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Chapter 6: Problem 13

A small barbershop, operated by a single barber, has room for at most two customers. Potential customers artive at a Poisson rate of three per hour, and the successive service times are independent exponential random variables with mean \(\frac{1}{4}\) hour. What is (a) the average number of customers in the shop? (b) the proportion of potential customers that enter the shop? (c) If the barber could work twice as fast, how much more business would he do?

Short Answer

Expert verified
In conclusion, for the given barbershop with a single barber and a maximum capacity of two customers: (a) The average number of customers in the shop is approximately 1.1794. (b) The proportion of potential customers that enter the shop is approximately 57.81%. (c) The barber would not do more business if he worked twice as fast; he would actually serve fewer customers.

Step by step solution

01

Identify the arrival rate (位) and service rate (渭)

The problem states that potential customers arrive at a Poisson rate of 3 per hour, so 位 = 3. The service time is exponentially distributed with a mean of 1/4 hour, therefore the service rate is the reciprocal of the mean service time: 渭 = 1/(1/4) = 4.
02

Calculate the utilization factor (蟻)

The utilization factor (蟻) represents the proportion of potential customers who enter the shop. It is calculated by dividing the arrival rate (位) by the service rate (渭): 蟻 = 位/渭 = 3/4 = 0.75
03

Calculate the probability of staying in the shop P_2

The probability of staying in the shop P_2 is calculated by the formula: P_2 = 蟻^3 * (1-蟻) / (1-蟻^3) = (0.75^3 * (1-0.75)) / (1-0.75^3) 鈮� 0.4219
04

Calculate the average number of customers in the shop (L)

The average number of customers (L) in the shop can be calculated using the formula: L = 蟻^2 + P_2 = 0.75^2 + 0.4219 鈮� 1.1794 Thus, the average number of customers in the shop is approximately 1.1794.
05

Calculate the proportion of potential customers that enter the shop

The proportion of potential customers that enter the shop is 1 - P_2: 1 - 0.4219 鈮� 0.5781 So, approximately 57.81% of potential customers actually enter the shop.
06

Calculate the increased service rate (渭') and customer increase if the barber could work twice as fast

If the barber could work twice as fast, the new service rate (渭') would be: 渭' = 2 * 渭 = 2 * 4 = 8 Now, we need to calculate the new utilization factor (蟻'): 蟻' = 位/渭' = 3/8 = 0.375 Next, we need to calculate the new probability of staying in the shop P_2': P_2' = 蟻'^3 * (1-蟻') / (1-蟻'^3) = (0.375^3 * (1-0.375)) / (1-0.375^3) 鈮� 0.1469 Now, we calculate the new average number of customers in the shop (L'): L' = 蟻'^2 + P_2' = 0.375^2 + 0.1469 鈮� 0.2899 Finally, we calculate the increase in the number of customers: Increase = L' - L 鈮� 0.2899 - 1.1794 鈮� -0.8895 Since the increase in the number of customers is negative, the barber would actually serve fewer customers if he worked twice as fast, as the shop would have a larger capacity, and fewer potential customers would enter. In conclusion: (a) The average number of customers in the shop is approximately 1.1794. (b) The proportion of potential customers that enter the shop is approximately 57.81%. (c) The barber would not do more business if he worked twice as fast; he would actually serve fewer customers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson Process
A Poisson process is a statistical model that describes a series of events occurring randomly over a fixed period of time or space. This process is particularly useful for modeling scenarios where events happen independently of one another and at a constant average rate.

In the context of the barbershop example from the exercise, potential customers arriving at the shop can be modeled as a Poisson process. Here, customers arrive randomly but with a predictable average rate, which in this case is three customers per hour. This arrival rate, denoted by the symbol \(\boldsymbol{\lambda}\) (lambda), is a key parameter in analyzing and understanding queueing systems like the barbershop.
Exponential Distribution
The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, which in our example refers to the service times between customers being serviced in the barbershop.

This distribution is characterized by a single parameter, the rate \(\boldsymbol{\mu}\) (mu), which is the reciprocal of the mean service time. So, if the mean service time is \(\frac{1}{4}\) hour, the service rate \(\boldsymbol{\mu}\) is 4 per hour. The memoryless property of the exponential distribution implies that the likelihood of a service being completed is the same at any moment, regardless of how long the service has already been in progress.
Service Rate
The service rate, symbolized as \(\boldsymbol{\mu}\), represents the capability of the service system鈥攊n this case, the barber鈥攖o complete a service in a given time frame. In simple terms, it's the number of customers the barber can serve per hour.

Knowing the service rate is essential for predicting how quickly customers can be processed through the system. If the mean service time is \(\frac{1}{4}\) hour, then the service rate would be the inverse, 4 customers per hour. It is imperative for queueing theory analysis to determine the balancing point between the customer arrival rate and the service rate to predict queue behavior and system performance.
Utilization Factor
The utilization factor, often represented by \(\boldsymbol{\rho}\) (rho), is a measure of how much of the system's capacity is being used. In other words, it's the fraction of time the service provider is actively serving customers.

In our barbershop scenario, \(\boldsymbol{\rho}\) is calculated by dividing the arrival rate (\(\boldsymbol{\lambda}\)) by the service rate (\(\boldsymbol{\mu}\)), resulting in a utilization factor of 0.75. This means that the barber is occupied 75% of the time on average. When the utilization factor is close to 1, the system is nearly at full capacity, leading to longer wait times and potential for increased queue length.

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Most popular questions from this chapter

Potential customers artive 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?

If \([X(t)]\) and \([Y(t))\) are independent continuous-time Markov chains, both of which are time reversible, show that the process \([X(t), Y(t)]\) is also a time reversible Markov chain.

There are \(N\) individuals in a population, some of whom have a certain infection that spreads as follows. Contacts between two members of this population occur in accordance with a Poisson process having rate \(\lambda .\) When a contact occurs, it is equally likely to involve any of the \(\left(\begin{array}{c}N \\ 2\end{array}\right)\) pairs of individuals in the population. If a contact involves an infected and a noninfected individual, then with probability \(p\) the noninfected individual becomes infected. Once infected, an individual remains infected throughout. Let \(X(t)\) denote the number of infected members of the population at time \(t\). (a) Is \(\\{X(t), t \geq 0\) a continuous-time Markov chain? (b) Specify the type of stochastic process. (c) Starting with a single infected individual, what is the expected time until all members are infected?

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed. At that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

After being repaired, a machine functions for an exponential time with rate \(\lambda\) and then fails. Upon failure, a repair process begins. The repair process proceeds sequentially through \(k\) distinet phases. First a phase 1 repair must be performed, then a phase 2, and so on. The times to complete these phases are independent, with phase \(I\) taking an exponential time with rate \(\mu_{i}, i=1, \ldots, k\).
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