Network Analysis-Prince Dudhatra

8/7/2019 Network Analysis-Prince Dudhatra

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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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A

C

B

D

G

E

F

Given Data :


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A

C

B

D

G

HE

F

Given Data :


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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


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 : 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 : [email protected]

Phone No. 0261-2255225 (O)

Documents

Network Analysis-Prince Dudhatra