Network analysis.pptxbbbbbbbbbbbbbbbbbbbbbbb

2024 Sourabh Saini 1
• Network : We represent activities of a project through networks. It
Network: We represent activities of a project through networks. It
arrange the activities in logical sequence, following the precedence –
succedence relationship.
• es care of precedence relationships.

CPM
•1.The time durations are deterministic (MBA degree).
•2.The duration of the project is fixed. And for a fixed duration it gives the most economical
schedule.
•3.Looping and probabilistic events are not allowed in the network.
PERT
•1.The time durations are probabilistic (Ph. D. degree, DRDO,ISRO, CSIR Labs)
•2.There is expected duration of the project.
•3.Simulation can be used to PERT network.

Network terminology:
• Activity: An activity is an identifiable job that has a beginning and an end. An activity
consumes time, manpower, and material resources.
• Example- excavation of foundation, placing of foundation concrete, construction of walls,
construction of roofing, wiring and electrification work.
• And activity represented by (AOA)
Activity Symbol of the activity Activity time Symbolic representation
Excavation of
foundation
Placing of foundation
concrete
A
B
15 Days
6 Days
Construction of walls C 21 Days
A
A
15
B
6
C
21

• Event: An event is the beginning and end of an activity. It does not consume
time, manpower, and materials. It represents a specific point in time. It is
represented by a circle.
• Event is symbolically represented by numbers or alphabets written inside the
circle.
• As seen, event 1 stands for the starting of foundation excavation, and Event 2
stands for the completion of foundation excavation. Thus, each activity has
two events.
Start excavation
for the foundation
Complete excavation
for the foundation
A
15
1 2

AOA Vs. AON
The same mini-project is shown with activities on arc…
C
E
D
B F
E
C
D
B F
…and activities on node.

i j
5
5
1. Each activity must be represented by one and only one arrow, an activity (i-j ) “ i” is the staring node
and “j” is terminal node.
2. No two activities should not have the same initial and same terminal nodes.
2
1
6
5 A
B
2
3
1
3
2
1
5
6
5
6
0 Dummy
0 Dummy
Dummy activity: Imaginary
activity in a network. It does not
consume time and resource
(manpower and material resources).

2 3
Lay foundation
Order material
(a) Incorrect precedence
relationship
(b) Correct precedence
relationship
3
4
2
Dummy
Lay
foundation
Order material
1
2 0
No two activities should have the same initial and same terminal nodes.

Job Immediate predecessors
A -
B -
C A
D A,B
E C,D
Ex-1. Develop AOA Network

Job Immediate predecessors
A -
B -
C A
D A,B
E C,D
1
3
4
6
5
C
A
B D
E
Ex1-Types of dummy activities (Activity on arrow).
3-4 is a logical dummy activity

3. The arrow heads should not form a close loop/ consistency in project network
A
C B

Hints for drawing networks:
Example 2
-‘A’ precedes ‘B’ and ‘C’ or ‘B’ and ‘C’ follow ‘A’ or ‘B ‘and ‘C’ depend on ‘A’
- ‘D’ follows ‘B’
- E follows C
- D and E are terminal activities.

Example 2.
-‘A’ precedes ‘B’ and ‘C’ or ‘B’ and ‘C’ follow ‘A’ or ‘B ‘and ‘C’ depend on ‘A’
- ‘D’ follows ‘B’
- E follows C
- D and E are terminal activities.
A
B
C E
D

Example 3.
-‘A’ and ‘B’ start immediately
- ‘C’ follows both ‘A’ and ‘B’
- D follows A
- E follows D
- The project is complete when C and E are done

Example 3.
-‘A’ and ‘B’ start immediately
- ‘C’ follows both ‘A’ and ‘B’
- D follows A
- E follows D
- The project is complete when C and E are done
A
B
E
D
C

Example 4.
-‘A’ starts immediately
- ‘B’ , ‘C’ and ‘D’ follow ‘A’
- E follows B, C, and D

A
E
C
D
B
D1
D2
D3
Example 4.
-‘A’ starts immediately
- ‘B’ , ‘C’ and ‘D’ follow ‘A’
- E follows B, C, and D

4. The arrows representing activities should be straight not curved.
- A, B and C start immediately
- D follows A
- E follows B and D
- C and E are terminal activities
D
A
B
C
E

4. The arrows representing activities should be straight not curved.
- A, B and C start immediately
- D follows A
- E follows B and D
- C and E are terminal activities
E
B
C
A D
D
A
B
C
E

5. Avoid crossover of activities whenever possible
.
D
B
C
A
E
E
F
A
B
D
F
C

Example 5.- Listed in the table are the activities and sequencing necessary for the completion of a recruitment procedure for
management trainees (MT) in an organization.

Ex-6. Find Critical Path of the Project

Critical Path
• Out of the seven paths, C -> J -> M -> L has the longest duration and this
is the critical path. The project duration is the duration of the critical path,
which is equal to 50 days.
Path Duration
1. A -> D -> G -> K 10+12+20+6 = 48 days
2. A -> D -> H -> L 10+12+3+18 = 43 days
3. A -> E -> I -> L 10+4+9+18 = 41 days
4. B -> I -> L 9+9+18 = 36 days
5. C -> J -> N 12+13+9 = 34 days
6. C -> J -> M -> L 12+13+7+18 = 50 days
7. C -> F -> I -> L 12+8+9+18 = 47 days

Forward Pass computation
• This is a method of computation of starting time of events. This computation
begins from the initial events and moves towards the final events.
• This series of computation is done to arrive at the at the earliest start time all the
events.
• Earliest expected time / Earliest start time (Te) : the time when an event can be
expected to start as early as possible. It is computed by adding the te’s of the
activity paths leading to that event.

Ex-7. Find Critical path
A
D
1
7
7
5
4 11
8
5
11
8
12
2
6 8
7
4
2
5
3

Solution:
Event-1: This is the start event or initial event. Hence, its Earliest Start Time is Zero. ie. TE (of
event-1) = 0
Event-2: Event 2 represents the end of activity 'A' and the beginning of activity 'C' no difference between these two
moments, i.e., activity 'C' can start at the y moment when activity 'A' ends.
Hence, the Earliest Start Time of Event 2 is given by
TE (of event-2 = TE (of event-1) + duration of activity-A
=0+5
= 5
Event-3: Activities D, E & F can start at the very same moment activity 'B' comes to Therefore, the Earliest Start Time
TE of Event 3 is given by
TE (of event-3) = TE (of event-1) + duration of activity-B
=0+4
= 4
-Forward computation method.

• Event-4: At event-4, two arrows terminate, and one arrow emerges. This means the completion of two activities
(activities C & D) and the start of (activity G).
Let us calculate the earliest completion time of the two activities C and D.
Earliest completion time of activity-C = TE (of event-2) + duration of activity-C
=5+7
= 12
Earliest completion time of activity D
= TE (of event-3) + duration of activity-D
=4+7
= 11
• Activity G can start only after both activities C & D are completed. While Activity D can be completed 11 days
after the start of Event 1, activity C can be completed only 12 days after the start of Event 1. Hence, activity G can
start only after 12 days. Thus, the Earliest Start Time (TE) of event 4 is equal to the maximum time of the two
activities, C & D.
Therefore, TE (of event-4) = 12

• Event-5: There is only one activity (activity-F) that enters the event-5. Hence,
TE (of event-5) = TE (of event-3) + duration of activity-F, =4+5
= 9
• Event-6: There are two activities (activities E & I) entering node-6 and one activity (activity-H) leaving node-6
(events are also sometimes referred to as 'nodes’).
• Activity-H can start only after both activities E & I are completed. Hence the Earliest Start Time (TE) of event-6 is
given by the maximum of the following two time estimates a & b.
(a) TE (of event-3) + duration of activity-E, = 4+8
= 12
(b) TE (of event-5) + duration of activity-l, =9+2
= 11
• Therefore activity-H can start only after 12 days from the commencement of the project. Hence, the Earliest start
time (TE) of event 6 is 12 days. i.e., TE (of event-6) = 12
• Event-7: Working on the similar lines as explained for event-6, TE (of event-7) is the maximum of the two time
estimates a & b.

(a) TE (of event-6) + duration activity-H
i.e., = 12 + 11
= 23
(b) TE (of event-5) + duration of activity-J
= 9+ 12
= 21
Therefore TE (of event-7) = 23
• Event-8: This is the final event. Since two activities are entering this node, TE for this event is the
maximum of the two-time estimates, a & b.
(a) TE (of event-4) + duration activity-G
= 12 +8 = 20
(b) TE (of event-7) + duration activity-K
= 23+11 = 34

A
D
1
7
7
5
4 11
8
5
11
8
12
2
6 8
7
4
2
5
3
TE= 0
TE= 5 TE= 12
TE= 34
TE= 23
TE= 4
TE= 9
TE= 12

Backward pass computation
• This is a method of computation of finishing time of events.
• This computation begins from the final events and moves towards the initial
event.
• Latest finish time ( : is the latest time by which an event should be completed so
that the project completion should not delayed.

• Event-7: There is only one activity emanating from event-7 (activity-K). Activity-K takes 11 days
for completion. Since the Latest Finish Time (TL) for event-8 is 34, event-7 can occur as late as on
the 23rd day (34-11). If the Latest Finish Time of event-7 is delayed beyond 23, the project
completion time will also extend beyond 34.
• Hence TL (of event-7) = T (for event-8) - duration of activity-K
= 34 -11 = 23.
• Event-6: Again, there is only one activity emanating from event-6 (Activity-H), which takes 11
days for completion. As per the logic applied to event-7, the Latest Finish Time of event-6 is given
by,
TL(of event-6) = TL(of event-7) - duration of activity-H
= 23 -11 = 12
• Event-5: Two activities are emanating from event-5. Hence there will be time estimates for finish
times out of which the minimum of the two time estimates will represent the Latest Finish Time.

• Activity-J takes 12 days for completion. Since the Latest Finish Time of event-7 is 23, event-5 can start as
late as on the 11th day (23-12).
• Activity-I takes 2 days for completion. Since the Latest Finish Time of event-6 is 12, event- 5 can start as
late as on the 10th day (12-2).
• The Latest Finish Time of event-5 is given by the minimum of the two time estimates, viz., 10.
T (of event-5) = 10.
₁
• Event-4: There is only one activity emanating from event-4. Therefore the Latest Finish Time of event-4 is
given by
TL (of event-4) = TL (of event-8) - duration of activity-G
= 34-8 = 26
• Event-3: There are three activities emanating from event-3 (activities-D, E & F). Hence three time
estimates of finish times are available out of which the minimum time represents the Latest Finish Time of
event-3.
TL (of event-3) is the minimum of
(a) TL(of event-4) duration of activity-D =26-7 = 19
(b) TL (of event-6) duration of activity-E =12-8 =4
(c) TL (of event-5) duration of activity-F = 10-5 =5
Therefore, TL (of event-3) = 4 (the minimum of the three time estimates).

• Event-2: There is only one activity emanating from event-2.
Therefore,
TL (of event-2) = TL (of event-4) - duration of activity-C
=26-7 = 19
• Event-1: There are two activities emanating from event-1. Hence TL of event-1 is
the minimum of the following two time estimates
(a) TL (of event-2) - duration of activity- A
= 19-5 = 14
(b) TL (of event-3) - duration of activity-B
= 4-4 = 0
Therefore TL (of event-1) = 0 If the backward pass computations are done
correctly, the Latest Finish Time of the start event (event-1) will be equal to its
Earliest Start Time.

A
D
1
7
7
5
4 11
8
5
11
8
12
2
6 8
7
4
2
5
3
TE= 0
TL= 0
TE= 5
TL= 0
TE= 12
TL= 26
TE= 34
TL= 34
TE= 23
TL= 23
TE= 4
TL= 4
TE= 9
TL= 10
TE= 12
TL= 12

• Slack time and critical path: We have calculated the Earliest Start Time (TE) of all events by doing forward
pass computations. We have also calculated the Latest Finish Time (TL) of all events by doing backward pass
computations. Thus TE & TL of all the events of the network is known. Slack Time (or Slack) of an event is
the difference between the Latest Finish Time (TL) and the Earliest Start Time (TE) of that event.
• The path connecting events with zero slack is the critical path.
• The following table reproduces the TL, TE & Slack of all events of the network that we have considered in
our illustration. (Refer Fig. 8.38).
• We find that events 1, 3, 6, 7 & 8 have zero slack. The path connecting these events is the critical path.
Critical path is denoted in the network by bold line. Fig. 8.39 indicates critical path in the given network.
Event TL TE Slack (TL - TE)
1 0 0 0
2 9 5 14
3 4 4 0
4 26 12 14
5 10 9 1
6 12 12 0
7 23 23 0
8 34 34 0

Ex-8. For the network shown below, workout the total float, free float, independent float of all
the activities.

• Calculation of Total float:
• Activities B, E, H, and K have zero total floats. Hence, all these four activities will lie on
the critical path.
• As we have already seen, the critical path for the given network is 1-3-6-7-8, which
confirms that these four activities are critical the activities and they lie on the critical path.
Activity (TL)j (TE)i tij Total float
A
B
C
D
E
F
G
H
I
J
K
19
4
26
26
12
10
34
23
12
23
34
0
0
5
4
4
4
12
12
9
9
2
5
4
7
7
8
5
8
11
2
12
11
(19-0-5)=14
(4-0-4)=0
(26-5-7)=14
(26-4-7)=15
(12-4-8)=0
(10-4-5)=1
(34-12-8)=14
(23-12-11)=0
(12-9-2)=1
(23-9-12)=2
(34-23-11)=0

• Calculation of Free Float
• Calculation of Independent Float
Activity (TE)j (TE)i tij Free Float
A
B
C
D
E
F
G
H
I
J
K
5
4
12
12
12
9
34
23
12
23
34
0
0
5
4
4
4
12
12
9
9
23
5
4
7
7
8
5
8
11
2
12
11
(5-0-5) = 0
(4-0-4) = 0
(12-5-7) = 0
(12-4-7) = 1
(12-4-8) = 0
(9-4-5) = 0
(34-12-8) = 14
(23-12-11) = 0
(12-9-2) = 1
(23-9-12) = 2
(34-23-11) = 0
Activity (TE)j (TL)i tij Independent Float
A
B
C
D
E
F
G
H
I
J
K
5
4
12
12
12
9
34
23
12
23
34
0
0
19
4
4
4
26
12
10
10
23
5
4
7
7
8
5
8
11
2
12
11
(5-0-5) = 0
(4-0-4) = 0
(12-19-7) = 0*
(12-4-7) = 1
(12-4-8) = 0
(9-4-5) = 0
(34-26-8) = 0
(23-12-11) = 0
(12-10-2) = 0
(23-10-12) = 1
(34-23-11) = 0

Ex-9. Find out the Critical path and project duration using forward pass and backward pass
calculations.
A
D
1
4
6
3
4 11
2
7
11
9
7
2
6 8
7
4
2
5
3

Solution
• Forward pass:
• The Earliest Start Times of events (TE) are indicated in the network shown in (Fig.
8.48)
Event Earliest Start Time (TE)
1 0
2 0+3=3
3 0+4=4
4 Higher of (a) 3+6=9 and (b) 4+4= 8, i.e., = 9
5 4+7=11
6 Higher of (a) 4+26 and (b) 11 + 2 = 13, i.e., = 13
7 Higher of (a) 13 + 11 = 24 and (b) 11+ 7 = 18, i.e., = 24
8 Higher of (a) 9+918 and (b) 24 + 11 = 35, i.e., = 35

A
D
1
4
6
3
4 11
2
7
11
9
7
2
6 8
7
4
2
5
3
TE= 0
TE= 3 TE= 9
TE= 35
TE= 24
TE= 4
TE= 11
TE= 13

• Backward Pass:
• The Latest Finish time of all events (along with their respective Earliest Start times) are given
in the network shown in fig. 8.49
Event Latest Finish Time (TL)
8 35 (Kept equal to TE)
7 35-11 = 24
6 24–11 = 13
5 Smaller of and (a) 13-2 = 11 (b) 24-7 = 17, i.e., = 11
4 35-9 = 26
3 Smaller of (a) 26-4= 22 (b) 13-2= 11 (c) 11-7=4 1, i.e., = 4
2 26-620
1 Smaller of (a) 20-3 = 17 (b) 4-4 = , i.e., = 0

A
D
1
4
6
3
4 11
2
7
11
9
7
2
6 8
7
4
2
5
3
TE= 0
TL= 0
TE= 3
TL= 0
TE= 9
TL= 26
TE= 35
TL= 35
TE= 24
TL= 24
TE= 4
TL= 4
TE= 11
TL= 11
TE= 13
TL= 13

We have already seen that all the events lying on the critical path have zero slack. However in the given
example there are three paths connecting zero slack events.
(a) 1-3-5-7-8: 29 days
(b) 1-3-5-6-7-8: 35 days
(c) 1-3-5-7-8: 28 days
Out of the above three paths, the critical path is the one that takes the longest time i.e., 1-3-5-6-7-8 is the
critical path. The project duration is 35 days.

Important Observation:
• "Though a critical path passes through events connecting zero-slacks (TE = TL), all the paths connecting
events of zero-slacks are not necessarily critical paths”
• Under Activity Float Analysis, we have seen that the Total float of activities that lie on the critical path is zero. We shall consider only
those activities for which TE and TL of head and tail events are equal.
• Calculation of Total Float:
• Critical path is that path connecting activities having Total Float Equal to zero.
Activity-ij (TL)j (TE)i tij Total float
(TL)j - (TE)i - tij
B 4 0 4 (4-0-4) = 0
E 13 4 2 (13-4-2) = 7
F 11 4 7 (11-4-7) = 0
I 13 11 2 (13-11-2)= 0
H 24 13 11 (24-13-11) = 0
J 24 11 7 (24-11-7) = 6
K 35 24 11 (35-24-11) = 0

Floats for activities:
• Slack is the maximum delay possible for an event without affecting its
overall duration.
• Floats for activities are the same as slacks are for events.
• So, we can define float as maximum delays possible for an activity
without changing the project duration.

There are three types of floats:
1. Total float: It is the maximum delay possible for an activity without considering
any delay in its precedence or succeeding activity.
2. Free float: It is the maximum delay possible for activities which will not affect
the float of the successor activity.
3. Independent float: It is the maximum delay possible for an activity with used
floats of preceding activities and will not affect the floats of succeeding activities.

The characteristics of float are:
Independent float <=Free float <=Total float
Only independent float can be negative, the rest two floats are
always positive or zero.
Activities with all floats = 0 are critical activities.

The applications of floats are as follows:
1.It identifies the critical activities as well as quantify maximum delays possible
for all not critical activities.
2.It is very important in crashing of a network (reducing the time and/or cost of
overall project).
3.It helps in resource allocation and smoothing.

Float: We can define the following for a given activity (i-j).
Earliest start time ( Tei ): This is the earliest occurrence time for the event from which the
activity arrow originates.
Earliest finish time : Tei + tij
Latest finish time: The latest occurrence time for the node at which the activity arrow
terminates, Tlj
Latest start time : Tlj - t Ij
h i j k
thi tij
tjk
Tei Tli Tej Tlj

Maximum time available = Tlj – Tei
Total float:= Total float for job i-j is the difference between maximum time available and
the actual time it takes.
TF =Tlj – Tei – tij
(It is the maximum delay possible for an activity without considering any delay in its precedence or
succeeding activity)
Free float: This is based on the possibility that all events occur at their earliest times, i.e.
all activities start as early as possible. It is the difference between earliest finish time and
earliest start time.
FF= Tej-Tei – tij
(It is the maximum delay possible for activities which will not affect the float of the successor activity)

Independent float: Let the preceding job h-i finish at its latest possible time Tli and the
succeeding job j-k start at its earliest possible time , which is Tej.
IF= Tej-Tli-tij
(It is the maximum delay possible for an activity with used floats of preceding activities and will not
affect the floats of succeeding activities)
h i j k
Tei Tli Tej Tll
thi
tij
tjk

j
i
tij
FF
TF
IF
Tej
Tli
Tei Tlj
TF =Tlj – Tei – tij
FF= Tej-Tei – tij
IF= Tej-Tli-tij

Ex-5. A construction project involves the following activities. Draw the network
diagram and mark the critical path. What is project completion time?
Activity 1-2 1-3 2-4 3-5 4-6 5-6 6-7 5-7
Proceeding
Activity
-- -- 1-2 1-3 1-3 &
2-4
3-5 4-6 &
5-6
3-5
Duration
(Days)
3 16 5 4 8 4 17 11

Activity 1-2 1-3 2-4 3-5 4-6 5-6 6-7 5-7
Proceeding
Activity
-- -- 1-2 1-3 1-3 &
2-4
3-5 4-6 &
5-6
3-5
Duration
(Days)
3 16 5 4 8 4 17 11
The network diagram for the given problem is-
Dummy
Activity
4
E
11
The activities have been named A to H for the sake of convenience

• The activities have been named A to H for the sake of convenience
• [Activity 1-2 is named 'A', activity 1-3 is named 'B' etc.]
Forward pass Computations Backward pass Computations
Event TE Event TL
1 0 7 41
2 0+3=3 6 41-17 = 24
3 0+16=16 5 Minimum of (a) 24-2 = 22 and (b) 41-
11 = 30, i.e., = 22
4 Maximum of (a) 3+5 = 8 and (b) 16+0 =
16, i.e., = 16
4 24-8 = 16
5 16+4=20 3 Minimum of (a) 16-0= 16 and (b) 22-2
= 18, l.e., = 16
6 Maximum of (a) 16 +8 = 24 and (b) 20 + 2
= 22 i.e., = 24
2 16-5 = 11
7 Maximum of (a) 24 + 17 = 41 (b) 20+11=
31., = 41
1 Minimum of (a) 11-3=8 (b) 1616 = 0,
i.e., = 0

• Nodes 1, 3, 4, 6, 7 have zero slacks.
• There is only one critical path, which is B→ (Dummy) →D→ G.
• The project completion time is (16+0+8+17) = 41 days. [Observation: Critical path can also pass
through dummy activity.]
• Let us find out the total floats of all the activities lying on the critical path and check up if they are
all zero.
• As expected, the total floats of all the activities lying on the critical path is zero.
Activity-ij (TL)j (TE)i tij Total float: (TL)j-(TE)i -tij
B
Dummy
D
G
16
16
24
41
0
16
16
24
16
0
8
17
16-0-16=0
16-16-0=0
24-16-8=0
41-24-17=0
TE= 0
TL= 0
TE= 24
TL= 24
TE= 16
TL= 16
TE= 3
TL= 11
TE= 20
TL= 20
TE= 16
TL= 16
TE=41
TL= 41

CPM/PERT using Excel
Sequence of
activities
No of activities
Objective /Total Time
Decision Variable
Constraints
=Sumproduct(..)
Ti = Time at which activity i begins,
ti = Normal activity time of activity i
Constraints, Tj-Ti ti OR TB-TA tA
Ending node
Starting node

Network solution using excel
Steps-
1. Identify the total no. of activities
2. Draw the network diagram based on
the precedence relationship
3. Determine the sequence of the
activities using network diagram.
5. Inset data on excel
6. Use solver to determine the total time
of projects

Network analysis.pptxbbbbbbbbbbbbbbbbbbbbbbb

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