# exponentials

• Mar 7th 2010, 04:33 PM
Anonymous1
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.
• Mar 7th 2010, 06:00 PM
Random Variable
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 $T$ = the total amount of time Bob spends in the hardware store

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

Let $T_{2}$ = the amount of time it takes for Bob to get his goods from server 1

Let $T_{3}$ = the amount of time Bob spends waiting in line for Al to pay server 2 for his goods

and Let $T_{4}$ = the amount of time it takes for Bob to pay server 2 for his goods

Then $T= T_{1}+T_{2}+T_{3}+T_{4}$

and $E[T] = E[T_{1}]+E[T_{2}]+E[T_{3}]+E[T_{4}]$

$E[T_{1}] = 6$ because of memoryless property of the exponential distribution

$E[T_{2}] = 6$

To find $E[T_{3}]$, condition on the availability of server 2 after Bob is done with server 1

$E[T_{3}] = E[T_{3}|\text{server 2 is free}]P(\text{server 2 is free}) + E[T_{3}|\text{server 2 is not free}]$ $P(\text{server 3 is not free})$

$= 0 + 3 \cdot P(\text{Bob is done with server 1 before Al is done with server 2})$

$= 3 \cdot \frac{1/6}{1/6+1/3} = 1$ (I used the probability than one exponential random variable is smaller than another exponential random variable.)

and $E[T_{4}] = 3$

so E[T] = 6+6+1+3= 16 minutes
• Mar 7th 2010, 06:04 PM
Anonymous1
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.
• Mar 7th 2010, 06:34 PM
Random Variable
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