Asymmetrical Multi-path Selection Game for Wireless Overlay Networks

Abstract

In order to transfer the increasing big data we need to make use of multiple wireless paths. Overlay networks provide the possibility of taking advantage of multiple available routing paths to realize the bandwidth aggregating. We present a game-theoretic study of the selfish strategic collaboration of multiple heterogeneous overlays when they are allowed to use massively-multipath transfer. Overlays are modeled as players in this multipath selection game model, we discuss the asymmetric case where all overlays have the different round trip times (RTT) and different wastefulness level, and demonstrate the existence and uniqueness of Nash equilibrium (NE). Then we find overlays differing only in their RTTs still receive equal throughput shares and utilities at the NE. However, if overlays differ only in their wastefulness levels, a more wasteful overlay has a larger utility and a larger throughput (bandwidth) share than a less wasteful overlay.

Appendices

Appendix 1

The appendix provides the proof of Theorem 1. First, we transform the objective function into a simpler form: Φ = C/(1 − p). It is easy to see that we need to find the minimal feasible p to minimize Φ.

Note that p is a function of n (see (9)). If we take T _0,i  = 4R _i , as recommended in [22], and let $\overline{\phi } (p) = \mu \sqrt p + 4\nu \left( {p^{3/2} + 32p^{7/2} } \right)$. (9) can be rewritten as

$$ F(p,n) = (1 - p)\left( {\sum\limits_{j = 1}^{m} {n_{i} /R_{j} } - C\overline{\phi } } \right) = 0. $$

(20)

Solving for p given a specific n is actually equivalent to solving the above equation. First note that, $\mathop {\lim }\limits_{p \to 0} F = \sum\nolimits_{j = 1}^{m} {n_{j} /R_{j} } ;\mathop {\lim }\limits_{p \to 1} F = - C \cdot \overline{\phi } (1) < 0$. Furthermore, F(p,n) is a strictly monotonic decreasing function of p since ${{\partial {\text{F}}} \mathord{\left/ {\vphantom {{\partial {\text{F}}} {\partial {\text{p}}}}} \right. \kern-0pt} {\partial {\text{p}}}} = - \sum\nolimits_{j = 1}^{m} {n_{j} /R_{j} - C\overline{\phi } < 0}$.

Thus, there must be a unique solution p in F (p, n) = 0 for any given feasible n. That is, p as a function of n is implicitly defined in F(p, n) = 0. Note that p is an increasing function of n, thus, minimal p _opt (satisfying F (p, n) = 0) is uniquely achieved at n _opt  = (1, 1, …, 1). Then, we see that the system cost expressed in (7) uniquely achieves minimum value at n _opt with

$$ \varPhi_{opt} = C/(1 - p_{opt} ). $$

(21)

Appendix 2

The appendix provides the proof of Theorem 2. Each player i tries to solve for its optimal strategy ${\hbox{n}}_{i}^{*}$, as a response to the strategies of all other players. Thus, if there is an interior point NE n _ne  = $\left( {n_{1}^{*} , \ldots, n_{m}^{*} } \right)$, then it must be true that ∀i, ∂U _i /∂ ${\hbox{n}}_{i}^{*}$ = 0, and $n_{i}^{*} = R_{i} \arg \max_{{n_{i} \in S_{i} }} U_{i} \left( {n_{1}^{*} ,n_{2}^{*} , \ldots ,n_{i}^{*} , \ldots ,n_{m}^{*} } \right)$. If we denote $z_{i} = n_{i} /R_{i}$, utility function (6) is rewrited as:

$$ U_{i} (z_{i} ) = \left[ {\frac{{Cz_{i} }}{{\left( {z_{i} + \sum\nolimits_{k = 1,k \ne i}^{m} {z_{k} } } \right)}}} \right] - \beta z_{i} B_{i} $$

(22)

In the following, we first introduce a fact indicating that the stationary point satisfying ∂U _i /∂z _i  = 0 is actually the maximum point if it is in [1, ∞). Then we show that there is a unique n _ne satisfying ∂U _i /∂ $z_{i}^{*}$ = 0, ∀i.

First, we need to seek all vectors n _ne satisfying a set of m equations

$$ \partial U_{i} /\partial z_{i} = 0,\forall_{i} \in [1, \, 2, \, 3, \ldots ,m]. $$

(23)

We first prove that if n _ne exists, $n_{i}^{*} /n_{j}^{*} = R_{i} /R_{j}$, ∀ _i,j . Then, we show that n _ne is actually unique by proving that there is only one $p^{*}$ such that $n_{i}^{*} /n_{j}^{*} = R_{i} /R_{j}$, ∀i, j.

To simplify calculation formula, first we need to set $\bar{\phi } = \mu \sqrt p + 4\nu (p^{3/2} + 32p^{7/2} )$, thus ${\text{B}}_{\text{i}} = 1/R_{i} \bar{\phi }$, $\phi_{i} = R_{i} \bar{\phi }$ and $\varphi_{i} = R_{i} \overline{\varphi }$. As ${\text{B}}_{\text{i}} = C/R_{i} ( 1- {\text{p}})\sum\nolimits_{i = 1}^{m} {{{n_{i} } \mathord{\left/ {\vphantom {{n_{i} } {R_{i} }}} \right. \kern-0pt} {R_{i} }}}$, we have $C\bar{\phi } = \left( {1 - {\text{p}}} \right)\sum\nolimits_{i = 1}^{m} {{{n_{i} } \mathord{\left/ {\vphantom {{n_{i} } {R_{i} }}} \right. \kern-0pt} {R_{i} }}}$, and utility function (Eq. 22) is further rewrited as:

$$ U_{i} (n_{i} ) = [Cz_{i} /(z_{i} + z_{ - i} )] - \beta_{i} z_{i} /\bar{\phi } = [Cz_{i} /(z_{i} + z_{ - i} )] - C\beta_{i} z_{i} /\left( {1 - p} \right)(z_{i} + z_{ - i} ) $$

For an arbitrary player i, we have:

$$ \frac{{\partial U_{i} }}{{\partial z_{i} }} = \frac{{Cz_{ - i} }}{{(z_{i} + z_{ - i} )^{2} }} - \frac{\beta }{{\bar{\phi }}} + \frac{{\beta n_{i} \frac{{\partial \bar{\phi }}}{\partial p}}}{{\bar{\phi }^{2} }}\frac{\partial p}{{\partial z_{i} }} = \frac{{Cz_{ - i} }}{{(z_{i} + z_{ - i} )^{2} }} - \frac{\beta }{{\bar{\phi }}} - \frac{{\beta z_{i} C\bar{\varphi }}}{{(z_{i} + z_{ - i} )^{2} [(p - 1)\bar{\varphi } - \bar{\phi }]}} $$

(24)

where ${{\bar{\phi } = \phi_{i} /R_{i} = \mu \sqrt p + 4\nu \left( {p^{3/2} + 32p^{7/2} } \right) = 1} \mathord{\left/ {\vphantom {{\bar{\phi } = \phi_{i} /R_{i} = \mu \sqrt p + 4\nu \left( {p^{3/2} + 32p^{7/2} } \right) = 1} {R_{i} B_{i} }}} \right. \kern-0pt} {R_{i} B_{i} }}$

$$ \bar{\varphi } = \varphi_{i} /R_{i} = \frac{\mu R}{2\sqrt p } + T_{0} \nu \left( {\frac{3}{2}\sqrt p + 112p^{5/2} } \right) $$

And $z_{ - i} = \sum\nolimits_{k = 1,k \ne i}^{m} {z_{k} }$ and $\bar{\varphi } = d\bar{\phi }/dp$, $\mu = \sqrt { 2 {\text{b}}/ 3}$, $\nu = 3/ 2\sqrt { 3 {\text{b}}/ 2}$, b = 1 or 2, and T _0,i = 4R _i .

Fact 1: Best response of a player is unique, which is the stationary point if the stationary point is in [1, ∞). First we need to show that for any given z _-i , there is only one unique maximal point for U _i . In fact, player i needs to solve the following equations to get a candidate for a maximal point z ^m _i :

$$ \beta z_{i} - z_{ - i} \left( {1 - p - \beta } \right)[\bar{\varphi }\left( {1 - p} \right)/\bar{\phi } + 1] = 0 $$

(25)

$$ C\bar{\phi } - (z_{i} - z_{ - i} )(1 - p) = 0 $$

(26)

where (Eq. 25) is a simplification of ∂U _i /∂z _i  = 0. We can regard z ^m _i and p as the implicit functions of z _-i . Note that for any given z _-i , there exists a unique pair of (z ^m _i , p) as the solution to (Eq. 25) and (Eq. 26). In fact, the unique stationary point z ^m _i obtained from this implicit function is indeed a maximal point. We can enlarge the domain of U _i to be (0, ∞), and notice that z ^m _i is also a unique stationary point for this enlarged domain. Since U _i (0, z _−i ) = 0 and $\mathop {\lim }\limits_{{z_{i} \to \infty }} U_{i} = - \infty$, the values of two end points are not larger than U _i (z ^m _i , z _−i ) given that z ^m _i is indeed an interior point. Then we can conclude z ^m _i is indeed a maximal point in domain (0, ∞). If z ^m _i is still a stationary interior point in [1, ∞), then it also must be a maximal point. We can show that z ^m _i  = f _i (z _−i ) and p = f _p (z _−i ) are continuous functions on domain z _-i  ∈ (0,∞), in which f _i (z _-i ) is referred to as the best response function. In addition, from implicit function theorem, we know that they are continuously differentiable.

Now, we go on to prove the existence and uniqueness of NE. Consider two arbitrary players i and j, and let $\delta_{i} z_{i} = \sum\nolimits_{k = 1,k \ne i}^{m} {z_{k} }$;$\delta_{j} z_{j} = \sum\nolimits_{k = 1,k \ne j}^{m} {z_{k} }$.

When ∂U _i /∂z _i  = ∂U _j /∂z _j  = 0, we have:

$$ (1 - p)[\delta_{i} + \beta \bar{\varphi }/((1 - p)\bar{\varphi } + \bar{\phi })]/(1 + \delta_{i} ) - \beta = 0 $$

(27)

$$ (1 - p)[\delta_{j} + \beta \bar{\varphi }/((1 - p)\bar{\varphi } + \bar{\phi })]/(1 + \delta_{j} ) - \beta = 0 $$

(28)

Let $\Delta = \beta \bar{\varphi }/((1 - p)\bar{\varphi } + \bar{\phi })$, then combining (Eq. 27) and (Eq. 28) leads to

$$ (\delta_{i} /(1 + \delta_{i} ) - \delta_{j} /(1 + \delta_{j} )) + \Delta (1/(1 + \delta_{i} ) - 1/(1 + \delta_{j} )) = 0 $$

(29)

For (Eq. 29) to be true, we need either ∆ = 1 or δ _i  = δ _j . We can show that ∆ = 1 cannot be true. We prove this by contradiction. Assume that it is true, then we can substitute it into (Eq. 27), and get β = 1−p. Substituting β = 1−p into ∆ = 1, we get φ = 0. We know that φ = 0 is impossible given that p ∈ (0, 1), thus ∆ ≠ 1. Thus, the only possible solution is δ _i  = δ _j , ∀i, j. This implies that $z_{i}^{*} = z_{j}^{*}$, i.e., $n_{i}^{*} /n_{j}^{*} = R_{i} /R_{j}$ at NE n _ne if it exists.

In the following, we prove that, when $n_{i}^{*} /n_{j}^{*} = R_{i} /R_{j}$, there exists a unique solution $p^{*}$ for (Eq. 23). Then we can conclude that there is one unique n _ne .

Since at NE all players have the same number of paths, from (Eq. 24), we obtain:

$$ (m - 1)/\beta - m/(1 - p) + \bar{\varphi }/((1 - p)\bar{\varphi } + \bar{\phi }) = 0 $$

(30)

Let F(p) denote the left-hand-side of (Eq. 30). Ideally, solving equation (Eq. 30) with p as unknown, we can get the loss rate at NE $p^{*}$. Then, substituting $p^{*}$ back into (Eq. 30), we can get n _ne as the number of paths of all players at NE. However, (Eq. 30) contains several powers of p such as 7/2 and 5/2, thus it is impossible to get an algebraic solution of p. Thus, in the following, we examine several properties of F(p), from which we make an inference about the behavior of NE. For an exact value of $p^{*}$ and n _ne when given a network setting, we can use Matlab to numerically solve for them.

First, we prove that (Eq. 30) has only one solution for p in (0, 1). Note that F (p) is a continuous function, and the domain of F (p) is p ∈ (0, 1), and lim _p→0 F (p) > 0 and lim_p→1 F (p) < 0. We claim that F (p) is a strictly monotonic decreasing function. If this claim is true, then there must be a single solution $p^{*}$ for F (p) = 0. In the following, we prove this claim.

Consider the derivative

$$ \frac{\partial F}{\partial p} = \frac{ - m}{{(1 - p)^{2} }} + \frac{{\bar{\varphi }^{\prime}\bar{\phi }}}{{[(1 - p)\bar{\varphi } + \bar{\phi }]^{2} }} < \frac{ - 1}{{(1 - p)^{2} }} + \frac{{\bar{\varphi }^{\prime}\bar{\phi }}}{{[(1 - p)\bar{\varphi } + \bar{\phi }]^{2} }} = \frac{{ - \overline{\phi }^{2} - 2(1 - p)\bar{\varphi }\bar{\phi } - (1 - p)^{2} (\bar{\varphi }^{2} - \bar{\varphi }^{\prime}\bar{\phi })}}{{(1 - p)^{2} [(1 - p)\bar{\varphi } + \bar{\phi }]^{2} }} $$

(31)

Thus, to prove that ∂F/∂p < 0, we only need to prove that $\bar{\varphi }^{ 2}$ > $\bar{\varphi }^{\prime}$ $\bar{\phi }$. This can be easily proved.

Then, substituting $p^{*}$ back into (Eq. 30), we can get $n_{i}^{*}$ as the number of paths of the ith player at NE. Therefore, we conclude that there is only one NE for this game and it is symmetric. That is, $z_{1}^{*} = z_{i}^{*} = \arg \max_{{n_{i} \in S_{i} }} U_{i} (z_{1}^{*} ,z_{2}^{*} , \ldots ,z_{i}^{*} , \ldots ,z_{m}^{*} )$ and $z_{i}^{*} = z_{j}^{*}$, thus $n_{i}^{*} /n_{j}^{*} = R_{i} /R_{j}$, ∀i at NE n _ne if it exists.

Therefore, F (p, m) is a monotonic decreasing function of p, and F ($p^{*}$ , m) = 0. Since F (p ₀ , m) < 0, $p^{*}$ must be smaller than p ₀ . When p < p ₀ , we get dF/dm = 1/β − 1/(1 − p) = 1/(1 − p0) − 1/(1 − p) > 0. Thus, as m increases, F(p,m) is strictly monotonic increasing. Since F(p,m) is a monotonic decreasing function of p, we see that as m increases, for any given F(p), p will be strictly increasing towards p ₀ . Then, it must be also true that for F (p ^*) = 0, as m increases, p approaches to p ₀ .

Recall that at NE, we must have $(1 - p^{*} )/R_{i} \bar{\phi }^{*} = C/mn_{i}^{*}$. Since all players must have at least one path, i.e., $n_{i}^{*}$ ≥ 1, we must make sure that $m(1 - p^{*} )/R_{\text{i}} \bar{\phi }^{*} \le C$. We know that as m increases, $p^{*}$ → p ₀  = 1−β, then $R_{i} \bar{\phi }^{*}$ (as a function of $p^{*}$) also increases to $R_{i} \bar{\phi }_{0}^{*}$ (function of p ₀ ). Thus, $(1 - p^{*} )/R_{i} \bar{\phi }^{*}$ is bounded below by $(1 - {\text{p}}_{0} )/R_{\text{i}} \bar{\phi }_{0}^{*}$. So, as m becomes larger and larger, eventually, $m(1 - p^{*} )/R_{\text{i}} \bar{\phi }^{*}$ will be larger than C, and NE is no longer an interior point. Supposing R ₁ is the minimum RTT, and $m(1 - p^{*} )/R_{ 1} \bar{\phi }^{*}$ is the largest value for all m(1 − $p^{*}$)/$R_{\text{i}} \bar{\phi }^{*}$, ∀i. Let m ₀ denote this threshold, then it is the largest m satisfying $m(1 - p^{*} )/R_{ 1} \bar{\phi }^{*} = m(1 - p^{*} )B_{1}^{*} \le {\text{C}}$. Since it is difficult to obtain an explicit expression of p as a function of m, we rely on the numerical method to identify m ₀ .