Question

Problem #2

PERT Analysis: The following represents a project with four activities. All times are in weeks.

Activity	Immediate Predecessor	Optimistic Time	Most Likely Time	Pessimistic Time
A	-	3	8	14
B	-	8	8	9
C	A	6	9	18
D	B	6	11	17

According to the data in the above Table:

What is the critical path?

What is the minimum expected completion time for the project?

According to the above Table, there are four activities in the project. Assume the normal distribution is appropriate to use to determine the probability of finishing by a particular time. If you wished to find the probability of finishing the project in 20 weeks or fewer, it would be necessary to find the variance and then the standard deviation to be used with the normal distribution. What is the probability of completing this project in 20 weeks or fewer?

Answer 1

EXPECTED TIME = (OPTIMISTIC TIME + (4 * MOST LIKELY TIME) + PESSIMISTIC TIME) / 6

VARIANCE = ((PESSIMISTIC TIME - OPTIMISTIC TIME) / 6)^2

ACTIVITY	EXPECTED TIME	VARIANCE
A	(3 + (4 * 8) + 14) / 6 = 8.17	((14 - 3) / 6)^2 = 3.3611
B	(8 + (4 * 8) + 9) / 6 = 8.17	((9 - 8) / 6)^2 = 0.0278
C	(6 + (4 * 9) + 18) / 6 = 10	((18 - 6) / 6)^2 = 4
D	(6 + (4 * 11) + 17) / 6 = 11.17	((17 - 6) / 6)^2 = 3.3611

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	8.17	0	8.17	1.17	9.34	1.17
B	8.17	0	8.17	0	8.17	0
C	10	8.17	18.17	9.34	19.34	1.17
D	11.17	8.17	19.34	8.17	19.34	0

ES = MAX(EF OF ALL PREDECESSORS); 0 FOR FIRST ACTIVITY

EF = ES + DURATION

LF = MIN(LS OF ALL SUCCESSOR ACTIVITIES); MAX EF AS LF FOR LAST ACTIVITY

LS = LF - DURATION

SLACK = LF- EF

CRITICAL PATH = LONGEST PATH WITH 0 SLACK:

CRITICAL PATH = B---D

DURATION OF PROJECT = 19.34

STANDARD DEVIATION OF CRITICAL PATH = SUM OF VARIANCE OF CRITICAL ACTIVITIES = SQRT(3.3889)

EXPECTED TIME = 19.34

DUE TIME = 20

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (20 - 19.34) / SQRT(3.3889) = 0.36

PROBABILITY FOR A Z VALUE OF 0.36 = NORMSDIST(0.36) = 0.6406

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