Question

In a PERT network for an activity Pessimistic, Most likely, and Optimistic times are 8 days, 6 days and 4 days respectively. The expected duration of the activity is _______

• A. 6 days
• B. 2 days
• C. 8 days
• D. 9 days

Hint:

Notice that the "most likely" time is given the most weight (a factor of 4) in the PERT formula.
This pulls the expected time towards the most probable outcome.
In this case, since the time estimates are symmetric around the most likely time (6-2=4, 6+2=8), the expected time is equal to the most likely time.

Answer 1

The Correct Option is A

Step 1: Recall why PERT uses a weighted average.
PERT recognises that most activities do not have a single fixed duration, so instead of guessing one number, it combines three estimates, an optimistic time $t_o$, a most likely time $t_m$, and a pessimistic time $t_p$, giving the most likely estimate four times the weight of the other two since it is considered the most probable outcome.
Step 2: Write down the given values.
Here $t_p = 8$ days, $t_m = 6$ days, and $t_o = 4$ days.
Step 3: Apply the weighted average formula.
\[ t_e = \frac{t_o + 4t_m + t_p}{6} = \frac{4 + 4(6) + 8}{6} = \frac{4 + 24 + 8}{6} = \frac{36}{6} \] Step 4: Simplify and conclude.
\[ t_e = 6 \text{ days} \] Notice this comes out equal to the most likely time in this particular case, since the optimistic and pessimistic estimates happen to be symmetric around it.
\[ \boxed{6\ \text{days}} \]