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Chapter 6: Problem 14

Potential customers arrive at a full-service, one-pump gas station at a Poisson rate of 20 cars per hour. However, customers will only enter the station for gas if there are no more than two cars (including the one currently being attended to) at the pump. Suppose the amount of time required to service a car is exponentially distributed with a mean of five minutes. (a) What fraction of the attendant's time will be spent servicing cars? (b) What fraction of potential customers are lost?

Short Answer

Expert verified
a) The attendant spends \(\frac{15}{21}\) of their time servicing cars. b) \(\frac{25}{18}\) of potential customers are lost.

Step by step solution

01

Understand the given information

We are given that potential customers arrive at the gas station at a Poisson rate of 20 cars per hour. Also, there can only be 2 cars at the pump at any given time. The service time for each car is exponentially distributed with a mean of 5 minutes, or \(1/12\) hours since there are 12 5-minute intervals in 1 hour.
02

Find key parameters for the model

First, we need to find the arrival rate and service rate. A Poisson process is determined by its rate parameter, in this case, the arrival rate is given as 20 cars per hour, or \(\lambda = 20\). The service rate parameter can be calculated as the inverse of the mean service time, which is given as \(1/12\) hours. So, the service rate is \(\mu = 12\).
03

Utilize Little's formula to find the probability the pump is occupied

We can use the \(P0\) formula for calculating the probability that the pump is not occupied: \(P0 = \frac{1}{1 + \frac{\lambda}{\mu} + \frac{\lambda^2}{2\mu^2}}\) Plugging in the values for \(\lambda\) and \(\mu\): \(P0 = \frac{1}{1 + \frac{20}{12} + \frac{20^2}{2*12^2}} = \frac{1}{1 + \frac{5}{3} + \frac{25}{18}} = 6/21\) The probability that the pump is occupied is: \(P_\text{occupied} = 1 - P0 = 1- 6/21 = 15/21\)
04

Calculate the fraction of the attendant's time spent servicing cars

The fraction of the attendant's time spent servicing cars is the probability that the pump is occupied: Fraction of time servicing cars = \(P_\text{occupied} = 15/21\)
05

Calculate the fraction of potential customers lost

We need to find the probability that thet there are more than 2 cars at the pump, which would cause a potential customer to leave. The probability that there are more than 2 cars in the system is: \(P_\text{lost} = \frac{\lambda^2}{2\mu^2}\) Plugging in the values for \(\lambda\) and \(\mu\): \(P_\text{lost} = \frac{20^2}{2*12^2} = \frac{25}{18}\) Thus, the fraction of potential customers lost = \(\frac{25}{18}\) To summarize: a) The fraction of the attendant's time spent servicing cars is \(15/21\). b) The fraction of potential customers lost is \(\frac{25}{18}\).

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Most popular questions from this chapter

Consider a birth and death process with birth rates \(\lambda_{i}=(i+1) \lambda, i \geqslant 0\), and death rates \(\mu_{i}=i \mu, i \geqslant 0\) (a) Determine the expected time to go from state 0 to state 4 . (b) Determine the expected time to go from state 2 to state 5 . (c) Determine the variances in parts (a) and (b).

In the \(M / M / s\) queue if you allow the service rate to depend on the number in the system (but in such a way so that it is ergodic), what can you say about the output process? What can you say when the service rate \(\mu\) remains unchanged but \(\lambda>s \mu ?\)

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Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right) \operatorname{arcs}(i, j), i \neq j\), \(i, j,=1, \ldots, n .\) (See Section \(3.6 .2\) for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, j)\) according to independent Poisson processes with rates \(\lambda_{i j} .\) An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j, i, j=1, \ldots, n\). Let \(P_{j}\) denote the proportion of time that the particle is at node \(j\). Show that $$ P_{j}=\frac{1}{n} $$ Hint: Use time reversibility.

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