On the collaborations of multiple selfish overlays using multi-path resources

Abstract

Overlay networks provide the possibility of taking advantage of multiple available paths to realize the bandwidth aggregating through multipath transfer. We present a game-theoretic study of the selfish strategic collaboration of multiple overlays when they are allowed to use multipath transfer, which is referred as the multipath selection game. Then we consider the equilibrium in this multipath selection game model where selfish players distribute their overlay traffic. Maximization of the utility functions for each overlay is the criterion of optimality. We adopt the objective of throughput maximization to capture the most typical overlay behaviors, and use the usual TCP as the basis of our analysis. We show analytically the existence and uniqueness of Nash equilibria in these games. Furthermore, we find that the loss of efficiency of Nash equilibria can be arbitrarily large if overlays do not have resource limitations.

Appendix A

The appendix provides the proof of Theorem 2. Each player i tries to solve for its optimal strategy n ^* _i , as a response to the strategies of all other players. Thus, if there is an interior point NE n _ne  = (n ^*₁ , …,n ^* _m ), then it must be true that $\forall i,\partial {U}_i/\partial {n_i}^{*}=0,{{\mathrm{and}\ \mathrm{n}}_{\mathrm{i}}}^{*}= \arg { \max}_{n_i\in {S}_i}{U}_i\left({n}_1^{*},{n}_2^{*},\dots, {n}_i^{*},\dots, {n}_m^{*}\right)$.

In the following, we first introduce Theorem 3 indicating that the stationary point satisfying ∂U _i /∂n _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 /∂n _i * = 0, ∀ i.

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

$$ \partial {U}_i/\partial {n}_i=0,\forall i\in \left[1,2,3\dots, m\right]. $$

(15)

We first prove that if n _ne exists, n _i * = n _j *, ∀ i, j. Then, we show that such n _ne is actually unique by proving that there is only one p _ne for which n _i * = n _j *, ∀ i, j.

For an arbitrary player i, we have

$$ \frac{\partial {U}_i}{\partial {n}_i}=\frac{C{n}_{-i}}{{\left({n}_i+{n}_{-i}\right)}^2}-\frac{\beta }{\phi }-\frac{\beta {n}_i C\varphi}{{\left({n}_i+{n}_{-i}\right)}^2\left[\left(p-1\right)\varphi -\phi \right]} $$

(16)

where

$$ \phi =1/B=\mu R\sqrt{p}+{T}_0\nu \left({p}^{3/2}+32{p}^{7/2}\right) $$

(17)

$$ \varphi =\frac{\mu R}{2\sqrt{p}}+{T}_0\nu \left(\frac{3}{2}\sqrt{p}+112{p}^{5/2}\right) $$

(18)

and n_−i = ∑  ^m _{k = 1,k ≠ i} n _k and φ = dϕ/dp.

Theorem 3

Best response of a player is unique and it is the stationary point if the stationary point is in [1, ∞).

First we need to show that for any given n _−i , there is only one unique maximal point for U _i . In fact, player i need to solve the following equations to get a candidate for a maximal point n ^m _i :

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

(19)

$$ C-\left({n}_i+{n}_{-i}\right)\left(1-\mathrm{p}\right)=0 $$

(20)

where (19) is a simplification of ∂U_i/∂n_i = 0. We can think of n ^m _i and p are implicit functions of n _-i . We note that for any given n _-i , there is a unique pair of (n ^m _i ,p) as the solution to (19) and (20). We can check that the unique stationary point n _i ^m obtained from this implicit function is indeed a maximal point. We can enlarge the domain of U_i to be (0, ∞), and notice that n_i ^m is also a unique stationary point for this enlarged domain. Since U_i (0, n_−i) = 0 and lim_ni→∞U_i = −∞, they are not larger than U_i(n ^m _i ,n_− i) given that n_i is indeed an interior point. Then we can conclude nm is indeed a maximal point in domain (0, ∞). If it is still a stationary interior point in [1, ∞), then it also must be a maximal point. We can show that n ^m _i  = f _i (n _− i ) and p = f_p(n _−i ) are continuous functions on domain n _-i ∈[1,∞). 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 δ _i n _i  = ∑  ^m _{k = 1,k ≠ i} n _k ; δ _j n _j  = ∑  ^m _{k = 1,k ≠ j} n _k . When ∂U _i /∂n _i  = ∂U _j /∂n _j =0, we get

$$ \left(1-\mathrm{p}\right)\left[{\updelta}_{\mathrm{i}}+\upbeta \upvarphi /\left(\left(1-\mathrm{p}\right)\upvarphi +\phi \right)\right]/\left(1+{\updelta}_{\mathrm{i}}\right)-\upbeta =0 $$

(21)

$$ \left(1-\mathrm{p}\right)\left[{\updelta}_{\mathrm{j}}+\upbeta \upvarphi /\left(\left(1-\mathrm{p}\right)\upvarphi +\phi \right)\right]/\left(1+{\updelta}_{\mathrm{j}}\right)-\upbeta =0 $$

(22)

Let Δ = βφ/((1 − p) φ + ϕ), then combining (21) and (22) leads to

$$ \left({\updelta}_{\mathrm{i}}/\left(1+{\updelta}_{\mathrm{i}}\right)-{\updelta}_{\mathrm{j}}/\left(1+{\updelta}_{\mathrm{j}}\right)\right)+\varDelta \left(1/\left(1+{\updelta}_{\mathrm{i}}\right)-1/\left(1+{\updelta}_{\mathrm{j}}\right)\right)=0 $$

(23)

For (23) to be true, we need either ∆ = 1 or δi = δj . We can show that ∆ = 1 cannot be true. We prove this by contradiction. Assume that it is true, then we can substitute it into (21), 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 = 2β/(1 − p − β), ∀ i, j. This implies that n ^* _i  = n ^* _j at NE n _ne if it exists.

In the following, we will prove that, when n ^* _i  = n ^* _j , there exists a unique solution p _ne for (15). Then we can conclude that there is one unique n _ne .

Since at NE all players have the same number of paths, from (16), we obtain

$$ \left(\mathrm{m}-1\right)/\upbeta -\mathrm{m}/\left(1-\mathrm{p}\right)+\upvarphi /\left(\left(1-\mathrm{p}\right)\upvarphi +\phi \right)=0 $$

(24)

Let F (p) denotes the right-hand-side of (24). Ideally, solving Eq. (24) with p as unknown, we can get loss rate at NE p*. Then, substituting p* back into (9), we can get n* as the number of paths of all overlays at NE. However, (24) contains several powers of p such as 7/2 and 5/2, which makes it impossible to get an algebraic solution of p. Thus, in the following, we examine several properties of F (p), and based on these properties make an inference about the behavior of NE. For an exact value of p* and n* when given a network setting, we can use Matlab to numerically solve for them.

First, we will prove that (24) has only one solution for p in (0, 1). We 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

$$ \frac{\partial F}{\partial p}=\frac{-m}{{\left(1-p\right)}^2}+\frac{\varphi \prime \phi }{{\left[\left(1-p\right)\varphi +\phi \right]}^2}<\frac{-1}{{\left(1-p\right)}^2}+\frac{\varphi \prime \phi }{{\left[\left(1-p\right)\varphi +\phi \right]}^2}=\frac{-{\phi}^2-2\left(1-p\right)\varphi \phi -{\left(1-p\right)}^2\left({\varphi}^2-\varphi \hbox{'}\phi \right)}{{\left(1-p\right)}^2{\left[\left(1-p\right)\varphi +\phi \right]}^2} $$

(25)

Thus, to prove that ∂F/∂p < 0, we only need to prove that φ ² > φ ′ ϕ. This can be easily proved.

Then, substituting p _ne back into (9), we can get n _ne as the number of paths of all overlays at NE. Therefore, we conclude that there is only one NE for this game and it is symmetric. That is, ${\mathrm{n}}_i^{*}= \arg { \max}_{n_i\in {S}_i}{U}_i\left({n}_1^{*},{n}_2^{*},\dots, {n}_i^{*},\dots, {n}_m^{*}\right)$ and n ^* _i  = n _ne , ∀ i.