Optimal booking control in revenue management with two substitutable resources

Abstract

This paper studies optimal booking policies for capacity control models in revenue management with two substitutable resources. Our model covers a broader class of problems than previous works including (i) flexible demand and opaque selling for (ii) both dynamic and static demand settings. We provide a unifying characterization of the structure of optimal booking control by exploiting concavity, submodularity, and subconcavity of the value function. Our characterization is based on the notion of optimal “booking paths” formalizing the idea that an optimal allocation of a demand batch decomposes into a sequence of optimal single-request allocations. In addition, we examine the relationship between our booking path-based and a switching curve-based policy, which has been known previously for the case with dynamic demand. We show that both these characterizations describe an optimal policy. Computationally, there is no advantage of implementing either switching curves or booking paths in the dynamic setting. In the static setting, however, one can resort to the simple criteria which we propose in order to construct the optimal booking paths, thereby accelerating the evaluation of the value function.

Appendices

Appendix

A Proof of Lemma 1

Lemma 1

Let any $x\in \mathbb {Z}^2_+$ be given. The following statement holds:

$$\begin{aligned} \mathop {H^{d}_{}}(V)(x) = \max _{u\in \mathcal {U}(d)}\Big \{ r^\top u+ V(x+ u)\Big \} \quad \forall d\in \mathbb {Z}_+. \end{aligned}$$

Proof

For demands $d\in \{0,1\}$, the result is trivial. For demands $d>1$, we prove the result by induction over the demand levels. Assume that the result holds for $d-1$. Then, we have

The first equality is the definition (7), the second equality follows from the induction assumption, and the third equality follows from simple rearrangement. To see the last equality, we define for vectors v and for sets $\mathcal {M}$ of vectors that $\mathcal {M}+v=\{m+v:m\in \mathcal {M}\}$. Noting the equality

$$\begin{aligned} \mathcal {U}(d) = \big (\mathcal {U}(d-1)+e_1\big ) \cup \big (\mathcal {U}(d-1)+e_2\big )\cup \mathcal {U}(d-1) \end{aligned}$$

completes the proof. $\square$

B Proof of Corollary 1

Corollary 1

Let $V_t,t\in \mathcal {T}$ be defined by either (RM-d) or (RM-s). In both cases, we have that

• (i) $V_t$ is multimodular for all $t\in \mathcal {T}$;

• (ii) $\varDelta _iV_t(x)$ is nondecreasing in $t$ for all $x\in \mathcal {X}$ and $i\in \mathcal {I}$.

Proof

The Lemmata B.7 and B.10 of Sect. 2 show, respectively, parts (i) and (ii) of Corollary 1 for (RM-d). The Lemmata B.12 and B.13 of Sect. 3 yield, respectively, parts (i) and (ii) of Corollary 1 for (RM-s). $\square$

1.1 B.1 Equivalent conditions for submodularity and subconcavity

Observe that the linear operator $\varDelta _i$ satisfies

$$\begin{aligned} \varDelta _i\varDelta _h f(x) = \varDelta _h\varDelta _i f(x) \quad \forall i,h\in \{1,2\}. \end{aligned}$$

(11)

Using the above equality and the definition of submodularity, we next state equivalent conditions for submodular functions (see also Chapter 10 in Sundaram 1996, for a similar discussion in terms of supermodular functions).

Lemma B.1

The following statements are equivalent:

• 1. The function $f:\mathbb {Z}^2_+ \rightarrow \mathbb {R}$ is submodular.

• 2. $\varDelta _i\varDelta _h f(x)\le 0$ for all $x \in \mathbb {Z}^2_+$ and $i,h\in \{1,2\}, i\ne h$.

• 3. $f(x) - f(x + e_i)$ nondecreasing in $x_h$ for all $x \in \mathbb {Z}^2_+$ and $i,h\in \{1,2\}, i\ne h$.

Proof

Given any $i,h\in \{1,2\},i \ne h$, it follows from the definition of submodularity (ii) using Eq. (11) that for any $x\in \mathbb {Z}^2_+$

$$\begin{aligned} \Leftrightarrow \quad \varDelta _i f(x)\le {}&\varDelta _i f(x+e_h)\\ \Leftrightarrow \quad 0\ge {}&\varDelta _i f(x) - \varDelta _i f(x+e_h) = \varDelta _i \varDelta _h f(x), \end{aligned}$$

which is part 2. Moreover, we have

$$\begin{aligned} \Leftrightarrow \quad \varDelta _i f(x)\le {}&\varDelta _i f(x+e_h)\\ \Leftrightarrow \quad f(x) - f(x+e_i) \le {}&f(x+e_h) - f(x+e_i+e_h). \end{aligned}$$

This shows part 3. and completes the proof.

Lemma B.2

The following statements are equivalent:

• 1. The function $f:\mathbb {Z}^2_+ \rightarrow \mathbb {R}$ is subconcave.

• 2. $\varDelta _1 f(x+e_2) - \varDelta _2 f(x +e_2) \le \varDelta _1 f(x) - \varDelta _2 f(x) \le \varDelta _1 f(x +e_1)-\varDelta _2 f(x+e_1)$ for all $x \in \mathbb {Z}^2_+$.

• 3. $f(x + e_{1}) - f(x + e_{2})$ nondecreasing in $x_{2}$ for all $x \in \mathbb {Z}^2_+$.

• 4. $f(x + e_{2}) - f(x + e_{1})$ nondecreasing in $x_{1}$ for all $x \in \mathbb {Z}^2_+$.

Proof

Given any $i,h\in \{1,2\},i \ne h$, it follows that for all $x\in \mathbb {Z}^2_+$

$$\begin{aligned}&\varDelta _i\varDelta _i f(x)\le {}\varDelta _i\varDelta _h f(x)\\&\quad \Leftrightarrow \varDelta _i f(x) - \varDelta _i f(x+e_i)\le {}\varDelta _h f(x) - \varDelta _h f(x+e_i)\\&\quad \Leftrightarrow \varDelta _i f(x) - \varDelta _h f(x)\le {}\varDelta _i f(x+e_i) - \varDelta _h f(x+e_i)\\&\quad \Leftrightarrow f(x+e_h) - f(x+e_i) \le {}f(x+e_i+e_h) - f(x+2e_i). \end{aligned}$$

Taking $i=1,h=2$ and $h=1,i=2$, we obtain part 2. from the third inequality and both parts 3. and 4. result from the last inequality. This completes the proof. $\square$

1.2 B:2 Monotonicity properties of the dynamic model

Lemma B.3

Let any $j\in \mathcal {J}$ and $t\in \mathcal {T}$ be given, and let $V_{t-1}$ be defined by (RM-d). If $V_{t-1}$ is multimodular, then $\mathop {H^{1}_{j}}(V_{t-1})$ is multimodular.

Proof

By Definition 2, this lemma builds on the fact that $\mathop {H^{1}_{j}}(V_{t-1})$ preserves the (i) submodularity and (ii) subconcavity which are both induced by $V_{t-1}$. The proof of (i) and (ii) is moved to Lemmata B.5 and B.6, respectively. $\square$

We verify the submodularity and subconcavity of $\mathop {H^{1}_{j}}(V_{t-1})$ in a case-by-case fashion making use of a result of Zhuang and Li (2010) (see their Lemma 2 on p. 463) in order to reduce the number of cases that need to be checked. We briefly restate their result in the next lemma.

Lemma B.4

Let $\{a_i,b_i,\tilde{a}_i,\tilde{b}_i:i=1,\dots ,m\}$ be four arbitrary real sequences. Assume that there exist $k_1,k_2\in \{1,\dots ,m\}$ such that

$$\begin{aligned} a_i - b_{k_1} \le \tilde{a}_{k_2} - \tilde{b}_j \quad \forall i,j\in \{1,\dots ,m\}. \end{aligned}$$

(12)

Then the following inequality holds,

$$\begin{aligned} \max \{a_1,\dots ,a_m\}-\max \{b_1,\dots ,b_m\}\le \max \{\tilde{a}_1,\dots ,\tilde{a}_m\}-\max \{\tilde{b}_1,\dots ,\tilde{b}_m\}. \end{aligned}$$

(13)

Proof

Using (12), we can state that

$$\begin{aligned} a_1 - b_{k_1}&\le \tilde{a}_{k_2} - b_j\\&\,\,\, {\vdots }\\ a_m - b_{k_1}&\le \tilde{a}_{k_2} - b_j \end{aligned}$$

hold for $j\in \{1,\dots ,m\}$ giving us the inequality

$$\begin{aligned} \max \{a_1,\dots ,a_m\} - b_{k_1} \le \tilde{a}_{k_2} - b_j \quad \forall j\in \{1,\dots ,m\}. \end{aligned}$$

(14)

(14) immediately leads us to

$$\begin{aligned} \max \{a_1,\dots ,a_m\} - b_{k_1} \le \tilde{a}_{k_2} - \max \{b_1,\dots ,b_m\}. \end{aligned}$$

(15)

Since $b_{k_1}\le \max \{b_1,\dots ,b_m\}$, it follows that $\max \{a_1,\dots ,a_m\} - b_{k_1}\ge \max \{a_1,\dots ,a_m\} - \max \{b_1,\dots ,b_m\}$. Thus, we can replace $b_{k_1}$ by $\max \{b_1,\dots ,b_m\}$ on the left hand side of (15) yielding

$$\begin{aligned} \max \{a_1,\dots ,a_m\} - \max \{b_1,\dots ,b_m\} \le \tilde{a}_{k_2} - \max \{b_1,\dots ,b_m\}. \end{aligned}$$

(16)

Likewise, it is possible to substitute $\tilde{a}_{k_2}$ by $\max \{\tilde{a}_1,\dots ,\tilde{a}_m\}$ on the right hand side of (16) in order to obtain the inequality (13). This completes the proof. $\square$

Remark 1

The main benefit of the above lemma is that verifying (12) requires to check $m^2$ inequalities, while verifying (13) requires to check $m^4$ inequalities.

Remark 2

Note that Lemma B.4 assumes $k_1$ and $k_2$ solving all inequalities in (12) simultaneously. In general, it might be easier to find $k^1_{ij}$ and $k^2_{ij}$ such that

$$\begin{aligned} a_i - b_{k^1_{ij}} \le \tilde{a}_{k^2_{ij}} - \tilde{b}_j \quad \forall i,j\in \{1,\dots ,m\}. \end{aligned}$$

In this case, the above result still follows if we use the definition $b_{k_1}=\max _{ij}\{b_{k^1_{ij}}\}$ and $\tilde{a}_{k_2}=\max _{ij}\{\tilde{a}_{k^2_{ij}}\}$.

To ease the subsequent discussion, we simplify the notation $\mathop {H^{1}_{j}}(V_{t-1})(x)$ to

$$\begin{aligned} \mathop {H_{}}(f)(x)=\max \left\{ r^1+ f(x +e_1), r^2+ f(x +e_2), f(x) \right\} \quad \forall x\in \mathbb {Z}^2_+. \end{aligned}$$

(17)

This notational simplification is without loss of generality because the results of both Lemmata B.5 and B.6 neither depend on the time index $t$ nor on the product index $j$.

Lemma B.5

If f is multimodular, then the operator $\mathop {H_{}}(f)$ is submodular.

Proof

By part 3. of Lemma B.1, $\mathop {H_{}}(f)$ is submodular if

$$\begin{aligned} \mathop {H_{}}(f)(x) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1) - \mathop {H_{}}(f)(x +e_1+ e_2). \end{aligned}$$

(18)

In what follows, we construct a case for each pair of values for $\mathop {H_{}}(f)(x)$ and $\mathop {H_{}}(f)(x+e_1+e_2)$. Each case is referred to as a pair of the identifiers shown in the first and third column of Table 5a.

By assumption, f is multimodular and hence implies for all $x\in \mathbb {Z}^2_+$ that

$$\begin{aligned} f(x)-f(x+e_2)&\le f(x+e_1)-f(x+e_1+e_2) \end{aligned}$$

(19)

$$\begin{aligned} f(x) - f(x+e_1)&\le f(x+e_1)-f(x+2e_1) \end{aligned}$$

(20)

$$\begin{aligned} f(x) - f(x+e_2)&\le f(x+e_2)-f(x+2e_2), \end{aligned}$$

(21)

where the first inequality is the submodularity of f (see also part 3. of Lemma B.1). Both inequalities (20) and (21) follow from the concavity of f. Starting with case (3.a), note that

$$\begin{aligned}&f(x) - \mathop {H_{}}(f)(x+e_2)\nonumber \\&\quad \le f(x) - f(x + e_2) \le f(x+e_1) - f(x+e_1+e_2)\nonumber \\&\quad \le r^1+f(x+2e_1) - f(x+2e_1+e_2)-r^1\nonumber \\&\quad \le \mathop {H_{}}(f)(x +e_1) - f(x+2e_1+e_2)-r^1, \end{aligned}$$

(22)

where the first inequality is true because $\mathop {H_{}}(f)(x+e_2) \ge f(x + e_2)$, the second and third inequalities follow from (19), and the last inequality holds because $\mathop {H_{}}(f)(x +e_1) \ge r^1+f(x+2e_1)$. Using the inequalities (19) and (21), we get case (3.b):

$$\begin{aligned}&f(x) - \mathop {H_{}}(f)(x+e_2)\nonumber \\&\quad \le f(x) - f(x + e_2) \le f(x+e_1) - f(x+e_1+e_2)\nonumber \\&\quad \le r^2+ f(x+e_1+e_2) -r^2-f(x+e_1+2e_2)\nonumber \\&\quad \le \mathop {H_{}}(f)(x +e_1) -r^2-f(x+e_1+2e_2). \end{aligned}$$

(23)

The case (3.c) can be established similarly with inequality (19), i.e., we obtain

$$\begin{aligned} f(x) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1) - f(x+e_1+e_2). \end{aligned}$$

(24)

Lemma B.4 allows us to combine the results (22), (23), and (24) to

$$\begin{aligned} f(x) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1) - \mathop {H_{}}(f)(x +e_1+ e_2). \end{aligned}$$

(25)

Likewise, the case (1.a) is constructed using (19), while the case (1.b) requires (21). Noting that case (1.c) is trivial because $r^1+ f(x +e_1) - r^1- f(x +e_1+e_2) = f(x +e_1) - f(x +e_1+e_2)$, we can conclude from the last three cases that

$$\begin{aligned} r^1+ f(x+e_1) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1) - \mathop {H_{}}(f)(x +e_1+ e_2). \end{aligned}$$

(26)

The case (2.c) is trivial because $r^2+ f(x+e_2) - f(x+e_2)=r^2+ f(x+e_1+e_2) - f(x+e_1+e_2)$. The cases (2.a) and (2.b) are derived using (20) and (19), respectively. Hence,

$$\begin{aligned} r^2+f(x +e_2) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1) - \mathop {H_{}}(f)(x +e_1+ e_2). \end{aligned}$$

(27)

Finally, combining (25), (26), and (27) yields the result (18) and completes the proof. $\square$

Table 5 Definition of identifiers (ID) for the cases of Lemmata B.5 and B.6

Lemma B.6

If f is multimodular, then the operator $\mathop {H_{}}(f)$ is subconcave.

Proof

By part 3. of Lemma B.2, $\mathop {H_{}}(f)$ is subconcave if

$$\begin{aligned} \mathop {H_{}}(f)(x+e_1) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1+ e_2) - \mathop {H_{}}(f)(x + 2e_2). \end{aligned}$$

(28)

We construct the subsequent cases by considering $\mathop {H_{}}(f)(x+e_1)$ and $\mathop {H_{}}(f)(x+2e_2)$, where each case is referred to as a pair of the identifiers depicted in columns one and three of Table 5b. Note that f is multimodular by assumption which implies for all $x\in \mathbb {Z}^2_+$:

$$\begin{aligned} f(x+e_1)-f(x+e_2)&\le f(x+e_1+e_2)-f(x+2e_2) \end{aligned}$$

(29)

$$\begin{aligned} f(x+e_1)-f(x+e_2)&\ge f(x+2e_1)-f(x+e_1+e_2) \end{aligned}$$

(30)

$$\begin{aligned} f(x) - f(x+e_2)&\le f(x+e_2)-f(x+2e_2), \end{aligned}$$

(31)

where the first inequality is part 3. of Lemma B.2 and the second inequality follows from multiplying part 4. of Lemma B.2 by $-1$. Inequality (31) is the definition of concavity in $x_2$.

Starting with case (4.f), note that

$$\begin{aligned} \begin{aligned}&r^1+ f(x+2e_1) - \mathop {H_{}}(f)(x+e_2) \le r^1+ f(x+2e_1) - r^1- f(x +e_1+ e_2) \\&\quad \le f(x+e_1+e_2) - f(x+2e_2) \le \mathop {H_{}}(f)(x +e_1+e_2) - f(x+2e_2), \end{aligned} \end{aligned}$$

(32)

where the first and last inequalities are true because $\mathop {H_{}}(f)(x+e_2) \ge r^1+ f(x + e_1+ e_2)$ and $\mathop {H_{}}(f)(x +e_1+e_2)\ge f(x+e_1+e_2)$, respectively, and the second inequality holds by (29) and (30). Likewise, we obtain case (4.d) using (29) and case (4.e) using (29) and (30). Hence, we can conclude that

$$\begin{aligned} r^1+ f(x+2e_1) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1+e_2) - \mathop {H_{}}(f)(x +2e_2). \end{aligned}$$

(33)

The cases (5.d) and (5.f) are readily available, while case (5.e) is achieved by applying inequality (30). The cases (5.d)–(5.f) lead to

$$\begin{aligned} r^2+ f(x+e_1+e_2) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1+e_2) - \mathop {H_{}}(f)(x +2e_2). \end{aligned}$$

(34)

The cases (6.e) and (6.f) are obtained with (29) and case (6.d) with (31), and together they imply

$$\begin{aligned} f(x+e_1) - \mathop {H_{}}(f)(x+e_2) \le \mathop {H_{}}(f)(x +e_1+e_2) - \mathop {H_{}}(f)(x +2e_2). \end{aligned}$$

(35)

Finally, we achieve the result (28) by combining (33), (34), and (35). This completes the proof. $\square$

Remark 3

Even though a multimodular function is the precondition of both Lemmata B.5 and B.6, the cases of Lemma B.5 are in fact established by using only concavity and submodularity. Similarly, the minimum requirements for the function f of Lemma B.6 are in fact concavity and subconcavity.

Lemma B.7

Let $V_t,t\in \mathcal {T}$ be defined by (RM-d). $V_t$ is multimodular for all $t\in \mathcal {T}$.

Proof

The proof is by induction over the time periods. For $t=0$, the boundary conditions (4) imply the submodularity

$$\begin{aligned} V_0(x) - V_0(x+e_i) - V_0(x+e_h) + V_0(x+e_i + e_h) = 0 \end{aligned}$$

and the subconcavity

$$\begin{aligned} V_0(x+e_i) - V_0(x+e_h) -V_0(x+e_i+e_h) + V_0(x+2e_h) ={}&{\left\{ \begin{array}{ll} -\bar{r}&{}\text { if }x_h = c_h - 1\\ 0&{}\text { otherwise} \end{array}\right. } \end{aligned}$$

for all $x\in \mathcal {X}$ and $i,h\in \{1,2\},i\ne h$. This shows that $V_0$ is multimodular, see Definition 2.

Assume that the result holds for $t-1$, i.e., $V_{t-1}$ is multimodular and herewith $\mathop {H^{0}_{j}}(V_{t-1})$. Applying Lemma B.3 to Eq. (6) yields that $V_t$ is a nonnegative weighted sum of multimodular functions, hence, multimodular. Repeating this argument for all $t\in \mathcal {T}$ completes the proof. $\square$

The two following Lemmata B.8 and B.9 provide lower and upper bounds which will be helpful to analyze the time monotonicity in Lemma B.10. To this end, we introduce two new operators $\mathop {T_{1}}(f)(x)$ and $\mathop {T_{2}}(f)(x)$, i.e., we define for $x\in \mathbb {Z}^2_+$

$$\begin{aligned} \mathop {T_{1}}(f)(x)= & {} \max \left\{ r^1+ f(x + e_1), f(x) \right\} \text { and } \mathop {T_{2}}(f)(x)\\= & {} \max \left\{ r^2+ f(x + e_2), f(x) \right\} . \end{aligned}$$

Now, the operator $\mathop {H_{}}(f)(x)$ as defined in (17) can be written in terms of $\mathop {T_{1}}(f)(x)$ und $\mathop {T_{2}}(f)(x)$, i.e.,

$$\begin{aligned} \mathop {H_{}}(f)(x)={}&\max \left\{ \mathop {T_{1}}(f)(x), \mathop {T_{2}}(f)(x) \right\} \quad \forall x\in \mathbb {Z}^2_+. \end{aligned}$$

Lemma B.8

Let $f:\mathbb {Z}^2_+\mapsto \mathbb {R}$ be a real-valued function. The following inequalities hold for $x,y\in \mathbb {Z}^2_+$:

$$\begin{aligned}&\min \left\{ \mathop {T_{1}}(f)(x)-\mathop {T_{1}}(f)(y),\mathop {T_{2}}(f)(x)-\mathop {T_{2}}(f)(y)\right\} \nonumber \\&\quad \le \mathop {H_{}}(f)(x)-\mathop {H_{}}(f)(y) \le \max \left\{ \mathop {T_{1}}(f)(x)-\mathop {T_{1}}(f)(y),\mathop {T_{2}}(f)(x)-\mathop {T_{2}}(f)(y)\right\} .\nonumber \\ \end{aligned}$$

(36)

Proof

Choose any $x,y\in \mathbb {Z}^2_+$. Consider the case $\mathop {H_{}}(f)(y)=\mathop {T_{1}}(f)(y)$. We have

$$\begin{aligned} \mathop {H_{}}(f)(x) - \mathop {H_{}}(f)(y) \ge \mathop {T_{1}}(f)(x) - \mathop {T_{1}}(f)(y) \end{aligned}$$

because $\mathop {H_{}}(f)(x)\ge \mathop {T_{1}}(f)(x)$. The case $\mathop {H_{}}(f)(y)=\mathop {T_{2}}(f)(y)$ leads to

$$\begin{aligned} \mathop {H_{}}(f)(x) - \mathop {H_{}}(f)(y) \ge \mathop {T_{2}}(f)(x) - \mathop {T_{2}}(f)(y) \end{aligned}$$

because $\mathop {H_{}}(f)(x)\ge \mathop {T_{2}}(f)(x)$. Both inequalities above imply the first “$\le$” in (36). Similarly, the second “$\le$” in (36) follows from the analysis of the cases $\mathop {H_{}}(f)(x)= \mathop {T_{1}}(f)(x)$ and $\mathop {H_{}}(f)(x)= \mathop {T_{2}}(f)(x)$. The proof is complete as our choice of x, y was arbitrary. $\square$

In the next lemma, we establish bounds for the single-mode operators $\mathop {T_{1}}(f)$ and $\mathop {T_{2}}(f)$.

Lemma B.9

Let $f:\mathbb {Z}^2_+\mapsto \mathbb {R}$ be a real-valued function. For all $i\in \{1,2\}$ and $x,y\in \mathbb {Z}^2_+$, we have

$$\begin{aligned}&\min \{f(x)-f(y),f(x+e_i) - f(y+e_i)\}\nonumber \\&\quad \le \mathop {T_{i}}(f)(x)-\mathop {T_{i}}(f)(y)\le \max \{f(x)-f(y),f(x+e_i) - f(y+e_i)\}. \end{aligned}$$

(37)

Proof

Similar to the proof of Lemma B.8. $\square$

In what follows, we use the shorthand notation $[x]^+=\max \{x,0\}$.

Lemma B.10

Let $V_t,t\in \mathcal {T}$ be defined by (RM-d). The following statement holds: $\varDelta _iV_t(x)$ is nondecreasing in $t$ for all $x\in \mathcal {X}$ and $i\in \mathcal {I}$.

Proof

The condition of this lemma can be equivalently stated as:

$$\begin{aligned} \varDelta _iV_t(x) - \varDelta _iV_{t-1}(x)\ge 0\quad \forall i\in \mathcal {I},t\in \mathcal {T},x\in \mathcal {X}. \end{aligned}$$

(38)

Let $\mathop {F_{1}}(V)(x)=\big [r^1-\varDelta _1 V(x)\big ]^+$ and $\mathop {F_{2}}(V)(x)=\big [r^2-\varDelta _2 V(x)\big ]^+$ for all $x\in \mathbb {Z}^2_+$. Moreover, we define the identity operator $\mathop {G^{0}_{}}(V_{t-1})=\varDelta V_{t-1}$ and the operator

$$\begin{aligned} \mathop {G^{1}_{}}(V)(x) ={}&\max \left\{ r^1-\varDelta _1 V(x), r^2- \varDelta _2 V(x), 0\right\} \nonumber \\ ={}&\max \left\{ \mathop {F_{1}}(V)(x),\mathop {F_{2}}(V)(x)\right\} \quad \forall x\in \mathbb {Z}^2_+. \end{aligned}$$

(39)

It is straightforward to show that the right hand side of (RM-d) can be expressed as

$$\begin{aligned} V_t(x) ={}&V_{t-1}(x) + \sum _{j\in \mathcal {J}}p_{j1t}\max \{r^1_j- \varDelta _1V_{t-1}(x), r^2_j- \varDelta _2V_{t-1}(x), 0\} \end{aligned}$$

(40)

$$\begin{aligned} ={}&V_{t-1}(x) +\sum _{j\in \mathcal {J}}p_{j1t}\mathop {G^{1}_{j}}(V_{t-1})(x) \quad \forall t\in \mathcal {T},x\in \mathcal {X}, \end{aligned}$$

(41)

where the second equality follows by setting $\mathop {G^{1}_{}}=\mathop {G^{1}_{j}},V=V_{t-1}$, and $r=r_j$. Using the definition of $\varDelta _i$ and the equality (41), the left hand side of (38) can be rewritten as

$$\begin{aligned}&\varDelta _iV_t(x) - \varDelta _iV_{t-1} (x)\\&\quad ={} \sum _{j\in \mathcal {J}}p_{j1t}\bigg [\mathop {G^{1}_{j}}(V_{t-1})(x) -\mathop {G^{1}_{j}}(V_{t-1})(x+e_i) \bigg ] \quad \forall i\in \mathcal {I},t\in \mathcal {T},x\in \mathcal {X}. \end{aligned}$$

It remains to show that the difference within brackets is nonnegative. Setting $\mathop {G^{1}_{}}=\mathop {G^{1}_{j}},V=V_{t-1}, r=r_j$, and $y=x+e_i$, it follows that

$$\begin{aligned}&\mathop {G^{1}_{}}(V)(x) -\mathop {G^{1}_{}}(V)(x+e_i) \\&\quad \ge {} \min \Big \{\mathop {F_{1}}(V)(x)-\mathop {F_{1}}(V)(y), \mathop {F_{2}}(V)(x)-\mathop {F_{2}}(V)(y)\Big \} \\&\quad ={} \min \Big \{\mathop {T_{1}}(V)(x)-\mathop {T_{1}}(V)(y), \mathop {T_{2}}(V)(x)-\mathop {T_{2}}(V)(y) \Big \} -V(x) +V(y) \\&\quad \ge {}\min \Big \{V(x)-V(y),V(x+e_1)-V(y+e_1), V(x+e_2)-V(y+e_2) \Big \}\\&\qquad -V(x) + V(y) \\&\quad ={} \min \Big \{0,\varDelta _1 V(y) - \varDelta _1 V(x),\varDelta _2 V(y) - \varDelta _2 V(x)\Big \} \\&\quad ={} \min \Big \{0,\varDelta _1 V(x+e_i) - \varDelta _1 V(x),\varDelta _2 V(x+e_i) - \varDelta _2 V(x)\Big \} \\&\quad \ge {} 0 \qquad \forall i\in \{1,2\},x\in \mathbb {Z}^2_+, \end{aligned}$$

where the first inequality follows from the first inequality of Lemma B.8, the first equality results from $\mathop {F_{i}}(V)(x)=\mathop {T_{i}}(V)(x)-V(x)$, the second inequality is a consequence of the first inequality of Lemma B.9, the second equality holds by simple rearrangement. The last equality is obvious and the last inequality can be derived whenever $V$ is concave and submodular, see Definition 1. Since Lemma B.7 has shown that $V_t$ is multimodular for all $t\in \mathcal {T}$, the proof is complete. $\square$

1.3 B.3 Monotonicity properties of the static model

Lemma B.11

Let any $j\in \mathcal {J}$ and $t\in \mathcal {T}$ be given, and let $V_{t-1}$ be defined by (RM-s). If $V_{t-1}$ is multimodular, then $\mathop {H^{d}_{j}}(V_{t-1})$ is multimodular for all $d\in \mathbb {Z}_+$.

Proof

We show the result by induction over the demand levels. $\mathop {H^{0}_{}}(V_{t-1})=V_{t-1}$ is multimodular by assumption. Consider $d\ge 1$ and assume that the result holds for $d-1$. Using $\mathop {H^{d}_{}}(V)=\mathop {H^1_{}(\mathop {H^{d-1}_{}}(V))}$ and that $\mathop {H^{1}_{}}$ preserves multimodularity (see Lemma B.3) completes the proof.

Lemma B.12

Let $V_t,t\in \mathcal {T}$ be defined by (RM-s). $V_t$ is multimodular for all $t\in \mathcal {T}$.

Proof

Applying Lemma 1, the proof is the same as that of Lemma B.7 except that here we apply Lemma B.11 to the equivalent value function (8). $\square$

Lemma B.13

Let $V_t,t\in \mathcal {T}$ be defined by (RM-s). The following statement holds: $\varDelta _iV_t(x)$ is nondecreasing in $t$ for all $x\in \mathcal {X}$ and $i\in \mathcal {I}$.

Proof

We need to verify that $\varDelta _iV_t(x) - \varDelta _iV_{t-1}(x) \ge 0$ for all $i\in \mathcal {I},t\in \mathcal {T}$, and $x\in \mathcal {X}$. Using the notation defined in the proof of Lemma B.10, we generalize (39) by defining the operator

$$\begin{aligned} \mathop {G^{d}_{}}(V)(x)={}&\max \Big \{ r^1- \varDelta _1\mathop {H^{d-1}_{}}(V)(x), r^2- \varDelta _2\mathop {H^{d-1}_{}}(V)(x), 0\Big \}\\ ={}&\max \{\mathop {F_{1}}(\mathop {H^{d-1}_{}}(V)),\mathop {F_{2}}(\mathop {H^{d-1}_{}}(V))\} \qquad \forall x\in \mathbb {Z}^2_+,d\ge 1. \end{aligned}$$

It is easy to show that the right hand side (RHS) of the equivalent value function (8) can be expressed as

(42)

where the second equality follows from setting $\mathop {G^{d}_{}}=\mathop {G^{d}_{j}},V=V_{t-1}$, and $r=r_j$, and the third equality follows from formulating the difference (inner brackets) as a telescoping series. Now, we use Eq. (42) and the definition of $\varDelta _i$ in order to rewrite $\varDelta _iV_t(x) - \varDelta _iV_{t-1}(x)$ as

$$\begin{aligned}&V_t(x) - V_t(x+e_i) - \varDelta _iV_{t-1}(x) \nonumber \\&\quad ={}\sum _{d=1}^\infty p_{kjdt}\Bigg [\sum _{z=1}^{d-1}\Big [\varDelta _i\mathop {H^{z}_{j}}(V_{t-1})(x) - \varDelta _i\mathop {H^{z-1}_{j}}(V_{t-1})(x)\Big ] \nonumber \\&\quad \quad +\Big [\mathop {G^{d}_{j}}(V_{t-1})(x) - \mathop {G^{d}_{j}}(V_{t-1})(x+e_i) \Big ] \Bigg ] \quad \forall i\in \mathcal {I},t\in \mathcal {T},j=j(t),x\in \mathcal {X}. \end{aligned}$$

(43)

To analyze the nonnegativity of the RHS of (43), we first rewrite the difference $\varDelta _i\mathop {H^{d}_{j}}(V_{t-1})(x) - \varDelta _i\mathop {H^{d-1}_{j}}(V_{t-1})(x)$ using the definition of $\varDelta _i$ as

$$\begin{aligned}&\mathop {H^{d}_{j}}(V_{t-1})(x) - \mathop {H^{d}_{j}}(V_{t-1})(x+e_i) - \Big [ \mathop {H^{d-1}_{j}}(V_{t-1})(x) - \mathop {H^{d-1}_{j}}(V_{t-1})(x+e_i)\Big ]\nonumber \\&\quad =\mathop {H^{d}_{j}}(V_{t-1})(x) - \mathop {H^{d-1}_{j}}(V_{t-1})(x) - \Big [ \mathop {H^{d}_{j}}(V_{t-1})(x+e_i) - \mathop {H^{d-1}_{j}}(V_{t-1})(x+e_i)\Big ]\nonumber \\&\quad \begin{aligned}&\quad =\max \Big \{ r^1_j-\varDelta _1\mathop {H^{d-1}_{j}}(V_{t-1})(x), r^2_j- \varDelta _2 \mathop {H^{d-1}_{j}}(V_{t-1})(x), 0 \Big \} \nonumber \\&\quad - \max \Big \{ r^1_j- \varDelta _1 \mathop {H^{d-1}_{j}}(V_{t-1})(x+e_i), r^2_j- \varDelta _2 \mathop {H^{d-1}_{j}}(V_{t-1})(x+e_i), 0\Big \} \end{aligned}\nonumber \\&\quad =\mathop {G^{d}_{j}}(V_{t-1})(x) - \mathop {G^{d}_{j}}(V_{t-1})(x+e_i) \qquad \forall i\in \mathcal {I},t\in \mathcal {T},j=j(t),x\in \mathcal {X},d\ge 1, \end{aligned}$$

(44)

where the second equality is obvious, the third equality follows from the definition of $\mathop {H^{d}_{}}$, and the last inequality follows from the definition of $\mathop {G^{d}_{}}$ with $\mathop {G^{d}_{}}=\mathop {G^{d}_{j}},V=V_{t-1}$, and $r=r_j$. Consequently, if we verify the nonnegativity of the RHS of (44), this implies the nonnegativity of the RHS of (43) and completes the proof. With the equality $\mathop {G^{d}_{j}}(V_{t-1})(x)=\mathop {G^{1}_{j}}(\mathop {H^{d-1}_{}}(V_{t-1}))(x)$, the RHS of (43) reads as

$$\begin{aligned}&\mathop {G^{1}_{j}}(\mathop {H^{d-1}_{}}(V_{t-1}))(x) - \mathop {G^{1}_{j}}(\mathop {H^{d-1}_{}}(V_{t-1}))(x+e_i)\\&\quad \ge 0 \quad \forall i\in \mathcal {I},t\in \mathcal {T},j=j(t),x\in \mathbb {Z}^2_+,d\ge 1, \end{aligned}$$

where the inequality can be derived similar to the proof of Lemma B.10 using that $\mathop {H^{d}_{j}}$ preserves multimodularity (see Lemmata B.11 and B.12). $\square$

C Proofs of Sect. 4

1.1 C.1 Proof of Theorem 1

We introduce the notation $R(u)=r_j^\top u + V_{t-1}(x+u) - V_{t-1}(x)$ to simplify the following discussion.

Theorem 1

Let any state $x\in \mathcal {X}$ at time $t\in \mathcal {T}$ and any demand $d\in \mathbb {Z}_+$ for product $j\in \mathcal {J}$ be given, and let $P_{jt}$ be a booking path according to Algorithm 1. Define $\ell =\min \{d,L\}$. The booking decision $u^{(\ell )}$ is optimal in both the dynamic model (RM-d) and the static model (RM-s).

Proof

Proof by contradiction, i.e., let $u^*\ne u^{(\ell )}$ be an optimal booking decision with $R(u^*)>R(u^{(\ell )})$. In particular, $R(u^*)\ge R(u)$ for all $u\in \mathcal {U}(d)$. Without loss of generality, we also assume that $u^*$ is chosen with smallest possible norm $||u^*||_1$. Let $x^P=x+u^{(\ell )}$ denote the state that results from following $P_{jt}$ up to the $\ell$th state and let $x^*=x+u^*$ denote the state resulting from $u^*$. In the following, we analyze eight possible cases, for clarity depicted in Table 6:

Table 6 Cases in the proof of Theorem 1

Case (I) $u^{(\ell )}_1>u^*_1$and $u^{(\ell )}_2=u^*_2$   Suppose that $x_{(l)}=x^*$ for some $l\in \{0,\dots ,L-1\}$. From the construction of $P_{jt}$ follows that $r^1(x^*)$ is positive and, hence,

$$\begin{aligned} R(u^*+ e_1)&={}r^\top (u^*+ e_1) + V_{t-1}(x+u^*+ e_1) - V_{t-1}(x)\\&=R(u^*)+ r^1(x+u^*)>R(u^*), \end{aligned}$$

which contradicts the optimality of $u^*$.

If $x^*\ne x_{(l)}$ for all $l\in \{0,\dots ,L-1\}$, consider the unique state $x_{(l)}$ with $x_{1l}=x^*_1$ having the largest $l$. Then, $e_{i(l+1)}=e_1$ by the definition of $l$ and $P_{jt}$, i.e., $r^1(x_{(l)})$ is positive and greater or equal than $r^2(x_{(l)})$. We define $K=x^*_2-x_{2l}$ and use subconcavity to state that

$$\begin{aligned} 0<r^1(x_{(l)})-r^2(x_{(l)}) ={}&r^1_j- r^2_j+ \varDelta _2V_{t-1}(x_{(l)}) -\varDelta _1V_{t-1}(x_{(l)}) \nonumber \\ \le {}&r^1_j- r^2_j+ \varDelta _2V_{t-1}(x_{(l)}+e_2) -\varDelta _1V_{t-1}(x_{(l)}+e_2)\nonumber \\&\vdots \nonumber \\ \le {}&r^1_j- r^2_j+ \varDelta _2V_{t-1}(x_{(l)}+(K-1)e_2)\nonumber \\&-\varDelta _1V_{t-1}(x_{(l)}+(K-1)e_2)\nonumber \\ ={}&r^1(x^*-e_2)-r^2(x^*-e_2). \end{aligned}$$

(45)

Hence, $R(u^*+e_1-e_2)=R(u^*)+r^1(x^*-e_2)-r^2(x^*-e_2)>R(u^*)$ is true and contradicts the optimality of $u^*$.

Case (II) $u^{(\ell )}_1=u^*_1$and $u^{(\ell )}_2>u^*_2$   The contradiction is obtained by interchanging the roles of $i=1$ and $i=2$.

Case (III) $u^{(\ell )}_1<u^*_1$and $u^{(\ell )}_2=u^*_2$   If $r^1(x^P)>0$, then $u^{(\ell )}_1+u^{(\ell )}_2=d$ follows from the definition of $u^{(\ell )}$. Since $u^*_1+u^*_2>u^{(\ell )}_1+u^{(\ell )}_2$, $u^*$ is infeasible ($u^*\notin \mathcal {U}(d)$) and herewith not optimal.

Suppose that $r^1(x^P)\le 0$ and set $K=x^*_1-x^P_1$. Concavity of $V$ implies that

$$\begin{aligned}&0\ge r^1_j-\varDelta _1V_{t-1}(x^P)\ge r^1_j-\varDelta _1V_{t-1}(x^P+e_1)\ge \cdots \\&\quad \ge r^1_j-\varDelta _1V_{t-1}(x^P+(K-1)e_1)=r^1(x^*-e_1). \end{aligned}$$

The relation $R(u^*-e_1)=R(u^*)-r^1(x^*-e_1)\ge R(u^*)$ means that $u^*-e_1$ is another optimal solution which, however, contradicts the smallest norm assumption of $u^*$.

Case (IV)   $u^{(\ell )}_1=u^*_1$and $u^{(\ell )}_2<u^*_2$. The contradiction is obtained by interchanging the roles of $i=1$ and $i=2$.

Case (V) $u^{(\ell )}_1>u^*_1$and $u^{(\ell )}_2<u^*_2$   Consider the unique state $x_{(l)}$ with $x_{1l}=x^*_1$ and largest $l$. Then, $e_{i(l+1)}=e_1$ by the definition of $l$ and $P_{jt}$. Hence, $r^1(x_{(l)})>0$ and $r^1(x_{(l)})\ge r^2(x_{(l)})$, and we obtain the result in (45) with $K=x^*_2-x_{2l}$ yielding the contradiction $R(u^*+e_1-e_2)>R(u^*)$.

Case (VI) $u^{(\ell )}_1<u^*_1$and $u^{(\ell )}_2>u^*_2$   Again, the contradiction is obtained by interchanging the roles of $i=1$ and $i=2$.

Case (VII)   $u^{(\ell )}_1<u^*_1$and $u^{(\ell )}_2<u^*_2$. As in Cases (III) and (IV), $r^1(x^P)>0$ or $r^2(x^P)>0$ implies $u^{(\ell )}_1+u^{(\ell )}_2=d$, $u^*_1+u^*_2>u^{(\ell )}_1+u^{(\ell )}_2$, $u^*$ is infeasible ($u^*\notin \mathcal {U}(d)$) and herewith not optimal.

Therefore, $r^1(x^P)\le 0$ and $r^2(x^P)\le 0$. Set $K_1=x^*_1-x^P_1$ and $K_2=x^*_2-x^P_2$. Submodularity of $V$ implies that

$$\begin{aligned} 0\ge r^1_j-\varDelta _1V_{t-1}(x^P)\ge r^1_j-\varDelta _1V_{t-1}(x^P+e_2)\ge \cdots \ge r^1_j-\varDelta _1V_{t-1}(x^P+K_2e_2) \end{aligned}$$

and concavity of  $V$ implies that

$$\begin{aligned} 0&\ge r^1_j-\varDelta _1V_{t-1}(x^P+K_2e_2)\ge r^1_j-\varDelta _1V_{t-1}(x^P+e_1+ K_2e_2) \\&\ge \cdots \ge r^1_j-\varDelta _1V_{t-1}(x^P+(K_1-1)e_1+ K_2e_2) = r^1(x^*-e_1). \end{aligned}$$

The relation $R(u^*-e_1)=R(u^*)-r^1(x^*-e_1)\ge R(u^*)$ means that $u^*-e_1$ is another optimal solution which, however, contradicts the smallest norm assumption of $u^*$.

Case (VIII) $u^{(\ell )}_1>u^*_1$and $u^{(\ell )}_2>u^*_2$   If $x_{(l)}=x^*$ for some $l\in \{0,\dots ,L-1\}$, then by construction of $P_{jt}$ we know that $r^1(x^*)$ or $r^2(x^*)$ is positive and, hence, either

$$\begin{aligned} R(u^*+ e_1)&={}r^\top (u^*+ e_1) + V_{t-1}(x+u^*+ e_1) - V_{t-1}(x)\\&=R(u^*)+ r^1(x^*)>R(u^*), \end{aligned}$$

or

$$\begin{aligned} R(u^*+ e_2)&={}r^\top (u^*+ e_2) + V_{t-1}(x+u^*+ e_2) - V_{t-1}(x)\\&=R(u^*)+ r^2(x^*)>R(u^*), \end{aligned}$$

which contradicts the optimality of $u^*$.

Otherwise, consider the unique state $x_{(l)}$ with $x_{1l}\le x^*_1$ and $x_{2l}\le x^*_2$ having the largest $l$. If $e_{i(l+1)}=e_1$, the definition of $l$ and $P_{jt}$ implies that $r^1(x_{(l)})$ is positive and greater or equal than $r^2(x_{(l)})$. As in Case (I), the inequality (45) then yields $R(u^*+e_1-e_2)>R(u^*)$, which contradicts the optimality of $u^*$. The alternative $e_{i(l+1)}=e_2$ leads to the symmetric contradiction as covered by Case (II). This completes the proof. $\square$

1.2 C.2 Proofs of Sect. 4.2

The first lemma of this section explains why using either of the switching curves $\mathop {S^{}_{1}}$ and $\mathop {S^{}_{2}}$ to control the booking process will end up in the same acceptance decision.

Lemma C.1

Let $V$ be concave and submodular and $x\in \mathbb {Z}^2_+$. Then,

• (i) $x_1 \ge \mathop {S^{}_{1}}(x_2)$ implies $\max \{r^1- \varDelta _1V(x_1,x_2),r^2- \varDelta _2V(x_1,x_2)\}\le 0$,

• (ii) $x_2 \ge \mathop {S^{}_{2}}(x_1)$ implies $\max \{r^1- \varDelta _1V(x_1,x_2),r^2- \varDelta _2V(x_1,x_2)\}\le 0$,

• (iii) $x_1 < \mathop {S^{}_{1}}(x_2)$ if and only if $x_2 < \mathop {S^{}_{2}}(x_1)$.

Proof

(i) By definition of $S_1$, we have $r^1-\varDelta _1 V(S_1(x_2),x_2) \le 0$. Concavity of V implies $r^1-\varDelta _1 V(x_1,x_2) \le 0$ for all $x_1\ge S_1(x_2)$. By definition of $S_1$, we also know that $r^2-\varDelta _2 V(S_1(x_2),x_2) \le 0$. Submodularity of V implies $r^2-\varDelta _2 V(x_1,x_2) \le 0$ for all $x_1\ge S_1(x_2)$. Both inequalities together imply $\max \{r^1-\varDelta _1 V(x_1,x_2),r^2-\varDelta _2 V(x_1,x_2) \}\le 0$, i.e., the statement.

(ii) Follows from (i) with the roles of $i=1$ and $i=2$ interchanged.

(iii) Now, the condition $x_1<\mathop {S^{}_{1}}(x_2)$ is equivalent to the condition $\varDelta _1V(x_1,x_2)-r^1> 0\; \textit{or}\;\varDelta _2V(x_1,x_2)-r^2> 0$ which is again equivalent to $x_2<\mathop {S^{}_{2}}(x_1)$ by the definitions of $S_1$ and $S_2$. $\square$

Theorem 2

Let the switching curves $S_1=\mathop {S_{1 jt}}$ and $S_2=\mathop {S_{2 jt}}$ for $V=V_{t-1}$ and $r=r_j$, products $j\in \mathcal {J}$, and periods $t\in \mathcal {T}$ be given. Then, the acceptance regions in the models (RM-d) and (RM-s) are

$$\begin{aligned} \mathcal {X}_{jt}^{acc} = \big \{(x_1,x_2)^\top \in \mathcal {X}: x_1< S_{1jt} (x_2) \big \} = \big \{(x_1,x_2)^\top \in \mathcal {X}: x_2< S_{2jt}(x_1) \big \} \end{aligned}$$

for all products $j\in \mathcal {J}$ and periods $t\in \mathcal {T}$. Moreover, $\mathcal {X}_{jt}^{acc}\subseteq \mathcal {X}_{jt-1}^{acc}$ for all $t\in \mathcal {T}$.

Proof

By the definition of the acceptance region and of switching curve $S_1$, a state $x\in \mathcal {X}_{jt}^{acc}$ is equivalent to the condition $\varDelta _1V(x_1,x_2)-r^1> 0\,or\,\varDelta _2V(x_1,x_2)-r^2> 0$. This is the first equality of the theorem, while the second equality is a direct consequence of the definition of $S_2$ and parts (ii) and (iii) of Lemma C.1. The fact that the acceptance region increases as time approaches the end of the booking horizon follows from applying time monotonicity results of Lemmata B.10 and B.13 for the models (RM-d) and (RM-s), respectively. This completes the proof. $\square$

Similar to Lemma C.1, the next lemma clarifies that using either switching curves $Q_1$ or $Q_2$ to control the booking process yields equal assignment decision.

Lemma C.2

Let $V$ be concave and submodular and $x\in \mathbb {Z}^2_+$. Then,

• (i) $x_1 \ge Q_1(x_2)$ implies $r^1- \varDelta _1V(x_1,x_2)<r^2- \varDelta _2V(x_1,x_2)$,

• (ii) $x_2 \ge Q_2(x_1)$ implies $r^1- \varDelta _1V(x_1,x_2)\ge r^2- \varDelta _2V(x_1,x_2)$,

• (iii) $x_1 < Q_1(x_2)$ if and only if $x_2 \ge Q_2(x_1)$.

Proof

(i) By definition of $Q_1$, we have $\varDelta _2V(Q_1(x_2),x_2)-\varDelta _1V(Q_1(x_2),x_2)<r^2-r^1$. Subconcavity of $V$ implies $\varDelta _2V(x_1,x_2)-\varDelta _1V(x_1,x_2)$ is non-increasing in $x_1$. Therefore,

$$\begin{aligned} \varDelta _2V(x_1,x_2)-\varDelta _1V(x_1,x_2)\le & {} \varDelta _2V(Q_1(x_2),x_2)-\varDelta _1V(Q_1(x_2),x_2)\\&<r^2-r^1\quad \forall x_1\ge Q_1(x_2), \end{aligned}$$

which shows (i).

(ii) By definition of $Q_2$, we have $\varDelta _2V(x_1,Q_2(x_1))-\varDelta _1V(x_1,Q_2(x_1))\ge r^2-r^1$. Subconcavity of $V$ implies $\varDelta _2V(x_1,x_2)-\varDelta _1V(x_1,x_2)$ is non-decreasing in $x_2$. Therefore,

$$\begin{aligned} \varDelta _2V(x_1,x_2)-\varDelta _1V(x_1,x_2)\ge & {} \varDelta _2V(x_1,Q_2(x_1))-\varDelta _1V(x_1,Q_2(x_1))\\\ge & {} r^2-r^1\quad \forall x_2\ge Q_2(x_1), \end{aligned}$$

which shows (ii).

(iii) The equivalence is a direct consequence of (i) and (ii). $\square$

Theorem 3

Let the switching curves $Q_1=Q_{1jt}$ and $Q_2=Q_{2jt}$ for $V=V_{t-1}$, revenue vector $r=r_j$, products $j\in \mathcal {J}$, and periods $t\in \mathcal {T}$ be given. Then, the assignment regions in the models (RM-d) and (RM-s) are

$$\begin{aligned} \mathcal {X}_{1jt}^{ass} ={}&\big \{(x_1,x_2)^\top \in \mathcal {X}: x_1<Q_{1jt}(x_2) \big \} = \big \{(x_1,x_2)^\top \in \mathcal {X}: x_2 \ge Q_{2jt}(x_1) \big \}\\ \mathcal {X}_{2jt}^{ass} ={}&\big \{(x_1,x_2)^\top \in \mathcal {X}: x_1\ge Q_{1jt}(x_2) \big \} = \big \{(x_1,x_2)^\top \in \mathcal {X}: x_2 < Q_{2jt}(x_1) \big \} \end{aligned}$$

for all products $j\in \mathcal {J}$ and periods $t\in \mathcal {T}$.

Proof

This proof works similar to the proof of Theorem 2. It follows from combining the definition of the assignment regions, $Q_1$, and $Q_2$ with the results provided in Lemma C.2. $\square$