1
Material Management
Class Note # 5-A ( in review )( in review )
Project Scheduling & Management
Prof. Yuan-Shyi Peter Chiu Feb. 2011
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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
◇◇
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 : 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
◇◇
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 P 5 : Network ◇◇
8
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
◇◇
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 : P 6 : Common Questions Common Questions about Projectabout Project
◇◇
10
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 ]◇◇
11
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 ]◇◇
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
(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
→◇◇
17
§ 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
◇◇
20
§ 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 ”
◇◇
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
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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!
◇◇
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 ]
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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 ]
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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
◇◇
35
§ 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
36
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
◇◇
40
[ 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
◇◇
41
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 ]
◇◇
43
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 ]
◇◇
45
(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 ]
◇◇
46
§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
48
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
49
§ 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 ?
﹛
50
§.§. 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
52
Advanced P.O.M.Advanced P.O.M.
Project Management
• Preview : Chap. 9 [ pp.503~537 ]Preview : Chap. 9 [ pp.503~537 ]
22. . Home Work due :Home Work due :
Problem:Problem:
# 3 # 3 a,b,ca,b,c p.514 p.514
# 5 a,b p.515# 5 a,b p.515
# 8 a,b p.521# 8 a,b p.521 # 19 a,b,c p.536# 19 a,b,c p.536
53
8.07 40 )E(T : BDFHI
7.68 41)E(T : ACEGI
2
1
(How to compute above?)
0.6537
(0.8554)(0.7642)
0.3554)(0.50.2642)(0.5
}Z { P0.72}Z { P 8.07
40-43P{Z
7.6841-43
P{Z
} 43P{T} 43P{T
} 43 T , 43P{T
} 43 ) T , max(T { P } 43 T P{
) T ,T ( max T
21
21
21
2 1
06.1
}}
■ Solution to Eg. 9-8
§P 12 : Path Independence §P 12 : Path Independence [ Eg. 9-8 ]
◇◇
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 §P 12 : Path Independence
[ Eg. 9-8 ]
◇◇
55
(B) What date of project completion can be assured by 90% chance ?
44.76 4084)(1.645)(2. t45.56 4177)(1.645)(2. t
1.645 Z ) error& trial ?t ( ..
.}2.84
40-t{zP}
2.7741-t
{zP
0.90 t}{TPt}{TP?t ; 90% }t {TP
2
1
r r
2r1r
r
950900
900
}
45.12 2
tt
21
§P 12 : Path Independence §P 12 : Path Independence
[ Eg. 9-8 ]
◇◇
56
9043.0
8984.0
)9641.0)(9319.0(
)4641.05.0()4319.05.0(
)803.1)487.1
))
.9664)(0.9357)(0
1.83) Z (P1.52) Z (P
45.2 t
Z (PZ (P2.84
40-45.12 Z (P
2.7741-45.12
Z (P
2 r1r
2r 1r
2r1r
• Assuming path independent in this problem more accurate than one CP.
• But Assumption of path independent may be inaccurate ?!?
[ Eg. 9-8 ] ◇◇
57
1
4
3
2 6
5 7
8 9
A
B
C
D
E
F
G
H
I
(3, 0.11)
(3, 0.13)
(8, 1.0) (7, 1.15)(15, 4.4)
(8, 1.0)
(15, 1.0)
(4, 0.86)(10, 3.1)
Network for PERT / Eg. 9-8
§P 12 : Path Independence §P 12 : Path Independence
[ Eg. 9-8 ]
◇◇
58
1
4
3
2
6
5
7 8A
B
C
D
E
F
G
H
I
P2
P1
5 PATHS
A - B - D - I 23.5 8.36A - C - P1 - D - I 20.8 6.58A - C - E - G - I 26.0* 7.16A - C - E - P2 - H - I 23.0 6.80A - C - F - H - I 23.2 6.80
E(T) Var(T)
﹛
§P 12 : Path Independence §P 12 : Path Independence
[ Eg. 9-8x ]
◇◇
59
• If we assume 5 paths are independent
0.0016
} 6.823.2 - 22 Z{ Pr
} 6.8
23 - 22 Z{ Pr } 7.16
26 - 22 Z{ Pr
} 6.5820.8- 22 Z{ Pr }
8.3623.5 - 22 Z{ Pr
) 22T , 22 T , 22T , 22T , 22T ( Pr ) 22 T ( Pr 54321
.}.26-22
{ZP22) (TP 22) T (P , CP One rr r 3 06680167
0.0668] [0.0016, 22}T {P of value True r
• Assuming path independence can have a very significant effect on the probabilities.
§P 12 : Path Independence §P 12 : Path Independence [ Eg. 9-8x ]
◇◇
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