Question

For the network shown in the following figure, the time estimates in days for each activity are indicated. Determine the probability ( in %) of completing the project in 36 days. The non exceedance probability (P) for various values of the probability factor Z is as follows. The critical path is shown indicated below.

$\begin{array}{ccccc}\text{Z}& \text{1.0}& \text{1.1}& \text{1.2}& \text{1.3}\\ \text{( Pr(\%))}& \text{84.13}& \text{86.43}& \text{88.49}& \text{90.32}\end{array}$

• A. 89.94

Answer 1

The correct option is A 89.94
$t_E=\frac{t_0+4t_L+t_P}{6}$

$\sigma ={\left(\frac{t_p-t_o}{6}\right)}^2$

Variance calculation is given in the table below :
$\begin{array}{cccc}\text{Activity}& t_0-t_m-t_p& t_E& {\sigma }^2\\ \text{1-2}& \text{6-9-18}& \text{10}& \text{4}\\ \text{2-4}& \text{4-7-22}& \text{9}& \text{9}\\ \text{4-5}& \text{4-10-22}& \text{11}& \text{9}\end{array}$

Critical path = 1 - 2- 4-5

$T_E=30days;T_S=36days$

$Z=\frac{T_s-T_E}{\sigma }$

To calculate $\sigma$ :-

${\sigma }^2$ of critical path

${\sigma }^2=4+9+9=22$

$\sigma =\sqrt{22}=4.69$

$Z=\frac{36-30}{4.69}=1.2793$

From the table

$Z=1.2793\Rightarrow$ Probability = 89.94 %