Renewal Reward Process Problem

Question

People arrive at a college admissions office at rate 1 per minute. When k people have arrive a tour starts. Student tour guides are paid $20 for each tour they conduct. The college estimates that it loses 10 cents in good will for each minute a person waits. What is the optimal tour group size?

To optimize the tour group size, we want to maximize the gaining from tour. So we want to calculate $\frac{R(t)}{t}$, the reward over time $t$. Using the law of large number, we have $$\frac{R(t)}{t} \xrightarrow[t\rightarrow\infty]{} \frac{\mathbb{E}[R_i]}{\mathbb{E}[t_i]}$$

I know $\mathbb{E}[t_i] = k$, but I am not sure how to compute $\mathbb{E}[R_i]$ . Intuitively, I guess $$\mathbb{E}[R_i] = 20 - 0.1 \times k \times \mathbb{E}[\text{waiting time}]$$ $$\mathbb{E}[\text{waiting time}] = \frac{k}{2} \times k = \frac{k^2}{2}$$

Is it correct? How to rigorously calculate the expected interval reward?

Answer 1

I don't see any randomness in the question. The arrival rate seems to be fixed at exactly one per minute. I think you are expected to compute the average cost per person to give a tour, counting the cost of the guide and the cost of waiting time, then choose $k$ to minimize that.

If we wait for $k$ people to show up, the number of person-minutes waiting is $\frac 12(k-1)k$ because the last person doesn't wait at all and the first one waits $k-1$ minutes. The cost of the tour is then $20+\frac 1{20}(k-1)k$ and the cost per guest is $\frac {20}k+\frac 1{20}(k-1)$. Taking the derivative and setting to zero we get $k=20$