Abstract—In this work, we consider a distributed wireless network where many transmitters communicate with a common receiver. Having the choice of their power control policy, transmitters are concerned with energy constraints : instantaneous energy-efficiency and long-term energy consumption. The individual optimization of the average energy-efficient utility over a finite horizon is studied by using control theory and a coupled system of Hamilton-Jacobi-Bellman-Fleming equations is obtained. As the corresponding stochastic differential game is difficult to analyze when the number of transmitters is large (in particular, the Nash equilibrium analysis becomes hard and even impossible), the problem is approximated by a mean-field game. The complete mean-field equilibrium analysis is then conducted (existence, uniqueness, and determination) and the efficiency of the corresponding power control policies is assessed numerically.

I. I NTRODUCTION Power control has always been recognized as an important problem for multiuser communications [1], [2]. With the appearance of new paradigms such as ad hoc networks [3], unlicensed band communications, and cognitive radio [4], [5], designing distributed power control policies has become especially relevant; in such networks, terminals can freely choose their power control policies and do not need to follow orders from central nodes. More recently, the need for building green communication networks has appeared to be stronger and stronger [6]. The goal is to manage energy consumption both at the mobile terminals and network infrastructure sides [7]. The work reported in this paper precisely falls into this framework that is, the design of green distributed power control policies in multiuser networks. More precisely, we consider a network which comprises many transmitters and one receiver. Each transmitter chooses its power control policy in order to maximize its average energy-efficiency (measured in bit per Joule). Note that a similar framework was analyzed in [8]. In [8], the problem is modeled by a static game: for each transmission block, the transmitters choose their power levels strategically but independently from past blocks. In [9], it has been shown that much more efficient control policies can be obtained by exploiting long-term interactions, which is done by using the models of repeated games. But, as the channel gains associated with the different communication links vary over time, there The authors would like to thank Hamidou Tembine from École Supérieure d’Électricité, Supélec for his many constructive comments ([email protected]).

is generally a loss of optimality when using repeated games. Indeed, in [10], [11], it is shown that the model of stochastic games is more appropriate and can lead to better policies. The problem is that, even though some special ad hoc policies can be found [11], stochastic games are not fully characterized in general; in particular, Folk theorems are only available in special cases. Additionally, the problem of characterizing the performance of distributed networks modeled by stochastic games becomes very hard and even impossible when the number of players becomes large. The same statement holds for determining individually control strategies. This is where mean-field games come into play. Mean-field games [12] represent a way of approximating a stochastic differential (or difference) game, by a much more tractable model. Typically, instead of depending of the actions and states of all the players, the mean-field utility of a player only depends on its own action and state, and depends on the others through a meanfield. It seems that the only work where mean-field games has been used for power control is [13]. Compared to the latter reference, the present work is characterized by a different utility function (no linear quadratic control assumptions is made here), the fact that the battery level of a transmitter is considered as part of a terminal state, and the existence and uniqueness analysis for of mean-field equilibria is conducted. At last, numerical results are provided. II. P ROBLEM STATEMENT In a wireless network with a set of transmitters N = {1, . . . , n} and one receiver, the received signal is y(t) =

n X

hi (t)ai (t) + z(t)

(1)

i=1

with hi (t) ∈ C the channel coefficient between transmitter i and the receiver at time t, ai (t) the symbols sequence sent by transmitter i during time-slot t and z(t) is a Gaussian noise with variance σ 2 . We denote pi (t) = |ai (t)|2 the transmitting power of transmitter i during time-slot t. We consider that for each transmitter, the dynamics of their channel coefficient is a Wiener process, meaning that ∀i ∈ N dhi (t) = ηdWi (t)

(2)

where ∀i ∈ N , Wi (t) are mutually independent Wiener processes of dimension 2 and η is the variance of hi (η < +∞).

As we consider that transmitters are concerned about their transmission rate but also about the energy they spend to reach this rate, we use energy-efficient utility as introduced by Mandayam and Goodman [8]. For each transmitter, the instantaneous utility is ui (p(t), h(t)) =

Rf (SIN Ri (p(t), h(t))) pi (t)

(3)

where R in bits/s is the transmitting rate of transmitter i (as we do not use this rate as a control in our work, we consider that it is constant for all the transmitters), p(t) = (p1 (t), . . . , pn (t)) and h(t) = (h1 (t), . . . , hn (t)). f is an efficiency function which represents the transmission success rate at the receiver. It takes its values in [0, 1] and depends on the SIN R of transmitter i. More details about f can be found in [8]. The SIN R writes pi (t)|hi (t)|2 , (4) SIN Ri (p(t), h(t)) = Pn 2 2 j6=i pj (t)|hj (t)| + σ with |hi (t)|2 the channel gain between transmitter i and the receiver. Here we consider that the manufacturer of the mobile transmitter wants its device to be able to transmit during a finite horizon which represents the desired battery life. We denote E0 the initial available energy for each transmitter. Note that the dynamics of Ei (t), the energy left for transmitter i at time t, is directly linked to the transmitting power by the following equation dEi (t) = −pi (t)dt.

(5)

0=

sup pi (T →T 0 )

n X ui (X(t), p(t)) − pj (t)∂Ej vj (t, X(t)) j=1

η2 2 +∂t vi (t, X(t)) + ∂hh vi (t, X(t)). 2

(9) With a similar reasoning as in [15], we can show that finding optimal power profiles p∗ (T → T 0 ) = (p∗1 (T → T 0 ), . . . , p∗n (T → T 0 ))

(10)

such that ∀i ∈ N , p∗i (T → T 0 ) ∈ n X pj (t)∂Ej vj (t, X(t)) arg max ui (X(t), p(t)) − pi (u→T )

j

(11) requires to solve ∀i ∈ N , ∀t ∈ [T, T 0 ] f 0 (γi (t))γi (t)−f (γi (t))=γi (t)2

∂E vi (t,X(t) i R

P 2 ∗ σ2 + n j6=i |hj | (t)pj (t) |hi |2 (t)

2

(12) with γi (t) = SIN Ri (t). Note that we consider that ∂Ei Vi (t, X(t)) ≥ 0, otherwise the optimal power p∗i (t) → ∞. In other words, we consider that the more energy we have in our battery, the better the payoff can be. III. N UMERICAL RESOLUTION A. Existence of a solution Here we propose to solve this equation with − a e γi (t) if γi (t) > 0, f (γi (t)) = 0 if γi (t) = 0.

(13) Each transmitter wants to maximize its utility during a finite 0 horizon T → T while taking into account the dynamics of an efficiency function introduced by Belmega et al. in [16]. the system, then the problem can be written Z T 0 Equation (12) can be rewritten vi (T ) = sup E ui (p(t), h(t))dt + q(E(T 0 ), h(T 0 )) a − γ a(t) i 0 e − 1 = pi (T →T ) T γi (t) (14) 2 Pn dhi (t) = ηdWi (t) σ + j6=i |hj |2 (t)p∗j (t) 2 2 ∂Ei vi (t, X(t) γi (t) . dEi (t) = −pi (t)dt R |hi |2 (t) (6) where vi (T ) is the Bellman function, the expectation of the or 2 Pn 2 continuous sum of utility for the optimal control path, E(t) = σ + j6=i |hj |2 (t)p∗j (t) 2 − y1 1 2 a ∂Ei vi 0 0 ( − 1) = y , e (E1 (t), . . . , En (t)), and q(E(T ), h(T )) is the instantaneous y R |hi |2 (t) utility value for the final state. (15) We denote X(t) = (E(t), h(t))T the state of the system at with y = γia(t) . Note that in (15) the right term being positive, time t. Then a solution may only exist if y ≤ 1. Thus if a solution γi∗ (t) exists, dX(t) = (−p(t)dt, ηdW(t))T , (7) γi∗ (t) ≤ a. (16) with W(t) a 2n-dimension Wiener process. (6) can be rewritThe existence of a non-zero solution depends on the term ten 2 Pn 2 Z T 0 2 ∗ a2 ∂Ei vi (t, X(t)) σ + j6=i |hj | (t)pj (t) 0 α= . (17) vi (T, X) = sup E ui (X(t), p(t))dt + q(X(T )) R |hi |2 (t) pi (T →T 0 ) T dX(t) = (−p(t)dt, ηWt )T (8) For each transmitter, the Hamilton-Jacobi-BellmanFleming [14] (HJBF) equation can be written

Heuristically, we can see that there exists a threshold αmax such that if α < αmax , there exists a global maximizer different from 0 and if α ≥ αmax , 0 is the global maximizer as illustrated on figure 1.

1 0.9

•

0.8 0.7

γmax i

0.6

IV. A PPROXIMATING THE STOCHASTIC DIFFERENTIAL POWER CONTROL GAME BY A MEAN - FIELD GAME

0.5 0.4

The previous system of equations is hard to solve because for each transmitter, the associated partial differential equation depends on all the channel coefficients and all the power profiles of the other transmitters. Of course, if the number of transmitters in the system increases greatly, it is true that the complexity of the resolution increases as well. But an interesting fact is that if the number of transmitters is large enough, then for one single transmitter point of view, it becomes equivalent to consider all the other transmitters as a continuum. Therefore only the distribution of the other transmitters states is needed for one transmitter to take into account the other transmitters, which highly simplifies the formulation of the problem. This is what is developed in the present section. Naturally, in a MAC network, the interaction between the different transmitters is expressed in the interference term. If we consider that CDMA is used in the network, then the interference seen by the receiver is

0.3 0.2 0.1 0

0

0.1

Figure 1.

0.2

0.3

0.4

0.5

0.6

α

0.7

0.8

0.9

1

Instantaneous global maxmizer depending on α.

0.5 0.45

Left term Right term

0.4

Left and Right terms

maximisers but by definition of αmax , y2 is the global maximizer. If α ≥ αmax , 0 is the global maximizer, by definition of αmax .

0.35 0.3 0.25 0.2 0.15 0.1

n

0.05 0

Ii (t) = 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 X pj (t)|hj (t)|2 . n−1

(21)

j6=i

y

Figure 2. Left and right terms of equation (20) depending on y for α = 0.5.

We consider homogeneous control in own state feedback form ∀i ∈ N pi (t) = α(t, si (t)), (22)

B. Implicit function theorem

with si (t) = (hi (t), Ei (t)). Then n

Consider

1 nX pj (t)|hj (t)|2 n−1n j6=i X n n 1 pj (t)|hj (t)|2 − α(t, si (t))|hi (t)|2 = n − 1 n j=1 Z n = |h|2 α(t, s)Mtn (ds) n−1 α(t, si (t))|hi (t)|2 − n−1 n 1 = Ant − Bn. n−1 n−1 t (23)

Ii (t) = g : [0, αmax [×]0, 1] → R 1 1 (α, y) → e− y ( − 1) − αy 2 . y

(18)

g is C ∞ , then if g(α0 , y0 ) = 0, there exists ϕ : R → R such that y0 = ϕ(α0 ). ϕ is C ∞ and ∂ϕ (y0 ) = ∂α

y02 1

e− y0

1 y03

−

2 y02

(19)

− 2α0 y0

(The fact that this expression is defined remains to prove.) with

C. Discussion about Uniqueness •

If α = 0, {0, 1} are the only solutions of 1 1 e− y ( − 1) = αy 2 , y

•

n

Mtn = (20)

and y = 1 is the global maximizer. For every α ∈ [0, αmax [, there exist three solutions {0, y1 , y2 |0 < y1 < y2 } for equation (20) as illustrated on figure 2 with α = 0.5. In this case, 0 and y2 are local

1X δs (t) . n j=1 j

(24)

If the number of transmitters becomes very large (n → ∞), we can consider that we have a continuum of transmitters. The convergence of the interference term when n → ∞ needs to be proven. Using admissible control, E[Btn ] < ∞, then Btn = 0. n→∞ n − 1 lim

(25)

n = 1, to prove Ii (t) converges weakly, it As limn→∞ n−1 suffices to prove Ant converge weakly. A sufficient condition is the weak convergence of the process Mtn . As stated by Kotelenez and Kurtz in [17], if the (sj (t)) are exchangeable, then there exists a distribution mt such that

mt = lim Mtn .

We expand the first term. Take the difference between the two HJBF equations and multiply by m ˆ 2 (s) − m ˆ 1 (s).

Then, we set Z

|h|2 α(t, s)mt (ds)

(27)

The SINR now writes 2 \Ri (si (t), mt ) = pi (t)|hi (t)| SIN σ 2 + Iˆi (t, mt )

(28)

(37)

˜ ˜ ∂t vˆ2 − ∂t vˆ1 =H(s(t), ∂E vˆ1 , m1 ) − H(s(t), ∂E vˆ2 , m2 ) 2 2 η 2 η 2 vˆ1 − ∂hh vˆ2 + ∂hh 2 2 (38) Z

We denote

s

\Ri (si (t), mt )) Ri f (SIN u ˆi (t) = =: rˆ(si (t), pi (t), mt ) pi (t) (29) We formulate the mean-field response problem Z T0 vˆT = sup E q(s(T 0 )) + rˆ(s(t), p(t), m∗t )dt (30) p(T →T 0 )

where

η2 2 ∂ vˆ2 = 0 2 hh

which gives Iˆi (t, mt ) =

m∗t

(36)

˜ ∂t vˆ2 + H(s(t), ∂E vˆ2 , m2 ) +

(26)

n→∞

η2 2 ˜ ∂t vˆ1 + H(s(t), ∂E vˆ1 , m1 ) + ∂hh vˆ1 = 0 2

T

is the mean-field optimal trajectory and 0 η 0 ds(t) = dt + dWt . −p(t) 0 0

(31)

We can write −∂ v ˆ = sup r ˆ (s(t), p(t), m ) − p(t)∂ v ˆ + t t t E t p∈PE vˆT = q(s(T )) ˜ ∂t mt + ∂E (mt ∂u H(s(t), ∂E vˆt , mt )) =

η2 2 2 ∂hh mt

η2 2 ˆt 2 ∂hh v

(32)

η2 2 ˜ m1 = 0 (40) ∂t m ˆ 1 + ∂E (m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 )) − ∂hh 2 ˜ ∂t m ˆ 2 + ∂E (m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 )) −

(42)

Z

with (34)

p

To prove the uniqueness of the solution, we suppose that there exists two solutions (ˆ v1 , m ˆ 1 ), (ˆ v2 , m ˆ 2 ) of the above system. We want to find a sufficient condition under which the quantity R (ˆ v (s) − v ˆ (s))( m ˆ (s) − m ˆ (s))ds is monotone in time, 2 1 2 1 s which is not possible. Compute the time derivative Z d (ˆ v2 (s) − vˆ1 (s))(m ˆ 2 (s) − m ˆ 1 (s))ds St = dt s Z (35) = (∂t vˆ2 (s) − ∂t vˆ1 (s))(m ˆ 2 (s) − m ˆ 1 (s))ds sZ + (ˆ v2 (s) − vˆ1 (s))(∂t m ˆ 2 (s) − ∂t m ˆ 1 (s))ds s

η2 2 η2 2 m2 − ∂hh m1 + ∂hh 2 2

(33)

˜ u, m) = sup{ˆ H(s, r(s, p, m) + hp, −ui}

η2 2 ∂ m2 = 0 (41) 2 hh

˜ ∂t m ˆ 2 − ∂t m ˆ 1 = − ∂E (m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 )) ˜ + ∂E (m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 ))

η2 2 2 ∂hh mt

(39)

We expand the second term. Take the difference between the two FPK equations and multiply by vˆ2 − vˆ1 .

A. Uniqueness of mean field solution We recall the mean field response problem 2 2 ˜ ∂E vˆt , mt ) + η2 ∂hh vˆt −∂t vˆt = H(s(t), vˆT = q(s(T )) ∂ m + ∂ (m ∂ H(s(t), ˜ ∂E vˆt , mt )) = t t E t u

(∂t vˆ2 (s) − ∂t vˆ1 (s))(m ˆ 2 (s) − m ˆ 1 (s))ds = Z ˜ H(s(t), ∂E vˆ1 , m1 )(m ˆ2 −m ˆ 1 )ds sZ ˜ − H(s(t), ∂E vˆ2 , m2 )(m ˆ2 −m ˆ 1 )ds s Z 2 η 2 ∂hh vˆ1 (m ˆ2 −m ˆ 1 )ds + s 2 Z 2 η 2 − ∂hh vˆ2 (m ˆ2 −m ˆ 1 )ds s 2

(∂t m ˆ 2 − ∂t m ˆ 1 )(ˆ v2 − vˆ1 )ds = Z ˜ − ∂E (m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 ))(ˆ v2 − vˆ1 )ds Zs ˜ + ∂E (m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 ))(ˆ v2 − vˆ1 )ds s Z 2 η 2 + ∂hh m2 (ˆ v2 − vˆ1 )ds s 2 Z 2 η 2 − ∂hh m1 (ˆ v2 − vˆ1 )ds s 2 (43) By integration by parts, Z Z ∂E kφ = − k∂φ (44) s

s

s

AAA This remains to explain! ZZZ

Thus Z (∂t m ˆ 2 − ∂t m ˆ 1 )(ˆ v2 − vˆ1 )ds = s Z ˜ (m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 ))(∂E vˆ2 − ∂E vˆ1 )ds sZ ˜ − (m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 ))(∂E vˆ2 − ∂E vˆ1 )ds s Z 2 η 2 ∂hh m2 (ˆ v2 − vˆ1 )ds + s 2 Z 2 η 2 − ∂hh m1 (ˆ v2 − vˆ1 )ds s 2 (45) The full derivative writes Z ˜ St = H(s(t), ∂E vˆ1 , m1 )(m ˆ2 −m ˆ 1 )ds Zs ˜ − H(s(t), ∂E vˆ2 , m2 )(m ˆ2 −m ˆ 1 )ds s Z ˜ − m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 )(∂E vˆ2 − ∂E vˆ1 )ds Zs (46) ˜ + m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 )(∂E vˆ2 − ∂E vˆ1 )ds s Z η2 2 2 (−∂hh m1 + ∂hh m2 )(ˆ v2 − vˆ1 )ds + 2 s Z η2 2 + (∂ 2 vˆ1 − ∂hh vˆ2 )(m ˆ2 −m ˆ 1 )ds 2 s hh

Introduce an auxiliary integral Z ˜ Cλ = H(s(t), ∂E vˆ1 , m1 )(m ˆλ −m ˆ 1 )ds Zs ˜ − H(s(t), ∂E vˆλ , mλ )(m ˆλ −m ˆ 1 )ds s Z ˜ − m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 )∂E vˆλ − ∂E vˆ1 )ds Zs ˜ + m ˆ λ ∂u H(s(t), ∂E vˆλ , mλ )(∂E vˆλ − ∂E vˆ1 )ds s

Cλ = λ

Z Zs

− Zs − Zs +

˜ H(s(t), ∂E vˆ1 , m1 )(m ˆ2 −m ˆ 1 )ds ˜ H(s(t), ∂E vˆλ , mλ )(m ˆ2 −m ˆ 1 )ds (52) ˜ m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 )(∂E vˆ2 − ∂E vˆ1 )ds ˜ m ˆ λ ∂u H(s(t), ∂E vˆλ , mλ )(∂E vˆ2 − ∂E vˆ1 )ds

s

Cλ =0 λ→0 λ lim

s

Z

2 2 (∂hh m1 − ∂hh m2 )(ˆ v2 − vˆ1 )ds = s Z η2 2 (∂ 2 vˆ1 − ∂hh vˆ2 )(m ˆ2 −m ˆ 1 )ds. 2 s hh

(47)

Then Z St = Zs − Zs − Zs +

˜ H(s(t), ∂E vˆ1 , m1 )(m ˆ2 −m ˆ 1 )ds ˜ H(s(t), ∂E vˆ2 , m2 )(m ˆ2 −m ˆ 1 )ds

(53)

Z d Cλ ˜ = − ∂λ H(s(t), ∂E vˆλ , mλ ) (m ˆ2 −m ˆ 1 )ds dλ λ Z s ˜ + ∂λ (m ˆ λ ∂u H(s(t), ∂E vˆλ , mλ ))(∂E vˆ2 − ∂E vˆ1 )ds (54)

For the same reason as the previous integration by part η2 2

(51)

d Cλ = dλ λ Z ˜ − ∂u H(s(t), ∂E vˆλ , mλ )(∂E vˆ2 − ∂E vˆ1 )(m ˆ2 −m ˆ 1 )ds s Z ˜ − ∂m H(s(t), ∂E vˆλ , mλ )(m ˆ2 −m ˆ 1 )2 ds Zs ˜ + (∂u H(s(t), ∂E vˆλ , mλ ))(m ˆ2 −m ˆ 1 )(∂E vˆ2 − ∂E vˆ1 )ds s Z 2 ˜ + m ˆ λ ∂uλ H(s(t), ∂E vˆλ , mλ )(∂E vˆ2 − ∂E vˆ1 )ds s

(55)

(48) However

˜ m ˆ 1 ∂u H(s(t), ∂E vˆ1 , m1 )(∂E vˆ2 − ∂E vˆ1 )ds

2 ˜ ∂uλ H(s(t),∂E vˆλ , mλ ) = 2 ˜ ∂uu H(s(t), ∂E vˆλ , mλ )(∂E vˆ2 − ∂E vˆ1 )

˜ m ˆ 2 ∂u H(s(t), ∂E vˆ2 , m2 )(∂E vˆ2 − ∂E vˆ1 )ds

s

2 ˜ + ∂um H(s(t), ∂E vˆλ , mλ )(m ˆ2 −m ˆ 1)

We introduce

(56)

=0

m ˆ λ = (1 − λ)m ˆ 1 + λm ˆ2 = m ˆ 1 + λ(m ˆ2 −m ˆ 1 ).

(49)

The same way vˆλ = (1 − λ)ˆ v1 + λˆ v2

(50)

Then Z d Cλ ˜ = − ∂m H(s(t), ∂E vˆλ , mλ )(m ˆ2 −m ˆ 1 )2 ds, dλ λ s (57) which can be proven to be greater than 0.

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