 125/day.Table 7.2: Data for Example 7.1ActivityPreceded byNormalCrashDuration (day) Cost (LE) Duration (day) Cost (LE)ABCDEFGHI-AABBCE, CFD, G, H128152355201312700050004000500010003000600025003000106122344151110720053004600500010503300630025803150SolutionThe cost slope of each activity is calculated. Both the crashability and the cost slope are shownbeneath each activity in the precedence diagram. The critical path is A-C-G-I and the contractduration in 59 days.Dr. Emad Elbeltagi
20 43D (23)24 470 12 12 20 20 25A (12) B (8)E (5)0 12 14 22 22 272@100 2@150 1@5027 47G (20)27 475@6047 59I (12)47 592@7512 27C (15)12 273@20027 32F (5)29 341@30032 45H (13)34 472@401. The activity on the critical path with the lowest cost slope is G, this activity can becrashed by 5 days, but if it is crashed by more than 2 days another critical path will begenerated. Therefore, activity G will be crashed by 2 days only. Then adjust timing of theactivities.20 43D (23)22 450 12 12 20 20 25A (12) B (8)E (5)0 12 14 22 22 272@100 2@150 1@5027 45G (18)27 453@6045 57I (12)45 572@7512 27C (15)12 273@20027 32F (5)27 321@30032 45H (13)32 452@40A new critical path will be formed, A-C-F-H-I.New contract duration is 57 days.The cost increase is 2 x 60 = LE 120.2. At this step the activities that can be crashed are listed below:Either A at cost LE 100/dayOr C at cost LE 200/dayOr I at cost LE 75/dayOr F & G at cost LE 360/dayOr H & G at cost LE 100/ dayDr. Emad Elbeltagi
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Project Time-Cost Trade-Off 7.1 Introduction In the previous chapters ...

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