Question

Question 1

A bank wishes to decide how many staff to schedule during the lunch period. During this period customers arrive at a rate of 9 per hour, according to a Poisson distribution. The enquiries that customers have (such as opening new accounts, arranging loans, etc.) take on average 15 minutes to deal with, these are exponential service times.

The bank manager wants to answer the following questions:

1.1 What is the arrival rate and the service rate for the single server model?                                         (2)

1.2 What service rate would be required so that customers averaged only 8 minutes in the system in the single server model?                                                                                                                                                 (4)

1.3 For that service rate, what is the probability of having more than four customers in the system for the single server model?                                                                                                                                 (5)

1.4 What service rate would be required to have only a 10 percent chance of exceeding four customers in the system for the single server model?                                                                                        (7)

1.5 The bank manager feels that four staff should be on duty during the lunch period but wants to make sure that the customers do not wait more than 3 minutes on average before they are served. Verify if four staff will be enough during the lunch period. Note that this is a multiple server model.

(7)

[25 marks]

Answer 1

1.1

Arrival rate, $\lambda$ = 9 per hour

Service rate, $\mu$ = 60/15 = 4 per hour

1.2

Target time in system (W) = 8 minutes = 8/60 = 0.133 hour

Average time in system, W = 1/($\mu$-$\lambda$)

1/($\mu$-9) = 0.133

$\mu$ = 1/0.133+9 = 16.5

Target service time = 16.5 per hour

1.3

Utilization rate, $\rho$ = $\lambda$/$\mu$ = 9/16.5 = 0.5455

Probability of having n customers in system, Pn = (1-$\rho$)*$\rho$ⁿ

P0 = (1-$\rho$)*$\rho$⁰ = (1-0.5455)*0.5455⁰ = 0.4545

P1 = (1-$\rho$)*$\rho$¹ = (1-0.5455)*0.5455¹ = 0.2479

P2 = (1-$\rho$)*$\rho$² = (1-0.5455)*0.5455² = 0.1352

P3 = (1-$\rho$)*$\rho$³ = (1-0.5455)*0.5455³ = 0.0738

P4 = (1-$\rho$)*$\rho$⁴ = (1-0.5455)*0.5455⁴ = 0.0402

Probability of having more than four customers in the system = 1 - (P0+P1+P2+P3+P4)

= 1 - (0.4545+0.2479+0.1352+0.0738+0.0402)

= 0.0483

1.4

With the service rate of 16.5, chance of exceed four customers in the system is only 4.83 %

By trial and error, the service rate can be reduced upto 14.3 per hour,

Utilization rate, $\rho$ = $\lambda$/$\mu$ = 9/14.3 = 0.629

Probability of having more than four customers in the system = 1 - (P0+P1+P2+P3+P4)

= 1 - (0.3706+0.2333+0.1468+0.0924+0.0582)

= 0.09875

~ 9.88 %