Concurrent Channel Probing and Data Transmission in Full-duplex MIMO Systems

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1 Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems Zhenzh Qan The Oho State Unversty 2015 Nel Ave. Columbus, Oho qan.209@osu.edu Fe Wu The Oho State Unversty 2015 Nel Ave. Columbus, Oho wu.1973@osu.edu Zzhan Zheng Tulane Unversty 6823 St. Charles Ave. New Orleans, Lousana zzheng3@tulane.edu ABSTRACT Kannan Srnvasan The Oho State Unversty 2015 Nel Ave. Columbus, Oho kannan@cse.oho-state.edu An essental step for achevng multplexng gan n MIMO downlnk systems s to collect accurate channel state nformaton (CSI) from the users. Tradtonally, CSIs have to be collected before any data can be transmtted. Such a sequental scheme ncurs a large feedback overhead, whch substantally lmts the multplexng gan especally n a network wth a large number of users. In ths paper, we propose a novel approach to mtgate the feedback overhead by leveragng the recently developed Full-duplex rados. Our approach s based on the key observaton that usng Full-duplex rados, when the base-staton (BS) s collectng CSI of one user through the uplnk channel, t can use the downlnk channel to smultaneously transmt data to other (non-nterferng) users for whch CSIs are already known. By allowng concurrent channel probng and data transmsson, our scheme can potentally acheve a hgher throughput compared to tradtonal schemes usng Half-duplex rados. The new flexblty ntroduced by our scheme, however, also leads to fundamental challenges n achevng throughout optmal schedulng. In ths paper, we make an ntal effort to ths mportant problem by consderng a smplfed group nterference model. We develop a throughput optmal schedulng polcy wth complexty O((N /I ) I ), where N s the number of users and I s the number of user groups. To further reduce the complexty, we propose a greedy polcy wth complexty O (N log N ) that not only acheves at least 2/3 of the optmal throughput regon, but also outperforms any feasble Half-duplex solutons. We derve the throughput gan offered by Full-duplex under dfferent system parameters and show the advantage of our algorthms through numercal studes. Permsson to make dgtal or hard copes of all or part of ths work for personal or classroom use s granted wthout fee provded that copes are not made or dstrbuted for proft or commercal advantage and that copes bear ths notce and the full ctaton on the frst page. Copyrghts for components of ths work owned by others than ACM must be honored. Abstractng wth credt s permtted. To copy otherwse, or republsh, to post on servers or to redstrbute to lsts, requres pror specfc permsson and/or a fee. Request permssons from permssons@acm.org. Mobhoc 17, July 10-14, 2017, Chenna, Inda 2017 Assocaton for Computng Machnery. ACM ISBN /17/07... $ Ness B. Shroff The Oho State Unversty 2015 Nel Ave. Columbus, Oho shroff.11@osu.edu CCS CONCEPTS Networks Network protocol desgn; Network control algorthms; KEYWORDS Schedulng, Full-duplex, Near-optmal throughput ACM Reference format: Zhenzh Qan, Fe Wu, Zzhan Zheng, Kannan Srnvasan, and Ness B. Shroff Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems. In Proceedngs of Mobhoc 17, Chenna, Inda, July 10-14, 2017, 10 pages. 1 INTRODUCTION Moble data traffc s expected to ncrease at rate of 53% per year by 2020 [1]. Mult-user MIMO (MU-MIMO), whch can potentally ncrease the network capacty lnearly wth the number of users, has been consdered as an mportant technque to confront ths data traffc challenge. Theoretcally, n a system wth M transmt and receve antennas, the throughput usng MU-MIMO can be M tmes of the throughput usng a sngle transmt and receve antenna par [20], where M s commonly referred as the spatal multplexng gan. In ths paper, we consder one mportant applcaton of MU- MIMO,.e., the downlnk wreless cellular network consstng of one Base Staton (BS) equpped wth many antennas and many users each equpped wth one antenna. In such systems, the BS could utlze MU-MIMO to transmt multple data streams to multple users smultaneously. Nevertheless, to take the advantage of MU- MIMO n practce, t s prerequste for the transmtter to learn the accurate channel state nformaton (CSI) of the users [14]. Note that n tradtonal wreless networks, rados can only operate n Halfduplex (HD) mode,.e., a rado cannot transmt and receve packets on the same frequency at the same tme. As a result, tradtonal schemes to harness the multplexng gan of MU-MIMO, e.g., [18, 24], requre that the channel state nformaton (CSI) of the users have to be learned frst before any data can be transmtted. Such a sequental channel learnng scheme ncurs a large overhead when there are a large number of users, whch would n turn substantally lmt the multplexng gans of MU-MIMO, especally f the channel

2 Mobhoc 17, July 10-14, 2017, Chenna, Inda Z. Qan et al. coherence tme s relatvely short [18, 24], the large channel learnng overhead has been a long-standng open problem whch lmts the achevable throughput of MU-MIMO n practce. Recently, Full-duplex (FD) rados [5, 6, 9] have been developed, whch allow smultaneous transmsson and recepton on the same frequency. The avalablty of Full-duplex provdes sgnfcant flexblty n desgnng wreless resource allocaton algorthms. For example, t has been shown that n some cases [22], Full-duplex can almost double the throughput and effectvely mprove spectrum effcency. Ths leads to the followng natural and mportant queston: Is t possble to leverage Full-duplex to address the feedback overhead challenge n Mult-user MIMO downlnk systems? In ths paper, we answer ths queston n the affrmatve. By usng a Full-duplex BS, we are able to break the boundary between the channel learnng phase and the data transmsson phase. As shown n Fg. 1, the BS receves the channel probng sgnal from Alce n round 1 and measures the downlnk channel to Alce assumng channel s recprocal 1. Then n round 2, the BS uses Full-duplex capablty to send data to Alce and receve the probng sgnal from Bob smultaneously, assumng Bob does not nterfere wth Alce. After the BS measures all downlnk channels, the BS operates n MU-MIMO mode n round 3. Compared to Half-duplex systems, once the BS knows the downlnk channel to Alce, t can start transmsson mmedately rather than watng untl the end of the channel learnng phase. Henceforth, we wll refer to ths concept as concurrent channel probng and data transmsson. Alce Base Staton Round 1 Bob Alce Base Staton Round 2 Bob Alce Base Staton Round 3 Fgure 1: Concurrent channel probng and data transmsson. Due to the nterference between users, the performance of concurrent channel probng and data transmsson scheme depends hghly on the set of users selected to send probng sgnals and the orderng of these users. Therefore, the followng mportant queston remans: How do we desgn a low-complexty schedulng polcy that acheves provably good throughput performance under the concurrent channel probng and data transmsson? Whle the desgn of hgh performance schedulng polces have been extensvely studed n tradtonal wreless systems [13], relatvely few efforts [23] have focused on the schedulng problem n Full-duplex systems. In partcular, t s much more challengng to consder ths problem under concurrent channel probng and data transmsson. The reason s that: 1) The orderng of users sendng probng sgnal matters. A user that sends a probng sgnal earler also starts transmsson earler. 2) Wthn one channel coherence tme, the schedulng decsons are coupled n terms of tme and nterference relatons. The rate receved by a certan user depends on 1 Measurng downlnk channel to a user through channel probng from the user s standard n a tme dvson duplex (TDD) system [18, 24]. Bob what tme t transmts the probng sgnal as well as the nterference relatons wth the users scheduled to send probng sgnals later. These two facts make the schedulng problem more complcated and classcal schedulng polces do not apply here. In ths paper, we am to develop a throughput near-optmal schedulng polcy and nvestgate the Full-duplex gan for a varous of network settngs. The key contrbutons of ths paper are summarzed as follows: We develop a schedulng polcy that acheves the optmal throughput regon under concurrent channel probng and data transmsson. Compared to Brute-Force search, the complexty has been decreased from O(N!) to O ((N /I ) I ). To further reduce the schedulng complexty n large systems, we desgn a low-complexty greedy polcy wth complexty O(N log N ) that not only acheves at least 2/3 of the optmal throughput regon but also outperforms any feasble Half-duplex solutons. We conjecture that the real performance of the greedy polcy s very close to the optmal, whch s confrmed by smulatons. We derve the throughput gan offered by Full-duplex under dfferent system parameters and use smulatons to valdate our theoretcal results. The rest of the paper s organzed as follows. We dscuss related works n Secton 2. In Secton 3, we descrbe the system model and problem formulaton. In Secton 4, we develop a throughput optmal polcy whch stablzes the system under any feasble arrval rates. In Secton 5, we desgn a low-complexty greedy polcy and provde provable performance guarantees. In Secton 6, we derve the Full-duplex gan under dfferent network settngs and system parameters. We conduct smulatons to valdate our theoretcal results n Secton 7 and make concludng remarks n Secton 8. 2 RELATED WORK In-band Full-duplex, as an emergng technology n wreless communcaton, was mplemented by combnng RF and baseband nterference cancellaton [5, 6, 9], enablng smultaneous b-drectonal transmsson between a par of nodes. Full-duplex has now been wdely studed n a number of wreless communcaton scenaros. Full-duplex WF-PHY based MIMO rados was frst mplemented n [4], and experments showed that the theoretcal doublng of throughput s practcally acheved. Whle t s hard to make Fullduplex MIMO rados ft n small personal devces, t s feasble to buld a Full-duplex MIMO Base Staton due to bgger sze and more powerful computatonal ablty [11]. In [7, 8], the authors proposed the contnuous feedback channel, whch enables sequental beamformng that update weghts whle also performng downlnk transmsson. The authors showed that the system outperforms ts Half-duplex counterpart and reduced the control overhead at the same tme. Ths work can be vewed as an prelmnary attempt of the dea of concurrent channel probng and data transmsson. However, the authors assumed that users are symmetrc and dd not consder the schedulng problem, whch s the focus of our study here. In addton to the research efforts focused on mplementaton and experments, there have also been several theoretcal works on Full-duplex systems. Although Full-duplex s expected to double the capacty n sngle par of nodes, [21] showed that the nter-lnk

3 Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems Mobhoc 17, July 10-14, 2017, Chenna, Inda nterference and spatal reuse substantally reduces network-level Full-duplex gan, makng t less than 2 n typcal cases. In order to deal wth the ncreasng nter-lnk nterference, [17] presented a new nterference management strategy to acheve a larger rate gan over Half-duplex systems. The capacty regon of mult-channel Full-duplex lnks was characterzed n [15] and rate gan s llustrated for varous channel and cancellaton scenaros. The authors n [22] also nvestgated the achevable throughput performance of MIMO, Full-duplex and ther varants that allow smultaneous actvaton of two RF chans. The schedulng problem n Full-duplex cut-through transmsson was consdered n [23], where the authors characterzed the nterference relatonshp between lnks n the network wth cut-through transmsson and desgned a Q-CSMA type of schedulng algorthm to leverage the flexblty of Full-duplex cut-through transmsson. In contrast to the aforementoned works, ths s the frst work that consders the schedulng problem under concurrent channel probng and data transmsson and provdes analytcal framework to characterze the network-level Full-duplex gan. 3 SYSTEM MODEL We consder the downlnk phase of a sngle-cell Full-duplex MIMO system. There are N users n ths system and each of them s equpped wth only one antenna. The Base Staton (BS) has multple antennas and Full-duplex capablty. In addton, we assume tme s slotted and we consder a dscrete-tme system. We use N denote the set of all users n the system. 3.1 Channel Model We consder a block fadng channel, where the channel state remans the same wthn each tme-slot, but may vary from tme-slot to tme-slot. We assume channel state nformaton (CSI) s only avalable at the user sde at the begnnng of each tme-slot. In order to fully acheve the multplexng gan of MU-MIMO, the BS needs to collect CSI va feedback through the uplnk channel. We assume that channels are recprocal, n whch case a user could send a probng sgnal on ts sngle antenna and the BS, by measurng on ts antennas, learns the downlnk CSI. Any CSI expres by the end of the current tme-slot, and t has to be learned agan n the next tme-slot. In practce, collectng CSI from multple users takes tme and ts overhead s lnear wth respect to the number of the correspondng users. We assume that n one tme-slot, the transmtter can collect CSI from at most K users. Therefore, each tme-slot can be further dvded nto K mn-slots and t takes one mn-slot to learn each CSI. The BS can only transmt one packet per mn-slot to each user whose channel nformaton s already known. In tradtonal Half-duplex systems, CSI collecton and data transmsson must be separated n tme to avod nterference. Data transmsson phase starts only f all desred CSIs are collected. Full-duplex systems, on the other hand, allows data transmsson mmedately after each CSI s collected. 3.2 User Groups Full-duplex capablty does not always offer free lunch, ts performance suffers from complex nterference patterns. One way to characterze nterference s usng user groups whch guarantee no nter-group nterference. Thus, we can break the schedulng problem nto two steps: 1) Gven N users, how to dvde them nto dfferent user groups. 2) Gven group nformaton, how to fnd a schedulng polcy that acheves good throughput performance. Dvdng users nto groups s not easy due to the conflct between nterference constrants and the desre to have more groups and less users n each group. We focus on the second step n ths work and leave the jont problem as the future work. The problem s stll challengng even when the group nformaton s already gven. Assume N users are splt nto I user groups, whch guarantees no nter-group nterference. For example, suppose user u and u j are from dfferent groups, the uplnk stream of user u does not nterfere wth the downlnk stream of user u j. Based on each user s geographcal statstcs, the group nformaton wll be determned once over a much larger tme scale. The group nformaton s assumed to be statc and remans the same n a tme-slot. Fg. 2 s an llustraton of a downlnk system wth 2 user groups. We use д(u) to denote the group ndex of user u, and let G д(u) denote the set of users n group д(u). Alce Base Staton Bob Bob Group 1 Group 2 Fgure 2: A downlnk system wth 2 user groups, the BS receves probng sgnal from Alce and transmts data packets to Bob (channel s already known) smultaneously. 3.3 Traffc Model The BS mantans a queue Q u to store packets requested by each user u. The arrval process to each queue s assumed to be statonary and ergodc. We assume packet arrval and departure both occur at the begnnng of each tme-slot. Let A u [t] denote the number of packet arrvals to queue Q u n tme-slot t. Let R u [t] denote the downlnk rate to queue Q u n tme-slot t. The queue-length Q u [t] evolves as: Q u [t + 1] = max {Q u [t] + A u [t] R u [t], 0}. (1) 3.4 Schedulng Polcy In each tme-slot t, a schedulng polcy P determnes the schedule based on the system state, e.g., queue-length and delay. Such schedule can be descrbed as a schedulng vector f = (u 1,,u K ), whch ndcates that user u sends a probng sgnal n the th mnslot. u = 0 mples that the BS s only transmttng, not learnng any channel n the th mn-slot. 0 element s also consdered as a dummy user from a dummy group wth zero queue-length. Due to nterference constrants, once the BS chooses to learn user u s channel durng the th mn-slot, t wll block all other users n G д(u) from recevng any packet. However, the BS can transmt data packets to users from other groups snce there s no nterference

4 Mobhoc 17, July 10-14, 2017, Chenna, Inda Z. Qan et al. between these groups. We use Ru f to denote the downlnk rate to user u under schedulng vector f. For all = 1,..., K, R u [t] = Ru f f schedulng vector f s adopted n tme-slot t. From now on, we omt the subscrpt [t] when lookng nto the schedule made n a certan tme-slot t. Note that Ru f s the number of mn-slots from + 1 to K such that the group of the scheduled user s dfferent from group д(u ),.e., Ru f = K j=+1 1 {д(u ) д(u j ) }. For example, f f = (u a,u b,u c, 0,, 0) and д(u a ) = д(u b ) д(u c ). From the second mn-slot to the K th mn-slot, there are K 2 users n f such that ts group s other than д(u a ). Thus, Ru f a = K 2. Smlarly, we have Ru f b = K 2 and Ru f c = K 3. Denote the set of feasble schedulng polces as Π. In ths paper, we manly focus on the throughput performance of the system. Frst we defne the optmal throughput regon for any gven system parameters N and K. As n [3, 12], a stochastc queueng network s sad to be stable f t behaves as a dscretetme countable Markov chan and the Markov chan s stable n the followng sense: 1) The set of postve recurrent states s non-empty. 2) It contans a fnte subset such that wth probablty one, ths subset s reached wthn fnte tme from any ntal state. When all the states communcate, stablty s equvalent to the Markov chan beng postve recurrent [16]. The throughput regon Λ P of a schedulng polcy P s defned as the set of arrval rate vectors for whch the network remans stable under ths polcy. Defnton 3.1. (Optmal throughput regon) The optmal throughput regon s defned as the unon of the throughput regons of all possble schedulng polces, whch s denoted by Λ,.e., Λ = Λ P. (2) P Π Defnton 3.2. (Throughput optmal polcy) A schedulng polcy s throughput-optmal f t can stablze any arrval rate vector strctly nsde Λ. 4 OPTIMAL SCHEDULING POLICY In ths secton, we propose a throughput-optmal schedulng polcy to the concurrent probng and transmsson problem. We frst observe that the followng classc result apples to our settng as well. Theorem 4.1. Any polcy that maxmzes the weght w (f) = Q u Ru f n each tme-slot, a.k.a., the MaxWeght schedulng polcy, u N s throughput-optmal. Proof. Please refer to the proof n [19]. From the theorem, t suffces to fnd a schedulng vector f such that the weght w(f) s maxmzed n each tme-slot,.e., f = arg max Q u Ru f. (3) f u N However, t s not trval to fnd a MaxWeght schedule wth low complexty. We note that for tradtonal wreless schedulng under 1-hop nterference, MaxWeght schedulng bols down to fndng a maxmum weghted matchng n each tme-slot, whch can be done n O (N 3 ) where N s the number of nodes. Ths result does not apply to our settng, however, snce the orderng of users sendng probng sgnal matters. A Brute-Force search enumerates all possble permutatons of users, leadng to a hgh complexty of O(N!), whch s nfeasble when N s large. Thus, an nterestng queston s how to fnd a MaxWeght schedule n our settng n a more effcent way. To ths end, we propose the followng algorthm wth complexty O((N /I ) I ) (polynomal when I s a constant regardless of N ). In the algorthm, m ndcates the number of users to be chosen from group, 1 I, and m = (m 1,,m I ) s the user-selecton vector. Algorthm 1 wll be appled to each tme-slot to generate the MaxWeght schedule. Algorthm 1 Search algorthm for MaxWeght Schedule Input: For all u N, group д(u) and queue-length Q u. Output: Schedulng vector f 1: Intalzaton: User-selecton vector m = (0, 0,, 0), ŵ = 0, f = (0, 0,, 0). 2: for all m such that m K do 3: Set schedulng vector f = (0, 0,, 0). 4: Set scheduled user set U = 5: for, 2,, I do 6: Add m users wth longest queue-length from group to U. 7: Fll n schedulng vector f wth users n U, followng the Longest Queue-length Frst order. 8: f w(f) > ŵ then 9: ŵ = w (f) 10: f = f 11: return f For a gven user-selecton vector m, Algorthm 1 pcks m users from group wth longest queue-length, for all = 1, 2,, I. It then generates a canddate schedulng vector f by fllng n users followng the Longest Queue-length Frst (LQF) order. The weght w(f) s evaluated for all possble user-selecton vectors m and ts resultng schedulng vector, Algorthm 1 returns the schedulng vector f that has the maxmum weght. Theorem 4.2. The schedule f returned by Algorthm 1 maxmzes weght w (f). Proof. We dvded the proof nto two steps. For the frst step, we show that the LQF order maxmzes the weght for a gven scheduled user set. Then for the user-selecton part, we show that t s suffcent to evaluate all possble user-selecton vectors m and ts resultng scheduled user set by addng m users wth longest queue-length from each group. The proof detals are provded n our techncal report [2]. Gven m, the schedule yelds maxmum weght s determned by: (1) For each group, add m users wth longest queue-length nto the scheduled user set U (m). (2) Schedule the users from U (m) followng the LQF order. Thus, traversng all possble m wll return the MaxWeght schedule. And ths proves the optmalty of Algorthm 1. Remark Applyng LQF to the set of all users does not guarantee the maxmum. Snce LQF s a myopc rule, t always gves hgher prorty to users wth longer queue-length regardless of ther nterference relatons. In fact, queue-length and nterference relatons both

5 Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems Mobhoc 17, July 10-14, 2017, Chenna, Inda play a key role n ths problem, and we need to do user-selecton to get a good balance between these two factors. 5 A LOW-COMPLEXITY GREEDY POLICY Although Algorthm 1 returns throughput optmal polcy n polynomal tme, the complexty O((N /I ) I ) grows very hgh when the number of groups I s large. It s nterestng to see whether there s any low-complexty polcy that acheves provably good throughput. In ths secton, we propose a greedy algorthm whch ncrementally adds users to the schedule and prove that t acheves at least 2/3 of the optmal throughput regon. In addton, our proposed greedy polcy always acheves a larger throughput regon than any schedulng polces under Half-duplex. 5.1 Greedy Algorthm Descrpton Defnton 5.1. (Margnal Gan) Gven a schedule f = (u 1,,u Ω, 0,, 0) and a user u that s a canddate user to be consdered n j th mn-slot (when evaluatng user u, the frst j 1 scheduled users have already been determned n f), the margnal gan u f, j s defned to be the weght dfference caused by addng user u as the j th element of f, assumng there are no future scheduled users,.e., u f, j = w ( (u 1,,u j 1,u, 0,, 0) ) w ( (u 1,,u j 1, 0,, 0) ). To evaluate the margnal gan of addng user u to the schedule f, we must consder the beneft as well as the cost. The beneft s obvous, we have one more user and t keeps transmttng packets untl the end of the current tme-slot,.e., receves a rate of K j. Hence ts weght contrbuton s Q u (K j). On the other hand, f we schedule user u n j th mn-slot, t wll block the transmsson of the prevously scheduled users that are from the same group д(u). Thus, the weght loss s j 1 Q u 1 {д(u )=д(u) }. Therefore, we have: u f, j j 1 = Q u (K j) Q u 1 {д(u )=д(u) }. (4) A postve margnal gan means that by addng a new user, the weght wll not be decreased. Margnal gan consders queue-length as well as the group nformaton and s able to dscrmnate dfferent cases (e.g., long queue-length & strong nterference v.s. short queuelength & weak nterference). Although the margnal gan s not the actual gan of user u j snce we do not know the future scheduled users, t s stll a good metrc to evaluate the potental gan of addng one canddate user to the current schedule. Moreover, as we wll soon see, the Margnal Gan-based Greedy (MGG) Algorthm acheves good throughput performance. The MGG Algorthm, nspred by Secton 4, we frst sort users accordng to ther queue-lengths, and then start from the user that has the longest queue-length n the system, the MGG Algorthm teratvely evaluates the user u wth next longest queue-length. The MGG Algorthm wll add user u f ts margnal gan s postve, otherwse skp user u and contnue to evaluate the user wth the next longest queue-length untl K users have been scheduled or all N users are all evaluated. The complexty of Algorthm 2 s at most O(N log N ) (comes from the sortng operaton), regardless of the value I takes. Compared to Algorthm 1, Algorthm 2 uses LQF and margnal gan to effcently select valuable users. Agan, applyng LQF only would Algorthm 2 Margnal Gan-based Greedy Algorthm Input: user u N, group д(u) and queue-length Q u. Output: Schedulng vector f G 1: Intalzaton: f G = (0,, 0) 2: Intalzaton: ndex = 1 3: Sort queue-length, assume Q u1 Q u2 Q un 4: for all from 1 to N do 5: f ndex K then 6: f fg,ndex u 0 then 7: Add user u to f G as the ndex th element 8: ndex = ndex + 1 9: return f G work poorly, snce t only gves hgher prorty to those users wth longer queue-length rather than large margnal gan. In fact, the nter-user nterference s very mportant and should not be gnored. 5.2 Performance Analyss The MGG Algorthm s smple, however t sacrfces some throughput performance. In ths secton, we am to provde a theoretcal worst-case lower bound on ts throughput performance. Theorem 5.2. The Greedy Algorthm 2 stablzes at least 2/3- fracton of the arrval vector on the optmal throughput regon,.e., acheves 2/3 of the optmal throughput regon. Proof. From [10], we know that t suffces to show thatw(f G ) 2/3w (f ), where f s the MaxWeght schedule. Consder the users selected by f G and f. Let A denote the set of users shared by both schedules, let B denote the set of users only scheduled n f and let C denote the set of users only scheduled n f G. Remark The MaxWeght schedule s not necessarly unque, but these schedules have the same weght. We can choose any of these schedules to be schedule f here. Remark In practce, users n B could nterfere wth users n A. Here n the proof, we am to show a stronger clam whch assumes that n the MaxWeght schedule, users from B do not nterfere wth users n A and B tself. Defnton 5.3. (Extra weght) Extra weght ϵ s defned to be the weght loss n the MGG schedule caused by nterference from users n C. That s to say, the total weght w (f G ) + ϵ s calculated as f there s no nterference caused by users n C, addng each user n C does not block the downlnk transmsson of all the scheduled users whch are from the same group. We dvde the proof nto two parts, for the frst part, we show that w(f G ) + ϵ w(f ). Then we show that ϵ 1/2w (f G ). Combnng both parts, we know w (f G ) 2/3w (f ), whch concludes the proof. Part 1 In ths part, we want to show that w (f G ) + ϵ w (f ), whch means the weght of the MGG schedule by gnorng the nterference caused by users n C s greater than the weght of the MaxWeght schedule. The followng lemmas llustrate the relatonshp between the MGG schedule and MaxWeght schedule, and these results wll be used later.

6 Mobhoc 17, July 10-14, 2017, Chenna, Inda Z. Qan et al. Lemma 5.4. Consder the MaxWeght schedule f = (u1,,u Ω, 0,, 0). For each 1 Ω, the margnal gan f, u 0. Proof. Please see our techncal report [2]. Remark Smlar to the MGG schedule generated by Algorthm 2, the MaxWeght schedule adds a user only f the margnal gan s non-negatve. The only dfference s that the MGG schedule wll gve hgher prorty to users wth longer queue-length, whereas the MaxWeght schedule may skp some users wth long queue lengths and choose other users wth large margnal gan. In the MaxWeght schedule, for each user u A B, we use t 1 (u) to denote the mn-slot that user u s scheduled. In the MGG schedule, for each user u N we defne t 2 (u) to be the mn-slot that ts margnal gan s evaluated (ether schedule u or skp u n t 2 (u) th mn-slot), f u has never been consdered as a canddate, t 2 (u) = K. Lemma 5.5. In the MaxWeght schedule, for each b B, consder user d whch has the longest queue-length among all users n group д(b) that are not scheduled n the MGG schedule. We have: t 1 (b) < t 2 (d),.e., b s scheduled earler n the MaxWeght schedule than the tme that d s skpped n the MGG schedule. Proof. Please see our techncal report [2]. Defne N B (t) and N C (t) to be the number of users n B and C scheduled n the MaxWeght and MGG schedule from the frst mn-slot to t th mn-slot. We have the followng lemma: Lemma 5.6. For each b B, whch s scheduled n t 1 (b) th mnslot, we have N B (t 1 (b)) N C (t 1 (b)). Proof. Please see our techncal report [2]. From Lemma 5.6, we can fnd a mappng h : B C, th user b n B corresponds to th user c n C, such that c s always scheduled earler than b,.e., t 1 (b ) t 2 (c ). For each user b, consder user d whch has the longest queue-length among all users n group д(b ) that are not scheduled n the MGG schedule. Note that users from group д(b ) only belongs to A or B, user d has the longest queue-length among all users n B G д(b ), thus Q d Q b. From Lemma 5.5, we know t 1 (b ) < t 2 (d ) and thus t 2 (c ) < t 2 (d ). Then Q c Q d due to the LQF order of evaluatng users n the MGG polcy. Therefore, Q c Q b. Lemma 5.7. The MGG schedule wll schedule more users than the MaxWeght schedule,.e., B C. Proof. Please see our techncal report [2]. Now we are ready to prove the result of part 1. Comparew(f G )+ϵ wth w(f ), we have two knds of losses. A loss: For each user a A, a wll be scheduled no earler n the MGG schedule than that n the MaxWeght schedule,.e., t 1 (a) t 2 (a) (corollary of Lemma 5.6). Each user a n the MGG schedule wll receve lower or equal rate than that n the MaxWeght schedule. B loss: In the MGG schedule, there s no weght contrbuted by users n B. If the total weght of the users n C can be used to cover A and B losses, then w (f G ) + ϵ w (f ) holds. Frst, we consder A loss: let Loss a denote the weght loss on user a. Loss a = Q a (K t 1 (a ) {a A a s scheduled after a n f } ) Q a (K t 2 (a ) {a A a s scheduled after a n f G } ) = Q a (t 2 (a ) t 1 (a )) 0. (5) Smlarly, we use Loss b to denote the weght loss on user b : Loss b = Q b (K t 1 (b )) 0. (6) The weght dfference w(f G ) + ϵ w (f ) s the total weght of C mnus A loss and B loss: w (f G ) + ϵ w (f ) C A B = Q c (K t 2 (c )) Loss a Loss b C A = Q c (K t 2 (c )) Q a (t 2 (a ) t 1 (a )) B Q b (K t 1 (b )) (d ) B C Q c (t 1 (b ) t 2 (c )) + Q c (K t 2 (c )) = B +1 (e) = A Q a (t 2 (a ) t 1 (a )) C A Q c (t 1 (b ) t 2 (c )) Q a (t 2 (a ) t 1 (a )). (7) where nequalty (d) comes from the property of mappng h and equaton (e) s derved by settng t 1 (b ) = K for any dummy user b, B < C. Note that for each, t 1 (b ) t 2 (c ) 0 and t 2 (a ) t 1 (a ) 0. Lemma 5.8. The R. H. S. of (7) s non-negatve. Proof. Please see APPENDIX A. The result of Lemma 5.8 concludes the proof of part 1. Part 2 In ths part, we want to show that ϵ 1/2w (f G ),.e., the extra weght s upper bounded by one half of the weght of the MGG schedule. We use ϵ and w (f G ) to denote the extra and actual weght from group. It suffces to show a stronger (per-group) clam: For each group, we have ϵ 1/2w (f G ). For each group, note that we only need to consder the worst case where all the users from group are n C. Otherwse, assume there are some users n A, then w (f G ) remans the same whle ϵ s smaller. Lemma 5.9. Assume n the MGG schedule, we havem users (u 1,, u m, wth queue-length Q u1 Q um ) from group, defne T m to be the smallest rate of the last scheduled user such that the MGG schedule s feasble (margnal gan s always non-negatve). Consder

7 Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems Mobhoc 17, July 10-14, 2017, Chenna, Inda the case K = K m T m + t 2 (u m ), we have ϵ K ( ) m 1/2w f G Km, where ϵ K ( ) m and w f G Km are extra weght and actual weght of f G from group under K m. Proof. Please see APPENDIX B. Note that K m s the smallest value of K such that the MGG schedule s feasble, for any K K m, extra weght ϵ wll be the same snce t s only related to u 1,,u m, however, w (f G ) wll ncrease wth K. ϵ K w (f K G ) ϵ K m w (f K G 1/2. (8) ) m Therefore, we know for every feasble MGG schedule, ϵ /w (f G ) s less than one half for any group = 1,, I. We fnsh the proof of part 2 and now we are able to show w (f G ) 2/3w(f ). Proposton The 2/3 worst-case lower bound s tght n terms of weght. Proof. Assume K = 2 r for some postve nteger r > 0. All the users have the same queue-length, and there are K 1 groups where each group has suffcently many users. Then the MaxWeght schedule wll serve K 1 users, one for each group, whch gves a total rate of K (K 1)/2, whle the MGG Algorthm serves K/2 users from group 1, K/4 users from group 2, and 1 user from group r, whch gves a total rate of (K 2 1)/3. As K, the effcency rato becomes arbtrarly close to 2/3. Theorem The throughput regon of the proposed MGG polcy s no smaller than the optmal throughput regon under Half-duplex. Proof. We frst prove the followng lemma, whch shows that the weght of MGG polcy domnates the weght of any Half-duplex polcy. Lemma The weght of the MGG polcy s no smaller than the maxmum weght under Half-duplex,.e., w (f G ) w H D, where w H D ( ) s the total weght calculated under Half-duplex. Proof. Please see our techncal report [2]. Now we need to show that the MGG polcy stablzes any arrval vector λ = (λ 1,, λ n ) wthn the optmal throughput regon under Half-duplex Λ H D. The followng lemma can be used to prove ths clam. Lemma Consder the capacty regon Λ H D under Half-duplex, wh D s the maxmum weght among all feasble schedulng polces under Half-duplex. If there exsts a Full-duplex schedulng polcy f G, such that w(f G ) wh D (f) for any queue-length vector, then polcy f G can stablze any arrval vector wthn Λ H D. Proof. Please see our techncal report [2]. Applyng Lemma 5.12 and 5.13, Theorem 5.11 follows. Remark Other promsng low-complexty algorthms, such as greedly select users wth the largest margnal gan or smply adopt certan amount of users from each group cannot work well ether n the comparson wth tradtonal Half-duplex schemes or under heterogeneous traffc arrvals. 6 CAPACITY GAIN OF FULL-DUPLEX OVER HALF-DUPLEX In ths secton, we wll dscuss the capacty gan of Full-duplex over Half-duplex. Let Λ F D and Λ H D denote the capacty regon under Full-duplex and Half-duplex mode, respectvely. To smplfy, we only evaluate the capacty magntude ν F D and ν H D along the (1,, 1) vector (e.g., (ν F D,,ν F D ) s the largest arrval vector such that all users have the same arrval rate and the queung system can be stablzed under Full-duplex mode). In addton, we assume all groups have the same sze,.e., N 1 = = N I = N /I. For Half-duplex, f the sum-rate s upper bounded by B H D, then the lowest servce rate s upper bounded by B H D /N. Accordng to the basc queung theory, ν H D B H D /N. The sum-rate s calculated by: N R H D = K I I m j m j. (9) where m j s the j th element n the user-selecton vector. If N K/2, the maxmum of the sum-rate s acheved by takng I m j = K/2, thus the upper bound B H D = K 2 4. Otherwse, f K s larger, the maxmum s acheved by schedulng all users n the system, B H D = (K N ) N. To sum up, ν H D = K 2 4N, K N, N K/2, otherwse. (10) Next, we wll look at the Full-duplex case, consder a randomzed polcy P whch uses random schedules from tme-slot to tme-slot, denote ts sum-rate as B F D. Snce the optmal throughput regon s the unon of the throuutghput regons of all possble schedulng polces, we have ν F D B F D /N. The sum-rate under f s calculated by: N I R f = m j m k + I I K m j m j. (11) k <j where m j s the j th element n the user-selecton vector m. The frst term of the R. H. S. of (11) calculates the total rate from the frst mn-slot to I m th j mn-slot, we only need to count the number of user pars (u,u j ) such that д(u ) д(u j ) and u s scheduled before u j. After I m th j mn-slot, all scheduled user wll have K I m j addtonal rate. The total rate from the remanng mn-slot s just ( K I m j ) I m j. To get the upper bound of the sum-rate, we need to solve the followng maxmzaton problem. I maxmze m j m m k + I I K m j m j k <j subject to m N /I,m N, for all = 1, 2,, I. If N /I I K +1 for all = 1, 2,, I, then the maxmum s acheved by takng m = I K +1 for all = 1, 2,, I. In ths case, B F D = I K 2 2(I +1). Otherwse, the maxmum s acheved by takng m = N /I for all.

8 Mobhoc 17, July 10-14, 2017, Chenna, Inda B F D = N (2I K N I N ) 2I. In a word, ν F D = I K 2 2N (I +1), N I K I +1, 2I K N I N 2I, otherwse. Defne Full-duplex gan G F D = ν F D ν H D, α = K/N. We have: (12) 2I I +1, α < I +1 I, G F D = 2(2Iα 1 I ) I α 2, I +1 I α < 2, (13) 1 + 2I (α I 1 1), α 2. Fx group number I = 10, Fg. 3 shows the Full-duplex gan G F D for dfferent α. As we can see n the fgure, f α s smaller than 1.1, 1.9 Full-duplex Gan n Sum-rate G FD α=1.0 α=1.5 α=3.0 Z. Qan et al Group Number I Fgure 4: Full-duplex gan versus group number I, when the K/N rato (α) s fxed. 1.8 Full-duplex Gan G FD K/N rato (α) Fgure 3: Full-duplex gan versus α, when the group number I = 10. Full-duplex gan G F D remans larger than 1.8. In ths regme, the number of users N s larger than (or comparable to) K, whch means the learnng phase takes as long as nearly K/2 mn-slots. Note that the Full-duplex gan comes from concurrent channel probng and data transmsson, the longer learnng phase takes, the larger G F D wll be observed. On the other hand, when α becomes larger, G F D decreases from 1.82 to Ths s because the learnng phase s neglgble compared to K, thus we don t have much gan compared to the tradtonal schemes. In general, when I becomes larger, the upper bound of the G F D becomes closer to 2, whch matches the expected potental of the Full-duplex gan. Fx α to be 1.0, 1.5 and 3, Fg. 4 shows how does the Full-duplex gan G F D change wth dfferent group number I. From Fg. 4, we can observe that the Full-duplex gan G F D keeps ncreasng as I becomes larger. The scheduler has more flexblty when gven more groups, thus a larger Full-duplex gan should be expected. Moreover, n many user regme (green and blue curve), G F D has mproved by 40% and 30% when I ncreases from 2 to 15. However, G F D does not mprove much n small user regme (red curve). The learnng phase only takes a small fracton of tme, thus G F D s always a lttle larger than 1.1, regardless of what value I takes. 7 NUMERICAL RESULTS In ths secton, we use smulatons to evaluate our proposed greedy polcy and compare ts performance wth tradtonal Half-duplex and Full-duplex MaxWeght Schedulng (MWS) schemes. 7.1 Smulaton Settngs We consder the downlnk system of a sngle-cell Full-duplex MIMO system. There are N users n ths system and each user s equpped wth only one antenna. The BS s assumed to have suffcently large number of antennas. Suppose all users are dvded nto I user groups such that users from dfferent group does not nterfere wth each other. Unlke the assumpton we make n Secton 6, each user group now could have dfferent group sze. In addton, we assume that each tme-slot has 15 mn-slots,.e., K = 15. We consder..d. arrval,.e., A u [t] = K, wth probablty λ, 0, otherwse. where λ s the scaled arrval rate of queue u, u N. 7.2 Performance of Greedy Polcy under Dfferent Regmes Fx group number I = 4, we then evaluate the performance of the proposed greedy polcy n three regmes whch represent three condtons of (13). Defne regme 1 as the many-user regme such that α In regme 1, we take N 1 = 8, N 2 = 5, N 3 = 6, N 4 = 1, wth sum N = 20 and α = Regme 2 denotes the moderate regme, where N s comparable wth K such that 1.25 α 2. In regme 2, N 1 = 3, N 2 = 2, N 3 = 2, N 4 = 3, wth sum N = 10 and α = 1.5. Regme 3 represents the small-user regme such that α 2. In regme 3, we take N 1 = 1, N 2 = 1, N 3 = 1, N 4 = 1, wth sum N = 4 and α = For all these three scenaros, we plot the average queue-length under dfferent arrval rate λ n Fg. 5. In all three regmes, the performance of the MGG polcy s very close to the Full-duplex MaxWeght polcy. Thus, the throughput performance of the MGG polcy s also very close to optmal. The Full-duplex gan s larger f α s small, meanng K s smaller compared to N. In ths case, the control overhead of sendng probng sgnals becomes the system bottleneck. Introducng Full-duplex reduces the control overhead and thus the throughput s mproved substantally. As α becomes larger, the control overhead no longer lmts the throughput, snce t only takes a small fracton of tme to send probng sgnals. As a result, Full-duplex gan decreases from 1.5 to 1.13 from as α ncreases from 0.75 to 3.75.

9 Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems Mobhoc 17, July 10-14, 2017, Chenna, Inda Average Queue-length 10 5 Half-duplex MWS MGG Polcy 10 4 Full-duplex MWS Average Queue-length 10 3 Half-duplex MWS MGG Polcy Full-duplex MWS Average Queue-length 10 3 Half-duplex MWS MGG Polcy Full-duplex MWS Arrval rate λ (as fracton of K) (a) Regme Arrval rate λ (as fracton of K) (b) Regme Arrval rate λ (as fracton of K) (c) Regme 3 Fgure 5: Average queue-length under dfferent arrval rate. 7.3 Performance of Greedy Polcy under Random Group Assgnments Gven N users, the way of assgnng users to dfferent groups affects the Full-duplex gan. In ths secton, we would lke to evaluate throughput performance under random group assgnments. Fx group number I = 4, number of users N = 10 and K = 15. Assume that each user has equal probablty to be assgned to each group, the followng fgure shows the emprcal CDF of the Full-duplex gan for samples of random group assgnments. CDF MGG Polcy Full-duplex MWS Full-duplex Gan Fgure 6: The emprcal CDF for Full-duplex gan compared to Half-duplex throughput optmal polcy From Fg. 6, we can observe that the Full-duplex gan of the MGG polcy and MaxWeght polcy have smlar dstrbutons. Although n theory there may exst scenaros n whch the MGG polcy s suboptmal, n typcal scenaros t acheves near-optmal throughput performance. The medan Full-duplex gan under the MaxWeght schedulng and the MGG polcy s around Although the lowest Full-duplex gan s around 1.3, n typcal scenaros (90% of all samples), the Full-duplex gan s larger than 1.44 (44% mprovement). 8 CONCLUSION In ths paper, we develop a throughput optmal schedulng polcy for concurrent channel probng and data transmsson scheme. To further reduce the complexty when there are a large number of groups, we propose a greedy polcy wth complexty O (N log N ) that not only acheves at least 2/3 of the optmal throughput regon but also outperforms any feasble Half-duplex solutons. Furthermore, we derve the Full-duplex gan for dfferent system parameters. Fnally, we use numercal smulatons to valdate our theoretcal results. ACKNOWLEDGMENTS Ths work has been supported n part by NSF grants CNS , CNS , CNS , CNS and CNS , Offce of Naval Research grant N , and from the Army Research Offce grant W911NF REFERENCES [1] Whte paper: Csco VNI Forecast and Methodology, http: // (2016). [2] Concurrent Channel Probng and Data Transmsson n Full-duplex MIMO Systems. (May. 2017). [3] Matthew Andrews, Krshnan Kumaran, Kavta Ramanan, Alexander Stolyar, Rajv Vjayakumar, and Phl Whtng Schedulng n a queung system wth asynchronously varyng servce rates. Probablty n the Engneerng and Informatonal Scences 18, 02 (2004), [4] Dnesh Bharada and Sachn Katt Full duplex MIMO rados. In USENIX NSDI [5] Dnesh Bharada, Emly McMln, and Sachn Katt Full duplex rados. ACM SIGCOMM Computer Communcaton Revew 43, 4 (2013), [6] Jung Il Cho, Mayank Jan, Kannan Srnvasan, Phl Levs, and Sachn Katt Achevng sngle channel, full duplex wreless communcaton. In ACM MOBICOM. ACM, [7] Xu Du, John Tadrous, Chrs Dck, and Ashutosh Sabharwal MIMO broadcast channel wth contnuous feedback usng full-duplex rados. In Aslomar Conference on Sgnals, Systems and Computers. IEEE, [8] Xu Du, John Tadrous, Chrs Dck, and Ashutosh Sabharwal MU-MIMO beamformng wth full-duplex open-loop tranng. In Internatonal Workshop on Sgnal Processng Advances n Wreless Communcatons (SPAWC). IEEE, [9] Melssa Duarte, Chrs Dck, and Ashutosh Sabharwal Experment-drven characterzaton of full-duplex wreless systems. IEEE Transactons on Wreless Communcatons 11, 12 (2012), [10] Atlla Erylmaz, Rayadurgam Srkant, and James R Perkns Stable schedulng polces for fadng wreless channels. IEEE/ACM Transactons on Networkng 13, 2 (2005), [11] Evan Everett and Ashutosh Sabharwal Spatal degrees-of-freedom n largearray full-duplex: the mpact of backscatterng. EURASIP Journal on Wreless Communcatons and Networkng 2016, 1 (2016), 286. [12] Bo J, Gagan R Gupta, Manu Sharma, Xaojun Ln, and Ness B Shroff Achevng optmal throughput and near-optmal asymptotc delay performance n multchannel wreless networks wth low complexty: a practcal greedy schedulng polcy. IEEE/ACM Transactons on Networkng 23, 3 (2015), [13] Xaojun Ln, Ness B Shroff, and Rayadurgam Srkant A tutoral on crosslayer optmzaton n wreless networks. IEEE Journal on Selected areas n Communcatons 24, 8 (2006), [14] Ja Lu, Atlla Erylmaz, Ness B Shroff, and Elzabeth S Bentley Understandng the mpact of lmted channel state nformaton on massve MIMO network performances. In ACM MOBIHOC [15] Jelena Maraševć and Gl Zussman On the Capacty Regons of Sngle- Channel and Mult-Channel Full-Duplex Lnks. arxv preprnt arxv: (2016). [16] Mchael J Neely Delay-based network utlty maxmzaton. IEEE/ACM Transactons on Networkng 21, 1 (2013),

10 Mobhoc 17, July 10-14, 2017, Chenna, Inda Z. Qan et al. [17] Achaleshwar Saha, Suhas Dggav, and Ashutosh Sabharwal On uplnk/downlnk full-duplex networks. In Aslomar Conference on Sgnals, Systems and Computers. IEEE, [18] Quentn H Spencer, Chrstan B Peel, A Lee Swndlehurst, and Martn Haardt An ntroducton to the mult-user MIMO downlnk. IEEE Communcatons Magazne 42, 10 (2004), [19] Leandros Tassulas and Anthony Ephremdes Stablty propertes of constraned queueng systems and schedulng polces for maxmum throughput n multhop rado networks. IEEE Trans. Automat. Control 37, 12 (1992), [20] Davd Tse and Pramod Vswanath Fundamentals of wreless communcaton. Cambrdge unversty press. [21] Xufeng Xe and Xnyu Zhang Does full-duplex double the capacty of wreless networks?. In IEEE INFOCOM [22] Yang Yang, Bo Chen, Kannan Srnvasan, and Ness B Shroff Characterzng the achevable throughput n wreless networks wth two actve RF chans. In IEEE INFOCOM [23] Yang Yang and Ness B Shroff Schedulng n wreless networks wth full-duplex cut-through transmsson. In IEEE INFOCOM [24] Anfu Zhou, Teng We, Xnyu Zhang, Mn Lu, and Zhongcheng L Sgnpost: Scalable MU-MIMO sgnalng wth zero CSI feedback. In ACM MOBIHOC. ACM, A PROOF OF LEMMA 5.8 Defnton A.1. (Avalable rate) Let S (t) denote the avalable rate n t th mn-slot: S (t + 1) = S (t) + t 1 (b j ) t 2 (c j ) f u G t = c j, (t 2 (a j ) t 1 (a j )) f u G t = a j. where the ntal value S (0) = 0 and ut G s the t th element n the MGG schedule f G. Start wth the frst scheduled user n f G, f we encounter wth a user from C, then the avalable rate wll be added t 1 (b j ) t 2 (c j ) more rates offered by users n C. Otherwse, the avalable rate wll be deducted by A loss rate t 2 (a j ) t 1 (a j ). In general, S(t + 1) s the sum of avalable rate of queue-length no smaller than Q u G. t The defnton of S (t) allows us to decouple the queue-length from ts rate, and to evaluate (7), we only need to compare the avalable rate" and A loss rate". If for any 1 t K, S (t) s always nonnegatve, then the R. H. S. of (7) s also non-negatve. Consder each t such that ut G A, S(t + 1) 0 means the sum of avalable rate receved by users wth queue-length hgher than Q u G s larger than t the A loss rate on user ut G. That s to say, for each a, there wll be suffcently many rate offered by users n C whch have longer queue-length than Q a. It s suffcent to show that the R. H. S. of (7) s non-negatve. On the other hand, we can rewrte the recurson formula of S(t) as: S (t + 1) = S(t) + t 1 (b j ) t 2 (c j ) f ut G = c j, t 1 (a j ) t 2 (a j ) f ut G = a j. Start from t = 1, for each user ut G, S(t) ncrements by the tme dfference of schedulng the same user or the correspondng user under mappng h. Thus, S (t) s actually the dfference between the sum of t dfferent tmestamps n MaxWeght schedule and the sum of t consecutve tmestamps from 1, 2, up to t. The later sum s the mnmum of the sum of t dfferent tmestamps, hence S(t) 0 holds for any 1 t K. B PROOF OF LEMMA 5.9 We use mathematcal nducton to prove ths lemma. Base Case: If m = 1, t s the trval case, snce ϵ K 1 = 0. If m = 2, we have: ϵ K 2 w (f K G ) = Q u1. (14) Q u1 (T 2 + t 2 (u 2 ) t 2 (u 1 ) 1) + Q u2 T 2 2 We know that Q u2 T 2 Q u1 and T 2 1, t 2 (u 2 ) t 2 (u 1 ) 1. ϵ K 2 Hence Q u1 (T 2 + t 2 (u 2 ) t 2 (u 1 ) 1) + Q u2 T 2 2Q u1 and w (fk G ) 2 1/2. Inductve hypothess: Assume the lemma holds for m users ϵ Km from group,.e., ( ) 1/2. w f G Km Inductve step: consder the case where we have m + 1 users from group (Q u1 Q u2 Q um+1 ). T m+1 must satsfy: Q um+1 T m+1 Q uj. (15) Q um+1 (T m+1 1) < Q uj. (16) User u 1,u 2,,u m wll determne T m : We then evaluate ϵ K m+1 /w (f G K m+1 ): ϵ K m+1 w ( f G Km+1 ) ϵ K m = ( ) w f G Km + Qum+1 T m+1 + m m 1 Q um T m Q uj. (17) m 1 Q um (T m 1) < Q uj. (18) + m Q uj. (19) Q uj (K m+1 K m 1) Gven the nductve hypothess, t suffces to show 2 Q uj Q um+1 T m+1 + Q uj (K m+1 K m 1). (20) From (15), we already know m Q uj Q um+1 T m+1. We only need to show m Q uj m Q uj (K m+1 K m 1), or equvalently, K m+1 K m 1 1. By defnton, K m+1 = t 2 (u m+1 ) + T m+1 t 2 (u m ) T m+1, K m = t 2 (u m ) + T m. The only thng left s to show T m+1 T m 1 (T m+1 > T m ). Suppose T m T m+1, from (18), we know: m 1 Q um T m < Q uj + Q um = Q uj. (21) Then, Q um+1 T m+1 Q um T m < Q uj. (22) (22) contradcts (15), therefore, T m+1 > T m, T m+1 T m 1, Lemma 5.9 holds.