Question

Problem 4 (designed to be used with the NORMDIST tutorial)

The following table represents a plan for a project:

JOB NO.	PREDECESSOR JOB(S)	a	m	b
A	-	5	6	7
B	A	3	4	5
C	A	6	7	14
D	A	4	5	12
E	B	1	4	7
F	C	2	3	4
G	D	6	9	15
H	E,F	4	6	8
I	H	3	7	17
J	G	4	5	6
K	I,J	6	8	10

1. Construct the appropriate network diagram.
2. Indicate the critical path.
3. What is the expected completion time for the project?
4. What is the probability the project will:
   take more than 43 days to complete;
   take less than 42 days to complete; and
   take between 37 and 42 days to complete?

USE AN EXCEL FILE

Answer 1

Answer a:

Answer b, c

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	5	6	7	6.00	0.11
B	3	4	5	4.00	0.11
C	6	7	14	8.00	1.78
D	4	5	12	6.00	1.78
E	1	4	7	4.00	1.00
F	2	3	4	3.00	0.11
G	6	9	15	9.50	2.25
H	4	6	8	6.00	0.44
I	3	7	17	8.00	5.44
J	4	5	6	5.00	0.11
K	6	8	10	8.00	0.44

Paths	Duration	Variance
ABEHIK	36.00
ACFHIK	39.00	8.333	CRITICAL PATH
ADGJK	35.50
Critical Path	Duration (Expected Completion time of the project)	Variance, σ2	Standard Deviation, σ (=sqrt(variance))
ACFHIK	39.00	8.333	2.887
Answer b	Answer c

Answer d (i)

we will find the variance of the tasks which lie on critical path	ACFHIK
mean project time (u) of critical path is=	39.00
Required completion time	43
standard deviation= sqrt(variance)=	2.887
because Z= (given completion time- u)/standard deviation	1.386
P(z)= using NORM.S.DIST(z,true)	0.9171
the probability of completing within 43 weeks, P(43) =	0.9171

Probability that the project will take more than 43 weeks to complete = 1- P(43) = 1-0.9171 = 0.0829 or 8.29%

Answer d (ii)

we will find the variance of the tasks which lie on critical path	ACFHIK
mean project time (u) of critical path is=	39.00
Required completion time	42
standard deviation= sqrt(variance)=	2.887
because Z= (given completion time- u)/standard deviation	1.039
P(z)= using NORM.S.DIST(z,true)	0.8507
the probability of completing within 42 weeks, P(42)=	0.8507
	or 85.07%

Answer d (iii)

we will find the variance of the tasks which lie on critical path	ACFHIK
mean project time (u) of critical path is=	39.00
Required completion time	37
standard deviation= sqrt(variance)=	2.887
because Z= (given completion time- u)/standard deviation	-0.693
P(z)= using NORM.S.DIST(z,true)	0.2442
the probability of completing within 37 weeks, P(37)=	0.2442

Probability that the project will take between 37 and 42 days to complete = P(42) - P(37) = 0.8507 - 0.2442 = 0.6065


=> Probability that the project will take between 37 and 42 days to complete =
0.6065 or 60.65%

Dear Student,
Please ask, if you have any doubts through the comment section. Do rate the answer.