Question

Estimated activity times and precedences are given below:

Activity	Optimistic	Most Likely	Pessimistic	Required Precedence
A	6	7	14
B	8	10	12
C	2	3	4
D	6	7	8	A
E	5	5.5	9	B, C
F	5	7	9	B, C
G	4	6	8	D, E
H	2.7	3	3.5	F

What is the probability that the project will be completed within:

A. 21 Days

B. 22 Days

C. 25 Days

Answer 1

Solution:

Activity	Optimistic time-a	Expected completion time-m	Pessimistic time-b	Expected time= (a+4*m+ b)/6	Variance, (sigma)^2= (b-a/6)^2
A	6	7	14	8.00	1.78
B	8	10	12	10.00	0.44
C	2	3	4	3.00	0.11
D	6	7	8	7.00	0.11
E	5	5.5	9	6.00	0.44
F	5	7	9	7.00	0.44
G	4	6	8	6.00	0.44
H	2.7	3	3.5	3.03	0.02

Path	Duration
ADG	21.00
BEG	23.00
BFH	20.03
CEG	15.00
CFH	13.03

Expected project duration is 23 days through BEG

Answer: the probability that the project will be completed within:

A. 21 Days: 0.0416

step 1	we will find the variance of the tasks which lie on critical path				1.33
step 2	mean project time (u) of critical path is=				23.00
step 3	Required completion time is				21
step 4	standard deviation= sqrt(variance)=				1.155
step 5	because Z= (given completion time- u)/standard deviation				-1.732
step 6	P(z)= using NORM.S.DIST(z,true)				0.0416

Answer: the probability that the project will be completed within:

B. 22 Days: 0.1932

step 1	we will find the variance of the tasks which lie on critical path				1.33
step 2	mean project time (u) of critical path is=				23.00
step 3	Required completion time is				22
step 4	standard deviation= sqrt(variance)=				1.155
step 5	because Z= (given completion time- u)/standard deviation				-0.866
step 6	P(z)= using NORM.S.DIST(z,true)				0.1932

Answer: the probability that the project will be completed within:

C. 25 Days: 0.9584

step 1	we will find the variance of the tasks which lie on critical path				1.33
step 2	mean project time (u) of critical path is=				23.00
step 3	Required completion time is				25
step 4	standard deviation= sqrt(variance)=				1.155
step 5	because Z= (given completion time- u)/standard deviation				1.732
step 6	P(z)= using NORM.S.DIST(z,true)				0.9584

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