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Project Management Chapter 3 part 3

Project Management Chapter 3 part 3. Crashing the Project (Time-Cost). Project Crashing (Time-Cost) (Reducing the length of time of a project). Estimated time is fixed... no more. Can reduce through additional resources Manpower, equipment Direct cost of activity is always increased.

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Project Management Chapter 3 part 3

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  1. Project ManagementChapter 3 part 3 Crashing the Project (Time-Cost)

  2. Project Crashing (Time-Cost)(Reducing the length of time of a project) • Estimated time is fixed... no more. • Can reduce through additional resources • Manpower, equipment • Direct costof activity is always increased

  3. Motivation • Need to reduce time of project because: • Requirement to complete in specified time frame • Economic advantage • Three kinds of costs • Crash costs (activity direct costs) • Can be given directly, or need to determine through linear interpolation • Administration costs (or project indirect costs) • Penalty costs • Incur crash costs to avoid administration and penalty costs

  4. CT=3 NT=7 Crash Time Total Cost = $9,000 Normal Time Total Cost = $3,000 Crash cost = $1,500 Interpolation Slope = (CC – NC)/(NT-CT) CC=Crash Cost NC=Normal Cost NT=Normal Time CT =Crash Time Time

  5. Procedure • Always crash one period at a time! • 1. Identify options – critical activities • 2. Select least expensive • 3. Savings? Implement if so. Update all paths • 4. Repeat 1-3 until no cost savings are possible Note: the critical path will change – can have two

  6. 5 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 ABCD 16 AXYZ 15 Cost Save Net Cumul What activity do we crash first?

  7. 5 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 ABCD 16 AXYZ 15 Cost Save Net Cumul

  8. 5 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 ABCD 16 AXYZ 15 Cost Save Net Cumul

  9. 5 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 ABCD 16 AXYZ 15 Cost Save Net Cumul

  10. 5 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 ABCD 16 AXYZ 15 Cost Save Net Cumul Choose B because it is the cheapest of the three alternatives (A, B, or C)

  11. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul400

  12. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 2 1 Admin Cost = $500 ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400 What activity(ies) do we crash next?

  13. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  14. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  15. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  16. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  17. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 1 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  18. 4 5 3 B C D 3 3 3 2 A 4 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B ABCD 16 15 AXYZ 15 15 Cost 100 Save 500 Net 400 Cumul 400

  19. 3 5 3 B C D 3 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 BBY ABCD 16 15 14 AXYZ 15 15 14 Cost 100 225 Save 500 500 Net 400 275 Cumul 400 675

  20. 3 5 3 B C D 3 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 BBY ABCD 16 15 14 AXYZ 15 15 14 Cost 100 225 Save 500 500 Net 400 275 Cumul 400 675 What activity(ies) do we crash next?

  21. 3 5 3 B C D 3 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY ABCD 16 15 14 AXYZ 15 15 14 Cost 100 225 Save 500 500 Net 400 275 Cumul 400 675

  22. 3 5 3 B C D 3 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY ABCD 16 15 14 AXYZ 15 15 14 Cost 100 225 Save 500 500 Net 400 275 Cumul 400 675

  23. 3 5 3 B C D 3 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 2 1 B BY ABCD 16 15 14 AXYZ 15 15 14 Cost 100 225 Save 500 500 Net 400 275 Cumul 400 675

  24. 3 5 3 B C D 2 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY A ABCD 16 15 14 13 AXYZ 15 15 14 13 Cost 100 225 250 Save 500 500 500 Net 400 275 250 Cumul 400 675 925

  25. 3 5 3 B C D 2 3 3 2 A 3 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY A ABCD 16 15 14 13 AXYZ 15 15 14 13 Cost 100 225 250 Save 500 500 500 Net 400 275 250 Cumul 400 675 925

  26. 3 4 3 B C D 2 3 3 2 A 2 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY A CY ABCD 16 15 14 13 12 AXYZ 15 15 14 13 12 Cost 100 225 250 325 Save 500 500 500 500 Net 400 275 250 175 Cumul 400 675 925 1100

  27. 3 4 3 B C D 2 3 3 2 A 2 5 3 2 X Y Z 2 2 1 Crash Costs A -$250 B-$100 C-$200 D- X-$350 Y-$125 Z-$325 Admin Cost = $500 B BY A CY ABCD 16 15 14 13 12 AXYZ 15 15 14 13 12 Cost 100 225 250 325 Save 500 500 500 500 Net 400 275 250 175 Cumul 400 675 925 1100

  28. 3 4 3 B C D 2 3 3 2 A 2 5 3 2 X Y Z 2 2 1 Admin Cost = $500 B BY A CYCZ We lose $$, so do not crash CZ ABCD 16 15 14 13 12 11 AXYZ 15 15 14 13 12 11 Cost 100 225 250 325 525 Save 500 500 500 500 500 Net 400 275 250 175 -25 Cumul 400 675 925 1100 1075

  29. 3 4 3 B C D 2 3 3 2 A 2 5 3 2 X Y Z 2 2 1 B BY A CY ABCD 16 15 14 13 12 AXYZ 15 15 14 13 12 Cost 100 225 250 325 Save 500 500 500 500 Net 400 275 250 175 Cumul 400 675 925 1100

  30. Conclusions • The most economical duration of the project is 12 weeks. • Achieved by crashing B twice, A once, C once and Y twice. • Spent $900 in increased direct costs. • Avoided $2000 in administrative costs. • Net savings $1100.

  31. Example 2 Network F A C E H B D G

  32. Critical Path • Four paths in the network: • Path 1: A – C – F – H: 9 weeks • Path 2: A – C – E – G – H: 15 weeks • Path 3: A – D – G – H: 13 weeks • Path 4: B – D – G – H: 14 weeks • Path 2 is critical

  33. Crashing Example 2 • Recall the project duration was 15 weeks • Suppose that project deadline is now 13 weeks • How should the project be crashed?

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