Answer: the probability that the critical path for this project will be completed within 210 days is 0.99725
Step-by-step explanation:
Given that;
Work break down is;
Time Estimates (days)
Activity Precedes Optimistic Most Likely Pessimistic
Start A,B - - -
A C,D 44 50 56
B D 45 60 75
C E 42 45 48
D F 31 40 49
E F 27 36 39
F End 58 70 82
we know that;
Expected Time = ( Optimistic Time + ( 4 × most likely time) + pessimistic time) / 6
and
Variance = [( pessimistic time - Optimistic time) / 6 ]²
so
ACTIVITY EXPECTED TIME VARIANCE
A (44 + (4 × 50) + 56) / 6 = 50 ((56 - 44) / 6)² = 4
B (45 + (4 × 60) + 75) / 6 = 60 ((75 - 45) / 6)² = 25
C (42 + (4 × 45) + 48) / 6 = 45 ((48 - 42) / 6)² = 1
D (31 + (4 * 40) + 49) / 6 = 40 ((49 - 31) / 6)² = 9
E
(27 + (4 * 36) + 39) / 6 = 35 ((39 - 27) / 6)² = 4
F (58 + (4 * 70) + 82) / 6 = 70 ((82 - 58) / 6)² = 16
ACTIVITY DURATION ES EF LS LF SLACK
A 50 0 50 0 50 0
B 60 0 60 30 90 30
C 45 50 95 50 95 0
D 40 60 100 90 130 30
E 35 95 130 95 130 0
F 70 130 200 130 200 0
FORWARD PASS:
ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT.
EF = ES + DURATION OF THE ACTIVITY.
CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.
CRITICAL PATH = ACEF
DURATION OF PROJECT = 200
Now the probability
VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL PATH) = 25
STANDARD DEVIATION OF CRITICAL PATH = SQRT(VARIANCE OF CRITICAL PATH) = 5
EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 200
DUE TIME = 210
Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)
Z = (210 - 200) / 5 = 2
Now form the Normal Distribution Table;
Z = 0.99725
Therefore, the probability that the critical path for this project will be completed within 210 days is 0.99725
