Question

!!! Que A small software development project at Krishna Dhir's firm has five major activities. The times are estimated and provided in the table below: Activity Immediate Predecessor a 2 الما COO m 5 6 7 5 3 b 8 9 10 14 3 4 2 3 a) The expected completion time for this project is (enter your response as a whole number),

2. what is the project variance value of completion?

Answer 1

EXPECTED TIME = (A + (4M) + B) / 6; WHERE A = OPTIMISTIC TIME, M = MOST LIKELY TIME, B = PESSIMISTIC TIME

VARIANCE = ((B - A) / 6)**2

ACTIVITY	EXPECTED TIME	VARIANCE
A	(2 + (4 * 5) + 8) / 6 = 5	((8 - 2) / 6)^2 = 1
B	(3 + (4 * 6) + 9) / 6 = 6	((9 - 3) / 6)^2 = 1
C	(4 + (4 * 7) + 10) / 6 = 7	((10 - 4) / 6)^2 = 1
D	(2 + (4 * 5) + 14) / 6 = 6	((14 - 2) / 6)^2 = 4
E	(3 + (4 * 3) + 3) / 6 = 3	((3 - 3) / 6)^2 = 0

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK(LF-EF)	CRITICAL
A	5	0	0 + 5 = 5	5 - 5 = 0	MIN LS(C) = 5	5 - 5 = 0	YES
B	6	0	0 + 6 = 6	9 - 6 = 3	MIN LS(D) = 9	9 - 6 = 3
C	7	MAX EF( A) = 5	5 + 7 = 12	12 - 7 = 5	MIN LS(E) = 12	12 - 12 = 0	YES
D	6	MAX EF( B) = 6	6 + 6 = 12	15 - 6 = 9	15	15 - 12 = 3
E	3	MAX EF( C) = 12	12 + 3 = 15	15 - 3 = 12	15	15 - 15 = 0	YES

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 = ACE

DURATION OF PROJECT = 15

VARIANCE OF PROJECT = VARIANCE OF CRITICAL PATH = 1 + 1 + 0 = 2

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