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Material Management Class Note # 6 Project Scheduling & Management. Prof. Yuan-Shyi Peter Chiu Feb. 2012. ◇. §. P 1: Introduction to Project Management. ■  Planning function at all levels of an organization ■ Poor project management  cost overruns

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Material Management Class Note # 6 Project Scheduling & Management


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    1. Material Management Class Note # 6 Project Scheduling & Management Prof. Yuan-Shyi Peter Chiu Feb. 2012

    2. §. P 1: Introduction to Project Management ■  Planning function at all levels of an organization ■Poor project management  cost overruns  delay ■ eg. Launching new products Organizing research projects Building new production facilities

    3. §. P 2: Two common techniques for Project Management ■  Critical Path Method (CPM) ~ deterministic problems  ■ Project Evaluation and Review Technique ( PERT ) ~ randomness allowed in the activity times.

    4. §. P 3 : Critical Path Analysis (1).Project definition ~ clear statement (2).Activity definition ~ project broken down into a set of indivisible tasks or activities (3).Activity relationships ~ precedence constraints (4).Project scheduling ~ starting & ending times (5).Project monitoring

    5. §. P 4 : Gantt Chart Fig.9-1 p.487 ~ Gantt Chartdoes not show the precedence constraints among tasks

    6. §. P 5 : Network • ■   can show the precedence • constraints • ■is a collection of nodes and • directed arcs •  arc : activity •  node : event ( start or • completion of a project) •   ■ Two conventional expressions: • Activity-on-arrow • Activity-on-node

    7. [ Eg. 9-2 ] Activity Predecessors A - B - C A D B E C , D Fig.9-2 p.488

    8. [ Eg. 9-3x ] Activity Predecessors A - B - C A D A , B E C , D P Fig.9-4 p.489

    9. §. P 6 : Common Questions about Project   (1)The minimum time required to complete the project ? (2) Starting & ending times for each activities ?    (3)  What activities can be delayed without delaying the entire project?

    10. [ Eg. 9-3x ] From Fig.9-4 there are 3 paths 1-2-4-5 A-C-E 1-2-3-4-5 A-P-D-E 1-3-4-5 B-D-E • The Minimum time required to • complete the project = Longest path Activity Time Predecessors A 1.5 - B 1.0 - C 2.0 A D 1.5 A , B E 1.0 C , D

    11. [ Eg. 9-3x ] A-C-E 4.5 A-P-D-E 4.0 B-D-E 3.5 *Critical path!! A-C-E Critical Path Critical activities ! Other activities have slack.

    12. [ Eg. 9-4 ] Task Time(in weeks) Immediate Predecessors A 3 - B 4 A C 2 A D 6 B , C E 5 C F 3 C G 7 E H 5 E , F I 8 D , G , H

    13. Network for Example 9-4 Fig.9-5 p.491

    14. § P7 : Finding the Critical Path (1) compute the earliest times for each activity ~ forward pass (2)  compute the latest times for each activity ~ backward pass

    15. § P7 : Finding the Critical Path (3) Critical Activity: ESi = LSi orEFi = LFi ESi : Earliest Starting time EFi : Earliest Finishing time LSi : Latest Starting time LFi : Latest Finishing time EFi = ESi + ti LSi = LFi –ti ESi EFi LSi LFi

    16. § P7.1 : Forward Pass → Fig.4

    17. § P7.2 : Backward Pass ← Fig.5

    18. § P 7.3 : Critical Path The set of critical activities and in proper order e.g. A – C – E – G – I

    19. §. P 7.4: Class Problems Discussion Chapter 9 : [# 3 a,b,c,4 a,b,c; 5 a,b;6 a,b,c,d]p. 496 Preparation Time : 25 ~ 30 minutes Discussion : 10 ~ 15 minutes

    20. § P 8 : Project cost & Alternatives Schedules  ■ Expediting costs ~ activity time can be reduced at additional cost ◆ Normal time ◆Expedited time ◆ One of the CPM cost-time relationship : “ linear model ”

    21. [ Eg. 9-5 ] p.499 Activity Normal Expedited Norm. Exp- Cost/wk Time(wks) TimeCost Cost A 3 1 1000 3000 1000 B 4 3 4000 6000 2000 C 2 2 2000 2000 - D 6 4 3000 6000 1500 E 5 4 2500 3800 1300 F 3 2 1500 3000 1500 G 7 4 4500 8100 1200 H 5 4 3000 3600 600 I 8 5 8000 12,800 1600 29,500 48,300

    22. [ Eg. 9-5 ] (A) Using expedited time : Fig.6

    23. [ Eg. 9-5 ] (B) CP solution ( when using expedited time ) : Fig.7 # CP=16 weeks *A-C-E-G-I *A-C-E-H-I

    24. • Normal Cost = $ 29,500 ; 25 weeks • Expedited Cost =$ 48,300 ; 16 weeks • Extra Cost = $18,800 • If Benefit = $1,500/week × 9 ( i.e. 25-16 ) = $13,500 (Saved) Spent $18,800 to save $13,500 ? →Not economy in expediting non-critical activities ! [ Answer to Eg. 9-5 ]

    25. § P 9 : Expediting Procedures [ Eg. 9-5 ] • List CP & its Normal and • Expedited Time and Cost. • CP: A – C – E – G – I (from previous …) Project CP CP Norm. Exped. Cost Time Activities Time Time / Week 25 A-C-E-G-I A 3 1 1000 C 2 2 - E 5 4 1300 G 7 4 1200 I 8 5 1600 Refer to Fig.9-7:Gantt Chart p.513 For a better picture!

    26. [ Eg. 9-5 ] (2) Pick the least expensive task without deriving a new CP ∴ to reduce A from 3 to 1, cost $2000 ∴ Next on G from 7 to 5 , cost $2400 (3) Repeat (1) & (2) until no more reduction in time are beneficial !

    27. [ Eg. 9-5 ] Project CP CP Norm. Expe. Cost Time Activities Time Time Red. Week 21 A-C-E-G-I A 1 1 - A-C-E-H-I C 2 2 - E 5 4 1300 G 5 4 1200 H 5 4 600 I 8 5 1600 ﹝ • Looks like on H but must reduce G together (Why?) ∴ G+H =$1800 / wk > $1500 ( X ) • Next on E from 5 to 4 cost $1300 / wk

    28. [ Eg. 9-5 ] Project CP CP Norm. Expe. Cost Time Activities Time Time Red. Week 20 A-C-E-G-I A 1 1 - A-C-E-H-I C 2 2 - E 4 4 - G 5 4 $1200 H 5 4 600 I 8 5 1600 ﹝ ◆ At this point no more reduction can be made for < $1500 / wk

    29. [ Eg. 9-5 ] ∴ From Original 25 weeks reduced to 20 weeks Cost Benefit 2000 2400 +1300 5700 $1500 × 5 7500 Make Sense!!

    30. [ Eg. 9-5 ] ◆ Final Solution ( in diagram ) Fig.8 CP: A-C-E-G-I A-C-E-P2-H-I

    31. [ Eg. 9-5 ] ◆ Final Solution ( in Gantt Chart ) Slack Fig.9

    32. The CPM Cost-Time Linear Model Fig.9-10 p.498 Fig.9-11 p.499

    33. §. P 9.1: Class Problems Discussion Chapter 9 : [ # 8 a,b ; 10 ]p. 502 [ # 30 a,b,c]p. 532 Preparation Time : 25 ~ 30 minutes Discussion : 10 ~ 15 minutes

    34. § P 10 : PERT ~ Introduction ■ Generalization of CPM, allows uncertainty in the activity time. ■ Terms : a : minimum activity time m : most likely activity time b : maximum activity time

    35. § P 10.1 : Beta Distribution a = 5 days , b = 20 , m = 17 X → 5 10 15 17 20

    36. § P 10.1 : Beta Distribution • Finite interval • Mode within interval • Used to describe the distribution of • individual activity times σ= μ=

    37. § P 10.2 : Uniform Distribution is a special case of the beta distribution. f(t) a b

    38. § P 10.3 : PERT ■ In PERT, one assumes : Total project time ~ Normal Distribution By Central Limit theorem ∵ T= , : Independent random var.

    39. §P 11: PERT ~ procedures (1). Estimates a , b , m , for all activities. (2). Using these estimates to compute μ and σ2 for all activities. (3). Using μ to find CP (critical path) (4). Total project time : E(T) = Var(T) = (5). Applications of E(T) ~ Normal Distribution

    40. §P 11: PERT ~ procedures [ Eg. 9-6 ]p.510 (1) (2) Act. MIN Likely MAX (a) (m) (b) σ2 μ A 2 3 4 B 2 4 10 C 2 2 2 D 4 6 12 E 2 5 8 F 2 3 8 G 3 7 10 H 3 5 9 I 5 8 18

    41. §P 11: PERT ~ procedures [ Eg. 9-6 ] (3) → CP = A-C-E-G-I

    42. §P 11: PERT ~ procedures [ Eg. 9.6 ] (4) → E(T) = 3+2+5+6.83+9.17=26 Var(T)=0.11+0+1.0+1.36+4.69=7.16 (5) →Total project time Normal(μ=26, σ= = 2.68)

    43. §P 11: PERT ~ procedures [ Eg. 9-7 ]p.510 Solution:

    44. §P 11: PERT ~ procedures [ Eg. 9-7 ] (1) (2) 0.2266 0.0681

    45. §P 11: PERT ~ procedures [ Eg. 9-7 ] (3) (3) 0.90

    46. §P 12 : Path Independence If CP E(T) =26 non-CP E(T) =25

    47. §P 12 : Path Independence ■In reality, there is a chance that non-critical path become critical !! ■ A-C-E-G-I A-C-F-H-I ■ Assuming independence of 2 or more paths – more accurate than assuming a single critical path. ﹜ Are they independent ?

    48. §P 12 : Path Independence [ Eg. 9-8 ] p.533 ﹛ A-C-E-G-I : 41 B-D-F- H-I : 40 • Almost independent

    49. § P13 : Something to think about ! ■ Activity Time ~ Randomness ~ Independence ■ Single Path ~ consists of independent activity ~ path completion time Normally Distribution ( Central Limit Theorem) ■ Pathmay not be independent . how to assume ? ■ When to assume one CP ? ﹛

    50. §. P 13.1: Class Problems Discussion Chapter 9 :[ #18,19,20,21 ] p. 515-7 The End