1. © J. Christopher Beck 20081 Lecture 6: Time/Cost Trade-off in Project Planning

2. © J. Christopher Beck 2008 2 Outline Project Planning with Time/Cost Tradeoffs Linear and non-linear

3. © J. Christopher Beck 2008 3 Readings P Ch 4.4-4.5 Slides borrowed from Twente & Iowa See Pinedo CD

4. © J. Christopher Beck 2008 4 Time/Cost Trade-Offs What if you could spend money to reduce the job duration More money  shorter processing time Run machine at higher speed

5. © J. Christopher Beck 2008 5 Linear Costs Money Processing time Marginal cost Cost to reduce job j by by 1 time unit

6. © J. Christopher Beck 2008 6 Problem Spend money to reduce processing times so as to minimize: “Overhead” cost Cost per activity

7. © J. Christopher Beck 2008 7 Solution Methods Objective: minimum cost of project Time/Cost Trade-off Heuristic Good schedules Works also for non-linear costs Linear programming formulation Optimal schedules Non-linear version not easily solved

8. © J. Christopher Beck 2008 8 Source (dummy) node Sink node Minimal cut set Cut set Sources, Sinks, & Cuts

9. © J. Christopher Beck 2008 9 Cuts Sets & Minimal Cut Sets Given a graph G(V,E) A cut set is a set of nodes such that in the graph formed by removing the nodes in S from G there is no path from the source to the sink A minimal cut set, R, is a cut set such that if you remove any node, a, from R, R \ {a} is not a cut set V is a set of nodes E is set of edges (arcs)

10. © J. Christopher Beck 2008 10 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

11. © J. Christopher Beck 2008 11 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

12. © J. Christopher Beck 2008 12 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

13. © J. Christopher Beck 2008 13 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

14. © J. Christopher Beck 2008 14 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

15. © J. Christopher Beck 2008 15 Is This a Cut Set? Is This a Minimal Cut Set? 1 3 69 11121413 247 10

16. © J. Christopher Beck 2008 16 Step 1: Set all processing times at their maximum Determine all critical paths Construct the graph G cp of critical paths Time/Cost Trade-off Heuristic

17. © J. Christopher Beck 2008 17 Step 2: Determine all minimal cut sets in G cp Consider those sets where all processing times are larger than their minimum If no such set STOP; otherwise continue to Step 3 Time/Cost Trade-off Heuristic

18. © J. Christopher Beck 2008 18 Time/Cost Trade-Off Heuristic Step 3: For each minimum cut set: Compute the cost of reducing all processing times by one time unit. Take the minimum cut set with the lowest cost If this is less than the overhead per time unit go on to Step 4; otherwise STOP

19. © J. Christopher Beck 2008 19 Time/Cost Trade-Off Heuristic Step 4: Reduce all processing times in the minimum cut set by one time unit Determine the new set of critical paths Revise graph G cp and go back to Step 2

20. © J. Christopher Beck 2008 20 Example 4.4.2 Jobs1234567891011121314 p max j 569127 106 97875 p min j 35795983764554 cajcaj 2025201530403525302025352010 cbjcbj 622126223127131851929122 cjcj 72434344452248 Error in text

21. © J. Christopher Beck 2008 21 Warning: Example 4.4.2 Unfortunately, the text is wrong for this example!

22. © J. Christopher Beck 2008 22 1. Step 1: Maximum Processing Times, Find G cp 1 2 3 69 58 47 1110121413

23. © J. Christopher Beck 2008 23 1. Step 1: Maximum Processing Times, Find G cp 1 2 3 69 58 47 1110121413 Cost = overhead + job costs = c o * C max + Σc b j = 6 * 56 + 224 = 560

24. © J. Christopher Beck 2008 24 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 Minimum cut set with lowest cost Cut sets: {1},{3},{6},{9}, {11},{12},{14}. Arbitrarily choose 12 Reduce p 12 from 8 to 7 1. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost

25. © J. Christopher Beck 2008 25 2. Step 4 & 1: Reduce p 12 by 1 1 2 3 69 58 47 1110121413 Cost = overhead + processing = c 0 * C max + Σjob costs = 6 * 55 + 226 = 556

26. © J. Christopher Beck 2008 26 2. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 13 Cut sets: {1},{3},{6},{9}, {11},{12,13},{14} Why isn’t {12} a cut set? What is the cost of {12, 13}? c 13 =4 Minimum cut set with lowest cost

27. © J. Christopher Beck 2008 27 3. Step 4 & 1: Reduce p 11 by 1 1 2 3 69 58 47 1110121413 Cost = overhead + processing = c 0 * C max + Σjob costs = 6 * 54 + 228 = 552

28. © J. Christopher Beck 2008 28 3. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 13 23 Min. cut sets! Where are they? Min cost min cut set: {2,11}: 4 Reduce Job 2 to 5 and Job 11 to 5 c 13 =4 Minimum cut set with lowest cost 247 10 c 2 =2c 4 =3c 7 =4 c 10 =4

29. © J. Christopher Beck 2008 29 4. Step 4 & 1: Reduce p 2 & p 11 by 1 1 2 3 69 58 47 1110121413 Cost = overhead + processing = c 0 * C max + Σjob costs = 6 * 53 + 232 = 550

30. © J. Christopher Beck 2008 30 4. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 13 Can’t reduce Job 2 anymore. Why? Min cost min cut set: {11,12}: 4 Reduce Job 11 to 4, Job 12 to 6 c 13 =4 Minimum cut set with lowest cost 247 10 c 2 =2c 4 =3c 7 =4 c 10 =4

31. © J. Christopher Beck 2008 31 5. Step 4 & 1: Reduce p 11 and p 12 by 1 1 2 3 69 58 47 1110121413 Cost = overhead + processing = c 0 * C max + Σjob costs = 6 * 52 + 236 = 548

32. © J. Christopher Beck 2008 32 5. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 13 Cutset: {6, 12}: 4 Reduce Job 6 to 11, Job 12 to 5 c 13 =4 247 10 c 2 =2c 4 =3c 7 =4 c 10 =4

33. © J. Christopher Beck 2008 33 6. Step 4 & 1: Reduce p 6 and p 12 by 1 1 2 3 69 58 47 1110121413 Cost = overhead + processing = c 0 * C max + Σjob costs = 6 * 51 + 241 = 547

34. © J. Christopher Beck 2008 34 6. Step 2 & 3: Min. Cut Sets in G cp & Lowest Cost 1 3 69 111214 c 1 =7 c 3 =4 c 6 =3c 9 =4 c 11 =2 c 12 =2 c 14 =8 13 All cut sets either can’t be reduced (a job at p min ) or has a cost ≥ 6. So we’re done. c 13 =4 247 10 c 2 =2c 4 =3c 7 =4 c 10 =4

35. © J. Christopher Beck 2008 35 Example 4.4.2 It would be very useful for you to work through this entire example on your own! Remember the book is wrong. You should use the slides.

36. © J. Christopher Beck 2008 36 Linear Programming Formulation The heuristic does not guarantee optimum See example 4.4.3 You are responsible for the LP formulation! see the text!

37. © J. Christopher Beck 2008 37 Can Also Have Non-linear Costs Arbitrary function c j (p j ) → cost of setting job j to processing time p j LP doesn’t work! See Section 4.5 A question I like: Given processing times and c j (p j ), which algorithm would you use (heuristic or LP)?