Dynamic Pricing Over Finite Horizons

Abstract

In this chapter, we first consider the problem of dynamically pricing one or more products that consume a single resource. Sales take place over a finite selling horizon, and the objective is to maximize the expected revenue that can be obtained from a finite inventory of the resource. We will assume that inventories cannot be replenished during the sales horizon. This problem setup holds for hotels, airlines, and seasonal merchandise including fashion retailing that have long procurement lead times. In this chapter, we focus on models that explicitly consider the stochastic and dynamic nature of demand. We use dynamic programming formulations to compute an optimal policy as a function of the remaining inventory and the time-to-go. In some cases, we are able to give closed-form solutions for the value function and the optimal pricing policy. In other cases, we resort to numerical solutions and to heuristic policies.

Appendix

Proof of Theorem 9.3

Notice that

$$\displaystyle \begin{aligned}\frac{\partial V(t,x)}{\partial t} = \mathcal{R}_t(\varDelta V(t,x)) \geq 0,\end{aligned}$$

with the inequality strict as long as d _t(p) > 0 for some p > ΔV (t, x), or equivalently if $\varDelta V(t,x) < \bar {r}_t$. This shows that V (t, x) is increasing in t. To show that V (t, x + 1) ≥ V (t, x), consider a sample path argument where the system with x + 1 units of inventory uses the optimal policy for the system with x units of inventory until either the system with x units runs out of stock or time runs out. If the system with x units of inventory runs out at time s, then the system with x + 1 units of inventory can still collect V (s, 1) ≥ 0. On the other hand, if time runs out, the two systems collect the same revenue. Consequently, the system with x + 1 units of inventory makes at least as much revenue resulting in V (t, x + 1) ≥ V (t, x).

Clearly ΔV (t, 1) ≤ ΔV (t, 0) = ∞. Assume as the inductive hypothesis that ΔV (t, y) is decreasing in y ≤ x for all t ≥ 0. We want to show that ΔV (t, x + 1) ≤ ΔV (t, x), or equivalently that

$$\displaystyle \begin{aligned} V(t,x+1) + V(t,x-1) \leq V(t,x) + V(t,x). \end{aligned} $$

(9.22)

We will use a sample path argument to construct the establish inequality (9.22). Consider four systems, one with x + 1 units of inventory, one with x − 1 units of inventory, and two with x units of inventory. Assume that we follow the optimal policy for the system with x + 1 and for the system with x − 1 that are on the left hand side of inequality (9.22). For the two systems on the right, we use the sub-optimal policies designed for x + 1 and x − 1 units of inventory, respectively. We follow these policies until one of the following events occurs: time runs out, the difference in inventories for the systems on the left drops to 1, or the inventory of the system with x − 1 units drops to zero. After that time, we follow optimal policies for all four systems.

To establish inequality (9.22), we will show that the revenues obtained for the systems in the right are at least as large as for the systems on the left, even though sub-optimal policies are used for the systems in the right. This is obviously true if we run out of time since the realized revenues of the two systems on the right are exactly equal to the realized revenues from the two systems on the left. Assume now that at time s ∈ (0, t), the difference in inventories of the two systems on the left hand side drops to 1, so that the states are (s, y + 1) and (s, y) for some y < x. This means that system on the left with x + 1 units of inventory had x − y units of sale and the system with x − 1 units of inventory had x − 1 − y units of sale. This implies that the system on the right that was following the policy designed for x + 1 reaches state (s, y), while the system that was using the policy designed for x − 1 reaches state (s, y + 1). Clearly, the additional optimal expected revenues over [0, s] for each pair of systems are V (s, y + 1) + V (s, y) = V (s, y) + V (s, y + 1), showing that the system on the right gets as much revenue as the system on the left even if sub-optimal policies are used for part of the horizon. Finally, if the inventory of the system with x − 1 units of inventory drops to 0 at some time s ∈ [0, t), so that state of the systems on the left are, respectively, (s, y) and (s, 0) for some y, such that 1 < y ≤ x, while the systems on the right are (s, y − 1) and (s, 1). From the inductive hypothesis, we know that ΔV (s, y) ≤ ΔV (s, 1) for all y ≤ x and all s ≤ t. Consequently,

$$\displaystyle \begin{aligned}V(s,y) + V(s,0) \leq V(s,y-1) + V(s,1),\end{aligned}$$

and once again the pair of systems on the right result in at least as much revenue even though sub-optimal policies are used for part of the sales horizon.

We now show that ΔV (t, x) is increasing in t. This is equivalent to

$$\displaystyle \begin{aligned}\frac{\partial V(t,x)}{\partial t} = \mathcal{R}_t(\varDelta V(t,x)) \geq \mathcal{R}_t(\varDelta V(t,x-1)) = \frac{\partial V(t,x-1)}{\partial t},\end{aligned}$$

but this is true on account of $\mathcal {R}_t(z)$ being decreasing in z and ΔV (t, x) being decreasing in x. We now show that P(t, x) = p _t(ΔV (t, x)) is decreasing in x. This follows because p _t(z) is increasing in z and ΔV (t, x) is decreasing in x.

Clearly, P(t, x) = p _t(ΔV (t, x)) is also increasing in t provided that p _t(z) is increasing in t since ΔV (t, x) is also increasing in t and p _t(z) is increasing in z. Clearly, p _t(z) = p(z) holds if d _t(p) = d(p) for all t. Moreover, if h _t(p) is decreasing in t, then p _t(z) is the root of the equation (p − z)h _t(p) = 1, and this is increasing in t.

We now show that V (t, x) is strictly increasing in t if $\varDelta V(t,x) < \bar {r}_t$. This follows because

$$\displaystyle \begin{aligned}\frac{\partial V(t,x)}{\partial t} = \mathcal{R}_t(\varDelta V(t,x)) >0,\end{aligned}$$

as any price $p \in (\varDelta V(t,x), \bar {r}_t)$ returns a positive profit. Notice that this is automatically true if p _t(z) is increasing in z because

$$\displaystyle \begin{aligned}\frac{\partial V(t,x)}{\partial t} = \mathcal{R}_t(\varDelta V(t,x)) \geq \mathcal{R}_t(\varDelta V(t,1)) = \mathcal{R}_t(V(t,1)) > 0,\end{aligned}$$

where the first inequality follows because $\mathcal {R}_t$ is decreasing and V (t, 1) = ΔV (t, 1) ≥ ΔV (t, x) for all x ≥ 1. The strict inequality follows because $V(t,1) < \bar {r}_t$, as otherwise the single unit must sell with probability one at the choke-off point, but by definition the demand at the choke-off point is zero. Consequently, there is a p _t > V (t, 1) with d _t(p) > 0 implying that $\mathcal {R}_t(V(t,1)) > 0$.

To show that V (t, x) is concave, notice that if $\mathcal {R}_t(z)$ is differentiable, then

$$\displaystyle \begin{aligned}\frac{\partial^2 V(t,x)}{\partial t^2} = \mathcal{R}^{\prime}_t(\varDelta V(t,x)) \frac{\partial \varDelta V(t,x)}{\partial t} \leq 0,\end{aligned}$$

follows since $\mathcal {R}^{\prime }_t(z) \leq 0$, on account of $\mathcal {R}_t(z)$ being decreasing in z, and from the fact that ΔV (t, x) is increasing in t.

Proof of Proposition 9.11

Recall that $\bar {V}(t,x) = \int _0^t \mathcal {R}_s(z(t,x))ds + xz(t,x)$. Since $\mathcal {R}_t(z)$ is continuously differentiable, either z(t, x) = 0 or z(t, x) > 0 is the root of $\int _0^td_s(p_s(z(t.x)))ds = x$. If z(t, x) = 0, then $\bar {V}(t,x) = \int _0^t \mathcal {R}_s(0)ds$, so

$$\displaystyle \begin{aligned}\frac{\partial \bar{V}(t,x)}{\partial t} = \mathcal{R}_t(0).\end{aligned}$$

On the other hand, if z(t, x) > 0, then

$$\displaystyle \begin{aligned}\frac{\partial \bar{V}(t,x)}{\partial t} = \mathcal{R}_t(z(t,x)) + [x - \int_0^td_s(p_s(z(t,x))dt] \frac{\partial z(t,x)}{\partial t} = \mathcal{ R}_t(z(t,x))\end{aligned}$$

since the last term cancels on account of $\int _0^td_s(p_s(z(t,x))dt = x$. Let ∂G(t, x)∕∂x = z(t, x). Then

$$\displaystyle \begin{aligned}\frac{\partial \bar{V}(t,x)}{\partial t} = \mathcal{R}_t\left (\frac{\partial G(t,x)}{\partial t}\right) = \frac{\partial G(t,x)}{\partial t}.\end{aligned}$$

Since the boundary conditions $\bar {V}(t,0) = G(t, 0) = \bar {V}(0,x) = G(0,x) = 0$, it follows that $\bar {V}(t,x) = G(t,x)$ as claimed.

Proof of Theorem 9.12

For ease of exposition, we will give the proof of Theorem 9.12 for the case of a single market with Poisson, rather than compound Poisson demands. However, the proof holds as stated. Clearly $V^h_b(T,c) \leq V_b(T,c) \leq \overline {V}_b(T,c)$. The idea of the proof is to show that the difference $\overline {V}_b(T,c) -V^h_b(T,c)$ becomes negligible relative to $V^h_b(T,c)$. First, notice that scaling arrival rates and demands does not change z(T, c) or the bid-price policy P(t, x) = p _t(Z(T, c)). To evaluate $V^h_b(T,c)$ let t = T − s be the time-to-go for each s ∈ [0, T]. The bid-price at time s is q _s = P _T−s(z(T, c)). The demand intensity at s is γ _s = d _T−s(P _T−s(z(T, c)). Let N _s be a Poisson process with intensity $\varGamma _s = \int _0^s\gamma _s ds$, and let τ _c be the first time the process reaches c, i.e. $\tau _c = \inf \{s \geq 0: N_s = c\}$. The bid-price policy earns revenues at rate q _s dN _s until time $\min (T,\tau _c)$, so

$$\displaystyle \begin{aligned}V^f_b(T,c) = \mathbb E\Bigg\{ \int_0^{\min(T,\tau_c)}q_sdN_s \Bigg\} = \mathbb E\Bigg\{\int_0^{\min(T,\tau_c)}q_s\gamma_sds \Bigg\}\end{aligned}$$

where the equality follows from Watanabe’s characterization of Poisson processes, see Bremaud (1980). Let $\varTheta _s = \int _0^s \theta _udu$, where θ _u = q _u γ _u, also let $\bar {\theta } = \max _{0 \leq u \leq T}\theta _u$. Then

$$\displaystyle \begin{aligned}V^f_b(T,c) = \varTheta_T - \mathbb E[\varTheta_T - \varTheta_{\min(T,\tau_c)}] \geq \varTheta_T - \bar{\theta} {\,} \mathbb E(T-\tau_c)^+.\end{aligned}$$

On the other hand, we can write

$$\displaystyle \begin{aligned}\bar{V}(T,c) = \varTheta_T - (\varGamma_T -c)z(T,c) .\end{aligned}$$

The result follows since the gap between $\bar {\theta } \mathbb E(T - \tau _c)^+ - (\varGamma _T - c)z(T,c)$ grows as the square root of b while the lower bound grows linearly with b.

Proof of Theorem 9.13

Our plan is to construct a heuristic with expected revenue V ^m(T, c) such that

$$\displaystyle \begin{aligned}\frac{V(T,c)}{\bar{V}(T,c)} \geq \frac{V^m(T,c)}{\bar{V}(T,c)} \geq \frac{1}{2}.\end{aligned}$$

Let

$$\displaystyle \begin{aligned}\bar{\pi} = \bar{V}(T,c) = \min_{z \geq 0}\left(\int_0^T \mathcal{R}_t(z)dt + zc\right),\end{aligned}$$

and set $z = 0.5\bar {\pi }/c$. Clearly

$$\displaystyle \begin{aligned}\bar{\pi} = \bar{V}(T,c) \leq \int_0^T \mathcal{R}_t(0.5\bar{\pi}/c)dt + 0.5 \bar{\pi},\end{aligned}$$

so

$$\displaystyle \begin{aligned}0.5 \bar{\pi} \leq \int_0^T \mathcal{R}_t(0.5\bar{\pi}/c)dt.\end{aligned}$$

Consider a heuristic policy that prices at $p_s(0.5 \bar {\pi }/c), 0 \leq s \leq T$, and let V ^m(T, x) be its expected revenue. We will show that $V^m(T,c) \geq 0.5 \bar {\pi }$. Suppose for a contradiction that $V^m(T,c) < 0.5 \bar {\pi }$. Then,

$$\displaystyle \begin{aligned}V^m(T,c) < 0.5 \bar{\pi} \leq \int_0^T \mathcal{R}_t(0.5\bar{\pi}/c)dt.\end{aligned}$$

We will show that in fact $V^m(T,c) \geq \int _{0}^{T} \mathcal {R}_{t}(0.5\bar {\pi }/c)dt$. Notice that

$$\displaystyle \begin{aligned}0.5 \bar{\pi} > V^m(T,c) = \sum_{x=1}^c \varDelta V^m(T,x),\end{aligned}$$

and since ΔV ^m(T, x) is decreasing in x, it follows that

$$\displaystyle \begin{aligned}\varDelta V^m(T,c) \leq V^m(T,c)/c < 0.5 \bar{\pi}/c.\end{aligned}$$

But then

$$\displaystyle \begin{aligned} \begin{array}{rcl} V^m(T,c) &\displaystyle = &\displaystyle \int_0^T R_t(p_t(0.5 \bar{\pi}/c), \varDelta V^m(t,c))dt \pm 0.5 \bar{\pi}/c \int_0^T d_t(p_t(0.5 \bar{\pi}/c))dt\\ {} &\displaystyle = &\displaystyle \int_0^T \mathcal{R}_t(0.5 \bar{\pi}/c)dt + \int_0^T[0.5 \bar{\pi}/c - \varDelta V^m(t,c)]d_t(p_t(0.5 \bar{\pi}/c))dt\\ {} &\displaystyle \geq &\displaystyle \int_0^T \mathcal{R}_t(0.5 \bar{\pi}/c)dt, \end{array} \end{aligned} $$

where the inequality follows because $0.5 \bar {\pi }/c > \varDelta V^h(t,c)$ for all t, since ΔV ^h(t, c) is increasing in t and the result holds at T. This shows that the assumption that $V^m(T,c) <0.5 \bar {V}(T,c)$ leads to a contradiction so it must be that $V^m(T,c) \geq 0.5 \bar {V}(T,c)$.

Proof of Theorem 9.14

Because d _t(p) = d(p) is continuous and o(1∕p), there exists a p _t(z) = p(z) for all z ≥ 0 that maximizes R _t(p, z) = (p − z)d _t(p) for all t ∈ [0, T]. Let $y = \int _0^T d_t(p(0))dt = T d(p(0))$. If y < c, then z(T, c) = 0 is optimal and the upper bound is $T\mathcal {R}(0) = y p(0)$. The bid-price heuristic would price at p _t = p(0) for all t ∈ (0, T] and would face Poisson demand with parameter y over [0, T], resulting in expected revenue $\min (c, \mbox{Poisson}(y))p(0)$. The ratio is therefore

$$\displaystyle \begin{aligned}\frac{V^h(T,c)}{V(T,c)} \geq \frac{V^h(T,c)}{\bar{V}(T,c)} = \frac{\mathbb E\min(c,\mbox{Poisson}(y))}{y}\end{aligned}$$

with y < c. On the other hand, if c ≥ y, then the assumed properties of d(p) guarantee the existence of a z = z(T, c) > 0 such that Td(p(z)) = c, so the upper bound yields cp(z) while the lower bound yields $\mathbb E\min (c,\mbox{Poisson}(c))p(z)$. Thus, the ratio is given by

$$\displaystyle \begin{aligned}\frac{V^h(T,c)}{V(T,c)} \geq \frac{V^h(T,c)}{\bar{V}(T,c)} = \frac{\mathbb E\min(c,\mbox{Poisson}(c))}{c}.\end{aligned}$$

It is well known that these ratios are bounded below by 1 − 1∕e.