Pricing schemes in processor sharing systems

Abstract

In this paper we study charging schemes for bandwidth or server usage under the processor sharing discipline. Specifically, we analyze post-payment and pre-payment (or payment on arrival) schemes in three charging frameworks: fixed-rate charging, Vickrey–Clarke–Groves based charging, and congestion based charging for users with logarithmic utilities. We show that in the absence of QoS constraints, the network operator can earn unbounded profits and thus there is a need to devise schemes where users are only charged if they are given a minimum rate. We obtain explicit characterizations for mean user payments and the operator’s mean revenue for these frameworks. We also analyze charge volatility via the second moments of the above implementations of arrival-based payments and post-payments. The volatility reflects the confidence in mean revenue for the operator and expected charges for a user. We present conditions under which a pre-payment mechanism is preferable over a post-payment mechanism. We also show that the same analysis can be applied to a scenario with admission control where each entering user is guaranteed a minimum service rate.

Appendix

Proof of Lemma 1

We have that $t(0) = \chi (\vec {0}) = \Phi (\vec {0})$. For $n \ge 1,$

$$\begin{aligned} t(n)&:=\sum _{\vec {x} : |\vec {x}| = n} \chi (\vec {x}) = \sum _{\vec {x} : |\vec {x}| = n} \frac{1}{C} \sum _{m=1}^K \Phi (\vec {x} - \vec {e}_m) \vec {\alpha }^{\vec {x}}\\&= \sum _{m=1}^K \rho _m \sum _{\vec {x} : |\vec {x}| = n-1} \Phi (\vec {x}) \vec {\alpha }^{\vec {x}} = \rho \cdot t(n-1) . \end{aligned}$$

Also, $\sum _{\vec {x} } \chi (\vec {x}) = \sum _{n=0}^\infty t(n) = \frac{\Phi (\vec {0})}{1 - \rho }.$ $\square$

Proof of Lemma 2

We start with

$$\begin{aligned} s_k(n)&= \sum _{\vec {x} : |\vec {x}| = n} \frac{x_k}{C} \sum _{m=1}^K \Phi (\vec {x} - \vec {e}_m) \vec {\alpha }^{\vec {x}} \nonumber \\&= \sum _{m=1}^K \rho _m \left[ \sum _{\vec {x} : |\vec {x}| = n-1} x_k \chi (\vec {x}) + \sum _{\vec {x} : |\vec {x}| = n-1} (\vec {e}_m)_k \chi (\vec {x}) \right] \nonumber \\&= \sum _{m=1}^K \rho _m s_k(n-1) + \sum _{m=1}^K \rho _m \sum _{\vec {x} : |\vec {x}| = n-1} (\vec {e}_m)_k \chi (\vec {x}) \nonumber \\&= \rho s_k(n-1) + \rho _k t(n-1) \nonumber \\&= \rho s_k(n-1) + \rho _k \rho ^{n-1} \Phi (\vec {0}) , \end{aligned}$$

(29)

where the last step follows from Lemma 1. It is easily shown that the result is a solution to the recursion in (29), starting with $s_k(0) = 0$. The first part of the result follows. Next,

$$\begin{aligned} \bar{s}_k(n) = \rho _k \Phi (\vec {0}) \sum _{m>n} m \rho ^{m-1} = \rho _k \Phi (\vec {0}) \frac{\rho ^n}{1-\rho } \left( n + \frac{1}{1-\rho } \right) . \end{aligned}$$

$\square$

Proof of Lemma 3

We have

$$\begin{aligned} s_{i,j}(n)&= \sum _{m=1}^K \frac{\alpha _m}{C} \sum _{\vec {x} : |\vec {x}|=n} x_i x_j \Phi (\vec {x} - \vec {e}_m) \vec {\alpha }^{\vec {x} - \vec {e}_m} \\&= \sum _{m=1}^K \rho _m \sum _{\vec {y} : |\vec {y}| = n-1} (\vec {y} + \vec {e}_m)_i (\vec {y} + \vec {e}_m)_j \Phi (\vec {y}) \vec {\alpha }^{\vec {y}} \\&= \rho s_{i,j}(n-1) + \rho _j s_i(n-1) + \rho _i s_j(n-1) \\&\quad + \mathbf 1_{(i=j)} \rho _i t(n-1) . \end{aligned}$$

Note that $s_{i,j}(0) = 0$ for any i, j and that $s_{i,j}(1)=0$ if $i \ne j$. The expression in (9) is the solution to this recursion. $\square$

Proof of Lemma 4

$$\begin{aligned} u(n)&= \sum _{\vec {x} : |\vec {x}| > n} \left( |\vec {x}| - \frac{1}{2} \right) \pi (\vec {x}) = \sum _{\vec {x} : |\vec {x}| > n} |\vec {x}| \pi (\vec {x}) - \frac{1}{2} \sum _{\vec {x} : |\vec {x}| > n} \pi (\vec {x})\\&= \sum _{\vec {x} : |\vec {x}| > n} \frac{(x_1 + \dots + x_K)\chi (\vec {x})}{\sum _{\vec {y}}\chi (\vec {y})} - \frac{1}{2} \sum _{\vec {x} : |\vec {x}| > n} \frac{\chi (\vec {x})}{\sum _{\vec {y}}\chi (\vec {y})} . \end{aligned}$$

Using $\sum _{\vec {x}} \chi (\vec {x}) = \sum _{n \ge 0} t(n)$,

$$\begin{aligned} u(n) = \frac{\sum _{k=1}^K \bar{s}_k(n)}{\sum _{m \ge 0} t(m)} - \frac{\sum _{m>n} t(m)}{2\sum _{m \ge 0} t(m)} . \end{aligned}$$

Using Lemma 1 and Lemma 2, we get

$$\begin{aligned} u(n) = \rho ^{n+1} \left( n + \frac{1}{1- \rho } \right) - \frac{\rho ^{n+1}}{2} . \end{aligned}$$

$\square$

Proof of Proposition 1

From (14), we have

$$\begin{aligned} R_V(\vec {x})&= |\vec {x}| ( |\vec {x}|-1)\sum _{n=1}^\infty \frac{1}{n|\vec {x}|^n} \\&= (|\vec {x}|-1)\left( 1 + \frac{1}{2|\vec {x}|} + \frac{1}{3|\vec {x}|^2} + \dots \right) \\&= |\vec {x}| - \frac{1}{2} - \sum _{m=2}^{\infty } \frac{1}{m (m+1) |\vec {x}|^{m-1} } < |\vec {x}| - \frac{1}{2}, \end{aligned}$$

which shows the required result. $\square$

Proof of Proposition 2

Let the system be in state $\vec {x}$. The mean revenue per unit time under fixed rate charging is given by

$$\begin{aligned} \bar{R}_F&= \sum _{\vec {x} : 1 \le |\vec {x}| \le n^*} \beta C \pi (\vec {x}) = \beta C \frac{1}{\sum _{\vec {x}} \chi (\vec {x})} \sum _{\vec {x} : 1 \le |\vec {x}| \le n^*} \chi (\vec {x})\\&= \beta C \frac{\sum _{n=1}^{n^*} t(n)}{\sum _{n=0}^\infty t(n)} . \end{aligned}$$

Using Lemma 1,

$$\begin{aligned} \bar{R}_F = \beta C \frac{1-\rho }{\Phi (\vec {0})} \frac{\Phi (\vec {0}) (\rho - \rho ^{n^*+1})}{1 - \rho } = \beta C \rho (1 - \rho ^{n^*}) . \end{aligned}$$

The mean revenue under VCG charging is

$$\begin{aligned} \bar{R}_V&= \sum _{\vec {x} : 2 \le |\vec {x}| \le n^*} \left( |\vec {x}| - \frac{1}{2}\right) \pi (\vec {x}) = u(1) - u(n^*) , \end{aligned}$$

where u(n) is given by Lemma 4. The result for $\bar{R}_V$ follows by simplification.

The mean revenue under congestion-based charging is given by

$$\begin{aligned} \bar{R}_L&= \sum _{\vec {x} : 1 \le |\vec {x}| \le n^*} \frac{|\vec {x}|^2}{C} \pi (\vec {x}) = \frac{1}{C} \sum _{\vec {x} : 1 \le |\vec {x}| \le n^*} |\vec {x}|^2 \frac{\chi (\vec {x})}{\sum _{\vec {y}}\chi (\vec {y})}\\&= \frac{1}{C} \frac{\sum _{1 \le n \le n^*} v(n)}{\sum _{n \ge 0} t(n)} . \end{aligned}$$

Using $v(n) = n^2 t(n)$ defined in (11) and Lemma 1,

$$\begin{aligned} \bar{R}_L = \frac{1-\rho }{C} \sum _{n=1}^{n^*} n^2 \rho ^n . \end{aligned}$$

$\square$

Proof of Proposition 3

To evaluate the mean payment by a class k user under fixed rate charging, consider the following integral where $A_k$ is the arrival process for class k users and $W_0^k$ is the random variable denoting the sojourn time of the class k arrival at time 0:

$$\begin{aligned} \bar{c}_k^F = \mathbb E_{A_k} \left[ \int _0^{W_0^k} \frac{\beta C}{|\vec {x}(t)|} \mathbf 1_{(1 \le |\vec {x}(t)| \le n^*)} dt \right] . \end{aligned}$$

Applying the Swiss Army formula (see [22]), equation (12), and Lemma 2,

$$\begin{aligned} \bar{c}_k^F&= \frac{\beta C}{\lambda _k} \mathbb E\left[ \frac{x_k}{|\vec {x}|} \mathbf 1_{( 1 \le |\vec {x}| \le n^*)} \right] = \frac{\beta C}{\lambda _k} \sum _{n=1}^{n^*} \sum _{\vec {x} : |\vec {x}| = n} \frac{x_k}{n} \pi (\vec {x}) \\&= \frac{\beta C}{\lambda _k} \frac{1}{\sum _{n=0}^\infty t(n)} \sum _{n=1}^{n^*} g_k(n) = \nu _k \beta \rho (1 - \rho ^{n^*} ) . \end{aligned}$$

Similarly, for VCG charging, the mean payment for a class k user is given by

$$\begin{aligned} \bar{c}_k^V = \mathbb E_{A_k} \left[ \int _0^{W_0^k} \left( 1 -\frac{1}{2 | \vec {x}(t) |} \right) \mathbf 1_{(2 \le |\vec {x}(t)| \le n^*)} dt \right] . \end{aligned}$$

Again, using the Swiss Army formula,

$$\begin{aligned} \bar{c}_k^V&= \frac{1}{\lambda _k} \mathbb E\left[ x_k \left( 1 - \frac{1}{2|\vec {x}|} \right) \mathbf 1_{(2 \le |\vec {x}| \le n^*)}\right] \nonumber \\&= \frac{1}{\lambda _k} \mathbb E\big [ x_k \mathbf 1_{(2 \le |\vec {x}| \le n^*)}\big ] - \frac{1}{\lambda _k} \mathbb E\left[ \frac{x_k}{2|\vec {x}|} \mathbf 1_{(2 \le |\vec {x}| \le n^*)} \right] . \end{aligned}$$

(30)

Let $J_1$ and $J_2$ be the first and the second term respectively in (30). Then,

$$\begin{aligned} J_1&= \frac{1}{\lambda _k} \sum _{n=2}^{n^*} \sum _{\vec {x} : |\vec {x}| = n} x_k \pi (\vec {x}) = \frac{1-\rho }{\lambda _k \Phi (\vec {0})} \sum _{n=2}^{n^*} s_k(n) \\&= \frac{\nu _k(1- \rho )}{C} \sum _{n=2}^{n^*} n \rho ^{n-1}, \end{aligned}$$

and,

$$\begin{aligned} J_2&= \frac{1}{2 \lambda _k} \sum _{n=2}^{n^*} \sum _{\vec {x} : |\vec {x}| = n} \frac{x_k}{|\vec {x}|} \pi (\vec {x}) = \frac{1 - \rho }{2\Phi (\vec {0}) \lambda _k} \sum _{n=2}^{n^*} g_k(n)\\&= \frac{\nu _k \rho }{2C} (1 - \rho ^{n^*-1}). \end{aligned}$$

Using the identity

$$\begin{aligned} \sum _{n=1}^m n \rho ^{n-1} = \frac{1 - \rho ^{m+1} - (m+1)(1 - \rho ) \rho ^m}{(1- \rho )^2}, \end{aligned}$$

and simplifying provides the required result. For congestion-based charging, the mean payment by a class k user is given by

$$\begin{aligned} \bar{c}_k^L = \mathbb E_{A_k} \left[ \int _0^{W_0^k} \frac{|\vec {x}(t)|}{C} \mathbf 1_{( 1 \le |\vec {x}(t)| \le n^*)} dt \right] . \end{aligned}$$

Applying the Swiss Army formula gives,

$$\begin{aligned} \bar{c}_k^L&= \frac{1}{\lambda _k C} \mathbb E [ x_k |\vec {x}| \mathbf 1_{(1 \le |\vec {x}| \le n^*)}] = \frac{1}{\lambda _k C} \sum _{n=1}^{n^*} \sum _{\vec {x} : |\vec {x}| = n} x_k n \pi (\vec {x}) \\&= \frac{1-\rho }{\lambda _k C \Phi (\vec {0})} \sum _{n=1}^{n^*} n s_k(n) = \frac{\nu _k ( 1 - \rho )}{C^2} \sum _{n=1}^{n^*} n^2 \rho ^{n-1} , \end{aligned}$$

which shows the required result. $\square$

Proof of Proposition 6

Suppose a class k arrival sees the system state as $\vec {x}$ on arrival. The fixed rate, pre-payment price is

$$\begin{aligned} p_k^F(\vec {x}) = \frac{\sigma _k^F}{|\vec {x} + \vec {e}_k|} \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) . \end{aligned}$$

It is required that the mean payment by a class k user equal $\bar{c}_k^F$, i.e.,

$$\begin{aligned} \mathbb E[ p_k^F(\vec {x})] = \bar{c}_k^F. \end{aligned}$$

(31)

Starting with the left hand side (LHS) of (31),

$$\begin{aligned} LHS&= \sum _{\vec {x}} \frac{\sigma _k^F}{|\vec {x} + \vec {e}_k|} \pi (\vec {x}) \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) \\&= \sigma _k^F (1-\rho ) \left[ \frac{A_{k,0}}{\rho } \log \frac{1}{1-\rho } + \left( \frac{ \frac{\rho }{1-\rho } - \log \frac{1}{1-\rho } }{\rho ^2} \right) \right. \\&\quad \times \left. \sum _{m=1}^K A_{k,m} \rho _m \right] . \end{aligned}$$

Equating this to $\bar{c}_k^F$ gives $\sigma _k^F$. Similarly, under VCG charging,

$$\begin{aligned} p_k^V(\vec {x}) = \sigma _k^V \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) , \end{aligned}$$

and

$$\begin{aligned} \mathbb E [p_k^V(\vec {x}) ]&= \sigma _k^V \sum _{\vec {x}} \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) \pi (\vec {x}) = \sigma _k^V \mathbb E[W_k] \\&= \sigma _k^V \frac{\nu _k}{C(1 - \rho )} . \end{aligned}$$

Equating this to $\bar{c}_k^V$ gives $\sigma _k^V$. Last, under congestion-based charging,

$$\begin{aligned} p_k^L (\vec {x}) = \sigma _k^L |\vec {x} + \vec {e}_k| \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) . \end{aligned}$$

Taking the expectation gives

$$\begin{aligned} \mathbb E[p_k^L(\vec {x}) ]&= \sigma _k^L \sum _{\vec {x}} |\vec {x} + \vec {e}_k| \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) \pi (\vec {x}) \\&= \frac{\sigma _k^L}{1-\rho }\left[ A_{k,0} + \frac{2}{1-\rho } \sum _{m=1}^K A_{k,m} \rho _m \right] , \end{aligned}$$

and equating this to $\bar{c}_k^L$ gives $\sigma _k^L$. $\square$

Proof of Proposition 7

The steps for deriving the second moment under congestion-based charging are outlined here. The proof for the other two charging models is similar.

$$\begin{aligned}&\mathbb E\big [(p_k^L (\vec {x}) )^2 \big ] = \sum _{\vec {x}} (p_k^L(\vec {x}))^2 \pi (\vec {x}) \\&\quad = (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 \left( A_{k,0} + \sum _{m=1}^K A_{k,m} x_m \right) ^2 \\&\qquad \times \,\pi (\vec {x}) (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 A_{k,0}^2 \pi (\vec {x}) \\&\quad \quad + (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 \sum _{m=1}^K A_{k,m}^2 x_m^2 \pi (\vec {x}) + (\sigma _k^L)^2 2 \\&\quad \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 A_{k,0} \sum _{m=1}^K A_{k,m} x_m \pi (\vec {x}) \\&\quad + (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 \sum _{i=1}^K \sum _{j=1 : j \ne i}^K A_{k,i} A_{k,j} x_i x_j \pi (\vec {x}) \\&\quad := S_1 + S_2 + S_3 + S_4\\ \end{aligned}$$

$$\begin{aligned}&S_1 = (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 A_{k,0}^2 \pi (\vec {x}) \\&\quad = \frac{(\sigma _k^L)^2 A_{k,0}^2 (1-\rho )}{\Phi (\vec {0})} \sum _{\vec {x}} (|\vec {x}|+1)^2 \chi (\vec {x})\\&\quad = \frac{(\sigma _k^L)^2 A_{k,0}^2 (1+\rho )}{(1-\rho )^2} . \\&S_2 = (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 \sum _{m=1}^K A_{k,m}^2 x_m^2 \pi (\vec {x}) \\&\quad = \frac{(\sigma _k^L)^2 (1-\rho )}{\Phi (\vec {0})} \sum _{m=1}^K A_{k,m}^2 \sum _{n=0}^\infty (n+1)^2 s_{m,m}(n) \\&\quad = 2 (\sigma _k^L)^2 \sum _{m=1}^K A_{k,m}^2 \frac{\rho _m(2 + 9\rho _m + 3 \rho _m \rho - \rho - \rho ^2)}{(1-\rho )^4} . \end{aligned}$$

$$\begin{aligned}&S_3 = (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 2 A_{k,0} \sum _{m=1}^K A_{k,m} x_m \pi (\vec {x}) \\&\quad = \frac{2 A_{k,0} (\sigma _k^L)^2 (1-\rho )}{\Phi (\vec {0})} \sum _{m=1}^K A_{k,m} \sum _{n=0}^\infty (n^2+2n+1) s_m(n) \\&\quad = 2 A_{k,0} (\sigma _k^L)^2 \frac{(2\rho +4)}{(1-\rho )^3} \sum _{m=1}^K A_{k,m} \rho _m .\\&S_4 = (\sigma _k^L)^2 \sum _{\vec {x}} |\vec {x} + \vec {e}_k|^2 \sum _{i=1}^K \sum _{j=1 : j \ne i}^K A_{k,i} A_{k,j} x_i x_j \pi (\vec {x}) \\&\quad = \frac{(\sigma _k^L)^2 (1-\rho )}{\Phi (\vec {0})} \sum _{i=1}^K \sum _{j=1 : j \ne i}^K A_{k,i} A_{k,j} \sum _{n=0}^\infty (n+1)^2 s_{i,j}(n) \\&\quad = \frac{6 (\sigma _k^L)^2 (3+\rho )}{(1-\rho )^4} \sum _{i=1}^K \sum _{j=1 : j \ne i}^K A_{k,i} A_{k,j} \rho _i \rho _j . \end{aligned}$$

Combining $S_1, S_2, S_3$ and $S_4$ gives the result. $\square$

Proof of Proposition 8

Let $\hat{\pi }(\vec {x})$ be the stationary distribution of the system under admission control. Noting that the underlying process is a truncation of a reversible process (see [23, Corollary 1.10]), the stationary distribution is given by

$$\begin{aligned} \hat{\pi }(\vec {x}) = \pi (\vec {x}) \frac{\sum _{\vec {y}} \pi (\vec {y})}{\sum _{\vec {y} : |\vec {y}| \le n^*} \pi (\vec {y}) } = \pi (\vec {x}) \left( \frac{1}{1-\rho ^{n^*+1}} \right) . \end{aligned}$$

The mean revenue is calculated as in Proposition 2 under $\hat{\pi }(\vec {x})$. $\square$

Proof of Proposition 9

Let $\hat{A}_k$ be the arrival process of class k users. Note that $\hat{A}_k$ is Poisson distributed for $|\vec {x}| < n^*$ and there is no new arrival if $|\vec {x}| = n^*$. The mean payment under the admission control system is

$$\begin{aligned} \hat{c}_k^{(\cdot )} = \mathbb E_{\hat{A}_k} \left[ \int _0^{W_k} c_k^{(\cdot )}(\vec {x}(t)) dt \right] = \frac{1}{\hat{\lambda }_k} \mathbb E \left[ x_k c_k^{(\cdot )} (\vec {x}) \right] . \end{aligned}$$

The second expectation above is under $\hat{\pi }$ instead of $\pi$ in Proposition 3 and the stochastic intensity $\hat{\lambda }_k$ of $\hat{A}_k$ is

$$\begin{aligned} \hat{\lambda }_k = \sum _{n=0}^{n^*-1} \lambda _k \pi (|\vec {x}| = n) = \lambda _k \left( \frac{1-\rho ^{n^*}}{1-\rho ^{n^*+1}} \right) . \end{aligned}$$

Thus, the mean payment under admission control is

$$\begin{aligned} \hat{c}_k^{(\cdot )} = \frac{1-\rho ^{n^*+1}}{1-\rho ^{n^*}} \frac{1}{1-\rho ^{n^*+1}} \bar{c}_k^{(\cdot )}. \end{aligned}$$

$\square$

Proof of Proposition 10

Let $P_k^F(\vec {x})$ be the random variable denoting the mean payment made by a given class k user when the state is $\vec {x}$. Let $Y(\vec {x})$ be the random variable indicating the payment made by this user in state $\vec {x}$ until the next event (arrival of a new user or departure of an existing user) occurs. Then,

$$\begin{aligned} P_k^F(\vec {x})&= \mathbf 1_{(\text {next event = arrival})} \sum _{m=1}^K \big [Y(\vec {x}) + P_k^F(\vec {x}+\vec {e}_m)\big ] \\&\quad + \,\mathbf 1_{(\text {next event = other user's departure})}\\&\quad \times \sum _{m=1}^K \big [Y(\vec {x}) + P_k^F(\vec {x} - \vec {e}_m)\big ] \\&\quad +\, \mathbf 1_{(\text {next event = tagged user's departure})} Y(\vec {x}) \\&\quad \text { for } |\vec {x}| < n^*. \end{aligned}$$

With a slight abuse of notation, let $Y(p; \vec {x})$ and $P_k^F(p; \vec {x})$ respectively denote the probability density function of $Y(\vec {x})$ and $P_k^F(\vec {x})$. Then,

$$\begin{aligned} Y(p; \vec {x}) = \hat{\omega }(\vec {x}) \frac{|\vec {x}|}{\beta C} e^{- \hat{\omega }(\vec {x}) \frac{|\vec {x}|}{\beta C} p}, \end{aligned}$$

and

$$\begin{aligned} P_k^F(p; \vec {x})&= \sum _{m=1}^K \frac{\lambda _m}{\hat{\omega }(\vec {x})} \int _0^p Y(q; \vec {x}) P_k^F(p-q; \vec {x}+\vec {e}_m) dq \\&\quad + \sum _{m=1}^K \frac{(\vec {x} - \vec {e}_k)_m C}{|\vec {x}| \nu _m \hat{\omega }(\vec {x})} \int _0^p Y(q; \vec {x}) P_k^F\\&\quad \times (p-q; \vec {x} - \vec {e}_m) dq \\&\quad + \frac{C}{|\vec {x}| \nu _k \hat{\omega }(\vec {x})} Y(p; \vec {x}) , \quad \text { for } |\vec {x}| < n^*. \end{aligned}$$

Let $P_k^F(s; \vec {x})$ be the Laplace–Stieltjis Transform (LST) of $P_k^F(p; \vec {x})$. Then, taking the LST of the above gives

$$\begin{aligned} \beta C P_k^F(s; \vec {x})&= \frac{1}{s + \frac{\hat{\omega }(\vec {x}) |\vec {x}|}{\beta C} } \left[ \sum _{m=1}^K \lambda _m(\vec {x}) P(s; \vec {x} + \vec {e}_m) \right. \\&\left. \quad + \sum _{m=1}^K \frac{(\vec {x} - \vec {e}_k)_m C}{\nu _m} P(s; \vec {x} - \vec {e}_m) + \frac{C}{\nu _k} \right] , \end{aligned}$$

for $|\vec {x}| < n^*$. Taking the derivative of the above once and twice and using

$$\begin{aligned}&\left. P_k^F(s; \vec {x}) \right| _{s=0} = 1, \left. -\left( P_k^F\right) '(s; \vec {x}) \right| _{s=0} = \eta _k^F(\vec {x}), \text { and }\\&\left. \left( P_k^F\right) ''(s; \vec {x})\right| _{s=0} = \xi _k^F(\vec {x}), \end{aligned}$$

gives (27) and (28). To obtain the second moment of payments by class k users, $\mathbb E [\xi _k^F(\vec {x})]$ is evaluated. Note that the state before arrival is $(\vec {x} - \vec {e}_k)$ when a payment of $P_k^F(\vec {x})$ is made.

For $|\vec {x}| = n^*$, the equations are similar except for $\lambda _m = 0$ since no arrivals take place in this state. $\square$