Question

Direction: Go through the following scenario and answer the THREE questions that follow.
To prepare a dish (e.g., Dosa- Sambhar, Idli-chutney, Rajma-Chawal, Mawa-Bati), the chef has to finish nine activities, some of which could be done simultaneously, while others could not be done simultaneously (see diagram). One of the challenges faced by the chef was to precisely calculate the preparation time of a dish and communicate the waiting time to the customers.
However, based on the past data, the chef had an idea about approximate time taken to complete each activity. He had noted down the best (optimistic), worst (pessimistic) and most likely (most commonly observed) time to finish each of the nine activities. Further, the chef realised that frequency of occurrence of most likely time was 66.666%, and the frequency of occurrence of pessimistic and optimistic times were 16.666% each. The diagram below shows the activities involved and the table shows the optimistic, pessimistic, and most likely times for each activity. Time is indicated in minutes in the table below.
Which of the following is the MOST volatile activity for the chef?

• A. H
• B. B
• C. G
• D. E
• E. C

Answer 1

The Correct Option is C

To determine the MOST volatile activity, we need to evaluate the variance of each activity using the given optimistic, pessimistic, and most likely times. This is often done using the Program Evaluation and Review Technique (PERT) formula.

Pert Formula for Variance:

The variance of an activity in PERT is calculated as:

\(\text{Variance} = \left(\frac{\text{Pessimistic} - \text{Optimistic}}{6}\right)^2\)

Let's calculate the variance for each activity:

• Activity A:
  • Optimistic = 1, Pessimistic = 1
  • Variance = \(\left(\frac{1 - 1}{6}\right)^2 = 0\)
• Activity B:
  • Optimistic = 2, Pessimistic = 3
  • Variance = \(\left(\frac{3 - 2}{6}\right)^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36}\)
• Activity C:
  • Optimistic = 10, Pessimistic = 13
  • Variance = \(\left(\frac{13 - 10}{6}\right)^2 = \left(\frac{3}{6}\right)^2 = \frac{1}{4}\)
• Activity D:
  • Optimistic = 5, Pessimistic = 5
  • Variance = \(\left(\frac{5 - 5}{6}\right)^2 = 0\)
• Activity E:
  • Optimistic = 3, Pessimistic = 6
  • Variance = \(\left(\frac{6 - 3}{6}\right)^2 = \left(\frac{3}{6}\right)^2 = \frac{1}{4}\)
• Activity F:
  • Optimistic = 1, Pessimistic = 1
  • Variance = \(\left(\frac{1 - 1}{6}\right)^2 = 0\)
• Activity G:
  • Optimistic = 5, Pessimistic = 10
  • Variance = \(\left(\frac{10 - 5}{6}\right)^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\)
• Activity H:
  • Optimistic = 9, Pessimistic = 13
  • Variance = \(\left(\frac{13 - 9}{6}\right)^2 = \left(\frac{4}{6}\right)^2 = \frac{4}{9}\)
• Activity I:
  • Optimistic = 5, Pessimistic = 5
  • Variance = \(\left(\frac{5 - 5}{6}\right)^2 = 0\)

By comparing the variances, Activity G has the highest variance of \(\frac{25}{36}\), making it the MOST volatile activity.

Conclusion: The correct answer is \(G\).

Answer 2

To determine the most volatile activity, we need to assess the variability of each activity using the following formula for the standard deviation (SD) in the PERT (Program Evaluation and Review Technique) method: SD = (Pessimistic time - Optimistic time) / 6. This provides a measure of uncertainty or volatility in the time estimate for each activity.

Activity	Optimistic Time (O)	Most Likely Time (M)	Pessimistic Time (P)
A	3	6	15
B	3	4	18
C	3	4	8
D	4	5	10
E	3	5	12
F	3	4	14
G	3	9	27
H	2	4	8
I	6	9	18

By applying the formula:

• SD(A) = (15 - 3) / 6 = 2
• SD(B) = (18 - 3) / 6 = 2.5
• SD(C) = (8 - 3) / 6 = 0.833
• SD(D) = (10 - 4) / 6 = 1
• SD(E) = (12 - 3) / 6 = 1.5
• SD(F) = (14 - 3) / 6 = 1.833
• SD(G) = (27 - 3) / 6 = 4
• SD(H) = (8 - 2) / 6 = 1
• SD(I) = (18 - 6) / 6 = 2

The highest standard deviation is for Activity G with a value of 4. Therefore, Activity G is the most volatile for the chef.