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I need some help to solve this problem:

There are two shops and each can serve at most one client per period. The number of clients that arrive in each period to each shop is

\begin{cases} 0 & \text{with probability} & p\\ 1 & \text{with probability} & 1-2p\\ 2 & \text{with probability} & p \end{cases}

where $p \in [0, 1/2]$. Clients arrivals to the two shops are independent.

Question 1: What is the expected number of clients the two shops combined serve per period?

Assume now the two shops can cooperate: they still can serve at most one client per period but if one of them is faced with 2 clients and the other has none, the overburdened shop can divert one of his clients to the other shop.

Question 2: Again, what is the expected number of clients they are able to serve per period?

What I come up with is the following:

Answer of question 1: Each shop serves $1$ client with probability $1-2p$ (when one comes) and $1$ client with probability $p$ (since when two comes, it can only serve one). This means each shop serves an expected number of clients equal to $1-2p + p = 1-p$. Since the two shops face the same arrival process, they combined serve on average $2(1-p)$ clients. Does it seem correct to you? If yes, do you have an alternative solution?

Answer of question 2: here I need your help! I have some ideas but I would like to see your opinion on how this can be solved!

Thanks!

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  • $\begingroup$ Your computation for the first part is correct. For the first part, independence is not relevant, but for the second it is, so you need to be clear about what you are assuming. Either way, case by case analysis should not be difficult. $\endgroup$
    – lulu
    Jun 30, 2020 at 15:47
  • $\begingroup$ Clients arrivals to the two shops are independent and each shop individually face the same arrival process that I have described. $\endgroup$
    – fennel
    Jun 30, 2020 at 16:00
  • $\begingroup$ You should edit your post to reflect that assumption. It's not obvious...I'd even say it was "unphysical" in the sense that, were this a real problem involving real shops, we'd expect "high/low volume times" for similar shops in similar locations to be highly correlated. $\endgroup$
    – lulu
    Jun 30, 2020 at 16:01
  • $\begingroup$ In any case, it's just a matter of case work, and there aren't very many cases. If you want to add in the answer you got, we could attempt to verify it for you. $\endgroup$
    – lulu
    Jun 30, 2020 at 16:02
  • $\begingroup$ yes you are right! My bad.. My try to answer question 2 is the following: Each shop serves $1$ client with probability $1-2p$ (when one comes), $1$ client with probability $p$ (when two comes) and $1$ client when the shop has $0$ clients but the other has $2$ with probability $p*p=p^2$. This means each shop serves an expected number of clients equal to $1-p+p^2$ and since they both face the same arrival process, they combined serve $2(1-p+p^2) $\endgroup$
    – fennel
    Jun 30, 2020 at 16:04

1 Answer 1

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Let $X_i$, $i=0,1$ be the number of customers that arrive at shop $i$ during some given period. Let $C_i$, $i=0,1$ be the number of customers served by shop $i$ during that period. For question 1, it is clear that $C_i = \min\{X_i, 1\}$, and hence \begin{align} \mathbb P(C_i=0) &= \mathbb P(X_i=0) = p\\ \mathbb P(C_i=1) &= \mathbb P(X_i>0) = 1-p. \end{align} It follows that $$ \mathbb E[C_i] = 0\cdot p + 1(1-p) = 1-p, $$ and so the expected number of customers served is $$ \mathbb E[C_1+C_2] = \mathbb E[C_1] + \mathbb E[C_2] = 2(1-p). $$ For question 2, we have $$ C_i = \min\{X_i,1\} + \mathsf 1_{\{X_i=0,X_{1-i}=2\}}, $$ and hence \begin{align} \mathbb P(C_i=0) &= p(1-p)\\ \mathbb P(C_i=1) &= 1-p(1-p). \end{align} So the expected number of customers served is $$ \mathbb E[C_1] + \mathbb E[C_2] = 2(1-p(1-p)). $$

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