Exponential Waiting times with two queues

Question

I am trying to understand the applications of exponential waiting times when there are two queues.

Let there be two counters in a mall, the first counter $X$ (where the order is placed), and the second counter $Y$ (where the item is collected). When a customer arrives, they first go to counter $X$ to order, and then go to counter $Y$ to collect the item.

Also, serving times at each counter are independent and Exponential, with parameter $λ1$ at counter $X$, and parameter $λ2$ at counter $Y$.

If someone is already being served at a counter, the next customer waits in a queue.

For simplicity, I start with the following. I assume that when I arrive, there is one customer at counter $X$, and no customer at counter $Y$ $($i. e., counter $Y$ is free when I arrive$)$. How can I calculate the expected time I have to wait before collecting my order?

I have worked out the following:

$E($time I have to wait to collect my order$)$

= $E($time taken to serve the previous customer who's at counter $X$ when I arrive$)$ $+$ $E[$max $($$Tx$, $Ty$$)$$]$ $+$ $E($time taken to serve me at counter $Y$$)$

= $E(Tx)$ $+$ $(1/$$λ1$$)$ $+$ $(1/$$λ2$$)$ $-$ $(1/$$($$λ1$ + $λ2$$)$$)$ $+$ $E(Ty)$

The $2$nd, $3$rd and $4$th terms in the above sum are what I get for $E[$max $($$Tx$, $Ty$$)$$]$. I am using the $E[$max $($$Tx$, $Ty$$)$$]$ because, when the previous customer is finished being served at counter $X$, the previous customer goes to counter $Y$, and I will start being served at counter $X$.

So here, there are 2 possibilities: $Tx$ > $Ty$, OR $Ty$ > $Tx$

$Tx$ ~ Exponential $($$λ1$$)$ and $Ty$ ~ Exponential $($$λ2$$)$.

Am I on the right track here? Or have I missed something or not understanding it correctly?

Any insights are most appreciated. Thank you so much!

Answer 1

For this particular question, there are some shortcuts using the memoryless property of exponential distributions.

• You have to wait an expected time $\frac{1}{\lambda_1}$ for the other person to finish being served at the first counter
• Then you have to wait an expected time $\frac{1}{\lambda_1}$ to be served yourself at the first counter
• Then you may or may not need to wait for the other person to finish being served at the second counter; you have to wait with probability $\frac{\lambda_1}{\lambda_1+\lambda_2}$ for an expected time $\frac{1}{\lambda_2}$, so combined an expected $\frac{\lambda_1}{\lambda_2(\lambda_1+\lambda_2)}$
• Finally you have to wait an expected time $\frac{1}{\lambda_2}$ to be served yourself at the second counter

Add these up: $\frac{2}{\lambda_1}+\frac{\lambda_1}{\lambda_2(\lambda_1+\lambda_2)} + \frac{1}{\lambda_2}$