Question

Submit your calculation/answer for Part 1 and MATLAB code/result/answer for Part 2 below:

Servicing Customers A supermarket you work part-time at has one express lane open from 5 to 6 PM on weekdays (Monday through Friday). This time of the day is usually the busiest since people tend to stop on their way home from work to buy groceries. The number of items allowed in the express lane is limited to 10 so that the average time to process an order is fairly constant at about 1 minute. The manager of the supermarket notices that there is frequently a long line of people waiting and hears customers grumbling about the wait. To improve the situation he decides to open additional express lanes during this time period. If he does, however, he will have to "pull" workers from other jobs around the store to serve as cashiers. Hence, he is reluctant to open more lanes than necessary. Knowing that you are a college student studying probability, your manager asks you to help him decide how many express lanes to open. His requirement is that there should be no more than one person waiting in line 95% of the time. With the task at hand, you set out to study the problem first. You start by counting the number of customer arrival in the express lane on a Monday from 5 to 6pm. There are a total of 81 arrivals. You repeat the experiment on the following four days (Tuesday through Friday) and note the total arrivals of 68, 72, 61 and 66 customers, respectively.

Part 1: Analysis (2% of final grade) In order to solve the problem, you decide to answer the following set of questions:

1)

2)Ans= 1.16

3) Ans = 67.71%

4) Ans= 96.53%

Part 2: Simulation (2% of final grade)

Before telling your manager your recommendation, you decide to simulate the problem first to verify your solution:

1) You decide to approximate the customer arrival process as follows. You treat each one-second interval as a Bernoulli trial. Assign it to be a one, if there is a customer arrives during that interval, zero if no customer arrives.

2) You count the number of customers arrives during a one-minute interval.

3) You count the total number of minutes out of a one-hour period that have two or fewer customers arrive. Does this number give your probability close to your calculation in Part 1 Prob 3?

4) Now based on your answer to Part 1 Prob 4, assign the arrivals in Part 2 Prob 1 with equal probabilities to the number of express lanes you recommend.

5) You count the number of customers arrives at each lane during a one-minute interval.

6) You count the total number of minutes out of a one-hour period that all lanes have two or fewer customers arrive. Does this number give you probability close to your calculation in Part 1 Prob 4?

Done in MatLab please.

Answer 1

• ANSWER FOR 1:

Average number of customer arrivals at the express lane from

1. to 6pm on weekdays = (81+68+72+61+66)/5 = 69.6

• ANSWER FOR 2:

          Average number of customer arrivals

• in a one-minute interval = 69.6/60 = 1.16

for the number of customers PMF arrived during a one-minute interval in this period

               P(X=k) = e-1.16 * 1.16k / (k!)

• ANSWER FOR 3:
• P(X<=2) = P(X=0) + P(X=1) + P(X=2)

                             = 0.3135 + 0.3636 + 0.2109

                             = 0.888

• for about 88.8% of the time, here TWO or fewer customers in queue

• P(X<=1) = P(X=0) + P(X=1)

                               = 0.3135 + 0.3636

                              = 0.6771

• there are 1 or lesser persons in queue about 67.71% of the time.
• This does not satisfy the manager's requirement of no more than one person waiting in line 95% of the time.

• ANSWER FOR 4:
• If there are 2 lanes, average number of persons in a one-minute interval = 1.16/2 = 0.58

• P(X<=1) = P(X=0) + P(X=1)

                        = 0.5599 + 0.3247

                       = 0.8846

• If there are 3 lanes, average number of persons in a one-minute interval = 1.16/3 = 0.3866

• P(X<=1) = P(X=0) + P(X=1)

= 0.6793 + 0.2627

= 0.942

• If there are 4 lanes, average number of persons in a one-minute interval = 1.16/4 = 0.29

• P(X<=1) = P(X=0) + P(X=1)
• = 0.7483 + 0.217
• = 0.9653

So about 96.53% of the time, there are 1 or fewer persons in queue

• So to satisfy the manager's requirement of of no more than one person waiting in line 95% of the time, there should be 4 lanes.