1. Problem 1
Activity Predecessor Successor
A - D,E
B - F
C - G,H
D A F
E A J
F B,D J
G C J
H C I
I H -
J E,F,G -
AOA Network
2
E
A
J
D 5 7
START
1 F
B I
3
C G 6
H
4

2. AON Network
A E
D
START
B F
J
END
G
C H I

3. Problem 2
Activity Predecessor Successor
a - c,f
b - d,e
c a h
d b h
e b h
f a i
g b i
h c,d,e i
i f,g,h -
AOA Network
6
f
h
2 i
c
a 5
g
7
1
d
3
b
e
4
AON Network
a f
c
START
d h END
i
e
b g

4. In the AOA Network representation a dummy task has to be
incorporated into the network to allow for two predecessors
of the task e. No such requirement is found in the AON
representation. It appears that an AON representation is
much easier than AOA.

5. Problem 7
Duration - 3 estimates
Most
Optimistic Likely Pessimistic
Activity Predecessor (months) (months) (months)
a - 2 4 9
b - 2 3 8
c b 4 7 14
d a 4 5 16
e b 5 7 12
f c,d 2 4 8
g c,d 6 8 14
h e,f 6 7 14
i g,h 2 3 9
Estimated
(a) Activity Time Variance S.D.
a 4.50 1.361 1.17
b 3.67 1.000 1.00
c 7.67 2.778 1.67
d 6.67 4.000 2.00
e 7.50 1.361 1.17
f 4.33 1.000 1.00
g 8.67 1.778 1.33
h 8.00 1.778 1.33
I 3.83 1.361 1.17
(b) AOA Network
a:4.50 2
d:6.67
1 g:8.67 i:3.83
4 6
c:7.67
f:4.33
b:3.67
3 5 h:8.00
e:7.50
Estimated
(c) Path Time Variance S.D.
3.83 1-2-4-6-7 23.67 8.500 2.92
0.17 1-2-4-5-6-7 27.33 9.500 3.08

6. 27.50 1-3-4-5-6-7 27.50 7.917 2.81
3.67 1-3-4-6-7 23.83 6.917 2.63
4.50 1-3-5-6-7 23.00 5.500 2.35
From the above table we see that the critical path is 1-2-4-
5-6-7 and the nearly critical path is 1-3-4-5-6-7
(d) Due date D = 36
Estimated time T = 27.50
Standard Deviation 3 2.81
Standard normal variate = 3.020979
Probability of completion in 40
months = 99.87
(e) Probability of completion = 0.8
Standard normal variate = 0.842
Estimated time T = 2 27.33
SD = 3.08
Due date D = 29.93
(f) For reduction in the project duration, activities on the
critical path can be reviewed for crashing options.The firm
must investigate methods that enables it to reduce the
variance of some of the activities on the critical path.
(g) The probability of completion in 36 months, on the near
critical path is :
Due date D = 36
Estimated time T = 27.33
Standard Deviation 3.08
Standard normal variate = 2.811838
Probability of completion in 36
months = 99.75
It is interesting to note that a near-critical path can
potentially bring down the probability of completion
compared to a critical path due to high path variance.
Therefore the notion of an official critical path in PERT
is not the same as in a CPM method.

7. What-if ??
Change the activity
durations to assess the
impact
i:3.83
7

8. ritical path is 1-2-4- This observation does not hold if you change the
1-3-4-5-6-7 activity durations.
months
months
%
months
months
activities on the
hing options.The firm
it to reduce the
the critical path.
nths, on the near
months
months
%
critical path can
of completion
h path variance.
itical path in PERT

9. Problem 8
Normal Crash
Duration Duration Normal Crash
Activity Predecessor (weeks) (weeks) Cost (Rs.) Cost (Rs.)
a - 6 5 10,000 15,000
b - 4 3 12,000 14,000
c a 5 n.a. 16,000 n.a.
d b 3 n.a. 18,000 n.a.
e c 4 2 11,000 17,000
f d 4 2 24,000 32,000
g c 4 3 12,000 18,000
h d 9 6 50,000 68,000
i e,f 2 n.a. 16,000 n.a.
j g,h,i 3 2 10,000 11,000
Indirect cost for the project per week (Rs.) 6,000
AON Network 4
g
a: c
e
6 5
START 4
f
b d 4
4 3
h
9
We evaluate our options by the
following table:
Crash Normal Crash Max.
Normal Time Time Cost NC Cost CC Crashing
Activity (Weeks) (Weeks) (Rs.) (Rs.) (Weeks)
a 6 5 10,000 15,000 1

10. b 4 3 12,000 14,000 1
c 5 n.a. 16,000 n.a. 0
d 3 n.a. 18,000 n.a. 0
e 4 2 11,000 17,000 2
f 4 2 24,000 32,000 2
g 4 3 12,000 18,000 1
h 9 6 50,000 68,000 3
i 2 n.a. 16,000 n.a. 0
j 3 2 10,000 11,000 1
Sum of the normal costs for all the activities in the above table con
the direct cost without crashing
The indirect cost is @ Rs. 6000 per week charged for the project dura
weeks
Crashable No. of
Activities on Activitiy weeks Direct Project
No. Critical path Crashed crashed Cost Duration
1 a,c,e,i,j None 0 179,000 20
2 a,c,e,i j 1 180,000 19
3 a,c,i e 2 186,000 17
4 c,i a 1 191,000 16
Therefore we find the optimum cost to be (Rs.)
Savings obtained by crashing activities (Rs.)
PREVIOUS SOLUTION

11. What-if ??
Change the cost
structure to assess the
impact
3
2
i j FINISH
Crashing
Cost/week
5000

12. 2000
n.a.
n.a.
3000
4000
6000
6000
n.a.
1000
in the above table constitutes
ed for the project duration of 20
Indirect Cost Total Cost
120,000 299,000
114,000 294,000
102,000 288,000
96,000 287,000
287,000
12,000
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