Threshold price for maximum return

Question

Q An online auction house charges seller of a product 10t units of money if the item is displayed for t days. You want to sell a product through this site. You are expecting 2 customers (on an average) per day. Each customer offers some price between 100 and 200. You have decided a threshold price of y, i.e., the product will be sold to the first customer offering a price of y or higher. What is good choice for y. What I tried : I let the threshold be y units. Then found i) E(profit|product remains on site for days and ii) Find expectation of (i) w.r.t t However I am not convinced if the above approach is right or not. I am getting y = 108.7 (approx)

Answer 1

Since you don't seem to be too interested in the specifics of the distributions of the customer arrivals and the prices, you might be satisfied with a model in which you pay the average of $5$ units per customer and the prices are uniformly distributed between $100$ and $200$. Then the expected return $r$ satisfies

$$ r = \frac{(200-y)\frac{200+y}2+(y-100)(r-5)}{100}\;. $$

Solving for $r$ yields

$$ r = \frac{y^2+10 y-41000}{2 (y-200)}\;, $$

and setting the derivative with respect to $y$ to $0$ then yields

$$ (2y+10)(y-200)=y^2+10y-41000 $$

with solutions $y=20(10\pm\sqrt{10})$, of which the smaller one, $y=20(10-\sqrt{10})\approx136.75$, lies in $[100,200]$ and is the desired threshold.

Another way to derive the expected return: When you eventually make the sale, the mean price will be $\frac{200+y}2$. You'll pay $5\frac p{1-p}$ for the extra customers, where $p=\frac{y-100}{100}$ is the probability to reject a customer. Together, this is

$$ \frac{200+y}2-\frac5{\frac{100}{y-100}-1}=\frac{y^2+10 y-41000}{2 (y-200)}\;. $$

(I've been assuming that every extra customer costs $5$ units; if the first customer also does, you can just subtract $5$ from the expected return without changing the optimal threshold.)