Question

There is an 82 percent chance the project below can be completed in X weeks or less. What is X?

Short answer

The value of X is 18 weeks.

Step-by-step solution

The network diagram

The network diagram for the given table is:

The expected time

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

Activity	Most Optimistic	Most Likely	Most Pessimistic	Expected time
A	2	5	11	=(B2+4*C2+D2)/6
B	3	3	3	=(B3+4*C3+D3)/6
C	1	3	5	=(B4+4*C4+D4)/6
D	6	8	10	=(B5+4*C5+D5)/6

Activity	Most Optimistic	Most Likely	Most Pessimistic	Expected time
A	2	5	11	5.5
B	3	3	3	3
C	1	3	5	3
D	6	8	10	8
E	4	7	10	7

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.

The value of X

We know that, to find the variance we can use

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

Activity	Most Optimistic	Most Likely	Most Pessimistic	Variance
A	2	5	11	=((D2-B2)/6)^2
B	3	3	3	=((D3-B3)/6)^2
C	1	3	5	=((D4-B4)/6)^2
D	6	8	10	=((D5-B5)/6)^2

Activity	Most Optimistic	Most Likely	Most Pessimistic	Variance
A	2	5	11	2.25
B	3	3	3	0
C	1	3	5	0.44444444
D	6	8	10	0.44444444
E	4	7	10	1

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

Value of 82% using NORMSINV function in excel:

=NORMSINV(0.82)

0.915365

We know that Expected time = X

Scheduled time = 16.5

Standard deviation = 2.69

Probability distribution = 0.915

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