Week 8 Network Scheduling

8/13/2019 Week 8 Network Scheduling

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Network Scheduling


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Some Definitions

Network: A diagram which shows the inter-relation ship and inter-dependency.

Activity: The performance of a task/work package/operation is known

ac

tivity. Activity consumes time and other resources.

Event: An event shows the start or completion of activities and doesnot consume time or any other resource. It is indicated in network by anumber enclosed in circle, square or triangle. It is also known as Nodeor Connector

Dummy Activities: It is an artificial activity shown by a dotted line in anetwork.

It does not consume time.

It is used to maintain logical sequence.

1

3

2 4

5

A

B

C

D

Dummy Activity


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Network Techniques Fundamental concept of a net work based planning and scheduling

is that the net work represents a time- oriented model of a system.

Net work planning and scheduling has immense scope in assisting

completion of complex civil engineering projects in time and within

reasonable cost.

There are two types of Net work used in most

• Activity-on-Arrow (A-O-A) Type or Arrow diagramming.

Activity shown by arrow and event shown by number in a geometrical

figure like circle

• Activity-on-node (A-O-N) Type or Precedence diagram.

Activity shown by boxes or circle and arrow represents inter-

relationship between activities


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AOA Project Network for House

3 2 01

3

1 1

11 2 4 6 7

3

5

Lay

foundation

Design house

and obtain

financing

Order and

receive

materials

Dummy

Finish

work

Select

carpet

Select

paint

Build

house

AON Project Network for House

13

2

2

43

31 5

1

61

71Start

Design house and

obtain financing

Order and receivematerials

Select paint

Select carpet

Lay foundations Build house

Finish work


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Situations in network diagram

AB

C

A must finish before either B or C can start

A

B

C both A and B must finish before C can start

D

C

B

A both A and C must finish before either of B

or D can start

A

C

B

D

Dummy

A must finish before B can start

both A and C must finish before D can start


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Network Examples

1. A Precedes B and B Precedes C

1 20

A B

3

C

B C A

2. A Precedes both B and C

1

2

0 A B

3

C

B

C

A

1

2

0 A

B3C

3. A and B Precede C

B

C

A


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1 2C

4. A Precedes C and B Precedes D

4 5D

A C

B D

0 A

3 B

5. A Precede C and D; B Precede D

2 31 A C

4D B

D

A

21B

C

2

0

1 A

B

4

3C

D

6. A Precedes B and C; B and C Precede D

B

D

A

C2

0 1 A

B

43

C D


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7. A project consist of 5 activities (A, B, C, D, E, F):

A (2 days), preceding B and C B (4 days), preceding D and E

C (5 days), preceding E D (6 days), preceding F

E (2 days), preceding F F (8 days)

2

0 1 A

B

4

3

C

D

2

4

5

6

2

8E

F5

B

4

E

2

C

5

D

6

A

2

F8


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Construction of Concrete Footing

1. Foundation Chain

A. Lay out of foundation

B. Dig foundation

C. Place formwork

G. Place steel reinforcementD. Place concrete

2. Steel Chain

E. Lay out of foundation

F. Dig foundation

G. Place steel reinforcement

D. Place concrete

3. Concrete Chain

H. Obtain Concrete

D. Place concrete

A B C G D

E F G D

H D S T A R T

F I N I S H

S E

A B C

G DF

H

F

0 2

1 3

4

5 6

A

B C

EF

H

D

G


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Network Schedule Analysis

• Objectives:

1. Find the critical set of activities that establishes the longest pathand defines the minimum duration of the project.

2. Calculate the early start times for each activity.

3. Calculate the late start times for each activity.

4. Calculate the float, or time, available for delay for each activity.


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• Path

• A connected sequence of activities leading from thestarting event to the ending event

• Critical Path• The longest path (time); determines the project duration

• Critical Activities

• All of the activities that make up the critical path

• Cannot be delayed without extending the projectduration.

• Float associated with a critical activity is zero.

• Critical activities lie along the longest path through thenetwork.

S E

A B C

G DF

H

F


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Logical Relationships

There are four types of logical relationship between activities:

Start to Start (SS)

This means activity starts at same time when predecessor activity starts.In figure Activity B has SS relation with activity A

Finish to Start (SF)

This mean activity start when predecessor activity is finished.

In figure Activity C has FS relation with activity B

Start to Finish (SF)This means B finishes with start of Activity A

In figure Activity C can start only after activity D is finished

Finish to Finish (FF)

This means activity finishes together at same time with predecessor activity

In figure Activity E has FF relation with activity B

Activity

A

Activity

B

Activity

C

Activity

D

Activity

E

Activity

C


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Forward Pass

• Earliest Start Time (ES)

• earliest time an activity can start

• ES = maximum EF of immediate predecessors

• Earliest finish time (EF)

• earliest time an activity can finish

• earliest start time plus activity time

Backward Pass

Latest Start Time (LS)

Latest time an activity can start without delaying

critical path timeLatest finish time (LF)

latest time an activity can be completed without

delaying critical path time

LS = minimum LS of immediate predecessors

EFT(I) = EST(I) + DUR(I)

LST(J) = LFT(J) - DUR(J)


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• Float

• Amount of duration by which activity can be delayed without affecting project

completion.


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Example on Activity-on-Node (A-O-N) or Precedence Diagram MethodA project consist of activities (A, B, C, D, E, F):

A (20 days), preceding B and C B (40 days), preceding D and E

C (50 days), preceding D and E D (60 days), preceding F

E (20 days), preceding F F (80 days)

B

40

E

20

C

50

D

60

A

20

F

80

0 20

0 20 20 70

20 70

20 60

30 70

70 130

70 130

70 90

110 130

130 210

130 210

Activity A

B

C

D

EF

Total Float (LF-EF)20 – 20 = 0

70 – 60 = 10

70 – 70 = 0

130 – 130 = 0

130 –

90 = 40210 – 210 = 0

FF (ESsucc -EF)20 – 20 = 0

70 – 60 = 0

70 – 70 = 0

130 – 130 = 0

130 –

90 = 40210 – 210 = 0

Interfering F (TF-FF)0 – 0 = 0

10 – 10 = 0

0 – 0 = 0

0 – 0 = 0

40 –

40 = 00 – 0 = 0


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ACTIVITY DURATION(DAY) PREDECESSOR

A 3 -

B 5 A

C 3 B

D 4 -

E 4 A

F 10 E,D

G 9 C,F

Example 1 on Activity-on-Arrow (A-O-A)


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A 3 -

B 5 A

C 3 BD 4 -

E 4 A

F 10 E,D

G 9 C,F

1

2 3

5

4

6

A

D

B

E

C

F

G3

5

3

4 10

4

9


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1

2 3

5

4

6

A

D

B

E

C

F

G

3

5

3

4 10

4

9

Early Start Box

Late Start Box Make forward pass through the network by adding durationtimes.

0

Forward Pass


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1

2 3

5

4

6

A

D

B

E

C

F

G3

5

3

4 10

4

9

0

0+3=3

38

If two or more activities terminate at a junction node, place the larger sum at ES box.

3+4=7

Larger=7

7


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1

2 3

5

4

6

A

D

B

E

C

F

G3

5

3

4 10

4

9

0

38

7

8+3=11

7+10=17

17 26

Early Start of 26 in the node 6 shows that it will take 26 days to complete the project.

Therefore, 26 days represents Late Finish of the project.


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1

2 3

5

4

6

A

D

B

E

C

F

G3

5

3

4 10

4

9

0

38

7

17

Enter 26 into Late Finish box at the node 6, and make a backward pass toestablish LF for each activity by deducting the durations.

26-9=17

17

14

26

26

Backward Pass


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In the case of junction nodes (with two or more activities),

place the smaller value in the LF box of that node.

1

2 3

5

4

6

A

D

B

E

C

F

G

3

5

3

4 10

4

9

0

38

7

17 26

2617

14

7

14-5=9

7-4=3

3Smaller=3

0


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1

2 3

5

4

6

A

D

B

E

C

F

G

3

5

3

410

4

9

0

38

7

17 26

2617

14

7

3

0


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Total Float (TF) =LF-D-ES

1

2 3

5

4

6

A

D

B

E

C

F

G3

5

3

4 10

4

9

0

3 8

7

17 26

2617

14

7

3

0

TFA = LF-D-ES

= 3-3-0

= 0

TFA=0

TFB=6

TFC=6

TFD=3

TFE=0

TFF=0

TFG=0


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AKTIVITY DURATION (DAY) PREDECESSOR

START 0 -

A 6 START

B 5 START

C 4 START

D 5 A

E 4 B

F (DUMMY) 0 C

G 7 C

H 2 D

I 3 D

J 5 E,F

K 4 G

L 5 I,J,K

M 2 H

FINISH 0 M,L



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1

2 7 8

3 6 9 10

4 5

A

6

B5

C

4

E

4

5

D

I

H

2

J

5

L

2M

G

7

K4

F

0

5

3


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1

2 7 8

3 6 9 10

4 5

A

6

B5

C

4

E

4

5

D

I

H

2

J

5

L

2M

G

7

K4

F

0

5

3

0

6 11

5

4

9

13

15

11

20

FORWARD PASS


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1

2 7 8

3 6 9 10

4 5

A

6

B5

C

4

E

4

5

D

I

H

2

J

5

L

2M

G

7

K4

F

0

5

3

0

6 11

5

4

9

13

15

11

20

0

7 12

6

4

10

18

15

11

20

BACKWARD PASS


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1

2 7 8

3 6 9 10

4 5

A

6

B5

C4

E

4

5

D

I

H

2

J

5

L

2M

G

7

K4

F

0

5

30 0

6 7 11 12

5 6

4 4

9 10

13 18

15 15

11 11

2020

Chapter 6 - Project Planning and Scheduling


TFA = LF-D-ES

= 7-6-0

= 1

TFA=1

TFB=1

TFC = LF-D-ES

= 4-4-0

= 0

TFC=0


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1

2 7 8

3 6 9 10

4 5

A

6

B5

C4

E

4

5

D

I

H

2

J

5

L

2M

G

7

K4

F

0

5

30 0

6 7 11 12

5 6

4 4

9 10

13 18

15 15

11 11

2020

FFA = EF-D-ES

= 6-6-0

= 0

FFA=0

Free Float (FF) =EF-D-ES

FFC = EF-D-ES

= 4-4-0

= 0

FFC=0


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TOTAL FLOAT (TF) = LF – ES - D

TF H = 18 –

11 - 2= 5

FREE FLOAT (FF) = EF – ES - D

FF H = 13 – 11 - 2

= 0

ACTIVITY H

TOTAL FLOAT (TF) = LF – ES - D

TFI

= 15 – 11 - 3

= 1

FREE FLOAT (FF) = EF – ES - D

FF I = 15 – 11 - 3

= 1

ACTIVITY I


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ACTIVITY DURATION ES EF LS LF FF TF

START 0

A 6

B 5

C 4

D 5

E 4

F 0

G 7

H 2

I 3

J 5

K 4

L 5

M 2

FINISH 0


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• Estimate the total project duration.

• Calculate the total float for each activity in the project

• Draw the bar chart according to your calculation

Activity Successor Duration

(week)

A B, C 2

B D 3

C E 2

D F 4

E G 5

F H 2

G H 3

H - 1


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Solution

A1 2

2

3

4

5

6

8 9

B

C

D

E

F

G

H3

2

4

5

2

3

1

7


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Solution

Total Project Duration = 13 weeks

A1

0

02

2

22

35

6

44

4

59

10

69

9

71

21

2

813

13

B

C

D

E

F

G

H3

2

4

5

2

3

1

12,112,3


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Solution

Critical Path : A C E G H

Event Activity

Duratio

n ES LS EF LF Total Float

(1) (2) (3) (4) (5) (6) (7) (8) = 7-3-4

1 - 2 A 2 0 0 2 2 0

2 - 3 B 3 2 2 5 6 1

2 - 4 C 2 2 2 4 4 0

3 - 5 D 4 5 6 9 10 1

4 - 6 E 3 4 4 9 9 0

5 - 7 F 2 9 10 12 12 1

6 - 7 G 3 9 9 12 12 0

7 - 8 H 1 12 12 13 13 0


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SolutionNo Activity week

1 2 3 4 5 6 7 8 9 10 11 12

1 A

2 B

3 C

4 D

5 E

6 F

7 G

8 H

13


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Example 4 on Activity-on-Arrow (A-O-

A)

Activity predecessor Duration

(week)

A - 2

B - 1

C - 3

D A 1

E B 3

F C 2

G D 4

H D, E 1

I D, E, F 2

J G 1

K H 2

L I 3

• Estimate the total project

duration.

• Calculate the total float foreach activity in the project

• Draw the bar chart according

to your calculation


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Solution


A

B

C

D

E

F

G

L

K

3

2

2

3

41

2

0

0

2

4

1

2

3

3

3

5

7

9

5

8

5

5

10

10

3

1

4

5

7

7

I

2

H

1

3

3

J

1

2

3

4

5

6

7

8

9

10

11


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Solution

1. Total Project Duration = 12 weeks

No Activity Total Floatweek

1 2 3 4 5 6 7 8 9 10

1 A 2

2 B 1

3 C 0

4 D 2

5 E 1

6 F 0

7 G 2

8 H 3

9 I 0

10 J 2

11 K 3

12 L 0


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Lag : A period of no activity that mustelapse between two events.

Lag (+ve)

Lead (-ve)

Waiting

period


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A 2 START

B 6 A

C 6 B

D 1 B

E 3 A

F 3 D,E

G 2 C(+2) ,F

EXAMPLE 01


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A

2

B

6

C

6

D

1

E

3

F3

G

2FS +2


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A

20 2

B

62 8

C

68 14

D

18 9

E

32 5

F39 12

G

216 18

16,12

FS +2

5,9

FORWARD PASS


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A

20 2

B

62 8

C

68 14

D

18 9

E

32 5

F39 12

G

216 18

16,12

FS +2

5,9

0 2 2 8 8 14

12 13

1816

1613

1310

8,122,10

BACKWARD PASS


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A

20 2

0 2

B

62 8

2 8

C

68 14

8 14

D

18 9

12 13

E

32 5

F

39 12

G216 18

16,12

FS +2

5,9

1816

1613

1310

8,122,10

COMPLETED


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A

20 2

0 2

B

62 8

2 8

C

68 14

8 14

D

18 9

12 13

E

32 5

F

39 12

G216 18FS +2

1816

1613

1310

TOTAL FLOAT

TFA = LF-EF

=2-2

=0

TFB = LF-EF

=8-8

=0

TFC = LF-EF

=14-14

=0

TFD = LF-EF

=13-9

=4

TFE = LF-EF

=13-5

=8

TFG = LF-EF

=18-18

=0

TFF = LF-EF

=16-12=4

00

4

8

0

4

0


=LF-EF


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A

20 2

0 2

B

62 8

2 8

C

68 14

8 14

D

18 9

12 13

E

32 5

F

39 12

G216 18FS +2

1816

1613

1310

FREE FLOAT

FFA = ESSUCC-EF-

Lag

=2-2-0

=0

00

4

8

0

4

0

FFB = ESSUCC-EF-

Lag

=8-8-0

=0

FFC = ESSUCC-EF-

Lag

=16-14-2

=0

FFD

= ESSUCC

-EF-

Lag

=9-9-0

=0

FFF = ESSUCC-EF-

Lag

=16-12-0

=4

FFA = ESSUCC-

EF- Lag

=18-18-

0

=0


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A

20 2

0 2

B

62 8

2 8

C

68 14

8 14

D

18 9

12 13

E

32 5

F

39 12

G216 18

16,12

FS +2

5,9

1816

1613

1310

8,122,10

CRITICAL PATH

00

4

8

0

4

0

Activities on critical path : A,B,C and G.


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Activity

ATFA = LF- EF

= 2 – 2

= 0

FFA = ESsucc - EF – Lag

= 2 – 2 – 0

= 0

Activity

BTFB = LF- EF

= 8 – 8

= 0

FFB = ESsucc - EF – Lag

= 8 – 8 – 0

= 0


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Activity

CTFC = LF- EF

= 14 – 14

= 0

FFC = ESsucc - EF – Lag

= 16 – 14 – 2

= 0

Activity

DTFD = LF- EF

= 13 – 9

= 4

FFD = ESsucc - EF – Lag

= 9 – 9 – 0

= 0


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Activity

ETFE = LF- EF

= 13 – 5

= 8

FFE = ESsucc - EF – Lag

= 9 – 5 – 0

= 4

Activity F

TFF = LF- EF

= 16 – 12

= 4

FFF = ESsucc - EF – Lag

= 16 – 12 – 0

= 4


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Activity

GTFG = LF- EF

= 18 – 18

= 0

FFG = ESsucc - EF – Lag

= 18 – 18 – 0

= 0


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SOLUTION


B

2

2

4

4

2

D

1

1

2

2

1

L = 0

L = 1E

2

2

3

3

1

C

5

5

6

6

1L = 1

L = 0 L = 2

A

0

0

2

2

2

0 0 0

0 0

Note

Total Float (TF):

Critical Path :

Tf


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No Activity week

1 2 3 4 5 6

1 A

2 B

3 C

4 D

5 E

L=1

L=1

L=2


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Example 3

No Activity Duration Relationship Lag

(week) (week)

1 A 3 A-B (S-S) 5

A-C (F-S) 2

2 B 1 B-D (F-S) 3

3 C 2 C-D (F-S) -

4 D 2 D-E (F-F) -

5 E 3 - -

1. Estimate the total project duration.

2. Calculate the total float for each activity in the project

3. Draw the bar chart according to your calculation


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C

5

7

7

9

2

B

5

5

6

6

1

L = 2

L = 5D

9

9

1

1

1

1

2

E

8

8

11

11

3

L = 3L = 2

A

0

0

3

5

3

2 2 0

0 0

Note

Total Float (TF):

Critical Path :

Tf

Activities on critical path : B, D, and E


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Week 8 Network Scheduling

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Transcript of Week 8 Network Scheduling