Transcript 投影片 1 - ndhu.edu.tw

Outline
 PERT/CPM
1
PERT/CPM
Program (or Project) Evaluation
and Review Technique /
Critical Path Method
2
PERT/CPM

Two related network based methods to schedule
activities for a (large) project

PERT: Program (or Project) Evaluation and Review
Technique

CPM: Critical Path Method

huge saving in projects such as the Manhattan Project,
the Apollo Project
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PERT/CPM

Two conventions

nodes for events and arcs for activities (our convention)
i

l
j
arcs for events and nodes for activities
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PERT/CPM - Example

A marine crop was going to launch an amphibious
operation. To do so, new amphibious landing
crafts were requested. Upon their arrivals, some
testing and tuning were necessary. At the same
time, the crop also needed to train new recruits for
landing operations. These new recruits were
training on the old landing crafts.

Assume that other than the precedent constraint,
the activities can be carried out in any order, with
or without other activities at the same time.
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PERT/CPM - Example



define activities
 A: the lead time to get new landing crafts
 B: testing and tuning of the new landing crafts
 C: training of the new recruits
define events
 1: the planning of the landing operation
 2: the arrival of landing crafts
 3: the landing operation was ready
the precedent relationship of the activities can be
represented by a network.
2
A
B
1
C
3
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PERT/CPM - Example A
1

2
B
C
3
Suppose that the duration of A = 10 weeks, of B = 5 weeks,
and of C = 20 weeks. How can the activities be carried out?
There are many ways.
Which way gives the shortest project completion time?
A
B
A
B
C
C

A
B
A
C
B
C
A
B
C
A
A
C
B
B
C
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PERT/CPM - Example

There are slack in activities A and B so that their
start times can be postponed without delaying the
completion time of the whole project.
A
B
A
C
A
C
B
C
B
A
B
C
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PERT/CPM - Example

Questions

What is an earliest start time of an activity?

What is the latest start time of an activity?
A
B
C
Too much delay in activity B so that the
completion time of the project is delayed.
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PERT/CPM - Example
 The
shortest completion time can be found
from the critical path.
 Let
ESi be the earliest start time of event i,
which is shown within the triangle beside
node i. For example, ES1 = 0.
2
A= 8
0
1
B=7
C = 20
3
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PERT/CPM - Example
2
A= 8
0
1
B=7
C = 20
8
3
ES2 = ES1 + 8 = 8
2
A= 8
0
1
B=7
C = 20
20
3
The earliest time to launch the
landing operation = 20 weeks
ES3
= max {ES1 + 20,
ES2 + 7}
= 20
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PERT/CPM - Example
 In
general, ESj = max ESi1 + li1 j , ..., ESin + lin j 
i1
.
.
.
in
li1 j
j
lin j
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PERT/CPM - Example
 Let
LSj be the latest start time of event j.
 With
the shortest project completion time
found, how to determine the latest start time
of each event?
 Box
the latest start time of an event beside it.
2
A= 8
1
B=7
C = 20
20
3
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PERT/CPM - Example
LS1
= min {LS2  8, LS3  20}
=0
0
1
13
LS2 = LS3  7 = 13
2
A= 8
B=7
C = 20
20
3
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PERT/CPM - Example
 In
general, LSi = min LS j1  lij1 , ..., LS jn  lijn 
lij1
i
lijn
j1
.
.
.
jn
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PERT/CPM - Example
 The
computation of the ES and LS can be
i
put in the same graph.
j
.
1
li1 j
i.n
lin j
.
Forward Pass
ES
ESj = max
ESi
1
LS
8
LSi = min
LSi
1
13
0
2
A= 8
0
1
20
C = 20
lij1
20
3
 lij1 , ..., LSin  lijn
j1
.
j.n
.
Backward Pass
i
B=7
+ li1 j , ..., ESin + lin j
lijn
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

PERT/CPM - Critical Path

Critical path: The path in which for all events ESi = LSi

path (1, 3) is critical

path (1, 2)-(2, 3) is not critical

Activities in a critical path do not have slack. Any delay of such
activities will delay the completion time of the project.
8
13
20
2
A= 8
0
0
1
B=7
C = 20
20
3
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PERT/CPM - Slack
ESi
LSi
 for
i
lij
ESj
j
LSj
activity (i, j):
slack, TSij = LSj ESi lij
 free slack, FSij = ESj ESi lij
 total
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PERT/CPM - Slack
TSij = LSj ESi lij
8
FSij = ESj ESi lij
13
20
2
A= 8
0
0
1
B=7
C = 20
20
3

TS12 = 5; FS12 = 0; TS23 = 5; FS23 = 5; TS13 = 0; FS13 = 0

Activity (1, 2) is not on the critical path, but its free slack is
equal to zero. Thus, any delay of the event delay the start
of subsequent activities by the same amount.
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PERT/CPM – Another Example
8
8
15
2
A= 8
0
0
1
B=7
C = 13
15
3

TS12 = 0; FS12 = 0; TS23 = 0; FS23 = 0; TS13 = 2; FS13 = 2

With FS13 = 2, it is possible to schedule the start time of
event (1, 3) within the time interval (0, 2) without delaying
the completion time of the project.
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PERT/CPM – Yet Another Example
30
2
30
A = 30
0
0
1
3
C=2
B=5
2
32
35
D=3
7
E=7
7
4
G =15
15
F = 25
I=2
10
5
H = 10
23
35
75
J =10
75
6
25
33
L= 30
8
9
45
K = 15
45
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PERT/CPM – Yet Another Example
 TSA
= 0, FSA = 0
 TSB
= 0, FSB = 0
 TSC
= 30, FSC = 0
 TSD
= 30, FSD = 30
 TSE
= 3, FSE = 0
 TSF
= 3, FSF = 3
 TSG
= 8, FSG = 0
 TSH
= 8, FSH = 0
 TSI
= 8, FSI = 8
 TSJ
= 0, FSJ = 0
 TSK
= 30, FSK = 30
 TSL
= 0, FSL = 0
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PERT/CPM – Yet Another Example
30
2
30
A = 30
0
0
1
3
C=2
B=5
2
32
35
D=3
7
E=7
7
4
G =15
15
F = 25
I=2
10
5
H = 10
23
35
75
J =10
75
6
25
33
L= 30
8
9
45
K = 15
45
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Two Comments on PERT/CPM

Sometimes dummy arcs of zero duration are
added to model the logical precedent relationship
among activities. （不少網路以沒有長度的虛擬
arcs表達按照邏輯activities間的關係）

In applying PERT/CPM in real life, the most
difficult part is the modeling part, i.e., to identify
events, to identify activities, and to determine the
precedent relationship. Calculation is simple.
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