Chapter 3, Problem 18PE

Explanation of Solution

Calculation of Expected Time (ET):

The ET is calculated using the following formula:

$\text{ET}\text{=}\frac{\text{Optimistic}\text{time}\text{+}\text{(4×Realistic Time)}\text{+}\text{Pesssimistic}\text{Time}}{\text{6}}$        (1)

For Task A,

Substitute, “3” for “Optimistic time”, “7” for “Realistic time” and “11” for “Pessimistic time” in the equation 1.

$\begin{array}{c}\text{ET}\text{=}\frac{\text{3}\text{+}\text{(4×7)}\text{+}\text{11}}{\text{6}}\\ \text{=}\text{7}\end{array}$

Therefore, the expected time for task A is “7”.

For Task B,

Substitute, “5” for “Optimistic time”, “9” for “Realistic time” and “13” for “Pessimistic time” in the equation 1.

$\begin{array}{c}\text{ET}\text{=}\frac{\text{5}\text{+}\text{(4×9)}\text{+}\text{13}}{\text{6}}\\ \text{=}\text{9}\end{array}$

Therefore, the expected time for task B is “9”.

For Task C,

Substitute, “1” for “Optimistic time”, “2” for “Realistic time” and “9” for “Pessimistic time” in the equation 1.

$\begin{array}{c}\text{ET}\text{=}\frac{\text{1}\text{+}\text{(4×2)}\text{+}\text{9}}{\text{6}}\\ \text{=}\text{3}\end{array}$

Therefore, the expected time for task C is “3”.

For Task D,

Substitute, “2” for “Optimistic time”, “3” for “Realistic time” and “16” for “Pessimistic time” in the equation 1.

$\begin{array}{c}\text{ET}\text{=}\frac{\text{2}\text{+}\text{(4×3)}\text{+}\text{16}}{\text{6}}\\ \text{=}\text{5}\end{array}$

Therefore, the expected time for task D is “5”.

For Task E,

Substitute, “2” for “Optimistic time”, “4” for “Realistic time” and “18” for “Pessimistic time” in the equation 1.

$\begin{array}{c}\text{ET}\text{=}\frac{\text{2}\text{+}\text{(4×4)}\text{+}\text{18}}{\text{6}}\\ \text{=}\text{6}\end{array}$

Therefore, the expected time for task E is “6”.

For Task F,

Substitute, “3” for “Optimistic time”, “4” for “Realistic time” and “11” for “Pessimistic time” in the equation 1