Power level management in wireless and DSL networks with transmission cost Текст научной статьи по специальности «Математика»

Eitan Altman1, Konstantin Avrachenkov1 and Andrey Garnaev2

1INRIA, France 2 St. Petersburg State University, Russia

Abstract. We study power level management in optimization and game frameworks with assumption that there is a transmission cost.

In the optimization framework there is a single decision-maker who assigns network resources and in the game framework players share the network resources according to Nash equilibrium. We study conditions for uniqueness of the Nash equilibrium. Besides we provide a closed form solution to the problems, which allows us to solve it in a finite number of operations, and we also consider a jamming plot of the game.

Keywords: Nash equilibrium, resource allocation, non-linear programming. Introduction

In wireless networks and DSL access networks the total available power for signal transmission has to be distributed among several resources. In the context of wireless networks the resources may correspond to frequency bands (e.g. as in OFDM), or they may correspond to capacity available at different time slots. In the context of DSL access networks the resources correspond to available frequency tones. This spectrum of problems can be considered in either optimization scenario or as a result of a non-cooperativie game scenario. The optimization scenario leads to “Water Filling Optimization Problem” [Heinzelman, 2000], [Goldsmith, 1997], [Tse, 2005] and the game scenario leads to “Water Filling Game” or “Gaussian Interference Game” [Lai, 2005], [Popescu, 2003], [Rose, 2004], [Yu, 2002]. In the optimization scenario, one needs to maximize a concave function (Shannon capacity) subject to power constraints. The Lagrange multiplier corresponding to the power constraint is determined by a non-linear equation. In the previous works [Heinzelman, 2000], [Goldsmith, 1997], [Tse, 2005] it was suggested to find the Lagrange multiplier by means of a bisection algorithm, where comes the name “Water Filling Problem”. Here

1 This work is partly supported by joint RFBR and NNSF Grant no.06-01-39005.

we show that the Lagrange multiplier and, hence, the optimal solution of the water filling problem can be found in explicit form with a finite number of operations. In the multiuser context one can view the problem in either cooperative or non-cooperative setting. If a centralized controller wants to maximize the sum of all users’ rates, the controller will face a non-convex optimization problem. On the other hand, in the non-cooperative setting, the power allocation problem becomes a game problem where each user perceives the signals of the other users as interference and maximizes a concave function of the noise to interference ratio. In [Lai, 2005], [Weissing, 1991] the spectrum of available resources was continuous, here as in [Popescu, 2003], [Song, 2002] we consider the discrete spectrum of available resources. A natural approach in the non-cooperative setting is the application of the Iterative Water Filling Algorithm (IWFA) [Yu, 2002]. Recently, the authors of [Luo, 2006] proved the convergence of IWFA under fairly general conditions. In the present work we analyze the case of symmetric water filling game with two users and transmission cost. We would like to note that even the symmetric scenario allows us to obtain several important conclusions about the Gaussian Interference Game. Our main result is an explicit form for the Nash equilibrium. In addition, to its mathematical beauty, the explicit solution allows one to find the Nash equilibrium in water filling game in a finite number of operations and to study limiting cases when the crosstalk coefficient is either small or large. As a by-product, we obtain an alternative simple proof of the convergence of the Iterative Water Filling Algorithm. Furthermore, it turns out that the convergence of IWFA slows down when the crosstalk coefficient is large. Using the closed form solution, we can avoid this problem. Finally, we compare the non-cooperative approach with the cooperative approach and conclude that the cost of anarchy is small in the case of small crosstalk coefficients. Applications that can mostly benefit from decentralized non-cooperative power control are ad-hoc and sensor networks with no predefined base stations [Goldsmith, 1997], [Lin, 1997], [Kwon, 1999]. An interested reader can find more references on non-cooperative power control in [Cover, 1991], [Lai, 2005]. We would like to mention that the water filling problem and jamming games with transmission costs have been analyzed in [Altman, 2007].

The structure of the paper is as follows: in the next Section 1, we recall the single decision-maker setup of the water filling optimization problem. Then, in Section 2 we provide its explicit solution. In Sections 3 and 4 we formulate two users symmetric water filling game and characterize its Nash equilibrium. In Section 5 we give an alternative simple proof of the convergence of the iterative water filling algorithm. In Section 6 we give the explicit form of the players’ strategy in the Nash equilibrium. In Section 7 we confirm our finding with the help of numerical example. In that section we also show that the cost of anarchy is small when the crosstalk coefficient is small. In Section 8 we consider a jamming plot of the game. We make conclusions in Section 9.

1. Single decision-maker

We consider the following power allocation problem in the case of a single decisionmaker. There is a single decision-maker (also called “Transmitter”) who wants to send information using n independent resources so as to maximize the Shannon capacity.

The strategy of Transmitter is T = (T,..., Tn), such that Ti > 0 for i G [1, n] and n=i Ti < T, where ni > 0 for i G [1, n] and T > 0. As the payoff to Transmitter we take the Shannon capacity minus expenses connected with transmission:

n / t \ n

V(T) = YJ^(i + ^)-cYJTl,

i= 1 i ' i=1

where N0 > 0 is the noise level in the sub-carrier i, and C > 0 is the cost of transmission of a unit signal. It is worth noting that the case C = 0 was investigated in [Avrachenkov, 2007]. Following the standard water-filling approach [Heinzelman, 2000], [Goldsmith, 1997], [Tse, 2005], which assumes application Kuhn-Tacker Theorem, we have the following result.

Theorem 1. Let 1/N0 = maxie[1n] 1/N0, and

T(w) = [1/(C + w) — №]+ for i G [1,n], and

H (w) = En=i Ti(w).

(i) If C > 1/N0, then T* = (0,..., 0) is the unique equilibrium with zero payoff.

(ii) If C < 1/N0, then T (w*) = (Ti(w* ),...,Tn(w*)) is the unique equilibrium with

zero payoff v(T(w*)), where

( 0 for H(0) < T,

w* = < the unique root

[ of the equation H(w) = T for H(0) > T.

Proof.

First, note that 9 < 0.

Thus, v is concave on Ti and the Kuhn-Tucker Theorem implies that T* is the optimal strategy if, and only if, there is a non-negative w, such that

1 rf=LV iolTi>°>

dTi N° + T* } <uj for T* = 0,

where

J

= 0 otherwise.

„ > , T; = T, (10)

So,

T(w)=[l/(C + w) - N0 + for * e [1,n],

which gives a form the optimal strategy has to have, and this form depends on a parameter w.

Now we have to find the optimal value of w. First, note that the function H(w) has the following properties:

(i) H(w) is non-negative and continuous in (0, to),

(ii) H(w) is positive and strictly decreasing in (0, [1/W0 — C] +),

(iii) H(w) = 0 for w e ([1/W0 — C] +, to).

Thus, the equation H(w) = T has a root, and this root is unique if, and only if, H(0) > T. This fact together with (21) imply the result.

It is interesting that the result of the previous theorem can be proved straight forward without application the Kuhn-Tacker Theorem. To complete the picture we also give this proof.

Let T* = (T*,. ..,Tn) be the optimal strategy. Let T* = (0,...,0). Then, there is a m e [1,n], such that T.. > 0. Let e be any enough small positive number and

k = m. Let T e'k = (Tl'k,..., Tl'k) be such that

It is clear that Te,k also is a strategy for any enough small positive e. Then, since T* is the optimal strategy, we have that v(T*) > v(Te,k). Thus,

T*m — e for * = m,

T£ + e for * = k,

T* for * e {m, k}.

ln

>

In ^1 H ^ - - e) + In ^1 +

So, putting e ^ 0 we have that

1

1

for any m = k.

T*m + N°m ~ T*k+N°k

Thus, there is a non-negative w, such that

for T * > 0, for T* = 0.

Then, further we just have to repeat word by word the previous proof.

2. Closed form solution for water filling problem

In the previous studies of the water filling problems it was suggested to use numerical (e.g., bisection) method to solve the equation (1). Here we propose an explicit form approach for the solution of equation (1).

Without loss of generality we can assume that the sub-carriers are arranged by the noise level as follows:

1 1 1 1

> pr > . . . > -r- >

N°~ N§----~ N°~ N°+1’

where N0+1 = to.

Theorem 2. The solution of the water filling optimization problem is given by: l

N°

(ii) if

(i) */ 4o < c, then T* = (0,..., 0),

k

T + E k= 1 N0

for an integer k, such that

№k+1- N°’ * )o,

(iii) otherwise

T + £(N0 - N°)

k

0

v=<——, ,„<k, d3)

0, if i > k,

where k can be found from the following conditions: vk < T < vk+1,

t

Vt = Y<(N0 - N0) for t e [1,n]. (14)

=1

Proof.

Consider separately two cases: (a) u* =0 and (b) u* > 0.

(a) Let u* =0. Consider separately two subcases:

(1) If C > 1/N0 then, by Theorem 1 (i), T* = (0,...,0) and (a) follows.

(2) If C < 1/N0 then, by Theorem 1 (ii), u* = 0 is the optimal one if

— - № c

< T. (15)

+

1

Let k be such that

Wi~c<W

Then, the optimal strategy T* given by (4) and (6) implies that i=1 ' ' i=1

So, (2) and (ii) follow.

(b) Let uj* > 0 and k G [1, n] be such that jL- > uj* + C > jJr—. Then,

k k+1

1 -N?

u* + C

\ ^ + c ~ Ni forie[l,A:],

I 0 for i € [k +1, n].

So

k ,

I -

u* + C

i=1

Since H(u*) = T, we have that

k

-C. (16)

k

0

i=1

Because of strictly decreasing H in (0,1/N°) we can find k from the following conditions:

H(1/N° - C) <T < H(1/N0+1 - C).

Since

k k+1

]T(N°+1 - N°) = ]T(N°+1 - N°) i=1 i=1

the switching sub-carrier k can be found from the following equivalent conditions:

t

¥k <T < yk+1, where <pt = ^2(№ - N°) for t € [1,n]. (17)

i=1

So, Theorem (1), (7) and (8) imply Theorem (2).

Let us demonstrate the closed form approach by a numerical example. Take n = 5, T =1, C = 0, N° = Ki-1, k = 1.7, ni = 1/5 for i € [1, 5]. Then, as the first step we calculate t for t € [1, 5]. In our case we get (0, 0.14, 0.616, 1.8298, 4.58108). Then, by (5), k = 3. Thus, by (4), the optimal water filling strategy is T* = (2.53,1.83,0.64,0, 0) with payoff 0.438.

+

3. Symmetric water filling game

In this section we consider game-theoretical formulation of the situation where a few users, namely, two users (Transmitters), try to send information through n resources so as to maximize the quality of the transmitted information. The strategy of Transmitter j is Tj = (Tj,. ..,T) with Tj > 0 and with

n

Y^Tj < Tj, (18)

=1

where Tj > 0 for j = 1, 2. The payoffs to Transmitters are given as follows:

n / rT11 \ n

AT\T-> = ±m(1+ Jf N?)-c,±Tl

=1 =1

where gi,ai,pi, i € [1,n],C1,C2 are some positive. These payoffs correspond to Shannon capacities. This is an instance of the Water-Filling or Gaussian Interference Game corresponding a particular case of OFDM wireless network and DSL access network.

In this work we restrict ourselves to the case of symmetric game with equal crosstalk coefficients, namely, ai = f3i, gi = g, N° = N°/ai for i € [1,n]. Then,

n / rp1 \ n

‘'1(T'T2, = S1"(1 + i5Fw)-c'ST-

n / rp2 \ n

,P(T\r>) = ^In (l + -jTT^) -C2Y.Tl

i=1 i i i=1

This situation can correspond, for example, to the scenario when the transmitters are situated at about the same distance from the base station.

We will assume that C1 > 0 and C2 > 0. The case C1 = C2 = 0 was investigated in [Avrachenkov, 2007].

We shall characterize a Nash Equilibrium of this problem. The strategies (T1*, T2*) constitute a Nash Equilibrium, if for any strategies (T1, T2) the following inequalities hold:

v1(T1, T2*) < v1 (T1*, T2*), v2(T1*, T2) < v2(T1*, T2*).

The goal of this paper is to find Nash Equilibrium. To perform this purpose, first, note that

d2v1(TSt2) _ 1

dT12 (N° + gT2 + T/)2

and

d2v2(T 2,T2)

dT22 (N0 + gTl + T2)2

< 0.

So, v1 and v2 are concave on T1 and T2 respectively, the Kuhn-Tucker Theorem implies the following result, describing the equilibrium structure.

Theorem 3. (T 1*,T2*) is a Nash equilibrium if, and only if, there are non-negative u1 and u2 (Lagrange multipliers) such that

1 -C'H fr^>0' (W)

gT22* + T0* + N0 < „2 for T0* = 0,

where

gT2* + T2* + N0 l< „2 for T2* =0,

„ > 0 for £”=2 t2* = T2, (21)

2 [=0 otherwise,

„2 /> 0 for KU ^ = T2 (22)

0 otherwise. Let us introduce the following sets:

1

1

r0 0(u\U2) = \ie [1, n] : ^1 | C1 < N0, -J—^<N0

/io(^2, w2) = {* € [l,n] : N° < and either N° > | ^

OT ^ < lo^ + C2 and lo^+C2 ~Ni ~ 9{-uj1+C1 ~ A

-foi^2^2) = {* € [l,n] : N0 < 2X 2 and either N0 > ^

I „2 + C 2 „2 + C2

1 1 1

I[2(„2,„2) = e [1,n] : 0 <

The next result characterizes the forms that the Nash equilibrium can take.

Lemma 1. Let (T2*,T2*) be a Nash equilibrium, then

(i) if T2* = 0 and T2* = 0 then i e I00(„2, „2),

(ii) ifT0* > 0 and Tf* = 0 then i G /(0(w1, lv2) and T0* = ui^_ci — N0,

(iii) ifT0* = 0 and T2* > 0 then i G /^(w1, lv2) and T2* = u2^C2 ~ N°,

(iv) if T1* > 0 and T2* > 0 then i G I[i(w1,^2) and

(-r+T^r - N?) -

T1*_ Vwi + Ci

1 _ £

T2* = (_z±çL~ N'] ~ g{^cT ~ N?)

1 - g2

Proof.

(i) follows directly from (10) where T0* = T2* = 0.

(ii) Let T1* > 0 and T2* = 0. Then, by (10), we have that

1 1 i ni

T1* + N0

w1 + C1.

Thus, 1-r > N° and Tl* = 1-r - N°.

w1 + C1 * * w1 + C1 *

Then, by (10), we have that

u,2 + C2> 1

9Tt + Ni g{^__No)+No

w + C

Thus,

g{ 1 1 1 - AT?) > 9 1 9 - AT?,

yVw1 + C’1 *'“w2+C2 ”

and the result follows.

(iii) can be proved similarly to (ii).

(iv) Let Tl* > 0 and T2* > 0. Then, by (10), we have that

^ > N° and „ ^ „ > N°.

(23)

1 uj2 + C2

Also, by (10), we have that T1* and T2* are given by (14). Then, since T1* > 0 and T2* > 0 we have that i G I'11(w1, w2). This completes the proof of the lemma.

Based on Lemma 1 and Theorem we straightforward obtain the following three theorems dealing with the situations where the transmission cost is too high either for both users or at least for one of them.

1

Theorem 4.

min{C’1, C2} > ^0

then there is unique equilibrium (T1*, T2*) and T1+ = T2* = (0,..., 0). Theorem 5. If

C2<h

then there is unique equilibrium (T 1*,T2*). Besides, T1* = (0,..., 0) and T2* T2*(w*) = (T?*(w*),...,T%*(w*)) where

Tf(w) = [1/(C2 + w) - N?]+ for i e [1,n]

and

"0 for H2(0) < T2,

the unique root

of the equation H2(w) = T2 for H2(0) > T2.

with

H 2(w)=^ T2(w).

Theorem 6. If

Cl<

then there is unique equilibrium (T1*, T2*). Besides, T2* = (0,..., 0) and T1* = T 1*(w*) = (T11*(w*),...,Tn1*(w*)) where

and

T1(u) = [1/(C1 + w) - N°] + for i e [1,n]

"0 for H 1(0) < T1,

the unique root

of the equation H 1(w) = T1 for H 1(0) > T1.

with H 1(w) = 7=1 T/(w). So, now we can restrict our attention to the case

max{Ci, C2} < ^0.

Although the game has symmetric nature there are some non-symmetric features impacted by the fact that the Lagrange multiplies are different as a rule. This difference will allow us to simplify the structure of the sets I' and the strategies. For this purpose, first, introduce the following auxiliary notations for positive w1 and w2:

(i) if w1 + C1 < w2 + C2, so 1/(w2 + C2) < 1/(w1 + C1) then let

I00 (w ,w ) = \ i e [1,n] :

1

< N

I10(w\w2) = |i e [1,n] :

I01 (w1,w2)

1 - g 1

I11 (w ,w ) = < i e [1,n] : N <

1-g

(ii) if w2 + C2 < w1 + C1, so 1/(w1 + C1) < 1/(w2 + C2) then let

I00(w1,w2) = <j i e [1,n] : I10 (w1,w2) = 9,

+ C2

N0

I01 (w1,w2) = {i e [1,n] :

. uj1 + C1 uj'z + C

1 - g

N 0 <

+ C*2 J’

(iii) if w2 + C2 = w1 + C1 then let

I00(w , w ) = {i e [1,n] :

+ C2

I10(w1,w2) = 9, I01 (w1, w2) = 9,

= {% G [l,n] : N° <

< n0},

}■

+ C2

The next lemma asserts that the sets I' coincide with the sets I.

Lemma 2. The following relations between sets I' and I hold for any non-negative w1 and w2.

(i) I00(w1,w2)= I00(w1,w ),

(ii) 110(w1,w2) = I10(w1,w2),

(iii) I01(w1,w2)= I01 (w1,w2),

(iv) I11(w1,w2)= Iu(w1,w2).

1

w

1

1

2

w

2

w

1

1

2

w

1

Proof.

Let, for example,

w1 + C1 < w2 + C2. (24)

(i) is obvious.

(ii) Let i e Thus, either ^2|c2 < N° < ul|cl or

1 g

w2 + C2 w1 ^ atO • I 1 1

< N < min

1 -g ..... lw1+C1’w2 + C2

Then, by (15), we obtain

1 1

+ C1 ’ LU2 + C2 J w2 + C2 and

w

+ C2

<

1 - g w2 + C2 '

Thus, I1 o(w1,w2) = I10(w1,w2).

(iii) Let i e I' 1 (w1,w2). Thus, either

1 < N° < 1

w

1 + C1 - * w2 + C2

+ C‘ ^ + C2 Ko < min

1 - g - ' { w1 + C 1 ’ w2 + C2

Then, by (15), we obtain

w

+ C1 cj2+ C2

>

1 - g w2 + C2 '

So, I' 1 (w1,w2) = 0 = I01 (w1, w2).

(iv) Let i e I' 1 (w1,w2). So,

ATo . • f^ + C1 w2+C2 tu2 + C2 w2+C2 uj1 + C1

TV, < mim 1 1 _ 1 '

1 - g 1 - g 1 - g

Thus, I11 (w1,w2) = 111 (w1,w2). This completes the proof of Lemma 2.

1

g

1

or

1

g

1

1

1

w

1

g

1

1

1

1

g

g

g

Now we introduce some strategies, which the Nash Equilibrium will have the form of. Namely, for non-negative j1 and j2, such that j1 + C1 < j2 + C2, and for i G [1,n] we introduce the following notations:

1

TTff

/ 1 \

g

i

g

UJL +CL uz + crz _No I if No < + C72 w1 + C1

1-3 * I * 1 - g

1______________g

if + C2 ^ + c1 < N0

O 1 — g I

/ 1 \

To>>2) —

uj1 +C1

-N0

TW2) —

1

TTff

CO2+c2_______cui + C1 _N0 I if N0 cu2 + C2 cu^C1

g 1 •» I 1 _

if

<72

1 - 3

_______

.1 I /'"'I

1 - 9

±C_ <

either in the following equivalent form as follows

if Ato < t2

if t2 < N0,

112 TO (w , W

^((l + ^ii-^-JVO) ifAT°<i t1- N0

0

if t2 < N0 < t1, if t1 < N0,

where

It is clear that

2 I /^2 1 I /'-»I

^.2 ________ Ce? H- O UJ ~\~ O

1 — g

t1 =

w1 + C

1

1

= t1

1

— gt1 + (1 — g)t2

(25)

(26)

w1 + C1 ’ w2 + C2

For the case w1 + C1 > w2 + C2, T1(w1,w2) and T2(w1,w2) can be defined by

symmetry.

The next result simplifies the form of the Nash equilibrium, given by Lemma 1,

and it shows that the strategies are not so symmetric as it could be expected and

their non-symmetric structure is motivated by difference in Lagrange multipliers and in the power of the signals the players have to transfer.

1

g

1

Theorem 7. Each Nash equilibrium is of the form (T 1(w1, w2),T2(w1,w2)) for some non-negative w1 and w2.

The next result shows that there is a monotonous dependence between the power of the signals the players have to transfer and Lagrange multipliers.

Corollary 1. Let (T 1(w1,w2),T2(w1,w2)) be a Nash equilibrium and w1w2 > 0. If

if1 > T2 (27)

then

w1 + C1 < w2 + C2. (28)

Proof.

Assume that (19) does not hold. Taen w1 + C1 > w2 + C2, and so

i g < i g

-- -------- V ----------- -

Thus,

Hence,

A + C1 w2 + C2 - w2 + C2 w1 + C1 ' Ti1(w1,w2) < T2(w1,w2) for i e [1,n].

if1 =£T1(w1,w2) < ^T2(w1,w2) = T2.

i=1 i=1

This contradiction to (18) completes the proof of Corollary 1.

To find the equilibrium strategies we have to find w1 and w2 such that the following conditions hold:

H 1(w1 ,w2)= T1, H2(w1,w2) = T2 for w1w2 > 0, (29)

or

H 1(w1, 0) = T1, H2(w1, 0) < T2 for w1 > 0, w2 =0, (30)

or

H 1(0, w2) < T1, H2(0, w2) < ;T2 for w1 =0, w2 > 0, (31)

or

H 1(0, 0) < f1, H2(0, 0) < T2 for w1 = w2 =0, (32)

where

n n

H 1(w1,w2) = ËT1(w1,w2) and H2(w1,w2) = ^T2(w1,w2).

=1 =1

It is clear that H 1(w1,w2) and H2(w1,w2) have the following properties, collected

in the next Lemma, which follow directly from the explicit formulas of the Nash

Equilibrium.

Lemma 3.

(i) H 1(w1,w2) and H2(w1,w2) are nonnegative and continuous,

(ii) H 1( w1,w2) is non-increasing on w1 and H2(w1 ,w2) is non-increasing on w2,

(iii) H 1(w1,w2) = 0 for enough big w1 and H3(w1,w2) = 0 for enough big w3, say, for w1 > 1/N0 and w2 > 1/N0 respectively,

(iv) H 1(w1,w2) is non-decreasing by w2 and H2(w1,w2) is non-decreasing by w1. Without loss of generality we can assume that

if1 > T2. (33)

Introduce the following notations:

H2{t2) = ]T n{t2-N°),

g {i:t2>N0}

H1(t\t2)= E TTiit1 - N°)+ E ((1+^t1 - gt2 - N°)

{i:t2<N0<t1} {i:N0<t2} g

where t2 <t1.

Lemma 4. Let (24) holds. Then the system of non-linear equations (20) has unique positive solution (w^,w^) if, and only if,

H2(1/C1) > T2 and H2(1/C 1,t*2) > T1

where t2 is the unique positive root of the equation:

H 2(t2 ) = T2. (34)

Proof.

Let (w1,w2) be the positive solution of (20). Then, by Corollary 1, w1 + C1 <

w2 + C2. Thus, instead of the system of equation (20) with variables

w1

and w2 we

can consider the following equivalent system of equation (26) with variables t1 and t2 where 0 < t2 < t1 < 1/C1:

H2(t2 )= T2, H 1(t1,t2)= T1. (35)

Consider the first equation of (26). It is clear that the function H2 has the following

properties:

(i) H2(-) is continuous in (0,1/C1),

(ii) Hi2(t) = 0 for t < N0,

(iii) H2(-) is strictly increasing in (N0,1/C1).

Thus, if H2 (1/C1) > T2 then there is the unique root t2 of (25).

Now we pass on to considering the second equation of (26). It is obvious that the function H1 has the following properties:

(i) H 1(-,t'2) is continuous and increasing in (t^ 1/C1),

(ii) H\tl ,t2) = H2(t2 ) = t2 < T1.

So, if H 1(1/C1,t2) > T2 there is the unique positive t\ such that

H 1(t1,t^) = T1. (36)

So, the system (26) has the unique solution t ,£1). Thus, (20) also has the unique solution and it can be found by (17). This completes the proof of Lemma 4.

Lemma 1 and proof of Lemma 4 imply the following main results describing the Nash equilibrium.

Theorem 8. Let (24) holds and

H2(1/C1) > T2 and H 1(1/C\tl) > T1.

The symmetric water filling game has the unique Nash equilibrium (T 1(w^, w^), T 2(w^, w^)) for g e (0, 1), where wl, w\ can be found through t\ and t2 from (16) which are the unique solution of the triangular system of equations (26).

Theorem 9. Let (24) holds and

H2(1/C1) > T2 and H 1(1/C 1,t2) < T1.

The symmetric water filling game has the unique Nash equilibrium (T 1(0,w2), T2(0, w^)) for g e (0,1), where w\ can be found through t\ = 1/C1 and t2 from (16).

Theorem 10. Let (24) holds and H2(1/C1) < T2. The symmetric water filling game

has the unique Nash equilibrium (T 1(0, 0), T2(0, 0)).

4. Ununiqueness of Nash Equilibrium

The assumption that g < 1 is essential for the uniqueness of Nash Equilibrium as it is shown in the following Proposition.

Proposition 1. For g =1 and and C1 = C2 = 0 the symmetric water filling game has a continuum of Nash equilibria.

Proof.

Suppose that (T 1*,T2*) be a Nash equilibrium. Then, similarly to Lemma 1, we have to consider three cases (i)-(iii) where at least one of the components of the vector (T 1+,T2*) is positive.

Power Level Management in Wireless and DSL Networks 31

(i) Let T/* > 0 and T,f* = 0. Then, by (10), we have that ^ ^0 = cj1. Thus,

\>N° and T1* = \-N°.

1 ^ ^ *i J. i 1

w1 w1

Then, by (10),

w2 > n---------------------- = w1.

Ti + Ni--------------------------№ + №

1 - i i w1

(ii) Let T2* > 0 and T1* = 0. Then, similarly to (ii), we have that

T2* = \- N° and \ > N°, iv1 > iv2.

i w2 i w2 i

(iii) Let T1* > 0 and T2* > 0. Then, by (10), we have that

1

T1* + t2* + N°

w1 = w2.

Assume that w1 > w2 then (i) does not hold, so T1* = 0 for each i which contradicts to (9). Similarly, the case w1 < w2 cannot hold.

Thus, w1 = w2 = w. So, T1* and T2*, i e [1, n] have to be any non-negative such that

n n

T1* + t2* = [1/w - N0]+ and ET1* = T 1^T2* = T2,

i=1 i=1

where w is the unique positive root of the equation

n

Y}1/w - n0]+ = t 1 + t 2.

i=1

It is clear that there is a continuum of such strategies. For example, if (T 1*,T2*) is the one of them, and let T1*, Tj2* > 0 and T^*, T* > 0 for some k and m. Then, it

is clear that the following strategies for any enough small positive e are also optimal:

T1* for i = k,m,

T1* = { T1* + e for i = k,

T1* - enk/nm for i = m,

T2* for i = k,m,

T2* = I T2* - e for i = k,

T2* + enk/nm for i = m.

This completes the proof of Proposition 1.

5. Convergence of an IWFA

In this section we describe a version of the water filling algorithm for finding the Nash Equilibrium for C1 = C2 =0 and supply a simple proof of its convergence based on some monotonicity properties. These properties give a simple proof of the convergence of the following water filling algorithm for finding the Nash Equilibrium:

Step 1. Let wQ and w2 be such that H ^w^w2) = H2(wq,w2) = 0, for example wQ = w0 = 1/Nq.

Step 2. Let wQ = wQ and define wQ such that H q(wq, wq) = Tq. Such wQ exists by Lemma 3 (i)-(iii).

Step 3. Then, by Lemma 3 (i), (v), H2(w Q,w2) = 0. Let wQ = wq and define wQ such that H2(wq, wQ) = T2.

Step 4. Then, by Lemma 3 (v), H 1 (wQ,wQ) < Tq, and so on.

So, we have non-increasing positive sequence (w Q,wQ). Thus, it converges to an (wq,wq) which produces a Nash Equilibrium.

6. Closed form solution for symmetric water filling game

In this section, based on the proof of Lemma 4, we propose the solution of the two players symmetric water filling game in the closed form for C 1 = C2 =0.

Without lost of generality we can assume that Tq > T2. Let k2 be such that N02+q > tQ > N0 . Then, since H2(tQ) = T2, we have that

(1 + g)T2 ^ti N0

(37)

k2

Since Hi2(-) is strictly increasing, k2 can be found from the condition

H2(N02) <T2 < H2(N02+1).

Hence, k2 can be found from the following equivalent conditions:

2 <T2 < ft2 + 1; (38)

where

1 k

Lp\ = ^(N0 - N0) for k < n and ^2+1 = 00.

+ g i=i

Since t1 is the root of the equation H 1(-,t'2) = T1, there is k1 > k2 such that N0i + i > t1 >N0!. So,

2

(i) if k1 > k2 then

k1 1 k 2

t1+ E N° + rr E(^+Ar°)

tl = --------------------p-^------------------, (39)

k1

(ii) if k1 = k2 then

1 k2

fl + YT~o^{gtl + N^ tl =----------------------------• (40)

Thus, k1 > k2 can be found as follows:

(i) if T1 < y>k2 + 1 then k1 = k2,

(ii) if T1 > ^k2+ 1 then k1 is given by the condition:

vki <T1 < vki+l, (41)

where

k 1 k2

fk= E (^k~^i)+ ^ ~ N° - gtl) for k G n]

i=k2 + 1 g i=1

and

^n+1 = ^.

We can summarize the obtained results in the following theorem.

Theorem 11. Let T1 >T2. Then, the Nash equilibrium strategies are given by

Ti* = s t1 - no if i e [k2 + 1,k1],

[0 if i e [k1 + 1,n], (42)

T2* = ilTgi1* ~ N°) */*€M2],

i [0 if i e [k2 + 1, n],

where k2, t"2, k1 and t1 are given by (29), (28), (32) and (30).

7. Numerical example

Let us demonstrate the closed form approach by a numerical example. Take n = 5, g = 0.9, T1 = 5, T2 = 0.5, N0 = Ki-1, k = 1.7, ni = 1/5 for i e [1, 5] and C1 = C2 =0. Then, as the first step we calculate ^ for t e [1, 5].

Fig. 1: Convergence of IWFA

In our case we get (0, 0.074, 0.324, 0.963, 2.411). Then, by (29), k2 = 3. Thus, by (28), t2 = 3.447. Then we calculate ^1 for t e [4, 5]. In our case we get (1.380, 4.131). So, by (32), k1 = 5. Using (30), we find t1 = 9.221. Thus, by (33) we have the following equilibrium strategies T1* = (7.062, 6.694, 6.067, 4.308, 0.869) and T2* = (1.288,0.919,0.293,0,0) with payoffs 0.909 and 0.062. We have run IWFA, which produced the same values for the optimal strategies and payoffs. However, we have observed that the convergence of IWFA is slow when g « 1. In Figure 1 we have plotted the total error in strategies HT1 - T 1*||2 + \ \T2 - T2*||2, where Tk are the strategies produced by IWFA on the k-th iteration and Ti* are the Nash equilibrium strategies. Our approach instantaneously finds the Nash equilibrium for all values of g. Also, it is interesting to note that by (33) the quantity of channels as well as the channels themselves used by weaker player (with smaller resources) is independent on behavior of the stronger player (with bigger resources) but of course each player allocating his/her resources among these channels take into account the opponent behaviour.

In Figure 2, we compare the non-cooperative approach with the cooperative approach. Specifically, we compare the transmission rates and their sum under Nash equilibrium strategies and under strategies obtained from the centralized optimization of the sum of transmitters’ rates. The main conclusions are: the cost of anarchy

'♦♦♦♦»fl**?*

*******_____

X X X Xxxxxxxxx A■ $8888880

0.8 •

0.6 -

0.4 -

0.2 ■

ffl ffl

Rate of Transmitter 1 (Game) □ Rate of Transmitter 2 (Game)

O Sum of Rates (Game)

& Sum of Rates (Optim.)

X Rate of Transmitter 1 (Optim.)

+ Rate of Transmitter 2 (Optim.)

0.2

0.4

0.6

0.8

Fig. 2 : Centralized Optimization vs. Game

0

0

g

is nearly zero for g G [0,1/4] and then it grows up to 22% when g grows from 1/4 to 1; the transmitter with more resources gains significantly more from the centralized optimization. Hence, the non-cooperative approach results in a more fair resource distribution.

In Table 1 we give strategies of both users obtained in the case of the centralized optimization for different values of the crosstalk coefficient g. First, we observe that when the crosstalk coefficient is large, the users share different resources. The user with the larger average power takes better resources. When the crosstalk coefficient is below 0.7, the users start to share the resources. As the value of the crosstalk coefficient decreases, the 2nd user with the smaller average power begins to occupy better resources. As expected, when the crosstalk coefficient is very small, the optimal strategies start to look like strategies which are optimal in the case of no interference.

8. A jamming plot

In this section we consider a plot where two users have even antagonistic objectives. Namely, let there are two mobile terminals and one base station. Since we use the framework of game theory, we shall use the terms mobiles and players in-terchangably. Player (user) 1 seeks to transmit information to the base station. We shall refer to it as “Transmitter”. Player (user) 2 has an antagonistic objective: to prevent or to jam the transmissions of Player 1 to the base station. Thus, we shall call Player 2 “Jammer”. Both players have in addition a transmission cost which prevents us from using zero-sum games to model our problem.

g

Table 1: Centralized optimization Users’ strategies

0.95

0.70

0.65

0.35

0.20

0.10

0.01

1st user 7.43 7.17 6.19 4.19 0.00

2nd user 0.00 0.00 0.00 0.00 2.50

1st user 7.83 7.22 5.98 3.96 0.00

2nd user 0.00 0.00 0.00 0.00 2.50

1st user 8.25 7.42 6.57 2.76 0.00

2nd user 0.00 0.00 0.00 0.83 1.67

1st user 8.27 7.73 4.77 3.33 0.90

2nd user 0.00 0.00 1.24 1.22 0.04

1st user 6.65 6.44 6.16 4.36 1.38

2nd user 0.91 0.99 0.60 0.00 0.00

1st user 6.27 6.65 6.47 4.53 1.08

2nd user 1.43 1.00 0.07 0.00 0.00

1st user 7.45 7.03 6.00 3.98 0.54

2nd user 1.58 0.92 0.00 0.00 0.00

The pure strategy of Transmitter is T = (Ti,..., Tn) where Ti > 0 for i G [1, n] and £n=i Ti < T where T > 0.

The pure strategy of Jammer is N = (N1,..., Nn) where Ni > 0 for i G [1, n] and n=i Ni < N where N > 0. The payoffs to Transmitter and Jammer are given as follows

Vt

vn

n / T \ n

(^> = I>{^jwTNf -C^T-

1=1 v b ' i=l

n / T \ n

(T, «I = - E1” (i + E N"

(43)

i=1

where N0 is the power level of the uncontrolled noise of the environement at state i, CT > 0 and CN > 0 are the costs of power usage for Transmiter and Jammer, and g > 0 is fading channel gains for Transmiter and Jammer when the environement is in state i.

We shall look for a Nash equilibrium, that is, we want to find (T*, N*) G A x B such that

vT(T, N*) < vT(T*, N*) for any T G A, vN(T*, N) < vN(T*, N*) for any N G B,

where A and B are the sets of all the strategies of Transmitter and Jammer, respectively.

In the special case when CT and CN are zero in (34), the game is zero-sum. As vT is convex in Ti and concave in N, we can apply Sion’s minimax Theorem to conclude that it has a saddle point.

Since vT and vN are concave in T and N, the Kuhn-Tucker Theorem implies the following theorem.

Theorem 12. (T*,N*) is a Nash equilibrium if, and only if, there are non-negative w and v such that

9 .. (T* AT*, 1______r j = u forT*> 0,

dTi t{- ’ > T*+gN*+N° T\<u; for T* = 0;

—v (T* N*)=________________^2________________C \ = V f0rN*>°

ON, ; (T* + gN* + N°)(gN* + N°) N forN*= 0

where

> 0 forE”=i T* = T,

= 0 for £”=1 T* <T;

> 0 for £n=i N* = NN,

= 0 for £n=i N* < NN.

(46)

For non-negative w and v let

/00(w, v) = /00(w) = {i e [1,n] : g/N0 < g(w + Ct)},

/10(w, v) = {i e [1,n] : g(w + Ct) < g/N0 < g(w + Ct) + v + Cn},

/11 (w, v) = {i e [1,n] : g(w + Ct) + v + Cn < g/N0},

7—1 .—f \ 1 1--- n x ^ + ^ for * G /11 (w, z/),

(w + Ct)g + v + Cn w + Ct v ’

Ti(ui,v) = \ Ct+ u ~ ^ ^°r * € ^10(w> z/)>

k0 for i e /00(w,v);

for i G /11 (w, ^), (48)

[0 for i e /00(w,v).

Theorem 13. Each Nash equilibrium is of the form (T(w,v),N(w,v)) for some nonnegative w and v.

Now we go on to finding optimal w and v. Let

n n

Ht(w,v) = ^ Ti(w,v), HN(w,v) = E Ni(w,v).

i=1 i=1

Then Theorem 13 implies that

ht(u, v) = J2 (kCt+uj - ^ + uj+Ct J2 (ijU+Ct) +v+Cn>

iElio iEln

1 No '

(w + Ct )g + v + Cn g

In the next lemma some monotonous properties of Ti(w, v) and Ni(w, v), Ht(w, v) and Hn (w, v) are obtained.

Lemma 5.

(i) For fixed w > 0 and 0 < vi < v2 we have:

(1) Ti(w,vi) < Ti(w,v2) where strict inequality holds if, and only if, i e Iio(w, vi),

(2) Ni(w, vi) > Ni(w, v2) where strict inequality holds if, and only if, i e Iio(w, vi),

(3) Ht(w, vi) < Ht(w,v2) where equality holds if, and only if, Iio(w, vi) = ty,

(4) Hn(w, vi) >Hn(w,v2) where equality holds if, and only if, Iio(w, vi) = ty.

(ii) For fixed v > 0 and 0 < wi < w2 we have:

(1) Ti(wi, v) < Ti(w2, v) where equality holds if, and only if, i e Ioo(wi, v),

(2) Ni(wi, v) > Ni(w2, v) where equality holds if, and only if, i e Iio(wi, v),

(3) Ht(wi,vi) > Ht(w2,v) where equality holds if, and only if, Ioo(w,vi) =

= [1,nL

(4) HN(wi,v) > Hn(w2,v) where equality holds if, and only if, Iio(wi,v) = 0.

(iii) Ht(w,v) and HN(w, v) are non-negative and continuous in [0,to) x [0, to).

(iv) If Hn(0, 0) < N then Hn(w, v) < N for w > 0 and v > 0.

Based on monotonous properties described in Lemma 5 we can establish the following result about the number of Nash equilibrium the game can have.

Theorem 14. There is at most one Nash equilibrium.

Note that

i£[i,nj: g(u + Cr )<g/N°<g(u + CT ) + CN

, c„ v i (49)

uj + Ct (u! + CT)g -\- CN

i£[i,nj: g(^ + C'T)+Cn<g/N0 K rJ

The following lemma supplying some properties of Ht (w, 0) follows straighforward from (40) and Lemma 5.

Lemma 6.

(i) Ht(■, 0) is non-negative and continuous in (0, x),

(ii) Ht(w, 0) = 0 for enough big w, namely, for w > imaxi{1/N° — Cn/g} — Ct,

(iii) Ht(w, 0) is strictly decreasing on w while Ht(w, 0) > 0.

Lemma 6 implies that if Ht(0,0) > T then there exists the unique positive wQ0 such that Ht (wQ0, 0) = N (indices 10 mean that at this moment we look for the optimal solution where w > 0 and v = 0). If Ht(0,0) < T then Ht(t, 0) < T for t > 0. Then, from Theorems 12 and 13 and Lemmas 5 (iv) and 6 we have the following theorem.

Theorem 15. Let Hn(0, 0) < N then

(a) if Ht(0, 0) < T then (T(0,0),N(0,0)) is Nash equilibrium,

(b) if Ht(0, 0) > T then (T(w *0, 0),N(w Q0, 0) is Nash equilibrium. By Lemma 5 the following Lemma holds.

Lemma 7.

(a) If Hn(0,0) > N then there is vQ 1 such that Hn (0,vQ 1) = N (subscript 01

signifies that we look for the optimal solution where w = 0 and v > 0) and there is w such that Hn (w, 0) = N.

(b) Hn (w,v) < N for each w > w and each non-negative v.

(c) For each w G (0, w] there is unique non-negative v(w) suchthat Hn (w,v(w)) = N.

(d) v(w) is continuous and strictly decreasing on w, v(0) = vQ 1 and v(w) = 0.

Thus, by Lemma 7 we can introduce the following notation:

Ht{uj)=Ht{uj,v{uj))= E ^ (c \u;- Ni^j +

i£ho(u,v(u)) V T + '

I v(^) + Cn x V ^_________________________9_i___________

uj + CT — 1 {lo + CT)g + {v{u) + CN)

i£iii(w,v(w))

Then by Lemma 5 Ht is continuous and strictly decreasing in (0, w). Thus, if Ht(0) < T then Ht(w) < T for w G (0, w). If Ht(w) > T then Ht(w) > T for w G (0, w). If Ht(w) < T and Ht(0) > T then there is unique wQ 1 G (0, w) such that Ht(wQ 1) = T (subscript 11 signifies that we look for the optimal solution where w,v > 0). Then, from Theorems 12 and 13 we have the following theorem.

Theorem 16. Let Hn(0,0) > N then

(a) if Ht(0) = Ht(0,vQi) < T then (T(0,vQi),N(0,vQi)) is Nash equilibrium,

(b) if Ht(0) = Ht(0,V)i) > T and Ht(w) = Ht(w, 0) > T then (T(w*o,0),

N(wf0, 0)) is Nash equilibrium,

(c) if Ht (0) = Ht (0,v* i) > T and Ht (w) = Ht (w, 0) < T then (T (w * i ,v(w * i)),

N (w * 1,v(w * 1 ))) is Nash equilibrium.

Theorems 14-16 imply the following main result.

Theorem 17. There is unique Nash equilibrium given by Theorems 15 and 16.

9. Conclusion

We have considered power level management problem for wireless and DSL access networks with transmission cost in optimization and game frameworks. Closed form solutions for the water filling optimization problem and two players symmetric water filling games have been provided. Namely, now one can calculate optimal/equilibrium strategies with a finite number of arithmetic operations. We have also provided a simple alternative proof of convergence for a version of iterative water filling algorithm. It had been known before that the iterative water filling algorithm converges very slow when the crosstalk coefficient is close to one. For our closed form approach possible proximity of the crosstalk coefficient to one is not a problem. We have shown that when the crosstalk coefficient is equal to one, there is a continuum of Nash equilibria. Finally, we have demonstrated that the price of anarchy is small when the crosstalk coefficient is small and that the decentralized solution is better than centralized with respect to fairness.

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