Question

Consider the following precedence chart: ACTIVITY PRECEDING ACTIVITIES OPTIMISTIC TIME (days) PESSIMISTIC TIME days) MOST LIKELY TIME (days) 10 10 10 B.C DE The total slack for activity D is O 2 O O O

Answer 1

1.SLACK OF D = A. 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 * 10) + 15) / 6 = 10	((15 - 5) / 6)^2 = 2.7778
B	(10 + (4 * 12) + 14) / 6 = 12	((14 - 10) / 6)^2 = 0.4444
C	(10 + (4 * 10) + 10) / 6 = 10	((10 - 10) / 6)^2 = 0
D	(2 + (4 * 4) + 6) / 6 = 4	((6 - 2) / 6)^2 = 0.4444
E	(4 + (4 * 8) + 12) / 6 = 8	((12 - 4) / 6)^2 = 1.7778
F	(4 + (4 * 8) + 12) / 6 = 8	((12 - 4) / 6)^2 = 1.7778
G	(10 + (4 * 12) + 14) / 6 = 12	((14 - 10) / 6)^2 = 0.4444
I	(4 + (4 * 8) + 12) / 6 = 8	((12 - 4) / 6)^2 = 1.7778
J	(2 + (4 * 4) + 6) / 6 = 4	((6 - 2) / 6)^2 = 0.4444

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	10	0	10	0	10	0
B	12	0	12	2	14	2
C	10	0	10	4	14	4
D	4	12	16	14	18	2
E	8	10	18	10	18	0
F	8	10	18	18	26	8
G	12	18	30	18	30	0
I	8	18	26	26	34	8
J	4	30	34	30	34	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 = AEGJ

DURATION OF PROJECT = 34

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