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

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 ◇

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 ◇

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 ◇

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

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 ◇

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

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

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 ◇

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 ]◇

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 ]◇

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 ]◇

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

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 ◇

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

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 ] 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 Activity Normal Expedited Norm. Exp- Cost/wk Time(wks) Time Cost Cost A B C － D E F G H I , ,500 48,300 [ Eg. 9-5 ] p.499

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

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 ]

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 ) = $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 Project CP CP Norm. Exped. Cost Time Activities Time Time / Week 25 A-C-E-G-I A C 2 2 － E G I (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!◇

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 ]

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 G H I 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 ]

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 I ﹝ ◆ At this point no more reduction can be made for < $1500 / wk [ Eg. 9-5 ]

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

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

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

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 ] 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

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 →

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

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

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 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◇

40 [ Eg. 9-6 ] p.510 Act. MIN Likely MAX (a) (m) (b) μ σ2σ2 A B C D E F G H I (1)(2) §P 11: PERT ~ procedures ◇

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

42 (4) → E(T) = =26 Var(T)= =7.16 (5) → Total project time Normal ( μ=26, σ= = 2.68 ) §P 11: PERT ~ procedures [ Eg. 9.6 ]◇

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

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

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

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

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

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

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 ? ﹛

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