Common Data for Question 01 & 02
Consider the following network
The optimistic time most likely time and pessimistic time of all the activities are given in the table below:
|
Activity
|
To
|
Tm
|
Tp
|
Te = $\frac{T_{o}+4T_{m}+T_{\rho }}{6}$
|
$\sigma = \frac{T_{\rho }-T_{o}}{6}$
|
$\sigma ^{2}=\left ( \frac{T_{\rho }-T_{o}}{6} \right )^{2}$
|
|
1 – 2
|
1
|
2
|
3
|
2
|
1/3
|
1/9
|
|
1 – 3
|
5
|
6
|
7
|
6
|
1/3
|
1/9
|
|
1 – 4
|
3
|
5
|
7
|
5
|
2/3
|
4/9
|
|
2 – 5
|
5
|
7
|
9
|
7
|
2/3
|
4/9
|
|
3 – 5
|
2
|
4
|
6
|
4
|
2/3
|
4/9
|
|
5 – 6
|
4
|
5
|
6
|
5
|
1/3
|
1/9
|
|
4 – 7
|
4
|
6
|
8
|
6
|
2/3
|
4/9
|
|
6 – 7
|
2
|
3
|
4
|
3
|
1/3
|
1/9
|
- The critical path duration of the network (in days) is
(a) 11 (b) 14
(c) 17 (d) 18
2. The standard deviation of the critical path is
(a) 0.33 (b) 0.55
(c) 0.77 (d) 1.66