Optimal pricing for a GI/M/k/N queue with several customer types and holding costs

Abstract

This paper studies optimal pricing for a GI/M/k/N queueing system with several types of customers. An arrival joins the queue if the price of service is not higher than the maximum amount that the arrival is willing to pay, and this maximum amount is defined by the customer type. A system manager chooses a price depending on the number of customers in the system. In addition, the system incurs holding costs when there are customers waiting in the queue for their services. Service times and holding costs do not depend on customer types. The holding costs are nondecreasing and convex with respect to the number of customers in the queue. This paper describes average reward optimal, canonical, bias optimal, and Blackwell optimal policies for this pricing problem.

Appendix

Proof of Lemma 4.6

(a) We prove statement (a) by contradiction. Assume that there exists some state $s\in R^f$ such that $U^f(s,\nabla w_s^f)< \max _{a\in A(s)}\{\alpha ^a[r_a-\nabla w^f_s]\}$. Consider a policy $g\in F$ such that $g_n=f_n$ for $n\ne s$ and $g_s$ is the action such that $\alpha ^{g_s}[r_{g_s}-\nabla w^f_s]=\max _{a\in A(s)}\{\alpha ^a[r_a-\nabla w^f_s]\}$. Let $e_n=\alpha ^{g_n}[r_{g_n}-\nabla w^f_n]-U^f(n,\nabla w^f_n)$ for $n=0,1,\ldots ,N$. Then $e_n$ is the same as $\gamma _n$ in formula (8.39) in [23] and, according to (8.39) in [23],

$$\begin{aligned} v^g-v^f=\sum _{n=0}^N\pi ^g_n e_n. \end{aligned}$$

(14)

Observe that $e_n=0$ for $n\ne s$ and $e_s=\alpha ^{g_n}[r_{g_n}-\nabla w^f_n]-U^f(s,\nabla w^f_s)>0$. According to (14), $v^g-v^f = \sum _{n=0}^N\pi ^g_n e_n = \pi ^g_s e_s$. Since $s\in R^g$ and $e_s>0$, $\pi ^g_s>0$ and, thus, $v^g-v^f=\pi ^g_s e_s>0$. We have a contradiction. Thus, if $f\in F^*$, then $U^f(n,\nabla w^f_n)=\max _{a\in A(n)}\{\alpha ^a[r_a-\nabla w^f_n]\}$ for $n\in R^f$.

(b) Let $f\in F^*$ and $\max _{a=1,2,\ldots ,m}\{\alpha ^a\times r_a\}=M$. Then $\pi ^f_n>0$ for $n=0,1,\ldots ,n^f$ and $v^f=v^*=\sum _{n=0}^{n^f}\pi _n^f R(n, f_n)$. Since $H_{n}\ge 0$ for $n=0,1,\ldots ,N$, $R(n, f_n)=\alpha ^{f_n} [r_{f_n}-H_{n+1}]-(1-\alpha ^{f_n})H_{n}\le M$ for $n=0,1,\ldots ,n^f-1$ and $R(n^f,f_{n^f})=0$. Then $v^f=\sum _{n=0}^{n^f}\pi ^f_n R(n,f_n)=\sum _{n=0}^{n^f-1}\pi ^f_n R(n,f_n)\le \sum _{n=0}^{n^f-1} \pi ^f_n M< \sum _{n=0}^{n^f}\pi ^f_n M= M$. To show $0<v^*$, consider a policy $g\in F$ such that $g_0= m$ and $g_n= 0$ for $n=1,2,\ldots ,N$. Then $\pi ^g_0>0$, $\alpha ^m=1$, $H_{0}=H_{1}=0$, and $R(0,g_0)=\alpha ^{m}[r_m -H_{1}]-(1-\alpha ^m)H_{0}=r_m >0$. Thus, $v^*\ge v^g=\pi ^g_0 R(0,g_0)>0$.

(c) Since $f\in F^*$, (6) and Statements (a) and (b) of this lemma imply $v^f=\alpha ^{f_0}[r_{f_0}-\nabla w_0^f]=\max _{a\in A(0)}\{\alpha ^a [r_a-\nabla w_0^f]\}<\max _{a=1,2,\ldots ,m}\{\alpha ^a\times r_a\}$. In addition, if $f_0= 0$, then $v^f=\pi ^f_0 R(0,f_0)=0$, which contradicts statement (b) of this lemma. Thus, $f_0\ne 0$. Since $v^f=\alpha ^{f_0}[r_{f_0}-\nabla w_0^f]=\max _{a\in A(0)}\{\alpha ^a [r_a-\nabla w_0^f]\}<\max _{a=1,2,\ldots ,m}\{\alpha ^a\times r_a\}$ and $f_0\ne 0$, the only possibility is $\nabla w_0^f>0$. Since $\nabla w^f_0>0$, $H_{0}=H_{1}=0$, and $p(1|1)>0$, (4) implies $\nabla y_0^f>0$.

(d) Since $f\in F^*$, $f^R\in F^*$. Consider a policy $g\in C^*$. Since $g\in C^*$, Lemma 4.1 implies $\nabla y^g_n=u_n$ and $\nabla w^g_n=z_n$ for $n=0,1,\ldots ,N-1$. From statement (c) of this lemma, $n^f>0$ and $n^g>0$. Consider two cases: (i) $0<n^g\le n^f$ and (ii) $0<n^f<n^g$.

(i) Let $0<n^g\le n^f$. Since $g\in C^*$, Proposition 4.1 implies $U^g(n,z_n)=\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}$ for $n=0,1,\ldots ,N$. Let $e_n=\alpha ^{f^R_n}[r_{f^R_n}-z_n]-U^g(n,z_n)$ for $n=0,1,\ldots ,N$. Then, $e_n= \alpha ^{f^R_n}[r_{f^R_n}-z_n]-\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}\le 0$ for $n=0,1,\ldots ,N$. Assume that there exists a state s, $0\le s\le n^f-1$, such that $e_s<0$. Observe that $\pi ^{f^R}_s>0$. According to (14), $v^{f^R}-v^g=\sum _{n=0}^N\pi ^{f^R}_n e_n=\pi ^{f^R}_s e_s<0$ and $f^R$ is not optimal. We have a contradiction. Thus, $e_n=0$ for $n=0,1,\ldots ,n^f-1$. Moreover, observe that $n^f\ge n^g$, $f^R_n=g_n= 0$ and, thus, $e_n=0$ for $n=n^f,n^f+1,\ldots ,N$. Since $e_n=\alpha ^{f^R_n}[r_{f^R_n}-z_n]-U^g(n,z_n)=0$ for all $n=0,1,\ldots ,N$, by substituting g with $f^R$ in (6), $\nabla y^{f^R}_n=u_n$ and $\nabla w^{f^R}_n=z_n$ for $n=0,1,\ldots ,N-1$. Since $\nabla w^{f^R}_n=z_n$ and $\alpha ^{f^R_n}[r_{f^R_n}-z_n]=U^g(n,z_n)$ for $n=0,1,\ldots ,N-1$, $U^{f^R}(n,\nabla w^{f^R}_n) = \alpha ^{f^R_n}[r_{f^R_n}-z_n]= U^g(n,z_n)=\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}$ for $n=0,1,\ldots ,N-1$. According to Proposition 4.1, $f^R\in C^*$.

(ii) Consider $0<n^f < n^g$. First, we show that $U^{f^R}(n,\nabla w_n^{f^R})=U^g(n,z_n)$ for $n=0,1,\ldots ,n^f-1$. Since $f^R\in F^*$ and $g\in C^{*}$, from (6), (7), and statement (a) of this lemma, $v^{f^R}=U^{f^R}(0,\nabla w_0^{f^R})=\max _{a\in A(0)}\{\alpha ^a[r_a-\nabla w^{f^R}_0]\}= v^g =U^g(0,z_0)=\max _{a\in A(0)}\{\alpha ^a[r_a-z_0]\}$. Assume $\nabla w^{f^R}_0< z_0$. Since $f^R_0\ne 0$ and $g_0\ne 0$, $\max _{a\in A(0)}\{\alpha ^a[r_a-z_0]\}=U^g(0,z_0)=\alpha ^{g_0}[r_{g_0}-z_0]<\alpha ^{g_0}[r_{g_0}-\nabla w^{f^R}_0]\le \max _{a\in A(0)}\{\alpha ^a[r_a-\nabla w^{f^R}_0]\}$. Similarly, $v^g>v^{f^R}$ if $\nabla w^{f^R}_0>z_0$. In either case we have a contradiction. Thus, $\nabla w^{f^R}_0=z_0$ and (4) implies $\nabla y^{f^R}_0=u_0$. Since $f^R\in F^*$ and $g\in C^*$, from (6), (7), and statement (a) of this lemma, $v^{f^R}=U^{f^R}(1,\nabla w^{f^R}_1)-H_{1}+p(0|1)\nabla y^{f^R}_0= \max _{a\in A(1)}\{\alpha ^a[r_a-\nabla w^{f^R}_1]\}-H_{1}+p(0|1)\nabla y^{f^R}_0 = v^g= U^g(1,z_1)-H_{1}+p(0|1)u_0= \max _{a\in A(1)}\{\alpha ^a[r_a-z_1]\}-H_{1}+p(0|1)u_0$. Since $\nabla y^{f^R}_0=u_0$, $U^{f^R}(1,\nabla w_1^{f^R})=\max _{a\in A(1)}\{\alpha ^a[r_a-\nabla w^{f^R}_1]\} = U^g(1,z_1)=\max _{a\in A(1)}\{\alpha ^a[r_a-z_1]\}$. Because $f^R_1\ne 0$ and $g_1\ne 0$, we can use the same arguments for proving $\nabla w^{f^R}_0 = z_0$ to show that $\nabla w^{f^R}_1=z_1$. Since $\nabla y^{f^R}_0=u_0$ and $\nabla w_1^f=z_1$, (4) implies $\nabla y^{f^R}_1=u_1$. By repeating the same argument $n^f-2$ times, $\nabla y^{f^R}_n = u_n$ and $\nabla w^{f^R}_n=z_n$ for $n=0,1,\ldots ,n^f-1$. Then statement (a) of this lemma and Proposition 4.1 imply $U^{f^R}(n,\nabla w_n^{f^R})=\max _{a\in A(n)}\{\alpha ^a[r_a-\nabla w_n^{f^R}]\}=\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}=U^g(n,z_n)$ for $n=0,1,\ldots ,n^f-1$.

Now we show that $U^g(n,z_n)=U^{f^R}(n,\nabla w_n^{f^R})$ for $n=n^f,n^f+1,\ldots ,N-1$. From the proven part of (ii), $\nabla y^{f^R}_n \nabla y^g_n= u_n$ for $n=0,1,\ldots ,n^f-1$. Then, according to (6) and (7), $v^{f^R} = -H_{n^f} + \sum _{j=0}^{n^f-1}p(j|n^f) [\sum _{i=j}^{n^f-1} u_i] = v^g = \alpha ^{g_{n^f}}[r_{g_{n^f}}-z_{n^f}]- H_{n^f} + \sum _{j=0}^{n^f-1}p(j|n^f) [\sum _{i=j}^{n^f-1}u_i]=\max _{a\in A(n^f)}\{\alpha ^a [r_a- z_{n^f}]\} - H_{n^f} + \sum _{j=0}^{n^f-1}p(j|n^f) [\sum _{i=j}^{n^f-1}u_i]$. Thus, $\alpha ^{g_{n^f}}[r_{g_{n^f}}-z_{n^f}]= \max _{a\in A(n^f)}\{\alpha ^a[r_a-z_{n^f}]\}=0$. Note that $n^f<n^g$ and $g_{n^f}\ne 0$. Because $\alpha ^{g_{n^f}}[r_{g_{n^f}}-z_{n^f}]=\max _{a\in A(n^f)}\{\alpha ^a[r_a-z_{n^f}]\}=0$ and $g_{n^f}\ne 0$, the only solution is $g_{n^f}=1$ and $r_{g_{n^f}}=z_{n^f}= r_1$. In addition, since $z_{n^f}= r_1$, (10) implies $z_n\ge r_1$ for $n=n^f+1,n^f+2,\ldots ,N-1$ and Proposition 4.1 implies $U^g(n,z_n)=\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}=0$ for $n=n^f+1, n^f+2,\ldots ,N$. Since $f^R_n=0$ for $n=n^f,n^f+1,\ldots ,N$, $U^g(n,z_n)=U^{f^R}(n,\nabla w_n^{f^R})=0$ for $n=n^f,n^f+1,\ldots ,N$.

Thus, $U^{f^R}(n,\nabla w_n^{f^R})=U^g(n,z_n)$ for $n=0,1,\ldots ,N$. Observe that $v^{f^R}=v^g=v^*$. By solving (6) for both policies $f^R$ and g, $\nabla y_n^{f^R}=u_n$ for $n=0,1,\ldots ,N-1$. According to (4), $\nabla w_n^{f^R}=z_n$ for $n=0,1,\ldots ,N-1$. Then $U^{f^R}(n,\nabla w_n^{f^R})=U^{f^R}(n, z_n)=U^g(n,z_n)=\max _{a\in A(n)}\{\alpha ^a[r_a-z_n]\}$ for $n=0,1,\ldots ,N$. Proposition 4.1 implies $f^R\in C^*$. $\square$