Study Guide
Network Model PERT CPM
-
University:
Kentucky State University -
Course:
COS 170 | Networking Fundamentals Academic year:
2024
-
Views:
456
Pages:
32
Author:
tenhentsjws
Network Model
PERT/CPM CPM
A network used to represent a project is called a project
network.
A project network consists of a number of nodes (events)
and a number of arcs (activities)
CPM (Critical Path Method)
Developed to coordinate maintenance projects in the chemical
industry
A complex undertaking, but individual tasks are routine (tasks’
duration = deterministic)
Time/cost/quantity
2 Two types of activity:
Real Activity: require time/cost/quantity to be
achieved.
2. Dummy activities: make network realistic
1.
3 Project Networks
A network used to represent a project is called a project
network.
A project network consists of a number of nodes and a
number of arcs.
Activity-on-arc (AOA): each activity is represented by an arc. A
node is used to separate an activity from its predecessors. The
sequencing of the arcs shows the precedence relationships. Example 1- A simple network
Consider the list of four activities for making a simple product:
Activity
Description
Immediate
predecessors
A
Buy Plastic Body
-
B
Design Component
-
C
Make Component
B
D
Assemble product
A,C
Immediate predecessors for a particular activity are the activities
that, when completed, enable the start of the activity in question. Sequence of activities
Can start work on activities A and B anytime, since
neither of these activities depends upon the completion
of prior activities.
Activity C cannot be started until activity B has been
completed
Activity D cannot be started until both activities A and
C have been completed.
The graphical representation (next slide) is referred to
as CPM network Network of Four Activities
Arcs indicate project activities
A
1
3
D
C
B
2
Nodes correspond to the beginning
and ending of activities
4 Example 2
Develop the network for a project with following activities and
immediate predecessors:
Activity
A
Immediate
predecessors
-
B
-
C
B
D
A, C
E
C
F
C
G
D,E,F
Try to do for the first five (A,B,C,D,E) activities Network of first five activities
A
1
D
3
4
E
B
C
2
5
We need to introduce
a dummy activity Network of Seven Activities
A
1
3
dummy
B
C
2
5
D
4
G
7
E
F
6
•Note how the network correctly identifies D, E, and F as the
immediate predecessors for activity G.
•Dummy activities is used to identify precedence relationships
correctly and to eliminate possible confusion of two or more
activities having the same starting and ending nodes
•Dummy activities have no resources (time, labor, machinery, etc) –
purpose is to PRESERVE LOGIC of the network EXAMPLES OF THE USE OF DUMMYACTIVITY
Network concurrent activities
a
a
2
1
2
Dummy
1
b
3
b
WRONG!!!
RIGHT
Activity c not
required for e
WRONG !
a
a
e
d
1
b
1
e
b
d
c
2
c
WRONG
!!!
RIGHT
RIGHT Scheduling with activity time
Activity
A
B
C
D
E
F
G
H
I
Immediate
predecessors
A
A
A
E
D,F
B,C
G,H
Completion
Time (week)
5
6
4
3
1
4
14
12
2
Total ……
51
This information indicates that the total time required to complete
activities is 51 weeks. However, we can see from the network that several
of the activities can be conducted simultaneously (A and B, for example). Earliest start & earliest finish time
We are interested in the longest path through the network, i.e.,
the critical path.
Starting at the network’s origin (node 1) and using a starting
time of 0, we compute an earliest start (ES) and earliest finish
(EF) time for each activity in the network.
Earliest Start (ES) = the earliest finish of the immediately
preceding activity
The expression EF = ES + t can be used to find the earliest finish
time for a given activity.
For example, for activity A, ES = 0 and t = 5; thus the earliest
finish time for activity A is
EF = 0 + 5 = 5 The Critical Path
A path through a network is one of the routes following the
arrows (arcs) from the start node to the finish node.
The length of a path is the sum of the (estimated) durations
of the activities on the path.
The (estimated) project duration equals the length of the
longest path through the project network.
This longest path is called the critical path. (If more than
one path tie for the longest, they all are critical paths.)
16-14 Arc with ES & EF time
EF = earliest finish time
ES = earliest start time
Activity
2
1
t = expected activity
time Network with ES & EF time
D[5,8]
3
2
5
7
4
1
6
3
Earliest start time rule:
The earliest start time for an activity leaving a particular node is equal to
the largest of the earliest finish times for all activities entering the node. Activity, duration, ES, EF, LS, LF
EF = earliest finish time
ES = earliest start time
Activity
3
2
LS = latest start time
LF = latest finish time Latest start & latest finish time
To find the critical path we need a backward pass calculation.
Starting at the completion point (node 7) and using a latest
finish time (LF) of 26 for activity I, we trace back through the
network computing a latest start (LS) and latest finish time
for each activity
The expression LS = LF – t can be used to calculate latest start
time for each activity. For example, for activity I, LF = 26 and t
= 2, thus the latest start time for activity I is
LS = 26 – 2 = 24 Network with LS & LF time
D[5,8]
3[7,10]
2
5
7
4
1
6
3
Critical Path: A,E,F,G,I
Latest finish time rule:
with 26
The latest finish time for an activity entering a particular node is equal to
the smallest of the latest start times for all activities leaving the node. Slack or Free Time or Float
Slack is the length of time an activity can be delayed without affecting the
completion date for the entire project.
For example, slack for C = 3 weeks, i.e Activity C can be delayed up to 3
weeks
(start anywhere between weeks 5 and 8).
2
ES
5
LS
8
EF
9
EF
12
LF-EF = 12 –9 =3
LS-ES = 8 – 5 = 3
LF-ES-t = 12-5-4 = 3
3 Activity schedule for our example
All activities on the critical path have zero slack
Activity
A
B
C
D
E
F
G
H
I
Earliest
Latest
Floating
start (ES) start (LS) Slack
(LS-ES)
0
0
5
5
5
6
10
9
24
0
6
8
7
5
6
10
12
24
0
6
3
2
0
0
0
3
0
Earliest
finish
(EF)
5
6
9
8
6
10
24
21
26
Latest
finish
(LF)
5
12
12
10
6
10
24
24
26
Floating
Slack
(LF-EF)
0
6
3
2
0
0
0
3
0
Critical
path
Yes
Yes
Yes
Yes
Yes IMPORTANT QUESTIONS
What is the total time to complete the project?
26 weeks if the individual activities are completed on schedule.
What are the scheduled start and completion times for each activity?
ES, EF, LS, LF are given for each activity.
What activities are critical and must be completed as scheduled in order to
keep the project on time?
Critical path activities: A, E, F, G, and I.
How long can non-critical activities be delayed before they cause a delay in
the project’s completion time
Slack time available for all activities are given. Importance of Float (Slack) and Critical Path
23
1.
Slack or Float shows how much allowance each activity has, i.e how
long it can be delayed without affecting completion date of project
2.
Critical path is a sequence of activities from start to finish with zero
slack. Critical activities are activities on the critical path.
3.
Critical path identifies the minimum time to complete project
4.
If any activity on the critical path is shortened or extended, project
time will be shortened or extended accordingly
Chapter 8 PERT (Program Evaluation and Review Technique
Many tasks pushed the boundaries of science & engineering (tasks’ duration =
probabilistic)
So far, times can be estimated with relative certainty, confidence
For many situations this is not possible, e.g Research, development, new products and
projects etc.
Use 3 time estimates
m= most likely time estimate, mode.
a = optimistic time estimate,
b = pessimistic time estimate, and
Expected Value (TE) = (a + 4m + b) /6
Variance
(V) = ( ( b – a) / 6 ) 2
Std Deviation (δ) = SQRT (V)
Expected critical path (the longest path)
Z = (Agreed Time –Expected critical path)/ Std Deviation
(return to the table)
Risk = 0.5±Z (+ if z positive, - if z is negative)
-0.5
Z
+0.5
Z Precedences And Project Activity Times
Immediate
Optimistic Most probable Pessimistic
Activity Predecessor(IP) Time (a)
Time (m)
Time (b)
EXP
Var S.Dev
TE
V
a
-
10
22
22
20
4
2
b
-
20
20
20
20
0
0
c
-
4
10
16
10
4
2
d
a
2
14
32
15
25
5
e
b,c
8
8
20
10
4
2
f
b,c
8
14
20
14
4
2
g
b,c
4
4
4
4
0
0
h
c
2
12
16
11
5.4
2.32
I
g,h
6
16
38
18
28.4 5.33
j
d,e
2
8
14
8
4
2 The complete network
2
1
B[0,20] 3
20[1,21]
4
D[20,35]
15[20,35]
6
F[20,34]
14[29,43]
H[10,21]
11[14,25]
7
5 Critical Path Analysis (PERT)
Activity
LS
ES
Slacks
Critical ?
a
0
0
0
Yes
b
1
0
1
c
4
0
4
d
20
20
0
e
25
20
5
f
29
20
9
g
21
20
1
h
14
10
4
i
25
24
1
j
35
35
0
Yes
Yes The complete network
2
1
B[0,20] 3
20[1,21]
4
D[20,35]
15[20,35]
6
F[20,34]
14[29,43]
H[10,21]
11[14,25]
7
5 Assume, PM promised to complete the project in the fifty days.
What are the chances of meeting that deadline?
Calculate Z, where
Z = (D-S) / V
Example,
D = 50; S (CRTICAL PATH) = 20+15+8 =43;
Z = (50 – 43) / 5.745
= 1.22 standard deviations.
V = (4+25+4) =33
The probability value of Z = 1.22, value from the table is 0.3888
Risk= 0.5+0.388= 0.888 88%
Z
1.22 Assume, PM promised to complete the project in the forty days.
What are the chances of meeting that deadline?
Calculate Z, where
Z = (D-S) / V
Example,
D = 40; S (CRTICAL PATH) = 20+15+8 =43;
Z = (40– 43) / 5.745
= -0.522 standard deviations.
V = (4+25+4) =33
The probability value of Z = -0.522, value from the table is 0.1985
Risk= 0.5-0.1985= 0.3015 30%
Z
= -0.522 Calculate the probability that the project can be
45,000 $ whereas the weekly cost is 1000$
Expected Cost: 43*1000$=43000$
Z=(45-43)/5.745 = 0.3481 from the table
0.1368
Risk=0.5+0.1368=0.6368 %64 Calculate the probability that the project can be
36,000 $ whereas the weekly cost is 1000$
Expected Cost: 43*1000$=43000$
Z=(36-43)/5.745 = -1.22 from the table: 0.3888
Risk=0.5-0.3888=0.12 12%
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