Question

Piotr Banasiewicz is a self-employed plumber who offers a 24-hour emergency service. For most of the year, calls arrive randomly at a rate of six a day. The time he takes to travel to a call and do the repair is randomly distributed with a mean of 90 minutes. Forecasts for February suggest cold weather and last time this happened Piotr received emergency calls at a rate of 18 a day. Because of repeat business, Piotr is anxious not to lose a customer and wants the average waiting time to be no longer in February than during a normal month. How many assistants should he employ to achieve this? • Assume a Single-Server model. • Ignore the stated Repair Time of 90 minutes. Instead, assume theRepair Time is 80 minutes. • Ignore the Arrival Rate of 6 per day. Instead, assume the Arrival Rateis 7 per day • In addition to the questions in the stated problem, determine the following: o Average server utilization o Average number of customers in the queue (Lq) o Average number of customers in the system (L) o Average waiting time in the queue (Wq) – express in days, hoursand minutes. o Average waiting time in the system (W) ) – express in days, hoursand minutes. o Probability there is 0 units in the system (P0) o Probability of "n" units in system with n=0 to up to n=20. o Probability of more than "n" units in system with n=0 to up to n=20. • Ignore the “How many assistants should he employ to achieve this?” question. • Instead, determine what the minimum Service Rate in customers per day should be for February, in order to maintain or improve the average waiting time (Wq) for the rest of the year. Please give step by step instructions on excel with formulas on how to solve.

Answer 1

Average server utilization = 7*80 / 24*60 = 38.89%

Average number of customers in the queue (Lq) = 7 - 1 = 6

Average number of customers in the system (L) = 7

Average waiting time in the queue (Wq) – express in days, hoursand minutes = 80/person

Average waiting time in the system (W) ) – express in days, hoursand minutes = 80*6 = 480 minutes

Probability there is 0 units in the system (P0) = 1/7 (when last person is not in the queue)

For the above question we will apply M/M/1 queuing model.

As it is stated that it it single server model.So he can't employ any worker. He should be only server.   

The following formula will be used

1. Average number of customers in the system

2. Average number of customers in the waiting line

3. Average waiting time in the system

4. Average waiting time in the queue

5. Utilization Factor

6. Probability of having n customers in the service system

Case 1)

Poisson call arrivals( = 6 per day) = = 0.0042 per minute .

Exponentially distributed Service time with rate( = 90 per minutes)= = 90

Case 2)

Poisson call arrivals( = 7 per day) = = 0.0049 per minute .

Exponentially distributed Service time with rate( = 90 per minutes)= = 80

Case 3) The minimum Service Rate in customers per day should be for February, in order to maintain or improve the average waiting time (Wq) for the rest of the year.

In february,Poisson call arrivals( = 18 per day) = = 0.0042 per minute , we have to estimate in order to maintain or improve . So,

The minimum Service rate should be 1.00212 in minutes.