Stabilizing policies for probabilistic matching systems

Abstract

In this work, we introduce a novel queueing model with two classes of users, where users wait in the system to match with a candidate from the other class, instead of accessing a resource. This new model is useful for analyzing the traffic in web portals that match people who provide a service with people who demand the service, for example, employment portals, matrimonial and dating sites, and rental portals. We provide a Markov chain model for these systems and derive the probability distribution of the number of matches up to some finite time given the number of arrivals. We prove that if no control mechanism is employed these systems are unstable for any set of parameters, and suggest four different classes of control policies to assure stability. Contrary to the intuition that the rejection rate should decrease as the users get more likely to match, we show that for certain control policies the rejection rate is insensitive to the matching probability. Even more surprisingly, we show that for reasonable policies the rejection rate may be an increasing function of the matching probability. We also prove insensitivity results related to the average queue lengths and waiting times.

Appendix

1.1 Proof of Theorem 1

Proof

(Proof of Theorem 1) When $q=1$, the result is trivial. When $0<q<1$, the probability of no matchings, i.e., $k=0$, when there are $m$ and $n$ class-1 and class-2 users, respectively, is $r^{mn}$. Also, we know that $k\le m \le n$ and $P_{k,m,n}^q=0$ for $k>m$. When $1\le k\le m$, conditioning on whether a specific class-1 user matches with any of the $n$ class-2 users or not, we get

$$\begin{aligned} P^q_{k,m,n}=r^nP^q_{k, m-1, n}+(1-r^n)P^q_{k-1, m-1, n-1}. \end{aligned}$$

(10)

It is clear that given $P^q_{0,m,n}=r^{mn}$ and $P^q_{k,m,n}=0$ when $k>m$, the solution to (10) is unique. We now use induction to prove that for $m\ge 1$

$$\begin{aligned} P^q_{1,m,n}=(1-r^n)r^{(m-1)n}\frac{1-r^m}{r^{m-1}(1-r)}. \end{aligned}$$

(11)

For $m=1$, $P^q_{1,1,n}=1-r^n$. Now, assume that (11) holds for $P^q_{k,m-1,n}$, where $2\le m \le n$. Then,

$$\begin{aligned} \begin{array}{rl} P^q_{1,m,n}&{}=r^nP^q_{k,m-1,n}+(1-r^n)P^q_{0,m-1,n-1}\\ &{}=\displaystyle r^n(1-r^n)r^{(m-2)n}\frac{1-r^{m-1}}{r^{m-2}(1-r)}+(1-r^n)r^{(m-1)(n-1)}\\ &{}=\displaystyle (1-r^n)r^{(m-1)n}\frac{1-r^m}{r^{m-1}(1-r)}, \end{array} \end{aligned}$$

and hence (11) holds for any $1\le m\le n$.

Now, we show that solving the three-dimensional recursion (10) can be reduced to solving a two-dimensional recursion. Suppose $\{a_{k,m}, k\ge 0, m\ge 0\}$ solves

$$\begin{aligned} a_{k,m}=\left\{ \begin{array}{ll} 1 &{}\quad k=0,\\ r^{k-m}a_{k-1,m-1}+a_{k,m-1}&{}\quad 1\le k\le m,\\ 0 &{}\quad k>m. \end{array}\right. \end{aligned}$$

(12)

Then,

$$\begin{aligned} P^q_{k,m,n}=a_{k,m}r^{(m-k)n}\prod _{i=0}^{k-1}(1-r^{n-i}) \end{aligned}$$

(13)

solves (10), where if $k=0$, $\prod _{i=0}^{-1}(1-r^{n-i})$ is assumed to be 1. To prove this statement, first observe that (13) implies $P^q_{0,m,n}=r^{mn}$. When $k=1$, $a_{1,0}=0, a_{1,1}=1$ and $a_{1,m}=r^{-m+1}+a_{1,m-1}$ when $m\ge 2$, which implies

$$\begin{aligned} a_{1,m}=\sum _{i=0}^{m-1}\left( \frac{1}{r}\right) ^i=\frac{r^{m}-1}{r^{m-1}(r-1)}. \end{aligned}$$

Now, fix $k,m$, and $n$ such that $n\ge m\ge 2$ and $k\ge 1$. Suppose that for $0\le n'<n, 0\le m'<n', 0\le k'\le m', 0\le m''<m, k''\ge 0$ and $0\le k'''\le k, P^q_{k',m',n'}, P^q_{k'',m'',n}$ and $P^q_{k''',m,n}$ given by (13) coincide with the solution of (10). Then, if $k+1\le m$,

$$\begin{aligned} P^q_{k+1, m,n}&= r^nP^q_{k+1,m-1,n}+(1-r^n)P^q_{k,m-1,n-1}\\&=a_{k,m}r^{(m-k-1)n}\prod _{i=0}^{k}(1-r^{n-i})+a_{k,m-1}r^{(m-k-1)(n-1)}\prod _{i=0}^{k}(1-r^{n-i})\\&=\left( a_{k+1,m-1}+r^{-m+k+1}a_{k,m-1}\right) r^{(m-k-1)n}\prod _{i=0}^{k}(1-r^{n-i})\\&=a_{k+1,m}r^{(m-k-1)n}\prod _{i=0}^{k}(1-r^{n-i}). \end{aligned}$$

This proves that if we can solve (12), (13) provides us with the solution of (10).

Now, we provide a solution to the recursion (12). Using (12) $m-k+1$ times, we get

$$\begin{aligned} a_{k,m}&= r^{-m+k} a_{k-1,m-1} + r^{-m+k+1} a_{k-1,m-2} +\cdots + \underbrace{ a_{k-1,k-1} }_1 + \underbrace{a_{k,k-1}}_{0}\nonumber \\&=\sum _{j=0}^{m-k} r^{-m+k+j} \cdot a_{k-1,m-1-j}. \end{aligned}$$

(14)

We now guess that for $0\le k\le m$, $a_{k,m}$ has the following form

$$\begin{aligned} a_{k,m}=\sum _{i=0}^{k} r^{-mi} \alpha _{k,i}. \end{aligned}$$

(15)

Then, $a_{0,m}=1$ implies $\alpha _{0,0}=1$. For $k\ge 1$, we plug (15) into both sides of (14) and we obtain

$$\begin{aligned} \sum _{i=0}^{k} r^{-mi} \alpha _{k,i}&=\sum _{j=0}^{m-k} r^{-m+k+j}\sum _{i=0}^{k-1}r^{-(m-1-j)i}\alpha _{k-1,i}\\&=\sum _{i=0}^{k-1} \alpha _{k-1,i}r^{ -m+k - mi+i}\sum _{j=0}^{m-k} \left( r^{(i+1)}\right) ^j\\&=\sum _{i=0}^{k-1} \alpha _{k-1,i} \frac{ 1 - r^{(i+1)(m-k+1) } }{ 1 - r^{(i+1)} } r^{ -m+k - mi+i }\\&=\sum _{i=0}^{k-1} \frac{ r^{ -m+k - mi+i } - r^{ -ki +2i +1 } }{ 1 - r^{i+1} } \alpha _{k-1,i}\\&=\sum _{i=0}^{k-1} r^{ -m(i+1) } \frac{ r^{ k +i } }{ 1 - r^{i+1} } \alpha _{k-1,i} + \sum _{i=0}^{k-1} \frac{ - r^{ -ki +2i +1 } }{ 1 - r^{i+1} } \alpha _{k-1,i}.\\ \end{aligned}$$

We then shift the index of the first sum on the right-hand side, and for $k\ge 1$, we get

$$\begin{aligned} \sum _{i=0}^{k} r^{-mi} \alpha _{k,i} = \sum _{i=1}^{k} r^{ -mi } \frac{ r^{ k +i-1 } }{ 1 - r^{i} } \alpha _{k-1,i-1} + \sum _{i=0}^{k-1} \frac{ - r^{ -ki +2i +1 } }{ 1 - r^{i+1} } \alpha _{k-1,i}. \end{aligned}$$

Now, comparing the coefficient of $r^{-mi}$ for $0\le i \le k$, we obtain

$$\begin{aligned} \alpha _{k,0} = \sum _{i=0}^{k-1} \frac{ - r^{ -ki +2i +1 } }{ 1 - r^{i+1} } \alpha _{k-1,i}, \end{aligned}$$

(16)

and for every $1\le i\le k$

$$\begin{aligned} \alpha _{k,i} = \frac{ r^{ k +i-1 } }{ 1 - r^{i} } \alpha _{k-1,i-1}. \end{aligned}$$

(17)

Repeating (17) $i$ times we have for $1\le i\le k$,

$$\begin{aligned} \alpha _{k,i}&=\frac{ r^{ k+i-1 } }{(1 - r^{i})} \frac{ r^{ k+i-3 } }{ (1 - r^{i-1}) } \frac{ r^{ k+i-5 } }{ (1 - r^{i-2}) } \cdots \frac{ r^{ k-i+5 } }{ (1 - r^{3}) } \frac{ r^{ k-i+3 } }{ (1 - r^{2}) } \frac{ r^{ k-i+1 } }{ (1 - r^{1}) } \alpha _{k-i,0}\nonumber \\&=\frac{ r^{ ki } }{ \prod _{j=1}^i (1 - r^{j}) } \alpha _{k-i,0}. \end{aligned}$$

(18)

Plugging (18) into (16) and substituting $\beta _k = \alpha _{k,0}$, we obtain a new recurrence:

$$\begin{aligned} \beta _k&= \sum _{i=1}^{k-1} \frac{ - r^{ -ki +2i +1 } }{ (1 - r^{i+1}) } \frac{ r^{ (k-1)i } }{ \prod _{j=1}^i (1 - r^{j}) } \beta _{k-1-i}+\frac{-r\beta _{k-1}}{1-r}\\&= \sum _{i=0}^{k-1} \frac{ - r^{ i + 1 }\beta _{k-1-i} }{ \prod _{j=1}^{i+1} (1 - r^{j}) }, \end{aligned}$$

and replacing $l=k-1-i$, we get

$$\begin{aligned} \beta _k = \sum _{l=0}^{k-1} \frac{ - r^{ k-l }\beta _{l} }{ \prod _{j=1}^{k-l} (1 - r^{j}) } . \end{aligned}$$

Thus, by now we have a recurrence of the form $\beta _0=1$ and for every $k\ge 1$

$$\begin{aligned} \beta _k = \sum _{l=0}^{k-1} \gamma _{k,l} \beta _{l}, \end{aligned}$$

(19)

where

$$\begin{aligned} \gamma _{i,j} = \frac{y_i}{y_j} z_{i-j}, \end{aligned}$$

(20)

with $y_i=r^i$ and $z_{i} = \frac{ -1 }{ \prod _{j=1}^{i} (1 - r^{j})}$. The recursion (19) is a 1-dimensional recurrence, and for $k\ge 1$ it has the general solution:

$$\begin{aligned} \beta _k&= \sum _{l>0} \left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=k} \left( \prod _{j=1}^{l} \gamma _{ i_{j}, i_{j-1} } \right) \right) \beta _0\nonumber \\&= \sum _{l>0} \left( \sum _{0=i_0<i_1<\cdots <i_{l-1}<i_l=k} \gamma _{i_l,i_{l-1}}\gamma _{i_{l-1},i_{l-2}} \cdots \gamma _{i_2,i_1} \gamma _{i_1,i_0} \right) . \end{aligned}$$

(21)

We prove this by induction.

• (i) when $k=1$, Eq. (21) implies

  $$\begin{aligned} \beta _1 = \sum _{l>0} \left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=1} \left( \prod _{j=1}^{l} \gamma _{ i_{j}, i_{j-1} } \right) \right) \beta _0=\gamma _{1,0}, \end{aligned}$$

  which satisfies Equation (19). Hence Eq. (21) holds for $\beta _1$.

• (ii) suppose Eq. (21) holds for all $\beta _n$, $1\le n\le k-1$. Then according to Equation (19),

  $$\begin{aligned} \beta _k&= \sum _{n=0}^{k-1} \gamma _{k,n} \beta _{n}\\&=\sum _{n=0}^{k-1} \gamma _{k,n} \left( \sum _{l>0} \left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=n} \left( \prod _{j=1}^{l} \gamma _{ i_{j}, i_{j-1} } \right) \right) \right) \beta _0\\&=\gamma _{k,0}+\gamma _{k,1}\gamma _{1,0}+\cdots +\sum _{l>0}\left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=n}\gamma _{k,n}\gamma _{n,i_{l-1}}\cdots \gamma _{i_1,0}\right) \\&\quad +\cdots +\sum _{l>0}\left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=k-1} \gamma _{k,k-1}\gamma _{k-1,i_{l-1}}\gamma _{i_{l-1},i_{l-2}}\cdots \gamma _{i_1,0}\right) \\&=\sum _{l>0} \left( \sum _{ 0=i_0<i_1<\cdots <i_{l-1}<i_l=k} \left( \prod _{j=1}^{l} \gamma _{ i_{j}, i_{j-1} } \right) \right) . \end{aligned}$$

  Therefore, if Eq. (21) holds for all $\beta _n$, $1\le n\le k-1$, then it also holds for $\beta _n, n=k$.

Now by adding Eq. (20) to (21), we have

$$\begin{aligned} \beta _k = \sum _{l>0} \left( \sum _{0=i_0<i_1<\cdots <i_{l-1}<i_l=k} \frac{ y_{k} }{ y_0 } \prod _{j=1}^{l} z_{ i_{j} - i_{j-1} }. \right) . \end{aligned}$$

Furthermore, using the substitution $d_j=i_{j}-i_{j-1} \ge 1$, where $d_1+d_2+\ldots +d_l=k$ we have:

$$\begin{aligned} \beta _k = \sum _{l>0} \left( \sum _{d_1+d_2+\cdots +d_l=k} \frac{ y_{k} }{ y_0 } \prod _{i=1}^{l} z_{ d_i } \right) , \end{aligned}$$

where indexes $d_1, d_2, \ldots$ are taken from $\mathbb {N}_+=\{1,2,3,\ldots \}$. Thus, for $k\ge 1$,

$$\begin{aligned} \beta _k = \sum _{l>0} \left( \sum _{d_1+d_2+\cdots +d_l=k} r^k \prod _{i=1}^{l} \frac{ -1 }{ \prod _{j=1}^{d_i} (1 - r^{j}) } \right) . \end{aligned}$$

(22)

Finally, (18) implies that for $1\le i\le k$,

$$\begin{aligned} \alpha _{k,i} = \frac{ r^{ ki } }{ \prod _{j=1}^i (1 - r^{j}) } \beta _{k-i}. \end{aligned}$$

Using (15), we have

$$\begin{aligned} a_{k,m}&=\sum _{i=0}^{k} r^{-mi} \alpha _{k,i} =\left( \sum _{i=0}^{k-1} r^{-mi} \alpha _{k,i} \right) + r^{-mk} \alpha _{k,k}\\&=\alpha _{k,0}+\left( \sum _{i=1}^{k-1} r^{-mi} \alpha _{k,i} \right) + r^{-mk} \alpha _{k,k}. \end{aligned}$$

As a result,

$$\begin{aligned} a_{k,m}&= \sum _{l>0} \left( \sum _{d_1+d_2+\cdots +d_l=k} r^k \frac{ (-1)^{l} }{ \prod _{i=1}^{l} \prod _{j=1}^{d_i} (1 - r^{j}) }\right) + r^{-mk} \frac{ r^{ k^2 } }{ \prod _{j=1}^k (1 - r^{j}) } \\&+ \sum _{i=1}^{k-1} r^{-mi} \frac{ r^{ ki } }{ \prod _{j=1}^i (1 - r^{j}) } \sum _{l>0} \left( \sum _{d_1+d_2+\cdots +d_l={k-i}} r^{k-i} \frac{ (-1)^{l} }{ \prod _{w=1}^{l}\prod _{j=1}^{d_w} (1 - r^{j}) } \right) , \end{aligned}$$

with indexes $d_1, d_2,\ldots$ taken from $\mathbb {N}_+=\{1,2,3,\ldots \}$. $\square$

1.2 Stationary probabilities under the ASQ policy

Proof

(Proof of Theorem 9) Under the ASQ policy, the rate balance equations can be written as follows:

$$\begin{aligned} (\lambda _1+\lambda _2)p_{0,0}&=\lambda _1(1-r)p_{0,1}+\lambda _2(1-r)p_{1,0},\end{aligned}$$

(23)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{i,i}&=\lambda _1r^ip_{i-1,i}+\lambda _2r^ip_{i,i-1}\nonumber \\&\quad +\lambda _1(1-r^{i+1})p_{i,i+1}+\lambda _2(1-r^{i+1})p_{i+1,i},\ i\ge 1,\end{aligned}$$

(24)

$$\begin{aligned} \lambda _2p_{i+1,i}&=\lambda _1r^ip_{i,i}+\lambda _1(1-r^{i+1})p_{i+1,i+1}, \ i\ge 0,\end{aligned}$$

(25)

$$\begin{aligned} \lambda _1p_{i,i+1}&=\lambda _2r^ip_{i,i}+\lambda _2(1-r^{i+1})p_{i+1,i+1}, \ i\ge 0,\end{aligned}$$

(26)

$$\begin{aligned} \sum _{i=0}^{\infty }\sum _{j=i-1}^{i+1}p_{i,j}&=1. \end{aligned}$$

(27)

The state space has a very special structure where the removal of a state in the form $(i,i), i>0$ disconnects the transition graph. This implies a rate balance for the transitions between states $\{(i, i-1),(i-1,i)\}$ and $(i,i)$, which implies the following detailed balance type equations:

$$\begin{aligned} (\lambda _1+\lambda _2)r^ip_{i,i}&=\lambda _1(1-r^{i+1})p_{i,i+1}+\lambda _2(1-r^{i+1})p_{i+1,i},\ i\ge 0,\end{aligned}$$

(28)

$$\begin{aligned} (\lambda _1+\lambda _2)(1-r^i)p_{i,i}&=\lambda _1r^ip_{i-1,i}+\lambda _2r^ip_{i,i-1}, \ i\ge 1, \end{aligned}$$

(29)

Equation (23) can obtained by setting $i=0$ in (28) and further, summing up (28) and (29) for a given $i\ge 1$ we have (24). Hence, any solution to the set of Eqs. (25)–(29) also solves (23)–(27) and hence should be unique. Further, Eqs. (25) and (26) imply $\frac{p_{i+1,i}}{p_{i,i+1}}=\frac{\lambda _1^2}{\lambda _2^2}.$ Hence, substituting $p_{i+1,i}=\frac{\lambda _1^2}{\lambda _2^2}p_{i,i+1}$ into (28) we obtain

$$\begin{aligned} p_{i,i+1}=\frac{\lambda _2}{\lambda _1}\frac{1-r^{i+1}}{r^i}p_{i,i}\ \mathrm{and}\ p_{i+1,i}=\frac{\lambda _1}{\lambda _2}\frac{1-r^{i+1}}{r^i}p_{i,i}. \end{aligned}$$

(30)

Then (30) and (29) together imply $p_{i+1,i+1}=\frac{r^i r^{i+1}}{(1-r^{i+1})^2}p_{i,i}, i\ge 0,$ and hence, for $i\ge 1$,

$$\begin{aligned} p_{i,i}&=\prod _{k=1}^{i}\frac{r^k r^{k-1}}{(1-r^k)^2}p_{0,0}=\frac{r^{i^2}p_{0,0}}{\left[ \prod _{k=1}^i(1-r^k)\right] ^2}. \end{aligned}$$

(31)

Substituting (31) into (30) and defining $\prod _{k=1}^0(1-r^k)=1$, for $i\ge 0$

$$\begin{aligned} p_{i,i+1}&=\frac{\lambda _2r^{i(i+1)}p_{0,0}}{\lambda _1\prod _{k=1}^{i}(1-r^k)\prod _{k=1}^{i+1}(1-r^k)}\\ p_{i+1,i}&=\frac{\lambda _1r^{i(i+1)}p_{0,0}}{\lambda _2\prod _{k=1}^{i}(1-r^k)\prod _{k=1}^{i+1}(1-r^k)}. \end{aligned}$$

Finally the result follows from plugging all $p_{i,j}$ back in (27). $\square$

1.3 Insensitivity proofs for functional threshold policies with $h(x)=x+d$

Lemma 20

Suppose that the functional threshold policy with a threshold function of the form $h(x)=x+d$, where $d\in \mathbb {N}$, is employed to stabilize a probabilistic matching system. For $(i,j)\in \mathbb {N}^2$, let $p_{i,j}$ be the stationary probability of being at state $(i,j)$. Now, define $a_l=\sum _{i=0}^\infty p_{i,i+l}$ and $a_{-l}=\sum _{j=0}^\infty p_{j+l,j}$ Then, if $\lambda _1\ne \lambda _2$,

$$\begin{aligned} a_{d+1}=\frac{1-\frac{\lambda _2}{\lambda _1}}{1-\left( \frac{\lambda _2}{\lambda _1}\right) ^{2d+3}} \text{ and } a_l=\left( \frac{\lambda _2}{\lambda _1}\right) ^{d+1-l}a_{d+1}, \text{ for } \ -d-1\le l\le d, \end{aligned}$$

and if $\lambda _1=\lambda _2$, $a_l=\frac{1}{2d+3},\ -d-1\le l \le d+1$.

Proof

When $q=1$, the process $\{X^{1,FT}(t)+d+1, t\ge 0\}$ is stochastically equivalent to an $M/M/1/2d+2$ system and the result follows. When $0<q<1$, the state space can be written as $\mathbb {S}=\{(i,i+l): i\in \mathbb {N}, -d-1\le l\le d+1, i+l\in \mathbb {N}\}$, hence the global balance equations are

$$\begin{aligned} (\lambda _1+\lambda _2)p_{0,0}&=\lambda _1(1-r)p_{0,1}+\lambda _2(1-r)p_{1,0}, \end{aligned}$$

(32)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{i,i}&=\lambda _1r^ip_{i-1,i}+\lambda _2r^ip_{i,i-1}+\lambda _1(1-r^{i+1})p_{i,i+1}\nonumber \\&\quad +\lambda _2(1-r^{i+1})p_{i+1,i},\ i\ge 1,\end{aligned}$$

(33)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{l,0}&=\lambda _1p_{l-1,0}+\lambda _1(1-r)p_{l+1,1}+\lambda _2(1-r^{l+1})p_{l+1,0},\nonumber \\&\qquad 1\le l\le d\end{aligned}$$

(34)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{i+l,i}&=\lambda _1r^ip_{i+l-1,i}+\lambda _1(1-r^{i+1})p_{i+l,i+1}+\lambda _2r^{i+l}p_{i+l,i-1}\nonumber \\&\quad +\lambda _2(1-r^{i+l+1})p_{i+l+1,i}, \ i\ge 1, 1\le l\le d, \end{aligned}$$

(35)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{0,l}&=\lambda _2p_{0,l-1}+\lambda _2(1-r)p_{1,l+1}+\lambda _1(1-r^{l+1})p_{0,l+1},\nonumber \\&\qquad 1\le l\le d\end{aligned}$$

(36)

$$\begin{aligned} (\lambda _1+\lambda _2)p_{i,i+l}&=\lambda _2r^ip_{i,i+l-1}+\lambda _2(1-r^{i+1})p_{i+1,i+l}+\lambda _1r^{i+l}p_{i-1,i+l}\nonumber \\&\quad +\lambda _1(1-r^{i+l+1})p_{i,i+l+1}, \ i\ge 1, 1\le l\le d, \end{aligned}$$

(37)

$$\begin{aligned} \lambda _2p_{i+d+1,i}&=\lambda _1r^ip_{i+d,i}+\lambda _1(1-r^{i+1})p_{i+d+1,i+1}, \ i\ge 0,\end{aligned}$$

(38)

$$\begin{aligned} \lambda _1p_{i,i+d+1}&=\lambda _2r^ip_{i,i+d}+\lambda _2(1-r^{i+1})p_{i+1,i+d+1}, \ i\ge 0,\end{aligned}$$

(39)

$$\begin{aligned} \sum _{i=0}^{\infty }\sum _{j=i-d-1}^{i+d+1}p_{i,j}&=1. \end{aligned}$$

(40)

We sum (33) for $i=1$ to $\infty$ and then add (32) to get

$$\begin{aligned} (\lambda _1+\lambda _2)a_0&=\lambda _1 a_{-1}+\lambda _2 a_{1}. \end{aligned}$$

(41)

Repeating the same procedures for pairs (34) and (35), (36) and (37), (38) and (39),

$$\begin{aligned} (\lambda _1+\lambda _2)a_l&=\lambda _2a_{l+1}+\lambda _1a_{l-1}, \ 1\le l \le d,\end{aligned}$$

(42)

$$\begin{aligned} (\lambda _1+\lambda _2)a_{-l}&=\lambda _1a_{-l-1}+\lambda _2a_{-l+1}, \ 1\le l\le d,\end{aligned}$$

(43)

$$\begin{aligned} \lambda _2a_{d+1}&=\lambda _1a_{d},\end{aligned}$$

(44)

$$\begin{aligned} \lambda _1a_{-d-1}&=\lambda _2a_{-d}. \end{aligned}$$

(45)

We notice that, similar to the case $q=1$, if we replace $b_l=a_{l-d}$ in (41)–(45), we obtain the global balance equations of an $M/M/1/2d+2$ system. Hence, the result follows. $\square$

Proof

(Proof of Theorem 12) Using PASTA property, $c_1=a_{d+1}$ and the result follows from Lemma 20. $\square$

Proof

(Proof of Theorem 14) Without lost of generality, assume $d\ge 0$ is an integer. The difference of average queue lengths can be written as

$$\begin{aligned} L_1-L_2&=\sum _{l=-d-1}^{d+1}la_l\\&=\frac{1-\frac{\lambda _2}{\lambda _1}}{1-(\frac{\lambda _2}{\lambda _1})^{2d+3}}\left( -\frac{\lambda _2}{\lambda _1}\right) ^{d+2}\sum _{l=-d-1}^{d+1}(-l)\left( \frac{\lambda _2}{\lambda _1}\right) ^{-l-1}. \end{aligned}$$

Using $\rho =\frac{\lambda _2}{\lambda _1}$,

$$\begin{aligned} \sum _{l=-d-1}^{d+1}(-l)\left( \frac{\lambda _2}{\lambda _1}\right) ^{-l-1}&=\sum _{l=-d-1}^{d+1}\frac{\partial }{\partial \rho } \rho ^{-l}\\&=\frac{\partial }{\partial \rho }\frac{\rho ^{d+1}(1-\rho ^{-2d-3})}{1-\rho ^{-1}}\\&=\!\frac{((d+2)\rho ^{d+\!1}\!+\!(d+1)\rho ^{-d-2})(\rho -1)\!-\!(\rho ^{d+2}\!-\!\rho ^{-d-1})}{(\rho -1)^2}. \end{aligned}$$

Hence,

$$\begin{aligned} L_1-L_2&=\frac{(d+2)\rho ^{2d+3}+d+1}{1-\rho ^{2d+3}} +\frac{(1-\rho )\rho ^{d+2}(\rho ^{d+2}-\rho ^{-d-1})}{(1-\rho ^{2d+3})(\rho -1)^2}. \end{aligned}$$

$\square$