Optimal Strategies in Jamming Resistant Uncoordinated Frequency Hopping Systems. Bingwen Zhang

Transcription

1 Optimal Stategies in Jamming Resistant Uncoodinated Fequency Hopping Systems by Bingwen Zhang A Thesis Submitted to the Faculty of the WORCESTER POLYTECHNIC INSTITUTE in patial fulfillment of the equiements fo the Degee of Maste of Science in Electical and Compute Engineeing by May 2013 APPROVED: Lifeng Lai, Assistant Pofesso Donald R. Bown III, Associate Pofesso Alexande Wyglinski, Associate Pofesso

2 Abstact Uncoodinated fequency hopping (UFH) has ecently emeged as an effective mechanism to defend against jamming attacks. Existing eseach focuses on the optimal design of the hopping patten, which implicitly assumes that the stategy of the attacke is fixed. In pactice, the attacke might adjust its stategy to maximize its damage on the communication system. In this thesis, we study the design of optimal hopping patten (the defense stategy) as long as the optimal jamming patten (the attack stategy). In paticula, we model the dynamic between the legitimate uses and the attacke as a zeo sum game, and study the popety of this game. We show that when the legitimate uses and the jamme can access only one channel at any time, the game has a unique Nash equilibium. In the Nash equilibium, the legitimate uses and Eve will access o jam only a subset of channels that have good channel quality. Futhemoe, the bette the channel, the lage the pobability that Eve will jam the channel and the smalle the pobability the legitimate uses will access this channel. We futhe extend the study to multiple access multiple jamming case and chaacteize the Nash equilibium. We also give numeical esults to illustate the analytical esults deived in this thesis.

3 iii Acknowledgements Fist and foemost, I shall geatly thank my eseach adviso, D. Lifeng Lai. He is not only a espectable and esponsible peson, but also povides valuable guidance and suppot to my eseach. Besides, I am gateful to him fo ecommending me as his PhD student. I hope I can have a vey pleasant collaboation with him on my PhD life. Thanks to my committee membes, D. Donald R. Bown III and D. Alexande Wyglinski fo shaing thei time on my thesis and pesentation. Thanks to my senio alumnus Jun Geng, my senio alumnus Heng Zhou and Ain Ul Aisha. They give me a lot of help on my eseach. Thanks to my family. My mothe Ziping Zhang and my fathe Zhenpeng Zhang, the most impotant pesons fo me, give me life, love and whateve I want unconditionally. The wok was suppoted by the National Science Foundation CAREER Awad unde gant CCF

4 iv Contents List of Figues vi 1 Intoduction Jamming Spead Spectum Techniques Uncoodinated Fequency Hopping(UFH) Existing Schemes to Impove Efficiency Leaing Based Appoach Coopeative Boadcasting Appoach Othe Reseach about UFH Summay of Thesis Backgound Zeo-sum Game Nash equilibium Model 20 4 Equal Channel Quality Case Poof Pove: E = {1, 2,..., N} Pove: E A, E B, A = B Detemine p t, p and p j Remak Geneal Channel Quality Case Poof Pove: if p j k > 0, then p j i > 0 fo all i > k. Hence, thee exists a numbe k such that E has the fom E = {k, k + 1,..., N} Pove: E A, E B and A = B Detemine p t, p and p j Remak

5 v 6 One Access Multiple Jamming Case Poof Remak Multiple Access One Jamming Case Poof Remak Multiple Access Multiple Jamming Case Poof Remak Numeical Simulation Equal Channel Quality Case Geneal Channel Quality Case One Access Multiple Jamming Case Multiple Access One Jamming Case Multiple Access Multiple Jamming Case Conclusion 92 Bibliogaphy 93

6 vi List of Figues 1.1 Fequency hopping Model of spead spectum system Cicula dependence Uncoodinated fequency hopping System model Channel Capacity (N=15) P t (= P ) (N=15) P j (N=15) Channel Quality (N = 15, k = 12) P t (= P ) (N = 15, k = 12) P j (N = 15, k = 12) Poof fo E = {k, k + 1,..., N} Poof fo E A Poof fo A = B Channel capacity Rewads of Alice and Bob fo accessing each channel Rewads of Eve fo accessing each channel k sepaates good channels and bad channels Channel Quality (N = 15, k = 12) P t (= P ) (N = 15, k = 12) P j (N = 15, k = 12) k sepaates good channels and bad channels Illustation of case 1 in one access multiple jamming case Illustation of case 2 in one access multiple jamming case Illustation of case 3 in one access multiple jamming case Channel Quality (N = 15, k = 13) Case 1: P t (= P ) (N = 15, k = 13) Case 1: P j (N = 15, k = 13) Case 2: P t (N = 15, k = 13)

7 vii 7.8 Case 2: P (N = 15, k = 13) Case 2: P j (N = 15, k = 13) Case 3: P t (N = 15, k = 13) Case 3: P (N = 15, k = 13) Case 3: P j (N = 15, k = 13) k sepaates good channels and bad channels Example of multiple access multiple jamming case Illustation of case 1 in multiple access multiple jamming case Illustation of case 2 in multiple access multiple jamming case Illustation of case 3 in multiple access multiple jamming case Channel Quality (N = 15, k = 12) Case 1: P t (= P ) (N = 15, k = 12) Case 1: P j (N = 15, k = 12) Case 2: P t (N = 15, k = 12) Case 2: P (N = 15, k = 12) Case 2: P j (N = 15, k = 12) Case 3: P t (N = 15, k = 12) Case 3: P (N = 15, k = 12) Case 3: P j (N = 15, k = 12) k sepaates good channels and bad channels Aveage data ate vs. Total numbe of channels Distibution of R, N = Distibution of R, N = R affected by Mj R affected by Mt R affected by Mt unde diffeent M j R affected by Mj unde diffeent M t (= M ) R affected by M unde diffeent M = Mt M, M = R unde the Nash equilibium in ou model and leaning based appoach.. 91

8 1 Chapte 1 Intoduction In this chapte, we fist intoduce the concept, consequences and categoies of jamming attacks in Section 1.1. The conventional anti-jamming method is intoduced in Section 1.2. The concept of uncoodinated fequency hopping (UFH) is intoduced in Section 1.3. The elated wok about UFH is shown and discussed in Section 1.4. In Section 1.5, we summaize the main contibutions of this thesis. 1.1 Jamming Wieless technology is becoming moe and moe popula [1] and is widely used by companies and individuals fo impotant communications, such as mobile e-commece tansactions, , and copoate data tansmissions [2]. As the esult, secuity issues become moe and moe impotant fo wieless netwoks. This is not a tivial poblem because wieless devices, including smat cellula phones and pesonal digital assistants (PDAs) with Intenet access, wee not oiginally designed with secuity as a top pioity [2]. Most of wieless netwok secuity poblems can be mitigated o fully addessed by changing wieless netwok secuity achitectues o using moe advanced cyptogaphic methods [3]. Howeve, thee ae still some theats that can not be addessed by these methods. Jamming is an impotant class of such theats [3]. Due to the openness of the wieless medium, attackes can easily implement jamming attacks to inject signals into the medium. Attackes can easily obseve communications be-

9 2 tween legitimate uses, and then make the tansmission in wieless netwoks fail by injecting false messages. The attackes can implement diffeent kinds of jamming attacks [4]: 1. Constant jamme: The constant jamme continually emits a adio signal. 2. Deceptive jamme: The deceptive jamme constantly injects egula packets to the channel. So legitimate uses will be deceived into believing the jamme is also a legitimate use in tansmitting state. 3. Random jamme: The andom jamme altenates between sleeping and jamming. 4. Reactive jamme: The eactive jamme only begins jamming when the jamming detects activity in the channel. Reactive jamming is the most impotant theat among the fou jammes [5]. The eason is that, while destoying the packets, the attacke minimizes its isk of being detected [5]. In fequency hopping, eactive jamme cannot complete the detection pocess if the hopping ate is high enough [6]. So it is can be seen that to mitigate jamming, the spead spectum techniques ae usually adopted [7]. 1.2 Spead Spectum Techniques Spead spectum techniques ae conventional anti-jamming methods [8]. The spead spectum signals usually have the chaacteistic that the bandwidth is much lage than the infomation ate which can be seen as edundancy. This kind of edundancy is added to the signal due to the signal is equied to ovecome sevee intefeence in the pocess of tansmission in the channel. The edundancy of the spead spectum signal can be chaacteize by bandwidth expansion facto which is usually much lage than one [7]. To intoduce edundancy to signal, we know that coding is an efficient method [7]. So how to code the signal to make it spead spectum is the fist key element in designing the spead spectum systems [7]. In the secuity aspect, in ode to avoid the attacke to demodulate the spead spectum signals, pseudoandomness is needed [7]. The pseudoandomness of the spead spectum

10 3 signals makes the signals seem to be andom noise to the attacke thus making it vey difficult fo the attacke to demodulate the signals [7]. This chaacteistic is actually elated to the pupose o application of these spead spectum signals [7]. In [7], the authos list the main puposes of the spead spectum signals: 1. To combat the effects of intefeence due to jamming, intefeence caused by othe uses of the channel and self-intefeence due to multipath popagation. 2. To hide a signal by tansmitting it at low powe, thus making it difficult fo an attacke to detect the signal in the pesence of backgound noise. 3. To achieve message pivacy in the pesence of attackes. 4. To obtain accuate ange (time delay) and ange ate (velocity) measuements in ada and navigation (this pupose is not diectly elated to communications). In combating the effects of intefeence of intentional jamming, the knowledge of the jamme is impotant [7]. If the jamme knows the chaacteistic of the tansmitting signal, it is easy fo the jamme to mimic this signal tansmitted by the tansmitte and confuse the eceive [7]. To pevent this to happen, the tansmitte intoduces the andomness (actually pseudoandomness) to the signal which is unpedictable fo the jamme while known to the eceive. So the only way fo the jamme to do jamming is to tansmit an intefeing signal without any pio knowledge about the pseudoandom patten [7]. Fequency-hopping spead spectum (FHSS), diect-sequence spead spectum (DSSS), time-hopping spead spectum (THSS), chip spead spectum (CSS), and combinations of these techniques ae foms of spead spectum [7]. Each of these techniques employs pseudoandom numbe sequences ceated using pseudoandom numbe geneatos to detemine and contol the speading patten of the signal acoss the allocated bandwidth [7]. Figue 1.1 shows the taditional fequency hopping (FH) which elies on secets shaed between the tansmitte and eceive. The shaed secet then detemines the hopping patten in FH. A thid paty who does not know the secet codes cannot pedict the tansmission [9]. Then the pobability of the tansmission being jammed is educed [9]. But the peequisite is a secet must be shaed by the tansmitte and the eceive [9].

11 4 Figue 1.1: Fequency hopping. Infom ation Sequence Channel encode M odulato channel dom odulato Chan nel decod e Output Data Pseud oand om Patten Geneato Pseudoandom Patten geneato Figue 1.2: Model of spead spectum system.

12 5 Figue 1.2 shows the model of spead spectum. Notice that we should have two identical pseudoandom patten geneatos, one at each side. In pactice, we equie the tansmitte and eceive have the same patten and we should have the pseudoandom patten geneato pefectly synchonized [7]. The poblem aises that if two nodes which have not communicated befoe but want to communicate in the pesence of jamme, the pseudoandom patten cannot be known by the othe side [9]. If we want the patten to be pe-stoed in each node in the netwok, scalability is a big issue [9]. If the netwok has N nodes, then fo each of them to communicate with othe nodes, N 1 pais of pe-shaed secets ae needed fo each node. The total numbe of pe-shaed secets in this netwok is N(N 1) 2. If N is lage, it is challenging to pe-distibute and futhe stoe N 1 pais of secets fo each node. In this context, the uncoodinated fequency hopping discussed below is poposed in [9] to solve this poblem. 1.3 Uncoodinated Fequency Hopping(UFH) The uncoodinated fequency hopping (UFH) to solve the poblem descibed above is oiginally poposed in [9], which can beak the cicula dependence of conventional spead spectum methods. Figue 1.3 descibes the cicula dependence poblem. In paticula, if two devices do not shae any secet keys o codes and want to execute a key establishment potocol in the pesence of a jamme, they have to use a jamming-esistant communication [9]. Howeve, known anti-jamming techniques such as fequency hopping and diect-sequence spead spectum ely on secet (speading) codes that ae shaed between the communication patnes pio to the stat of thei communication [9]. This ceates the cicula dependence. Figue 1.4 illustates the high level idea of UFH. In UFH, the tansmitte and eceive hop andomly between channels in an uncoodinated manne. The tansmission is successful when they ae in the same channel and the jamme is not in that channel. Figue 1.4 shows 3 diffeent scenaios of UFH: 1. In timeslot 1, both tansmitte and eceive ae in channel 5, while Eve is not in channel 5. The tansmission is successful.

13 6 Key establishment in the pesence of jamme Dependency cicle FH Fequency hopping sequence Figue 1.3: Cicula dependence. Figue 1.4: Uncoodinated fequency hopping.

14 7 2. In timeslot 2, the tansmitte and eceive ae not in the same channel. The tansmission is failed. 3. In timeslot 3, both tansmitte and eceive ae in channel 3, while Eve is also in channel 3. This tansmission is jammed, so it is failed. Thus UFH beaks the cicula dependence by not elying on the hopping patten [9] and by establishing a secet key when the tansmission is successful. In [9], it is shown that UFH scheme can be as esistant to jamming as coodinated fequency hopping. The authos assume the legitimate communication nodes have the ability to stoe a few megabytes of data and can pefom elliptic cuve cyptogaphy (ECC) based public key cyptogaphy. The attacke in this model is computationally bounded and also enegy constained. The goal of the attacke is to intefee the communication of the legitimate nodes by inseting messages, modifying messages o jamming messages. In thei scheme, a message M which is going to be sent by tansmitte is split into l fagments M 1, M 2,..., M l. And the behavio of tansmitting fagment M j does not elate to any channel and does not elate to the fagments sent befoe. The authos call this scheme andomized. At the eceive side, the fagments of message M should be eassembled. This pocess is to avoid the jamming attack. In this pape, the authos assume that the eceive switches channel less often than the tansmitte, thus educing the numbe of patially eceived fagments. The scheme of avoiding inseting messages and modifying messages is also designed using Hash link. Fo each fagment M i, some additional messages ae added to M i to fom a packet m i = id i M i h(m i+1 ). id is the message identifie, i is the fagment numbe, M i is the fagment of the message, and h(m i+1 ) is the hash value of the next packet. Fo the last packet, m l = id l M l h(m 1 ), so it foms an unbeakable hash link in the fagments of messages. This ensues the attacke cannot pefom effective inseting o modifying attack. The hash-linked packets ae tansmitted with a high numbe of epetition. The tansmission is successful when the tansmitte eceives the eply that the eassembling is successful and then the tansmission is finished. The tansmission can also be finished unsuccessfully with the numbe of epetitions has exceeded an theshold value. When the packets ae eceived, the eceive stats to compute the packets into a whole message. Fist, the eceive link the

15 8 packets accoding to the fagment numbe. Second, the eceive computes the hash value of the i + 1th packet h(m i+1 ) and compaes it with the hash value pat in the pevious packet. When all the packets ae linked successfully, the eceive sends the eply message. Fo the secuity analysis, given two consecutive packets m i 1 = id i 1 M i 1 h(m i ) and m i = id i M i h(m i+1 ) whee 2 < i < l, the attacke need to foge a m i. So the attacke needs to find h i = h(id i M i h i+1 ) = h(id i M i h i+1 ). But this means to find a collision of hash function h( ), so this is impossible fo the computationally bounded attacke to find a collision of hash function. The pocess that last packet in chain linking to the fist packet avoids inseting additional chain heads. The attacke needs to foge a m 1 = id 1 M 1 h 2 fo m 1 = id 1 M 1 h 2. So h(m 1 ) = h(m 1 ), which also means finding a collision fo h( ). This UFH scheme fo key establishment woks like this: fist, the tansmitte and eceive use a key establishment potocol to geneate a key and use the UFH scheme to communicate to agee on a key; second, the tansmitte and eceive use this key to find the hopping sequence. In this model, the authos show that fo all attacke types, jamming is the best stategy fo the attacke. The authos also state that thee is no pio wok that focuses on cicula dependence of anti-jamming establishment and they have not been able to find a scheme to tansfe abitay length messages without a pe-shaed key. Compaing to FH, UFH does not need shaed secets and synchonization. Howeve, UFH suffes fom low thoughput [6], as the tansmitte and eceive ae uncoodinated and hence most of them they ae opeating at the same channel. 1.4 Existing Schemes to Impove Efficiency As discussed above, since the hopping in UFH is uncoodinated, UFH often achieves a low efficiency [6]. To alleviate this poblem, thee have been some ecent woks attempting to incease the thoughput of UFH. They mainly focus on two aspects: leaning based appoach and coopeative boadcasting appoach, which will be discussed in Section and Section espectively. UFH. Section will discuss some othe eseach elated to

16 Leaing Based Appoach In [6], the authos develop an almost optimal and adaptive UFH-based anti-jamming scheme and give the thoough quantitative pefomance analysis fo this type of schemes. The UFH-based anti-jamming communication is fomulated as a non-stochastic Multiamed Bandit Poblem and online adaptive UFH algoithm against oblivious and adaptive jamme is poposed. The authos show that the pefomance diffeence between thei algoithm and the optimal one is no moe than O(k T n ln n) in T timeslots, whee k is the numbe of fequencies the eceive can eceive simultaneously and n is the total numbe of othogonal fequencies. A thoough quantitative pefomance unde vaious stategies of the sende, the eceive and the jamme is made. The authos also analyze the paamete selection to achieve the optimality. In the model of this pape, each node can tansmit and eceive ove k s and k channels espectively, whee k s n and k n. It is also assumed that the thee paties, i.e., the tansmitte, the eceive and the jamme have no pio knowledge of othes stategy. The authos do not conside message authentication and pivacy in this model because this can be achieved on the application laye. The authos assume the jamme can jam k j channels in one timeslot. The authos divide the jammes into two categoies: oblivious jamme and adaptive jamme. The oblivious jamme selects the jamming stategy independent of the past channel status he has obseved. The adaptive jamme can adaptively change his jamming stategy based on his past expeiences and obsevations. In this pape, the authos do not assume the channel quality can be estimated and known befoe o duing tansmission. So the algoithm poposed is tying to do online leaning the stategy of jamme and use the stategy of jamme to achieve optimality. In [10], the authos conside powe contol jointly with UFH poblem. The poposed appoach in this pape utilizes online leaning theoy to detemine both the hopping channels and the tansmitting powes based on the histoy of channel status. The sende in this model has a powe limited budget. Using the poposed appoach, the tansmitte can choose both tansmission powe of each channel and which channel to tansmit simultaneously. In [11], the authos discuss pimay use emulation (PUE) attack to fight ove channels with the seconday use in cognitive adio systems. In this scenaio, thee ae two paties

17 10 instead of thee paties in the UFH model. The authos model the PUE and andom hopping as a zeosum game between the attacke and seconday use. The Nash equilibium of this model is found. One impotant assumption hee is that the channel statistics ae known. In this pape, available pobability of each channel is known. In [12], the authos change known channel statistics model into unknown channel statistics model. In this model, the seconday uses need to face the challenge that how to addess the uncetainties in the channel statistics and the attacke s policy. The authos adopt advesaial bandit algoithm which is significantly modified in the context of blind dogfight in spectum. This is actually a way to lean the optimal defense stategy using past expeience and adaptively change the stategy to the channel dynamics and attack stategy. This key idea is simila as [6] which focuses on a diffeent poblem. The leaning based appoach implicitly assumes that the stategy of the jamme is fixed. What if the attacke is also intelligent so that the attacke can implement leaning method to lean the stategy of the legitimate uses and adjust its own stategy? Fo intelligent attackes, we can not use leaning based appoach and optimize the thoughput using the leaned stategy of the attacke since the stategy of the jamme is no longe fixed. This motivates us to think about the UFH poblem with intelligent jammes Coopeative Boadcasting Appoach In [13], [14], [15], [16], [17], [18] and [19], a collaboative UFH-based boadcast (CUB) scheme to achieve a highe communication efficiency is poposed and the main idea is to allow nodes that aleady eceive the message to help boadcast. The authos show that thei CUB scheme can achieve highe communication efficiency and is moe esistant to jamming attack than existing jamming esistant boadcast scheme. In this pape, the authos assume a souce node intends to tansmit a message to N nodes. The analysis is mainly focused on single hop. The message is split into M fagments of equal length, and each of them is tansmitted duing one time slot (fequency hop). The CUB scheme is an extension of pevious pai-wise UFH schemes. The authos name the staightfowad extension of UFH in the boadcast scenaio without coopeation as UFH-based Boadcast (NUB). In NUB, each node selects one of C channels in each time slot to eceive a packet, and epeat

18 11 this until the whole boadcast message is eceived. NUB does not have elay nodes in the boadcast pocess and in this pape it is shown that NUB takes longe time than paiwise tansmission to a single eceive. The authos mainly focus on jamming attack and assume the computation and tansmission capability of the attacke is bounded. The time slot is t p, and the jamming attack needs t p to successfully jam a packet, and it takes t s time to sense a channel. The authos categoize the jamming attacks into esponsive attacks and nonesponsive attacks, based on whethe the jamme senses the tansmission befoe implementing jamming attacks. Fo esponsive attacks, it is assumed that C J channels can be blocked simultaneously and t J time is needed to switch those channels. In one time slot, the numbe of channels can be jammed is n J C J and n J = t p t p+t J. Fo nonesponsive attacks, it is assumed that C s channels can be sensed simultaneously. In one time slot, the numbe of channels can be sensed is n s C s and n s = t p t p t J t s. The authos assume the attacke can implement esponsive and nonesponsive attacks simultaneously, which is a wost case, and they call it esponsive-sweep stategy. In this wost case, the attacke can jam n s C s + n J C J channels in one time slot, and each time slot is jammed with pobability n sc s +n J C J C. The authos popose thee elay channel selection stategies: Random Relay Channel selection (RRC), Sweep Relay Channel selection (SwRC), and Static Relay Channel selection (StRC). In RRC, each elay node selects andomly and independently one channel in all C channels to tansmit a packet. But RRC often esults in collision, which is two elay nodes select one same channel. SwRC is an idealized vesion of RRC. In SwRC, the fist node selects one channel in all C channels, and the second node selects one channel in the left C 1 channels and so on. SwRC avoids collision but it equies a lot of infomation exchange between elay nodes. In StRC, the selection of channels is no longe andom, the channels in boadcast pocess is fixed and nonovelapping. The authos assume the nodes have unique IDs and a suitable algoithm can make the pobability of channel collision is negligible. The authos mainly adopt a Random Receiving Channel Selection (RRxC) scheme which means the eceive hops andomly among the C channels. Specially, fo StRC, the authos design an Adaptive Receiving Channel selection (ARxC) stategy. As the StRC, each node in ARxC is assumed to know the elay channel list. If all the elay channels in the channel list ae jammed, the stategy is switched to RRxC. In CUB, the authos design the contol scheme of

19 12 tansmission duation, the tansmitte (souce node o elay node) stops tansmitting when a ACK signal is eceived o a maximum tansmission duation is eached. The authos also analyze the coopeation gain of RRC and StRC stategy. Simulations also indicate a significant impovement of pefomance of CUB compaed to noncollaboative UFH-based boadcast scheme. [20] investigates efficient Media Access Contol (MAC) stategies fo the UFH-based collaboative boadcast. To minimize the boadcast delay and to significantly educe enegy cost, the closed-fom expession of channel access pobabilities is given. This pape is based on [14]. The authos mainly conside two issues: boadcast delay and total enegy cost. The authos divide the synchonization among the elays and souce into two categoies. The fist categoy is pefect synchonization elays, all tansmittes ae synchonized both in time and tansmission content. The second categoy is asynchonous elays, which is moe ealistic one, whee two o moe tansmission ove the same channel fail. The boadcast delay is the time fom the beginning of the tansmission to the message is successfully eceived by the eceive. The enegy consumption is all the enegy consumed in this pocess. The authos give the minimal delay stategy and enegy efficient stategy. The authos show that if boadcast delay is the main concen, the elays should aggessively access the wieless media. As the netwok gows, to educe enegy consumption, the channel access pobability should be gadually educed. [21] addesses the poblem of anti-jamming boadcast communication among nodes that do not shae secet keys. This pape is based on [9]. Thee instances of Uncoodinated Spead Spectum (USS) ae pesented: Uncoodinated Fequency Hopping (UFH), Uncoodinated DSSS (UDSSS) and hybid UFH-UDSSS. UFH andomizes the selection of the fequency channels and UDSSS andomizes the selection of the speading codes. The feasibility and pacticability of the schemes poposed was demonstated by a USRP/GNU Radio based pototype implementation. Coopeative boadcasting appoach does not focus on impoving the thoughput of one hop in UFH. If the thoughput of one hop can be inceased by choosing appopiate stategy, the total thoughput of coopeative boadcasting can be inceased automatically. As the esult, we mainly focus on optimizing the thoughput of one hop in UFH.

20 Othe Reseach about UFH In [22], the authos popose a new USD-FH scheme fo Diffie-Hellman (DH) key establishment using UFH befoe the FH communication stats. This is based on [9], and ties to design a moe efficient scheme. The advantage of this scheme ove othes is that it does have to split the DH message into multiple packets. The basic idea of USD-FH is to tansmit each DH key establishment message using a one time pseudoandom hopping patten, and befoe the actual message tansmission, the seed of the pseudoandom patten is disclosed. Fo enegy bounded jamme, it is vey difficult to jam all the channels, so thee is always a chance that the eceive gets the seed of the pseudoandom hopping patten while the jamme does not. Since the jamme cannot jam the message tansmitted using pseudoandom hopping patten, so the eceive can eceive the DH message coectly. This pape uses UFH to establish key befoe FH communication. [23] mainly talks about the coodination between a seconday tansmitte and a seconday eceive in ode to use the same spectum white space. This is simila to the synchonization between the tansmitte and eceive in UFH. The authos build a new tansmission scheme within a famewok of fequency-hopping spead spectum (FHSS) tansmission with M-ay fequency-shift keying (M-FSK) modulation. When the white space detection eo is lage, which happens moe often duing the beginning stage of the seconday use tansmission, the speading gain is inceased to educe the intefeence to the pimay use. When the white space eo detection is small enough, which happens moe often afte beginning stage of tansmission, the white space in spectum is known, the speading gain is deceased to incease the data ate of the seconday uses. The pupose of using FHSS is to avoid intefeence and to impove secuity. In [24], the authos popose an appoximation fo the channel capacity of M-fequency T-use multiple access channel. In [25], the authos popose a detection scheme fo uncoodinated naow-band FH systems. The detection scheme can detect the existence and the numbe of the naowband signals colliding with the desied signal. These papes ae mainly focus on designing the UFH scheme in diffeent specific scena-

21 14 ios. In this thesis, the oiginal model poposed in [9] is used which will lead to moe geneal esults. 1.5 Summay of Thesis In this thesis, the goal is to optimize the thoughput of UFH assuming that the attacke is intelligent. Since the jamme is intelligent, it will also adapt its attack stategy. Hence, when we design ou optimal hopping stategy, we need to take the dynamics of the attacke into consideation. In this thesis, we model the inteaction of the legitimate uses and the jamme in UFH as a zeo-sum game, and study the optimal hopping stategy of the legitimate uses and coespondingly the optimal attack stategy of the jamme using game theoy [26]. In this thesis, we use the name Alice to denote the tansmitte, Bob to denote the eceive and Eve to denote the jamme. The oganization and main contibutions of the thesis ae the following: In Chapte 2, the backgound of the zeo-sum game and the Nash equilibium ae intoduced. In Chapte 3, the zeo-sum game model of UFH is intoduced. The stategies of the tansmitte, the eceive and the attacke ae defined. equilibium of ou game is also given. The definition of the Nash In Chapte 4, we study the case that all the channels have the same capacity. We fully chaacteize the Nash equilibium fo this case. In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. We show that Alice, Bob and Eve always access o jam all channels. Alice, Bob and Eve access o jam each channel with an equal pobability. The aveage thoughput is a deceasing function of N fo N 2, whee N is the total numbe of channels. In Chapte 5, we study the geneal channel quality case with R 1 R 2... R N, whee N is the total numbe of channels. We fully chaacteize the Nash equilibium fo this case. In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. Alice, Bob and Eve do not always access o jam all channels.

22 15 A new vaiable k is intoduced and it is fo Alice, Bob and Eve to decide on which channels they should take actions. k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. It is simple to veify that N k This implies that Alice and Bob will access at least two channels. Othewise, if they access only one channel, this channel will be jammed by the attacke with pobability 1. In Chapte 6, we extend the model into one access multiple jamming case, in which case Alice and Bob can access one channel in one timeslot, while Eve can jam multiple channels in one timeslot. We fully chaacteize the Nash equilibium fo this case. In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. Alice, Bob and Eve do not always access o jam all channels. k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob. When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. It is simple to veify that N k + 1 M j + 1, and M j is the numbe of channels Eve can jam in one timeslot. This implies that Alice and Bob will access at least M j + 1 channels. Othewise, if they access only M j channels, this channel will be jammed by the attacke with pobability 1. In Chapte 7, we extend the model into multiple access one jamming case, in which case Alice and Bob can access multiple channels in one timeslot, while Eve can jam one channels in one timeslot. In the Nash equilibium, Alice, Bob and Eve do not always opeate on the same set of channels. k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. But in some cases, Alice and Bob have to access channels

23 16 fom 1 to k 1. k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob. When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. It is simple to veify that N k This implies that Alice and Bob will access at least two channels. Othewise, if they access only one channel, this channel will be jammed by the attacke with pobability 1. In Chapte 8, we extend the model into multiple access multiple jamming case, in which case Alice and Bob can access multiple channels in one timeslot, and Eve can jam multiple channels in one timeslot. In the Nash equilibium, Alice, Bob and Eve do not always opeate on the same set of channels. k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. But in some cases, Alice and Bob have to access channels fom 1 to k 1. k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob. When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. It is simple to veify that N k + 1 M j + 1. This implies that Alice and Bob will access at least M j + 1 channels. Othewise, if they can access only M j channels, this channel will be jammed by the attacke with pobability 1. In Chapte 9, numeical simulations of ou optimal stategies ae shown. The popeties of optimal stategies ae shown, and we find that the waste case which should be avoided. The compaison between ou optimal stategy and the leaning appoach algoithm is shown. In Chapte 10, conclusions of this thesis ae given.

24 17 Chapte 2 Backgound In this chapte, we intoduce the backgound of stategic games, including zeo-sum game which will be used to model UFH in this thesis in Section 2.1. In Section 2.2, we intoduce and discuss the concept and implications of the Nash equilibium. 2.1 Zeo-sum Game A stategic game is a model used to model the inteaction of a set of decision makes [27]. The concept of stategic game theoy is widely used in economics, political science, and psychology, as well as logic and biology [27]. In game theoy, the decision makes of a game is efeed as playes [27]. In a stategic game model, the inteactions between playes ae affected not only by his own actions, but also by the actions taken by othe playes [27]. Thee is a clea distinction between a stategic game and a one paty optimization poblem [28]. The playes in a stategic game usually do not have complete contol of the esult of thei actions, which suits the case in UFH with intelligent jammes, but in a one paty optimization poblem, the esult can be completely contolled by his own actions, so in stategic game, we usually cannot get a global optimization esult [28]. In stategic games, each playe has his own action set [27], which is the set of actions the playe can take. The action set can be the same fo all playes, and can also be diffeent fo diffeent playes. Each playe in a game should has pefeences ove the action set, which means the playe may pefe some actions moe than othes because those actions can give

25 18 him moe ewad. Oveall, a stategic game consists of a set of playes {1,, N}, a action set Y i = {y i } fo each playe and pefeences ove the action set fo each playe [27]. And the pefeence is a function of stategy in the action set, and is usually called the payoff function u i, which depends not only on his own action but the actions of othe playes of the game. Fom the angle of coopeation, thee ae two types of games: coopeative games and noncoopeative games [27]. In noncoopeative games, zeo-sum game is one of the most impotant fom. Zeo-sum game is usually used to descibe the situation when the playes ae in competitive elationship. Playes in zeo-sum game do not coopeate and the gain of one playe will lead to loss to the othe playe [28]. As the name of zeo-sum game implies, in zeo-sum game the sum of total ewads of all the playes is identically zeo [27]: N u i (y 1,, y N ) = 0. (2.1) i=1 In some games, the sum of the total ewad is not zeo, but a nonzeo constant [27]. We may efe this kind of games including zeo-sum game as constant-sum game [27]. One popety of zeo-sum game which makes it widely used is that, nonzeo constant games can be easily tansfomed to zeo-sum games without changing the natue of the games [27]. So when we model a constant-sum game, we always choose zeo-sum game fo it is the same in natue as nonzeo constant-sum game [27]. 2.2 Nash equilibium As the pevious section stated, in stategic games, the global optimization usually cannot be eached due to the patial contol to the game of each playe [28]. This is due to the fact that the optimal action of one playe depends on the actions taken by othe playes [27]. So when choosing an action a playe must take account into the actions taken by othe playes [27]. So the belief of othe playes is vey impotant. This belief may be fom the past expeience of othe playes and this expeience is sufficient fo the playe to pedict what the opponents will behave [27]. Given the othe playes stategies, if the playe is ational, he can choose the optimal stategy. Unde this cicumstance, the optimal stategy

26 19 of each playe given othe playes stategies is impotant at the optimization aspect [27]. In game theoy, the Nash equilibium is a solution concept of a non-coopeative game involving two o moe playes, in which each playe is assumed to know the equilibium stategies of the othe playes, and no playe has anything to gain by changing only his own stategy unilateally [27]. Definition Let y i denote the stategy of playe i, and y i denote the set of stategies of playes except fo playe i. u i is the payoff function of playe i. y is said to be a Nash equilibium if fo each playe i and evey stategy y i taken by the playe, y is at least as good as the stategy (y i, y i ) in which playe i chooses y i while evey othe playe j chooses yj. That is, fo evey playe i and y i, u i (y ) u i (y i, y i). (2.2) This definition can be explained this way: if each playe has chosen a stategy and no playe can benefit by changing his o he stategy while the othe playes keep theis unchanged, then the cuent set of stategy choices and the coesponding payoffs constitute a Nash equilibium [27]. In the Nash equilibium, changing one s own stategy unilateally can not lead to a geate ewad [27] fo him. Eveyone is taking his best stategy while taking into account the decisions of the othes [27]. Fo the UFH poblem, if the stategy taken by the legitimate tansmitte and eceive is in the Nash equilibium, fom definition it can be concluded that the jamme cannot do bette even if he knows the stategy of the legitimate tansmitte and eceive. So algoithm of leaning the stategy taken by the legitimate tansmitte and eceive and then design a optimal stategy will not wok in the Nash equilibium. This also secues the UFH in the sense of data ate.

27 20 Chapte 3 Model In this chapte, we intoduce the basic model of ou UFH poblem. At the same time, the definition of paametes and the Nash equilibium in ou model ae given. We conside a time-slotted wieless system with N channels, each with channel capacity R i, i = 1,, N. Without loss of geneality, we assume R 1 R 2... R N. Hee, to assist the pesentation, we assume that all teminals can access o jam one channel at any given time slot. The moe geneal case in which the teminals can access o jam moe than one channel at each time slot will be consideed in Chapte 6, Chapte 7 and Chapte 8. In UFH, the tansmitte (Alice) and eceive (Bob) hop andomly though these N channels. We use p t i and p i to denote the pobabilities that Alice and Bob will access channel i at any time slot espectively. Futhemoe, we define p t [p t 1,, pt N ] and p [p 1,, p N ] with p t i = 1 and p i = 1. The jamme will jam channel i with a pobability pj i, and similaly we define p j [p j 1,, pj N ] with p j i = 1. We assume that if Eve chooses to jam a channel, then the communication between Alice and Bob though that channel will fail. The tansmission between Alice and Bob is successful when Alice and Bob use the same channel and at the same time and Eve is not jamming this channel. And we use A, B and E to denote the suppot set of channels, fo channels in the suppot set, Alice, Bob and Eve will access o jam with non-zeo pobabilities. Figue 3.1 shows 3 diffeent scenaios of UFH: 1. In timeslot 1, both tansmitte and eceive ae in channel 5, while Eve is not in

28 21 Figue 3.1: System model. channel 5. The tansmission is successful. 2. In timeslot 2, the tansmitte and eceive ae not in the same channel. The tansmission is failed. 3. In timeslot 3, both tansmitte and eceive ae in channel 3, while Eve is also in channel 3. This tansmission is jammed, so the it is failed. The aveage thoughput of UFH is R = N R i p t ip i i=1 ( ) 1 p j i. (3.1) Clealy, in UFH, Alice and Bob would like to maximize the aveage thoughput, while Eve would like to minimize it. We model this scenaio as a zeo-sum game, with Alice and Bob being one paty and Eve being the othe paty. In this game, the stategy of Alice and Bob is to choose p t and p, and the stategy of Eve is to choose p j. The ewad fo Alice and Bob is R and the ewad fo Eve is R. A stategy pai {(p t, p ), p j } is called a

29 22 Nash equilibium if ( R p t, p, p j ) R ( p t, p, p j ), p t, p, (3.2) R ( p t, p, p j) R ( p t, p, p j ), p j. (3.3) This implies that neithe paty will eceive a lage ewad by unilateally deviate fom this equilibium, hence they have no motivation to do so.

30 23 Chapte 4 Equal Channel Quality Case In this chapte, we study the case when all the channels have the same capacity, which is R 1 = R 2 =... = R N. The Nash equilibium is given at the beginning of this chapte, and the poof is given in Section 4.1. Remaks ae given in Section 4.2. We can denote Lemma In this case, the Nash Equilibium is: fo i {1, 2,..., N}. R i = R, i {1, 2,..., N}. (4.1) p t i = p i = p j 1 i = N, (4.2) This esult is intuitive which is illustated in Figue 4.1, Figue 4.2 and Figue 4.3. Figue 4.1 shows the channel capacity ae the same fo all channels. Figue 4.2 shows the stategy of Alice and Bob, they access all the channels with equal pobability. Figue 4.3 shows the stategy of Eve, she jams all the channels with equal pobability. 4.1 Poof The poof is oganized as: Let A, B and E denote the suppot set fo Alice, Bob and Eve espectively. In Section 4.1.1,we pove E = {1, 2,..., N}. In Section 4.1.2, we pove E A, E B, A = B. So A = B = E = {1, 2,..., N}. In Section 4.1.3, we detemine the Nash equilibium.

31 channel capacity channel capacity channel index Figue 4.1: Channel Capacity (N=15) P t * (=P * ) 0.06 pobability channel index Figue 4.2: P t (= P ) (N=15).

32 P j * 0.06 pobability channel index Figue 4.3: P j (N=15). Poof. The ewad of Alice and Bob accessing channel i is: R i (1 p j i ) ( ) = R 1 p j i. (4.3) The ewad of Eve jamming channel i is: whee S = N j=1 R jp t j p j. N j=1,j i Then we deive the Nash equilibium step by step. R j p t jp j = S + R i p t ip i, (4.4) Pove: E = {1, 2,..., N}. If E {1, 2,..., N}, then i {1, 2,..., N} s.t. p j i = 0. Case 1: p t i = p i = 1, then it is obvious fo Eve to incease his ewad by jamming channel i. So this is not a Nash equilibium. Case 2: p t i 1 and p i 1. Since Eve neve jams channel i, so Alice and Bob can always incease thei ewad by allocating moe pobability into channel i. But they cannot

33 26 achieve maximum ewad by setting p t i = p i = 1. So Alice and Bob can always incease thei ewad by changing thei stategy unilateally. This is not a Nash equilibium. Fom above, we can conclude E = {1, 2,..., N} Pove: E A, E B, A = B. If E A, then Eve is jamming some channel that is neve used by the tansmitte. So Eve can incease his ewad by jamming some othe channel. So E A 1. The poof is the same fo E B. If A B, then Alice is tansmitting on some channel that is can neve used by Bob o Bob is listening on some channel that Alice s message neve comes fom. This means Alice o Bob is wasting he o his esouces. So Alice and Bob can incease thei ewad by allocating thei pobability on the same set of channels. Thus, A = B. The above conclusion implies that if all the channel quality ae equal, then A = B = E = {1, 2,..., N} Detemine p t, p and p j Fist we will show that to achieve the Nash equilibium, 1 p j i = C0 and p t i p i = C 1, whee C 0 and C 1 ae constants independent of i. If 1 p j i C0, then l 1, l 2 {1, 2,..., N} s.t. whee i {1, 2,..., N}, 1 p j l 1 = max{1 p j i }, (4.5) R = R[p t l 1 p l1 ( 1 p j l 1 ) + p t l 2 p l2 ( 1 p j l 2 ) + 1 p j l 1 > 1 p j l 2, (4.6) N i=1 i l 1 i l 2 ( p t i p i 1 p j ) i ]. (4.7) So thee exist anothe stategy of Alice and Bob p t and p, which satisfy that fo i l 1 1 The symbol does not mean pope set in this thesis.

34 27 and i l 2, p t i = p t i, p i = p i, (4.8) p t l 1 > p t l1, p l1 > p l 1, (4.9) p t l 2 < p t l2, p t l2 < p t l2. (4.10) This implies p t l 1 + p t l2 = p t l1 + p t l2, (4.11) p l 1 + p t l 2 = p l1 + p t l 2. (4.12) We can always find stategy p t and p because we have poved A = B = E = {1, 2,..., N}, which means the pobability of accessing each channel in the game is nonzeo. It is obvious that the stategy p t and p can incease the ewad of Alice and Bob. So Alice and Bob can always incease thei ewad by moving thei pobability of accessing channel l 2 into accessing channel l 1. Thus this is not a Nash equilibium. Then 1 p j i = C0. This means that all the p j i ae all equal, so p j i = 1 N. Simila to the poof of 1 p j i = C0, if p t i p i C 1, then l 3, l 4 {1, 2,..., N} s.t. p t l 3 p l3 = max p t ip i, (4.13) p t l 3 p l3 > p t l 4 p l4. (4.14) So Eve can always incease his ewad by moving his pobability of jamming channel l 4 into jamming channel l 3. So this is not a Nash equilibium. Then in the Nash equilibium p t i p i = C 1. Unde the Nash equilibium, R = N i=1 ( C ) = (N 1) C 1. (4.15) N Since Alice and Bob fom one paty of the game, so the we can take the pai (p t, p ) as the stategy of Alice and Bob. So Alice and Bob want maximize R, that is, to maximize C 1. We build two vectos A 1 = [ p p t1, t2,..., p t N ], (4.16) A 2 = [ p 1, p 2,..., p N ], (4.17)

35 28 A 1 and A 2 ae two vecto in R N. Then A 1 = 1 and A 2 = 1. A 1 A 2 = N i=1 p t i p i = N C1 = N C 1, (4.18) i=1 so we can convet the poblem of maximizing C 1 into maximizing A 1 A 2. A 1 A 2 = A 1 A 2 cos θ A 1 A 2 = 1, (4.19) the equality holds when θ = 0, whee θ is the measue of angle between A 1 and A 2. So we can maximize C 1 by setting θ = 0. The two vectos A 1 and A 2 have the same length, and the same diection, so they ae equal. So Then we can detemine p t i = p i = C 1, i {1, 2,..., N}. (4.20) p t i = p i = 1, i {1, 2,..., N}. (4.21) N Fom the poof above, Alice and Bob cannot incease thei ewad by changing to anothe stategy unilateally. So in the equal channel quality scenaio, the Nash Equilibium is fo i {1, 2,..., N}. p t i = p i = p j 1 i = N, (4.22) 4.2 Remak Remak Since the Nash equilibium is obtained, the aveage thoughput is R = R N 1 N 2. (4.23) Remak In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. In paticula, A = B = E = {1,, N}. Remak Alice, Bob and Eve always access o jam all channels. Remak Alice, Bob and Eve access o jam each channel with equal pobability. Remak R is a deceasing function of N fo N 2. Notice R each maximum when N = 2.

36 29 Chapte 5 Geneal Channel Quality Case In this chapte, we study the a moe geneal case, R 1 R 2... R N. We chaacteize the Nash equilibium of the game fo this geneal channel quality case. Duing the deivation, we also study the popeties of the stategies that achieve this equilibium. The Nash equilibium is given at the beginning of this chapte, and the poof is given in Section 5.1. Remaks ae given in Section 5.2. Lemma The unique Nash equilibium of this game is p t i p j i = p i = = 1 1 Ri N l=k 1 Rl, (5.1) N k R i N l=k 1 R l, (5.2) { fo k i N, whee k = min k R k > N k N i=k 1 R i }, and p t i = p i = p j i = 0, i < k. (5.3) Figue 5.1, Figue 5.2 and Figue 5.3 give an example to illustate ou esults. Figue 5.1 shows the channel capacity. Figue 5.2 shows the stategy of Alice and Bob, we can see that Alice and Bob only access channels fom k to N, and when the channel quality is bette, Alice and Bob access this channel with a smalle pobability. Figue 5.3 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability.

37 channel capacity channel capacity channel index Figue 5.1: Channel Quality (N = 15, k = 12). 0.4 P t * (=P * ) 0.3 pobability channel index Figue 5.2: P t (= P ) (N = 15, k = 12).

38 P j * 0.3 pobability channel index Figue 5.3: P j (N = 15, k = 12). 5.1 Poof Poof. The poof is oganized as follows. In Section 5.1.1, we show that in the Nash equilibium if Eve jams channel k with a non-zeo pobability, it will jam all channels that have bette channel qualities with a non-zeo pobability. Hence, thee exists a numbe 1 k N, such that the jamming set E has the fom E = {k, k + 1,..., N}. In Section 5.1.2, we show that E A, E B and A = B. A = B = E and detemine the Nash equilibium. Befoe poceeding to the detailed poof, we have the following facts. 1. The ewad of Alice and Bob both accessing channel i is R i (1 p j i In Section 5.1.3, we show that ). (5.4) 2. The ewad of Eve jamming channel i is whee S N j=1 R jp t j p j. N j=1,j i R j p t jp j = S + R i p t ip i, (5.5)

39 32 Figue 5.4: Poof fo E = {k, k + 1,..., N} Pove: if p j k > 0, then p j i > 0 fo all i > k. Hence, thee exists a numbe k such that E has the fom E = {k, k + 1,..., N}. We will pove this by contadiction. Figue 5.4 shows the idea how to pove this. Suppose this claim is not tue, then in the Nash equilibium stategy of Eve p j, thee exists some k 1 > k such that p j k 1 = 0 and p j k > 0. That is, Eve gives up jamming some channel that is not the wost in he suppot set. In this case, the ewad of Alice and Bob accessing channel k 1 is Then we have R k1 ( 1 p j k 1 ) = R k1. (5.6) ( ) R k1 R k > R k 1 p j. k We then have the following two cases, each of which will lead to a contadiction. Case 1: If p t k = p k = 0, that is Alice and Bob neve use channel k. Then Eve can incease its ewad by educing the pobability of jamming channel k to zeo. This contadicts with the definition of the Nash equilibium. Case 2: If p t k > 0 o p k > 0, then Alice and Bob can incease thei ewads by tansfeing thei pobability fom channel k to channel k 1. definition of the Nash equilibium. This contadicts with the

40 33 Figue 5.5: Poof fo E A. Figue 5.6: Poof fo A = B. This completes the poof that in the Nash equilibium, E must have the fom E = {k, k + 1,..., N} with 1 k N Pove: E A, E B and A = B Figue 5.5 and Figue 5.6 show the idea how to pove this. If E A, namely E\A ϕ, then Eve is jamming some channels that ae neve used by the tansmitte. Hee E\A {k k E and k / A}, and ϕ is the empty set. So Eve can incease his ewad by moving jamming pobabilities fom channels in E\A to channels in A, which contadicts the definition of the Nash equilibium. Hence, in the Nash equilibium, we have E A 1. The poof of E B is the same. If A B, then Alice is tansmitting on some channel that 1 The symbol does not mean pope set in this thesis.

41 34 is can neve used by Bob o Bob is listening on some channel that Alice s message neve comes fom. This means Alice o Bob is wasting he o his esouces. So Alice and Bob can incease thei ewad by allocating thei pobability on the same set of channels. Thus, A = B. Now, we show A = B = {k 1, k 1 + 1,..., N} with k 1 k 1 k. Because we have poved E A, E B and A = B, so we have two cases: Case 1: A = B = E, so k 1 = k. Case 2: E A and E A. Since Eve neve jams channel 1 to k 1, so fo channel 1 to k 1, k 1 i=1 R i p t ip i R k 1 k 1 i=1 p t ip i R k 1 ( k 1 i=1 p t i ) ( k 1 i=1 p i ). (5.7) The equality holds when Alice and Bob access channel 1 to k 2 with pobability zeo. So Alice and Bob should put all thei pobability fom channel 1 to k 2 into channel k 1. Then Alice and Bob should set k 1 = k 1. So A = B = {k 1, k,..., N} Detemine p t, p and p j Figue 5.7, Figue 5.8 and Figue 5.9 show the high level idea how to detemine p t, p and p j. In the Nash equilibium, in the suppot set, the ewad should be equal to a constant. Intuitively, if this is not tue, the othe paty will find the channel with maximum of the ewad and use this channel. Next, the concete poof is povided. Fom the side of Eve, Eve is not going to jam channel k 1. So we have R k 1p t k 1 p k 1 R k pt k p k = R k +1p t k +1 p k +1 =... = R Np t Np N = C 1, (5.8) whee C 1 is a constant. We have R k 1p t k 1 p k 1 C 1, (5.9) because if this inequality does not hold, then Eve can incease he ewad by jamming channel k 1 and thus contadicts with the definition of the Nash equilibium. Fom the side of Alice and Bob, we have R k 1 = R k ) ) ) (1 p j k = R k +1 (1 p j k +1 =... = R N (1 p j N = C 0, (5.10)

42 channel capacity channel capacity channel index Figue 5.7: Channel capacity. Rewad of Alice and Bob fo accessing each channel 15 ewad channel index Figue 5.8: Rewads of Alice and Bob fo accessing each channel.

43 36 0 Rewad of Eve fo jamming each channel ewad channel index Figue 5.9: Rewads of Eve fo accessing each channel. whee C 0 is a constant. So this implies R k 1 < R k R k R N. (5.11) We can find p j fist. Fom the discussion above, we know R i (1 p j i ) = C 0 fo i {k, k + 1,..., N}. Then we have p j i = 1 C 0 R i fom k to N, we have Fom this, we have 1 = N N p j l = l=k l=k fo i {k, k + 1,..., N}. Summing p j i ( 1 C ) 0 R l = (N k + 1) C 0 N l=k 1 R l. C 0 = N k N l=k 1 R l, p j i = 1 N k R i N l=k 1 R l. So we have C 0 < R k R k R N. If R m satisfies R m > N m N i=m 1 R i, (5.12)

44 37 then fo m + 1, R m ( R m ( N N R i=m+1 i R m+1 ( ) 1 R i=m i > N m, ) > N m, N ) 1 R i=m+1 i R m+1 > > N (m + 1), N (m + 1) N i=m+1 1 R i, m + 1 also satisfies the inequality. The thid inequality { use the fact that R m+1 R m. By induction, we can conclude that if k is in the set k R k > N k N }, then all the numbes i=k 1 R i fom k to N ae also in that set. So fo { j < min k R k > } N k N i=k 1, (5.13) R i we have Next, we show that R j k = k m min { N j N i=j 1 R i. (5.14) k R k > } N k N i=k 1. (5.15) R i We show this by contadiction. Suppose k = k k m, we have the following two cases: 1. If k > k m, the ewad of Alice and Bob accessing channel i k is ( ) R i 1 p j i = N k N. (5.16) 1 l=k R l Fom the discussion above, we know that fo i < k, p j i Bob accessing channel k 1 is R k 1. We have k 1 k m, so ( ) N k 1 R k 1 > R k 1 N l=k 1 R l + 1 R k 1 = 0. The ewad of Alice and N l=k 1 > N k + 1, R k 1 > N k N l=k 1 R l = R k, 1 R l (1 p jk ).

45 38 The ight hand side tem is the ewad of Alice and Bob acessing channel k. So the ewad of Alice and Bob accessing channel k 1 is bette than accessing channel k. Hence, Alice and Bob has the motivation to deviate fom this stategy, which contadicts the definition of Nash equilibium. 2. If k < k m, then so p j k R k N k N l=k 1 R l, (5.17) = 1 N k R i N l=k 1 R l 0. This contadicts with ou assumption that E = {k, k +1,..., N} which means p j k > 0. So in the Nash equilibium, k = min { k R k > } N k N i=k 1, (5.18) R i p j i = 1 N k R i N l=k 1, R l i {k, k + 1,..., N}. (5.19) and In the following, we chaacteize p t and p. Fist we show that k 1 = k. If k 1 = k 1, then we have R k 1 = C 0 = R k R k 1. N k N i=k 1 R i

46 39 The total ewad of Alice and Bob is N R = R i p t ip i (1 p j i i=k 1 ) ( N ) = R k 1p t k 1 p k 1 + C 0 = C 0 ( C 0 [ p t k 1 p k 1 + pt k p k + i=k p t ip i N i=k +1 p t ip i (p t k 1 + pt k )(p k 1 + p k ) + N = C 0 [p t k p k + N i=k +1 p t ip i and the equality holds when p t k 1 = p k 1 = 0. ], ) i=k +1 p t ip i ] So the ewad of Alice and Bob will incease if they tansfe thei effot of accessing channel k 1 to accessing channel k. So k 1 = k 1 is not a Nash equilibium. So k 1 = k. Since k 1 = k, we can build two vectos A 1 and A 2, A 1 = [ p t 1, p t 2 p,..., t N ], A 2 = [ p 1, p 2,..., p N ], A 1 and A 2 ae two vecto in R N. Then A 1 = 1 and A 2 = 1. A 1 A 2 = = N p t i p i i=k N C1 i=k R i A 1 A 2 = 1, the equality holds when A 1 and A 2 have the same diection. A1 and A 2 also have same length, so they ae equal. Then C 1 = 1 ( N ) 2. 1 l=k Rl So p t i = p i = C1 R i = 1 Ri N. 1 l=k Rl

47 40 So the Nash equilibium is p t i p j i = p i = = 1 1 Ri N, 1 l=k Rl N k R i N l=k 1 R l, fo k i N, whee k = min{k R k > N k N i=k 1 R i }. 5.2 Remak Remak Since the Nash equilibium is obtained, the aveage thoughput is R = N k ( N l=k 1 Rl ) 2. (5.20) Remak In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. In paticula, A = B = E = {k,, N}. Remak Alice, Bob and Eve do not always access o jam all channels. k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. Figue 5.10 shows k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. Figue 5.10: k sepaates good channels and bad channels. Remak When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability.

48 41 Remak It is simple to veify that N k This implies that Alice and Bob will access at least two channels. Othewise, if they access only one channel, this channel will be jammed by the attacke with pobability 1.

49 42 Chapte 6 One Access Multiple Jamming Case In this chapte, we assume that Eve can jam moe than one channels simultaneously, while Alice and Bob can tansmit and eceive though only one channel each time. Let M j denote the numbe of channels Eve can jam simultaneously, 1 M j N. Let denote the set of channels Eve jams, whee has M j elements. So p j i = i p j. The Nash equilibium is given at the beginning of this chapte, and the poof is given in Section 6.1. Remaks about this case ae given in Section 6.2. Lemma The Nash Equilibium in this case is, p t i = p i = 1 Ri N l=k, (6.1) 1 Rl { fo k i N, whee k = min k R k > (N k+1) M j N i=k 1 R i }, and p j i = 1 (N k+1) M j R i N l=k 1 R l, (6.2) p t i = p i = p j i = 0, i < k. (6.3) Figue 6.1, Figue 6.2 and Figue 6.3 give an example to illustate ou esults with M j = 2. Figue 6.1 shows the channel capacity. Figue 6.2 shows the stategy of Alice

50 channel capacity 250 channel capacity channel index Figue 6.1: Channel Quality (N = 15, k = 12). and Bob, we can see that Alice and Bob only access channels fom k to N, and when the channel quality is bette, Alice and Bob access this channel with a smalle pobability. Figue 6.3 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. 6.1 Poof Poof. Similaly, the can pove that E = {k, k +1,..., N} and A = B = {k 1, k 1 +1,..., N}, k 1 k 1 k. The stategy of Eve in the Nash equilibium be found, then we can get A = B = E. Then we can get the Nash Equilibium using vecto. The ewad of Alice and Bob accessing channel i is R i 1 j p i, whee p j is the pobability fo Eve to choose jamming channel set.

51 P t * (=P * ) 0.3 pobability channel index Figue 6.2: P t (= P ) (N = 15, k = 12). 0.8 P j * 0.6 pobability channel index Figue 6.3: P j (N = 15, k = 12).

52 45 The ewad of Eve jamming channel i is R j p t jp j j / i whee S = j R jp t j p j. = i S + R j p t jp j, j Similaly, the can pove that E = {k, k + 1,..., N} and A = B = {k 1, k 1 + 1,..., N}, k 1 k 1 k. To achieve the Nash equilibium, R i 1 j p i S + i j R j p t j p j whee C 1, C 2 ae constants independent of. Fom we have N i=k R i (N k + 1) (N k + 1) 1 j p i = C 1, i k (6.4) = C 2, i k 1 (6.5) 1 j p i 1 j p i N p i=k i = C 1, = C 1 R i, N =, R i j ( N k M j 1) (N k + 1) ( N k +1 i=k C 1 = C 1 N i=k 1 R i, N 1 ) = C 1, R M j i=k i C 1 = (N k + 1) M j N i=k 1 R i, p j i = Ω i Ω p j = 1 (N k +1) M j R i N l=k 1 R l.

53 46 Similaly, we can pove k = min { } k R k > (N k + 1) M j N i=k 1. (6.6) R i Obviously, k N M j, that is, Eve has at least M j + 1 channels to jam, then Alice and Bob have at least M j + 1 channels to access. Fom S + i noting S is the same fo all diffeent i and j R j p t j { i } = p j = C 2, ( ) N k1 M j 1 (6.7) is independent of i, we can ewite equation 6.7 as whee C 3 is independent of i. R j p t j p j = C 3, (6.8) j i Let us conside two diffeent channels i and j, = = R m p t m p = R m m p t m p m, m m j R m p t m p m + R m p t m m m i i,j i,j i,j / m \{i} i,j / i/,j R m p t m p m + m R m p t m p m = m m \{j} R m p t m p m + R m p t m p m + i,j / i/,j p m R m p t m p m, m R m p t m p m, m i/,j t R i pi p i i,j / t R j pj p j, i/,j

54 47 Notice that m \{i} i,j / i/,j m \{j} whee Ω 1 is a M 1 set. So we have that is R i p i,j / R m p t m p m = R m p t m p m = t i p i = Ω1 i/ Ω 1,j / Ω 1 Ω1 i/ Ω 1,j / Ω 1 R j p i/,j R m p t m p m, m Ω 1 R m p t m p m, m Ω 1 t j p j, {Ω i Ω, j / Ω} R i p t i p i = {Ω i / Ω, j Ω} R j p t j p j. Because so we have {Ω i Ω, j / Ω} = {Ω i / Ω, j Ω} = R i p t i p i = R j p t j p j, ( ) N k 1, M j 2 thus R i p t i p i = C 4, fo all i, whee C 4 is a constant independent of i. Then we tansfom ou poblem into this: Fo Alice and Bob, find the optimal solution of j p i R i p t i p i R = i A = 1 R i p t i p i 1 j p i, (6.9) (N k +1) M j R i N i=k 1 R i, (6.10) = C 3. (6.11) This poblem has been solved in Chapte 5. k 1 = k. Then the Nash equilibium fo Alice and Bob is p t i = p i = Using the same method, we can pove 1 Ri N l=k 1 Rl, k i N.

55 48 So fo the multiple jamming case, the Nash equilibium is p t i = p i = 1 Ri N, 1 i=k Ri p j i = 1 (N k +1) M j R i N l=k 1 R l, { whee k = min k R k > (N k+1) M j N i=k 1 R i }, and (6.12) p t i = p i = p j i = 0, i < k. (6.13) Notice that is a M j -element set and the element ae chosen without eplacement in {k, k + 1,..., N}. So can take ( ) N k +1 M j values. In the Nash equilibium, we have (N k + 1) equations. It can be easily veified that N k So ( ) N k +1 M j > (N k + 1) fo M j 2, which means p j has infinite numbe of solutions. So Eve has infinite numbe of specific stategies in the Nash equilibium, but these stategies have to satisfy j p i (N k +1) Mj R = 1 i N l=k 1 R l. 6.2 Remak Remak Since the Nash equilibium is obtained, the aveage thoughput is R = N k + 1 M j ( N l=k 1 Rl ) 2. (6.14) Remak In the Nash equilibium, Alice, Bob and Eve always opeate on the same set of channels. In paticula, A = B = E = {k,, N}. Remak Alice, Bob and Eve do not always access o jam all channels. k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. Figue 6.4 shows k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1.

56 49 Remak k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob. Figue 6.4: k sepaates good channels and bad channels. Remak When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. Remak It is simple to veify that N k + 1 M j + 1. This implies that Alice and Bob will access at least M j + 1 channels. Othewise, if they access only M j channels, this channel will be jammed by the attacke with pobability 1. Remak Alice and Bob have a unique Nash equilibium, while Eve has infinite numbe of stategies in the Nash equilibium if M j 2. The Nash equilibium of Eve is given in the fom of maginal distibution.

57 50 Chapte 7 Multiple Access One Jamming Case In this chapte, we study the case the sende and eceive can access multiple channels in one time slot, but the attacke can only jam one channel at a time. We assume Alice and Bob can access M t and M channels espectively, whee 1 M t N and 1 M N. The stategy of Alice and Bob taken in a time slot is denoted by Ω A and Ω B. Obviously, Ω A and Ω B ae subset of A and B, and Ω A and Ω B ae M t set and M set espectively. Let p t i denote Ω A p t Ω A, and p i denote Ω B p Ω B. S is a subset of {1, 2,..., N}, then i Ω A i Ω B R S i S R i. And without loss of geneality, we can assume M t M. The Nash equilibium is given at the beginning of this chapte, and the poof is given in Section 7.1. Remaks about this case ae given in Section 7.2. Lemma The Nash equilibium in this case is given unde thee diffeent conditions: 1. Case 1:

58 51 If N k + 1 > M t, p t i = M t Ri N l=k 1 Rl, (7.1) p i = p j i = 1 M Ri N l=k 1 Rl, (7.2) N k R i N l=k 1 R l, (7.3) fo k i N, whee k = min { k R k > N k N i=k 1 R i }, and p t i = p i = p j i = 0, i < k. (7.4) 2. Case 2: If M t N k + 1 > M, p t i = 1, (7.5) fo N M t + 1 i N, p i = p j i = 1 M Ri N l=k 1 Rl, (7.6) N k R i N l=k 1 R l, (7.7) fo k i N, whee k = min { k R k > N k N i=k 1 R i }, and p t i = 0, i < N Mt + 1, (7.8) p i = p j i = 0, i < k. (7.9) 3. Case 3: If N k + 1 M, p t i = M t Ri N, 1 l=k t Rl k t i N

59 52 Figue 7.1: Illustation of case 1 in one access multiple jamming case. Figue 7.2: Illustation of case 2 in one access multiple jamming case. p i = Ri M N, 1 l=k t Rl k t i N { whee k = min k R k > N k p j R i = 1 i N l=k 1, k i N R l N k N i=k 1 R i }, k t = max { k M t Rk N i=k 1 1 Ri Figue 7.1, Figue 7.2 and Figue 7.3 illustate the thee cases espectively, and notice the backets denote the numbe of channels one can access o jam, not the stategy. We can see that case 1 is the case Alice and Bob cannot access all the channels fom k to N, }.

60 53 Figue 7.3: Illustation of case 3 in one access multiple jamming case. case 2 is the case only one of Alice and Bob can access all the channels fom k to N, and case 3 is the case both Alice and Bob can access all the channels fom k to N. Figue 7.4 shows the channel capacity. Figue 7.5 and Figue 7.6 give an example to illustate case 1 with M t = M = 2. Figue 7.5 shows the stategy of Alice and Bob, we can see that Alice and Bob only access channels fom k to N, and when the channel quality is bette, Alice and Bob access this channel with a smalle pobability. Figue 7.6 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. Figue 7.7, Figue 7.8 and Figue 7.9 give an example to illustate case 2 with M t = 4, M = 2. Figue 7.7 shows the stategy of Alice, we can see that Alice takes constant stategy. Figue 7.8 shows the stategy of Bob, we can see that Bob only access channels fom k to N, and when the channel quality is bette, Bob access this channel with a smalle pobability. Figue 7.9 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. Figue 7.10, Figue 7.11 and Figue 7.12 give an example to illustate case 2 with M t = 4, M = 4. Figue 7.10 and Figue 7.11 shows the stategy of Alice and Bob,

61 channel capacity 250 channel capacity channel index Figue 7.4: Channel Quality (N = 15, k = 13). 0.4 P t * (=P * ) 0.3 pobability channel index Figue 7.5: Case 1: P t (= P ) (N = 15, k = 13).

62 P j * 0.3 pobability channel index Figue 7.6: Case 1: P j (N = 15, k = 13). 1 P t * 0.8 pobability channel index Figue 7.7: Case 2: P t (N = 15, k = 13).

63 P * 0.3 pobability channel index Figue 7.8: Case 2: P (N = 15, k = 13). 0.4 P j * 0.3 pobability channel index Figue 7.9: Case 2: P j (N = 15, k = 13).

64 57 1 P t * 0.8 pobability channel index Figue 7.10: Case 3: P t (N = 15, k = 13). 1 P * 0.8 pobability channel index Figue 7.11: Case 3: P (N = 15, k = 13).

65 P j * 0.3 pobability channel index Figue 7.12: Case 3: P j (N = 15, k = 13). we can see that both Alice and Bob take constant stategy. Figue 7.12 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. 7.1 Poof Poof. The ewad of Alice and Bob accessing channel i is R ΩA Ω B \{m}p j m + R m/ ΩA Ω B p j m m Ω A Ω B m/ Ω A Ω B = = i Ω A,i Ω B i Ω A,i Ω B i Ω A,i Ω B (R ΩA Ω B R m ) p j m + R ΩA Ω B 1 m Ω A Ω B m Ω A Ω B R m ( 1 p j m ) m Ω A Ω B p j m (7.10)

66 59 The ewad of Eve jamming channel i is R ΩA Ω B \{i}p t Ω A p Ω B = R ΩA ΩB\{i}p i Ω A Ω B = = t Ω A p Ω B (R R ΩA ΩB i) p i Ω A Ω B R ΩA Ω B p t Ω A p Ω B + R i i/ Ω A Ω B t Ω A p Ω B To achieve the Nash equilibium, in suppot set E, R ΩA Ω B p t Ω A p ΩB + R i whee C 0 is a constant independent of i, notice is constant fo all i, so R i R ΩA Ω B p t Ω A i Ω A ΩB i/ Ω A Ω B t pω A p Ω B i Ω A Ω B p i Ω A Ω B p ΩB R ΩA Ω B p t Ω A p Ω B R ΩA Ω B p t Ω A p Ω B t Ω A p ΩB = C 0 p t Ω A p ΩB = C 1, (7.11) whee C 1 is a constant independent of i. And in suppot set A and B, i Ω A,i Ω B m Ω A Ω B R m whee C 2 is a constant independent of Ω A and Ω B. Fom R i ( 1 p j m ) = C 2, (7.12) p t Ω A p ΩB = C 1, fo and i Ω A ΩB p t i p i i Ω A p t Ω A i Ω B p t Ω B,

67 60 then we have R i p t Ω A p ΩB i Ω A ΩB = R i = R i p t i p i = C 1. i Ω A p t Ω A p Ω B i Ω B Fom i Ω A,i Ω B conside two diffeent channels i and j, = i Ω A,i Ω B i Ω A,i Ω B j Ω A,j Ω B Now we investigate = m Ω A Ω B R m m Ω A Ω B \{i} m Ω A Ω B \{i} i Ω A,i Ω B i Ω A,i Ω B j Ω A,j Ω B + i Ω A,i Ω B j Ω A,j / Ω B m Ω A Ω B \{i} m Ω A Ω B \{j} m Ω A Ω B \{i} m Ω A Ω B R m ( 1 p j m ) = ( 1 p j m ) = C 2, j Ω A,j Ω B m Ω A Ω B R m R m (1 ) ( p j m + R i 1 p j ) i R m (1 ) ( p j m + R j 1 p j j ), R m (1 ) p j m R m (1 p j m ) + R m (1 p j m ) + m Ω A Ω B \{i} i Ω A,i Ω B j / Ω A,j Ω B i Ω A,i Ω B j / Ω A,j / Ω B m Ω A Ω B \{i} ( 1 p j m ), R m (1 ) p j m R m (1 p j m ),

68 61 = m Ω A Ω B \{j} j Ω A,j Ω B j Ω A,j Ω B i Ω A,i Ω B + j Ω A,j Ω B i Ω A,i/ Ω B m Ω A Ω B \{j} m Ω A Ω B \{j} R m (1 ) p j m R m (1 p j m ) + R m (1 p j m ) + m Ω A Ω B \{j} j Ω A,j Ω B i/ Ω A,i Ω B j Ω A,j Ω B i/ Ω A,i/ Ω B m Ω A Ω B \{j} R m (1 ) p j m R m (1 p j m ). Let Ω A1 and denote a M t 1 set, so i Ω A,i Ω B j / Ω A,j Ω B j Ω A,j Ω B i/ Ω A,i Ω B m Ω A Ω B \{i} m Ω A Ω B \{j} R m (1 p j m ) = R m (1 p j m ) = Ω A1,Ω B i/ Ω A1,j Ω B j / Ω A1,i Ω B Ω A1,Ω B i/ Ω A1,j Ω B j / Ω A1,i Ω B m Ω A1 Ω B R m m Ω A1 Ω B R m ( 1 p j m ), ( 1 p j m ). So we have, j Ω A,j Ω B i/ Ω A,i Ω B Similaly, we can pove and i Ω A,i Ω B j Ω A,j / Ω B j Ω A,j Ω B i/ Ω A,i/ Ω B m Ω A Ω B \{j} m Ω A Ω B \{i} m Ω A Ω B \{j} R m (1 p j m ) = R m (1 p j m ) = R m (1 p j m ) = j Ω A,j Ω B i/ Ω A,i Ω B j Ω A,j Ω B i Ω A,i/ Ω B j Ω A,j Ω B i/ Ω A,i/ Ω B m Ω A Ω B \{j} m Ω A Ω B \{j} m Ω A Ω B \{j} R m (1 p j m ). R m (1 ) p j m R m (1 p j m ).

69 62 Notice and = = m Ω A Ω B \{i} i Ω A,i Ω B j Ω A,j Ω B i Ω A,i Ω B j Ω A,j Ω B m Ω A Ω B R m m Ω A Ω B \{j} j Ω A,j Ω B i Ω A,i Ω B j Ω A,j Ω B i Ω A,i Ω B m Ω A Ω B R m R m (1 ) p j m ( 1 p j m ) R m (1 ) p j m ( 1 p j m ) i Ω A,i Ω B j Ω A,j Ω B j Ω A,j Ω B i Ω A,i Ω B R i (1 p j i ), R j (1 p j j ), Hence we have, R i (1 p j i ) = R j (1 p j j ). i Ω A Ω B,j / Ω A Ω B j Ω A Ω B,i/ Ω A Ω B Simila in Chapte 6, R i (1 p j ) ( i = R j 1 p j j ), we have R i (1 p j i ) = C3, i k. Similaly, we can give the solution as p j i = 1 N k R i N i=k 1 R i, (7.13) { fo k i N, whee k = min k R k > N k N i=k 1 R i }. Now we can notice k is detemined by the channel quality, but M t and M ae detemined by uses. Fom discussion in Chapte 5 and Chapte 6, we notice that Alice and Bob have no motivation to access channels fom 1 to k 1. So we should discuss diffeent cases.

70 63 If N k + 1 > M t, fom conclusion in Chapte 5 and Chapte 6 we have k 1 = k, we can simplify ou poblem as, R i p t i p i N p t i i=k N p t i i=k = C 1, i k, = M t, = M. Using the same vecto method Chapte 5 and Chapte 6, we have p t i = Ri M t N, 1 l=k Rl p i = Ri M N, 1 l=k Rl But thee is still one question unansweed: Is p t i < 1 and p i < 1? Now we define we set: { } K 1 = k R k > N k N i=k 1, R i Fo k K 1, This is because so Then we have K 2 = N i=k k R k > N i=k N k N i=k R k R i > N k. R k R i 1, i k, Rk Ri Rk Ri R k R i. N i=k 1 Ri R k R i > N k.. So k is also in K 2. So min K 1 min K 2. So fo k = min{k R k > Rk > N k N l=k 1 Rl M t N. 1 l=k Rl N k N i=k 1 R i },

71 64 Figue 7.13: k sepaates good channels and bad channels. So p t i < 1. The poof is same fo p i < 1. If M t N k + 1 > M, Alice can cove all the channel fom k to N, while Bob cannot. C 3 = N k N R l=k 1 m whee m = 1, 2,..., k 1, and we know Alice and Bob have R l no motivation to access channels except channel k to N. So Alice have to constantly access channels fom N M t + 1 to N, and the stategy of Bob keeps andom. If M N k +1, both Alice and Bob can cove all the channel fom k to N. We know Alice and Bob have no motivation to access channels except channel k to N, howeve, both Alice and Bob have to access some othe channel to access M t and M channels espectively. So both Alice and Bob will access some channels below k. 7.2 Remak Remak Unde the Nash equilibium in this case, Alice, Bob and Eve do not always opeate on the same set of channels. Remak k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. Figue 7.13 shows k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. But in Case 2 and Case 3, Alice and Bob have to access channels fom 1 to k 1. Remak k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob.

72 65 Remak When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. Remak It is simple to veify that N k This implies that Alice and Bob will access at least two channels. Othewise, if they access only one channel, this channel will be jammed by the attacke with pobability 1. Remak Alice and Bob do not have a unique Nash equilibium if M t 2 and M 2, while Eve has unique stategy unde the Nash equilibium. The Nash equilibium of Alice and Bob is given in the fom of maginal distibution.

73 66 Chapte 8 Multiple Access Multiple Jamming Case In this chapte, we study the case Alice, Bob and Eve can access o jam moe than one channel simultaneously. Alice and Bob can access M t and M channels espectively, while Eve can jam M j channels, 1 M t N, 1 M N and 1 M j N. Let Ω A, Ω B and denote the set of channels Alice, Bob and Eve access o jam. Ω A is a M t set, Ω B is a M set and is a M j set. Let p t i denote Ω A p t Ω A, p i denote Ω B p Ω B and p j i denote i Ω A i Ω B p j. And without loss of geneality, we can assume M t M. Figue 8.1 shows an i example of multiple access multiple jamming case with M t = 2, M = 2 and M j = 2. The Nash equilibium is given at the beginning of this chapte, and the poof is given in Section 8.1. In the end, some emaks about this case ae given in Section 8.2. Lemma The Nash equiliium in this case is given unde thee diffeent conditions: 1. Case 1:

74 67 Figue 8.1: Example of multiple access multiple jamming case. If N k + 1 > M t, p t i = M t Ri N l=k 1 Rl, (8.1) p i = M Ri N l=k 1 Rl, (8.2) p j i = 1 N k +1 M j R i N l=k 1 R l, (8.3) { fo k i N, whee k = min k R k > N k+1 M j N i=k 1 R i }, and p t i = p i = p j i = 0, i < k. (8.4) 2. Case 2: If M t N k + 1 > M, p t i = 1, (8.5)

75 68 fo N M t + 1 i N, p i = M Ri N l=k 1 Rl, (8.6) p j i = 1 N k +1 M j R i N l=k 1 R l, (8.7) fo k i N, whee k = min { k R k > N k+1 M j N i=k 1 R i }, and p i = 0, i < N M t + 1, (8.8) p i = p j i = 0, i < k. (8.9) 3. Case 3: If N k + 1 M, p t i = M t Ri N, 1 l=k t Rl k t i N p i = Ri M N, 1 l=k t Rl k t i N p j i = 1 N k +1 M j R i N l=k 1 R l, k i N { { whee k = min k R k > N k+1 M j N i=k 1 R i }, k t = max k M t Rk N i=k 1 1 Ri Figue 8.2, Figue 8.3 and Figue 8.4 illustate the thee cases espectively, and notice the backets denote the numbe of channels one can access o jam, not the stategy. Fom the figues, we can see that case 1 is the case Alice and Bob cannot access all the channels fom k to N, case 2 is the case only one of Alice and Bob can access all the channels fom k to N, and case 3 is the case both Alice and Bob can access all the channels fom k to N. }. Figue 8.5 shows the channel capacity.

76 69 Figue 8.2: Illustation of case 1 in multiple access multiple jamming case. Figue 8.3: Illustation of case 2 in multiple access multiple jamming case.

77 70 Figue 8.4: Illustation of case 3 in multiple access multiple jamming case. 300 channel capacity 250 channel capacity channel index Figue 8.5: Channel Quality (N = 15, k = 12).

78 P t * (=P * ) 0.3 pobability channel index Figue 8.6: Case 1: P t (= P ) (N = 15, k = 12). 0.8 P j * 0.6 pobability channel index Figue 8.7: Case 1: P j (N = 15, k = 12).

79 72 1 P t * 0.8 pobability channel index Figue 8.8: Case 2: P t (N = 15, k = 12). Figue 8.6 and Figue 8.7 give an example to illustate case 1 with M t = M = 2 and M j = 2. Figue 8.6 shows the stategy of Alice and Bob, we can see that Alice and Bob only access channels fom k to N, and when the channel quality is bette, Alice and Bob access this channel with a smalle pobability. Figue 8.7 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. Figue 8.8, Figue 8.9 and Figue 8.10 give an example to illustate case 2 with M t = 6, M = 2 and M j = 2. Figue 8.8 shows the stategy of Alice, we can see that Alice takes constant stategy. Figue 8.9 shows the stategy of Bob, we can see that Bob only access channels fom k to N, and when the channel quality is bette, Bob access this channel with a smalle pobability. Figue 8.10 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. Figue 8.11, Figue 8.12 and Figue 8.13 give an example to illustate case 2 with M t = 6, M = 5 and M j = 2. Figue 8.11 and Figue 8.12 shows the stategy of Alice

80 P * 0.3 pobability channel index Figue 8.9: Case 2: P (N = 15, k = 12). 0.8 P j * 0.6 pobability channel index Figue 8.10: Case 2: P j (N = 15, k = 12).

81 74 1 P t * 0.8 pobability channel index Figue 8.11: Case 3: P t (N = 15, k = 12). 1 P * 0.8 pobability channel index Figue 8.12: Case 3: P (N = 15, k = 12).

82 P j * 0.6 pobability channel index Figue 8.13: Case 3: P j (N = 15, k = 12). and Bob, we can see that both Alice and Bob take constant stategy. Figue 8.13 shows the stategy of Eve, we can see that Eve only jams channels fom k to N, and when the channel quality is bette Eve jams this channel with a lage pobability. 8.1 Poof Poof. The oganization of poof is the same as Chapte 7.

83 76 The ewad of Alice and Bob access channel i is: = = = i Ω A Ω B i Ω A Ω B i Ω A Ω B i Ω A Ω B R ΩA Ω B \ p j Ω A Ω B = R Ω A Ω B 1 R Ω A Ω B R ΩA Ω B p j + Ω A Ω B Ω A Ω B Ω A Ω B p j + R ΩA Ω B \ p j Ω A Ω B R ΩE Ω A Ω B p j (R ΩA Ω B R ΩA Ω B ) p j In the Nash equilibium, i Ω A Ω B R Ω A Ω B Ω A Ω B whee C 0 is a constant independent of i. So we have = i Ω A Ω B j Ω A Ω B R Ω A Ω B R Ω A Ω B whee j is a channel index diffeent fom i. R ΩE Ω A Ω B p j = C 0, Ω A Ω B Ω A Ω B R ΩE Ω A Ω B p j R ΩE Ω A Ω B p j,

84 77 = = i Ω A Ω B R i i Ω A Ω B R i i Ω A Ω B + R Ω A Ω B ( 1 p j ) i + ( 1 p j ) i + Ω A Ω B i Ω A Ω B i Ω A Ω B j Ω A Ω B R ΩE Ω A Ω B p j R ΩA Ω B \{i} ΩE R ΩA Ω B \{i} ΩE R ΩA Ω B \{i} p j R ΩA Ω B \{i} p j i Ω A Ω B j / Ω A Ω B R ΩA Ω B \{i} ΩE R ΩA Ω B \{i} p j, Notice that R ΩA Ω B \{i} ΩE R ΩA Ω B \{i} p j = i Ω A Ω B j / Ω A Ω B j Ω A Ω B i/ Ω A Ω B R ΩA Ω B \{j} ΩE R ΩA Ω B \{j} p j, and R ΩA Ω B \{i,j} ΩE R ΩA Ω B \{i,j} p j = i Ω A Ω B j Ω A Ω B i Ω A Ω B j Ω A Ω B R ΩA Ω B \{i,j} ΩE R ΩA Ω B \{i,j} p j, so we have R i (1 p j ) i = R j (1 p j j ). i/ Ω A Ω B j Ω A Ω B i Ω A Ω B j / Ω A Ω B

85 78 Because so {(Ω A, Ω B ) i / Ω A Ω B, j Ω A Ω B } = {(Ω A, Ω B ) i Ω A Ω B, j / Ω A Ω B }, whee C 1 is a constant independent of i. R i (1 p j i ) = C 1, i A B, The ewad of Eve jamming channel i is: R ΩA Ω B \ p t Ω A p Ω B i = i = i Ω A Ω B In the Nash equilibium, R ΩA Ω B p t Ω A p Ω B R ΩA Ω B p t Ω A i (R ΩA Ω B R ΩA Ω B ) p t Ω A p Ω B + p ΩB Ω A Ω B Ω A Ω B whee C 2 is a constant independent of i. Because is the same fo all i, so i i Ω A Ω B R ΩA Ω B p t Ω A Ω A Ω B = R ΩA Ω B p t Ω A p Ω B. R ΩA Ω B p t Ω A p Ω B R ΩA Ω B p t Ω A p ΩB = C 2, p ΩB R ΩA Ω B p t Ω A p ΩB = C 3, whee C 3 is a constant independent of i. Then, R ΩA Ω B p t Ω A p ΩB i = i Ω A Ω B R p ΩA ΩB ΩE i Ω A Ω B t Ω A p ΩB + i i/ Ω A Ω B Ω A Ω B i E R ΩA Ω B p t Ω A p ΩB

86 79 = i + i = i = i + i j R i p i Ω A Ω B i/ Ω A Ω B Ω A Ω B R i p i Ω A Ω B R i p i Ω A Ω B Ω A Ω B t Ω A p ΩB + i R ΩA Ω B \{i}p t Ω A p ΩB t Ω A p ΩB + t Ω A p ΩB + t R ΩA ΩB ΩE\{i}pΩ A p ΩB i Ω A Ω B i Ω A Ω B i j / Conside a channel j diffeent fom i, notice i j / Ω A Ω B Ω A Ω B R ΩA Ω B \{i}p t Ω A p ΩB R ΩA Ω B \{i}p t Ω A p ΩB = j i/ and we can subtact the common tem R ΩA Ω B \{i}p t Ω A p ΩB R ΩA Ω B \{i}p t Ω A p ΩB Ω A Ω B R ΩA Ω B \{j}p t Ω A p ΩB, R ΩA Ω B \{i,j}p t Ω A p ΩB, i Ω A Ω B j then we have i j / R i p i Ω A Ω B t Ω A p ΩB = i/ j R j p j Ω A Ω B t Ω A p ΩB. Because { i, j / } = { i /, j }, so R i p i Ω A Ω B t Ω A p ΩB = R j p j Ω A Ω B t Ω A p ΩB.

87 80 t R i pω A p ΩB = C 4 i Ω A Ω B whee C 4 is a constant independent of i. And we can ewite it as R i p i p t i = C4 So fa we have R i (1 p j i ) = C 1, i A B R i p i p t i = C 4, i E = M t, i A p t i p i = M, i B i E p j i = M j, whee C 1 and C 4 ae constant independent of i. Fom the discussion in Chapte 6 and Chapte 7, we can give the esult same as the lemma. 8.2 Remak Remak Unde the Nash equilibium in this case, Alice, Bob and Eve do not always opeate on the same set of channels. Remak k is a vaiable that is fo Alice, Bob and Eve to decide on which channels they should take actions. Figue 8.14 shows k sepaates good channels and bad channels. Alice, Bob and Eve have no motivation to access o jam channels fom 1 to k 1. But in Case 2 and Case 3, Alice and Bob have to access channels fom 1 to k 1. Remak k is detemined by the channel capacity and M j, and it is not elated to Alice and Bob.

88 81 Figue 8.14: k sepaates good channels and bad channels. Remak When the channel quality is bette, Eve jams this channel with a lage pobability while Alice and Bob access this channel with a smalle pobability. Remak It is simple to veify that N k + 1 M j + 1. This implies that Alice and Bob will access at least M j + 1 channels. Othewise, if they can access only M j channels, this channel will be jammed by the attacke with pobability 1. Remak The Nash equilibium of Alice, Bob and Eve is given in the fom of maginal distibution.