Question

Question 3 1 pts Consider the network described in the table below. Activity Preceding Optimistic Probable Pessimistic 11 14 |A,B 7 11 13 G D, E

Answer 1

1. C. 4.14

EXPECTED TIME

FORMULA:

(a + (4m) + b) / 6

A = (5 + (4 * 11) + 14) / 6 = 11

B = (3 + (4 * 3) + 9) / 6 = 4

C = (6 + (4 * 10) + 14) / 6 = 10

D = (3 + (4 * 5) + 7) / 6 = 5

E = (4 + (4 * 6) + 11) / 6 = 7

F = (6 + (4 * 8) + 13) / 6 = 9

G = (2 + (4 * 4) + 6) / 6 = 4

H = (3 + (4 * 3) + 9) / 6 = 4

VARIANCE

FORMULA:

((b - a) / 6) ^ 2

A = (14 - 5) / 6) ^ 2 = 2.25

B = (9 - 3) / 6) ^ 2 = 1

C = (14 - 6) / 6) ^ 2 = 1.78

D = (7 - 3) / 6) ^ 2 = 0.44

E = (11 - 4) / 6) ^ 2 = 1.36

F = (13 - 6) / 6) ^ 2 = 1.36

G = (6 - 2) / 6) ^ 2 = 0.44

H = (9 - 3) / 6) ^ 2 = 1

CPM ANALYSIS

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	11	0	11	3	14	3
B	4	0	4	8	12	8
C	10	0	10	0	10	0
D	5	11	16	14	19	3
E	7	4	11	12	19	8
F	9	10	19	10	19	0
G	4	16	20	19	23	3
H	4	19	23	19	23	0

FORWARD PASS = ES = MAX EF OF THE PREDECESSOR

EF = ES + DURATION

BACKWARD PASS = LF = MIN LS OF ALL PREDECESSORS

LS = LF - DURATION

CRITICAL PATH = PATH WITH LONGEST DURATION AND 0 NET SLACK

CRITICAL PATH & DURATION

C + F + H = 23

VARIANCE OF CRITICAL PATH = 1.78 + 1.36 + 1 = 4.14