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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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ACTIVITY DURATION(DAY) PREDECESSOR
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
Example 2 on Activity-on-Arrow (A-O-A)
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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
Total Float (TF) =LF-D-ES
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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Example 3 on Activity-on-Arrow (A-O-A)
• 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
Total Project Duration = 10 weeks
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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ACTIVITY DURATION(DAY) PREDECESSOR
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
Total Float (TF) =LF-D-ES
=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
Total Project Duration = 6 weeks
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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Total Project Duration = 11 weeks
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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