A revisit to the markup practice of irreversible dynamic pricing

Abstract

We consider an irreversible dynamic pricing situation in which a firm uses real-time inventory information to decide the most opportune time to raise its sales prices. Feng and Xiao (Oper Res 48:332–343, 2000b) has studied this problem along with the opposite markdown case. In quite symmetric fashions, they established the optimality of threshold policies for both cases. Though the earlier work has made dramatic advances in dynamic pricing and at the same time pioneered with many relevant techniques, we believe its treatment of the markup case warrants some revision. In particular, we find it is in possession of an erstwhile-unknown complementarity property between price flexibility and inventory, whose counterpart is not true for the markdown case. This property is needed in the derivation leading to the optimality of a threshold policy. Our development also allows demand to be time-dependent in a product form, and naturally leads to an efficient policy-computing algorithm.

Appendices

Appendices

Proofs for the markup case

Proof of Proposition 2

We can use a sample-path argument to prove that

$$\begin{aligned} v^k_n(t)\ge \frac{v^k_{n-1}(t)+v^k_{n+1}(t)}{2}. \end{aligned}$$

(A.1)

We allow four pools of inventories, termed 1, 2, ${\bar{1}}$, and ${\bar{2}}$, to start at time t with the same price ${\bar{p}}^k$ and experience the same sample path over the interval [t, T]. These pools have different starting inventory levels and may exert different price controls, though. Pools 1 and 2 start with $n+1$ and $n-1$ initial items and pools ${\bar{1}}$ and ${\bar{2}}$ with n items.

Besides applying optimal time-increasing stopping-time pricing policies to pools 1 and 2, we apply the higher of the two prices for pools 1 and 2 to pool ${\bar{1}}$ and the lower of the two prices to pool ${\bar{2}}$, until the first moment, say s, when pool 1 is to have generated one more demand arrival than pool ${\bar{1}}$. After s, we let pool ${\bar{1}}$ follow pool 1’s decisions and pool ${\bar{2}}$ follow pool 2’s decisions. As both the minimum and maximum of two decreasing functions are decreasing functions themselves, pools ${\bar{1}}$ and ${\bar{2}}$ can be considered as adopting time-increasing stopping-time pricing policies as well.

Suppose the moment ever occurred, i.e., $s\in [t,T)$. Then, it has already been shown by Zhao and Zheng that the total sales revenue made by pools ${\bar{1}}$ and ${\bar{2}}$ amounts to the same as that by pools 1 and 2. Suppose the moment never occurred, i.e., $s=T$. Then, it has been shown by Zhao and Zheng that pools ${\bar{1}}$ and ${\bar{2}}$ can generate as high a total sales revenue as pools 1 and 2. So regardless, on every sample path, pools ${\bar{1}}$ and ${\bar{2}}$ can generate as high a total revenue as pools 1 and 2. Thus, we have proved (A.1). $\square$

Proof of Proposition 3

We use a sample-path approach to show the following: for any $n=0,1,\ldots ,N-1$,

$$\begin{aligned} v^k_{n+1}(t)-v^k_n(t)\ge v^{k+1}_{n+1}(t)-v^{k+1}_n(t),{\quad }\forall k=0,1,\ldots ,K-1,\;t\in [0,T]. \end{aligned}$$

(A.2)

We prove by induction on the inventory level n. Let us first prove

$$\begin{aligned} v^k_1(t)-v^k_0(t)\ge v^{k+1}_1(t)-v^{k+1}_0(t),{\quad }\forall k=0,1,\ldots ,K-1,\;t\in [0,T]. \end{aligned}$$

(A.3)

For every possible k and t, we introduce four pools of inventories, 1, 2, ${\bar{1}}$, and ${\bar{2}}$, that experience identical sample paths. At time t, pools 1 and ${\bar{1}}$ are with price index $k+1$, and pools 2 and ${\bar{2}}$ are with price index k. Also, pools 1 and ${\bar{2}}$ have one item, while pools ${\bar{1}}$ and 2 are out of stock. Apparently, pools ${\bar{1}}$ and 2 will continue to hold zero inventory. On the other hand, pool ${\bar{2}}$ can immediately raise its price to ${\bar{p}}^{k+1}$ and then match actions taken by pool 1. Hence, (A.3) is true.

Now, for some $n=1,2,\ldots N-1$, suppose it is true that, for any $m=0,1,\ldots ,n-1$,

$$\begin{aligned} v^k_{m+1}(t)-v^k_m(t)\ge v^{k+1}_{m+1}(t)-v^{k+1}_m(t),{\quad }\forall k=0,1,\ldots ,K-1,\;t\in [0,T]. \end{aligned}$$

(A.4)

We now prove that

$$\begin{aligned} v^k_{n+1}(t)-v^k_n(t)\ge v^{k+1}_{n+1}(t)-v^{k+1}_n(t),{\quad }\forall k=0,1,\ldots ,K-1,\;t\in [0,T]. \end{aligned}$$

(A.5)

For any possible k and t, we rely on pools 1, 2, ${\bar{1}}$, and ${\bar{2}}$ that experience identical sample paths. At time t, pools 1 and ${\bar{1}}$ are with price index $k+1$, and pools 2 and ${\bar{2}}$ are with price index k. Also, pools 1 and ${\bar{2}}$ both have $n+1$ items, while pools ${\bar{1}}$ and 2 both have n items. For $s\in [t,T]$, let us use $N_i(s)$ to denote the inventory level of pool i at time s. For instance, we have $N_{{\bar{1}}}(t)=n$ and $N_{{\bar{2}}}(t)=n+1$. We let pools 1 and 2 execute their respective optimal decisions. For a certain period, we let pool ${\bar{1}}$ follow pool 1’s decisions and pool ${\bar{2}}$ follow pool 2’s decisions. Note that prices adopted by all pools are increasing over time.

We let this certain period end at the first moment, say $s\in [t,T)$, when (a) pools 1 and 2 have reached the same price, (b) pools 2 and ${\bar{2}}$ have just experienced one more arrival than pools 1 and ${\bar{1}}$, or (c) pools 1 and ${\bar{2}}$ both have just one item left while pools ${\bar{1}}$ and 2 have no item left. We may denote the case where none of the above occurs by $s=T$. This is the case when by time T, at any moment the price taken by pool 2 has always been strictly lower than that taken by pool 1, yet all pools have admitted the same demand arrivals, and none of the pools have run out of stock. We caution that the opposite to (b) will not occur, since before its price “catches up” with that of pool 1, pool 2 always charges a strictly lower price than pool 1 and hence, by (S1’), has more chance to realize sales.

For all cases, pools ${\bar{1}}$ and ${\bar{2}}$ will have together generated the same revenue as pools 1 and 2 by time s. This also means that we are already done when $s=T$.

When (a) ever occurs, we may let pools ${\bar{1}}$ and ${\bar{2}}$ both execute optimal decisions from time s on. Then, within the time interval [s, T], pool ${\bar{1}}$ will behave exactly the same as pool 2, and pool ${\bar{2}}$ will behave exactly the same as pool 1. So, pools ${\bar{1}}$ and ${\bar{2}}$ will continue to together produce the same revenue as pools 1 and 2 do.

When (b) ever occurs, denote the price taken by pool 1 at time s by $k_1$ and the price taken by pool 2 at time s by $k_2$. We have $k_1>k_2$, and

$$\begin{aligned} n+1=N_1(t)\ge N_1(s)=N_{{\bar{1}}}(s)+1=N_{{\bar{2}}}(s)+1=N_2(s)+2. \end{aligned}$$

(A.6)

Hence, from the induction hypothesis (A.4), we have

$$\begin{aligned} v^{k_2}_{N_{{\bar{2}}}(s)}(s)-v^{k_2}_{N_2(s)}(s)\ge v^{k_1}_{N_{{\bar{2}}}(s)}(s)-v^{k_1}_{N_2(s)}(s). \end{aligned}$$

(A.7)

But from Proposition 2, we also have

$$\begin{aligned} v^{k_1}_{N_{{\bar{2}}}(s)}(s)-v^{k_1}_{N_2(s)}(s)\ge v^{k_1}_{N_1(s)}(s)-v^{k_1}_{N_{{\bar{1}}}(s)}(s). \end{aligned}$$

(A.8)

Combining (A.7) and (A.8), we obtain

$$\begin{aligned} v^{k_2}_{N_{{\bar{2}}}(s)}(s)-v^{k_2}_{N_2(s)}(s)\ge v^{k_1}_{N_1(s)}(s)-v^{k_1}_{N_{{\bar{1}}}(s)}(s). \end{aligned}$$

(A.9)

In view of the memorylessness property of the Poisson process, (A.9) means that, conditioned on the same sample path up to the provocation of (b), pools ${\bar{1}}$ and ${\bar{2}}$ will on average together earn more than pools 1 and 2 in the time interval [s, T].

When (c) ever occurs, denote the price taken by pool 1 at time s by $k_1$ and the price taken by pool 2 at time s by $k_2$. We have $k_1>k_2$, and

$$\begin{aligned} N_1(s)=N_{{\bar{1}}}(s)+1=N_{{\bar{2}}}(s)=N_2(s)+1=1. \end{aligned}$$

(A.10)

From the induction hypothesis (A.4), we have

$$\begin{aligned} v^{k_2}_{N_{{\bar{2}}}(s)}(s)-v^{k_2}_{N_2(s)}(s)\ge v^{k_1}_{N_1(s)}(s)-v^{k_1}_{N_{{\bar{1}}}(s)}(s). \end{aligned}$$

(A.11)

In view of the memorylessness property of the Poisson process, (A.11) means that, conditioned on the same sample path up to the provocation of (c), pools ${\bar{1}}$ and ${\bar{2}}$ will on average together earn more than pools 1 and 2 in the time interval [s, T].

In view of all the above, we see that (A.5) is true. We have hence completed the induction process. Therefore, (A.2) is true. $\square$

Proof of Theorem 1

The proof has use of the following lemma, which was originated in Karlin (1968) and also used in Feng and Xiao (2000b).

Lemma 1

Let $k>0$ and $\phi (t)=\int _t^{+\infty }\rho (s)\cdot exp(-k\cdot (s-t))\cdot ds$. Then, $\phi (t)$ will be decreasing in $t\ge 0$ if $\rho (s)$ is decreasing in $s\ge 0$.

We prove by induction on k. Let us focus on proving (a[K]) to (f[K]) first. Take $n=1,2,\ldots ,N$. From (3), we have

$$\begin{aligned} v^K_n(t)={\bar{p}}^K\cdot {\mathbb {E}}[(N^K(T)-N^K(t))\wedge n]={\bar{p}}^K\cdot \left( n-\sum _{m=0}^{n-1}(n-m)\cdot {\mathbb {P}}[N^K(t,T)=m]\right) , \end{aligned}$$

(A.12)

which, by (2), amounts to

$$\begin{aligned} v^K_n(t)={\bar{p}}^K\cdot n-{\bar{p}}^K\cdot \exp \left( -{\bar{\alpha }}^K\cdot {{\hat{\beta }}}(t,T)\right) \cdot \sum _{m=0}^{n-1}\frac{(n-m)\cdot \left( {\bar{\alpha }}^K\cdot {{\hat{\beta }}}(t,T)\right) ^m}{m!}. \end{aligned}$$

(A.13)

Taking derivative over t, we find that, for $t\in (0,T)$,

$$\begin{aligned} d_t v^K_n(t)=-{\bar{p}}^K{\bar{\alpha }}^K\cdot \beta (t)\cdot \exp \left( -{\bar{\alpha }}^K\cdot {{ \hat{\beta }}}(t,T)\right) \cdot \sum _{m=0}^{n-1} \frac{\left( {\bar{\alpha }}^K\cdot {{\hat{\beta }}}(t,T)\right) ^m}{m!}. \end{aligned}$$

(A.14)

Taking differences over n, we find that

$$\begin{aligned} v^K_n(t)-v^K_{n-1}(t)={\bar{p}}^K\cdot \left( 1-\exp \left( -{\bar{\alpha }}^K\cdot {{\hat{\beta }}}(t,T)\right) \cdot \sum _{m=0}^{n-1}\frac{\left( {\bar{\alpha }}^K \cdot {{\hat{\beta }}}(t,T)\right) ^m}{m!}\right) . \end{aligned}$$

(A.15)

From (A.14) and (A.15), it can be checked that

$$\begin{aligned} {{\mathcal {G}}}^K_n(t)\circ v^K(t)+{\bar{p}}^K{\bar{\alpha }}^K\cdot \beta (t)=0, {\quad }\forall t\in (0,T). \end{aligned}$$

(A.16)

Consulting (5) and (6), we obtain

$$\begin{aligned} v^K_n(t)={\bar{\alpha }}^K\cdot \int _t^T\beta (s)\cdot \left( {\bar{p}}^K+v^K_{n-1}(s)\right) \cdot \exp \left( -{\bar{\alpha }}^K\cdot {{\hat{\beta }}}(t,s)\right) \cdot ds. \end{aligned}$$

(A.17)

In view of the construction (9), we may see (c[K]), that

$$\begin{aligned} \begin{array}{l} u^K\equiv \left( u^K_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T]\right) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=v^K\equiv \left( v^K_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T]\right) . \end{array} \end{aligned}$$

(A.18)

From (A.16), we may confirm (d[K]) with the understanding that $\tau ^{K+1}_n=0$ for $n=1,2,\ldots ,N$. The convention for the $\tau ^{K+1}_n$’s also leads directly (e[K]). By (A.16) and the convention on $\tau ^{K+1}_n$ and $v^{K+1}_n(t)$, we may see that (a[K]) and (b[K]) are both true. To verify (f[K]), we can use the same method on (f[k]) for $k=K-1,K-2,\ldots ,0$, which is presented near the end of this proof.

Suppose for some $k=K-1,K-2,\ldots ,0$, we have (a$[k+1]$), that

$$\begin{aligned} ({{\mathcal {G}}}^{k+1}_n)^+(\tau ^{k+2}_n)\circ v^{k+2}+{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}\cdot \beta (\tau ^{k+2}_n)\le 0,{\quad }\forall n=1,2,\ldots ,N, \end{aligned}$$

(A.19)

(b$[k+1]$), that ${{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+2}/\beta (t)$ is increasing in t for $n=1,2,\ldots ,N$ and $t\in (0,T)$, (c$[k+1]$), that

$$\begin{aligned} u^{k+1}_n(t)=v^{k+1}_n(t),{\quad }\forall n=0,1,\ldots ,N,\;t\in [0,T], \end{aligned}$$

(A.20)

(d$[k+1]$), that, for $n=1,2,\ldots ,N$,

$$\begin{aligned} v^{k+1}_n(t)=v^{k+2}_n(t)\; \text{ and } \;{{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+1}+{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}\cdot \beta (t)\le 0,{\quad }\forall t\in (0,\tau ^{k+2}_n), \end{aligned}$$

(A.21)

and

$$\begin{aligned} {{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+1}+{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}\cdot \beta (t)=0,{\quad }\forall t\in (\tau ^{k+2}_n,T). \end{aligned}$$

(A.22)

(e$[k+1]$), that $\tau ^{k+2}_n$ is decreasing in n, and (f$[k+1]$), that $v^{k+1}_n(t)$ has decreasing differences between n and t.

Now we embark on showing (a[k]) to (f[k]). For $n=1,2,\ldots ,N$, by the definition of $\tau ^{k+1}_n$ through (11), we know that

$$\begin{aligned}&u^k_n(t)>v^{k+1}_n(t),{\quad }\forall t\in (\tau ^{k+1}_n,T), \end{aligned}$$

(A.23)

$$\begin{aligned}&u^k_n(\tau ^{k+1}_n)=v^{k+1}_n(\tau ^{k+1}_n), \end{aligned}$$

(A.24)

and

$$\begin{aligned} d^+_t u^k_n(\tau ^{k+1}_n)\ge d^+_t v^{k+1}_n(\tau ^{k+1}_n). \end{aligned}$$

(A.25)

By (5) and (6), we may see that the construction (10) renders

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ u^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0,{\quad }\forall t\in (\tau ^{k+1}_n,T). \end{aligned}$$

(A.26)

Our construction through (10)–(12) also guarantees that

$$\begin{aligned} u^k_{n-1}(\tau ^{k+1}_n)\ge v^{k+1}_{n-1}(\tau ^{k+1}_n). \end{aligned}$$

(A.27)

Combining (A.24)–(A.27), we obtain

$$\begin{aligned} ({{\mathcal {G}}}^k_n)^+(\tau ^{k+1}_n)\circ v^{k+1}+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (\tau ^{k+1}_n)\le 0. \end{aligned}$$

(A.28)

Thus we have (a[k]).

Note that (a$[k+1]$) means (A.19). This, (b$[k+1]$), and (d$[k+1]$) together lead to the fact that ${{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+2}/\beta (t)$ is increasing in t and below $-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}$. From the definition of $u^{k+1}_n(t)$ for $t\in [0,\tau ^{k+2}_n]$ through (12), (c$[k+1]$), and (e$[k+1]$), we may see that

$$\begin{aligned} \frac{{{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+1}}{\beta (t)}=\frac{{{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+2}}{\beta (t)},{\quad }\forall t\in (0,\tau ^{k+2}_n). \end{aligned}$$

(A.29)

Hence, in view of the above and (b$[k+1]$) again, we may see that ${{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+1}/\beta (t)$ is increasing in t and below $-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}$ when $t\in (0,\tau ^{k+2}_n)$, and is flat at $-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}$ for $t\in (\tau ^{k+2}_n,T)$. Now, note that

$$\begin{aligned} \frac{{{\mathcal {G}}}^k_n(t)\circ v^{k+1}-{{\mathcal {G}}}^{k+1}_n(t)\circ v^{k+1}}{\beta (t)}=({\bar{\alpha }}^k-{\bar{\alpha }}^{k+1})\cdot \left( v^{k+1}_{n-1}(t)-v^{k+1}_n(t)\right) , \end{aligned}$$

(A.30)

which, by (S1’) and (f$[k+1]$), is increasing in t. This and the just proved result together lead to the increase of ${{\mathcal {G}}}^k_n(t)\circ v^{k+1}/\beta (t)$ in t. Hence, we have (b[k]).

From (a[k]) and (b[k]), we obtain, for $n=1,2,\ldots ,N$,

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ v^{k+1}+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\le 0,{\quad }\forall t\in (0,\tau ^{k+1}_n). \end{aligned}$$

(A.31)

Our construction (12) and (c$[k+1]$) dictate that

$$\begin{aligned} u^k_n(t)=v^{k+1}_n(t),{\quad }\forall t\in [0,\tau ^{k+1}_n]. \end{aligned}$$

(A.32)

We now define ${\overline{\lambda }}$, so that

$$\begin{aligned} {\overline{\lambda }}={\bar{\alpha }}^0\cdot \sup _{t\in [0,T]}\beta (t). \end{aligned}$$

(A.33)

By the continuity of $\beta (\cdot )$ and the compactness of [0, T], we know that ${\overline{\lambda }}$ is a strictly positive and finite constant. By (S1’), we know that ${\overline{\lambda }}$ is the highest instantaneous arrival rate that could ever materialize. By its construction, $u^k_n(t)$ is uniformly bounded by $N{\overline{\lambda }} T$; it is also Lipschitz continuous in t with coefficient $N{\overline{\lambda }}$, and hence absolutely continuous in t. By (A.23), (A.26), (A.31), and (A.32), and as well as the fact that $v^{k+1}_n(T)=0$ for every n, we may see that $u^k\equiv (u^k_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T])$ satisfies the sufficient conditions (i) to (iv) stipulated in Proposition 1. Hence, we have shown (c[k]), that $u^k_n(t)=v^k_n(t)$ for every $n=0,1,\ldots ,N$ and $t\in [0,T]$.

From (A.26), (A.31), (A.32), and (c[k]), we easily have (d[k]).

For $n=1,2,\ldots ,N-1$, we have, from (11), (c$[k+1]$), and (c[k]),

$$\begin{aligned} v^k_n(t)-v^{k+1}_n(t)>0,{\quad }\forall t\in (\tau ^{k+1}_n,T). \end{aligned}$$

(A.34)

By Proposition 3, however, we have

$$\begin{aligned} v^k_{n+1}(t)-v^{k+1}_{n+1}(t)\ge v^k_n(t)-v^{k+1}_n(t). \end{aligned}$$

(A.35)

Combining (A.34) and (A.35), we obtain

$$\begin{aligned} v^k_{n+1}(t)-v^{k+1}_{n+1}(t)>0,{\quad }\forall t\in (\tau ^{k+1}_n,T). \end{aligned}$$

(A.36)

But in view of (11), (c$[k+1]$), and (c[k]), this leads to $\tau ^{k+1}_{n+1}\le \tau ^{k+1}_n$. Thus, we have (e[k]).

Let us now turn to the proof of (f[k]). For convenience, we denote $d_t v^k_n(t)/\beta (t)$ by $w^k_n(t)$. When $t\in (0,\tau ^{k+1}_n)$, which is $\emptyset$ when $k=K$, we have $w^k_n(t)=w^{k+1}_n(t)$ from (d[k]). Hence, following (f$[k+1]$), we know that $w^k_n(t)$ is decreasing in t. Let $t\in (\tau ^{k+1}_n,T)$ then. By (d[k]),

$$\begin{aligned} v^k_n(t)-v^k_{n-1}(t)={\bar{p}}^k+\frac{w^k_n(t)}{{\bar{\alpha }}^k}. \end{aligned}$$

(A.37)

By the boundary conditions $v^k_n(T)=v^k_{n-1}(T)=0$, we therefore have

$$\begin{aligned} w^k_n(T^-)=-{\bar{p}}^k{\bar{\alpha }}^k. \end{aligned}$$

(A.38)

Taking derivative on (A.37), it follows that

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ w^k=0,{\quad }\forall t\in (\tau ^{k+1}_n,T). \end{aligned}$$

(A.39)

In view of (5), (6), and (A.38), we can solve (A.39) to obtain, for $t\in (\tau ^{k+1}_n,T)$,

$$\begin{aligned} w^k_n(t)=-{\bar{p}}^k{\bar{\alpha }}^k\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,T))+{\bar{\alpha }}^k\cdot \int _t^T\beta (s)\cdot w^k_{n-1}(s)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s))\cdot ds, \end{aligned}$$

(A.40)

which, by the identity

$$\begin{aligned} {\bar{\alpha }}^k\cdot \int _t^T \beta (s)\cdot \exp \left( -{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s)\right) \cdot ds=1-\exp \left( -{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,T)\right) , \end{aligned}$$

(A.41)

results in

$$\begin{aligned} w^k_n(t)=-{\bar{p}}^k{\bar{\alpha }}^k+{\bar{\alpha }}^k\cdot \int _t^T \beta (s)\cdot \left( {\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1}(s)\right) \cdot \exp \left( -{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s)\right) \cdot ds. \end{aligned}$$

(A.42)

Since ${{\hat{\beta }}}(0,\cdot )$ is a strictly increasing function on [0, T], we can define strictly increasing function $\theta (\cdot )$ on $[0,{{\hat{\beta }}}(0,T)]$, so that

$$\begin{aligned} {{\hat{\beta }}}(t,\theta (y))={{\hat{\beta }}}(0,\theta (y))-{{\hat{\beta }}}(0,t)=y-{{\hat{\beta }}}(0,t),{\quad }\forall y\in [0,{{\hat{\beta }}}(0,T)]. \end{aligned}$$

(A.43)

This then leads to

$$\begin{aligned} d_y \theta (y)=\frac{1}{d_s{{\hat{\beta }}}(0,s)} \mid _{s=\theta (y)}=\frac{1}{\beta (\theta (y))}. \end{aligned}$$

(A.44)

In view of the above, we can bring a change of variables to the integral involved in (A.42), so that the latter becomes

$$\begin{aligned} w^k_n(t)=-{\bar{p}}^k{\bar{\alpha }}^k+{\bar{\alpha }}^k\cdot \int _{{{\hat{\beta }}}(0,t)}^{{{\hat{\beta }}}(0,T)} \left( {\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1}(\theta (y))\right) \cdot \exp \left( -{\bar{\alpha }}^k\cdot (y-{{\hat{\beta }}}(0,t))\right) \cdot dy. \end{aligned}$$

(A.45)

Suppose $w^k_{n-1}(t)$ is decreasing in t for $t\in (0,T)$, then since $\theta (\cdot )$ is increasing, we know $w^k_{n-1}(\theta (y))$ is decreasing in y for $y\in (0,{{\hat{\beta }}}(0,T))$. From (A.38) which also applies to $w^k_{n-1}(T^-)$, we may see that

$$\begin{aligned} {\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1}(\theta (y))\ge 0,{\quad }\forall y\in (0,{{\hat{\beta }}}(0,T)), \end{aligned}$$

(A.46)

and

$$\begin{aligned} {\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1}(\theta (({{\hat{\beta }}}(0,T))^-))=0. \end{aligned}$$

(A.47)

Hence, (A.45) can be rewritten as

$$\begin{aligned} w^k_n(t)=-{\bar{p}}^k{\bar{\alpha }}^k+{\bar{\alpha }}^k\cdot \int _{{{\hat{\beta }}}(0,t)}^{+\infty } ({\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1} (\theta (y)))^+\cdot \exp (-{\bar{\alpha }}^k\cdot (y-{{\hat{\beta }}}(0,t)))\cdot dy. \end{aligned}$$

(A.48)

By the decrease of $({\bar{p}}^k{\bar{\alpha }}^k+w^k_{n-1}(\theta (y)))^+$ in $y\ge 0$, the increase of ${{\hat{\beta }}}(0,t)$ in t, and Lemma 1, we can get the decrease of $w^k_n(t)$ in t on $(\tau ^{k+1}_n,T)$. But combining with the earlier result on the other half interval, we may get the decrease of $w^k_n(t)$ in t on the entire (0, T) from that of $w^k_{n-1}(t)$. As $w^k_0(t)=0$ for all $t\in (0,T)$, we can therefore use induction on n to prove the decrease of $w^k_n(t)$ in $t\in (0,T)$ for all $n=0,1,\ldots ,N$.

For $t\in (0,\tau ^{k+1}_n)$, which is $\emptyset$ when $k=K$ and a subset of $(0,\tau ^{k+1}_{n-1})$ by (e[k]), we have from (d[k]) that,

$$\begin{aligned} v^k_n(t)-v^k_{n-1}(t)=v^{k+1}_n(t)-v^{k+1}_{n-1}(t). \end{aligned}$$

(A.49)

Hence, $v^k_n(t)-v^k_{n-1}(t)$ is decreasing in t by (f$[k+1]$). For $t\in (\tau ^{k+1}_n,T)$, we can achieve the same property by (A.37) and the just proved decrease of $w^k_n(t)$ in t. Thus, we have shown (f[k]). Note that, when $k=K$, the proof has no involvement of any $[k+1]$-property.

We have now completed the induction process. Therefore, (a[k]) to (f[k]) are all true for $k=K,K-1,\ldots ,0$. From these, we see that, for any $k=K-1,K-2,\ldots ,0$ and $n=1,2,\ldots ,N$,

$$\begin{aligned} v^k_n(t)\left\{ \begin{array}{ll} =v^{k+1}_n(t),&{}{\quad }\forall t\in [0,\tau ^{k+1}_n],\\ >v^{k+1}_n(t),&{}{\quad }\forall t\in (\tau ^{k+1}_n,T). \end{array}\right. \end{aligned}$$

(A.50)

Hence, we may see that each $\tau ^{k+1}_n$ offers an optimal time by which the firm is to raise its price from ${\bar{p}}^k$ to ${\bar{p}}^{k+1}$ when it has n remaining items. $\square$

The markdown case

In the markdown case, the firm has to consecutively charge the sequence of decreasing prices ${\bar{p}}^K,{\bar{p}}^{K-1},\ldots ,{\bar{p}}^0$. Again, we show the optimality of a threshold policy $\tau \equiv (\tau ^k_n\mid k=1,2,\ldots ,K,\;n=1,2,\ldots ,N)\in (\Delta _N)^K$. Under it, the firm should lower its price from ${\bar{p}}^k$ to ${\bar{p}}^{k-1}$ when the threshold $\tau ^k_n$ corresponding to its current inventory level n is about to be passed.

Let $v^k_n(t)$ be the maximum revenue the firm can make in time interval [t, T] when it starts time t with price ${\bar{p}}^k$ and inventory level n. Since the lowest price ${\bar{p}}^0$ is the last price the firm can charge before running out of stock, we have

$$\begin{aligned} v^0_n(t)={\bar{p}}^0\cdot {\mathbb {E}}\left[ (N^0(T)-N^0(t))\wedge n\right] . \end{aligned}$$

(B.1)

When the firm is charging any other price ${\bar{p}}^k$ for $k=1,2,\ldots ,K$, it has yet to dynamically decide the time to switch to the next price ${\bar{p}}^{k-1}$. Hence, we have

$$\begin{aligned} v^k_n(t)=\sup _{\tau \in {{\mathcal {T}}}(t)}{\mathbb {E}}\left[ {\bar{p}}^k\cdot \{(N^k(\tau )-N^k(t))\wedge n\}+v^{k-1}_{(n-N^k(\tau )+N^k(t))^+}(\tau )\right] , \end{aligned}$$

(B.2)

where ${{\mathcal {T}}}(t)$ is again the set of stopping times later than t. Contrast this with (4), and we can see that in the markdown case, it is the next lower price ${\bar{p}}^{k-1}$ that will be tried after ${\bar{p}}^k$. Similar to Proposition 1, we have the following markdown counterpart.

Proposition 5

For any $k=1,2,\ldots ,K$, a function vector $u\equiv (u_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T])$ that is uniformly bounded and absolutely continuous in t for every n will be the value-function sub-vector $v^k\equiv (v^k_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T])$, if it satisfies the following:

• (i) $u_n(t)\ge v^{k-1}_n(t)$ for every $n=0,1\ldots ,N$ and $t\in [0,T]$,

• (ii) $u_n(T)=0$ for every $n=0,1,\ldots ,N$ and $u_0(t)=0$ for every $t\in [0,T]$,

• (iii) for $n=1,2,\ldots ,N$ and $t\in [0,T]$, $u_n(t)=v^{k-1}_n(t)$ implies ${{\mathcal {G}}}^k_n(t)\circ u+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\le 0$,

• (iv) for $n=1,2,\ldots ,N$ and $t\in [0,T]$, $u_n(t)>v^{k-1}_n(t)$ implies ${{\mathcal {G}}}^k_n(t)\circ u+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0$.

Proposition 5 offers a hint as to how the value functions $v^k_n(t)$ and threshold levels $\tau ^k_n$ can be constructed. First, due to (B.1), $v^0_n(t)$ can be constructed in an n-loop. Then, for any $k=1,2,\ldots ,K$, suppose $v^{k-1}_n(t)$ is known. We can then go through an n-loop to find all the $v^k_n(t)$’s. First, we can let $v^k_0(t)=0$ as suggested by (ii) of the proposition. Second, suppose $v^k_{n-1}(t)$ is known for some $n=1,2,\ldots ,N$. Then, starting with $t=T$, we can equate $v^k_n(t)$ to $v^{k-1}_n(t)$ for ever smaller t values until it is to occur that ${{\mathcal {G}}}^k_n(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)>0$. The latter is not allowed by (iii) and (iv) of the proposition. For time $t'$ earlier than this t, which we mark as $\tau ^k_n$, we let $v^k_n(t')$ be the solution to ${{\mathcal {G}}}^k_n(t')\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t')=0$.

According to (i) of the proposition, we still need $v^k_n(t')\ge v^{k-1}_n(t')$ for $t'<\tau ^k_n$ for the thus constructed $v^k_n(\cdot )$ to be the true value function. Nevertheless, let us go ahead with the construction thus outlined. Not knowing whether what shall be constructed are the true value functions, we call them u’s instead of v’s. Formally, here is the iterative procedure.

First, let

$$\begin{aligned} u^k_0(t)=0,{\quad }\forall k=0,1,\ldots ,K,\;t\in [0,T]. \end{aligned}$$

(B.3)

Then, for $n=1,2,\ldots ,N$ and $t\in [0,T]$, let

$$\begin{aligned} u^0_n(t)={\bar{\alpha }}^0\cdot \int _t^T \beta (s)\cdot \left( {\bar{p}}^0+u^0_{n-1}(s)\right) \cdot \exp \left( -{\bar{\alpha }}^0\cdot {{\hat{\beta }}}(t,s)\right) \cdot ds. \end{aligned}$$

(B.4)

Next, we go over an outer loop on $k=1,2,\ldots ,K$ and an inner loop on $n=1,2,\ldots ,N$. At each k and n, first let

$$\begin{aligned} \tau ^k_n=\inf \left\{ t\in [0,T]\mid {{\mathcal {G}}}^k_n(t)\circ u^{k-1}+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\le 0\right\} , \end{aligned}$$

(B.5)

with the understanding that $\tau ^k_n=0$ when the concerned inequality is always true and $\tau ^k_n=T$ when it is never true. Then, let

$$\begin{aligned} u^k_n(t)=u^{k-1}_n(t),{\quad }\forall t\in [\tau ^k_n,T], \end{aligned}$$

(B.6)

and when $t\in [0,\tau ^k_n)$,

$$\begin{aligned} u^k_n(t)=u^{k-1}_n(\tau ^k_n)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,\tau ^k_n))+{\bar{\alpha }}^k\cdot \int _t^{\tau ^k_n}\beta (s)\cdot ({\bar{p}}^k+u^k_{n-1}(s))\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s))\cdot ds. \end{aligned}$$

(B.7)

In obtaining (B.4) and (B.7), we have resorted to (5) and (6) for solving ${{\mathcal {G}}}^k_n(t)\circ u^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0$ for $k=0,1,\ldots ,K$. Like its markup counterpart, the ODE again boils down to (13). With the terminal condition $u^0_n(T)=0$ and the second half of (6), we will have (B.4). For $k=1,2,\ldots ,K$, with $u^k_n(\tau ^k_n)=u^{k-1}_n(\tau ^k_n)$ and again the second half of (6), we will have (B.7). Now, we proceed to show the optimality of the construction.

Proposition 6

For any fixed $k=0,1,\ldots ,K$ and $t\in [0,T]$, the value function $v^k_n(t)$ is concave in n: $v^k_{n+1}(t)-v^k_n(t)\le v^k_n(t)-v^k_{n-1}(t)$.

For convenience, we have let the yet undefined $\tau ^0_n=T$ for $n=1,2,\ldots ,N$ and $v^{-1}_n(t)=u^0_n(t)$ for $n=1,2,\ldots ,N$ and $t\in [0,T]$. Using Proposition 6, we can prove the following.

Theorem 2

The u and $\tau$ as constructed from (B.3) to (B.7) satisfy the following for $k=0,1,\ldots ,K$:

(a[k]):

${{\mathcal {G}}}^k_n(t)\circ v^{k-1}/\beta (t)$ is decreasing in t for $n=1,2,\ldots ,N$ and $t\in (0,T)$;

(b[k]):

$u^k_n(t)=v^k_n(t)$ for any $n=0,1,\ldots ,N$ and $t\in [0,T]$;

(c[k]):

for any $n=1,2,\ldots ,N$, we have ${{\mathcal {G}}}^k_n(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0$ for $t\in (0,\tau ^k_n)$ and ${{\mathcal {G}}}^k_n(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\le 0$ for $t\in (\tau ^k_n,T)$;

(d[k]):

$\tau ^k_n$ is decreasing in n;

(e[k]):

more than having decreasing differences between n and t, $v^k_n(t)$ satisfies the following for every $t\in (0,T)$:

$$\begin{aligned} d_t v^k_1(t)\le 0, \end{aligned}$$

and for $n=1,2,\ldots ,N-1$,

$$\begin{aligned} d_t v^k_{n+1}(t)-d_t v^k_n(t)\le {\bar{\alpha }}^k\cdot \beta (t)\cdot \left( v^k_{n-1}(t)-2v^k_n(t)+v^k_{n+1}(t)\right) , \end{aligned}$$

which is negative due to Proposition 6.

Proof

We prove by induction on k. Let us first focus on proving (a[0]) to (e[0]). Take $n=1,2,\ldots ,N$. First, we have

$$\begin{aligned} {{\mathcal {G}}}^0_n(t)\circ v^0(t)+{\bar{p}}^0{\bar{\alpha }}^0\cdot \beta (t)=0,{\quad }\forall t\in (0,T). \end{aligned}$$

(B.8)

and (b[0]), that

$$\begin{aligned} u^0\equiv \left( u^0_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T]\right) =v^0\equiv \left( v^0_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T]\right) . \end{aligned}$$

(B.9)

From (B.8), we may confirm (c[0]) with the understanding that $\tau ^0_n=T$ for $n=1,2,\ldots ,N$. The convention for the $\tau ^0_n$’s also leads directly (d[0]). From (B.1), we have

$$\begin{aligned} v^0_1(t)={\bar{p}}^0\cdot {\mathbb {E}}[N^0(t,T)\wedge 1]={\bar{p}}^0\cdot {\mathbb {P}}[N^0(t,T)\ge 1]<{\bar{p}}^0. \end{aligned}$$

(B.10)

As $v^0_0(t)=0$ for any $t\in [0,T]$, we can apply (B.8) at $n=1$ to get

$$\begin{aligned} d_t v^0_1(t)={\bar{\alpha }}^0\cdot \beta (t)\cdot (v^0_1(t)-{\bar{p}}^0),{\quad }\forall t\in (0,T). \end{aligned}$$

(B.11)

which is negative by (B.10). For $n=1,2,\ldots ,N-1$, we can apply (B.8) at both n and $n+1$ to obtain

$$\begin{aligned} d_t v^0_{n+1}-d_t v^0_n(t)={\bar{\alpha }}^0\cdot \beta (t)\cdot \left( v^0_{n-1}(t)-2v^0_n(t)+v^0_{n+1}(t)\right) ,{\quad }\forall t\in (0,T). \end{aligned}$$

(B.12)

So (e[0]) is satisfied as well. Finally, by our default definition of $v^{-1}_n(t)$ and (B.8), we know that (a[0]) is true.

Suppose for some $k=1,2,\ldots ,K$, we have (a$[k-1]$), that ${{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-2}/\beta (t)$ is decreasing in t for $n=1,2,\ldots ,N$ and $t\in (0,T)$, (b$[k-1]$), that

$$\begin{aligned} u^{k-1}_n(t)=v^{k-1}_n(t),{\quad }\forall n=0,1,\ldots ,N,\;t\in [0,T], \end{aligned}$$

(B.13)

(c$[k-1]$), that, for $n=1,2,\ldots ,N$,

$$\begin{aligned} {{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)=0,{\quad }\forall t\in (0,\tau ^{k-1}_n), \end{aligned}$$

(B.14)

and

$$\begin{aligned} {{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)\le 0,{\quad }\forall t\in (\tau ^{k-1}_n,T), \end{aligned}$$

(B.15)

(d$[k-1]$), that $\tau ^{k-1}_n$ is decreasing in n, and (e$[k-1]$), that, for $t\in (0,T)$,

$$\begin{aligned} d_t v^{k-1}_1(t)\le 0,{\quad }\forall t\in (0,T), \end{aligned}$$

(B.16)

and for $n=1,2,\ldots ,N-1$,

$$\begin{aligned} d_t v^{k-1}_{n+1}-d_t v^{k-1}_n(t)\le {\bar{\alpha }}^{k-1}\cdot \beta (t)\cdot \left( v^{k-1}_{n-1}(t)-2v^{k-1}_n(t)+v^{k-1}_{n+1}(t)\right) ,{\quad }\forall t\in (0,T). \end{aligned}$$

(B.17)

Now we embark on showing (a[k]) to (e[k]). Take $n=1,2,\ldots ,N$. From the definition of $\tau ^{k-1}_n$ through (B.5), (b$[k-1]$), and (c$[k-1]$), we may see that

$$\begin{aligned} {{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)=0<{{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-2}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t), \end{aligned}$$

(B.18)

for $t\in (0,\tau ^{k-1}_n)$, and

$$\begin{aligned} {{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)={{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-2}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)\le 0, \end{aligned}$$

(B.19)

for $t\in (\tau ^{k-1}_n,T)$. Therefore,

$$\begin{aligned} {{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)=[{{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-2}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\cdot \beta (t)]\wedge 0, \end{aligned}$$

(B.20)

and hence by the strict positivity of $\beta (\cdot )$,

$$\begin{aligned} \frac{{{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}}{\beta (t)}=\left[ \frac{{{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-2}}{\beta (t)}+{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}\right] \wedge 0-{\bar{p}}^{k-1}{\bar{\alpha }}^{k-1}. \end{aligned}$$

(B.21)

Note that the above is even true for $k=1$ due to the default definitions. Combining (a$[k-1]$) and (B.21), we obtain the decrease of ${{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}/\beta (t)$ in t. Now, note that

$$\begin{aligned} \frac{{{\mathcal {G}}}^k_n(t)\circ v^{k-1}-{{\mathcal {G}}}^{k-1}_n(t)\circ v^{k-1}}{\beta (t)}=({\bar{\alpha }}^{k-1}-{\bar{\alpha }}^k)\cdot \left( v^{k-1}_n(t)-v^{k-1}_{n-1}(t)\right) , \end{aligned}$$

(B.22)

which, by (S1’) and (e$[k-1]$), is decreasing in t. This and the just proved result together lead to the decrease of ${{\mathcal {G}}}^k_n(t)\circ v^{k-1}/\beta (t)$ in t. Hence, we have (a[k]).

By the definition of $\tau ^k_n$ through (B.5) and (b$[k-1]$), we have

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ v^{k-1}+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)>0,{\quad }\forall t\in (0,\tau ^k_n). \end{aligned}$$

(B.23)

Due to the strict positivity of $\beta (\cdot )$ and (b$[k-1]$), the threshold $\tau ^k_n$ also satisfies

$$\begin{aligned} \tau ^k_n=\inf \left\{ t\in [0,T]\mid \frac{{{\mathcal {G}}}^k_n(t)\circ v^{k-1}}{\beta (t)}+{\bar{p}}^k{\bar{\alpha }}^k\le 0\right\} . \end{aligned}$$

(B.24)

But this and (a[k]) will lead to

$$\begin{aligned} \frac{{{\mathcal {G}}}^k_n(t)\circ v^{k-1}}{\beta (t)}+{\bar{p}}^k{\bar{\alpha }}^k\le 0,{\quad }\forall t\in (\tau ^k_n,T). \end{aligned}$$

(B.25)

Combining (B.23), (B.25), and the strict positivity of $\beta (\cdot )$, we arrive to

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ v^{k-1}+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\left\{ \begin{array}{ll} >0, &{} \;\;\;\;\;\;\forall t\in (0,\tau ^k_n),\\ \le 0, &{} \;\;\;\;\;\;\forall t\in (\tau ^k_n,T). \end{array}\right. \end{aligned}$$

(B.26)

Our construction (B.6) and (b$[k-1]$) dictate that

$$\begin{aligned} u^k_n(t)=v^{k-1}_n(t),{\quad }\forall t\in [\tau ^k_n,T]. \end{aligned}$$

(B.27)

Also, we may see that the construction (B.7) renders

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ u^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0,{\quad }\forall t\in (0,\tau ^k_n). \end{aligned}$$

(B.28)

From the first half of (B.26), we have, for any $n=1,2,\ldots ,N$ and $t\in [0,\tau ^k_n)$,

$$\begin{aligned} \begin{array}{l} v^{k-1}_n(t)<v^{k-1}_n(\tau ^k_n)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,\tau ^k_n))\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+{\bar{\alpha }}^k\cdot \int _t^{\tau ^k_n}\beta (s)\cdot ({\bar{p}}^k+v^{k-1}_{n-1}(s))\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s))\cdot ds. \end{array} \end{aligned}$$

(B.29)

Using (B.7) and (B.29), as well as the fact that $u^k_0(t)=v^{k-1}_0(t)=0$ for any $t\in [0,T]$, we can use induction over n to prove that

$$\begin{aligned} u^k_n(t)>v^{k-1}_n(t),{\quad }\forall n=1,2,\ldots ,N\; \text{ and } t\in [0,\tau ^k_n). \end{aligned}$$

(B.30)

By its construction, $u^k_n(t)$ is uniformly bounded by $N{\overline{\lambda }} T$; it is also Lipschitz continuous in t with coefficient $N{\overline{\lambda }}$, and hence absolutely continuous in t. By (B.26)–(B.28), and (B.30), as well as the fact that $v^{k-1}_n(T)=0$ for every n, we may see that $u^k\equiv (u^k_n(t)\mid n=0,1,\ldots ,N,\;t\in [0,T])$ satisfies the sufficient conditions (i) to (iv) stipulated in Proposition 5. Hence, we have shown (b[k]), that $u^k_n(t)=v^k_n(t)$ for every $n=0,1,\ldots ,N$ and $t\in [0,T]$.

From the second half of (B.26), (B.27), (B.28), and (b[k]), we easily have (c[k]).

Now take $n=1,2,\ldots ,N-1$. By (S1’), (e$[k-1]$), and Proposition 6, we have

$$\begin{aligned} \begin{array}{ll} d_t v^{k-1}_{n+1}(t)-d_t v^{k-1}_n(t)&{}\le {\bar{\alpha }}^{k-1}\cdot \beta (t)\cdot (v^{k-1}_{n-1}(t)-2v^{k-1}_n(t)+v^{k-1}_{n+1}(t))\\ &{}\le {\bar{\alpha }}^k\cdot \beta (t)\cdot (v^{k-1}_{n-1}(t)-2v^{k-1}_n(t)+v^{k-1}_{n+1}(t))\le 0. \end{array} \end{aligned}$$

(B.31)

We therefore have the negativity of

$$\begin{aligned} \begin{array}{l} {{\mathcal {G}}}^k_{n+1}(t)\circ v^{k-1}-{{\mathcal {G}}}^k_n(t)\circ v^{k-1}\\ \;\;\;\;\;\;\;\;\;\;\;\;=d_t v^{k-1}_{n+1}(t)-d_t v^{k-1}_n(t)+{\bar{\alpha }}^k\cdot \beta (t)\cdot \left( 2v^{k-1}_n(t)-v^{k-1}_{n-1}(t)-v^{k-1}_{n+1}(t)\right) . \end{array} \end{aligned}$$

(B.32)

But by the definition of $\tau ^k_n$ and $\tau ^k_{n+1}$, this leads to $\tau ^k_{n+1}\le \tau ^k_n$. Therefore, we have (d[k]).

Let us turn to the proof of (e[k]). When $t\in (\tau ^k_1,T)$, we have

$$\begin{aligned} d_t v^k_1(t)=d_t v^{k-1}_1(t), \end{aligned}$$

(B.33)

which is negative by (B.16). When $t\in (0,\tau ^k_1)$, we have

$$\begin{aligned} \begin{array}{ll} d_t v^k_1(t)&{}={\bar{\alpha }}^k\cdot \beta (t)\cdot (v^k_1(t)-{\bar{p}}^k)\\ &{}={\bar{\alpha }}^k\cdot \beta (t)\cdot [v^{k-1}_1(\tau ^k_1)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,\tau ^k_1))\\ &{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+{\bar{p}}^k{\bar{\alpha }}^k\cdot \int _t^{\tau ^k_1}\beta (s)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s))\cdot ds-{\bar{p}}^k]\\ &{}={\bar{\alpha }}^k\cdot \beta (t)\cdot \exp (-{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,\tau ^k_1))\cdot (v^{k-1}_1(\tau ^k_1)-{\bar{p}}^k), \end{array} \end{aligned}$$

(B.34)

where the first equality is due to (c[k]), the second equality is from (b[k]) and (B.7), and the last equality is by the following result from integration by parts:

$$\begin{aligned} {\bar{\alpha }}^k\cdot \int _t^{\tau ^k_1}\beta (s)\cdot \exp \left( -{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,s)\right) \cdot ds=1-\exp \left( -{\bar{\alpha }}^k\cdot {{\hat{\beta }}}(t,\tau ^k_1)\right) . \end{aligned}$$

(B.35)

But from (B.1), (B.2), and the fact that ${\bar{p}}^k>{\bar{p}}^{k-1}>\cdots >{\bar{p}}^0$, it is obvious that $v^{k-1}_1(\tau ^k_1)\le {\bar{p}}^{k-1}<{\bar{p}}^k$. So we know that $d_t v^k_1(t)\le 0$ for $t\in (0,\tau ^k_1)$ as well.

Now let $n=1,2,\ldots ,N-1$. For $t\in (\tau ^k_n,T)$, we have $t\in (\tau ^k_{n+1},T)$ as well due to (d[k]). Thus, we have

$$\begin{aligned} v^k_n(t)=v^{k-1}_n(t)\; \text{ and } \;v^k_{n+1}(t)=v^{k-1}_{n+1}(t), \end{aligned}$$

(B.36)

and hence

$$\begin{aligned} \begin{array}{ll} d_t v^k_{n+1}(t)-d_t v^k_n(t)&{}=d_t v^{k-1}_{n+1}(t)-d_t v^{k-1}_n(t)\\ &{}\le {\bar{\alpha }}^{k-1}\cdot \beta (t)\cdot (v^{k-1}_{n-1}(t)-2v^{k-1}_n(t)+v^{k-1}_{n+1}(t))\\ &{}\le {\bar{\alpha }}^k\cdot \beta (t)\cdot (v^{k-1}_{n-1}(t)-2v^{k-1}_n(t)+v^{k-1}_{n+1}(t))\\ &{}\le {\bar{\alpha }}^k\cdot \beta (t)\cdot (v^k_{n-1}(t)-2v^k_n(t)+v^k_{n+1}(t)), \end{array} \end{aligned}$$

(B.37)

where the first inequality is by (e$[k-1]$), the second inequality is by (S1’) and Proposition 6, and the last inequality is from (B.36) and the fact that $v^k_{n-1}(t)\ge v^{k-1}_{n-1}(t)$. For $t\in (0,\tau ^k_n)$, we have, by (c[k]),

$$\begin{aligned} {{\mathcal {G}}}^k_{n+1}(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)\le 0={{\mathcal {G}}}^k_n(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t). \end{aligned}$$

(B.38)

This leads to

$$\begin{aligned} d_t v^k_{n+1}(t)-d_t v^k_n(t)\le {\bar{\alpha }}^k\cdot \beta (t)\cdot (v^k_{n-1}(t)-2v^k_n(t)+v^k_{n+1}(t)). \end{aligned}$$

(B.39)

Therefore, we have (e[k]).

We have thus completed the induction process. Therefore, (a[k]) to (e[k]) are all true for $k=0,1,\ldots ,K$. From these, we see that, for any $k=1,2,\ldots ,K$ and $n=1,2,\ldots ,N$,

$$\begin{aligned} v^k_n(t)\left\{ \begin{array}{ll} >v^{k-1}_n(t),&{}{\quad }\forall t\in [0,\tau ^k_n),\\ =v^{k-1}_n(t),&{}{\quad }\forall t\in [\tau ^k_n,T]. \end{array}\right. \end{aligned}$$

(B.40)

Hence, we may see that each $\tau ^k_n$ offers an optimal time beyond which the firm is to drop its price from ${\bar{p}}^k$ to ${\bar{p}}^{k-1}$ when it has n remaining items. $\square$

For the thus constructed threshold levels, it may be tempting to conjecture that $\tau ^k_n$ is decreasing in k as well. We have the following relevant result.

Proposition 7

Suppose $\tau ^k_n>0$ for some $k=1,2,\ldots ,K-1$ and $n=1,2,\ldots ,N$. Then, we have $\tau ^{k+1}_n\le \tau ^k_n$ if and only if

$$\begin{aligned} v^k_n(\tau ^k_n)-v^k_{n-1}(\tau ^k_n)\le \frac{{\bar{p}}^k{\bar{\alpha }}^k-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}}{{\bar{\alpha }}^k-{\bar{\alpha }}^{k+1}}. \end{aligned}$$

Proof

If $\tau ^k_n=T$, we have both $\tau ^{k+1}_n\le T=\tau ^k_n$ and, due to (S1),

$$\begin{aligned} v^k_n(T)-v^k_{n-1}(T)=0\le \frac{{\bar{p}}^k{\bar{\alpha }}^k-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}}{{\bar{\alpha }}^k-{\bar{\alpha }}^{k+1}}. \end{aligned}$$

(B.41)

Now suppose $\tau ^k_n\in (0,T)$. From Theorem 2, we know that

$$\begin{aligned} {{\mathcal {G}}}^k_n(t)\circ v^k+{\bar{p}}^k{\bar{\alpha }}^k\cdot \beta (t)=0,{\quad }\forall t\in (0,\tau ^k_n]; \end{aligned}$$

(B.42)

we have $\tau ^{k+1}_n\le \tau ^k_n$ if and only if

$$\begin{aligned} {{\mathcal {G}}}^{k+1}_n(\tau ^k_n)\circ v^k+{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}\cdot \beta (\tau ^k_n)\le 0. \end{aligned}$$

(B.43)

Due to (S1’) and (B.42), the above (B.43) is equivalent to

$$\begin{aligned} v^k_n(\tau ^k_n)-v^k_{n-1}(\tau ^k_n)\le \frac{{\bar{p}}^k{\bar{\alpha }}^k-{\bar{p}}^{k+1}{\bar{\alpha }}^{k+1}}{{\bar{\alpha }}^k-{\bar{\alpha }}^{k+1}}. \end{aligned}$$

(B.44)

This completes our proof $\square$

In the last inequality in Proposition 7, the left-hand side is independent of ${\bar{p}}^{k+1}$ or ${\bar{\alpha }}^{k+1}$; yet, the right-hand side is dependent on both, and can be arbitrarily small. Hence, we may see that $\tau ^k_n$ is not necessarily decreasing in k. Thence, there exists a possibility for the following “leapfrog” phenomenon: Right after the time has gone past $\tau ^k_n$, it has also passed the time $\tau ^{k-1}_n$, and so on and so forth. Therefore, when the time passes beyond $\tau ^k_n$ and it is currently charging ${\bar{p}}^k$ while with n items, the firm should ultimately switch to some price ${\bar{p}}^{{\tilde{k}}^-(k,n)}$, where ${\tilde{k}}^-(k,n)$ is not necessarily $k-1$. For each $n=1,2,\ldots ,N$, we can use the following iterative procedure to find $({\tilde{k}}^-(k,n)\mid k=1,2,\ldots ,K)$:

One of our computational studies shall confirm that a threshold policy for the markdown case is not necessarily k-monotone.

We can adapt the recursive procedure described from (B.3) to (B.7) to the following algorithm Markdown.

The algorithm’s time complexity is apparently O(KNQ).

Other algorithms

For the markup case, there is always the brute-force method Markup2.

In this algorithm, each $p^k_{nq}$ indicates the new price taken by the firm when it has n items and is charging price ${\bar{p}}^k$ right before time $q\cdot \Delta T$.

For the markdown case, there is also the brute-force method Markdown2.

Here, $p^k_{nq}$ means the same as in Markup2.

For the reversible-pricing case, we can translate Zhao and Zheng’s (2000) method in Section 4.2 into the following algorithm Reversible.

For this case, there is again the brute-force method Reversible2.

In the algorithm, each $p_{nq}$ stores the price to be taken at time $q\cdot \Delta T$ when the firm has n items at that time.

Now we introduce the three algorithms needed in computation. First, the following is algorithm Markup3.

Next comes algorithm Markdown3.

Finally, the following is algorithm Reversible3.