Question

The following represents a project that should be scheduled using CPM:

	IMMEDIATE PREDECESSORS	TIMES (DAYS)
ACTIVITY		a	m	b
A	—	1	4	7
B	—	1	3	5
C	A	2	5	11
D	A	1	8	9
E	B	1	2	3
F	C,D	1	6	11
G	D,E	1	2	3
H	F,G	2	3	3

b. What is the critical path?

	B-E-G-H
	A-C-F-H
	A-D-F-H
	A-D-G-H

c. What is the expected project completion time? (Round your answer to 3 decimal places.)

Project completion time             days

d. What is the probability of completing this project within 22 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.)

Probability           

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	(1 + (4 * 4) + 7) / 6 = 4	((7 - 1) / 6)^2 = 1
B	(1 + (4 * 3) + 5) / 6 = 3	((5 - 1) / 6)^2 = 0.4444
C	(2 + (4 * 5) + 11) / 6 = 5.5	((11 - 2) / 6)^2 = 2.25
D	(1 + (4 * 8) + 9) / 6 = 7	((9 - 1) / 6)^2 = 1.7778
E	(1 + (4 * 2) + 3) / 6 = 2	((3 - 1) / 6)^2 = 0.1111
F	(1 + (4 * 6) + 11) / 6 = 6	((11 - 1) / 6)^2 = 2.7778
G	(1 + (4 * 2) + 3) / 6 = 2	((3 - 1) / 6)^2 = 0.1111
H	(2 + (4 * 3) + 3) / 6 = 2.83	((3 - 2) / 6)^2 = 0.0278

USING EXPECTED DURATION TO FIND THE CRITICAL PATH OF THE PROJECT

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	4	0	4	0	4	0
B	3	0	3	10	13	10
C	5.5	4	9.5	5.5	11	1.5
D	7	4	11	4	11	0
E	2	3	5	13	15	10
F	6	11	17	11	17	0
G	2	11	13	15	17	4
H	2.83	17	19.83	17	19.83	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 = ADFH

DURATION OF PROJECT = 19.83

PROBABILITY OF FINISHING THE PROJECT WITHIN 22 DAYS:

VARIANCE = SIGMA(VARIANCE OF CRITICAL ACTIVITIES) = 1 + 1.7776 + 2.7778 + 0.0278 = 5.5834

STDEV = SQRT(VARIANCE)

EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 19.83

DUE TIME = 22

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

Z = (22 - 19.83) / SQRT(5.5836) = 0.92

THE PROBABILITY DISTRIBUTION CORRESPONDING TO A Z VALUE OF 0.92 = 0.8212

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