Scheduling and Resource Allocation in OFDMA Wireless Systems

Transcription

1 Schedulng and Resource Allocaton n OFDMA Wreless Systems Janwe Huang, Vjay Subramanan, Randall Berry, and Rajeev Agrawal February 2009 Book chapter n Orthogonal Frequency Dvson Multple Access Fundamentals and Applcatons. 1 Introducton Schedulng and resource allocaton are essental components of wreless data systems. Here by schedulng we refer the problem of determnng whch users wll be actve n a gven tme-slot; resource allocaton refers to the problem of allocatng physcallayer resources such as bandwdth and power among these actve users. In modern wreless data systems, frequent channel qualty feedback s avalable enablng both the scheduled users and the allocaton of physcal layer resources to be dynamcally adapted based on the users channel condtons and qualty of servce QoS requrements. Ths has led to a great deal of nterest both n practce and n the research communty on varous channel aware schedulng and resource allocaton algorthms. Many of these algorthms can be vewed as gradent-based algorthms, whch select the transmsson rate vector that maxmzes the projecton onto the gradent of the system s total utlty [1 4, 8, 9, 25, 28, 29]. One example s the proportonally far rule [3,4] frst proposed for CDMA 1xEVDO based on a logarthmc utlty functon of each user s throughput. A larger class of throughput-based utltes s consdered n [2] where effcency and farness are allowed to be traded-off. The Max Weght polcy e.g. [6 8] can also be vewed as a gradent-based polcy, where the utlty s now a functon of a user s queue-sze or delay. Compared to TDMA and CDMA technologes, OFDMA dvdes the wreless resource nto non-overlappng frequency-tme chunks and offers more flexblty for resource allocaton. It has many advantages such as robustness aganst ntersymbol 1

2 nterference and multpath fadng as well as and lower complexty of recever equalzaton. Owng to these OFDMA has been adopted the core technology for most recent broadband wreless data systems, such as IEEE WMAX, IEEE a/g Wreless LANs, and LTE for 3GPP. Ths chapter dscusses gradent-based schedulng and resource allocaton n OFDMA systems. Ths bulds on prevous work specfc to the sngle cell downlnk [28] and uplnk [25] settng e.g., Fg. 1. The key contrbuton of the book chapter s provdng a general framework that ncludes each of these as specal cases and also apples to multple cell/sector downlnk transmssons e.g., Fg. 2. In partcular, several mportant practcal constrants are ncluded n ths framework, namely, 1 nteger constrants on the tone allocaton,.e., a tone can be allocated to at most one user; 2 constrants on the maxmum SNR.e., rate per tone, whch models a lmtaton on the avalable modulaton and codng schemes; 3 self-nose on tones due to channel estmaton errors e.g., [11] or phase nose [24]; and 4 user-specfc mnmum and maxmum rate constrants. We not only provde the optmal algorthm for solvng the optmzaton problem correspondng to the generalzed model, but also provde low complexty heurstc algorthms that acheve close to optmal performance. Most prevous work on OFDMA systems focused on solvng the resource allocaton problem wthout jontly consderng the problem of user schedulng. We wll brefly survey ths work n the next secton. Then we descrbe our general formulaton together wth the optmal and heurstc algorthms to solve the problem. Fnally, we wll summarze the chapter and outlne some future research drectons. Base Staton Base Staton User 1 User K User 1 User K User 2 User 2 Sngle Cell Downlnk Communcatons Sngle Cell Uplnk Communcatons Fgure 1: Example of a sngle cell downlnk left and uplnk scenero rght. 2

3 User 1 User 1 User 2 User 2 Base Staton 2 Base Staton 1 User K2 User K1 Base Staton 3 User 1 User K3 User 2 Fgure 2: Example of a multple cell/sector downlnk scenero the dfferent base statons could represent dfferent sectors of the same base staton shown by the crcle. 3

4 2 Related Work on OFDMA resource allocaton A number of formulatons for sngle cell downlnk OFDMA resource allocaton have been studed e.g., [12 21]. In [13, 14], the goal s to mnmze the total transmt power gven target bt-rates for each user. In [14], the target bt-rates are determned by a far queueng algorthm, whch does not take nto account the users channel condtons. A number of papers ncludng [15 18, 20, 21] have studed varous sumrate maxmzaton problems, gven a total power constrant. In [16 18] there s also a mnmum bt-rate per user that must be met. [21] consders both mnmum and maxmum rate targets for each user and also takes nto account several constrants that arse n Moble WMax. In [20], certan delay senstve users are modeled as havng fxed target bt-rate.e. ther maxmum and mnmum rates are the same, whle other best effort users have no bt-rate constrants. Thus the scheduler attempts to maxmze the sum-rate of the best effort users whle meetng the rate-targets of the delay senstve ones. In [12, 19], weghted sum-rate maxmzaton s consdered. Ths s a specal case of the resource allocaton problem we study here for a gven tmeslot but does not account for constrants on the SNR per carrer, rate constrants, or self-nose. In [12], a suboptmal algorthm wth constant power per tone was shown n smulatons to have lttle performance loss. Other heurstcs that use a constant power per tone are gven n [15 17]; we wll brefly dscuss a related approach n Secton 4. In [19], a dual-based algorthm smlar to ours s consdered, and smulatons are gven whch show that the dualty gap of ths problem quckly goes to zero as the number of tones ncreases. In [22], the nformaton theoretc capacty regon of a sngle cell downlnk broadcast channel wth frequency-selectve fadng usng a TDM scheme s gven; the feasble rate regon we consder, wthout any maxmum SNR and rate constrants, can be vewed as a specal case of ths regon. None of these papers consder self-nose, rate constrants or per user SNR constrants. Moreover, most of these papers optmze a statc objectve functon, whle we are nterested n a dynamc settng where the objectve changes over tme accordng to a gradent-based algorthm. It s not a pror clear f a good heurstc for a statc problem appled to each tme-step wll be a good heurstc for the dynamc case, snce the optmalty result n [1 3,6 8,29] s predcated on solvng the weghted-rate optmzaton problem exactly n each tme-slot. Smulaton results n [28] show that ths does hold for the heurstcs presented n Secton 4. Resource allocaton for a sngle cell OFDMA uplnk has been presented n [32 39]. In [32], a resource allocaton problem was formulated n the framework of Nash Barganng, and an teratve algorthm was proposed wth relatvely hgh complexty. The authors of [33] proposed a heurstc algorthm that tres to mnmze each user s transmsson power whle satsfyng the ndvdual rate constrants. In [34], the author consdered the sum-rate maxmzaton problem, whch s a specal case of the problem 4

5 consdered here wth equal weghts. The algorthm derved n [34] assumes Raylegh fadng on each subchannel; we do not make such an assumpton here. In [35], an uplnk problem wth multple antennas at the base staton was consdered; ths enables spatal multplexng of subchannels among multple users. Here, we focus on sngle antenna systems where at most one user can be assgned per sub-channel. The work n [36 39] s closer to our model. The authors n [36] also consdered a weghted rate maxmzaton problem n the uplnk case, but assumed statc weghts. They proposed two algorthms, whch are smlar to one of the algorthms descrbed n ths chapter. We propose several other algorthms that outperform those n [36] wth smlar or slghtly hgher complexty. Paper [37] generalzed the results n [36] by consderng utlty maxmzaton n one tme-slot, where the utlty s a functon of the nstantaneous rate n each tme-slot. Another work that focused on per tme-slot farness s [39]. Fnally, [38] proposed a heurstc algorthm based on Lagrangan relaxaton, whch has hgh complexty due to a subgradent search of the dual varables. Resource allocaton and nterference management of mult-cell downlnk OFDMA systems were presented n [42 49]. A key focus of these works s on nterference management among multple cells. Our general formulaton ncludes the case where resource coordnaton leads to no nterference among dfferent cells/sectors/stes. In our model, ths s acheved by dynamcally parttonng the subchannels across the dfferent cells/sectors/stes. In addton to beng easer to mplement, the nterference free operaton assumed n our model allows us to optmze over a large class of achevable rate regons for ths problem. If the nterference strength s of the order of the sgnal strength, as would be typcal n the broadband wreless settng, then ths parttonng approach could also be the better opton n an nformaton theoretc sense [31]. 1 3 OFDMA Schedulng and Resource Allocaton 3.1 Gradent-based Wreless Schedulng and Resource Allocaton Problem Formulaton Let us consder a network wth a total of K users. In each tme-slot t, the schedulng and resource allocaton decson can be vewed as selectng a rate vector r t = r 1,t,..., r K,t from the current feasble rate regon Re t R K +. If a user s not scheduled hs rate s smply zero. Here e t ndcates the tme-varyng channel state 1 We note that our dscussons do not drectly apply to the case of frequency reuse, where dfferent non-adjacent cells may use the same frequency bands. In practce, frequency reuse s typcally consdered together wth fxed frequency allocatons, whle here we consder dynamc frequency allocatons across dfferent cells. 5

6 nformaton of all users avalable at the scheduler at tme t. The decson on the rate vector s made accordng to the gradent-based schedulng framework n [1 3,29] that s bascally a stochastc verson of the condtonal gradent/frank-wolfe algorthm [26]. Namely, an r t Re t s selected that has the maxmum projecton onto the gradent of the system s total utlty functon UW t := K U W,t, 1 =1 where U s an ncreasng concave utlty functon that measures user s satsfacton for dfferent values of throughput, and W,t s user s average throughput up to tme t. In other words, the schedulng and resource allocaton decson s the soluton to max UW t T r t = r t Re t max r t Re t K U W,t r,t, 2 where U s the dervatve of U. As a concrete example, t s useful to consder the class of commonly used so-elastc utlty functons gven n [2, 5], =1 U W,t = { c α W,t α, α 1, α 0, c logw,t, α = 0, 3 where α 1 s a farness parameter and c s a QoS weght. In ths case, after takng dervatves, 2 becomes c W,t α 1 r,t. 4 max r t Re t Wth equal class weghts c = c for all, settng α = 1 results n a schedulng rule that maxmzes the total throughput durng each slot. For α = 0, ths results n the proportonally far rule, and as α ncreases wthout bound, we get closer to a max-mn far soluton. Thus, ths famly of utlty functons yelds a flexble class of polces: the α parameter allows for the choce of an approprate farness objectve whle the c parameter allows one to dstngush relatve prortes wthn each farness class. However, more generally, we consder the problem of max r t Re t w,t r,t, 5 where w,t 0 s a tme-varyng weght assgned to the th user at tme t. In the case of 4, we let w,t = c W,t α 1. In 4 these weghts are gven by the gradents of throughput-based utltes; however, other methods for generatng the weghts possbly dependng upon queue-lengths and/or delays [6 8] are also possble. We note 6

7 that 5 must be re-solved at each schedulng nstance because of changes n both the channel state and the weghts e.g., the gradents of the utltes. Whle the former changes are due to the tme-varyng nature of wreless channels, the latter changes are due to new arrvals and past servce decsons. 3.2 General OFDMA rate regons The soluton to 5 depends on the channel state dependent rate regon Re, where we suppress the dependence on tme for smplcty. We consder a model approprate for general OFDMA systems ncludng sngle cell downlnk and uplnk as well as multple cell/sector/ste downlnk wth frequency sharng; related sngle cell downlnk and uplnk models have been consdered n [12, 22, 25, 28]. In ths model, Re s parameterzed by the allocaton of tones to users and the allocaton of power across tones. In a tradtonal OFDMA system at most one user may be assgned to any tone. Intally, as n [13, 14], we make the smplfyng assumpton that multple users can share one tone usng some orthogonalzaton technque e.g. TDM. 2 In practce, f a schedulng nterval contans multple OFDMA symbols, we can mplement such sharng by gvng a fracton of the symbols to each user; of course, each user wll be constraned to use an nteger number of symbols. Also, wth a large number of tones, adjacent tones wll have nearly dentcal gans, n whch case ths tme-sharng can also be approxmated by frequency sharng. The two approxmatons becomes tght as the number of symbols or tones ncreases, respectvely. We dscuss the case where only one user can use a tone n Secton 4. Let N = {1,..., N} denote the set of tones 3 and K = {1, 2,..., K} the set of users. For each j N and user K, let e j be the receved sgnal-to-nose rato SNR per unt transmt power. We denote the transmt power allocated to user on tone j by p j, and the fracton of that tone allocated to user by x j. As tones are shared resources, the total allocaton for each tone j must satsfy x j 1. For a gven allocaton, wth perfect channel estmaton, user s feasble rate on tone j s r j = x j B log 1 + p je j x j whch corresponds to the Shannon capacty of a Gaussan nose channel wth bandwdth x j B and receved SNR p j e j /x j. 4 Ths SNR arses from vewng p j as the 2 We focus on systems that do not use superposton codng and successve nterference cancellaton wthn a tone, as such technques are generally consdered too complex for practcal systems. 3 In practce, tones may be grouped nto subchannels and allocated at the granularty of subchannels. As dscussed n [28], our model can be appled to such settngs as well by approprately redefnng the sub-channel gans {e j } and nterpretng N as the set of sub-channels. 4 To better model the achevable rates n a practcal system we can re-normalze e j by γe j, where, 7

8 energy per tme-slot user uses on tone j; the correspondng transmsson power becomes p j /x j when only a fracton x j of the tone bandwdth s allocated. Smlarly ths can also be explaned by tme-sharng as follows: a channel of bandwdth B s used only a fracton x j of the tme wth average power p j whch leads to the power durng channel usage to be p j /x j. Wthout loss of generalty we set B = 1 n the followng Self-nose In a realstc OFDMA system, mperfect carrer synchronzaton and channel estmaton may result n self-nose e.g. [11, 24]. We follow a smlar approach as n [11] to model self-nose. Let the receved sgnal on the jth tone of user be gven by y j = h j s j + n j, where h j, s j and n j are the complex channel gan, transmtted sgnal and addtve nose, respectvely, wth n j CN 0, σ 2. 5 Assume that h j = h j + h j,δ, where h j s recever s estmate of h j and h j,δ CN 0, δj. 2 After matched-flterng, the receved sgnal wll be z j = h jy j resultng n an effectve SNR of h j 4 p j Eff-SNR = σj 2 h j 2 + δj 2 p j h j = p je j, β j p j e j where p j = E s j 2, β j = δ2 j and e h j 2 j = h j 2 σj 2. 6 Here, β j p j e j s the self-nose term. As n the case wthout self-nose β j = 0, the effectve SNR s stll ncreasng n p j. However, t now has a maxmum of 1/β j. In general, β j may depend on the channel qualty e j. For example, ths happens when self-nose arses prmarly from estmaton errors. The exact dependence wll depend on the detals of channel estmaton. As an example, usng the model n [23, Secton IV] t can be shown that when the plot power s ether constant or nversely proportonal to channel qualty subject to maxmum and mnmum power constrants modelng power control, β s nversely proportonal to the channel condton for large e. On the other hand β j = β s a constant when self-nose s due to phase nose as n [24]. For smplcty of presentaton, we assume constant β j = β n the remander of the paper except n Fg. 4 where we we allow βe 1/e to llustrate the mpact γ [0, 1] represents the system s gap from capacty. 5 We use the notaton x CN 0, b to denote that x s a 0 mean, complex, crcularly-symmetrc Gaussan random varable wth varance b := E x 2. 6 Ths s slghtly dfferent from the Eff-SNR n [11] n whch the sgnal power s nstead gven by h j 4 p j ; the followng analyss works for such a model as well by a smple change of varables. For the problem at hand, 6 seems more reasonable n that the resource allocaton wll depend only on h j and not on h j. We also note that 6 s shown n [23] to gve an achevable lower bound on the capacty of ths channel. 8

9 of self-nose on the optmal power allocaton. The analyss s almost dentcal f users have dfferent β j s. We assume that e j s known by the scheduler for all and j as s β. For example, n a frequency dvson duplex FDD downlnk system, ths knowledge can be acqured by havng the base staton transmt plot sgnals, from whch the users can estmate ther channel gans and feedback to the base staton. In a tme dvson duplex TDD system, these gans can also be acqured by havng the users transmt uplnk plots; for the downlnk case, the base staton can then explot recprocty to measure the channel gans. In both cases, ths feedback nformaton would need to be provded wthn the channel s coherence tme. Wth self-nose, user s feasble rate on tone j becomes r j = x j log 1 + p j e j pj e j =: x j f x j + βp j e j x j, 7 where agan x j models tme-sharng of a tone and the functon f s gven by 1 fs = log 1 +, β 0. 8 β + 1/s More generally, we assume that a user s rate on channel j s gven by pj e j r j = x j f, 9 for some functon f : R + R + that s non-decreasng, twce contnuously dfferentable and concave wth f0 = 0, wthout loss of generalty 7 f 0 := df 0 = ds fs fs df lm s 0 = sup s s>0 = 1, and lm s t + t = 0. We also assume by contnuty8 ds that xfp/x s 0 at x = 0 for every p 0. From the assumptons on the functon f t follows that xfp/x s jontly concave n x, p; ths can be easly proved by showng that the Hessan s negatve semdefnte [26, 27]. It s easy to verfy that f gven by 8 satsfes the above propertes. We should, however, pont out that usng the theory of subgradents [26,27], our mathematcal results easly extend to a general 7 Usng the dea that Shannon capacty log1 + s s a natural upper bound for fs, t follows that 0 < df ds 0 1. Therefore, f f 0 1, then we can solve the problem usng a scaled verson of functon,.e., fs = fs/ df ds 0, after scalng the rate constrants by the same amount; the power and subchannel allocatons wll be the same n the two cases. The Shannon capacty upper bound df also yelds that 0 lm t + ds t lm s + fs log1+s s lm s + s = 0, as concavty of f and f0 = 0 mply that df ft ds t t for all t > 0. 8 Usng the Shannon capacty functon, log1 + s, upper bound, we have for p > 0, that ft log1+t lm x 0 xfp/x = p lm t + t p lm t + t = 0. For p = 0, we drectly get the property from f0 = 0. 9 x j

10 f that s only non-decreasng and concave. For nstance, t can be easly proved from frst prncples that xfp/x s jontly concave n x, p f f s merely concave. We conscously choose the smpler settng of twce contnuously dfferentable functons to keep the level of dscusson smple, but to ad a more nterested reader, we wll strve to pont out the loosest condtons needed for each of our results. Before proceedng we should pont out that, operatonally, f s a functon of the receved sgnal-to-nose rato, and thus, abstracts the usage of all possble sngle-user decoders, ncludng the optmal decoder that yelds Shannon capacty General power constrant - sngle cell downlnk, uplnk and multcell downlnk wth frequency sharng Let {K m } M m=1 be non-empty subsets of the set of users K that form a coverng,.e., M m=1k m = K. We assume that there s a vector of non-negatve power budgets {P m } M m=1 assocated wth these subsets, so that K m j p j P m for each m. Ths condton ensures that there s no user who s unconstraned n ts power usage. Ths provdes a common formulaton of the sngle cell downlnk and uplnk schedulng problems as descrbed n [28] and [25], respectvely. For the sngle cell downlnk problem M = 1 and K 1 = K, and for the sngle cell uplnk problem M = K and K = {} for K. More generally, f {K m } M m=1 s a partton,.e., mutually dsjont, then we can vew the transmtters for users K m as co-located wth a sngle power amplfer. For example, such a model may arse n the downlnk case where M := {1, 2,..., M} represents sectors or stes across whch we need to allocate common frequency/channel resources, but whch have ndependent power budgets. A key assumpton, however, s that we can make the transmssons from the dfferent sectors/stes non-nterferng by tme-sharng or by some other sutable orthogonalzaton technque Capacty Regon - max SNR and mn/max rate constrants Under these assumptons, the rate regon can be wrtten as { Re = r : r = pj e x j f j x j and R mn j p j P m, m, j K m r R max,, } x j 1, j, x, p X, 10 where X := { x, p 0 : x j 1, p j x js j e j }, j. 11 Here and n the followng, a boldfaced symbol wll ndcate the vector of the correspondng scalar quanttes, e.g. x := x j and p := p j. Also, any nequalty such 10

11 as x 0 should be nterpreted componentwse. The lnear constrant on x j, p j n 11 usng s j models a constrant on the maxmum rate per subchannel due to a lmtaton on the avalable modulaton and codng schemes; f user can send at a maxmum rate of r j on tone j, then s j = f 1 r j. We have also assumed that each user K has maxmum and mnmum rate constrants R max and R mn, respectvely. In order to have a soluton we assume that the vector of mnmum rates {R mn } K s feasble. For the vector of maxmum rates, t s more convenent to assume that {R max } K s nfeasble. Otherwse the optmzaton problem assocated wth feasblty see Secton 3.5 wll yeld an optmal soluton. Typcally we wll set R mn = 0 and R max to be the tme-varyng buffer occupancy. However, wth tght mnmum throughput demands one can magne usng a non-zero R mn to guarantee ths. 3.3 Optmal Algorthms From 5 and 10, the optmal schedulng and resource allocaton problem can be stated as: max V x, p := pj e w x j f j x,p X x j P2 j pj e j subject to: x j f R mn K η j x j pj e j x j f j x j R max K γ x j 1 j N µ j p j P m m = 1, 2,..., M λ m K m j where set X s gven n 11. As a rule, varables at the rght of constrants wll ndcate the dual varables that we wll use to relax those constrants whle constructng the dual problem later. One mportant pont to note s that as descrbed above, the optmzaton problem P2 s not convex and so we can not appeal to standard results such as Slater s condtons to guarantee that s has zero dualty gap [26, 27]. In partcular, note that the maxmum rate constrants have a concave functon on the left sde. To show that we stll have no dualty gap, we wll consder a related convex problem n hgher dmensons that has the same prmal soluton and the same dual. The new 11

12 optmzaton problem P1 s gven by max w r subject to: r pj e j x j f, K α x j j x j 1, j N µ j p j P m, m = 1, 2,..., M λ m K m R mn j r R max, K x, p X. P1 Ths problem s easly seen to be convex due to the jont concavty of xfp/x as a functon of x, p and also wll satsfy Slater s condton. 9 Hence, t wll have zero dualty gap [26, 27]. The problem P1 can be practcally motvated as follows: the physcal PHY layer gves the scheduler at the MAC layer a maxmum rate that t can serve per user based upon power and subchannel allocatons, and the scheduler then drans from the queue an amount that obeys the mnmum and maxmum rate constrants mposed by the network layer and the maxmum rate constrant from the PHY layer output. If the scheduler chooses not to use the complete allocaton gven by the PHY layer, then the fnal packet sent by the MAC layer s assumed to be constructed usng an approprate number of padded bts. However, we wll now show that at the optmal, there s no of loss optmalty n assumng that the scheduler never sends less than what the PHY layer allocates,.e., the frst constrant n Problem P1 s always made tght at an optmal soluton. Ths pont of vew s exemplfed n schematc shown n Fgure 3. Assume that there s an optmzer of P1 at whch for some user, r < j x jf p je j x j. We wll now construct another feasble soluton that wll satsfy the above relatonshp wth equalty. Let γ [0, 1] and set p j := γp j. Note that by convexty, both the power and subchannel constrants are satsfed for every value of γ. Now j x jfγ p je j x j s a non-decreasng and contnuous functon of γ takng values 0 at γ = 0 and j x jf p je j x j at γ = 1. Therefore, there exsts a γ 0, 1 such that r = j x jfγ p j e j x j as desred. Ths procedure can be followed for every user for whom r < j x jf p je j x j, so that at the end we satsfy r = j x jf p je j x j for a feasble x, p. Therefore both the optmal value and an optmzer of problem P1 9 More precsely, Slater s condton wll be satsfed provded that the mnmum rate R mn are strctly n the nteror of rate-regon Re. If R mn = 0 for all ths wll trvally be true. 12

13 Network Layer QoS Constrants Buffer Sze Transmsson Rate MAC Layer Scheduler Channel Qualty Measurements Maxmum Transmsson Rate Transmsson Rate Physcal Layer Fgure 3: Schematc of a scheduler that has cross-layer vsblty. 13

14 concdes wth those for problem P2. The loosest condton needed for the above to hold s f beng non-decreasng and concave wth f0 = 0. Henceforth, we wll only work wth Problem P1. Before proceedng to solve the problem by dual methods, we frst defne some key notaton. For two numbers, x, y R we set x y := mnx, y, x y := maxx, y and x + = [x] + := x Dual of Problem We now proceed to derve a closed-form expresson for the dual functon for problem P1. The Lagrangan obtaned by relaxng the marked constrants of P1 usng the correspondng dual varables s gven by Lr, x, p, α, µ, λ = w α r + j µ j x j j m µ j + M pj e j α x j f x j λ m P m + m=1,j λ m p j. 12 The correspondng dual functon s then gven by maxmzng ths Lagrangan over r, x and p. Frst optmzng over rate r [R mn, R max ] and notng that the Lagrangan s lnear n r we get Lx, p, α, µ, λ = +,j w α + R max α x j f p je j x j j K m j α w + R mn + M µ j + λ m P m j m=1 µ j x j λ m p j. m The optmzng r s gven by the followng {R max } f α < w ; K, r {R } f α > w ; and [R mn K m, R max ] f α = w 13 Note that the last term of equaton 12 can be rewrtten as λ m p j = p j λ m = p jˆλ 14 m j,j m: K m,j where ˆλ := m: K m λ m. K m j 14

15 Now maxmzng the Lagrangan over power p requres us to maxmze pj e j α x j [f ˆλ ] p j e j x j α e j x j over p j for each, j. From the assumptons on the functon f, t s easy to check that the maxmzng p j wll be of the form p je j ˆλ = g s j, 16 x j α e j for some functon g : R + [0, ] wth gx = 0 for x f 0. Specfcally f df/ds s monotoncally decreasng, we may show that g = df 1 ds,.e., the nverse of the dervatve of f. Otherwse, snce df/ds s stll a non-ncreasng functon we can set gx = nf{t : df/dst = x}. Usng the non-ncreasng property of df/ds we can see that gx y = g x df y. Note that we have assumed df/ds0 = 1 ds and lm t + df/dst = 0 but we do not assume that lm s + fs = + e.g., see the self-nose example. In case f s not dfferentable, then we would defne the functon g usng the subgradents of f. In all cases, the key concluson from 16 s that the optmal value of p j s always a lnear functon of x j. Note that when f = log1 + 1, wth β 0, as gven by 8, then β+1/s where gx = q1/x 1 +, { z, f β = 0, qz = 2β ββ+1 z 1, f β > 0. 2ββ+1 2β+1 2 Fgure 4 shows p j n 16 as a functon of e j for the specfc choce of f from 8 wth three dfferent values of β = 0, 0.01, 0.1. When β = 0, 16 becomes a waterfllng type of soluton n whch p j s non-decreasng n e j. For a fxed β > 0, ths s not necessarly true,.e., due to self-nose, less power may be allocated to better subchannels. We also consder the case where β = 10/e to model the case where self-nose s due to channel estmaton error. Insertng the expresson for p j nto the Lagrangan yelds Lx, α, µ, λ = w α + R max α w + R mn + M µ j + λ m P m j m=1 + ˆλ x j [α f g s j ˆλ ] ˆλ g s j µ j, α,j e j e j α e j

16 β=0 β=0.01 Optmal power p j * β=10/e β= Channel condton e j db Fgure 4: Optmal power p j as a functon of the channel condton e j. Here x j = 1, α = 1, s j = +, and ˆλ = 15. whch s now a lnear functon of {x j }. Thus, optmzng over x j yelds the dual functon for P1, Lα, λ, µ = +,j = + j w α + R max [ α f g α e j w α + R max ˆλ s j α w + R mn ˆλ e j α w + R mn ] ˆλ [µ j α, α e j µ j + + µ j + λ m P m j m ] ˆλ g s j µ j α e j + λ m P m m + µ j, 18 + where µ j a, b := a f gb s j b gb s j. 16

17 Note that any choce such that ˆλ {1}, f µ j α, α e j > µ j, x ˆλ j [0, 1], f µ j α, α e j = µ j, 19 {0}, f µ j α, < µ j wll optmze the Lagrangan n Optmzng the Dual Functon over µ ˆλ α e j From the dualty theory of convex optmzaton [26, 27] the optmal soluton to problem P1 s gven by mnmzng the dual functon n 18 over all α, λ, µ 0. We do ths coordnate-wse startng wth the µ varables. The followng lemma characterzes ths optmzaton. Lemma 1 For all α, λ 0, Lα, λ := mn Lα, λ, µ µ 0 = w α + R max α w + R mn + λ m P m + m j µ jα, λ, 20 where for every tone j, the mnmzng value of µ j s acheved by µ jα, λ := max µ j ˆλ α,. 21 α e j The proof of Lemma 1 follows from a smlar argument as n [9]. Note that 21 requres searchng for the maxmum value of the metrcs µ j across all users for each tone j. Snce Lα, λ s the mnmum of a convex functon over a convex set, t s a convex functon of α, λ Optmzng the Dual Functon over α, λ Now we are ready to optmze the remanng varables n the dual functons, namely, α, λ. In the sngle cell downlnk case wth no rate constrants and thus no α varables, ths reduces to a one dmensonal problem n λ and hence, t can be mnmzed usng an terated one dmensonal search e.g., the Golden Secton method [26]. Snce there s no dualty gap, at λ = arg mn λ 0 Lλ, Lλ gves the optmal objectve 17

18 value of problem P1. Smlarly, n the absence of rate constrants, the multple stes/sectors problem wth a partton of the users {K m } M m=1 also leads to a one dmensonal problem wthn each partton. In general, however, one would need to use subgradent methods [26, 27] to numercally solve for the optmal α, λ. The followng lemma characterzes the set of subgradents of Lα, λ wth respect to α, λ. Lemma 2 About any α 0, λ 0 0, Lα, λ dα 0 α α 0 + m dλ 0 mλ m λ 0 m, 22 wth dλ m = P m p j = P m x j ˆλ g s j 23 e j α e j K m K m dα m = x ˆλ jf g s j r 24 α j e j where x js satsfy x j 1 and µ j α, λ 1 x j = 0; j, and satsfy the equaton 19 wth µ j = µ jα, λ as gven n equaton 21, and r satsfy equaton 13. Thus the subgradents dλ m and dα are parameterzed by r, x and are lnear n these varables. Moreover, the permssble values of r le n a hypercube and those of x n a smplex. Observe that the dual functon at any pont α, λ s obtaned by takng the maxmum of the Lagrangan over r, p, x satsfyng x j 1, j N, x, p X. In case r, p, x s unque, then the resultng Lagrangan s a gradent to the dual functon at α, λ. In case there are multple optmzers, the resultng Lagrangans are each a subgradent, and every subgradent can be obtaned by a convex combnaton of these subgradents so that the set of subgradents s convex. The lemma follows easly by substtutng for the optmal r, p, x. Havng characterzed the set of subgradents, a method smlar to that used n [25] for the sngle cell uplnk problem can be used to solve for the optmal dual varables α, λ numercally. In each step of ths method we change the dual varables along the drecton gven by a subgradent subject to non-negatvty of the dual varables. The convergence of ths procedure for a proper step-sze choce s once agan guaranteed by the convexty of Lα, λ see [26, Exer ], [25]. 18

19 3.3.4 Optmzng the dual functon over α Snce the dmenson of α equals the number of users and the dmenson of µ equals the number of tones, t may be computatonally better to optmze over α nstead of µ f the number of users s greater, and then use numercal methods to solve the problem. Next we detal the means to optmze over α before µ. The dual functon contans many terms that have defntons wth +, and therefore we would need to dentfy exactly when these terms are non-zero. For ths we need to solve a non-lnear equaton whch s guaranteed to have a unque soluton. We frst dscuss ths and then apply t to optmzng the dual functon over α. Gven y, z 0, defne by vy, z the unque soluton wth 1 x < + to xf 1 g x df ds where t s easy to show that xf g z 1 df x 1 g x df ds z = y, z g ds 1 df x z s a monotoncally ds ncreasng functon takng value 0 at x = 1 and ncreasng wthout bound as x +. If y fz/df/dsz z 0, then vy, z = y +z/fz where t s easy to verfy that vy, z z/fz 1/df/dsz 1/df/ds0 = 1 from the concavty of f and from f0 = 0. Otherwse we need to solve for the unque 1 x 1/df/dsz such that 1 1 xf g g = y. x x µj e For our results we wll be nterested n v j, s ˆλ j, usng whch we also defne ν j := ˆλ v µj e j ˆλ, s j s j ˆλ e j and ζ j := µ j + e j fs j where ν j = ζ j f µ je j s ˆλ j. Frst note that we can rewrte the functon n 18 as follows fs j dfs j ds, Lα, µ, λ = j µ j + m λ m P m + L, where L = w α + R max α w + R mn + [ ˆλ α e j ˆλ f g s j e j j ˆλ α e j 19 ] ˆλ g s j µ je j. α e j ˆλ +

20 Now usng the quanttes defned earler n ths secton, one can wrte L as follows L = [ ˆλ α e j 1 {0 α ζ e j } fs j s j µ je j + j j ˆλ ˆλ α e j ˆλ ] ˆλ 1 {ζj <α ν j } f g g µ je j ˆλ α e j α e j ˆλ + w α + R max α w + R mn. Mnmzng L over α 0 can now be accomplshed by a smple one dmensonal search; we defne the optmal vector of α s to be α λ, µ. Thereafter one would need to use a subgradent method [25, 26] to numercally mnmze over µ, λ. A subgradent of L wth respect to λ m s gven by P m K m p j where p j s taken from 16 where one substtutes x j from 19. A subgradent of L wth respect to µ j s gven by 1 x j where we substtute for x j from 19. Note, however, that t s mportant that we also meet the followng constrants for all, namely, R mn x jf p j x j f α < w, then j f α > w, then j x jf x jf R max ; p j x j p j x j = R max ; and = R mn. The proof of ths follows by retracng the steps of the proof of Lemma 2 wth the roles of α and µ beng swtched. 3.4 Prmal optmal soluton For the general OFDMA problem we presented two methods to solve for V : n the frst method we showed how to characterze the dual varables µα, λ and then we proposed numercally solvng for the optmal α, λ usng subgradent methods, whle n the second method followed the same strategy after swtchng the roles of µ and α. However, we stll need to solve for the values of the correspondng optmal prmal varables. Concentratng on the frst method, we know by dualty theory [26] that gven α, λ we need to fnd one vector from the set of r, x, p that also satsfes prmal feasblty and complementary slackness. These constrants can easly be seen to translate to the followng: dλ m 0, dλ mλ m = 0, m; 25 dα 0, dα α = 0,

21 From the lnearty of dλ m, dα n r, x t follows that the prmal optmal r, x, p are the soluton of a lnear program n r, x. For the sngle cell downlnk case wth no rate constrants, as we have prevously noted searchng for the dual optmal s a one dmensonal numercal search n λ. In that case, the search for prmal optmal soluton turns out to have addtonal structure as shown n [28]. 3.5 OFDMA Feasblty Next we turn to the correspondng feasblty problem, whch can be stated as: V = mn σ 27 subject to: R x j f p je j, α x j j x j 1 j µ j p j σ m λ m P m K m j x, p X. The vector of rates R s feasble f V 1,.e., all the power constrants wll also be satsfed by a vector x, p. As mentoned earler, we need to check that R = R mn s ndeed feasble; otherwse problems P1 and P2 are both nfeasble as well. Moreover, f R = R max s also feasble, then r = R max s the optmzer for problems P1 and P2. In whch case, the optmal soluton to the problem above wth R = R max wll also yeld an optmal soluton to the schedulng problem. Observe that problem 27 s convex and satsfes Slater s condtons. Fnally, we also note that other alternate formulatons of the feasblty problem are possble where one could ether apply the σ constrant also on the subchannel utlzaton or swtch the roles of subchannel and power utlzaton. All of these wll yeld the same concluson about feasblty although the actual solutons, n terms of x, p, would possbly be dfferent. The Lagrangan consderng the marked constrants s Lσ, x, p, α, µ, λ = σ 1 λ m µ j + α R m j + µ j x j pj e j α x j f + p j λ x j j j 21

22 where λ := λ m m: K m Pm. As before, mnmzng over p j yelds p j e j x j Substtutng ths n the Lagrangan, we get Lσ, x, α, µ, λ = α R µ j + σ 1 λ m j m [ x j,j = g λ α e j s j. α fg ˆλ α e j s j ˆλ e j g ˆλ α e j s j µ j ]. Mnmzng over 0 x j 1 yelds Lσ, α, µ, λ = L j µ j + σ 1 m λ m where L = α R j [ α fg ˆλ α e j s j ˆλ e j g ˆλ α e j s j µ j ] +. Next we mnmze L over all values of σ. Snce there are no constrants on σ, t follows that the resultng L s fnte only when m λ m = 1; for all other values we would get L =. Hereafter we wll assume that m λ m = 1. Thus Lσ, x, α, µ, λ = L j µ j. Note that as before, as a functon of α the problem s now separable. Therefore we only need to maxmze L over α 0. Smlarly we can wrte L as Lσ, x, α, µ, λ = j ˆL j + α R, where we have ˆL j = µ j + [ α fg ˆλ α e j s j ˆλ e j g ˆλ α e j s j µ j ] + As a functon of µ j the problem s now separable, and we only need to maxmze ˆL j over µ 0. 22

23 Thus, we could optmze frst over ether µ or α, once agan based upon whether the number of users or subchannels s smaller. In ether case, the methodology and the functons that appear are very smlar to the correspondng problem n the schedulng problem P1, and due to space constrants we do not elaborate on ths. Care must be take, however, whle evaluatng subgradents wth respect to λ. Here we propose usng a projected gradent method [26, 27] based upon the constrant m λ m = 1 to numercally solve for the optmal λ. 3.6 Power allocaton gven subchannel allocaton In many of the suboptmal schedulng algorthms that we wll dscuss, a central feature wll be a computatonally smpler but stll close to optmal method to provde a subchannel allocaton. Once the subchannel allocaton has been made, all that wll reman s the power allocaton problem, subject to the varous constrants that we dscussed earler. Here we dscuss how ths can be solved n an optmal manner. A smlar queston can also be asked about the feasblty problem, hence we also dscuss ths here. In all cases, we assume that we are gven a feasble subchannel allocaton. Snce we are gven a feasble subchannel allocaton x, the Lagrangan of the new schedulng problem power allocaton only can be easly derved by settng µ = 0. For ths we once agan use the formulaton based upon Problem P1. The optmal power allocaton s then gven by p j results from substtutng ths formula s = x j e j g ˆλ α e j sj. The Lagrangan that Lx, α, λ = λ m P m + w α + R max α w + R mn m + ˆλ α x j f g s j ˆλ x j ˆλ g s j. e j j α e j e j α Now t s easy to argue that f R mn = 0 and R max = + and f the K m s form a partton, then wthn each partton the λ m s can be solved for as n Secton 3.3. In any case, n ths settng solvng for the optmal α 0 s easer, but uses some of the functons descrbed at the end of Secton 3.3. However, after ths step we would stll need to solve for λ numercally; f the parttons assumpton holds, then t would only need a sngle dmensonal search wthn each partton. A fnte-tme algorthm for achevng the optmal λ has been gven n [25,28] under the assumpton that f represents the Shannon capacty as n 8 wth β = 0. 23

24 3.6.1 Feasblty check Under the assumpton that a feasble subchannel allocaton has already been provded, even the feasblty check problem becomes a lot easer. As before we can assume m λ m = 1, and that the optmal power allocaton s gven by p j = x j e j g λ sj e j α, and substtutng ths we get Lx, α, λ = α ˆR j λ x j [α f g s j λ ] λ g s j. e j α e j e j α Agan solvng for the optmal α s smpler. Once agan the λ vector would need to be computed numercally, subject to t beng a probablty vector,.e., m λ m = 1 and λ m 0 for each m. 4 Low Complexty Suboptmal Algorthms wth Integer Channel Allocaton There are two shortcomngs wth usng the optmal algorthm outlned n the prevous secton for schedulng and resource allocaton: the complexty of the algorthm n general s not computatonally feasble for even moderate szed systems; the soluton found may requre a tme-sharng channel allocaton, whle practcal mplementatons typcally requre a sngle user per sub-channel. One way to address the second pont s to frst fnd the optmal prmal soluton as n the prevous secton and then project ths onto a nearby nteger soluton. Such an approach s presented n [28] for the case of a sngle cell downlnk system M = 1 wthout any rate constrants. In that settng, after mnmzng the dual functon over µ, one optmzes the functon Lλ, whch only depends on a sngle varable. Ths functon wll have scalar subgradents whch can then be used to develop rules for mplementng such an nteger projecton. Moreover, n ths case snce Lλ s a one-dmensonal functon the search for the optmal dual values s greatly smplfed. However, n the general settng, ths type of approach does not appear to be promsng. 10 In ths secton we dscuss a famly of sub-optmal algorthms SOA s for the general settng that try to reduce the complexty of the optmal algorthm, whle sacrfcng lttle n performance. These algorthms seek to explot the problem structure revealed by the optmal algorthm. Furthermore, all of these sub-optmal algorthms enforce an nteger tone allocaton durng each schedulng nterval. In the followng we consder the general model from Secton 3.1 wth the restrcton that {K m } forms 10 See [25] for a more detaled dscusson of ths n the context of the uplnk scenaro. 24

25 a partton of the user groups.e. each user s n only one of these sets and that R mn = 0 for all. In a typcal settng both of these assumptons wll be true. In the optmal algorthm, gven the optmal λ and α, the optmal tone allocaton up to any tes s determned by sortng the users on each tone accordng to the ˆλ metrc µ j α, α e j cf. 19. Gven an optmal tone allocaton, the optmal power allocaton s gven by 16. In each SOA, we use the same two phases wth some modfcatons to reduce the complexty of computng λ, α and the optmal tone allocaton. Specfcally, we begn wth a subchannel Allocaton CA phase n whch we assgn each tone to at most one user. We consder two dfferent SOAs that mplement the CA phase dfferently. In SOA1, nstead of usng the metrc gven by the optmal λ and α, we consder metrcs based on a constant power allocaton over all tones assgned to a partton. In SOA2, we fnd the tone allocaton, once agan through a dual based approach, but here we frst determne the number of tones assgned to each user and then match specfc tones and users. In all cases we assgn the tones to dstnct parttons whch wll, n turn, yeld an nterference-free operaton. After the tone allocaton s done n both SOAs, we execute the Power Allocaton PA phase n whch each user s power s allocated across the assgned tones usng the optmal power allocaton n CA n SOA1: Progressve Subchannel Allocaton Based on Metrc Sortng In ths famly of SOAs, tones are assgned sequentally n one pass based on a per user metrc for each tone,.e., we terate N tmes, where each teraton corresponds to the assgnment of one tone. Let N n denote the set of tones assgned to user after the nth teraton. Let g n denote user s metrc durng the nth teraton and let l n be the tone ndex that user would lke to be assgned f he/she s assgned the nth tone. The resultng CA algorthm s gven n Algorthm 1. Note that all the user metrcs are updated after each tone s assgned. We consder several varatons of Algorthm 1 whch correspond to dfferent choces for steps 4 and 5. The choces for step 4 are: 4A: Sort the tones based on the best channel condton among all users. Ths nvolves two steps. Frst, for each tone j, fnd the best channel condton among all users and denote t by µ j := max e j. Second, fnd a tone permutaton {α j } j N such that µ α1 µ α2 µ αn, and set l n = α n for each user at the nth teraton. Each max operaton has complexty of OK, and the sortng operaton has a complexty of ON logn. The total complexty s O NK + N log N. We note that ths s a one-tme pre-processng that needs to done before the CA phase starts. Durng the tone allocaton teratons, the users just choose the tone ndex 25

26 Algorthm 1 CA Phase for SOA1 1: Intalzaton: set n = 0 and N n = for each user. 2: whle n < N do 3: n : Update tone ndex l n for each user. 5: Update metrc g n for each user. 6: Fnd n = arg max g n break tes arbtrarly. 7: f g nn 0 then 8: Assgn the nth tone to user n: { N n = N n 1 {l n}, f = n; N n 1, otherwse. 9: else 10: Do not assgn the nth tone. 11: end f 12: end whle from the sorted lst. 4B: Sort the tones based on the channel condtons for each ndvdual user. For each user at the nth teraton, set l n to be the tone ndex wth the largest gan among all unassgned tones,.e., l n = arg max j N \ N n 1 e j. Ths requres K sorts one per user; these also need to be performed only once snce each tone assgnment does not change a user s orderng of the remanng tones and can be done n parallel. The total complexty of the K sortng operatons s O KN log N, whch s hgher than that n 4A. Durng the nth teraton, let k n = j Km N j n denote the number of tones assgned to users n the group to whch user belongs,.e., m. The choces for Lne 5 are: 5A: Set g n to be the total ncrease n user s utlty f assgned tone l n, assumng the power for each user group s allocated unformly over the tones assgned 26

27 to that group,.e., g n = [ w j N n 1 {l n} f P e j s k n 1+1 j R max ],f k n 1 > 0; j Nn 1 f P e j s k n 1 j R max [ ] w j N n 1 {l n} f P e j s k n 1+1 j R max,otherwse. 28 5B: Set g n to be user s gan from only tone l n, agan assumng constant power allocaton wthn each group,.e. ] P e,l n g n = w [f k n s j R max. Compared wth 5A, ths metrc s smpler to calculate but gnores the change n user s utlty due to the decrease n power allocated to any tones n N n 1. It also does not accurately enforce the maxmum rate constrant, snce t only consders one tone at a tme. The complexty of ether of these choces over N teratons s ONK, and so the total complexty for the CA phase s O NK + N log N f 4A s chosen or O KN log N f 4B s chosen. Algorthms smlar to SOA1 wth 4B and 5B have been proposed n the lterature for both the sngle cell downlnk settng [12] 11 and the uplnk [36] wthout rate or SNR constrants. In the sngle cell downlnk case, the algorthm n [12] s shown va numercal examples to have near optmal performance. In the uplnk case, ths also performs reasonably well n smulatons [36], but [25] shows that better performance can be obtaned usng 4B and 5A nstead. 4.2 CA n SOA2: tone Number Assgnment & tone User Matchng SOA2 mplements the CA phase through two steps: tone number assgnment CNA and tone user matchng CUM. The algorthm s summarzed n Algorthm The man dfference wth the algorthm n [12] s that after each teraton n, t then checks to see f w r s ncreasng and f not t stops at teraton n 1. Such a step can be added to Algorthm 1; however, unless the system s lghtly loaded t s unlkely to have a large mpact on the performance. 27

28 Algorthm 2 CA Phase of SOA2 1: subchannel Number Assgnment CNA step: determne the number of tones n allocated to each user such that K n N. 2: subchannel User Matchng CUM step: determne the tone assgnment x j {0, 1} for all users and tones j, such that j N x j = n subchannel Number Assgnment CNA In the CNA step, we determne the number of tones n assgned to each user K. The assgnment s calculated based on the approxmaton that each user sees a flat wde-band fadng tone. Notce that here we do not specfy whch tone s allocated to whch user; such a mappng wll be determned n the CUM step. The CNA step s further dvded nto two stages: a basc assgnment stage and an assgnment mprovement stage. Stage 1, Basc Assgnment: Here, the assgnment s based on the normalzed SNR averaged over all tones. Specfcally, we model each user as havng a normalzed SNR e = 1 N j N e j, and then determne a tone number assgnment n for all by solvng: max {n 0, K} subject to: w n f K n N K n f Pm e j K m n j s Pm e j K m n j s R max. SOA2-CNA Here, we are agan assumng that power s allocated unformly over all the channels assgned to a gven user group. Unfortunately, n general the objectve n Problem SOA2-CNA s not concave. However, n the specal case of the uplnk K m = {} t wll be. 12 In the case of the sngle cell downlnk, f nfa/n s ncreasng for all a > 0 as n our general formulaton, then the problem can be re-formulated to have a concave objectve by notng that n ths case t must be that K n = N at any optmal soluton. Addtonally, due to the maxmum rate constrant, the constrant set may not be convex; ths can be accommodated by consderng a hgher dmensonal problem as n Secton Some care s requred at the pont where the SNR constrant becomes actve as the objectve s not dfferentable there; nevertheless, by evaluatng left and rght dervatves the concavty can be shown. 28

29 Next, we focus on solvng Problem SOA2-CNA n the uplnk settng wthout maxmum rate constrants. In ths case, the problem wll have a unque and possbly non-nteger soluton, whch we can agan use a dual relaxaton to fnd. Consder the Lagrangan Ln, λ := P e w n f s λ n N. K n Optmzng Ln, λ over n 0 for a gven λ s equvalent to solvng the followng K subproblems, n λ = arg max w P e n f s λn,. 29 n 0 n Problem 29 can be solved by a smple lne search over the range of 0, N]. Substtutng the correspondng results nto the Lagrangan yelds Lλ := w n P e λ f n K λ s λ n λ N, whch s a convex functon of λ [26]. The optmal value K K λ = arg mn Lλ 30 λ 0 can be found by a lne secton search over: [0, max w f P ē N/K ]13. For a gven search precson, the maxmum number of teratons needed to solve ether 29 or 30 s fxed. 14. Hence, the worst case complexty of the solvng each subproblem s ndependent of K or N. Snce there are K subproblems n 29, t follows that the complexty of the basc assgnment step s OK. If the resultant channel allocatons contan non-nteger values, we wll approxmate wth an nteger soluton that satsfes K n = N. 15 normalzed SNR e = 1 N Snce each user s allocated only a subset of the tones, the j N e j s typcally a pessmstc estmate of the averaged 13 The upperbound of the search nterval can be obtaned by examnng the frst order optmalty condton of For example, f we use b-secton search to solve 29 and stop when the relatve error of the soluton s less than N/2 10, then we only need a maxmum of ten search teratons. 15 One possble nteger approxmaton s the followng. Assume n s the unque optmal soluton of Problem SOA2-CNA. Frst, sort users n the descendng order of the mantssa of n, fr n = n n. That s, fnd a user permutaton subset {α k, 1 k N} such that fr n α 1 fr n α2 fr n αm. Second, for each user, let ñ = n. Thrd, calculate the number of unallocated tones, N A = N ñ. Fnally, adjust users wth large mantssas such that all the tones are allocated,.e., ñ α = ñ α + 1 for all 1 N A. The resultng {ñ } K gve the nteger approxmaton. 29