Instructions

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.

Question 33

The expected time to prepare the dish is the weighted average of optimistic, pessimistic and most likely time.

Which of the following is the expected wait time for the chef to communicate to the customers?

Solution

From the points given in the question, we can calculate the expected time for each activity.

The table is given below:

First activity A will be started, and then B, and C can be started simultaneously. E and H will be dependent on C.

Similarly, D, F, and G will be dependent on B.

D, F, and G will not depend on C, H, E, and vice versa.

I activity will be started after all the 8 activities are done.

A takes = 1 min

Then B, D, F, and G will take 16 minutes.

In these 16 minutes, C and E will also be completed simultaneously.

Hence, in the first (16+1) = 17 minutes, A, B, C, D, E, F, and G will be completed. Now H will take another 10.33 minutes. After that I will take 5 minutes.

Hence, the total time taken = (17+10.33+5) = 32.33 minutes


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