(23)

Example 9.14

End

(3,17)
B

O
(20,23)
( , )
(16,20)

(4,5)
(4 5)

M
(16,17)

E
(0,3)

(0)
Start

A

( , )
(3,4)
C

( , )
(4,7)
D

( , )
(7,8)
G

( , )
(8,9)
H

(9,12)

( , )
(12,14)

I

K

(4,6)

•

CP = A‐C‐D‐G‐H‐I‐K‐L‐M‐O

(9,11)

F

• Results:
• Critical Path is in Red
• Total duration is 23.

N

J

L
(14,16)

• Early Start/Early Finish ‐‐‐‐ Go forward
Example 9.15

•TS=6‐3=3
•FS=20‐17=3

(3,17)
B

•TS=6‐4=2
•FS=7‐5=2

(6,20)

Start

(0,3)
A
(0,3)

(16,20)

( , )
(3,4)

( , )
(4,7)

( , )
(7,8)

( , )
(8,9)

C
(3,4)

D
(4,7)

G
(7,8)

H

•TS=5‐4=1
•FS=7‐6=1

(20,23)
O
(20,23)
(16,20)
(16,17)
M
N

( , )
(4,5)
E
(6,7)
(0)

End

(23)

(4,6)
F
(5,7)

(8,9)

•TS=10‐9=1
•FS=12‐11=1

(9,12)
I
(9,12)
(9,11)
(9 11)
J
(10,12)

( , )
(12,14)
K
(12,14)

(19,20)
(14,16)
(14 16)
L
(14,16)

•TS=LS‐ES=19‐16=3
•FS=ES(Suc)‐EF
• FS=20‐17=3

• Late Start/ Late Finish ‐‐‐‐ Go Backward
• Results:
•Total Slack (TS) & Free Slack (FS) for CP =0
• For all other ‐‐ TS=FS
• B: TS=FS=3, E: TS=FS=2, F: TS=FS=1
• J: TS=FS=1, N: TS=FS=3

•Formula y •TS=Late Start – Early Start
•FS=Early Start(Successor) – Early
Finish

Example 9.19

• Early Start/Early Finish ‐‐‐‐ Go forward
(0,1)
(0 1)
B

(0)
Start

A
(0,5)
C

• Results:
• Critical Path is in Red
• Total duration is 22.
•

(12,14)

CP = C E F G H
= C‐E‐F‐G‐H

D
(5,11)

(11,12)

E

F

(14,22)

(22)

G

(0,3)

H

End

Example 9.19
• Late Start/ Late Finish ‐‐‐‐ Go Backward
(0,1)
(0 1)
B
(11,12)
(0)
Start

(0,3)
A
(0,5)

•TS=8‐0=8
•FS=3‐3=0

C

(12,14)

D

(11,12)

(5,11)

(22)

H

End

(11,12)

E

(14,22)

G

(3,4)
( , )
(8,11)

•TS=0
•FS=0

•TS=11‐0=11
•FS=12‐1=11

F

• Results:
•Total Slack (TS) & Free Slack (FS) for CP =0
• For Activities B &D ‐‐ TS=FS
• B: TS=FS=11, D: TS=FS=8,
• A: TS= 8 FS=0
TS= 8, FS=0

•TS=LS‐ES=11‐3=8
TS LS ES 11 3 8
•FS=ES(Suc)‐EF
• FS=12‐4= 8

Example 9.19

• Critical Path in the above examples are:
Critical Path in the above examples are:
• CP = C‐E‐F‐G‐H
• Any delay to any of these activities will delay the
Any delay to any of these activities will delay the entire project.
Critical Path is the longest path through the
• Critical Path is the longest path through the network. • The sum of the durations of critical tasks is the shortest time to complete the project.
• Critical activities have zero Total Slack (TS) or Free slack (FS). All activities on the critical path must start & finish on time in order to complete the project in a timely manner. project in a timely manner

Gantt Chart

Slack

Late Start

Example 9.20

Activity

Optimistic (a)

Most Likely (m)

Pessimistic (b)

Mean (µ)

SD (σ)

Var.

A

1

3

4

2.83

0.50

0.25

B

1

1

2

1.17

0.17

0.03

C

4

5

9

5.50

0.83

0.69

D

1

1

1

1.00

0.00

0.00

E

4

6

12

6.67

1.33

1.78

F

1

1

2

1.17

0.17

0.03

G

1

2

3

2.00
2 00

0.33
0 33

0.11
0 11

H

6

8

10

8.00

0.67

0.44

Formula :
Formula :
Mean µ(F) =(a+4m+b)/6
SD σ(F)=(b‐a)/6 σ2
Var(F)= σ2

Example 9.20

(Mean, Standard Deviation)
(1.17,0.17)
B
(2,0.33)

A

D

(5.5,0.83)

(6.67,1.33)

(1.17,0.17)

C

Start

(1,0)

E

F

• Critical Path is in Red
•

CP = C‐E‐F‐G‐H

(8,0.67)

G

(2.83,0.50)

H

End

Example 9.20

(Mean, Standard Deviation)
(1.17,0.17)
(0,1.17)
(0 1 17)
B

(0)
Start

(2,0.33)
(2 0 33)

(2.83,0.50)
(0,2.83)

(1,0)
(2.83,3.83)

(13.34,15.34)

A

D

G

(5.5,0.83)
(0,5.5)
C
• Critical Path is in Red
Critical Path is in Red
• CP = C‐E‐F‐G‐H
• Total duration 23.34

(1.17,0.17)
(6.67,1.33)
(5.5,12.17) (12.17,13.34)
E

F

(8,0.67)
(15.34,23.34)
H

End
(23.34)

Example 9.20
(1.17,0.17)

(Mean, Standard Deviation)
•TS=LS‐ES
• TS=12.17‐ 0= 12.17
•FS=ES(Suc)‐EF
• FS=13.34‐1.17= 12.17

(0,1.17)
(0 1 17)
B (12.17,13.34)

(0)
Start

(2.83,0.50)
(0,2.83)
A
(9.51,12.34
(5.5,0.83)
C
(0,5.5)

• Results:
•Total Slack (TS) & Free Slack
(FS) for CP =0
• For all other :
B: TS=FS=12.17,
B: TS=FS=12 17
D: TS=FS=9.51,
A: TS= 9.51,FS= 0

(1,0)
(2.83,3.83)

(2,0.33)

D
(12.34,13.34)
(6.67,1.33)
E
(5.5,12.17)

G
(13.34,15.34)

(1.17,0.17)
F
(12.17,13.34)

•TS=LS‐ES
• TS=9.51‐ 0= 9.51
•FS=ES(Suc)‐EF
FS ES(S ) EF
• FS=2.83‐2.83= 0

(8,0.67)

(23.34)

H

End

(15.34,23.34)

•TS=LS‐ES
• TS=12.34‐2.83= 9.51
•FS=ES(Suc)‐EF
•FS ES(Suc) EF
• FS=13.34‐3.83= 9.51

Example 9.20

• Calculation for critical path.
• No, Critical Path has not changed, it has the same path as shown on Exercise 9.19.
• Calculation for Mean, SD and Variance
Expected Value (mean) of the critical path
• µ(CP) = µ(C)+ µ(E)+ µ(F)+ µ(G)+ µ(H) µ(CP) = 5.5+6.67+1.17+2.0+8.0 = 23.34
Variance of the CP
• V(CP)=V(C)+V(E)+V(F)+V(G)+V(H)
• V(CP)=0.69+1.78+0.03+0.11+0.44 =3.06
Standard Deviation of CP
• σ(CP)=SQRT(V(CP)) =1.75

Example 9.20
d.1

• Probability of duration less than or equal to 22 weeks
•Formula :
•P(X ≤ τ) = P {Z ≤ τ‐ µ(X)} σ(X) •Where Z is the standard normal deviate
Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation
•Probability that the project will be completed in 22weeks or less : µ(X) = 23.34 , σ(X) = 1.75, and τ = 22
(X) 23 34 (X) 1 75 d
22
P(Z ≤22)= P(Z ≤(22‐23.34)/1.75 )= ‐0.77
P(Z ≤ ‐0.77) = 1‐0.7794 = 0.22 = 22%
(
)

Example 9.20
d.2 • Probability of duration obtained from the critical path is 23.34 weeks
•Formula :
•P(X ≤ τ) = P {Z ≤ τ‐ µ(X)}
P(X ) P {Z
(X)}
σ(X)
•Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation µ( )
( )
•Probability that the project will be completed in 23.34 weeks : µ(X) = 23.34 , σ(X) = 1.75, and τ = 23.34
P(Z ≤23.34) = P(Z ≤ (23.34‐23.34)/1.75) = 0
P(Z ≤ 0) = 0.5000 = 50%
P(Z ≤ 0) 0 5000 50%

d.3

• Probability of duration more than 26 weeks
•Formula :
•P(X ≤ τ ) = P {Z ≤ τ‐ µ(X)} σ(X) •Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation µ(X) is the mean and σ(X) is the standard deviation
•Probability that the project will takes more than 26 weeks : µ(X) = 23.34 , σ(X) = 1.75, and τ = 26
P(Z ≤ 26)= P(Z ≤ (26‐23.34)/1.75 )= 1.52
P(Z ≤ 1.52) = P(Z ≥ ‐1.52) = 1‐0.9357 = 0.0643 = 6.43%

Example 9.22

• Early Start/Early Finish ‐‐‐‐ Go forward
(0,30)
(0 30)

(30,33)
(30 33)

A
(0,15)
B

F

(25,32)

C

Start

(15,16)

(0,25)

(0)

D

E

• Critical Path is in Red
•

CP = C‐E‐I‐J‐K

(33,38)
(33 38)
G
(46,54)

(33,35)
H

(32,36)
I

K

(36,46)
J

(54)
End

Example 9.22

• Late Start/ Late Finish ‐‐‐‐ Go Backward
(0,30)
(0 30)

(30,33)
(30 33)

(33,38)
(33 38)

A
(1,31)

(15,16)

(33,35)

(46,54)

B
(16,31)
(16 31)

F
(31,32)
(31 32)

H
(34,36)
(34 36)

K

(0,25)

(25,32)

(32,36)

C

E

(0,25)

Start

G
(41,46)

(0,15)

(0)

D
(31,34)

(25,32)

I
(32,36)

• Critical Path is in Red
•

CP = C‐E‐I‐J‐K

(46,54)
(36,46)
J
(36,46)

(54)
End

Example 9.22
A ‐‐‐
• TS=1‐0 = 1
•FS= = 30‐30 = 0

• Total Slack (TS) & Free Slack (FS)
(0,30)
(0 30)

(30,33)
(30 33)

G ‐‐‐ TS=LS‐ES
• TS=41‐33 = 8
•FS=ES(Suc)‐EF
( )
• FS= = 46‐38 = 8

(33,38)
(33 38)

A
(1,31)

Start

TS FS 0
TS =FS =0

B ‐‐‐ TS=LS‐ES
• TS=16‐0 = 16
• FS=ES(Suc)‐EF
• FS= = 15‐15 = 0

G
(41,46)

(0,15)

(15,16)

(33,35)

(46,54)

B
(16,31)
(16 31)

F
(31,32)
(31 32)

H
(34,36)
(34 36)

K

(0,25)

(25,32)

(32,36)

C

E

(0,25)

(0)

D
(31,34)

(25,32)

I
(32,36)

D ‐‐‐ TS=LS‐ES
• TS=31‐30 = 1
•FS=ES(Suc)‐EF
• FS= = 33‐33 = 0

F ‐‐‐ TS=LS‐ES
• TS=31‐15 = 16
•FS=ES(Suc)‐EF
• FS= = 32‐16 = 16

(54)
End

(46,54)
(36,46)
J
(36,46)

H ‐‐‐ TS=LS‐ES
• TS=34‐33 = 1
TS=34 33 = 1
•FS=ES(Suc)‐EF
• FS= = 36‐35 = 1

Total Slack (TS) & Free
Slack (FS) for CP =0

Example 9.22

• Critical Path in the above examples are:
Critical Path in the above examples are:
• CP = C‐E‐I‐J‐K
• Any delay to any of these activities will delay the entire project.
• The critical path activities have no slack.
• The critical path is the longest path through
The critical path is the longest path through the network diagram.

Example 9.22

•Gantt Chart

Total Slack =8
Free Slack=8

Total Slack =16
T l Sl k 16
Free Slack=16

Example 9.23

Activity Optimistic Most Likely Pessimistic Mean (µ) SD (σ)
A
B
C
D
E
F
G
H
I
J
K

25
10
20
3
5
1
4
2
4
8
6

30
15
25
3
7
1
5
2
4
10
8

45
20
35
5
12
1
7
3
6
14
15

31.67
15.00
15 00
25.83
3.33
7.50
1.00
5.17
2.17
4.33
10.33
8.83

3.33
1.67
1 67
2.50
0.33
1.17
0.00
0.50
0.17
0.33
1.00
1.50

Var.
11.11
2.78
2 78
6.25
0.11
1.36
0.00
0.25
0.03
0.11
1.00
2.25

(Mean, Standard Deviation)
Example 9.23
(31.67,3.33)
(31 67 3 33)
A
(0)

(15,1.67)

Start

B

(25.83,2.50)
C
• Critical Path is in Red
•

CP = C‐E‐I‐J‐K = 56.82

(3.33,0.33)
(3 33 0 33)
D

(1,0)
F

(7.50,1.17)
E

(5.17,0.50)
(5 17 0 50)
G
(8.83,1.50) (56.82)

(2.17,0.17)
H

(4.33,0.33)
I

K

(10.33,1.0)
J

End

(Mean, Standard Deviation)
Example 9.23
(31.67,3.33)
(0.00,31.67)
(0 00 31 67)

(3.33,0.33)
(31.67,35.00)
(31 67 35 00)

A
(15,1.67)
(15 1 67)
(0,15)

(0)
Start

D
(1,0)
(1 0)
(15,16)

B

F

(25.83,2.50)
(0.00,25.83)
C

• Critical Path is in Red
•

CP = C‐E‐I‐J‐K = 56.82
C E I J K 56 82

(7.50,1.17)
(25.83,33.33)
E

(5.17,0.50)
(35.00,40.17)
(35 00 40 17)
G
(8.83,1.50)
(8 83 1 50)
(47.99,56.82) (56.82)

(2.17,0.17)
(2 17 0 17)
(35.00,37.17)
H
(4.33,0.33)
(33.33,37.66)
I

K
(10.33,1.0)
(37.66,47.99)
J

End

(Mean, Standard Deviation)
Example 9.23

A
(0.49,32.16)

D
(32.16,35.49)

G
(42.82,47.99)

(0)

(56.82)

Start

B
F
(17.33,32.33) (32.33,33.33)
(17 33 32 33) (32 33 33 33)

C
(0,25.83)

• Critical Path is in Red
•

CP = C‐E‐I‐J‐K = 56.82

E
(25.83,33.33)

H
(35.49,37.66)
(35 49 37 66)

I
(33.33,37.66)

K

End

(47.99,56.82)

J
(37.66,47.99)

Example 9.23
A ‐‐‐ TS=0.49‐0.00 = 0.49
•FS= = 31.67‐31.67 = 0
(0.00,31.67)
A
(0.49,32.16)
(0)
Start

(31.67,35.00)
D
(32.16,35.49)

(15,16)
(0,15)
B
F
(17.33,32.33) (32.33,33.33)
(17 33 32 33) (32 33 33 33)
(0.00,25.83)
C
(0,25.83)

B ‐‐‐ TS=17.33‐0.00 = 17.33
B TS=17 33 0 00 = 17 33
•FS= = 15‐15 = 0

• Critical Path is in Red
•

D ‐‐‐ TS=32.16‐31.67 = 0.49
•FS= = 35.00‐35.00 = 0

CP = C‐E‐I‐J‐K = 56.82

(25.83,33.33)
E
(25.83,33.33)

(35.00,40.17)
G
(42.82,47.99)
(35.00,37.17)
H
(35.49,37.66)
(35 49 37 66)
(33.33,37.66)
I
(33.33,37.66)

F ‐‐‐ TS=LS‐ES
• TS=32.33‐15.00 =
17.33
•FS=ES(Suc)‐EF
•FS=ES(Suc) EF
• FS= = 33.33‐16 =
17.33

G ‐‐‐ TS=LS‐ES
• TS=42.82‐35.00 =
7.82
•FS=ES(Suc)‐EF
• FS= = 47.99‐40.17
= 7.82
(47.99,56.82) (56.82)
K

End

(47.99,56.82)
(37.66,47.99)
J
(37.66,47.99)
H ‐‐‐ TS=LS‐ES
• TS=35.49‐35.00 =
0.49
•FS=ES(Suc)‐EF
•FS=ES(Suc) EF
• FS= = 37.66‐37.17
= 0.49

Example 9.23

• Calculation for critical path.
• No, Critical Path has not changed, it has the same path as shown on Exercise 9.22
• Calculation for Mean, SD and Variance
Expected Value (mean) of the critical path
• µ(CP) = µ(C)+ µ(E)+ µ(I)+ µ(J)+ µ(K) µ(CP) =25.83+7.50+4.33+10.33+8.83=56.82
Variance of the CP
• V(CP)=V(C)+V(E)+V(I)+V(J)+V(K)
• V(CP)=6.25+1.36+0.11+1.0+2.25 =10.97
Standard Deviation of CP
• σ(CP)=SQRT(V(CP)) =3.31

Example 9.23
• Probability of duration obtained from the critical path is 54 days.
•Formula :
•P(X ≤ τ ) P {Z
P(X
) = P {Z ≤ τ‐ µ(X)}
(X)}
σ(X)
•Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation µ( )
( )
•Probability that the project will be completed in 54 days or less : µ(X) = 56.82 , σ(X) = 3.31, and τ = 54
P(Z ≤ 54) =P (Z ≤ (54‐56.82)/3.31) = ‐0.852
P(Z ≤‐0.852) = 1‐0.8023 = 0.198 = 19.8%
P(Z ≤ 0 852) 1 0 8023 0 198 19 8%
• Probability of duration obtained from the critical path is 56.82 days.
•Formula :
F
l
•P(X ≤ τ ) = P {Z ≤ τ‐ µ(X)} σ(X) •Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation
•Probability that the project will be completed in 56.82 days : µ(X) = 56.82 , σ(X) = 3.31, and τ = 56.82 µ(X) = 56 82 σ(X) = 3 31 and τ = 56 82
P(Z ≤56.82) = P(Z ≤ (56.82‐56.82)/3.31) = 0.00
P(Z ≤0) = 0.50 = 50%

Example 9.23
• Probability of duration obtained from the critical path is 64 days or more.
•Formula :
•P(X ≤ τ ) P {Z
P(X
) = P {Z ≤ τ‐ µ(X)}
(X)}
σ(X)
•Where Z is the standard normal deviate
•µ(X) is the mean and σ(X) is the standard deviation µ( )
( )
•Probability that the project will be completed in 64 days or more : µ(X) = 56.82 , σ(X) = 3.31, and τ = 64
P(Z ≤64) = P(Z ≤ (64‐56.82)/3.31 )=7.18/3.31= 2.169
P(Z ≤64) P(Z ≤ (64 56 82)/3 31 ) 7 18/3 31 2 169
P(Z ≤2.17) = P(Z≥‐2.17) =1‐0.985 = 0.015 = 1.5%