Question

Question: To complete the wing assembly for and experimental aircraft, Jim Gilbert has laid out nine activities involved. These activities have been labeled A through I in the following table. The company is planning to finish the project in 34 weeks. Please answer the following questions.

1. Determine the expected (estimated) duration and variance for each activity. You can round the expected duration numbers without decimals.

Activity	Immediate Predecessor(s)	Optimistic	Probable	Pessimistic	Expected duration	Variance
A	--	5	9	13	Answer	Answer
B	--	3	3	9	Answer	Answer
C	--	6	9	15	Answer	Answer
D	A,C	5	7	12	Answer	Answer
E	B	4	6	14	Answer	Answer
F	C,B	3	4	8	Answer	Answer
G	D	4	4	4	Answer	Answer
H	E,F	2	4	9	Answer	Answer
I	G,H	7	10	13	Answer	Answer

2. Draw the network diagram in a page and then fill the below table:

Activity	Early Start	Early Finish	Late Start	Late Finish	Slack
A	Answer	Answer	Answer	Answer	Answer
B	Answer	Answer	Answer	Answer	Answer
C	Answer	Answer	Answer	Answer	Answer
D	Answer	Answer	Answer	Answer	Answer
E	Answer	Answer	Answer	Answer	Answer
F	Answer	Answer	Answer	Answer	Answer
G	Answer	Answer	Answer	Answer	Answer
H	Answer	Answer	Answer	Answer	Answer
I	Answer	Answer	Answer	Answer	Answer

3. What is the critical path? Draw the network carefully in a page to make the correct sequence of the critical path.

4.

A) What are the project variance and the project standard deviation?

B) What is the probability that the project will be finished in less than 40 weeks?

Answer 1

1.

Activity	Predecessor	Opt	Probable	Pessi	Expected	Variance
A	-	5	9	13	9	1.78
B	-	3	3	9	4	1.00
C	-	6	9	15	9.5	2.25
D	A,C	5	7	12	7.5	1.36
E	B	4	6	14	7	2.78
F	C,B	3	4	8	4.5	0.69
G	D	4	4	4	4	0.00
H	E,F	2	4	9	4.5	1.36
I	G,H	7	10	13	10	1.00

te = (to + 4tm + tp)/6

te = expected time

to = optimistic time

tm = probable time

tp = pessimistic time

variance = (tp - to)^2/36

The network diagram is drawn with following terminology:

ES	Sla	EF
	Act
LS	Dur	LF

where Act is activity, Dur is duration, ES and EF are earliest start and finish times and LS and LF are latest start and finish times and Sla is slack time calculated as LS-ES

The ES and EF are calculated in the forward path. For activities A, B and C , ES is 0 and EF is ES + duration i.e. 9, 4 and 9.5 respectively. Further, the ES for E is EF for B i.e. 4 and hence EF for B is 4+7=11. For F, the ES will be maximum EF of B and C i.e. 9.5 hence EF for F will be 14. Similarly we reach the last activity I with ES being maximum EF of G and H i.e 21. Hence EF for I will be 21+10 = 31 which is the project completion time

To calculate LS and LF we start backwards from I. Here LF will be 31 and LS will be LF-duration i.e. 21. Next LF for both G and H will be LS for I i.e. 21. Hence LS for G and H will be 17 and 16.5 respectively. Further, for activity C, the LF will be minimum LS of D and F i.e. 9.5. Hence LS for C will be 0.

Finally we calculate slack time for all using the formula LS-ES

2. The summary of ES, EF, LS, LF and slack is given below

Activity	ES	EF	LS	LF	Slack
A	0	9	0.5	9.5	0.5
B	0	4	5	9	5
C	0	9.5	0	9.5	0
D	9.5	17	9.5	17	0
E	4	11	9	16.5	5
F	9.5	14	12	16.5	2.5
G	17	21	17	21	0
H	14	18.5	16.5	21	2.5
I	21	31	21	31	0

3. From the diagram above, the path in which the activities have a slack of zero is the critical path. The same is highlighed in RED. i.e. path C-D-G-I that takes 31 weeks

4.

A. The project variance is the sum of variance of activities on the critical path i.e C,D,G and I = 2.25+1.36+0+1 = 4.61

Hence project standard deviation = sqrt of variance = 2.14735

B. For project completion in less than 40 weeks, we need to calculate z statistic

z = (x-u)/SD

where x = 40, u = 31 and SD = 2.14735

Hence z = 4.191213

To calculate the probability, we need to refer to standard normal table or use normsdist(z) function in excel that gives 0.9999

i.e. there is a 99.99% probability of the project finishing in less than 40 weeks