Download presentation

Presentation is loading. Please wait.

2
1 Material Management Class Note # 5-A ( in review ) Project Scheduling & Management Prof. Yuan-Shyi Peter Chiu Feb. 2011

3
2 ■ 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 §. P 1: Introduction to Project Management ◇

4
3 ■ Critical Path Method (CPM) ~ deterministic problems ■ Project Evaluation and Review Technique ( PERT ) ~ randomness allowed in the activity times. §. P 2: Two common techniques for Project Management ◇

5
4 (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 §. P 3 : Critical Path Analysis ◇

6
5 Gantt Chart ~ Gantt Chart does not show the precedence constraints among tasks §. P 4 : Gantt Chart Fig.9-1 p.487◇

7
6 ■ 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 §. P 5 : Network ◇

8
7 [ Eg. 9-2 ] Activity Predecessors A － B － C A D B E C, D Fig.9-2 p.488◇

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

10
9 (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? §. P 6 : Common Questions about Project ◇

11
10 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 (1)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 [ Eg. 9-3x ]◇

12
11 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. [ Eg. 9-3x ]◇

13
12 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 [ Eg. 9-4 ]◇

14
13 Network for Example 9-4 Fig.9-5 p.491◇

15
14 (1) compute the earliest times for each activity ~ forward pass (2) compute the latest times for each activity ~ backward pass § P7 : Finding the Critical Path ◇

16
15 ES i EF i LS i LF i § P7 : Finding the Critical Path ◇ (3) Critical Activity: ESi = LSi or EFi = LFi ESi : Earliest Starting time EFi : Earliest Finishing time LSi : Latest Starting time LFi : Latest Finishing time EF i = ES i + t i LS i = LF i – t i

17
16 § P7.1 : Forward Pass Fig.4 →◇

18
17 § P7.2 : Backward Pass Fig.5 ←◇

19
18 § P 7.3 : Critical Path The set of critical activities and in proper order e.g. A – C – E – G – I◇

20
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 ] 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 ◇

21
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 ” ◇

22
21 Activity Normal Expedited Norm. Exp- Cost/wk Time(wks) Time Cost 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 [ Eg. 9-5 ] p.499

23
22 Fig.6 (A) Using expedited time : [ Eg. 9-5 ]

24
23 Fig.7 ＊ A－C－E－G－I＊ A－C－E－H－I＊ A－C－E－G－I＊ A－C－E－H－I (B) CP solution ( when using expedited time ) : # CP=16 weeks [ Eg. 9-5 ]

25
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 ]

26
25 § P 9 : Expediting Procedures 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 (1)List CP & its Normal and Expedited Time and Cost. CP: A – C – E – G – I (from previous …) [ Eg. 9-5 ] Refer to Fig.9-7:Gantt Chart p.513 For a better picture!◇

27
26 (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 ! [ Eg. 9-5 ]

28
27 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 ﹝ [ Eg. 9-5 ]

29
28 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 [ Eg. 9-5 ]

30
29 ∴ From Original 25 weeks reduced to 20 weeks 2000 2400 +1300 5700 $1500 × 5 7500 Make Sense!! Cost Benefit [ Eg. 9-5 ]

31
30 Fig.8 CP:A － C － E － G － I A － C － E － P 2 － H － I ◆ Final Solution ( in diagram ) [ Eg. 9-5 ]

32
31 Fig.9 ◆ Final Solution ( in Gantt Chart ) Slack [ Eg. 9-5 ]

33
32 The CPM Cost-Time Linear Model Fig.9-10 p.498 Fig.9-11 p.499

34
33 §. P 9.1: Class Problems Discussion Chapter 9 : [ # 8 a,b ; 10 ] Chapter 9 : [ # 8 a,b ; 10 ] p. 502 [ # 30 a,b,c ] [ # 30 a,b,c ] p. 532 Preparation Time : 25 ~ 30 minutes Discussion : 10 ~ 15 minutes

35
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◇

36
35 § P 10.1 : Beta Distribution a = 5 days, b = 20, m = 17 X → 5 10 15 17 20

37
36 (1)Finite interval (2)Mode within interval (3)Used to describe the distribution of individual activity times μ= σ= § P 10.1 : Beta Distribution

38
37 § P 10.2 : Uniform Distribution § P 10.2 : Uniform Distribution is a special case of the beta distribution. a b f(t)

39
38 § P 10.3 : PERT ■ In PERT, one assumes : Total project time ~ Normal Distribution By Central Limit theorem ∵ T=, : Independent random var.◇

40
39 §P 11: PERT ~ procedures (1). Estimates a, b, m, for all activities. (2). Using these estimates to compute μ and σ 2 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◇

41
40 [ Eg. 9-6 ] p.510 Act. MIN Likely MAX (a) (m) (b) μ σ2σ2 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 (1)(2) §P 11: PERT ~ procedures ◇

42
41 (3) → CP = A － C － E － G － I §P 11: PERT ~ procedures [ Eg. 9-6 ]◇

43
42 (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 ) §P 11: PERT ~ procedures [ Eg. 9.6 ]◇

44
43 Solution: §P 11: PERT ~ procedures [ Eg. 9-7 ] p.510 ◇

45
44 (1) (2) 0.2266 0.0681 §P 11: PERT ~ procedures [ Eg. 9-7 ] ◇

46
45 (3) 0.90 §P 11: PERT ~ procedures [ Eg. 9-7 ] ◇

47
46 §P 12 : Path Independence If CP E(T) =26 non-CP E(T) =25

48
47 ■ 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 ? §P 12 : Path Independence

49
48 Almost independent ﹛ A － C － E － G － I : 41 B － D － F － H － I : 40 §P 12 : Path Independence [ Eg. 9-8 ] p.533

50
49 § P13 : Something to think about ! ■ Activity Time ~ Randomness ~ Independence ■ Single Path ~ consists of independent activity ~ path completion time Normally Distribution ( Central Limit Theorem) ■ Path may not be independent. how to assume ? ■ When to assume one CP ? ﹛

51
50 §. P 13.1: Class Problems Discussion Chapter 9 : [ # 18,19, 20, 21 ] Chapter 9 : [ # 18,19, 20, 21 ] p. 515-7 The End

52
51 The End

53
52

54
53 (How to compute above?) ■ Solution to Eg. 9-8 §P 12 : Path Independence [ Eg. 9-8 ]◇

55
54 ■ If consider only one CP, then P{T< 43} = 0.7642 ■ If consider both paths, then P{T< 43} = 0.6537 ■ For this network 0.6537 is far more accurate. (A) Discussion 1 : §P 12 : Path Independence [ Eg. 9-8 ]◇

56
55 (B) What date of project completion can be assured by 90% chance ? ｝ §P 12 : Path Independence [ Eg. 9-8 ]◇

57
56 Assuming path independent in this problem more accurate than one CP. But Assumption of path independent may be inaccurate ?!? [ Eg. 9-8 ]◇

58
57 Network for PERT / Eg. 9-8 §P 12 : Path Independence [ Eg. 9-8 ]◇

59
58 5 PATHS A － B － D － I 23.5 8.36 A － C － P1 － D － I 20.8 6.58 A － C － E － G － I 26.0* 7.16 A － C － E － P2 － H － I 23.0 6.80 A － C － F － H － I 23.2 6.80 E(T) Var(T) ﹛ §P 12 : Path Independence [ Eg. 9-8x ]◇

60
59 If we assume 5 paths are independent Assuming path independence can have a very significant effect on the probabilities. §P 12 : Path Independence [ Eg. 9-8x ]◇

61
60 The End

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google