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Chapter 3

Chapter 3. Project Management. Month. 0 2 4 6 8 10. | | | | |. Activity Design house and obtain financing Lay foundation Order and receive materials Build house Select paint Select carpet Finish work. 1 3 5 7 9. Month. A Gantt Chart. Node. 1. 2. 3. Branch. The Project Network.

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Chapter 3

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  1. Chapter 3 Project Management

  2. Month 0 2 4 6 8 10 | | | | | Activity Design house and obtain financing Lay foundation Order and receive materials Build house Select paint Select carpet Finish work 1 3 5 7 9 Month A Gantt Chart

  3. Node 1 2 3 Branch The Project Network Network consists of branches (arrows) & nodes We will stick to Activity on Arrow Representation

  4. 3 Lay foundation Dummy Build house Finish work 2 0 3 1 1 2 4 6 7 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 5 Project Network for a House

  5. Critical Path • A path is a sequence of connected activities running from start to end node in network • The critical path is the path with the longest duration in the network • Project cannot be completed in less than the time of the critical path

  6. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 1 2 4 6 7 5 The Critical Path A: 1-2-3-4-6-73 + 2 + 0 + 3 + 1 = 9 months B: 1-2-3-4-5-6-73 + 2 + 0 + 1 + 1 + 1 = 8 months C: 1-2-4-6-73 + 1 + 3 + 1 = 8 months D: 1-2-4-5-6-73 + 1 + 1 + 1 + 1 = 7 months

  7. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 1 2 4 6 7 5 The Critical Path A: 1-2-3-4-6-7 3 + 2 + 0 + 3 + 1 = 9 months B: 1-2-3-4-5-6-7 3 + 2 + 0 + 1 + 1 + 1 = 8 months C: 1-2-4-6-7 3 + 1 + 3 + 1 = 8 months D: 1-2-4-5-6-7 3 + 1 + 1 + 1 + 1 = 7 months The Critical Path

  8. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 3 Finish at 9 months Start at 5 months 2 0 1 3 1 1 2 4 6 7 1 2 4 6 7 3 1 1 Start at 8 months Start at 3 months 5 5 The Critical Path Activity Start Times

  9. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 1 2 4 6 7 5 Early Times • ES - earliest time activity can start • Forward pass starts at beginning of CPM/PERT network to determine ES times • EF = ES + activity time • ESij=maximum (EFi) • EFij=ESij- tij • ES12=0 • EF12 =ES12- t12=0 + 3 = 3months

  10. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 1 2 4 6 7 5 Computing Early Times • ES23 = max EF2 = 3 months • ES46 = max EF4 = max 5,4 = 5 months • EF46 = ES46 - t46 = 5 + 3 = 8 months • EF67 = 9 months, the project duration

  11. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet Early Start and Finish Times 3 3 (ES = 3, EF = 5) (ES = 5, EF = 5) 2 0 (ES = 5, EF = 8) 1 2 4 6 7 1 3 1 1 2 4 6 7 3 (ES = 0, EF = 3) (ES = 3, EF = 4) (ES = 8, EF = 9) 1 1 (ES = 5, EF = 6) (ES = 6, EF = 7) 5 5 Computing Early Times

  12. Late Times • LS - latest time activity can start & not delay project • Backward pass starts at end of CPM/PERT network to determine LS times • LF = LS + activity time • LSij=LFij-tij • LFij= minimum (LSj)

  13. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 1 2 4 6 7 5 Computing Late Times • LF67 = 9 months • LS67 = LF67 - t67 = 9 - 1 = 8 months • LF56 = minimum (LS6) = 8 months • LS56 = LF56 - t56 = 8 - 1 = 7 months • LF24 = minimum (LS4) = min(5, 6) = 5 months • LS24 = LF24 - t24 = 5 - 1 = 4 months

  14. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet Early and Late Start and Finish Times 3 3 ( ) ( ) ES = 3, EF = 5 LS = 3, LF = 5 ES = 5, EF = 5 LS = 5, LF = 5 ( ) ES = 5, EF = 8 LS = 5, LF = 8 2 0 1 2 4 6 7 1 3 1 1 2 4 6 7 3 ( ) ( ) ( ) ES = 0, EF = 3 LS = 0, LF = 3 ES = 3, EF = 4 LS = 4, LF = 5 ES = 8, EF = 9 LS = 8, LF = 9 1 1 5 ( ) ( ) 5 ES = 5, EF = 6 LS = 6, LF = 7 ES = 6, EF = 7 LS =7, LF = 8 Computing Late Times • LF67 = 9 months • LS67 = LF67 - t67 = 9 - 1 = 8 months • LF56 = minimum (LS6) = 8 months • LS56 = LF56 - t56 = 8 - 1 = 7 months • LF24 = minimum (LS4) = min(5, 6) = 5 months • LS24 = LF24 - t24 = 5 - 1 = 4 months

  15. Activity Slack • Activities on critical path have ES = LS & EF = LF • Activities not on critical path have slack • Sij = LSij - ESij • Sij = LFij - EFij • S24 = LS24 - ES24 = 4 - 3 = 1 month

  16. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 Activity LS ES LF EF Slacks *1-2 0 0 3 3 0 *2-3 3 3 5 5 0 2-4 4 3 5 4 1 *3-4 5 5 5 5 0 4-5 6 5 7 6 1 *4-6 5 5 8 8 0 5-6 7 6 8 7 1 *6-7 8 8 9 9 0 * Critical path 1 2 4 6 7 5 Activity Slack Data

  17. Lay foundation Dummy Build house Finish work 2 0 3 1 3 1 Design house and obtain financing Order and receive materials 1 1 Select paint Select carpet 3 3 Activity Slack Activity LS ES LF EF Slacks *1-2 0 0 3 3 0 *2-3 3 3 5 5 0 2-4 4 3 5 4 1 *3-4 5 5 5 5 0 4-5 6 5 7 6 1 *4-6 5 5 8 8 0 5-6 7 6 8 7 1 *6-7 8 8 9 9 0 * Critical path S = 0 S = 0 1 2 4 6 7 1 2 4 6 7 2 0 S = 0 1 3 1 S = 0 S = 1 S = 0 3 5 5 1 1 S = 1 S = 1 Activity Slack Data

  18. Probabilistic Time Estimates • Reflect uncertainty of activity times • Beta distribution is used in PERT

  19. a + 4m + b 6 Mean (expected time): t = 2 b - a 6 Variance: 2 = where a = optimistic estimate m = most likely time estimate b= pessimistic time estimate Probabilistic Time Estimates • Reflect uncertainty of activity times • Beta distribution is used in PERT

  20. P(time) P(time) a m t b a t m b Time Time P(time) a m = t b Time Example Beta Distributions Figure 6.11

  21. E 2 6 Equipment testing and modification Equipment installation Final debugging A D Dummy L System development F I M 1 3 5 7 9 B Manual Testing System Training System changeover Position recruiting Job training System Testing C G J K Dummy H 4 8 Orientation Southern Textile Company

  22. 2 6 1 3 5 7 9 4 8 TIME ESTIMATES (WKS) ACTIVITYa m b 1 - 2 A 6 8 10 1 - 3 B 3 6 9 1 - 4 C 1 3 5 2 - 5 D 0 0 0 2 - 6 E 2 4 12 3 - 5 F 2 3 4 4 - 5 G 3 4 5 4 - 8 H 2 2 2 5 - 7 I 3 7 11 5 - 8 J 2 4 6 8 - 7 K 0 0 0 6 - 9 L 1 4 7 7 - 9 M 1 10 13 Activity Estimates

  23. 2 6 1 3 5 7 9 4 8 TIME ESTIMATES (WKS) MEAN TIME VARIANCE ACTIVITYa m b t2 1 - 2 6 8 10 8 0.44 1 - 3 3 6 9 6 1.00 1 - 4 1 3 5 3 0.44 2 - 5 0 0 0 0 0.00 2 - 6 2 4 12 5 2.78 3 - 5 2 3 4 3 0.11 4 - 5 3 4 5 4 0.11 4 - 8 2 2 2 2 0.00 5 - 7 3 7 11 7 1.78 5 - 8 2 4 6 4 0.44 8 - 7 0 0 0 0 0.00 6 - 9 1 4 7 4 1.00 7 - 9 1 10 13 9 4.00 Activity Estimates

  24. 2 6 1 3 5 7 9 4 8 6 + 4(8) + 10 6 a + 4m + b 6 2 b - a 6 4 9 10 - 6 6 t = = = 8 weeks  2 = = = week 2 Early and Late Times For Activity 1-2 a= 6, m = 8, b = 10

  25. 2 6 1 3 5 7 9 4 8 ACTIVITYtES EF LS LF S 1 - 2 8 0.44 0 8 1 9 1 1 - 3 6 1.00 0 6 0 6 0 1 - 4 3 0.44 0 3 2 5 2 2 - 5 0 0.00 8 8 9 9 1 2 - 6 5 2.78 8 13 16 21 8 3 - 5 3 0.11 6 9 6 9 0 4 - 5 4 0.11 3 7 5 9 2 4 - 8 2 0.00 3 5 14 16 11 5 - 7 7 1.78 9 16 9 16 0 5 - 8 4 0.44 9 13 12 16 3 8 - 7 0 0.00 13 13 16 16 3 6 - 9 4 1.00 13 17 21 25 8 7 - 9 9 4.00 16 25 16 25 0 Early and Late Times

  26. ES = 8, EF = 13 LS = 16, LF = 21 2 6 5 ES = 0, EF = 8 LS = 1, LF = 9 ES = 8, EF = 8 LS = 9, LF = 9 ES = 13, EF = 17 LS = 21, LF = 25 8 0 4 ES = 0, EF = 6 LS = 0, LF = 6 ES = 6, EF = 9 LS = 6, LF = 9 ES = 9, EF = 16 LS = 9, LF = 16 9 1 3 5 7 9 6 3 7 ES = 16, EF = 25 LS = 16, LF = 25 ES = 3, EF = 7 LS = 5, LF = 9 4 0 3 ES = 0, EF = 3 LS = 2, LF = 5 ES = 9, EF = 13 LS = 12, LF = 16 ES = 13, EF = 13 LS = 16, LF = 16 2 4 8 ES = 3, EF = 5 LS = 14, LF = 16 Southern Textile Company

  27. ES = 8, EF = 13 LS = 16, LF = 21 2 6 5 Total project variance ES = 0, EF = 8 LS = 1, LF = 9 ES = 8, EF = 8 LS = 9, LF = 9 ES = 13, EF = 17 LS = 21, LF = 25 8 0 4 2 = 2 + 2 + 2 + 2 = 1.00 + 0.11 + 1.78 + 4.00 = 6.89 weeks^2 13 35 57 79 ES = 0, EF = 6 LS = 0, LF = 6 ES = 6, EF = 9 LS = 6, LF = 9 ES = 9, EF = 16 LS = 9, LF = 16 9 1 3 5 7 9 6 3 7 ES = 16, EF = 25 LS = 21, LF = 25 ES = 3, EF = 7 LS = 5, LF = 9 4 0 3 ES = 0, EF = 3 LS = 2, LF = 5 ES = 9, EF = 13 LS = 12, LF = 16 ES = 13, EF = 13 LS = 16, LF = 16 2 4 8 ES = 3, EF = 5 LS = 14, LF = 16 Southern Textile Company

  28. x -   Z = Probabilistic Network Analysis Determine probability that project is completed within specified time where  = tp = project mean time  = project standard deviation x = proposed project time Z = number of standard deviations xis from mean

  29. Probability Z  = tp x Time Normal Distribution Of Project Time

  30. What is the probability that the project is completed within 30 weeks? Southern Textile Example

  31. What is the probability that the project is completed within 30 weeks? P(x 25 weeks)  = 25 x = 30 Time (weeks) Southern Textile Example

  32. What is the probability that the project is completed within 30 weeks? Z = = = 1.91 P(x 25 weeks)  2 = 6.89 weeks  = 6.89  = 2.62 weeks x -   30 - 25 2.62  = 25 x = 30 Time (weeks) Southern Textile Example

  33. What is the probability that the project is completed within 30 weeks? Z = = = 1.91 P(x 25 weeks)  2 = 6.89 weeks  = 6.89  = 2.62 weeks x -   30 - 25 2.62  = 25 x = 30 Time (weeks) From Excel, a Z score of 1.91 corresponds to a probability = normsdist(1.91) = 0.9719 or use norm.dist(30,25,2.62,TRUE) Southern Textile Example

  34. What is the probability that the project is completed within 22 weeks? P(x 22 weeks) 0.3729 x = 22  = 25 Time (weeks) Southern Textile Example

  35. What is the probability that the project is completed within 22 weeks? Z = = = -1.14  2 = 6.89 weeks  = 6.89  = 2.62 weeks P(x 22 weeks) x -   0.3729 22 - 25 2.62 x = 22  = 25 Time (weeks) Southern Textile Example

  36. What is the probability that the project is completed within 22 weeks? Z = = = -1.14  2 = 6.89 weeks  = 6.89  = 2.62 weeks P(x 22 weeks) x -   0.3729 22 - 25 2.62 x = 22  = 25 Time (weeks) From Excel, a Z score of -1.14 corresponds to a probability = normsdist(-1.14) = 0.1271 or use norm.dist(22,25,2.62,TRUE) Southern Textile Example

  37. Project Crashing • Crashing is reducing project time by expending additional resources • Crash time is an amount of time an activity is reduced • Crash cost is the cost of reducing the activity time • Goal is to reduce project duration at minimum cost

  38. 3 8 0 12 4 12 4 1 2 4 6 7 4 4 5 Housebuilding Network

  39. $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – 3 8 0 12 4 12 4 1 2 4 6 7 4 4 Normal activity 5 Normal cost Normal time | | | | | | | 0 2 4 6 8 10 12 14 Weeks Housebuilding Network

  40. $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – 3 Crash cost 8 0 Crashed activity 12 4 12 4 1 2 4 6 7 4 4 Normal activity 5 Normal cost Crash time Normal time | | | | | | | 0 2 4 6 8 10 12 14 Weeks Housebuilding Network

  41. $7,000 – $6,000 – $5,000 – $4,000 – $3,000 – $2,000 – $1,000 – – Total crash cost $2,000 Total crash time 5 3 = = $400 per week Crash cost 8 0 Crashed activity 12 4 12 4 1 2 4 6 7 Slope = crash cost per week 4 4 Normal activity 5 Normal cost Crash time Normal time | | | | | | | 0 2 4 6 8 10 12 14 Weeks Housebuilding Network Figure 6.16

  42. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 Normal Activity and Crash Data

  43. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 3 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 $500 8 0 12 4 12 4 1 2 4 6 7 $400 $3,000 $7,000 $7,000 4 4 $200 $200 5 Normal Activity and Crash Data

  44. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 3 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 $500 8 0 12 4 12 4 1 2 4 6 7 $400 $3,000 $7,000 $7,000 4 4 $200 $200 5 Normal Activity and Crash Data

  45. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 3 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 $500 8 0 7 4 12 4 1 2 4 6 7 $3,000 $7,000 $7,000 4 4 $200 $200 5 Normal Activity and Crash Data Crash cost = $2,000

  46. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 3 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 $500 8 0 7 4 12 4 1 2 4 6 7 $3,000 $7,000 $7,000 4 4 $200 $200 5 Normal Activity and Crash Data Crash cost = $2,000

  47. 3 1 2 4 6 7 5 TOTAL NORMAL CRASH ALLOWABLE CRASH TIME TIME NORMAL CRASH CRASH TIME COST PER ACTIVITY (WEEKS) (WEEKS) COST COST (WEEKS) WEEK 3 1-2 12 7 $3,000 $5,000 5 $400 2-3 8 5 2,000 3,500 3 500 2-4 4 3 4,000 7,000 1 3,000 3-4 0 0 0 0 0 0 4-5 4 1 500 1,100 3 200 4-6 12 9 50,000 71,000 3 7,000 5-6 4 1 500 1,100 3 200 6-7 4 3 15,000 22,000 1 7,000 $75,000 $110,700 $500 7 0 7 4 12 4 1 2 4 6 7 $3,000 $7,000 $7,000 4 4 $200 $200 5 Normal Activity and Crash Data Crash cost = $2,000 + $500 = $2,500

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