This document discusses the Critical Path Method (CPM) for calculating the schedule of activities in a network diagram. It defines key CPM terms like earliest start/finish times, latest start/finish times, critical path. It then performs forward and backward pass calculations on a sample network to determine the schedule, highlighting the critical path with a duration of 12 days. Finally, it asks students to explain CPM and calculations.

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Unit No.4.
NETWORK ANALYSIS
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean, EDP & Associate Professor MBA
1
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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302-DECISION SCIENCE
Unit No.4. Network Analysis
4.9 CPM Network
Calculations
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean EDP & Associate Professor MBA
2
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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NETWORK ANALYSIS
❑ At the End of the Session Student will be able to
understand-
A. CPM Network Calculations

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CPM NETWORK CALCULATIONS
❑ After drawing the Network and getting the time
estimates for all the activities, we can analyze the
network in order to schedule the performance of the
activities and find out the flexibility in scheduling the
activities.
❑ This we can do by CPM where only perfect time
estimates will be considered or
❑ We can do this by PERT whereby three different time
estimates along with Time and Cost trade off is
considered.
❑ To do so, let’s first study the various notations in it-

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CPM NETWORK CALCULATIONS
Aij - where A is the Activity which starts at i (Tail Node)
and ends at j (Head Node)
Ei - Earliest Occurrence time of event i
Li - Latest Occurrence time of event i
tij – Time duration of activity A(i,j)
ESij - Earliest Starting time for activity A(i,j)
EFij - Earliest Finishing time for activity A(i,j)
LSij - Latest Starting time for activity A(i,j)
LFij - Latest Finishing time for activity A(i,j)

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CPM NETWORK CALCULATIONS
1
3
2 64 7
5
A (2)
B (3) E (5)
I (2)
H
F (3)
G (2)
C (4)
D(1)

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FORWARD PASS CALCULATIONS
In FPC we begin with initial node with beginning time as 0
so that,
E1 = 0
For any Activity,
ESij = Ei hence
EFij = ESij + tij OR EFij = Ei + tij
And
Ej = Maximum of EFij
Thus, Ej = Maximum of [Ei + tij]

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BACKWARD PASS CALCULATIONS
In BPC we begin with final node by assuming Latest
Occurrence time for terminal event (say N) Ln as equal to its
Earliest Occurrence time En i.e.
Ln = En
For any Activity,
LFij = Lj hence
LSij = LFij - tij OR LSij = Lj - tij
And
Li = Minimum of LSij
Thus, Li = Minimum of [Lj - tij]

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CPM NETWORK CALCULATIONS
Do the FPC and BPC for the Network as given below-
1
3
2 64 7
5
A (2)
B (3) E (5)
I (2)
H
F (3)
G (2)
C (4)
D(1)

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FORWARD PASS CALCULATIONS
In the beginning, E1 = 0
For any Activity,
Ej = Maximum of [Ei + tij] Thus,
E2 = [E1 + t12] = 0 + 2 = 2
E3 = [E2 + t23] = 2 + 3 = 5
E4 = Maximum of [E3 + t34] & [E2 + t24]
E4 = Maximum of [5 + 1] & [2 + 4] = Max of [6, 6] = 6
E5 = [E4 + t45] = 6 + 2 = 8
E6 = Max of [E3 + t36], [E4 + t46] & [E5 + t56] = Max of [10, 9, 8]
E6 = 10
E7 = E6 + t67 = 10 + 2 = 12

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BACKWARD PASS CALCULATIONS
In the beginning, Ln = En i. e. L7 = E7 which gives L7 = 12
For any Activity,
Li = Minimum of [Lj - tij] Thus,
L6 = [L7 – t67] = 12 - 2 = 10
L5 = [L6 – t56] = 10 - 0 = 10
L4 = Minimum of [L6 - t46] & [L5 - t45]
L4 = Minimum of [10 - 3] & [10 - 2] = Min of [7, 8] = 7
L3 = Minimum of [L6 – t36] & [L4 - t34]
L3 = Minimum of [10 - 5] & [7 - 1] = Min of [5, 6] = 5
L2 = Minimum of [L3 – t23] & [L4 - t24]
L3 = Minimum of [5 - 3] & [7 - 4] = Min of [2, 3] = 2
L1 = [L2 – t12] = 2 - 2 = 0

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NETWORK CALCULATIONS
Activity Duration
tij
EST
ESij = Ei
EFT
EFij = ESij + tij
LST
LSij = LFij - tij
LFT
LFij = Lj
A: 1-2 2 0 = 0 + 2 = 2 = 2 - 2 = 0 2
B: 2-3 3 2 = 2 + 3 = 5 = 5 - 3 = 2 5
C: 2-4 4 2 = 2 + 4 = 6 = 7 - 4 = 3 7
D: 3-4 1 5 = 5 + 1 = 6 = 7 - 1 = 6 7
E: 3-6 5 5 = 5 + 5 = 10 = 10 - 5 = 5 10
F: 4-6 3 6 9 7 10
G: 4-5 2 6 8 8 10
H: 5-6 0 8 8 10 10
I: 6-7 2 10 12 10 12

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CRITICAL PATH METHOD
1
0 0
A (2)
B (3) E (5)
I (2)
H
F (3)
G (2)
C (4)
D(1)
2
2 2
3
5 5
4
6 7
5
8 10
6
10 10
7
12 12

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CRITICAL PATH METHOD
Path Path Duration
tij
A-B-E-I 2 + 3 + 5 + 2 = 12 Days (Critical Path)
A-B-D-F-I 2 + 3 + 1 + 3 +2 = 11 Days
A-B-D-G-H-I 2 + 3 + 1 + 2 + 0 +2 = 10 Days
A-C-F-I 2 + 4 + 3 + 2 = 11 Days
A-C-G-H-I 2 + 4 + 2 + 0 + 2 = 10 Days

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EXERCISE
❑ State and Explain Critical Path Method with Network
Calculations.

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For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in

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Thank You