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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