Congestion-based leadtime quotation and pricing for revenue maximization with heterogeneous customers

Abstract

This paper studies a queuing model where two customer classes compete for a given resource and each customer is dynamically quoted a menu of price and leadtime pairs upon arrival. Customers select their preferred pairs from the menu and the server is obligated to meet the quoted leadtime. Customers have convex–concave delay costs. The firm does not have information on a given customer’s type, so the offered menus must be incentive compatible. A menu quotation policy is given and proven to be asymptotically optimal under traditional large-capacity heavy-traffic scaling.

Appendices

Appendix A: Technical proofs in Sects. 3 and 4

Proof of Lemma 1

First, observe that (c ₁-c ₂)(⋅) is increasing on $(0,\underline{x}_{2}\vee d_{1})$. To see this, note that

$$(c_{1}-c_{2})^{\prime}(x)=c_{2}^{\prime}(x+d_{2}-d_{1})-c_{2}^{\prime}(x).$$

For x≤d ₁, we have x+d ₂-d ₁≤d ₂. Thus, (c ₁-c ₂)′(x)>0 for x≤d ₁. Next, suppose without loss of generality that $\underline{x}_{2}>d_{1}$ and $x\in(d_{1},\underline{x}_{2})$. Then, since $c_{2}^{\prime}(x)<\nobreak c$ for $x\leq\underline{x}_{2}$, and $c_{2}^{\prime}(x+d_{2}-d_{1})\geq c$, we conclude that (c ₁-c ₂)′(x)>0. Thus, (c ₁-c ₂) is increasing on $[0,\underline{x}_{2}\vee d_{1})$. Then observe that $(c_{1}-c_{2})^{\prime}(x)=c_{2}^{\prime }(x+d_{2}-d_{1})-c_{2}^{\prime}(x)$ is decreasing over $(\underline{x}_{2}\vee d_{1},d_{2})$ since the first term on the right hand side is decreasing while the second one is increasing. Also note that $(c_{1}-c_{2})^{\prime }(\underline{x}_{2}\vee d_{1})\geq0$ whereas (c ₁-c ₂)′(d ₂)<0. Thus, there exists a unique $d^{\ast}\in(\underline{x}_{2}\vee d_{1},d_{2})$ such that (c ₁-c ₂)′(d ^∗)=0. To conclude that d ^∗ is the unique maximizer note that (c ₁-c ₂)′(x)<0 for $x\in\lbrack d_{2},\overline{x}_{2}]$, while (c ₁-c ₂)′(x)=0 for $x>\overline{x}_{2}$, which also proves that (c ₁-c ₂)(⋅) is constant on $[\overline{x}_{2},\infty)$.

Now if x ₂≤d ^∗, then (2) follows immediately because (c ₁-c ₂)(⋅) is increasing on [0,d ^∗]. If x ₂>d ^∗ then to prove (2), it suffices to show that

$$ (c_{1}-c_{2}) (\overline{x}_{2})\geq (c_{1}-c_{2}) (\underline{x}_{1}).$$

(31)

This is because $(c_{1}-c_{2})(x_{2}) \ge (c_{1}-c_{2})(\overline{x}_{2})$ by the first part of the lemma. Moreover, because $(c_{1}-c_{2})(\underline{x}_{1}) \ge (c_{1}-c_{2})(x_{1})$ (again by the first part of the lemma and because $\underline{x}_{1}<d^{\ast}$) it suffices to show (31). To this end, we consider the following two cases: Case (i) $c_{1}^{\prime}(0)\geq c$; Case (ii) $c_{1}^{\prime}(0)<c$. In Case (i), we have $\underline{x}_{1}=0$. Thus, proving (2) reduces to checking $(c_{1}-c_{2})(\overline{x}_{2})\geq0$, which follows from Assumption 1(vi).

In Case (ii), we have

where we use the fact that 0< $\underline{x}_{1}$ to conclude that $c_{1}^{\prime}(x)=c_{2}^{\prime}(x+d_{1}-d_{2})<c$ for $x<\underline{x}_{1}$. □

Proof of Proposition 1

Given a feasible policy for which there exists an interval (t ₁,t ₂) such that Δ₁(t)>0, we can improve the objective by modifying Δ₁(⋅) on (t ₁,t ₂) such that Δ₁(t)=0. This also relaxes the constraint (12). Thus, without loss of optimality, we have Δ₁(t)=0 for all t. Similarly, it follows that Δ₂(t)=(c ₁-c ₂)(τ ₁(t)) for all t. □

Proof of Proposition 3

First, we verify that q _i (⋅) is monotone. To that end, note that if $W>\lambda_{1} \underline{\tilde{x}}_{1}+\lambda_{2} \underline{x}_{2}$, then $q_{1}(W)=\underline{\tilde{x}}_{1}$, and $q_{2}(W)=W-\underline{\tilde{x}}_{1}$, and the result follows. Otherwise, i.e., $W\le \lambda_{1}\underline{\tilde{x}}_{1}+\lambda_{2} \underline{x}_{2}$, then a necessary condition for optimality is that

$$ \biggl \vert \tilde{h}_1^\prime \biggl(\frac{q_1}{\lambda_1} \biggr)-h_2^\prime \biggl(\frac{q_2}{\lambda_2} \biggr)\biggr \vert $$

(32)

is minimized over q ₁,q ₂≥0 such that q ₁+q ₂=W.

Let W ₂>W ₁ and suppose that q _i (W ₂)<q _i (W ₁) for some i∈{1,2}, then we must have q _j (W ₂)>q _j (W ₁)+(W ₂-W ₁) for j≠i. Notice that setting q _i (W ₂)=q _i (W ₁) and q _j (W ₂)=q _j (W ₁)+W ₂-W ₁ decreases the difference in (32) by the strict convexity of $\tilde{h}_{1}$ on $(0, \underline{\tilde{x}}_{1})$ and h ₂ on $(0, \underline{x}_{2})$ (which follows from strict convexity of c _i on [0,α _i ]). Thus, q _i (⋅) is monotone. That is, for W ₂>W ₁,

(33)

(34)

Putting q ₁(W)=W-q ₂(W) in (34) gives

$$q_2(W_2)-q_2(W_1)\le W_2-W_1. $$

(35)

Then combining (33) and (35) gives the Lipschitz continuity of q ₂(⋅):

$$q_2(W_2)-q_2(W_1)\le |W_2-W_1|.$$

The Lipschitz continuity of q ₁(⋅) follows similarly. Finally, we note by the minimality of q ₁(⋅) (among multiple optimal solutions) that

$$\frac{q_1(W)}{\lambda_1}\le \frac{q_2(W)}{\lambda_2} \quad \mbox{and}\quad \frac{q_1(W)}{\lambda_1}\le \underline{\tilde{x}}_1,$$

from which we conclude that q ₂(W)≥Wλ ₂/(λ ₁+λ ₂). □

Appendix B: Proofs of auxiliary results in Sect. 6.2

Proof of Lemma 2

It suffices to show that for i=1,2

$$\mathbb{P} \biggl(\bigl \vert A_i^n\bigl(t_k^n\bigr)-A_i^n\bigl(t_{k-1}^n\bigr)-\lambda_i^n\bigl(t_k^n-t_{k-1}^n\bigr) \bigr \vert >\frac{\zeta}{2}n^\beta \biggr) \le \frac{C_1}{2}e^{-C_2\zeta^2n^{2\beta-1+\alpha_1}}.$$

By Theorem 2.1 of Csorgo and Horvath [13], there exists a standard Brownian motion $\tilde{B}_{i}$ and an error process $\tilde{\varepsilon}_{i}^{n}$ for each n such that

$$A_i^n(t)=\lambda_i nt+\sqrt{\lambda_i n}\tilde{B}_i(t)+\tilde{\varepsilon}_i^n(t)\quad \mbox{for }t\ge 0,$$

where for positive constants $\tilde{C}_{1}, \tilde{C}_{2}$

$$\mathbb{P} \Bigl(\sup_{0\le t\le T}\bigl \vert \tilde{\varepsilon}^n(t)\bigr \vert >\tilde{C}_1 \log (\lambda_inT)+x \Bigr)\le \tilde{C}_1 e^{-\tilde{C}_2x}.$$

Using this result, we write for n sufficiently large that

Note by a straightforward application of Markov’s inequality that

$$\mathbb{P} \bigl(\bigl \vert \tilde{B}_i(1)\bigr \vert >x \bigr)\le 2\exp\bigl\{-x^2/2\bigr\} \quad \mbox{for } x>0.$$

Then for i=1,2,

where the last inequality follows for n sufficiently large because $\frac{1-\alpha_{1}}{2}<\beta<1-\alpha_{1}$.

Then letting

$$C_1=2(2+\tilde{C}_1) \quad \mbox{and}\quad C_2=\frac{1}{32(\lambda_1+\lambda_2)},$$

the result follows. □

Appendix C: Proofs of technical results in Sect. 6.3

Proof of Lemma 3

Fix T>0, and note that , which provides the induction basis. As the induction hypothesis assume that

for j=0,1,…,k-1. Then it suffices to show that

(36)

To this end, note that

(37)

Consider the first term on the right-hand side:

Next, we consider each term on the right-hand side:

(38)

for sufficiently large n, where the first inequality follows since $q_{2} ( \widehat{Q}^{n}(t_{k}^{n}) ) \geq 0$, and

$$Q_{2}^{n}\bigl(t_{k-1}^{n}\bigr)\leq Q^{n}\bigl(t_{k-1}^{n}\bigr)\leq W^{n}\bigl(t_{k-1}^{n}\bigr)+\varepsilon \leq \frac{2\delta }{3n^{\varepsilon }}\sqrt{n}$$

for sufficiently large n. The third inequality follows from the fact that

$$\widehat{Q}_{2}^{n}\bigl(t_{k}^{n}\bigr)-\widehat{Q}_{2}^{n}\bigl(t_{k-1}^{n}\bigr)\leq \frac{A_{2}^{n}(t_{k}^{n})-A_{2}^{n}(t_{k-1}^{n})}{\sqrt{n}}.$$

The fourth inequality follows independence of the increments of the Poisson process, while the last inequality follow from Lemma 2 for n sufficiently large where the right-hand side $\delta \sqrt{n}/(3n^{\varepsilon })$ was replaced by its half to account for centering of the left-hand side.

Similarly,

Note that the second term on the right-hand side is zero for n sufficiently large since

$$q_2 \bigl(\widehat{Q}^n \bigl(t_k^n\bigr) \bigr)\le \widehat{Q}^n\bigl(t_k^n\bigr)\le \widehat{W}^n \bigl(t_k^n\bigr) +\frac{3}{\sqrt{n}} \le \frac{2\delta}{3n^\varepsilon}+\frac{3}{\sqrt{n}}<\frac{\delta}{n^\varepsilon}.$$

Moreover, for sufficiently large n,

(39)

Then combining (38)–(39) we conclude that

(40)

Next, consider

and note that

where

First, consider $p_{1}^{n}$ and recall that on we have

$$\bigl \vert \widehat{Q}_{2}^{n}\bigl(t_{k-1}^{n}\bigr)-q_{2} \bigl( \widehat{Q}^{n}\bigl(t_{k-1}^{n}\bigr) \bigr) \bigr \vert \leq \frac{\delta }{n^{\varepsilon }}.$$

Thus, on the event of interest we have

$$\widehat{q}_{2} \bigl( \widehat{Q}^{n}\bigl(t_{k-1}^{n}\bigr) \bigr) <\widehat{Q}_{2}^{n}\bigl(t_{k-1}^{n}\bigr)\leq q_{2} \bigl(\widehat{Q}^{n}\bigl(t_{k-1}^{n}\bigr) \bigr)+\frac{\delta }{n^{\varepsilon }}.$$

Also note that

Since $\widehat{W}_{n}(t_{k-1}^{n})>\frac{\delta}{3n^{\varepsilon}}$ on the event of interest, for sufficiently large n we have

$$W^n\bigl(t_{k-1}^n\bigr)>\frac{ \delta n^{0.5-\varepsilon}}{3}>\mu^n\kappa^n.$$

Thus, the incremental idleness during [t _k-1,t _k ] will be zero. Then

$$W^n\bigl(t_k^n\bigr)-W^n\bigl(t_{k-1}^n\bigr)=A_1^n\bigl(t_k^n\bigr)+A_2^n\bigl(t_k^n\bigr)-A_1^n\bigl(t_{k-1}^n\bigr)-A_2^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n.$$

Moreover, since |Q ⁿ(t)-W ⁿ(t)]|≤3 for all t≥0, we write

$$\bigl \vert Q^n\bigl(t_k^n\bigr)-Q^n\bigl(t_{k-1}^n\bigr)\bigr \vert \le \bigl \vert A_1^n\bigl(t_k^n\bigr)+A_2^n\bigl(t_k^n\bigr)-A_1^n\bigl(t_{k-1}^n\bigr)-A_2^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n\bigr \vert +6,$$

and therefore,

by Lipschitz continuity of q ₂(⋅). Then we can write

(41)

Also note that since (by Proposition 3)

we have $Q_{2}^{n}(t)>0$ for all $t\in \lbrack t_{k-1}^{n},t_{k}^{n}]$ for n sufficiently large.

Moreover, since $\widehat{Q}_{2}^{n}(t_{k-1}^{n})>q_{2} (\widehat{Q}^{n}(t_{k-1}^{n}) )$, the flexible server works on class 2 during $[t_{k-1}^{n}, t_{k}^{n}]$. Then

(42)

Combining (41)–(42), we see that a necessary condition for the event of interest is that

On the event of interest, we have $Q_{2}^{n}(t_{k-1}^{n})\leq \sqrt{n}q_{2} ( \widehat{Q}_{2}^{n}(t_{k-1}^{n}) ) +\frac{\delta \sqrt{n}}{n^{\varepsilon }}$. Thus, we conclude that

Then for sufficiently large n,

Then we conclude by Lemma 2 that

(43)

Next, consider $p_{4}^{n}$ and note that

(44)

We first bound the first term on the right-hand side of (44):

where the first term on the right is zero for n sufficiently large. To bound the second term on the right note that we can argue as in bounding $p_{1}^{n}$ that

Then for n sufficiently large

by Lemma 2. Thus we have the following bound for the first term on the right-hand side of (44).

(45)

For bounding the second term on the right-hand side of (44) note that, the flexible server gives priority to class 1 on $[t_{k-1}^{n}, t_{k}^{n}]$, and since

$$\widehat{Q}_2^n\bigl(t_{k-1}^n\bigr)\le q_2 \bigl(\widehat{Q}^n\bigl(t_{k-1}^n\bigr) \bigr)-\frac{\delta}{2n^\varepsilon},$$

we have

$$\widehat{Q}_1^n\bigl(t_{k-1}^n\bigr)>q_1 \bigl(\widehat{Q}^n\bigl(t_{k-1}^n\bigr) \bigr)+\frac{\delta}{2n^\varepsilon}\ge \frac{\delta}{2n^\varepsilon} >\frac{\mu^n\kappa^n}{\sqrt{n}}$$

for n sufficiently large, which in return implies that $Q_{1}^{n}(t)>0$ for all t∈[t _k-1,t _k ]. Therefore, the flexible server cannot serve any class 2 jobs during [t _k-1,t _k ].

Note that on the event of interest we have

$$q_2 \bigl(\widehat{Q}^n\bigl(t_{k-1}^n\bigr) \bigr)-\frac{\delta}{2n^\varepsilon}\le q_2 \bigl(\widehat{Q}\bigl(t_{k-1}^n\bigr) \bigr)-\frac{\delta}{2n^\varepsilon}.$$

Also note that

$$W^n\bigl(t_k^n\bigr)=W^n\bigl(t_{k-1}^n\bigr)+A^n\bigl(t_k^n\bigr)-A^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n+L^n\bigl(t_k^n\bigr)-L^n\bigl(t_{k-1}^n\bigr).$$

However, since $\widehat{W}^{n}(t_{k-1}^{n})>\frac{\delta}{3n^{\varepsilon}}$, we have $L^{n}(t_{k}^{n})-L^{n}(t_{k-1}^{n})=0$ for n sufficiently large. Hence

$$W^n\bigl(t_k^n\bigr)-W^n\bigl(t_{k-1}^n\bigr)=A^n\bigl(t_k^n\bigr)-A^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n.$$

Moreover, since |Q ⁿ(t)-W ⁿ(t)|≤3 for all t, we write

$$\bigl \vert Q^n\bigl(t_k^n\bigr)-Q^n\bigl(t_{k-1}^n\bigr)\bigr \vert \le \bigl \vert A_1^n\bigl(t_k^n\bigr)+A_2^n\bigl(t_k^n\bigr)-A_1^n\bigl(t_{k-1}^n\bigr)-A_2^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n\bigr \vert +6,$$

and therefore,

(46)

In particular,

(47)

Furthermore, since the flexible server exerts no effort on class 2, we have

(48)

Combining (47) and (48), a necessary condition for $\widehat{Q}_{2}^{n}(t_{k}^{n})<q_{2} (\widehat{Q}^{n}(t_{k}^{n}) )-\frac{\delta}{n^{\varepsilon}}$ is that

Since $\widehat{Q}_{2}^{n}(t_{k-1}^{n})>q_{2} ( \widehat{Q}^{n}(t_{k-1}) ) -\frac{\delta }{n^{\varepsilon }}$ on the event of interest for sufficiently large n,

We can thus argue as in the case of bounding $p_{1}^{n}$ that for sufficiently large n,

(49)

Then combining (45) and (49), we conclude that for sufficiently large n

(50)

Next consider $p_{2}^{n}$, and note that a necessary condition on for $\widehat{Q}_{2}^{n}(t_{k}^{n})>q_{2} ( \widehat{Q}^{n}(t_{k}^{n}) )+\frac{\delta }{n^{\varepsilon }}$ and $\widehat{Q}_{2}^{n}(t_{k-1}^{n})<q_{2} ( \widehat{Q}^{n}(t_{k-1}^{n}) )$ is that at least one of the following holds:

$$\widehat{Q}_{2}^{n}\bigl(t_{k}^{n}\bigr)-\widehat{Q}_{2}^{n}\bigl(t_{k-1}^{n}\bigr)>\frac{\delta }{3n^{\varepsilon }},\qquad q_{2} \bigl( \widehat{Q}^{n}\bigl(t_{k-1}^{n}\bigr) \bigr) -q_{2} \bigl(\widehat{Q}^{n}\bigl(t_{k}^{n}\bigr) \bigr) >\frac{\delta }{3n^{\varepsilon }}.$$

Otherwise,

which is a contradiction. Therefore,

Note that for sufficiently large n,

$$\mathbb{P} \biggl(A_2^n\bigl(t_k^n\bigr)-A_2^n\bigl(t_{k-1}^n\bigr)>\frac{\delta}{3n^\varepsilon}\sqrt{n} \biggr) \le C_1 \exp\biggl \{-C_2\frac{\delta^2}{36}n^{\alpha_1-2\varepsilon}\biggr\}.$$

Moreover, since $W^{n}(t_{k-1}^{n})>\frac{\delta \sqrt{n}}{3n^{\varepsilon}}$, for sufficiently large n, we have

$$W^n\bigl(t_k^n\bigr)-W^n\bigl(t_{k-1}^n\bigr)=A_1^n\bigl(t_k^n\bigr)+A_2\bigl(t_k^n\bigr)-A_1^n\bigl(t_{k-1}^n\bigr)-A_2^n\bigl(t_{k-1}^n\bigr)-\mu^n\kappa^n.$$

Thus,

Note that for sufficiently large n,

Therefore, for sufficiently large n

(51)

Similarly, note that

Note that for sufficiently large n

(52)

Also note that, as in the case of bounding $p^{n}_{2}$,

(53)

Then combining (52) and (53), we have

(54)

Then combining (40), (43), (50), (51), and (54) we conclude for sufficiently large n that

where C ₃=13C ₁ and $C_{4}=\frac{C_{2}z_{2}^{2}}{16(L_{2}+1)^{2}}$, and the last inequality follows since ε<0.5-α ₂. □

Proof of Lemma 4

First, note that on the set of (for sufficiently large n) the number of class 2 jobs arriving in each period exceeds $\mu_{2a}^{n}$ but is less than $\mu_{2a}^{n}+3\mu_{2b}^{n}$. Next, recall that $Q_{2a}^{n}(0)=Q_{2b}^{n}(0)=0$. So, k=0 constitutes an induction basis. As the induction hypothesis assume that (29) holds for k=1,…,j, and consider k=j+1. If $Q_{2a}^{n}(t_{j+1}^{n})<\lfloor \lambda_{2} \underline{x}_{2}\sqrt{n} \rfloor$, then there are two possibilities during $[t_{j}^{n}, t_{j+1}^{n}]$ under the proposed policy: Either there were not enough arrivals during $[t_{j}^{n}, t_{j+1}^{n}]$ to have $Q_{2a}^{n}(t_{j+1}^{n})=\lfloor \lambda_{2} \underline{x}_{2} \sqrt{n}\rfloor$, or the server pool 2a received help from other server pools during $[t_{j}^{n}, t_{j+1}^{n}]$ . In the latter case, (since the other server pools cannot help class 2a before helping class 2b when they are idle) it must be that $Q_{2b}^{n}(t) = 0$ for some $t\in [t_{j}^{n}, t_{j+1}^{n}]$. Thus, on the set , in this case

$$Q_{2b}^n \bigl(t_{j+1}^n\bigr)\le A_2^n \bigl(t_{j+1}^n\bigr)-A_2^n\bigl(t_j^n\bigr)-\mu_{2a}^n\le 3\mu_{2b}^n.$$

If the server pool 2a did not receive help during $[t_{j}^{n}, t_{j+1}^{n}]$ , i.e., the former case, then at most $\mu_{2b}^{n}$ jobs are routed to class 2b during $[t_{j}^{n}, t_{j+1}^{n}]$. Then we have two further subcases to consider.

Case (i). $Q_{2a}^{n}(t_{j}^{n})=\lfloor \lambda_{2}\underline{x}_{2}\sqrt{n}\rfloor$. This is not possible because with no help to server pool 2a we would have $Q_{2a}^{n}(t_{j+1}^{n})=\lfloor \lambda_{2}\underline{x}_{2}\sqrt{n}\rfloor$ on the set , which is a contradiction.

Case (ii). $Q_{2a}^{n}(t_{j}^{n})<\lfloor \lambda_{2}\underline{x}_{2}\sqrt{n}\rfloor$, which is the only remaining possibility. Then by the induction hypothesis $Q_{2b}^{n}(t_{j}^{n})\le 3\mu_{2b}^{n}$. Thus

$$Q_{2b}^n\bigl(t_{j+1}^n\bigr)\le \bigl[Q_{2b}^n\bigl(t_j^n\bigr)-\mu_{2b}^n \bigr]^++\mu_{2b}^n \le 3\mu_{2b}^n.$$

 □