÷ÈÓ°Ö±²¥

The network diagram for the given table is:

$\text{Expected} \text{time} \text{=} \frac{\text{a + 4m + b}}{\text{6}}$

Now, we can find the expected path durationas follow:

$$\begin{gathered} \text{A - B - D} \text{=} \text{5.5 + 3 + 8} \text{=} \text{16.5} \\ \text{A - C - E} \text{=} \text{5.5 + 3 + 7} \text{=} \text{15.5} \end{gathered}$$

The longest time of 16.5 weeks belongs to the path A-B-D and this path would be the critical pathof the project.

We know that, to find the variance we can use

$Variance={\left(\frac{b-a}{6}\right)}^2$

$$\begin{gathered} \text{Total} \text{variance} \text{=} \text{2.25 + 0 + 0.44} \\ \text{=} \text{2.69} \end{gathered}$$

Value of 82% using NORMSINV function in excel:

We know that Expected time = X

Scheduled time = 16.5

Standard deviation = 2.69

Probability distribution = 0.915

$$\begin{gathered} \text{Now,} \text{probability} \text{distribution} \text{=} \frac{\text{Expected} \text{time}-\text{Scheduled} \text{time}}{\text{Standard} \text{deviation}} \\ \text{0.915} \text{=}\frac{\mathrm{χ}-16.5}{\sqrt{2.69}} \\ \text{X-16.5=0.915x1.64} \\ \mathrm{χ}=1.5+16.5 \\ \mathrm{χ}=18 \end{gathered}$$