Thread: exponentials

In a hardware store you must first go to server 1 to get your goods and then to server 2 to pay for them. Suppose that the times for the times for the two activities are exponentially distributed with means 6 min and 3 min. Compute the average amount of time it takes Bob to get his goods and pay if when he comes in there is a customer named Al with server 1 and no one at server 2.

I remember doing this problem in a stochastic models class (although I don't think it was a hardware store and I don't think the customers had names).

Here's what I did:


Let = the total amount of time Bob spends in the hardware store

Let = the amount of time Bob spends waiting in line for Al to get his goods from server 1

Let = the amount of time it takes for Bob to get his goods from server 1

Let = the amount of time Bob spends waiting in line for Al to pay server 2 for his goods

and Let = the amount of time it takes for Bob to pay server 2 for his goods


Then

and


because of memoryless property of the exponential distribution





To find , condition on the availability of server 2 after Bob is done with server 1





(I used the probability than one exponential random variable is smaller than another exponential random variable.)


and


so E[T] = 6+6+1+3= 16 minutes

Thanks so much. I tried to construct something similar, but got lost around E(T3). And your final answer matches with the back of my book btw.

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