Math Is Fun Forum

rkruz

Member

Registered: 2009-07-06

Posts: 1

Simple exponential dist problem......need help

problem:
Service time is exponentially distributed with mean 30 minutes.
1st customer arrived at t=0 min.
2nd customer arrives at t=5 min
what is expected duration from arrival of 1st to departure of the 2nd customer?

Can you help provide some hints on how to solve this?

thank you.

juriguen

Member

Registered: 2009-07-05

Posts: 59

Re: Simple exponential dist problem......need help

I hope this can help, you can search for Queueing Theory in google or, directly, wikipedia:

"A Poisson process models random events (such as a customer arrival, a request for action from a web server, or the completion of the actions requested of a web server) as emanating from a memoryless process. That is, the length of the time interval from the current time to the occurrence of the next event does not depend upon the time of occurrence of the last event."

In Queueing Theory the customer inter-arrival time, as well as the service time, both can usually be modeled using the Exponential Distribution, which satisfies the memoryless property.

In any case, an Exponential Distribution is characterized by the following probability density function:

For this pdf you can calculate the mean, which is

In the problem you post, they already tell you what is the service mean time, 30 minutes. Since the inter-arrival time is independent from the service time, you don't need the information regarding when the customers arrive (0 and 5 minutes), but only to know there are in fact 2 customers in the server's queue, waiting to be attended.

The expected time between the departure of the first customer and of the second is equal to the mean of the exponential distribution, hence 30 minutes.

Jose

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“Make everything as simple as possible, but not simpler.” -- Albert Einstein

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Simple exponential dist problem......need help

It's a bit trickier than that, because the 2nd customer arrives 5 minutes after the first one.
If they both arrived at the same time then they'd both have an expected service time of 30 minutes and so the expected service time for the two of them is 1 hour.

However, the actual answer is slightly higher than this because the first customer could be served very quickly and in that case there would be a wait before the second customer arrives.

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Why did the vector cross the road?
It wanted to be normal.

juriguen

Member

Registered: 2009-07-05

Posts: 59

Re: Simple exponential dist problem......need help

You're right I was suspecting my solution was too easy!

Would then this formulation be correct?

The total time from the arrival of the 1st customer to the departure of the 2nd can be written as the following random variable:

Where T1 and T2 characterize the serving time of the 1st and 2nd customers.

Then we would have:

Hope it is correct now!

Jose

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“Make everything as simple as possible, but not simpler.” -- Albert Einstein