Handoff calls’ joining behavior and incentive mechanism in wireless cellular networks with retrial orbit

Abstract

In wireless cellular networks, handoff is a key element in providing quality of service and supporting mobility for users. There is an interaction between wireless cellular networks and handoff calls which strategically act to achieve their own objectives. In order to reasonably control the access of handoff calls and achieve the optimization of social network resources, we present a game-theoretic queueing model to investigate the strategic behavior of handoff calls in wireless cellular networks. The noncooperative joining behavior of handoff calls that maximizes their expected net benefit selfishly is considered. Moreover, the socially optimal joining behavior of handoff calls that maximizes their social total welfare is analyzed. The equilibrium and socially optimal strategies of handoff calls are obtained. Counter-intuitively, we find that the socially optimal joining probability is larger than the equilibrium joining probability. In order to eliminate the difference of joining probabilities, an appropriate incentive-payment is proposed to attract handoff calls to join the system actively. In doing so, it is feasible to induce individuals to behave in a socially best way.

Appendices

Appendix 1

In order to simplify the matrices with complicated construction, we define some symbols as following: $\mu_{1} = r_{1} + h_{1} ,\;\mu_{2} = r_{2} + h_{2}$.

• 1) The block ${\mathbf{B}}_{0}$ is a $(c + 1)(c + 2)/2$ dimensional matrix with its elements given by:

  $${\mathbf{B}}_{0} = \left( {\begin{array}{*{20}c} {{\mathbf{B}}_{00} } & {{\mathbf{B}}_{01} } & {} & {} & {} & {} \\ {{\mathbf{B}}_{10} } & {{\mathbf{B}}_{11} } & {{\mathbf{B}}_{12} } & {} & {} & {} \\ {} & {{\mathbf{B}}_{21} } & {{\mathbf{B}}_{22} } & {{\mathbf{B}}_{23} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {{\mathbf{B}}_{c - 1,c - 2} } & {{\mathbf{B}}_{c - 1,c - 1} } & {{\mathbf{B}}_{c - 1,c} } \\ {} & {} & {} & {} & {{\mathbf{B}}_{c,c - 1} } & {B_{c,c} } \\ \end{array} } \right).$$

  (25)

  Furthermore, its elements are shown as follows:

  $$ {\mathbf{B}}_{0,0} = \left( {\begin{array}{*{20}c} { - \varphi_{0} } & {\lambda_{2} \beta_{0} } & {} & {} & {} & {} \\ {\mu_{2} } & { - \varphi_{1} } & {\lambda_{2} \beta_{1} } & {} & {} & {} \\ {} & {2\mu_{2} } & { - \varphi_{2} } & {\lambda_{2} \beta_{2} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {(c - 1)\mu_{2} } & { - \varphi_{c - 1} } & {\lambda_{2} \beta_{c - 1} } \\ {} & {} & {} & {} & {c\mu_{2} } & { - \varphi_{c} } \\ \end{array} } \right)_{(c + 1) \times (c + 1)} , $$

  (26)

  where $\varphi_{k} = - (\lambda_{1} q\alpha_{k} + \lambda_{2} \beta_{k} + k\mu_{2} ),k = 0,1 \cdots c - 1$, $\varphi_{c} = - (\lambda_{1} q\omega + c\mu_{2} )$.

  $$ {\mathbf{B}}_{i,i} = \left( {\begin{array}{*{20}c} { - \tau_{i} } & {\lambda_{2} \beta_{i} } & {} & {} & {} & {} \\ {\mu_{2} } & { - (\tau_{i + 1} + \mu_{2} )} & {\lambda_{2} \beta_{i + 1} } & {} & {} & {} \\ {} & {2\mu_{2} } & { - (\tau_{i + 2} + 2\mu_{2} )} & {\lambda_{2} \beta_{i + 2} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {(c - i - 1)\mu_{2} } & { - (\tau_{c - 1} + (c - i - 1)\mu_{2} )} & {\lambda_{2} \beta_{c - 1} } \\ {} & {} & {} & {} & {(c - i)\mu_{2} } & { - (\tau_{c} + (c - i)\mu_{2} )} \\ \end{array} } \right)_{(c + 1 - i) \times (c + 1 - i)} , $$

  (27)

  where $\tau_{k} = - (\lambda_{1} q\alpha_{k} + \lambda_{2} \beta_{k} + i\mu_{1} ),k = i, \cdots ,c - 1$, $\tau_{c} = - (\lambda_{1} q\omega + i\mu_{1} )$. $B_{cc} = - (\lambda_{1} q + c\mu_{1} )$.

  $$ {\mathbf{B}}_{i,i + 1} = \left( {\begin{array}{*{20}c} {\lambda_{1} q\alpha_{i} } & {} & {} & {} \\ {} & {\lambda_{1} q\alpha_{i + 1} } & {} & {} \\ {} & {} & \ddots & {} \\ {} & {} & {} & {\lambda_{1} q\alpha_{c - 1} } \\ 0 & 0 & 0 & 0 \\ \end{array} } \right)_{(c + 1 - i) \times (c - i)} i = 0,1, \cdots ,c - 1. $$

  (28)
  $$ {\mathbf{B}}_{i + 1,i} = \left( {\begin{array}{*{20}c} {(i + 1)\mu_{1} } & {} & {} & {} & 0 \\ {} & {(i + 1)\mu_{1} } & {} & {} & 0 \\ {} & {} & \ddots & {} & 0 \\ {} & {} & {} & {(i + 1)\mu_{1} } & 0 \\ \end{array} } \right)_{(c - i) \times (c + 1 - i)} i = 0,1, \cdots ,c - 1. $$

  (29)

• 2) The block ${\mathbf{A}}_{0}$ is a $(c + 1)(c + 2)/2$ dimensional matrix with its elements given by:

  $$ {\mathbf{A}}_{0} = \left( {\begin{array}{*{20}c} {{\mathbf{D}}_{0} } & {} & {} & {} & {} \\ {} & {{\mathbf{D}}_{1} } & {} & {} & {} \\ {} & {} & {{\mathbf{D}}_{2} } & {} & {} \\ {} & {} & {} & \ddots & {} \\ {} & {} & {} & {} & {{\mathbf{D}}_{c} } \\ \end{array} } \right), $$

  (30)

  where ${\mathbf{D}}_{i} = \left( {\begin{array}{*{20}c} 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & {\lambda_{1} q\omega } \\ \end{array} } \right)_{(c + 1 - i) \times (c + 1 - i)} i = 0,1, \cdots ,c$.

• 3) The block ${\mathbf{A}}_{2}$ is a $(c + 1)(c + 2)/2$ dimensional matrix with its elements given by:

  $$ {\mathbf{A}}_{2} = \left( {\begin{array}{*{20}c} 0 & {{\mathbf{E}}_{0} } & 0 & \cdots & 0 \\ 0 & 0 & {{\mathbf{E}}_{1} } & 0 & 0 \\ 0 & 0 & 0 & \ddots & \vdots \\ \vdots & \vdots & \vdots & \vdots & {{\mathbf{E}}_{c - 1} } \\ 0 & 0 & 0 & \cdots & 0 \\ \end{array} } \right), $$

  (31)

  where ${\mathbf{E}}_{i} = \left( {\begin{array}{*{20}c} \delta & 0 & \cdots & 0 \\ 0 & \delta & \cdots & 0 \\ 0 & 0 & \ddots & 0 \\ \vdots & \vdots & \vdots & \delta \\ 0 & 0 & 0 & 0 \\ \end{array} } \right)_{(c + 1 - i) \times (c - i)} i = 0,1, \cdots ,c - 1$.

• 4) The block ${\mathbf{A}}_{1}$ is a $(c + 1)(c + 2)/2$ dimensional matrix with its elements given by:

  $$ {\mathbf{A}}_{1} = \left( {\begin{array}{*{20}c} {{\mathbf{A}}_{00} } & {{\mathbf{A}}_{01} } & {} & {} & {} & {} \\ {{\mathbf{A}}_{10} } & {{\mathbf{A}}_{11} } & {{\mathbf{A}}_{12} } & {} & {} & {} \\ {} & {{\mathbf{A}}_{21} } & {{\mathbf{A}}_{22} } & {{\mathbf{A}}_{23} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {{\mathbf{A}}_{c - 1,c - 2} } & {{\mathbf{A}}_{c - 1,c - 1} } & {{\mathbf{A}}_{c - 1,c} } \\ {} & {} & {} & {} & {{\mathbf{A}}_{c,c - 1} } & {A_{c,c} } \\ \end{array} } \right). $$

  (32)

Its elements are shown as follows.

$$ {\mathbf{A}}_{0,0} = \left( {\begin{array}{*{20}c} { - \varphi_{10} } & {\lambda_{2} \beta_{0} } & {} & {} & {} & {} \\ {\mu_{2} } & { - \varphi_{11} } & {\lambda_{2} \beta_{1} } & {} & {} & {} \\ {} & {2\mu_{2} } & { - \varphi_{12} } & {\lambda_{2} \beta_{2} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {(c - 1)\mu_{2} } & { - \varphi_{1c - 1} } & {\lambda_{2} \beta_{c - 1} } \\ {} & {} & {} & {} & {c\mu_{2} } & { - \varphi_{1c} } \\ \end{array} } \right)_{(c + 1) \times (c + 1)} , $$

(33)

where $\varphi_{1k} = - (\lambda_{1} q\alpha_{k} + \lambda_{2} \beta_{k} + k\mu_{2} + \delta ),k = 0,1 \cdots c - 1$, $\varphi_{1c} = - (\lambda_{1} q\omega + c\mu_{2} )$.

$${\mathbf{A}}_{i,i} = \left( {\begin{array}{*{20}c} { - \tau_{1i} } & {\lambda_{2} \beta_{i} } & {} & {} & {} & {} \\ {\mu_{2} } & { - ( - \tau_{1i + 1} + \mu_{2} )} & {\lambda_{2} \beta_{i + 1} } & {} & {} & {} \\ {} & {2\mu_{2} } & { - ( - \tau_{1i + 1} + 2\mu_{2} )} & {\lambda_{2} \beta_{i + 2} } & {} & {} \\ {} & {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {} & {(c - i - 1)\mu_{2} } & { - ( - \tau_{1c - 1} + (c - i - 1)\mu_{2} )} & {\lambda_{2} \beta_{c - 1} } \\ {} & {} & {} & {} & {(c - i)\mu_{2} } & { - ( - \tau_{1c} + (c - i)\mu_{2} )} \\ \end{array} } \right)_{(c + 1 - i) \times (c + 1 - i)},$$

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where $\tau_{1k} = - (\lambda_{1} q\alpha_{k} + \lambda_{2} \beta_{k} + i\mu_{1} + \delta ),k = i, \cdots ,c - 1,$, $\tau_{1c} = - (\lambda_{1} q\omega + i\mu_{1} )$, $i = 1, \cdots ,c - 1$. $A_{cc} = - (\lambda_{1} q + c\mu_{1} )$.

$$ {\mathbf{A}}_{i,i + 1} = \left( {\begin{array}{*{20}c} {\lambda_{1} q\alpha_{i} } & {} & {} & {} \\ {} & {\lambda_{1} q\alpha_{i + 1} } & {} & {} \\ {} & {} & \ddots & {} \\ {} & {} & {} & {\lambda_{1} q\alpha_{c - 1} } \\ 0 & 0 & 0 & 0 \\ \end{array} } \right)_{(c + 1 - i) \times (c - i)} i = 0,1, \cdots ,c - 1. $$

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$$ {\mathbf{A}}_{i + 1,i} = \left( {\begin{array}{*{20}c} {(i + 1)\mu_{1} } & {} & {} & {} & 0 \\ {} & {(i + 1)\mu_{1} } & {} & {} & 0 \\ {} & {} & \ddots & {} & 0 \\ {} & {} & {} & {(i + 1)\mu_{1} } & 0 \\ \end{array} } \right)_{(c - i) \times (c + 1 - i)} i = 0,1, \cdots ,c - 1. $$

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Appendix 2

2.1 Proof of Theorem 2

We mark a type-1 call which arrives in the system, analyze the equilibrium strategy and discuss it in three cases.

Case 1: $R < C \cdot e^{{\varepsilon E(T_{1} (0))}}$. In this case, even if no other type-1 call joins, the expected net benefit of a type-1 call which joins is negative. Therefore, the unique pure-strategy equilibrium is balking, i.e., $q_{e} = 0$. Moreover, balking is a dominant strategy.

Case 2: $R > C \cdot e^{{\varepsilon E(T_{1} (1))}}$. In this case, even if all type-1 calls join, they all enjoy a positive expected net benefit. Therefore, the unique pure-strategy equilibrium is joining, i.e., $q_{e} = 1$. Moreover, joining is a dominant strategy.

Case 3: $C \cdot e^{{\varepsilon E(T_{1} (0))}} \le R \le C \cdot e^{{\varepsilon E(T_{1} (1))}}$. In this case, we can find a unique mixed-strategy equilibrium $q_{ee} \in [0,1]$ which can be derived by solving equation $R - C \cdot e^{{\varepsilon E(T_{1} (q_{ee} ))}} = 0$. Then, any strategy between joining and balking is a best response. Therefore, $q_{e} = q_{ee}$. We complete the proof.

Appendix 3

3.1 Proof of Theorem 3

We need to know the monotonicity of the function $S(q)$ before analyzing the problem of maximizing $S(q)$ for $q \in [0,1]$.

Taking the derivative of function $S(q)$ with respect to $q$, we get

$$S^{\prime}(q) = {\lambda_{1}}^{e} R - C \cdot e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q)).$$

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According to Remark 2, we think that $E(N_{1} (q)$ should be convex in $q$. Both $e^{{\varepsilon E(N_{1} (q))}}$ and $E^{\prime}(N_{1} (q))$ are increasing in $q$. Both $C$ and $\varepsilon$ are positive. Hence, $S^{\prime}(q)$ is decreasing in $q$.

Next, we consider the following three cases:

Case 1: ${\lambda_{1}}^{e} R > C \cdot e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q))$ for $q \in [0,1]$. In this case, we observe that $S^{\prime}(q) > 0$ for any $q \in [0,1]$, which implies that $S(q)$ is strictly increasing in $q$. The unique maximum is attained at $q = 1$. Thus, the best response of type-1 calls is to join the system with probability $q^{*} = 1$. That is, $q^{*} = 1$ is socially optimal joining strategy.

Case 2: ${\lambda_{1}}^{e} R < C \cdot e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q))$ for $q \in [0,1]$. In this case, we observe that $S^{\prime}(q) < 0$ for any $q \in [0,1]$, which implies that $S(q)$ is strictly decreasing in $q$. The unique maximum is attained at $q = 0$. Thus, the best response of type-1 calls is to join the system with probability $q^{*} = 0$. That is, $q^{*} = 0$ is socially optimal joining strategy.

Case 3: There is a $q^{\prime}$ which satisfies ${\lambda_{1}}^{e} R = C \cdot e^{{\varepsilon E(N_{1} (q^{\prime}))}} \cdot \varepsilon E^{\prime}(N_{1} (q^{\prime}))$. In this case, we know that $S^{\prime}(q) \ge 0$ for $q \in [0,q^{\prime}]$ and $S^{\prime}(q) < 0$ for $q \in (q^{\prime},1]$. The unique maximum is attained at $q^{\prime}$. Thus, the best response of type-1 calls is to join the system with probability $q^{*} = q^{\prime}$. That is, $q^{*} = q^{\prime}$ is a socially optimal joining strategy.

We complete the proof.

Appendix 4

4.1 Proof of Lemma 2

Let

$$Y(q) = e^{{\varepsilon E(T_{1} (q))}} ,\;L(q) = ({\lambda_{1}}^{e} )^{ - 1} e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q)).$$

According to (20) and (22), we obtain that $Y(0) < Y(1)$, $L(0) < L(1)$, $Y(0) > L(0)$ and $Y(1) > L(1)$. Since the relationship of the above inequalities, we only need to compare $Y(0)$ and $L(1)$ in the following two cases.

1.If $Y(0) > L(1)$, we will have the following five subcases.

Case1a: if $R/C < L(0)$, then $q^{*} = q^{\prime} = 0$.

Case1b: if $L(0) \le R/C \le L(1)$, then $q^{*} = q^{\prime} \in [0,1]$, and $q_{e} = 0$. Therefore, the inequality $q^{*} \ge q_{e}$ holds.

Case1c: if $L(1) < R/C < Y(0)$, $q^{*} = 1$, and $q_{e} = 0$. Thus, we obtain $q^{*} > q_{e}$.

Case1d: if $Y(0) \le R/C \le Y(1)$, $q_{e} = q_{ee}$, and $q^{*} = 1$. We get $q^{*} \ge q_{e}$.

Case1f: if $R/C > Y(1)$, $q^{*} = q_{e} = 1$.

2. $Y(0) \le L(1)$, the five subcases are given as follows.

Case2a: if $R/C < L(0)$, then $q^{*} = q_{e} = 0$.

Case2b: if $L(0) \le R/C \le Y(0)$, then $q^{*} = q^{\prime} \in [0,1]$, and $q_{e} = 0$. Therefore, the inequality $q^{*} \ge q_{e}$ holds.

Case2c: if $Y(0) < R/C < L(1)$, $q^{*} = q^{\prime}$, and $q_{e} = q_{ee}$. The above numerical experiments imply this result, as shown in Figs. 12–13. Thus, we can obtain $q^{*} > q_{e}$.

Case2d: if $L(1) \le R/C \le Y(1)$, $q_{e} = q_{ee}$, and $q^{*} = 1$. We get $q^{*} \ge q_{e}$.

Case2f: if $R/C > Y(1)$, $q^{*} = q_{e} = 1$.

Appendix 5

5.1 Proof of Theorem 4

The system administrator can give an incentive-payment $P$ to the type-1 calls that decide to join the system. The reward of type-1 calls that have been served becomes $R + P$. But the socially objective function is the same with (21) since the incentive-payment $P$ is not included in the socially objective function with a transfer of income from one social group. Thus, the resulting socially optimal joining probability remains the same with (22). However, the equilibrium of joining with probability is given by the following theorem.

In the considered wireless cellular network system, an equilibrium strategy “joining with probability $q_{e}$” is given by

$$ q_{e} = \left\{ {\begin{array}{*{20}l} {0,\;\;\;{\text{if }}R + P < C \cdot e^{{\varepsilon E(T_{1} (0))}} ,} \hfill \\ {q_{ee} {,}\;{\text{if }}C \cdot e^{{\varepsilon E(T_{1} (0))}} \le R + P \le C \cdot e^{{\varepsilon E(T_{1} (1))}} ,} \hfill \\ {1,\;\;\;\;{\text{if }}R + P > C \cdot e^{{\varepsilon E(T_{1} (1))}} ,} \hfill \\ \end{array} } \right. $$

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where $q_{ee}$ is the solution to equation $R + P - C \cdot e^{{\varepsilon E(T_{1} (q))}} = 0$.

We will discuss the following three cases:

Case 1: When ${\lambda_{1}}^{e} R < C \cdot e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q))$ for $\forall q \in [0,1]$, the socially optimal joining probability is $q^{*} = 0$. If we let $P = C(e^{{\varepsilon E(T_{1} (0))}} - ({\lambda_{1}}^{e} )^{ - 1} \varepsilon \cdot e^{{\varepsilon E(N_{1} (0))}} \cdot E^{\prime}(N_{1} (0)))$, then $R + P\;{ < }\;{\text{C}} \cdot e^{{\varepsilon E(T_{1} (0))}}$, which is the condition that the equilibrium joining probability is 0. Therefore,

$$P = C(e^{{\varepsilon E(T_{1} (0))}} - ({\lambda_{1}}^{e} )^{ - 1} \varepsilon \cdot e^{{\varepsilon E(N_{1} (0))}} \cdot E^{\prime}(N_{1} (0)))$$

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is the appropriate incentive-payment that makes the equilibrium joining probability coincide with the socially optimal joining probability, which gives the first branch of (24).

Case 2: When $\lambda_{1}^{e} R > C \cdot e^{{\varepsilon E(N_{1} (q))}} \cdot \varepsilon E^{\prime}(N_{1} (q)) \,$ for $\forall q \in [0,1]$, the socially optimal joining probability is $q^{*} = 1$. We need to get

$$P = C(e^{{\varepsilon E(T_{1} (1))}} - ({\lambda_{1}}^{e} )^{ - 1} \cdot e^{{\varepsilon E(N_{1} (1))}} \cdot \varepsilon E^{\prime}(N_{1} (1)))$$

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to make the equilibrium joining probability coincide with the socially optimal joining probability. So it follows the third branch of (24).

Case 3: When $C \cdot e^{{\varepsilon E(N_{1} (0))}} \cdot \varepsilon E^{\prime}(N_{1} (0)) \le {\lambda_{1}}^{e} R \le C \cdot e^{{\varepsilon E(N_{1} (1))}} \cdot \varepsilon E^{\prime}(N_{1} (1))$, the socially optimal joining probability $q^{\prime}$ which satisfies ${\lambda_{1}}^{e} R = C \cdot e^{{\varepsilon E(N_{1} (q^{\prime}))}} \cdot \varepsilon E^{\prime}(N_{1} (q^{\prime}))$. Then, let $q_{ee} = q^{\prime}$, and we have that

$$ P = C \cdot e^{{\varepsilon E(T_{1} (q^{\prime}))}} - R. $$

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Thus, we can obtain that the equilibrium joining probability is the same as socially optimal joining probability. We complete the proof.