1. Chapter 8 - Project Management1 Lecture 2 Today’s lecture covers the followings: 1.To study “project crashing” concept 2.LP formulation for project management problem 3.The use of QM (Try it yourself!) Tutorial: Chapter 8: 8 th Ed:Q26 and Q30 9tj Ed: Q21 and Q23 (to p2) (to p26)

2. Chapter 8 - Project Management2 Project Crashing Basic Concept In last lecture, we studied on how to use CPM to determine solution for a project problem There, we determine its critical path and completion time. Question: Can we cut short its project completion time? If so, how! (to p3)

3. Chapter 8 - Project Management3 Project Crashing Solution! Yes, the project duration can be reduced by assigning more resources to project activities But, doing this would somehow increase our project cost! How do we strike a balance? (to p4)

4. Chapter 8 - Project Management4 Trade-off concept Here, we adopt the “Trade-off” concept ie, we attempt to “crash” some “critical” events by allocating more sources to them, and also to maintain a balance that the shortening time is not less than the normal activities How to do that: Question: What criteria should it be based on when deciding to crashing critical times? (to p11) (to p5)

5. Chapter 8 - Project Management5 Example – crashing (1) The critical path is 1-2-3, the completion time =11 How? Path: 1-2-3 = 5+6=11 weeks Path: 1-3 = 5 weeks Now, how many days can we “crash” it? 1 3 2 5 (1) 6(3) 5(0) Normal weeks Max weeks can be crashed (to p6)

6. Chapter 8 - Project Management6 Example – crashing (1) 1 3 2 5 (1) 6(3) 5(0) The maximum time that can be crashed for: Path 1-2-3 = 1 + 3 = 4 Path 1-3 = 0 Total weeks can be crashed = 4 + 0 = 4 Are we to use up all these 4 weeks? (to p7)

7. Chapter 8 - Project Management7 Example – crashing (1) 1 3 2 5 (1) 6(3) 5(0) If we used all 4 days, then path 1-2-3 has (5-1) + (6-3) = 7 completion weeks Now, we need to check if the completion time for path 1-3 has lesser than 7 weeks (why?) Now, path 1-3 has (5-0) = 5 weeks Since path 1-3 still shorter than 7 weeks, we used up all 4 crashed weeks Question: What if path 1-2 has, say 8 week completion time? 4(0) 3(0) (to p8)

8. Chapter 8 - Project Management8 Example – crashing (1) 1 3 2 5 (1) 6(3) 8(0) Such as Now, we cannot use all 4 days (Why?) Because path 1-2-3 will not be critical path anymore as path 1-3 would now has longest hour to finish Rule: When a path is a critical path, it will stay as a critical path So, we can only reduce the path 1-2-3 completion time to the same time As path 1-2. (HOW?) (to p9)

9. Chapter 8 - Project Management9 Example – crashing (1) 1 3 2 5 (1) 6(3) 8(0) Solution: We can only reduce total time for path 1-2-3 = path 1-2, that is 8 weeks If the cost for path 1-2 and path 2-3 is the same then We can random pick them to crash so that its completion Time is 8 weeks (to p10)

10. Chapter 8 - Project Management10 Example – crashing (1) 1 3 2 5 (1) 6(3) 8(0) Solution: 1 2 3 5 (1)6(3) 8(0) OR 4(0)4(1) 3(0) Now, paths 1-2-3 and 1-3 are both critical paths (to p4)

11. Chapter 8 - Project Management11 Time-cost Trade-off In this subject, the decision for “crashing” the project is based on the trade-off between “time and cost” The method is called “Time-cost Trade-off” How it works? –We determine an average crash cost for each event How to do that? –Procedural step. (to p13) (to p12)

12. 12 Project Crashing and Time-Cost Trade-Off Example Problem (1 of 3) Table 8.5 Normal Activity and Crash Data for the Network in Figure 8.16 AB C D E=(A-B) F= (D-C)/(A-B) Note: A,B,C,D are given Note: we will use F values to decide We need to compute E and F which path to crash! F (to p11)

13. Chapter 8 - Project Management13 Time-Cost Trade-Off Steps: 1. use “normal cost” to determine the critical path 2. for each event, compute their average crash cost 3. for each section of critical path, crash their maximum time by retaining this section be part of the “critical” path. 4. compute total crashing costs and completion time Example (to p14)

14. Chapter 8 - Project Management14 Example: trade-off Consider the same example as show in below Step 1 determine it critical path Step 2 determine all average unit crash cost Step 3 crashing events with minimum costs Step 4 compute crashed weeks and costs (to p15) (to p16) (to p21) (to p20) (to p17) More example!

15. Chapter 8 - Project Management15 Step 1 Using CPM, the critical path is 1-2-3-4-6-7 (to p14)

16. Chapter 8 - Project Management16 Step 2 (to p14)

17. 17 First, we cluster each segment of critical path into sections that can be crashed and to consider to crash them one section at a time Step 3: Section1 Section 2 Section 3 Section 4 (to p18)

18. 18 We now add the normal and crashed time and cost to each segment Step 3: Section1 Section 2 Section 3 Section 4 12(5) $400 8(3) $500 12(3) $$7000 4(1) $7000 4(1) $3000 4(3) $$200 (to p19)

19. 19 We now crashed them one section at a time as follows: Step 3: Section1 Section 2 Section 3 Section 4 12(5) $400 8(3) $500 12(3) $$7000 4(1) $7000 4(1) $3000 4(3) $$200 7(0) 5(0) 9(0) 3(0) (to p14)

20. 20 We now crashed them one section at a time as follows: Step 4: 12(5) $400 8(3) $500 12(3) $$7000 4(1) $7000 4(1) $3000 4(3) $$200 7(0) 5(0) 9(0) 3(0) Total crash cost =( 5*$400)+(3*$500)+(3*$7000)+(1*$7000)= 31,000 Total crashed weeks= 5+3+3+1=12 Note: critical path is 1-2-3-4-6-7 Completion time = 7+5+0+9+3 = 24 (to p14)

21. Chapter 8 - Project Management21 Crashed cost 1 2 3 4 10(5)5(4) 4(2) 4(1) How to solve this problem? (to p22)

22. Chapter 8 - Project Management22 Further detail steps 1.Determine the critical path 2.Crash the critical path to the level where other non-critical paths become a critical one 3.Consider for further crashing until all possible crashing resources were consumed! (to p23)

23. Chapter 8 - Project Management23 Critical path 1 2 3 4 10(5)5(4) 4(2) 4(1) The critical path is 1-3-4, completion time is 10+5 = 15 (to p24)

24. Chapter 8 - Project Management24 Crash to a level to which other non- critical path is introduced 1 2 3 4 10(5)5(4) 4(2) 4(3) The non-critical path is 1-2-4, has the processing time = 4+4 = 8 So, we try to reduce the critical path to this level ! 5(0)3(2) Both critical Paths = 8 (to p25)

25. Chapter 8 - Project Management25 Crash all resources until no further can be reduced! 1 2 3 4 10(5)5(3) 4(2) 4(3) Stop, since no more resources can be reduced in path 1-3-4 5(0)3(1) Both critical Paths = 7 3(2) 2(0) (to p1)

26. Chapter 8 - Project Management26 Formulating the CPM/PERT Network as a Linear Programming Model - The objective is to determine the earliest time the project can be completed (i.e., the critical path time). normal CPM crashing model (to p27) (to p30) (to p1)

27. Chapter 8 - Project Management27 LP formulation General linear programming model is: minimize Z =  c i x i subject to x j - x i  t ij for all activities i  j x i, x j  0 where x i = earliest event time of node i x j = earliest event time of node j t ij = time of activity i  j LP formulation for the project management (to p28)

28. Chapter 8 - Project Management28 LP for the CPM Let consider a simple problem as outlined as follows: Let x i be denote as each node i And segment of say path 1-2 as x 2 -x 1 Then (to p29)

29. Chapter 8 - Project Management29 Objective is Minimize Z = x1 + x2 + x3 + x4 + x5 + x6 + x7 Subject to x2 - x1  12 (for path 1-2) x3 - x2  8(for path 2-3) x4 - x2  4(for path 2-4) x4 - x3  0(for path 3-4) x5 - x4  4(for path 4-5) x6 - x4  12(for path 4-6) x6 - x5  4(for path 5-6) x7 - x6  4(for path 6-7) xi, xj  0 Do you know how to read the results from the LP output? (to p26)

30. Chapter 8 - Project Management30 General concept All formulation of CPM is used, except we need one more variable to represent the crashed cost per unit of each path Example! (to p31)

31. Chapter 8 - Project Management31 Consider again the following crashed cost as an example - Objective is to reduce the project duration from 36 to 30 weeks at the minimum possible crash cost. We now y to represent these Our objective is to min these How? (to p32)

32. Chapter 8 - Project Management32 Min 400y 12 + 500y 23 + 3000y 24 + 0y 34 + 200y 45 + 7000y 46 + 200y 56 + 7000y 67 And all y ij <= their total allowance crash time A complete model is shown in next slide (to p33)

33. Chapter 8 - Project Management33 The CPM/PERT Network as a Linear Programming Model Example Problem Project Crashing - Model Formulation x i = earliest event time of node i x j = earliest event time of node j y ij = amount of time by which activity i  j is crashed (i.e., reduced) minimize Z = $400y 12 + 500y 23 + 3000y 24 + 200y 45 + 7000y 46 + 200y 56 + 7000y 67 subject to y 12  5y 12 + x 2 - x 1  12 y 23  3y 23 + x 3 - x 2  8 y 24  1y 24 + x 4 - x 2  4 y 34  0y 34 + x 4 - x 3  0 y 45  3y 45 + x 5 - x 4  4 y 46  3y 46 + x 6 - x 4  12 y 56  3y 56 + x 6 - x 5  4 y 67  1x 67 + x 7 - x 6  4 x 7  30 x j, y ij  0 New set of equations Max crashing time for critical path i.e. total allowable crashed time CPM value (to p26)