Question

A supermarket has two exponential checkout counters, each operating at rate \(\mu .\) Arrivals are Poisson at rate \(\lambda\). The counters operate in the following way: (i) One queue feeds both counters. (ii) One counter is operated by a permanent checker and the other by a stock clerk who instantaneously begins checking whenever there are two or more customers in the system. The clerk returns to stocking whenever he completes a service, and there are fewer than two customers in the system. (a) Find \(P_{n}\), proportion of time there are \(n\) in the system. (b) At what rate does the number in the system go from 0 to \(1 ?\) From 2 to \(1 ?\) (c) What proportion of time is the stock clerk checking? Hint: Be a little careful when there is one in the system.

Short answer

The proportion of time there are \(n\) customers in the system is given by \(P_n\). To calculate these probabilities, we use the following relationships: - \(P_1 = \frac{λ}{µ}P_0\) - \(P_2 = \frac{λ}{µ}(P_1 + P_0)\) - \(P_{n+1} = \frac{λ}{2µ}P_n, n\ge 2\) After normalizing, we find \(P_0 = \left[\frac{1 - \frac{λ}{2µ}}{1 - \frac{λ}{µ}}\right]^{-1}\), and we can calculate all the \(P_n\)s using this value. The rate at which the number of customers in the system goes from 0 to 1 is given by \(λP_0\), while the rate from 2 to 1 is \(2µP_2\). The proportion of time the stock clerk is checking is calculated as the sum of probabilities when there are two or more customers in the system, resulting in \(\sum_{n=2}^{\infty} P_n = 1 - P_0 - P_1\).

Step-by-step solution

Analyze Queue System Dynamics

Let's first describe the queue dynamics through the following probabilities and rates: - \(p_n\): the probability there are \(n\) customers in the system - \(λ\): arrival rate of customers (Poisson) - \(µ\): service rate of each counter (exponential) For \(n \ge 2\), we have two counters operating. Therefore, the service rate for each of these states is \(2µ\). Now we can analyze the probabilities in a step-by-step fashion.

The probability of No Customers in the system (\(P_0\))

In a stable system, the arrival rate equals the service rate. When there are no customers, the only possible state change is an arrival. Thus, - \(λP_0 = µP_1\) - \(P_1 = \frac{λ}{µ}P_0\)

The probability of One Customer in the system (\(P_1\))

When there is one customer in the system, the permanent checker provides the service, and the stock clerk goes back to stocking. The state changes can only occur due to the arrival of a new customer or the departure of the one being served. Thus, - \(λP_1 = µP_2 + (λ - µ)P_1\) - \(P_2 = \frac{λ}{µ}(P_1 + P_0)\)

The probability of Two or More Customers in the system (\(P_n, n\ge 2\))

When there are two or more customers in the system (\(n\ge 2\)), both the checkers provide service, so the service rate becomes \(2µ\). Now we can set up the following balance equations: - \(λP_n = 2µP_{n+1} + (λ - 2µ)P_n\) for \(n\ge 2\) - \(P_{n+1} = \frac{λ}{2µ}P_n, n\ge 2\)

Calculate the Rate of Transitions

(a) We need to use the normalization condition \(\sum_{n=0}^{\infty}P_n = 1\). After substituting the equations from step 2, 3 and 4, - \(P_0 \left(1 + \frac{λ}{µ} + \frac{λ^2}{µ^2} + \frac{λ^3}{(2µ)^2} + \frac{λ^4}{(2µ)^3} + \cdots\right) = 1\) - \(P_0 \left(\frac{1}{1 - \frac{λ}{µ}} + \frac{λ^2}{µ^2}\frac{1}{1 - \frac{λ}{2µ}}\right) = 1\) - \(P_0 = \left[\frac{1 - \frac{λ}{2µ}}{1 - \frac{λ}{µ}}\right]^{-1}\) Now, we can calculate all the \(P_n\)s by substituting \(P_0\). (b) Rate at which the number of customers in the system goes from 0 to 1: \(λP_0\) - Rate at which the number of customers in the system goes from 2 to 1: \(2µP_2\) (c) Finally, the proportion of time the stock clerk is checking is given by the sum of probabilities when there are two or more customers (\(n\ge 2\)): - \(\sum_{n=2}^{\infty} P_n = 1 - P_0 - P_1\)