University:
California State University, Northridge
Course:
MGT 360 | Management and Organizational Behavior
Academic year:

2024
Views:

388

Pages:

18
Author:

vas3s1ziuzy

Project Crashing Crashing       Crashing: shortening duration of activities. Because  Some activities were delayed.  Client is willing to pay more for earlier completion. Crashing changes the schedule for remaining activities. It has impact on schedules for all the subcontractors. Often introduces unanticipated problems. The faster an activity is completed, the more it costs. There is always a lower bound on task duration. 2 Linear Time / Cost Tradeoff Slope (bj) = increase in cost from reducing task duration by one time unit Cost Crash cost Crash point Ccj Slope = Crash Cost - Normal Cost Crash Time - Normal Time Normal point Normal cost CN j c tj Crash time tN j Time Normal time 3 Crashing Algorithm Assume each task can be crashed one day at a time (simplifying assumption, but not necessary) Crash only critical activities. Crashing other activities can only increase cost without changing project duration To decrease project duration by one day, the critical path or paths must decrease by one day. 1. Find the critical path or paths 2. If there is no other critical activity which could still be reduced, and shorten the critical path. Stop. 3. Crash the cheapest critical activity (or combination of activities) to shorten the critical path (or paths) by one day. 4. Go to 1 4 Crashing 21 Days to 20 Days Network 6 60 6 A 5 60 30 90 6 100 Start 7 30 7 50 7 B 4 50 33.33 150 5 70 6 C 4 7 D 7 5 E 4 100 30 160 9 40 9 F 7 40 40 120 30 ACF ADG BEG A 6 60 1 1 B 7 50 1 C 6 100 1 D 7 30 1 E 5 70 1 F 9 40 1 G 7 50 1 1 Cost 400 21 20 19 End 30 70 15 85 7 50 7 G 4 50 60 230 Activities A,C, and F are on Critical Path. A and C are the least-cost choice. We crash A because it affects two paths. Lower A’s normal time by one day. It now equals the crash time and cannot be shortened further. The critical path is unchanged. The critical time will be lowered to 20 days. The cost of the project will be $400+30(A)= $430. 5 Crashing 20 Days to 19 Days Network 1 5 90 6 A 5 60 30 90 6 100 Start 7 30 7 50 7 B 4 50 33.33 150 5 70 6 C 4 7 D 7 5 E 4 100 30 160 9 40 9 F 7 40 40 120 30 ##### 30 70 15 85 ACF ADG BEG A 5 90 1 1 B 7 50 C 6 100 1 1 D 7 30 1 E 5 70 1 F 9 40 1 G 7 50 1 1 Cost 430 20 19 19 End 7 50 7 G 4 50 60 230 Activities A,C, & F are still on the critical path. A cannot be crashed any more. C is the least-cost choice. Lower C’s normal time by one day. The critical path is unchanged The critical time will be lowered to 19 days The cost of the project will be $400+ 30(A) + 30(C) = $460 6 Crashing 19 Days to 18 Days Network 1 5 90 6 A 5 60 30 90 Start 1 5 130 0 7 30 0 7 50 7 B 4 50 33.33 150 0 5 70 6 C 4 7 D 7 5 E 4 100 30 160 0 9 40 9 F 7 40 40 120 30 ##### 30 70 15 85 ACF ADG BEG A 5 90 1 1 B 7 50 1 C 5 130 1 D 7 30 1 E 5 70 1 F 9 40 1 G 7 50 1 1 Cost 460 19 19 19 End 0 7 50 7 G 4 50 60 230 All activities are now critical. 3 paths; ACF,EDG, and BEG. A cannot be crashed any more. The only way to crash ACF is to crash C or F. C is cheaper. On ADG path, A and D cannot be crashed. The only way to crash ADG is to crash G. Crashing G crashes BEG path too. The critical time will be lowered to 18 days The cost of the project will be $400+ 30(A) + 30(C2) + 60(G) = $550 7 Crashing 18 Days to 17 Days Network 1 5 90 6 A 5 60 30 90 Start 2 4 160 0 7 30 0 7 50 7 B 4 50 33.33 150 0 5 70 6 C 4 7 D 7 5 E 4 100 30 160 0 9 40 9 F 7 40 40 120 30 ##### 30 70 15 85 ACF ADG BEG A 5 90 1 1 B 7 50 1 C 4 160 1 D 7 30 1 E 5 70 1 F 9 40 1 G 6 110 1 1 Cost 550 18 18 18 End 1 6 110 7 G 4 50 60 230 All activities are critical. There are three paces ACF, ADG, and BEG. Crash F and G by 1 at cost of 60 +40 = 100 The critical time has been lowered to 17 days The cost of the project is $400+ 30(A) + 30(C2) + 60(G2)+40(F) = $650 8 Crashing 17 Days to 16 Days Network 1 5 90 6 A 5 60 30 90 Start 2 4 160 0 7 30 0 7 50 7 B 4 50 33.33 150 0 5 70 6 C 4 7 D 7 5 E 4 100 30 160 2 7 120 9 F 7 40 40 120 30 ##### 30 70 15 85 ACF ADG BEG A 5 90 1 1 B 7 50 1 C 4 160 1 D 7 30 1 E 5 70 1 F 7 120 1 G 4 230 1 1 Cost 750 16 16 16 End 3 4 230 7 G 4 50 60 230 We can shorten the project to 16 days by crashing F and G by another day. The cost of the project is $400+ 30(A) + 30(C2) + 60(G3)+40(F2) = $750 Activities A,C,F, and g have been crashed to their limits. No further crashing will help so B,D, and E remain at their normal times and costs. 9 Time-Cost Tradeoff Project Crashing 750 Project Cost 700 650 600 Example 1 550 500 450 400 15 16 17 18 19 20 21 Project Duration 10 Crash the Following Network. Prepare Cost-time Curve 11 Crash the Following Network. Prepare Cost-time Curve A 0 2 20 Start 0 3 40 3 A 2 40 40 80 2 B 1 20 60 80 0 2 20 2 C 2 20 ##### 20 0 4 30 4 D 1 30 30 120 0 3 10 3 E 1 10 35 80 B 3 40 End ABE AC AD 1 1 1 C 2 20 D 2 20 E 4 30 1 3 10 1 1 1 Cost 120 8 5 7 12 Crash the Following Network. Prepare Cost-time Curve A 0 2 20 Start 0 3 40 3 A 2 40 40 80 2 B 1 20 60 80 0 2 20 2 C 2 20 ##### 20 0 4 30 4 D 1 30 30 120 1 2 45 3 E 1 10 35 80 B 3 40 End ABE AC AD 1 1 1 C 2 20 D 2 20 E 4 30 1 2 45 1 1 1 Cost 155 7 5 7 13 Crash the Following Network. Prepare Cost-time Curve A 0 2 20 Start 1 2 80 3 A 2 40 40 80 2 B 1 20 60 80 0 2 20 2 C 2 20 ##### 20 0 4 30 4 D 1 30 30 120 1 2 45 3 E 1 10 35 80 B 2 80 End ABE AC AD 1 1 1 C 2 20 D 2 20 E 4 30 1 2 45 1 1 1 Cost 195 6 4 6 14 Crash the Following Network. Prepare Cost-time Curve A 0 2 20 Start 1 2 80 3 A 2 40 40 80 2 B 1 20 60 80 0 2 20 2 C 2 20 ##### 20 1 3 60 4 D 1 30 30 120 2 1 80 3 E 1 10 35 80 B 2 80 End ABE AC AD 1 1 1 C 2 20 D 2 20 E 3 60 1 1 80 1 1 1 Cost 260 5 4 5 15 Crash the Following Network. Prepare Cost-time Curve A 1 1 80 Start 1 2 80 3 A 2 40 40 80 2 B 1 20 60 80 0 2 20 2 C 2 20 ##### 20 2 2 90 4 D 1 30 30 120 2 1 80 3 E 1 10 35 80 B 2 80 End ABE AC AD 1 1 1 C 1 80 D 2 20 E 2 90 1 1 80 1 1 1 Cost 350 4 4 4 16 Time-Cost Curve $400 All crash e(2)+a+d(2)+b 300 e(2)+a+d 200 e+a e e a,e e(2),a,d e(2),a,d(2),b e(2),a,d(3),b 120 120+35=155 155+40=195 195+35+30= 260 260+30+60=350 350+30=380 8 7 6 5 4 4 100 All normal 0 2 4 6 8 Total Duration (Days) 10 17 Practice The same example as we solved. Activity c either 0 or 2 but not 1. Solve it in the easiest way. Prepare Time-Cost Trade-off Curve Activity Predecesor a b [c] a d a e b f c g d,e T(N) 6 7 6 7 5 9 7 Time(C) Cost(N) Cost(C) C/T 5 60 90 30 4 50 150 33.3 4 100 160 60 7 30 30 4 70 85 15 7 40 120 40 4 50 230 60 Left 1 3 2 1 2 3 18