Minimax and Maxmin Solutions to Gaussian

Jamming in Block-Fading Channels under Long

Term Power Constraints

George T. AmariucaiDepartment of ECE

Louisiana State UniversityBaton Rouge, LA 70803

Shuangqing WeiDepartment of ECE

Louisiana State UniversityBaton Rouge, LA 70803

Rajgopal KannanDepartment of CS

Louisiana State UniversityBaton Rouge, LA 70803

Abstract-Without assuming any knowledge on source's code-book and its output signals, we formulate a Gaussian jammingproblem in block fading channels as a two-player zero-sumgame. The outage probability is adopted as an objective function,over which transmitter aims at minimization and jammer aimsat maximization by selecting their power control strategies.Under long term power constraints, the optimal power controlstrategies of each player for both maxmin and minimax problemsare obtained. Numerical results demonstrate a sharp differencebetween the resulting outage probabilities of the minimax andmaxmin solutions, which thus implies the non-existence of Nashequilibria.

I. INTRODUCTION

The problem of jamming in wireless networks started toattract interest in the 80's when several works [1], [2] studiedsimple, point-to point communication systems affected byintelligent jammers. The jammer was assumed to have accessto either a noise-distorted version of the transmitter's output[1], or the transmitter's input message [2]. The mean-squarederror was considered as a performance metric.The saddle-point policy of the jamming game formulated in

[1] consists of an amplifying transmitter and ajammer that per-forms a linear transformation of the transmitter's output signal.A deterministic problem (shown to display no saddle point)and a probabilistic one are investigated in [2]. It is interestingto note that for the probabilistic formulation, the saddle pointis attained when the jammer ignores its information about thetransmitter's output.

Similar results were obtained in [3] for correlated jammerssuffering from phase/time jitters at acquisition or at transmis-sion. Channel capacity was used as the objective function.

Relatively few papers on this subject followed in sequel untilrecently when several extensions to more complex, multi-userchannels with fading were derived in [4], [5], [6], [7], [8].

It is shown in [5] that, in the absence of channel stateinformation (CSI) at both transmitter and jammer, an equilib-rium point is obtained when the jammer completely ignoresits information about the encoder's output.

'This work was funded in part by NSF under grants ITR-0312632 andIIS-0329738, and in part by the Board of Regents of Louisiana under grantsLEQSF(2004-07)-RD-A- 17.

Broadcast and multiple access channels (MAC) are investi-gated in [6] under a complete CSI and uncorrelated jammerassumption. The sum-rate is used as a performance indicatorfor the broadcast channel, while this role is played by anarbitrary weighting of the user's rates for the MAC. Proofsof existence of multiple Nash equilibria and conditions foruniqueness are provided.

Similar results for the multiple access channel are presentedin [7]. The paper covers all possible cases in terms of CSI andcorrelation of the jammer with transmitter's output, for a two-transmitter, one jammer scenario.The general tendency seems to be in favor of an assumption

that jammer has access to either the transmitter's output orinput and consequently is able to produce correlated jammingsignals. Uncorrelated jammers are often studied only as aparticular case. We, however, argue that the correlation as-sumption is sometimes inappropriate because of the effect ofcausality. Although the jammer can acquire information aboutthe transmitted data, there are significant time delays betweenthe original signal and the jamming signal at receiver, whichis not taken into consideration in previous works.

In addition, most recent works adopt ergodic capacity as acommon objective function over which transmitter and jammerfight against each other [7], [5], which is not a suitable metricif delay constraint is considered.

In this paper, we take a look at a constant-rate wirelesssystem with power and delay constraints. Without the presenceof a jammer, our scenario would be similar to the one studiedin [9]. We also consider a block fading channel with additivewhite Gaussian noise (AWGN). Each codeword in a frame oflength n = MN is assumed to span M blocks each of whichconsists of N channel uses. Channel fading variable remainsconstant over each block and varies independently acrossdifferent blocks. Moreover, all M coefficients of a frame areassumed available to both transmitter and jammer non-causally[9]. However, the jammer has no knowledge about the outputof the transmitter, or the codebook the transmitter uses.Our channel model as depicted in Figure 1 is very similar

to that in [7]. The difference, however, is that we investigatethe jamming in delay constrained block fading channels, and

1-4244-1037-1/07/$25.00 C2007 IEEE. 312

hence the probability of outage is an objective function,where the outage probability is defined as the probabilitythat the instantaneous mutual information of the channel islower than the transmission rate [9]. While [7] assumes nodelay constraint and employs ergodic capacity as an objectivefunction.

This difference brings more complexity to our problemwhich is formulated as a two-player, zero-sum game, whereonly pure strategies (no randomized strategies) are considered.The power constraint in a block fading channel can be man-ifested in terms of either short term or long term [9]. In thispaper, we devote our efforts exclusively to elaborate on thelong-term case as the solution to this case is more involvedthan the short-term one [10]. In particular, only the resultsfor M = 1 case are presented as this special case enables usto develop intuition and provides the mechanism needed forsolving the more general M > 1 case [10].

II. CHANNEL MODEL AND PROBLEM FORMULATION

The channel model is given in Figure 1. Transmission ina block fading channel is depicted in Figure 2, where eachcodeword spans a concatenation of M blocks, each of whichhas N channel uses. As assumed in [9], we let N -> Xo inorder to average out the impact of Gaussian noise.

Over a given frame, the transmitter (Tx) allocates power Pmto block m, 0 < m < M -1, while the jammer (Jx) investspower Jm in jamming the same block with the worst possiblejamming signal which is not correlated with the transmitter'soutput and white Gaussian distributed [11].The channel squared fading coefficient hm is constant over

the length of one block. The vector h = [ho, h,,...,,hm1]of channel coefficients over a whole frame is assumed to beperfectly known to transmitter, receiver and jammer beforetransmission begins. This condition does not imply non-causality if we think of modeling a multi-carrier system [9].

Fig. 1. Channel model

Frame I Frame 2 Frame 3

hii *il ..h h2 h, I...h hhi||

(N < oCBok

Channeluses

Fig. 2. Frames, blocks and channel uses

Then the mutual information over a sub-channel m whentransmitter uses Gaussian codebook is given

I(hm,Pm, Jm) = l/2log(1 + hh 'PI )2N + Jrn

where o-2 is the variance of the AWGN.In the sequel, the following denotations will be repeatedly

used:. Power allocated by transmitter over a frame:PM =

UNi 1' PM. Power allocated by jammer over a frame:

Tl 1 YM-1 ;J1MVI/-m =o. Instantaneous mutual information for a frame:IM = MmO I(hm, Pm:Jm)

Note that PM is a function of the channel realization h, sowe often write PM (h) when this relation needs to be explicitlyemphasized. PM (h) can also be interpreted as the functiongiving the power distribution across different frames. We usePM (h) and JM (h) to denote inter-frame power allocation forthe case M = 1, since in this case a frame only contains oneblock.The probability of outage, defined as P0,t = Pr(IM < R)

with R denoting the fixed transmission rate of source, isadopted as an objective function over which both transmitterand jammer vary their transmission powers based on the CSIabout h. The transmitter wants to achieve reliable communi-cation and minimize the outage probability, while the jammerwants to induce outage and maximize the outage probability,which makes it a two-player zero-sum game. Both of themseek only optimal pure strategies.

Therefore, the long-term power constrained jamming gamecan be formulated as:

Tx Minimize Pr(IM(h, {Pm}: {Jm}) < R) 1Subject to E[PM(h)] <P (

Jx Maximize Pr(IM(h, {Pm}: {J.n}) <R)x Subject to E[JM(h)] <j (2)

where expectation is with respect to the vector of channelcoefficients h = (ho,h...h hm- 1) C R4M, and P and J arethe upper-bounds on average transmission power of the sourceand jammer, respectively.The two players need to make decisions on whether or not

to transmit, given a fixed opponent's strategy, as well as therequired power for a player to achieve its objective over aframe. In the sequel, we look at both the maximin and minimaxsolutions to the above two player game in order to find outwhether a Nash Equilibrium exists, as well as the associatedpower control functions PM (h) and JM (h).

Let m denote the probability measure introduced by theprobability density function (p.d.f.) of h, i.e., for a set v C

R4m, we have m(vc) = fa f (h)dh. Integrating with respectto this measure is equivalent to computing an average withrespect to the p.d.f. given by f (h), i.e., dm(h) = f (h)dh.Both transmitter and jammer have to plan considering theprobability distribution of the channel coefficient vector, andtheir opponent's strategy.

313

III. MAXIMIN SOLUTION FOR M = 1

The maximin solution is defined as the set of optimalstrategies when the jammer plays first. No matter what strategythe jammer employs, the transmitter will take advantage ofits weaknesses, and try to obtain a minimum of the outageprobability. The best option for the jammer is to maximizethe minimum achievable by the transmitter.

Note that in a real situation, neither of the two players willplay first, hence none of them will gain access to its opponent'sstrategy. This is why the maximin solution should be thoughtof as the worst case scenario for the jammer, and the best casescenario for the transmitter.

Let X c R+ denote the set of channel realizations overwhich the jammer invests non-zero power. The transmitter willnot transmit any power over a frame that is going to end up inoutage. Therefore, it will start allocating enough power to theframes over which reliable transmission is easiest to achieve,and go on with this technique until the average power reachesthe limit set by the power constraint.The jammer needs to find the best choice of the set 9,

and its optimal power allocation strategy JM(h) over Y,such that when the transmitter employs its optimal strategy,the probability of outage is maximized.

For the case of M = 1, the transmitter can trans-mit reliably over a frame with minimum power expense ifIM (h, PM (h), JM (h)) = R, or equivalently, PM (h) = [(c-1)/h](JM(h) + usf), where c = exp(2R) is a given constant.It is easy to see that the larger JM (h) is for a particular h,the more difficult it is for the transmitter to achieve reliablecommunication. Therefore, the transmitter prefers the framesover which the required PM (h) is relatively small.

Let 2 c R+ denote the set of channel realizations overwhich the transmitter uses non-zero power. The jammer shoulddeploy some JM (h) over X such that the required PM (h) isconstant over the whole interval X. The purpose of jammerbeing active over X \ 5 is to sort of intimidate the transmitterto such an extent to maximize the outage probability. Thetransmitter plays second, and hence takes advantage of thejammer's weaknesses. It always chooses to be active on thesubset of X on which the required PM (h) is least. This iswhy the optimal jammer strategy is to display no weakness,i.e. to make PM(h) constant over X.

Note that the power needed by a player to achieve its ob-jective (reliable communication for transmitter, and outage forjammer) over some frame, given a fixed opponent's behavior,will be denoted as "required power". Depending on its optimalstrategy, this power may or may not be actually employed bythe player.

These considerations are formalized in Theorem 1 below.Before proving the theorem, we provide some insightfulcomments.

Consider the set 5 of channel realizations over whichthe transmitter allocates non-zero power. Let K denote themaximum level of power required for the transmitter tocommunicate reliably on 5.

Because of the way the transmitter chooses the set Y, thisimplies that the required transmitter power outside 5 is atleast K.On the other hand, the required transmitter power over X9

5 should be no greater than K, and hence equal to K, becauseotherwise the jammer would be wasting power.The next Proposition deals with the optimum way of allo-

cating the jammer's power over Y. Think of 5 as a linearlyordered subset of the positive real line.

Proposition 1: The jammer should adopt such a strategy asto make the transmitter's best choice of 5 intersect ( onthe left-most part of 5, and the required transmitter powerequal to some constant K on a n/Y and to (c- 1)2Ih onY\

Proof: Consider that the jammer picks a certain strategyJM(h). Because the transmitter's strategy is predictable, thejammer knows exactly what is the set Y, as well as themaximum level of required transmitter power that will bematched by transmitter. Denote this level by K.We assume the existence of two sets X, _ c Y, of non-

zero m-measure, such that h1 < h2, Vhl C X, h2 C 6, andPM(h) < K on X, and PM(h) > (c- 1)u2h on X, wherePM(h) is the power required for the transmitter to establishreliable communication over the block characterized by thechannel realization h.We show that if this is the case, then for the same K, the

jammer can improve its strategy., i.e. there is a different choiceof JM(h), which maintains the level K, but requires morepower from the transmitter to achieve reliable communicationover the same Y. But since the transmitter already uses allof its power for the first JM(h), and the required transmitterpower outside 5 is larger than K, the transmitter has nochoice but to reduce the set 5 to one of its proper subsets.This is equivalent to increasing the probability of outage.

Denote the necessary transmitter power allocation functionsover 1 and 6 by PM,l(h) and PM,2(h) respectively. Thepower invested by the jammer over the interval W U 6 is

-=[J hPM, (h)f (h)dh + hPM,2(h)f(h)dh]-c

(3)X 4N2f (h)dh,

while the power invested by the transmitter for obtainingreliable communication on the two intervals is

Thy = J PM, 1 (h) f (h)dh + J PM,2 (h)f (h)dh (4)

Next, we attempt to decrease the transmitter's necessarypower on 6 and increase the transmitter's necessary poweron X, while keeping the power invested by jammer constant.There are two cases to be considered as illustrated in Figure3 for an intuitive description of the technique.

Case I. Assume we can take enough jamming power fromX, to make the required transmitter power on v equal to K.

314

Case II

(ii)

PI

Q

Fig. 3. Improving the jammer's strategy

That is, assume there exists a positive PM,3(h) > h o-

such that

J h[K-PM,,(h)]f (h)dh

f h[PM,2(h) -PM,3(h)] f(h)dh (5)

and denote I (h) = h[K -PM,l (h)] and r(h) = h[PM,2 (h)-PM,3 (h)].

Then, allocating the jammer power such that the requiredtransmitter power functions are K instead of Pl (h) on X,and P3(h) instead of P2(h) on 6, the power invested by thejammer stays the same. However, the power invested by thetransmitter in achieving reliable communication is now:

PWy = paM +

replaced by PM,4(h) and it stops jamming on X, the powerneeded at the transmitter is now

' = P +

+ (I -l(h)f(h)dh J -r(h)f(h)dh).-h0

and the arguments under Case I can be used.This argument implies that for any pair of sets X, 6, of

positive measure, with the property that h1 < h2, Vhl CQ, h2 C X, we should have either PM,l(h) = PM,2(h) C{K, (c- 1)X /h}, or PM,l(h) = K and PM,2(h) = (c-1)ufjh. Taking sets of infinitely small measure, we have thedesired result of the Proposition.

We can now formulate the main result of this section.

Theorem 1: Transmitting JM (h), satisfying the power con-straint with equality, such that the transmitter power requiredfor reliable communication is PM(h) = K, Vh C [h*, h*],and PM(h) = (c- 1)j2h, Vh C [O,xo) \ (h*,h*], for someh* < heC R+ and some constant K C R+ UJ{ci} is anoptimal jammer strategy for the maximin problem. Note thatPM (h) should be continuous at h*1.The values of h*, h* and K that maximize the outage

probability can be found by solving the following problemnumerically:

Find min f (h)dh, whereh1 hI 0

ho is given by X Kf (h)dh + h 'Nf(h)dhho ~~~h2

cK 12K is givenbyK =~

+ (J -l(h)f(h)dh J -r(h)f(h)dh). (6)h* is given by 1h2 (h

Obviously, fa /jl(h)f(h)dh > fe1 r(h)f (h)dh, since Vh1 c

X, h2 C X, we have h1 < h2, and, dividing both terms byfe r(h)f(h)dh = fa I (h)f(h)dh, one gets the mean values

of 1/h over the sets a and 6, under two different probabilitymeasures.

Case II: Now suppose the jamming power on 6 is notenough to make the required transmitter power on a equal toK. Then the jammer can just stop transmitting over 6, anduse the saved power to increase the required transmitter poweron v from PM,3(h) to some PM,4(h) < K.

Thus, we can find some power allocation functionPM,4(h) <K such that

hh[PM,4(h) - PM,1 (h)] f (h) dh

[PM,2(h) oNjf(h)dh.Denote I(h) = h[PM,4(h) -PM,l(h)]f(h) and r(h) =

h[PM,2(h) c1 UN]f (h). While the power used by the

jammer does not change if it transmits such that PM,1 (h) is

Proof: The feasible solution above (unique up to sets ofmeasure zero) follows from the system of statements below.

By Proposition 1, the only optimal power allocation forthe jammer is such that the required transmitter power on

Ynn should be equal to some positive K (there is norestriction on the finiteness of K), and the intersectionY5 n should fill in the left-most part of 5.

If the maximum value of the transmitter's required poweron n a is K, then the required transmitter power on

\ should also be equal to K, because otherwiseeither the jammer (if > K), or the transmitter (if < K)would be wasting power;

Since the transmitter chooses the regions with the lowestrequired power first, the required transmitter power on

R+ \ (Y U 9) should be larger than K (and hencecontinuity of PM (h) in h* follows);Knowing that the transmitter's required power withoutjammer is a decreasing function of h, and that it is easierfor the jammer to impose a power K to the transmitter

315

Case I

\ PM 2 (h+).,,-J

K

K

IPI

ransmitter'srequired powerwithout jammer

I

rr

ransmitjmerreurdloe

(8)

2,

C72 Tf(h)dh J.

at lower h, the set . \ X should be situated on theright-most side of the positive real line R+.

This shows that an optimal jamming strategy is to make therequired PM (h) equal to K over some interval [hr*, h*] andto set JM (h) = 0 over [0, ox) \ (hr, h ].

Under these conditions, the transmitter's strategy is totransmit over the set [h*, o) U(5 nx), where yncan be any subset of 9 with the measure resulting from thetransmitter's power constraint. For simplicity, the numericalproblem above considers 5 xn to be the right-most sideof 2, i.e. an interval of the form [ho, h*]. However, this isnot mandatory. The transmitter can choose any part of X (aslong as it stays within the limits of its own power constraints),without influencing the probability of outage. This is becausethe power invested by the transmitter on any subset of 9equals a constant K times the m-measure of that subset. Hencethe transmitter will always pick subsets of equal measure, andthe probability of outage is determined only by this measure.

Recall that the transmitter only transmits if it can achievereliable communication. Therefore, outage will only occuroutside of Y. a

The numerical problem is described in Figure 4.

o5Q-

au

U)

----- Tx power required without jammerTx power required with jammer - case 1

..........-Tx power required with jammer - case ;

\ ~~~~~~~~tranlsmitter tranlsmits

jammertransThits

_ E~~~~~~........ ..... ...........

0 h,h hh hSquared channel coefficienth

Fig. 4. Maximin problem - power distribution between frames

Picking some hl, we can determine K, h2 and ho (in thisorder), and then find the probability of outage as P,tt(hl) =1-M[(ho, oc)]. The optimum hi, denoted by h*, is the oneminimizing the m-measure of the set (ho, oc).

IV. MINIMAX SOLUTION FOR M 1

The minimax solution is defined as the set of optimalstrategies when the transmitter plays first. The jammer willalways take advantage of the transmitter's weaknesses, and tryto obtain a maximum probability of outage. The best optionfor the transmitter is to minimize the maximum achievable bythe jammer. The minimax solution should be thought of asthe worst case scenario for the transmitter, and the best casescenario for the jammer.

Let R5 c R1+ denote the sets of channel realizations(frames) over which the jammer and transmitter use non-zeropowers, respectively.

The jammer will not transmit any power over a frame, ifoutage is not going to be induced, or if the transmitter is notpresent, i.e. 9 c 5.The jammer will start allocating power to the frames over

which outage is easiest to induce, and go on with thistechnique until the average power reaches the limit set by itspower constraint.From the transmitter's point of view, we need to find the

best choice of the set 5, and the best power allocation PM (h)over 5, satisfying the power constraints, such that, whenjammer employs its optimal strategy, the outage probabilityis minimized.

For the case M 1, the jammer can induce out-age over a frame, with minimum power expense, ifIM(h, PM(h), JM(h)) = R, or equivalently, JM(h) =

hPM(h)/(c-1)- o, with c exp(2R). Hence, the jammerprefers the frames for which the required JM(h) is less.The optimal transmitter's strategy is to allocate its power

such that the required JM (h) is constant on the whole set 5,and hence to display no weakness.

These considerations are formalized in Theorem 2 below.Before proving the theorem, we provide some insightfulcomments.

Let K denote the maximum value of the required jammingpower over X. Then the required jamming power on 5 \ Xcannot be less than K, because then the jammer's choice ofX is not optimal, and cannot be larger than K, because thenthe transmitter would be wasting power.The next Proposition deals with the optimum way of allo-

cating the transmitter's power over the set X.Proposition 2: The transmitter's optimal way to allocate its

power is to make the required jamming power remain equalto some constant K on all of X.

Proof: The proof is similar to that of Proposition 1 [10]and is omitted here due to the space constraint. UThe main result of this section is the following theorem.Theorem 2: Transmitting PM (h), satisfying the power con-

straint with equality, such that the required JM (h) equals Kfor h C [h*,oo), and JM(h) = VVh C [0,h*), for someheC R+, is an optimal strategy for the minimax problem.The values of h* and K that minimize the outage probabilitycan be found by solving the following problem numerically:

Find maxj f(h)dh, whereh*x J X ( )0

Ihoho is given by Kf (h)dh = J,

xay

and where K is given by X0 ' 1)(A N) f(h)dh P.

Proof: The feasible solution above (unique up to sets ofmeasure zero) follows the observations listed below:

. By Proposition 2, the required jamming power on Xshould be equal to some positive K (there is no restrictionon the finiteness of K);

316

r x r T r X 1 T

2

. The required jamming power on . \ X should be equalto K, because otherwise either the jammer (if < K) orthe transmitter (if > K) would be wasting power;

. Since it is easier for the transmitter to impose a certainjamming power K at higher h, the set 5 should besituated on the right-most side of the positive real lineR+.

The jammer's strategy is to transmit on any subset of 5, aslong as it satisfies its own power constraints. For simplicity, thenumerical problem above considers that the jammer transmitson the left-most part of Y, i.e. on an interval of the form[hz, h*]. However, this is not mandatory. The jammer canchoose any part of 5 (as long as it stays within the limits ofits own power constraints), without influencing the probabilityof outage. This is because the power invested by the jammeron any subset of 5 equals a constant K times the m-measureof that subset. Hence the jammer will always pick subsets ofequal measure, and the probability of outage is determinedonly by this measure.

Recall that the jammer only transmits if it can induceoutage. Therefore, outage will occur with certainty on S andoutside of Y. a

The numerical problem is described in Figure 5.

rT r T r~~ Jx required power-case 1

..Jxrequired power -case 2_

hxhh hosquared channel coefficient h

Fig. 5. Minimax problem - power distribution between frames

Picking some hx, we can determine K and ho (in thisorder), and then find the probability of outage as Pott(hi) =1 -m[(ho, oc)]. The optimum hx, denoted by h*, is the onemaximizing the m-measure of the set (ho, oc).

Using the two numerical methods presented above, we havecomputed the two solutions for a fixed jamming power upper-bound J and a sequence of transmitter power upper-boundsP. The channel coefficient h is assumed to be exponentiallydistributed with parameter A = 1/6. The significant differencebetween the outcomes of maxmin and minimax solutions interms of outage probability versus P demonstrate the non-existence of Nash Equilibria in this two-player zero-sum game.

V. CONCLUSIONS

We have formulated the jamming problem over a constant-rate, block fading channel, with long-term power constraintsin terms of a two-person zero-sum game.The maximin and minimax solutions turn out to be generally

different, as demonstrated by both Theorem 1 and Theorem 2,

P ,,, vs. P for M=1, J=10, 62 =5, R=1, exponential fading withX=1/6

0.9

0.

co

O 0.Q3

).5

0.4

0.3

0.25 10 15 20 25 30

Average transmitter power P

Fig. 6. Outage probability vs. P - maximin and minimax cases

as well as numerical results. This implies that under long-termpower constraints Nash equilibria don't exist.

For both solutions, whoever plays first is clearly at somedisadvantage. The second player will always try to exploitthe first player's weakness. Our results show that the optimalstrategy for the first player is to exhibit no weakness, i.e. topresent the second player with an indifferent choice space. Thesecond player will then pick a set out of the choice space, overwhich it will allocate non-zero power.The extension to more general cases with M > 1, as well

as the two-player zero-sum game under a short-term powerconstraint will be presented in detail in [10].

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