Float Slack

1

• Float (Slack)
• Float (Slack ) refers to the amount of time by
  which a particular event or an activity can be
  delayed without affecting the time schedule of
  the network.
• Float (Slack)
• Float (Slack) is defined as the difference
  between latest allowable and the earliest
  expected time.
• Event Float/Slack LS ES
• Where LS Latest start time
• ES Early start time.

2
Earliest start Denoted as ES Earliest start
time is the earliest possible time by which the
activity can be started. Early finish time
Denoted as EF Early finish time is the earliest
possible time by which the activity can be
completed.
3
Latest start time Denoted as LS Latest start
time is the latest possible time by which the
activity can be started Late finish time
Denoted as LS Late finish time is the latest
possible time by which the activity can be
completed.
4
Total float (TF) / Total slack (TS) Total float
of the job is the differences between its Late
start and Early start or Late finish and Early
finish i.e. TF( CA) LS (CA) - ES (CA) Or TF(
CA) LF (CA) - EF (CA) CA Current activity
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Free float (FF) Free float is the amount of time
a job can be delayed without affecting the Early
start time of any other job. FF(CA) ES(SA) EF
(CA) CA Current Activity SA Succeeding
Activity
6
Independent Float (IF) Independent Float is the
amount of time that can be delayed without
affecting either predecessor or successor
activities. IF ES(SA) LF(PA) - Duration
of CA ES Early Start LF Late Finish SA
Succeeding Activity PA Preceding Activity CA
Current Activity
7
Example Construct the Network for the following
Project and determine the following i) Critical
Path ii) ES,EF,LS,LF iii) TF,FF

8

9
Construction of the Network and Determination
Critical Path
4
C
B
3
F
3
6
5
10
1
G
0
E
4
A
14
D
3
2
7
10
Determination of ES and EF
4
C
B
3
F
3
6
5
10
1
G
0
E
4
A(0,14)
14
D
3
2
7
11
Determination of ES and EF
4
C
B
3
F
3
6
5
10
1
G
0
E
4
A(0,14)
14
D(14,21)
3
2
7
12
Determination of ES and EF
4
C
B
3
F
3
6
5
10
1
0
G
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
13
Determination of ES and EF
4
C
B
3
F(25,35)
3
6
5
10
1
G
0
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
14
Determination of ES and EF
4
C
B(0,3)
3
F(25,35)
3
6
5
10
1
0
G
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
15
Determination of ES and EF
4
C
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10
1
0
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
16
Determination of ES and EF
4
C(14,17)
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10
1
0
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
17
Determination of LS and LF
4
C(14,17)
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10(25,35)
1
E(21,25)
4
A(0,14)
14
D(14,21)
3
2
7
18
Determination of LS and LF
4
C(14,17)
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10(25,35)
1
E(21,25)
4(21,25)
A(0,14)
14
D(14,21)
3
2
7
19
Determination of LS and LF
4
C(14,17)
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10(25,35)
1
E(21,25)
4(21,25)
A(0,14)
14
D(14,21)
3
2
7(14,21)
20
Determination of LS and LF
4
C(14,17)
B(0,3)
3
F(25,35)
3
G(14,14)
6
5
10(25,35)
1
E(21,25)
4(21,25)
A(0,14)
14 (0,14)
D(14,21)
3
2
7(14,21)
21
Determination of LS and LF
4
C(14,17)
B(0,3)
3(22,25)
F(25,35)
3
G(14,14)
6
5
10(25,35)
1
E(21,25)
4(21,25)
A(0,14)
14 (0,14)
D(14,21)
3
2
7(14,21)
22
Determination of LS and LF
4
C(14,17)
B(0,3)
3(22,25)
F(25,35)
3
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)
14 (0,14)
D(14,21)
3
2
7(14,21)
23
Determination of LS and LF
4
C(14,17)
B(0,3)
3(22,25)
F(25,35)
3(19,22)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)
14 (0,14)
D(14,21)
3
2
7(14,21)
Job (ES,EF)
Key
Duration (LS,LF)
24
Determination of TF and FF
4
C(14,17)
B(0,3)
3(22,25)
F(25,35)
3(19,22)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)(0)
14 (0,14)(0)
D(14,21)
3
2
7(14,21)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
25
Determination of TF and FF
4
C(14,17)
B(0,3)(11)
3(22,25)
3(19,22)(19)
F(25,35)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)(0)
14 (0,14)(0)
D(14,21)
3
2
7(14,21)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
26
Determination of TF and FF
4
C(14,17)(8)
B(0,3)(11)
3(22,25)(8)
3(19,22)(19)
F(25,35)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)(0)
14 (0,14)(0)
D(14,21)
3
2
7(14,21)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
27
Determination of TF and FF
4
C(14,17)(8)
B(0,3)(11)
3(22,25)(8)
3(19,22)(19)
F(25,35)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)
4(21,25)
A(0,14)(0)
14 (0,14)(0)
D(14,21)(0)
3
2
7(14,21)(0)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
28
Determination of TF and FF
4
C(14,17)(8)
B(0,3)(11)
3(22,25)(8)
F(25,35)
3(19,22)(19)
G(14,14)
6
5
10(25,35)
0(22,22)
1
E(21,25)(0)
4(21,25)(0)
A(0,14)(0)
14 (0,14)(0)
D(14,21)(0)
3
2
7(14,21)(0)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
29
Determination of TF and FF
4
C(14,17)(8)
B(0,3)(11)
3(22,25)(8)
F(25,35)(0)
3(19,22)(19)
G(14,14)
6
5
10(25,35)(0)
0(22,22)
1
E(21,25)(0)
4(21,25)(0)
A(0,14)(0)
14 (0,14)(0)
D(14,21)(0)
3
2
7(14,21)(0)
TF( CA) LS (CA) - ES (CA)
Job (ES,EF)(FF)
Key
FF(CA) ES(SA) EF (CA)
Duration (LS,LF)(TS)
IF ( ES(SA) LF(PA)) - Duration of CA
30
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Example Construct the Network for the following
Project and determine the following i) Critical
Path ii) ES,EF,LS,LF iii) TF,FF

32
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5
(5 ,9)(0)
(9,17)(0)
4(8, 12)(3)
8(12,20)(3)
3
8
(2, 5)(0)
(13, 20)(0)
(5, 6)(7)
(0, 2)(0)
3(5, 8)(3)
1(12, 13)(7)
7(13, 20)(0)
1
2
6
(9,13)(0)
2(0, 2)(0)
4(16, 20)(7)
(7,13) (0)
(2, 7)(0)
6(7, 13)(0)
5(2, 7)(0)
(7, 9)(0)
4
7
2(14,16)(7)
(ES,EF)(FF)
Key
Duration (LS,LF)(TS)
34
PERT Model
Historical Evolution. Before 1957 there was no
generally accepted procedure to aid the
management of a project. In 1958 PERT was
developed by team of engineers working on a
Polaris Missile programme of the navy. This was
a large project involved 250 prime contractors
and about 9000 job contractors. It had about 19
million components. In such projects it is
possible that a delay in the delivery of a small
component might hold the progress of entire
project. PERT was used successfully and the total
time of completion was reduced from 7 years to 5
years.
35
Differences between PERT CPM
36
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• In PERT model, 3 time values are associated
• with each activity. They are
• Optimistic time to
• Pessimistic time tp
• Most likely time tm
• These three times provide a measure of
  uncertainty associated with that activity

38
Optimistic Time This is the shortest possible
time in which the activity can be finished. It
assumes that every thing goes well. Pessimistic
Time This is the longest time that an activity
could take. It assumes that every thing goes
wrong. Most likely Time It is the estimate of
the normal time that an activity would take. This
assumes normal delays.
39
Expected Time ( te) te can be calculated by
the following formula te (to 4tm tp) /
6 Example. If a job has to 5 days, tp 17
days, tm 8 days Then Expected time for the job
would be te (to 4tm tp) / 6 (5 4 x
8 17) / 6 9 days
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• Variability of activity times
• Standard deviation and Variance are commonly used
  in statistics to measure the variability of
  number.
• In PERT model, to measure the variability of an
  activity time duration standard deviation and
  variance are used.
• A large standard deviation represents high
  variability and vice-versa.

41

• Calculation of Standard Deviation and Variance
• Variance (Standard deviation )2
• Standard deviation (t p t o) / 6
• Expected length of the Critical Path te of all
  the activities along the Critical Path

42
Probability of completing the project within a
given date Z (TS TE ) / s Where TS
Scheduled time for project completion
TE Expected time for the project completion
s Standard deviation for the Network
43
sNetwork vSum of variances along the
Critical Path v
(sNetwork )2
44
Example Construct the Network for the following
project and calculate the probability of
completing the project in 25 days
45
1.Construction of the Network
2 - 3 - 4
2
4
2- 6 -10
6 - 4 - 10
2 - 4 - 6
0 - 0 - 0
1
6
1 - 3 - 5
4 - 8 - 12
3 - 6 - 9
3
5
46
2. Calculation of Expected time for all the
activities
2 - 3 - 4
2
4
2- 6 -10
3
6 - 4 - 10
6
10
4
2 - 4 - 6
0 - 0 - 0
1
6
0
1 - 3 - 5
4 - 8 - 12
3
8
3 - 6 - 9
3
5
6
Expected Time ( te) te can be calculated by
the following formula te (to 4tm tp) / 6
47
3. Determination of Critical Path
2 - 3 - 4
2
4
2- 6 -10
3
6 - 4 - 10
6
10
4
2 - 4 - 6
0 - 0 - 0
1
6
0
1 - 3 - 5
4 - 8 - 12
3
8
3 - 6 - 9
3
5
6
to - tm - tp
Key
te
Expected Duration of the project Te 20 days
48
S s2 4.00
49
sNetwork vSum of variances along the
Critical Path v
(sNetwork )2 v 4 2
50
Probability of completing the project within a
given date Z (TS TE ) / s Where TS
Scheduled time for project completion
TE Expected time for the project completion
s Standard deviation for the Network
(25 20) / 2 2.5
51
From the Normal distribution Table, we get the
probability of completing the project in 25 days
is 99.4
52
Example. The following table lists the jobs of a
network along with their time estimates.
53
a) Draw the project network. b) What is the
probability that the job will be completed in 35
days? c) What due date has 90 chance of being
met?
54
1.Construction of the Network
2
6 -12 - 30
4 - 19 - 28
3 - 6 - 15
2 - 5 - 8
3
1
6
6 -12 - 30
3 - 9 - 27
3 - 9 - 27
4
5
55
2. Calculation of Expected time for all the
activities
2
6 -12 - 30
4 - 19 - 28
18
14
3 - 6 - 15
2 - 5 - 8
3
1
6
7
5
6 -12 - 30
3 - 9 - 27
14
3 - 9 - 27
11
11
1 - 4 - 7
4
5
4
Expected Time ( te) te can be calculated by
the following formula te (to 4tm tp) / 6
56
3. Determination of Critical Path
2
6 -12 - 30
4 - 19 - 28
18
14
3 - 6 - 15
2 - 5 - 8
3
1
6
7
5
6 -12 - 30
3 - 9 - 27
14
3 - 9 - 27
11
11
1 - 4 - 7
4
5
4
to - tm - tp
Key
te
Expected Duration of the project Te 32 days
57
As there are two Critical Paths, the path which
gives more variance(s2) is taken as Critical
Path Path A
S s2 32.00
58
Path B
S s2 36.00
v 36 6
v
s
S s2
Therefore the Critical Path is 1 - 3 - 5 - 6
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b) Probability of completing the project within a
given date Z (TS TE ) / s Where TS
Scheduled time for project completion
TE Expected time for the project completion
s Standard deviation for the Network
(35 32) / 6 0.5
60
From the Normal distribution Table, we get the
probability of completing the project in 35 days
is 69.15
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c) The due date for 90 chance of being
met. Probability of completing the project within
a given date Z (TS TE ) / s The value of Z
from the table for a 90 probability is 1.28 TS
? (to be calculated) ,TE 32, s 6 i.e. 1.28
(TS 32) / 6 TS 39.68 days
s
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