4.1 The non-cooperative game
We consider the simple form of the interaction among mobile users, where they only can choose the amount of throughput to keep (i.e., x
i
) and thereafter the amount of throughput to inject in the system (i.e., 1−x
i
). Power control is assumed to be run by the base station. Each mobile user will then face the problem of deciding the appropriate amount of throughput to allocate to others. We propose a non-cooperative game where each user (selfishly) decides the amount of throughput to yield to others in order to satisfy its throughput requirement d
i
.
We denote the VMIMO game on its normal form as \(\mathcal {G}=[\Omega,\mathcal {A},\mathbf {\Theta }, \mathbf {D}]\), where Ω={1,..,m} is the set of players (mobile users), \(\mathcal {A}=[0,1]^{m}\) is the space of strategies set for all users, and Θ=[Θ
1,Θ
2,⋯,Θ
m
] is the vector of payoff functions. In our setting, the average throughput will be considered as the utility function. D=(d
1,d
2,⋯,d
m
) denotes the throughput demand vector which defines the set of objectives (QoS) to meet by mobile users.
Each mobile user is looking to experience an average throughput greater than or equal to its throughput demand required by its running services; otherwise, the service would not discard. Then, the interaction among mobile users can be clearly explained as follows:
-
Step 1 : Mobile user i receives an average throughput \({T_{i}^{C}}\) from the cellular networks. This received throughput may be greater than, less than, or equal to its required throughput demand d
i
.
-
Step 2 : Mobile user i decides the amount of throughput x
i
to keep for itself. Then 1−x
i
represents the amount of throughput to re-inject into the system through WLAN interface. We refer to this as a virtual throughput since it will be shared according to some sharing rule by the other mobile users.
-
Step 3 : Each mobile user will fine-tune unilaterally the amount of throughput to keep (alternatively to re-inject into the WLAN) until its average total throughput is equal to its throughput demand.
4.2 Feasible strategy and satisfactory solution
When one imposes constraints over the payoff functions that each mobile user obtains or over the action that a player can choose in the game \(\mathcal {G}\), it becomes plausible to replace the Nash equilibrium concept by a constrained (Nash) equilibrium. Here, in the presence of QoS constraints, the set of individual actions reduces to the set of actions which verifies the constraints given the actions adopted by the other mobile users.
Definition
1.
Let X
−i
=(x
1,x
2,…,x
i−1,x
i+1,…,x
m
)be the strategy vector of all users but the ith user. a strategy \(x_{i}^{*}\) of mobile user i is said to be a feasible strategy iff
$$ \mathbf \Theta_{i}(x_{i}^{*},\mathbf X_{-i}) \geq d_{i}. $$
((7))
From now on, each mobile user will seek to achieve its throughput demand by fine-tuning the amount of throughput to re-inject in the WLAN subsystem.
Definition
2.
A strategy profile \(\mathbf X^{*}=(x_{1}^{*},x_{2}^{*},\ldots,x_{m}^{*})\) is a satisfaction Nash equilibrium point (SNEP) iff
-
(a)
All individual throughput demand are met. Namely,
$$ \mathbf \Theta_{i}(\mathbf X^{*}) \geq d_{i}, \qquad\forall i, $$
((8))
and
-
(b)
X
∗ is a feasible strategy profile.
In the remainder, we will be interested in the case of strict equality. This is not a restriction but a plausible point that has sense if the required demand is equal to the rate at which packets are generated. Moreover, energy efficiency considerations can also be a good reason to only address the case of strict equality.
4.3 Existence and uniqueness of a SNEP
We turn now to prove the existence/uniqueness of the SNEP point and to characterize it for the one-shot version of \(\mathcal {G}\). When a SNEP (if exists) \(\mathbf {X}^{*}=\left (x_{1}^{*},x_{2}^{*},\ldots,x_{m}^{*}\right)\) is achieved, each mobile user should met its own throughput demand, see Definition 2. Namely, we have
$$ x_{i}^{*} {T_{i}^{C}}+\sum_{j\neq i} \left(1-x_{j}^{*}\right) {T_{j}^{W}}\pi_{j,i}(\mathbf{D})=d_{i}, \qquad\forall i, $$
((9))
Let X=(x
1,x
2,⋯,x
m
) be the column vector of the unknowns. Thus, we can write the non-cooperative VMIMO problem as the following non-homogeneous linear system:
$$ \mathbf{A\cdot X=\underline{D}}, $$
((10))
where A is a square matrix of size #
m, and which is given explicitly by the following
$${}{\begin{aligned} \mathbf A= \left(\begin{array}{cccc} {T_{1}^{C}} & -{T_{2}^{W}}\pi_{1,2}(\mathbf{D}) & \ldots & -{T_{m}^{W}}\pi_{1,m}(\mathbf{D})\\ -{T_{1}^{W}} \pi_{2,1}(\mathbf{D}) & {T_{2}^{C}} & \ldots & -{T_{m}^{W}}\pi_{2,m}(\mathbf{D})\\ \vdots & \vdots & \vdots & \vdots\\ -{T_{1}^{W}} \pi_{m,1}(\mathbf{D}) & -{T_{2}^{W}}\pi_{m,2}(\mathbf{D}) & \ldots & {T_{m}^{C}}\\ \end{array} \right) \end{aligned}} $$
and
$$\mathbf{\underline{D}}=\left(\begin{array}{c} d_{1}-\sum\limits_{j\neq 1}{T_{j}^{W}}\pi_{1}(\mathbf{D})\\ d_{2}-\sum\limits_{j\neq 2}{T_{j}^{W}}\pi_{2}(\mathbf{D})\\ \vdots \\ d_{K}-\sum\limits_{j\neq K}{T_{j}^{W}}\pi_{m}(\mathbf{D}) \end{array} \right) $$
Lemma
1.
If A is a Cramer invertible matrix and \(\mathbf {A}^{-1}\cdot \underline {\mathbf {D}} \in [\!0, 1]^{m}\). Then, the game \(\mathcal {G}\) has a unique pure satisfaction equilibrium given by \(\mathbf {X}^{*}=\mathbf {A}^{-1}\underline {\mathbf {D}}\).
Proof.
It is straightforward that the column and the row vectors of the matrix A are non-collinear, therefore it is a Cramer matrix with non-zero determinant. Henceforth, the system has a unique solution given by \(\mathbf {X}^{*}=\mathbf {A}^{-1}\underline {\mathbf {D}}\).
4.4 A sufficient condition for existence of a symmetric SNEP
Now, we seek to derive a sufficient condition for existence of a satisfaction equilibrium. For that, we imagine an equivalent network with symmetric setting, with same radio conditions and with same parameters for all mobile users. This symmetric network has the same number of users and will only be used to derive an upper bound on the total demand such that an satisfaction equilibrium still exists. We set for the equivalent network R
i
=R
j
=R, L
i
=L
j
=L, d
i
=d
j
=d, ρ
i
=ρ
j
=ρ, f(γ
i
)=f(γ
j
)=f(γ), and π
i,j
(·)=π
j,i
(·)=π(·)∀i,j. Here, one consider the worst case where all mobile users ask for a throughput demand \(d=\max \limits _{i} d_{i}\). Note that the total demand generated by the symmetric network is m·d.
The idea behind derivation of our sufficient condition is the following simple reasoning: A unique satisfaction equilibrium will exist if and only if the total throughput demand \(\sum \limits _{i=1}^{i=m}d_{i}\) does not exceed the achievable throughput (total capacity) of the network. We have the following result:
Proposition
1.
Existence of a satisfaction equilibrium is guaranteed if the total throughput demand \(\sum \limits _{i=1}^{i=m}d_{i}\) verifies
$$ \sum_{i=1}^{i=m}d_{i} \le md \le \frac{mLR}{M}f(\gamma)\cdot\mathbf{min}\bigg(\left(m-1\right)\rho\pi(d),1\bigg). $$
((11))
Proof.
Let \(\mathbf \Xi \triangleq \sum \limits _{i=1}^{i=m}\mathbf \Theta _{i}\) be the total average throughput experienced by mobile users altogether. From Eq. (6), we have:
$$ {}{\begin{aligned} \mathbf{\Xi}=\sum_{i=1}^{i=m}\left[x_{i}\frac{L_{i}}{M_{i}}R_{i}f(\gamma_{i})+\sum_{j\neq i} (1-x_{i}) \rho_{j}\frac{L_{j}}{M_{j}}R_{j}f(\gamma_{j})\pi_{j,i}(\mathbf{D})\right] \end{aligned}} $$
((12))
After some algebra, the total experienced throughput for the symmetric equivalent network writes:
$$ \begin{aligned} \mathbf{\Xi}_{\text{sym}}&=\frac{mLR}{M}f(\gamma)\left[x\left[1-(m-1)\rho\pi(d)\right]\right.\\ & \quad \left.+(m-1)\rho\pi(d)\right] \end{aligned} $$
((13))
It is straightforward from Eq. (13) that the total throughput is an affine function on the amount of throughput x to be kept by a given user. Thus, it suffices to analyze the sign of the term (1−(m−1)ρ
π(d)). We have two cases:
-
Case 1 : (1−(m−1)ρπ(d))≥0
-
In this case Ξ
sym
reaches its maximum when x is equal to 1, i.e., all the users will satisfy their throughput demand from the base station. Here, no throughput re-injection is required to achieve the satisfaction equilibrium. Replacing x by 1 in (13) yields
$$ \Xi_{\text{sym}}^{1}=\frac{mLR}{M}f(\gamma). $$
((14))
-
Case 2 : (1−(m−1)ρπ(d))≤0
-
Now the total throughput Ξ
sym is maximized when x is set to 0. All mobile users will satisfy their whole throughput demand from the WLAN subsystem, i.e., after re-injecting their whole throughput received first from the cellular subsystem. We have the following
$$ \Xi_{\text{sym}}^{0}=\frac{m(m-1)LR}{M}\rho f(\gamma)\pi(q). $$
((15))
Combining (14) and (15), and in order to ensure the existence of a satisfaction equilibrium, the total demand must verify:
$$ md \leq \mathbf{min}\left(\Xi_{\text{sym}}^{0}, \, \Xi_{\text{sym}}^{1}\right), $$
((16))
which completes the proof.
