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Lecture
onNetwork Analysis
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Topics be covered in this lecture
1. Introduction - Activity, Event & Network
3. Installation of a New Machine-Network
2. Errors in Network
4. Fulkersons rule5. Types of data
6. How to draw Network ?
7. Analysis of Network
E, L, EST, EFT, LST, LFT
Critical Path, Slack, Float
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INTRODUCTION
In Network Analysis :
Two aspects : CPM & PERT
CPM - Critical Path Method
PERT - Programme Evaluation &
Review Technique
Events & Activities
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itij
i - j
Network, Activity, Event
i-j - an activity, i & j are events
tij - duration of activity i-j
i - starting event of activity i-j
j - end event of activity i-j
j
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2
31
Example of a Network
AC
B
Activities : A, B, C
Events : 1, 2, 3
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Definition
1. Network - Pictorial representation
of the interrelationship of all
events & activities.
2. Event - instant of time whencertain activities have been
started or completed.
3. Activity - The work content
required to be achieved to
accomplish an event.
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Errors in Network
1.Looping2.
3.Parallel activity
Dangling
2
31
AC
B
Looping
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Dangling
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A
B
dummy
Parallel activity
1 2
A
B
1
2
3
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Project : Installation of a new machine
A - Awaiting Delivery of new machine (10)B - Installation of new machine (3)
C - Remove existing machine (2)
D - Dispose existing machine (5)
E - Install power supply (4)
F - Connect machine to power (1)
G - Test complete installation (2)
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C D
A
E
B F
G
Networkforgiven project
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C D
A
E
B FG
1X
X
X
543
2
6
XX
X X
Numbering Events using
Fulkersons rule
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C D
A
E
B FG
1 543
2
6
1. Activity 1-2 1-3 1-4 2-3 2-6 3-4 4-5 5-6
Duration 2 10 4 0 5 3 1 2
Types ofData :
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C D
A
E
B FG
1 543
2
6
Types ofData :
2. Activity A B C D E F G
Imm. Pred. - A,C - C - B,E F
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C D
A
E
B FG
1 543
2
6
Types ofData :
3. C < B, D
A < B
B, E < F
F C
B > A, C
F > B, E
G > F
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How to draw network ?
Activity A B C D E F G H
Imm.Pred - - - A B,D C B F,E
A
B
C
D
G
F
E
Case : I - Given Data
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Activity A B C D E F G H
Imm.Pred - - - A B,D C B F,E
A
C
B
D
G
E
F
Given Data :
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Activity A B C D E F G H
Imm.Pred - - - A B,D C B F,E
A
C
B
D
G
HE
F
Given Data :
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Activity A B C D E F G H
Imm. Pred - - - A B,D C B F,E
A D
G
EC
F
H
B
Given Data :
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How to draw network ?
A< F, G ; B< D ; C < E ; D < H ; E, G < I ;
E, F, G, H < J
A
B
C
D
G
FE
Case : II - Given Data
H
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A< F, G ; B < D ; C < E ; D < H ; E, G < I ;
E, F, G, H < J
A
B
C
D
G
F
E
Given Data :
H
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A< F, G ; B < D ; C < E ; D < H ; E, G < I ;
E, F, G, H < J
A
B
C
D
G
F
E
Given Data :
H
I
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A< F, G ; B< D ; C < E ; D < H ; E, G < I ;
E, F, G, H < J
A
B
C
D
G
F
E
Given Data :
H
I
J
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A< F, G ; B< D ; C < E ; D < H ; E, G < I ;
E, F, G, H < J
A
B
C
D
G
F
E
Given Data :
H
I
J
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Activity: A B C D E F G H I J K L
Im.Pre. : - - A B B C,D E E G H I,J F
A
B
C
D
G
F
E
H
I
J
K
L
Case : III - Given Data
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Activity: A B C D E F G
Im.Pre. : - A A B,C C D E,F
A
C
D
G
FB
Case : IV - Given Data
E
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Recapitulation
Introduction - Activity, Event & Network
Installation of a New Machine - Network
Errors in Network
Fulkersons rule
T
ypes of data How to draw Network ?
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ANALYSIS OF NETWORK
How to locate Critical Path ?
For Event : E, L, Slack, HS, TS
For Activity : EST, EFT, LST, LFT
For Activity : Floats - F, FF, IF
Relationship between Float & Slack
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B
A FJ
1
5
4
3
2
6
7
E
Network for Analysis
D
I
H
8C
8
2
15
9
38
90
41
A - J : Activities
1 - 8 Events
G
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B
A FJ
1
5
4
3
2
6
7
G
E
D
I
H
8
E=14E=9
E=1
E=0C
82
1
5
9
3
89
04
1
E=4
E=(4, 12)
E=(15,13, 12)
E=(17,20)
E = Earliest Occurrence
time of Event
= 12
= 20
= 15
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B
A FJ
1
5
4
3
2
6
7
G
E
D
I
H
8
L=17L=12
L=9
L= ( 3,0,8 )C 8
2
1
5
9
3
8
9
0
41
L=( 4,9 )
L=( 12,18 )
L=18
L=20
L = Latest Occurrence
time of Event
= 12
= 4
= 0
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B
A F
J
1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0C
82
1
5
9
3
89
04
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20
L=204
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0
C
8
2
1
5
9
3
8
9
0
4
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
For Events 1,3,6 and 8 E = L
Hence 1, 3, 6 and 8 are Critical
Events
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0
C8
2
1
5
9
3
8
9
0
4
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
Hence1 - 3 - 6 - 8 is a Critical Path
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Critical Events : E=L : 1, 3, 6, 8
Critical pathis 1 - 3 - 6 - 8
A network can have more than
one critical path
Critical Path is a path on which allEvents will be Critical.
Critical path is a Longest path.
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0
C8
2
1
5
9
3
8
9
0
4
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
If for activity F , duration ischanged from 9 to 14 then ?
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0C
8
2
1
5
9
3
8
14
0
4
1
E=4, L==4
E=(4, 12)=12
L=(18,12)=12
E=(15,18, 12)=18
L=18
E=(14,20)=20
L=20
1 - 3 - 6 - 8 and
Event 7 also becomes Critical ,hence
1 - 3- 7 - 8 are Critical Paths
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CPM Analysis
Ej , LjEi , Litij
i j
For Event : Ej , Lj , Slack, HS, TS
Slack = Difference between L & E of an Event
TS = Tail Slack = Difference of L & E of i
With respect to i-j activity :
HS = Head Slack =D
ifference of L & E of j
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CPM Analysis
Ej, LjEi , Litij
i j
For Event : Ei , Li, Ej , Lj , Slack, HS, TS
Slack = (L-E)
TS = (Li-Ei)
HS = (Lj-Ej)
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=9
L=12
E=1
L=9
E=0
L=0
C8
2
1
5
9
3
8
9
0
4
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
For Event 2
E = 9 , L = 12
Slack = 3
For Event 5
E = 14 , L = 17
Slack = 3
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CPM Analysis
Ej, LjEi , Litij
i j
For Activity : EST, EFT, LST, LFT
EST = Earliest Start Time
LST = Latest Start Time
EFT = Earliest Finish Time
LFT
= Latest FinishT
ime
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CPM Analysis
Ej, LjEi , Litij
i j
To find value of : EST, EFT, LST, LFT
EST = Ei
LFT = Lj
EFT = Ei + tij
LST
= Lj - tij
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=1
L=9
E=0
L=0
C8
2
1
5
9
3
8
9
0
4
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
For Activity 2 - 5
EST = 9
EFT = 14
LFT = 17
LST = 12
E=9
L=12
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CPM Analysis
Ej, LjEi , Litij
i j
For Activity : Types of Floats
F = Total Float
FF = Free Float
IF = Independent Float
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CPM Analysis
Ej, LjEi , Litij
i j
For Activity : Formulae for Floats
F = ( Lj - Ei ) - tij
FF = ( Ej - Ei ) - tij
IF = ( Ej - Li ) - tij
Negative IF has to be considered as Zero
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Floats : Graphical Presentation
Lj
Ej
Li
Ei
tij
F
tij
tij
FF
IF
F = (Lj - Ei) - tijFF = (Ej - Ei) - tij
IF = (Ej - Li) - tij
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B
A F
J1
5
4
3
2
6
7
G
E
D
I
H
8
E=14
L=17
E=1
L=9
E=0
L=0
C8
2
1
5
9
3
8
9
04
1
E=4, L=(9,4)=4
E=(4, 12)=12
L=(18,12)=12
E=(15,13, 12)=12
L=18
E=(14,20)=20L=20
For Activity 2 - 5
F = 3, FF= 0, IF = 0
E=9
L=12
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Floats : Graphical Presentation
Lj
Ej
Li
Ei
tij
F
tij
tij
FF
IF
Relationship between Slack & Float
F = FF + HS
FF = IF + TS
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Recapitulation
Analysis of a selected Network
E, L, Critical Path
For Event : E, L, Slack, HS, TS
For Activity : EST, EFT, LST, LFT
For Activity : Floats - F, FF, IF
Relationships between Float & Slack
GATE 2002
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GATE - 2002
Job A B C D E F G H I
Imm. Pred. - - A A B,C B,C F F D,E,G
Duration 2 4 6 8 6 4 2 8 6
(a)
(b)
(c)
Drawnetworkandnumbertheevents.
Determine all critical path(s) andtheirduration(s).
What arethe floats forjobs B and D ?
Questions :
E=14E=2
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B
E I1
5
3
2
4
G
F
D
H
6
E=14E=2
E=8
E=0C
8
2
8
2
4
64
E=12E=20
Critical Paths :
1-2-3-5-6
1-2-3-4-5-6
1-2-3-4-6
For B : F=4, FF=4, IF=4
D : F=4, FF=4, IF=4
L=0
L=2
6
L=8 L=12
L=14
6
(2)
(3)
(1)
L=20
GATE 2003
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GATE - 2003(1)
(A)
A Projectconsistsof activitiesAto M showninthenetinthe following figurewiththedurationofthe activities
markedindays
The projectcan becompleted
between 18, 19 days
(C) between 24, 26 days
(B) between 20, 22 days
(D) between 60, 70 days
A, 2
C, 4
E, 5
B, 8
F, 9
G, 6
D, 10
H, 10
K, 3
L, 3
M, 8
GATE 2003
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GATE - 2003
Solution :
Hence,the projectcan becompleted
(C)between 24, 26 days
A, 2
C, 4
E, 5
B, 8
F, 9
G, 6
D, 10
H, 10
K, 3
L, 3
M, 8
1
6
5
4
3
2
70
2
4
13
19
16
24
GATE 2000
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GATE - 2000
Job A B C D E F G
Imm. Pred. - - A,B A C,D A A,B
Duration 3 5 5 4 4 4 6
(a)
(b)
Drawnetworkandnumbertheevents.
Determinecritical path anditsduration.
Questions :
GATE 2000
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GATE - 2000
Solution :
B-C-E is Critical Path. Duration = 14.
A, 2
B, 5
D, 4
C, 5
G, 6
F, 4
E, 41 4
3
2
50,0
3,6
5,5
10,10
14,14
0
GATE PROBLEM
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Job A B C D E F G
Imm. Pred. - - A A B D,E C,F
Duration 2 2 7 12 10 3 4
(a)
(b)
Constructnetwork. Findcritical path.
Estimate projectduration.
Questions :
GATE PROBLEM
GATE PROBLEM
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Solution :
A-D-F-G is Critical Path. Duration = 21.
A, 2
B, 2
D, 12
C, 7G, 4
F, 3
E, 10
1 4
3
2
5 6
GATE PROBLEM
GATE QUESTION
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GATE QUESTION
Adummy activity is usedinnetworkto
describe :
(a) Resourcerestriction
(b) Precedence relationship
(c) Necessary timedelay
(d) Resourceidleness
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Thank youThank youFor any Query or suggestion :Contact :Dr.D.B. NaikProfessor, Training & PlacementSardar Vallabhbhai National InstituteofTechnology, SuratIchchhanath, Surat 395 007
Email : dbnaik_svr@yahoo.com
Phone No. 0261-2255225 (O)
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