Consider the following work breakdown structure: 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 What is the probability that the critical path for this project will be completed within 210 days

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