How Many Copies of a Magazine Should a Drugstore Order to Maximize Profit?

[Usagi]

The owner of a small drugstore is to order copies of a news magazine for the n potential readers among his customers. Customers act independently and each one of them will actually express interest in buying the news magazine with probability p. Suppose that
the store owner actually pays B for each copy of the news magazine, and the price to customers is C. If magazines left at the end of the week have no salvage value, what is the optimum number of copies to order?

So I'm not too sure on how to approach this question. I have several ideas but I am not sure on how to complete the question.

I think that the optimum number of copies to order would be the number such that the owner's expected profit is maximised, ie maximising E(C-B).

Now the owner has a probability p of making C and has probability of (1-p) of losing B since the customer won't buy anything.

Where can I go from here?

EDIT: Actually let's assume that the owner buys N copies and sells i copies thus he would want to maximise E(iC-NB) = iE(C) - NE(B), now how do I go about finding E(C) and E(B)?

 

[chisigma]

The owner of a small drugstore is to order copies of a news magazine for the n potential readers among his customers. Customers act independently and each one of them will actually express interest in buying the news magazine with probability p. Suppose that
the store owner actually pays B for each copy of the news magazine, and the price to customers is C. If magazines left at the end of the week have no salvage value, what is the optimum number of copies to order?

So I'm not too sure on how to approach this question. I have several ideas but I am not sure on how to complete the question.

I think that the optimum number of copies to order would be the number such that the owner's expected profit is maximised, ie maximising E(C-B).

Now the owner has a probability p of making C and has probability of (1-p) of losing B since the customer won't buy anything.

Where can I go from here?

EDIT: Actually let's assume that the owner buys N copies and sells i copies thus he would want to maximise E(iC-NB) = iE(C) - NE(B), now how do I go about finding E(C) and E(B)?

A very interesting question!... If n are the bought copies and k the sold copies, the the gain is $\displaystyle G(n)=K\ C - n\ B$. The expected number of k is...

$\displaystyle E\{k\}= (1-p)^{n}\ \sum_{k=1}^{n} n\ (n-1)...(n-k+1)\ \frac{1}{(k-1)!}\ (\frac{p}{1-p})^{k} $ (1)

... so that we obtain...

$\displaystyle G(n)= C\ (1-p)^{n}\ \sum_{k=1}^{n} n\ (n-1)...(n-k+1)\ \frac{1}{(k-1)!}\ (\frac{p}{1-p})^{k} - B\ n$ (2)

Now we have to maximize the discrete function G(n) respect to n and that requires a little more study...

Kind regards

$\chi$ $\sigma$

 

[Usagi]

Thanks for that chisigma, however could you expand a bit on how you arrived at (1)?

Cheers.

 

[chisigma]

Thanks for that chisigma, however could you expand a bit on how you arrived at (1)?

Cheers.

The probability of k events with probability p in n trials is...

$\displaystyle P_{n,k}= \frac{n\ (n-1)... (n-k+1)}{k!} (\frac{p}{1-p})^{k}\ (1-p)^{n}$ (1)

... so that is...

$\displaystyle E\{k\}= \sum_{k=1}^{n} k\ P_{n,k}= (1-p)^{n}\ \sum_{k=1}^{n} n\ (n-1)...(n-k+1)\ \frac{1}{(k-1)!}\ (\frac{p}{1-p})^{k}$ (2)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

A little more thinking and one remembers that for a binomial distribution is $\displaystyle E\{k\}= n\ p$ so that the function to be maximized is...

$\displaystyle E\{G\}= n\ (C\ p-B)$ (1)

It is clear now that the discrete function (1) cannot be maximized respect to n. It is possible however, because C and p are not independent, maximize (1) respect to C...

Kind regards

$\chi$ $\sigma$

 

[mfb1]

Let P(n,p,k) be the probability that k customers want to buy the magazine. We know that it is a binominal distribution and P(n,p,k)=\frac{(n-k)!\,k!}{n!} p^k (1-p)^{n-k}

Now, compare the situation "x magazines printed" and "x+1 magazines printed".
- If there are not more than x potential customers (let's call this probability Q), the sold copies are just limited by the number of customers. Printing one more costs B.
- If there are x+1 or more potential customers (probability 1-Q), the sold copies are just limited by the number of available magazines. Printing one more gives C-B.

Neglecting the discrete nature of the problem, the optimal point is Q*B=(1-Q)*(C-B) => Q=1-B/C and (1-Q)=B/C
Q=1-B/C is the optimal probability that not all magazines are sold.
(1-Q)=B/C is the optimal probability that all magazines are sold.
Given p and n, it is possible to sum over the individual cases (0, 1, ... customers) until this probability Q is reached. As far as I know, there is no closed form for this sum.

Check:
If B=0, buy everything you can get: Q=1, buy n magazines => correct
If B>C, do not buy anything: Q<0, buy 0 magazines => correct

 

[chisigma]

Let P(n,p,k) be the probability that k customers want to buy the magazine. We know that it is a binominal distribution and P(n,p,k)=\frac{(n-k)!\,k!}{n!} p^k (1-p)^{n-k}

Now, compare the situation "x magazines printed" and "x+1 magazines printed".
- If there are not more than x potential customers (let's call this probability Q), the sold copies are just limited by the number of customers. Printing one more costs B.
- If there are x+1 or more potential customers (probability 1-Q), the sold copies are just limited by the number of available magazines. Printing one more gives C-B.

Neglecting the discrete nature of the problem, the optimal point is Q*B=(1-Q)*(C-B) => Q=1-B/C and (1-Q)=B/C
Q=1-B/C is the optimal probability that not all magazines are sold.
(1-Q)=B/C is the optimal probability that all magazines are sold.
Given p and n, it is possible to sum over the individual cases (0, 1, ... customers) until this probability Q is reached. As far as I know, there is no closed form for this sum.

Check:
If B=0, buy everything you can get: Q=1, buy n magazines => correct
If B>C, do not buy anything: Q<0, buy 0 magazines => correct

Some minor details...

a) the probability to sell k copies over n is $\displaystyle P_{k,n}= \frac{n!}{k!\ (n-k)!}\ p^{k}\ (1-p)^{n-k}$ ...

b) the quantity to be maximized is $\displaystyle G=n\ (C\ p-B)$. The owner of the drugstore can only modify C and n [p is strongly dependent from c, if C=0 then p is close to 1 and if C=10.000 dollars then p is close to 0...]. You can do that however only if You know how C and p are related. Of course G cannot be maximized operating on n...

Kind regards

$\chi$ $\sigma$