Question

A network consists of the following list. Times are given in weeks.

Activity	Preceding	Optimistic	Probable	Pessimistic
A	--	5	11	14
B	-	3	3	9
C	--	6	10	14
D	A, B	3	5	7
F	C	6	8	13
G	D, E	2	4	6
H	F	3	3	9

Calculate the probability that the project will be completed in less than 25 weeks.

NOTE

NO need to draw the network diagram on this page,  

It is NOT sufficient, for example, to write the “probability is = __x %.”

Show your work how you got the probability.

Answer 1

PROBABILITY = 24.2 %

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

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

ACTIVITY	EXPECTED TIME	VARIANCE
A	(5 + (4 * 11) + 14) / 6 = 10.5	((14 - 5) / 6)^2 = 2.25
B	(3 + (4 * 3) + 9) / 6 = 4	((9 - 3) / 6)^2 = 1
C	(6 + (4 * 10) + 14) / 6 = 10	((14 - 6) / 6)^2 = 1.7778
D	(3 + (4 * 5) + 7) / 6 = 5	((7 - 3) / 6)^2 = 0.4444
E	(6 + (4 * 8) + 13) / 6 = 8.5	((13 - 6) / 6)^2 = 1.3611
F	(2 + (4 * 4) + 6) / 6 = 4	((6 - 2) / 6)^2 = 0.4444
G	(3 + (4 * 3) + 9) / 6 = 4	((9 - 3) / 6)^2 = 1

CPM ANALYSIS

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	10.5	0	10.5	3	13.5	3
B	4	0	4	9.5	13.5	9.5
C	10	0	10	0	10	0
D	5	10.5	15.5	13.5	18.5	3
E	8.5	10	18.5	10	18.5	0
F	4	18.5	22.5	18.5	22.5	0
G	4	22.5	26.5	22.5	26.5	0

FORWARD PASS: ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT. EF = ES + DURATION OF THE ACTIVITY.

BACKWARD PASS: LF = MINIMUM LS OF ALL SUCCESSOR ACTIVITIES; COMPLETION TIME OF THE PROJECT IF NO SUCCESSORS ARE PRESENT. LS = LF - DURATION OF THE ACTIVITY.

SLACK = LF - EF, OR, LS - ES

CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.

CRITICAL PATH = CEFG

DURATION OF PROJECT = 26.5

PROBABILITY

VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL PATH) = 4.5833

STANDARD DEVIATION OF CRITICAL PATH = SQRT(VARIANCE OF CRITICAL PATH) = 2.14

EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 26.5

DUE TIME = 25

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

Z = (25 - 26.5) / 2.14 = -0.7

PROBABILITY FOR A Z VALUE OF -0.7 = NORMSDIST(-0.7) = 0.242

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