OPERATIONS
RESEARCH
CPM / PERT
Pert Case Study
Duration (Weeks)
S.No. Activity Predecessor (s)
Optimistic Most Pessimistic
time likely time time
1 A - 5 6 7
2 B - 1 3 5
3 C - 1 4 7
4 D A 1 2 3
5 E B 1 2 9
6 F C 1 5 9
7 G C 2 2 8
8 H E,F 4 4 10
9 I D 2 5 8
10 J H,G 2 2 8
Q.1 Construct the project network.
Q.2 Find the expected duration and variance of each activity
Q.3 Find the critical path and expected project completion time
Q.4 What is the probability of completing the project on or before 22 weeks
NETWORK DIAGRAM
2
A
B
1 3
C
4
NETWORK DIAGRAM
D
2 5
A
B E
1 3 6
NETWORK DIAGRAM
D
2 5
A
B E
1 3 6
F
C H
4
G 7
NETWORK DIAGRAM
D
2 5 I
A
B E
1 6 8
3
F
C H
4 G J
7
te – Expected Time
te = to + 4tm + tp
6
Variance
2
σ= tp - to
6
EXPECTED DURATION
Duration (Weeks)
Mean
S.No. Activity Predecessor (s)
Duration
Optimistic Most Pessimistic
time likely time time
1 A - 5 6 7 6
2 B - 1 3 5 3
3 C - 1 4 7 4
4 D A 1 2 3 2
5 E B 1 2 9 3
6 F C 1 5 9 5
7 G C 2 2 8 3
8 H E,F 4 4 10 5
9 I D 2 5 8 5
10 J H,G 2 2 8 3
VARIANCE
Duration (Weeks)
Mean
s.No. Activity Predecessor (s) Variance
Duration
Optimistic Most Pessimistic
time likely time time
1 A - 5 6 7 6 0.11
2 B - 1 3 5 3 0.44
3 C - 1 4 7 4 1.00
4 D A 1 2 3 2 0.11
5 E B 1 2 9 3 1.78
6 F C 1 5 9 5 1.78
7 G C 2 2 8 3 1.00
8 H E,F 4 4 10 5 1.00
9 I D 2 5 8 5 1.00
10 J H,G 2 2 8 3 1.00
CPM – CRITICAL PATH METHOD
D
2 5 I
A
B E
1 6 8
3
F
C H
4 G J
Duration – We will consider the Mean Duration out of three
CPM – CRITICAL PATH METHOD
D(2)
2 5 I(5)
A(6)
B(3) E(3)
1 6 8
3
F(5)
C(4) H(5)
4 J(3)
G(3) 7
CPM – CRITICAL PATH METHOD
D(2)
2 5 I(5)
A(6)
B(3) E(3)
1 6 8
3
F(5)
C(4) H(5)
4 J(3)
G(3) 7
To find critical path and completion time we will have to
find the
2- Latest Completion Time – Backward Pass
1- Earliest Start Time – Forward Pass
CPM – CRITICAL PATH METHOD
6 8
D(2)
2 5 I(5)
0 A(6) 17
0
3 E(3)
B(3) 9 8
1 3 6
F(5)
C(4) H(5)
4 J(3)
G(3) 7
4 14
For forward pass we will consider the maximum value
CPM – LATEST COMPLETION TIME
10 12
6 8
D(2)
0 2 5 I(5) 17
0 A(6) 17
0 6 9
3 E(3)
B(3) 9 8
1 3 6
F(5)
C(4) H(5)
4 J(3)
4 G(3) 7 14
4 14
For Backward pass we will consider the minimum value
CPM – THE LONGEST PATH IN THE NETWORK
10 12
6 8
D(2)
0 2 5 I(5) 17
0 A(6) 17
0 6 9
3 E(3)
B(3) 9 8
1 3 6
F(5)
C(4) H(5)
4 J(3)
4 G(3) 7 14
4 14
Three conditions to be met
1- Earliest starting time of each starting node should be equal to latest
completion time each starting
2- Earliest start time of ending node must be equal to latest completion of
ending node
3- Earliest start time completion node – earliest start time of starting node =
Duration = Latest completion of completing node – latest completion of starting
node
EXPECTED PROJECT COMPELITION TIME
10 12
6 8
D(2)
0 2 5 I(5) 17
0 A(6) 17
0 6 9
3 E(3)
B(3) 9 8
1 3 6
F(5)
C(4) H(5)
4 J(3)
4 G(3) 7 14
4 14
1 4 6 7 8
4 + 5+ 5+ 3 = 17 weeks
PROBABILITY OF COMPLETING THE
PROJECT ON OR BEFORE 22 WEEKS
Mean
S.No. Activity Variance
Duration
3 C 4 1
6 F 5 1.78
8 H 5 1
10 J 3 1
17 4.78
P(x <= 22) = p x – mu <= 22-17 = 2.28
σ 2.19
Z value = 2.28 see z table normal distribution
P [z<= 2.28 ] = .09887 = 98.87%
