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Material Management

Class Note # 6

Project Scheduling & Management

Prof. Yuan-Shyi Peter Chiu Feb. 2012

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■  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 P 1: Introduction to Project ManagementProject Management

◇◇

3

■  Critical Path Method (CPM)

~ deterministic problems

 ■ Project Evaluation and Review

Technique ( PERT )

~ randomness allowed in the activity

times.

§.§. P 2: P 2: Two common techniques Two common techniques

for Project Management for Project Management

◇◇

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(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 : P 3 : Critical Path AnalysisCritical Path Analysis

◇◇

5

B

C

D

E

12:00 1 2 3 4 5 6:00

A

~ Gantt ChartGantt Chart does not show the precedence constraints among tasks

§.§. P 4 : Gantt Chart P 4 : Gantt Chart

Fig.9-1 p.487

◇◇

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■   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 P 5 : Network ◇◇

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1

2

4

3

5

A

B

C

D

E

[ Eg. 9-2 ]

Activity Predecessors

A － B － C A D B E C , D

Fig.9-2 p.488

◇◇

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Activity Predecessors

A － B － C A D A , B E C , D

1

2

4

3

5

A

B

C

D

E

P

[ Eg. 9-3x ]

Fig.9-4 p.489

◇◇

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    (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 : P 6 : Common Questions Common Questions about Projectabout Project

◇◇

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From Fig.9-4 there are 3 paths

1－ 2－ 4－ 5 A－ C－ E1－ 2－ 3－ 4－ 5 A－ P－ D－ E1－ 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 ]◇◇

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A－ C－ E 4.5A－ P－ D－ E 4.0B－ D－ E 3.5

＊Critical path!!

A－ C－ E Critical Path

Critical activities !

Other activities have slack.

[ Eg. 9-3x ]◇◇

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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 ]◇◇

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1 2

3

45

6

7 8 A

B

C

P1

E

F

G

P2 H

I

D

Network for Example 9-4

Fig.9-5 p.491

◇◇

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(1) compute the earliest times

for each activity

~ forward pass

(2)  compute the latest times

for each activity

~ backward pass

§ P7 : § P7 : Finding the Critical PathFinding the Critical Path

◇◇

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ESi EFi

LSi LFi

§ P7 : § P7 : Finding the Critical PathFinding the Critical Path◇◇

(3) Critical Activity:

ESi = LSi or EFi = LFi

ESi : Earliest Starting time EFi : Earliest Finishing time

LSi : Latest Starting timeLFi : Latest Finishing time

EFi = ESi + ti

LSi = LFi – ti

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§ P7.1 : Forward Pass§ P7.1 : Forward Pass

30

73

137

53

55

105

1710

2517

85

1010

1 2

3

4

5

6

7 8

B=4

C=2

P1=0

E=5

F=3

G=7

P2=0 H=5

I=8

D=6

A=3

1510

Fig.4

→◇◇

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§ P7.2 : Backward Pass§ P7.2 : Backward Pass

30

117

1711

53

1111

105

1710

2517

129

1212

1 2

3

4

5

6

7 8

B=4

C=2

P1=0

E=5

F=3

G=7

P2=0 H=5

I=8

D=6

A=3

1712

Fig.5

←

◇◇

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§ P 7.3 : Critical Path§ P 7.3 : Critical Path

The set of critical activities and in proper order

e.g. A – C – E – G – I

◇◇

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§.§. P 7.4: P 7.4: Class ProblemsClass Problems Discussion Discussion

Chapter 9 :Chapter 9 : [ [ # 3 # 3 a,b,ca,b,c,, 4 4 a,b,ca,b,c;; 5 a,b; 5 a,b; 6 6 a,b,c,da,b,c,d ]] p. 496

Preparation Time : 25 ~ 30 minutesDiscussion : 10 ~ 15 minutes

◇◇

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§ P 8 : Project cost & § P 8 : Project cost & Alternatives Schedules 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 ”

◇◇

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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

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1 2

3

4

5

6

7 8

B=3

C=2

P1=0

E=4

F=2

G=4

P2=0 H=4

I=5

D=4

A=1

Fig.6

(A) Using expedited time :

[ Eg. 9-5 ]

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1 2

3

4

5

6

7 8

B=3

C=2

P1=0

E=4

F=2

G=4

P2=0 H=4

I=5

D=4

A=1

10

10

74

41

117

84

31

31

77

33

73

73

117

117

1611

1611

75

53

77

77

117

117

Fig.7

＊ A － C － E － G － I＊ A － C － E － H － I

(B) CP solution ( when using expedited time ) :

# CP=16 weeks

[ Eg. 9-5 ]

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• 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 ]

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§ P 9 : Expediting Procedures§ 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!

◇◇

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(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 ]

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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 ]

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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 ]

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∴ From Original 25 weeks reduced

to 20 weeks

2000 2400+1300 5700

$1500× 57500

Make Sense!!

Cost Benefit

[ Eg. 9-5 ]

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1 2

3

4

5

6

7 8

B=4

C=2

P1

E=4

F=3

G=5

P2 H=5

I=8

D=6

A=1

10

10

1 5

2 6

126

115

31

31

66

33

73

73

127

127

2012

2012

74

63

77

77

127

127

Fig.8

CP: A － C － E － G － IA － C － E － P2 － H －

I

◆ Final Solution ( in diagram )

[ Eg. 9-5 ]

31

0 10 15 205

ABCDEFGHI

Fig.9

◆ Final Solution ( in Gantt Chart )

Slack

[ Eg. 9-5 ]

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Expedited cost

Normal cost

Expeditedtime

Normaltime

The CPM Cost-Time Linear Model

Fig.9-10 p.498

Fig.9-11 p.499

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§.§. P 9.1: P 9.1: Class ProblemsClass Problems Discussion Discussion

Chapter 9 : Chapter 9 : [ # 8 [ # 8 a,ba,b ; 10 ] ; 10 ] p. 502

[ [ # 30 # 30 a,b,ca,b,c ]] p. 532

Preparation Time : 25 ~ 30 minutesDiscussion : 10 ~ 15 minutes

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§ P 10 : PERT ~ Introduction§ 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

◇◇

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§ P 10.1 : Beta Distribution§ P 10.1 : Beta Distribution

f(X)

a = 5 days , b = 20 , m = 17

X →5 10 15 17 20

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(1) Finite interval

(2) Mode within interval

(3) Used to describe the distribution of

individual activity times

μ= 6

4 bma σ= 6ab

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a-b

2

2

§ P 10.1 : Beta Distribution§ P 10.1 : Beta Distribution

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§ P 10.2 : Uniform Distribution§ P 10.2 : Uniform Distribution

is a special case of

the beta distribution.

a b

f(t)

2ba

2ba

m

12

a-b

2

2

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§ P 10.3 : PERT§ P 10.3 : PERT

■ In PERT, one assumes :

Total project time ~

Normal Distribution

By Central Limit theorem

∵ T= ,

iT

iT : Independent random var.

◇◇

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§P 11: PERT ~ procedures§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

k 2 1 ... 222 ... 2 1 k

◇◇

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[ Eg. 9-6 ] p.510

Act. MIN Likely MAX

(a) (m) (b)

6

b4ma

36

a-b

2

μ σ2

A 2 3 4B 2 4 10C 2 2 2D 4 6 12E 2 5 8F 2 3 8G 3 7 10H 3 5 9 I 5 8 18

(1) (2)

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

◇◇

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1 2

3

4

5

6

7 8

B=

C=

P1=0

E=

F=

G=

P2=0 H=

I=

D=

A=

(3) →

CP = A － C － E － G － I

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

[ Eg. 9-6 ]

◇◇

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(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)

16.7

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

[ Eg. 9.6 ]

◇◇

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90% )t (TP (3)

? 28) (TP (2)

? 22) (TP (1)

r

r

r

Solution:

0.0681 0.4319-0.5

1.49)- (ZP 2.68

26-22 Z {P

-22 Z { P ) 22 T {P

r

r

rr

}

})1(

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

[ Eg. 9-7 ] p.510

◇◇

44

0.2266 0.2734-0.5

Z (P

26-28 Z { P

? ) 28 T {P

r

r

r

)75.0

}68.2

)2(

16 22 26 36 16 26 28 36

(1) (2)

0.22660.0681

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

[ Eg. 9-7 ]

◇◇

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(3)

weeks29.43

8)(1.28)(2.626 t

1.28 Z -t

90% )t T (P

0.90

r

28.1

16 26 30 36

(3) 0.90

§P 11: PERT ~ procedures§P 11: PERT ~ procedures

[ Eg. 9-7 ]

◇◇

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§P 12 : Path Independence §P 12 : Path Independence

If CP E(T) =26 non-CP E(T) =25 503

682

.

.

CP

25 26

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■ 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 §P 12 : Path Independence

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1

4

3

2 6

5 7

8 9

A

B

C

D

E

F

G

H

I

• Almost independent ﹛A － C － E － G － I : 41B － D － F － H － I : 40

? wks)43 withincompleted(Project Pr

§P 12 : Path Independence §P 12 : Path Independence

[ Eg. 9-8 ] p.533

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§ P13 : § P13 : Something to think aboutSomething 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 ?

﹛

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§.§. P 13.1: P 13.1: Class Class ProblemsProblems DiscussionDiscussion

Chapter 9 :Chapter 9 : [ #[ # 1818,19,,19, 2020,, 2121 ] ] p. 515-7

The EndThe End