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 ABC is critical path=30 Crash costCrash per weekwks avail A5002 B8003 C5,0002 D1,1002 C 10 B 10 A 10 D 8 Cheapest way to gain 1 Week is to cut A

7. Time-Cost Example ABC is critical path=29 Crash costCrash per weekwks avail A5001 B8003 C5,0002 D1,1002 C 10 B 10 A 9 D 8 Cheapest way to gain 1 wk Still is to cut A Wks IncrementalTotal GainedCrash $Crash $ 1500500

8. Time-Cost Example ABC is critical path=28 Crash costCrash per weekwks avail A5000 B8003 C5,0002 D1,1002 C 10 B 10 A 8 D 8 Cheapest way to gain 1 wk is to cut B Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000

9. Time-Cost Example ABC is critical path=27 Crash costCrash per weekwks avail A5000 B8002 C5,0002 D1,1002 C 10 B 9 A 8 D 8 Cheapest way to gain 1 wk Still is to cut B Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800

10. Time-Cost Example Critical paths=26 ADC & ABC Crash costCrash per weekwks avail A5000 B8001 C5,0002 D1,1002 C 10 B 8 A 8 D 8 To gain 1 wk, cut B and D, Or cut C Cut B&D = $1,900 Cut C = $5,000 So cut B&D Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600

11. Time-Cost Example Critical paths=25 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0002 D1,1001 C 10 B 7 A 8 D 7 Can’t cut B any more. Only way is to cut C Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500

12. Time-Cost Example Critical paths=24 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0001 D1,1001 C 9 B 7 A 8 D 7 Only way is to cut C Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500

13. Time-Cost Example Critical paths=23 ADC & ABC Crash costCrash per weekwks avail A5000 B8000 C5,0000 D1,1001 C 8 B 7 A 8 D 7 No remaining possibilities to reduce project length Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500 75,00014,500

14. Time-Cost Example C 8 B 7 A 8 D 7 No remaining possibilities to reduce project length Wks IncrementalTotal GainedCrash $Crash $ 1500500 25001,000 38001,800 48002,600 51,9004,500 65,0009,500 75,00014,500 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

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) Sum of critical path activity times, t Project variance (V) Sum of critical path activity variances, v

18. Example ActivityambE[T]variance A2484.331 B36.111.56.482 C48107.671 Project18.54 C C B B A A 4.33 6.48 7.67

19. Sum of 3 Normal Random Numbers 102030405060 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

20. Back to the Example: Probability of <= 21 wks 18.5 21 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%

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