Question

In a PERT network, the time estimate of an activity is as per the following: Optimistic time is \(4\) days, most likely time is \(6\) days and pessimistic time is \(14\) days. The expected time and standard deviation of the activity (in days) are respectively

• A. \(6.0\) & \(2.76\)
• B. \(6.0\) & \(3.78\)
• C. \(7.0\) & \(1.67\)
• D. \(7.0\) & \(2.76\)

Hint:

PERT formulas: \[ \boxed{ t_e=\frac{a+4m+b}{6} } \] \[ \boxed{ \sigma=\frac{b-a}{6} } \] \[ \boxed{ \text{Variance}=\sigma^2=\left(\frac{b-a}{6}\right)^2 } \]

Answer 1

The Correct Option is C

Step 1: Use the PERT expected time formula. The expected time is \[ t_e=\frac{a+4m+b}{6}, \] where \[ a=4,\qquad m=6,\qquad b=14. \] Thus, \[ t_e = \frac{4+4(6)+14}{6} = \frac{42}{6} = 7\ \text{days}. \]

Step 2: Calculate the standard deviation. The standard deviation is \[ \sigma=\frac{b-a}{6}. \] Therefore, \[ \sigma = \frac{14-4}{6} = \frac{10}{6} = 1.67\ \text{days}. \] Hence, \[ \boxed{t_e=7.0\ \text{days},\qquad \sigma=1.67\ \text{days}} \] is the correct answer. Therefore, \[ \boxed{(C)} \] is the correct answer.