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Waiting line problem: Poisson and Exponential distributions

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Speedy oil provides a single channel automobile oil change and lubrication service. Customers provide an arrive rate of 2.5 cars per hour. The service rate is 5 cars per hour. Assume that arrivals follow a Poisson probability distribution and that service times follow an exponential probability distribution.

a) What is the average time of cars in the system ?

b) What is the average time that a car waits for oil and lubrication service to begin?

c) What is the average time a car spend in the system?

d) What is the probability that an arrival has to wait for service?

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Solution Summary

We solve a waiting line problem in which the arrivals follow a Poisson distribution and the service times follow an exponential distribution. No knowledge of waiting time theory is needed to understand the solution, everything is derived in the solution itself.

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The service time probability distribution is of the form:

P(t) = 1/tau Exp(-t/tau)

This means that the probability that the service time is between t and t + dt is P(t)dt for infinitesimal dt. The parameter tau is the average time it takes to service one car. You can verify this easily by calculating the integral of t*P(t)dt from t=0 to infinity. The service rate is 1/tau.

In this case the average time of cars in the system is thus 12 minutes.

The Poisson distribution has the form:

Q(n)= lambda^n/n! Exp(-lambda)

Here Q(n) denotes the probability of n events in some time interval and lambda is the average number of events in that time interval. In this problem we take the events to be the arrivals of customers. Let's denote the rate at which the customers are arriving by R. In this problem it is given that R = 2.5 hour^(-1). Long after the lubrication service starts operating a dynamical equilibrium will be achieved. The average rate at which cars are arriving will equal the rate at ...

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