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Project Management

Project Management. Dr. Ron Lembke Operations Management. Can We Go Faster?. MacArthur Maze. Key SF artery I-80 west to 880 south April 29, 2007, 8,600 gal gas Day 8 – CC Meyers gets job Day 25 – opened to traffic $876 k bid, $2.5m cost $200,000 per day incentive $5,000,000 bonus.

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Project Management

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  1. Project Management Dr. Ron Lembke Operations Management

  2. Can We Go Faster?

  3. MacArthur Maze • Key SF artery • I-80 west to 880 south • April 29, 2007, 8,600 gal gas • Day 8 – CC Meyers gets job • Day 25 – opened to traffic • $876 k bid, $2.5m cost • $200,000 per day incentive • $5,000,000 bonus

  4. China Coal Plant Official Deadline: June 29 Internal Goal: March 31 Delivered ON TIME Lost Profits: $400 million US 1995: Pledged $100m to Princeton

  5. Time-Cost Models 1. Identify the critical path. 2. Find cost per day to expedite each node on critical path. 3. For cheapest node to expedite, reduce it as much as possible, or until critical path changes. 4. Repeat 1–3 until no feasible savings exist.

  6. Time-Cost Example D 8 A 10 B 10 C 10 • ABC is critical path=30 Crash cost Crash per week wks avail A 500 2 B 800 3 C 5,000 2 D 1,100 2 Cheapest way to gain 1 Week is to cut A

  7. Time-Cost Example D 8 A 9 B 10 C 10 • ABC is critical path=29 Crash cost Crash per week wks avail A 500 1 B 800 3 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 Cheapest way to gain 1 wk Still is to cut A

  8. Time-Cost Example D 8 A 8 B 10 C 10 • ABC is critical path=28 Crash cost Crash per week wks avail A 500 0 B 800 3 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 Cheapest way to gain 1 wk is to cut B

  9. Time-Cost Example D 8 A 8 B 9 C 10 • ABC is critical path=27 Crash cost Crash per week wks avail A 500 0 B 800 2 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 Cheapest way to gain 1 wk Still is to cut B

  10. Time-Cost Example D 8 A 8 B 8 C 10 • Critical paths=26 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 1 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 To gain 1 wk, cut B and D, Or cut C Cut B&D = $1,900 Cut C = $5,000 So cut B&D

  11. Time-Cost Example D 7 A 8 B 7 C 10 • Critical paths=25 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 2 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 Can’t cut B any more. Only way is to cut C

  12. Time-Cost Example D 7 A 8 B 7 C 9 • Critical paths=24 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 1 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 Only way is to cut C

  13. Time-Cost Example D 7 A 8 B 7 C 8 • Critical paths=23 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 0 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 7 5,000 14,500 No remaining possibilities to reduce project length

  14. Time-Cost Example D 7 A 8 B 7 C 8 • Now we know how much it costs us to save any number of weeks. • Customer says he will pay $2,000 per week saved. • Only reduce 5 weeks. • We get $10,000 from customer, but pay $4,500 in expediting costs. • Increased profits = $5,500 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 7 5,000 14,500 No remaining possibilities to reduce project length

  15. What about Uncertainty?

  16. PERT Activity Times • 3 time estimates • Optimistic times (a) • Most-likely time (m) • Pessimistic time (b) • Follow beta distribution • Expected time: t = (a + 4m + b)/6 • Variance of times: v = (b - a)2/36 

  17. Project Times • Expected project time (T) • Sumofcritical path activity times, t • Project variance (V) • Sum of critical path activity variances, v

  18. A B C 6.48 4.33 7.67 Example Activitya m bE[T]variance A 2 4 8 4.33 1 B 3 6.1 11.5 6.48 2 C 4 8 10 7.67 1 Project 18.5 4

  19. Sum of 3 Normal Random Numbers Average value of the sum is equal to the sum of the averages Variance of the sum is equal to the sum of the variances Notice curve of sum is more spread out because it has large variance 10 20 30 40 50 60

  20. Back to the Example: Probability of <= 21 wks Average time = 18.5, st. dev = 2 21 is how many standard deviations above the mean? 21-18.5 = 2.5. St. Dev = 2, so 21 is 2.5/2 = 1.25 standard deviations above the mean Book Table (p. 498) says area between Z=1.25 and –infinity is 0.8944 Probability <= 21 wks = 0.8944 = 89.44% 18.5 21

  21. Conclusion • Determined most cost-effective way to “crash” a project • Cheapest way to crash a given number of weeks • Stop crashing when marginal cost exceeds marginal benefit • Computed project probabilities • Use probabilities of each activity • Can talk about likelihood of finishing project in a given amount of time

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