Problem 1

Activity Predecessor SuccessorA - D,EB - FC - G,HD A FE A JF B,D JG C JH C II H -J E,F,G -

AOA Network

1

2

4

START

3

7

C

A

D

E

G

J

IBF

6

5

H

AON Network

START

A

C

D

B

G

E

F

H

ENDJ

I

Problem 2

Activity Predecessor Successora - c,fb - d,ec a hd b he b hf a ig b ih c,d,e ii f,g,h -

AOA Network

AON Network

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.

1

2

4

START

a

b

c

d

e

f

h

g

END

5

3

6

7

b

a

f

h

c

d

e

g

i

i

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.

Problem 7

Duration - 3 estimates

Activity Predecessora - 2 4 9b - 2 3 8c b 4 7 14d a 4 5 16e b 5 7 12f c,d 2 4 8g c,d 6 8 14h e,f 6 7 14i g,h 2 3 9

(a) Activity Variance S.D.a 4.50 1.361 1.17b 3.67 1.000 1.00c 7.67 2.778 1.67d 6.67 4.000 2.00e 7.50 1.361 1.17f 4.33 1.000 1.00g 8.67 1.778 1.33h 8.00 1.778 1.33I 3.83 1.361 1.17

(b) AOA Network

(c) Path Variance S.D.3.83 1-2-4-6-7 23.67 8.500 2.920.17 1-2-4-5-6-7 27.33 9.500 3.08

27.50 1-3-4-5-6-7 27.50 7.917 2.813.67 1-3-4-6-7 23.83 6.917 2.63

Optimistic (months)

Most Likely

(months)Pessimistic

(months)

Estimated Time

Estimated Time

2

3

4 6

5

1

a:4.50

b:3.67

d:6.67

c:7.67

e:7.50

f:4.33

g:8.67

h:8.00

i:3.83

4.50 1-3-5-6-7 23.00 5.500 2.35

(d) Due date D = 36Estimated time T = 27.50Standard Deviation 3 2.81Standard normal variate = 3.020979

99.87

(e) Probability of completion = 0.8Standard normal variate = 0.842Estimated time T = 2 27.33SD = 3.08Due date D = 29.93

(f)

(g)

Due date D = 36Estimated time T = 27.33Standard Deviation 3.08Standard normal variate = 2.811838

99.75

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

Probability of completion in 40 months =

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.

The probability of completion in 36 months, on the near critical path is :

Probability of completion in 36 months =

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.

What-if ??

Change the activity durations to assess the

impact

7

monthsmonths

%

months

months

monthsmonths

%

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

This observation does not hold if you change the activity durations.

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.

The probability of completion in 36 months, on the near

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

Problem 8

Activity Predecessora - 6 5 10,000 15,000b - 4 3 12,000 14,000c a 5 n.a. 16,000 n.a.d b 3 n.a. 18,000 n.a.e c 4 2 11,000 17,000f d 4 2 24,000 32,000g c 4 3 12,000 18,000h d 9 6 50,000 68,000i 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

Activitya 6 5 10,000 15,000 1b 4 3 12,000 14,000 1c 5 n.a. 16,000 n.a. 0

Normal Duration (weeks)

Crash Duration (weeks)

Normal Cost (Rs.)

Crash Cost (Rs.)

We evaluate our options by the following table:

Normal Time (Weeks)

Crash Time (Weeks)

Normal Cost NC

(Rs.)

Crash Cost CC

(Rs.)

Max. Crashing (Weeks)

START

b

a:

d

c

h

f

g

e6 5

4 3

9

4

4

4

d 3 n.a. 18,000 n.a. 0e 4 2 11,000 17,000 2f 4 2 24,000 32,000 2g 4 3 12,000 18,000 1h 9 6 50,000 68,000 3i 2 n.a. 16,000 n.a. 0j 3 2 10,000 11,000 1

No.1 a,c,e,i,j None 0 179,000 202 a,c,e,i j 1 180,000 193 a,c,i e 2 186,000 174 c,i a 1 191,000 16

Therefore we find the optimum cost to be (Rs.)Savings obtained by crashing activities (Rs.)

Sum of the normal costs for all the activities in the above table constitutes the direct cost without crashing

The indirect cost is @ Rs. 6000 per week charged for the project duration of 20 weeks

Crashable Activities on Critical path

Activitiy Crashed

No. of weeks

crashedDirect Cost

Project Duration

PREVIOUS SOLUTION

What-if ??

50002000n.a.

Change the cost structure to assess the impact

Crashing Cost/week

i j FINISH

23

n.a.3000400060006000n.a.

1000

Indirect Cost Total Cost120,000 299,000114,000 294,000102,000 288,00096,000 287,000

287,00012,000

Sum of the normal costs for all the activities in the above table constitutes

The indirect cost is @ Rs. 6000 per week charged for the project duration of 20

NEXT SOLUTION