Consider the network described in the table below.
Activity |
Immediate Predecessor(s) |
Pessimistic |
Probable |
Optimistic |
J |
-- |
15 |
10 |
8 |
K |
-- |
9 |
8 |
7 |
L |
J |
10 |
6 |
5 |
M |
J |
3 |
3 |
3 |
N |
K,M |
9 |
5 |
1 |
O |
K,M |
10 |
7 |
4 |
P |
L,N |
10 |
8 |
3 |
a. Calculate the expected duration of each activity.
b. Calculate the expected duration and variance of the critical path.
c. Calculate the probability that the project will be completed in fewer than 30 time units.
d. What time corresponds to when the top 30% of those complete the project?
a. Expected activity duration = pessimistic duraion + 4x probable duration + optimistic duration /6
Duration of J = 15+40+8/6 = 10.5
Duration of K =9+32+7 /6 =8
similarly the durations of activities
Activity | duration |
J | 10.5 |
K | 8 |
L | 6.5 |
M | 3 |
N | 5 |
O | 7 |
P | 7.5 |
b.
Critical path is JMNP
Variance = ( optimistic time -pessimistic time/6)^{2}
Variance of J = ( 15-8/6)^{2} = 1.36
Variance of M =0
Variance of N = (1-9/6)^{2} = 1.77
Variance of P = (3-10/6)^{2} = 1.36
Total variance of critical path = 1.36+0+1.77+1.36 =4.49
Standard deviation = ( variance)^{1/2} = 2.11
c.
Probability of completion of project before 30 time units
z = x-mean / sd
z =30-26/2.11 =1.89
which corresponds to 97.06% probability
d. Let this time be t
now z = t-26/2.11
z corresponding to 30% is -0.52
-0.52 = t-26 /2.11
which gives t = 24.9
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