Revenue Optimization with Poisson Demand

Worked Solution

How to Think About It: This is a classic price-quantity tradeoff. Raising the price increases revenue per unit but decreases demand. With Poisson demand, the expected quantity equals the rate parameter, so expected revenue is $p \cdot \lambda_0 e^{-\beta p}$ -- a hump-shaped function of $p$. The unconstrained optimum is where the marginal revenue from a price increase exactly offsets the lost demand. The vacancy constraint adds a floor on demand: if you price too high, the apartment sits empty too often.

Quick Estimate: Expected revenue is $R(p) = p \lambda_0 e^{-\beta p}$. Setting the derivative to zero: $\lambda_0 e^{-\beta p}(1 - \beta p) = 0$, giving $p^{*} = 1/\beta$. For example, if $\beta = 0.1$ (each dollar of price reduces log-demand by 0.1), the optimal price is $\$10$. Revenue at the optimum is $\lambda_0 \cdot 10 \cdot e^{-1} \approx 3.68 \lambda_0$.

Approach: Differentiate expected revenue for part (i). For part (ii), use Lagrangian optimization with the vacancy constraint.

Formal Solution:

Part (i): Unconstrained optimization.

Since $Q \sim \text{Poisson}(\lambda)$ with $\lambda = \lambda_0 e^{-\beta p}$, we have $E[Q] = \lambda = \lambda_0 e^{-\beta p}$.

Expected revenue: $$R(p) = p \cdot E[Q] = p \lambda_0 e^{-\beta p}$$

Take the derivative: $$R'(p) = \lambda_0 e^{-\beta p} - \beta p \lambda_0 e^{-\beta p} = \lambda_0 e^{-\beta p}(1 - \beta p)$$

Setting $R'(p) = 0$: since $\lambda_0 e^{-\beta p} > 0$ for all $p$, we need $1 - \beta p = 0$, so: $$p^{*} = \frac{1}{\beta}$$

Second derivative check: $R''(p) = \lambda_0 e^{-\beta p}(-\beta - \beta + \beta^2 p) = \lambda_0 e^{-\beta p}(\beta^2 p - 2\beta)$. At $p = 1/\beta$: $R''(1/\beta) = \lambda_0 e^{-1}(\beta - 2\beta) = -\beta \lambda_0 e^{-1} < 0$. Confirmed maximum.

Optimal revenue: $R(p^{*}) = \frac{\lambda_0}{\beta} e^{-1} = \frac{\lambda_0}{\beta e}$.

Part (ii): Vacancy-constrained optimization.

The vacancy probability is: $$P(Q = 0) = e^{-\lambda} = e^{-\lambda_0 e^{-\beta p}}$$

The constraint $P(Q = 0) \leq v_{\max}$ becomes: $$e^{-\lambda_0 e^{-\beta p}} \leq v_{\max}$$ $$-\lambda_0 e^{-\beta p} \leq \ln v_{\max}$$ $$\lambda_0 e^{-\beta p} \geq -\ln v_{\max} = \ln(1/v_{\max})$$ $$e^{-\beta p} \geq \frac{\ln(1/v_{\max})}{\lambda_0}$$ $$p \leq \frac{1}{\beta} \ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right)$$

Define the maximum feasible price: $$\bar{p} = \frac{1}{\beta} \ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right)$$

This requires $\lambda_0 > \ln(1/v_{\max})$ for $\bar{p} > 0$ (otherwise the constraint is infeasible at any positive price).

The constraint binds when $p^{*} > \bar{p}$, i.e., when: $$\frac{1}{\beta} > \frac{1}{\beta} \ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right)$$ $$1 > \ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right)$$ $$e > \frac{\lambda_0}{\ln(1/v_{\max})}$$ $$\lambda_0 < e \cdot \ln(1/v_{\max})$$

So the constrained solution is: $$p^{*}_{\text{constrained}} = \begin{cases} 1/\beta & \text{if } \lambda_0 \geq e \cdot \ln(1/v_{\max}) \\ \bar{p} = \frac{1}{\beta}\ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right) & \text{if } \lambda_0 < e \cdot \ln(1/v_{\max}) \end{cases}$$

When $\lambda_0$ is large (high base demand), the unconstrained optimum already keeps vacancy low enough. When $\lambda_0$ is small, you must price lower than the revenue-optimal level to maintain sufficient occupancy.

Answer: (i) $p^{*} = 1/\beta$. (ii) The vacancy constraint binds when $\lambda_0 < e \cdot \ln(1/v_{\max})$, in which case the optimal price is $\bar{p} = \frac{1}{\beta}\ln\left(\frac{\lambda_0}{\ln(1/v_{\max})}\right)$.