On Tractability Aspects of Optimal Resource Allocation in OFDMA Systems
aa r X i v : . [ c s . N I] D ec On Tractability Aspects of Optimal ResourceAllocation in OFDMA Systems
Di Yuan , Jingon Joung , Chin Keong Ho , and Sumei Sun Department of Science and Technology, Link¨oping University, Sweden Institute for Infocomm Research (I R), A ∗ STAR, Singapore
Emails: [email protected], { jgjoung; hock; sunsm } @i2r.a-star.edu.sg Abstract
Joint channel and rate allocation with power minimization in orthogonal frequency-division multiple access (OFDMA) has attracted extensive attention. Most of the researchhas dealt with the development of sub-optimal but low-complexity algorithms. In this pa-per, the contributions comprise new insights from revisiting tractability aspects of com-puting optimum. Previous complexity analyses have been limited by assumptions of fixedpower on each subcarrier, or power-rate functions that locally grow arbitrarily fast. Theanalysis under the former assumption does not generalize to problem tractability with vari-able power, whereas the latter assumption prohibits the result from being applicable towell-behaved power-rate functions. As the first contribution, we overcome the previouslimitations by rigorously proving the problem’s NP-hardness for the representative loga-rithmic rate function. Next, we extend the proof to reach a much stronger result, namelythat the problem remains NP-hard, even if the channels allocated to each user is restrictedto a consecutive block with given size. We also prove that, under these restrictions, thereis a special case with polynomial-time tractability. Then, we treat the problem class wherethe channels can be partitioned into an arbitrarily large but constant number of groups,each having uniform gain for every individual user. For this problem class, we present apolynomial-time algorithm and prove optimality guarantee. In addition, we prove that therecognition of this class is polynomial-time solvable.
Keywords : orthogonal frequency-division multiple access, resource allocation, tractability. Introduction
In orthogonal frequency division multiple access (OFDMA) systems, resource allocation amountsto finding the optimal assignment of subcarriers to users, and, for each user, the allocation ofrate or power over the assigned subset of sub-carriers. In this paper, we focus on the problem ofminimizing the total transmit power, subject to delivering specified data rates, by power assign-ment and subcarrier allocation. The popularity of OFDMA for wireless communications hasled to an intense research effort in this area. A majority of the works has focused on heuristicand thus sub-optimal solutions, see, for example, [1–7], and the references therein. For globaloptimality, a branch-and-bound algorithm is developed in [8].We address a complementary but fundamental aspect of the resource allocation problem: Towhat extent is it tractable? In contrast to the significant amount of research on algorithms, worksalong the line of tractability analysis are few [9, 10]. The edge of fundamental understandingof problem tractability is formed in respect of the following limitations. In [10], the power oneach subcarrier is assumed to be given, i.e., it can be arbitrarily set to any fixed value for thepurpose of proving complexity. Thus the result does not apply to the problem where powersare optimization variables. In [9], the resource allocation problem is shown to be NP-hard forone particular type of rate function r mn ( P ) ∈ C inc , where r mn ( P ) is the rate as a function ofpower P for the m th user and n th subcarrier, and C inc is the set of all increasing functions suchthat r mn (0) = 0 . By this result, there exists some (but possibly ill-behaved ) function in C inc for which the problem is NP-hard. However, the result does not carry over to well-behavedsubclasses of functions in C inc . In fact, for any r mn ( P ) ∈ C linear , where C linear is the set of linearfunctions, the problem is solvable in polynomial time (see Section 3). Thus, whereas dealingwith C inc is in general intractable for the problem in question, there exists some subclass in C inc that admits global optimality at low complexity.For the (wide) class of functions in C inc but not in C linear , the tractability of the resource al- Indeed, the proof in [9] relies on a power function (i.e., the inverse of the rate function) growing arbitrarilyfast for arbitrarily small rate increase, meaning that the function is not locally Lipschitz continuous. C concave , for which C linear ⊂ C concave ⊂ C inc holds. Indeed, it is very commonly assumed that the rate function is given by r mn ( P ) =log(1 + g mn P ) , with g mn > . We will thus investigate the complexity for this representa-tive case of C concave , and, if the problem is NP-hard for the function, examine to what extentrestrictions on input structure (e.g., assuming the number of subcarrier for each user is part ofthe input) will admit better tractability. The specific contributions are as follows. • We rigorously prove that given the representative logarithmic rate function r mn ( P ) =log(1 + g mn P ) with g mn > , the resource allocation problem is NP-hard. Since this ratefunction is in C concave , it follows that in the hierarchy C linear ⊂ C concave ⊂ C inc , the hardnessresult holds except for the linear case. The contribution leads a significant refinement ofthe result of tractability. • We extend the NP-hardness analysis to arrive at a much stronger result. Namely, theproblem remains NP-hard, even if the following two restrictions are jointly imposed: 1)the subcarriers allocated to each user form a consecutive block, and 2) the block size isgiven as part of the problem’s input. We also prove that, with these two restrictions, thereis a special case admitting polynomial-time tractability. • We identify a tractable problem subclass, with the structure that the channels can bepartitioned into an arbitrarily many but constant number of groups, possibly with varyinggroup size, and the channels within each group have uniform gain for each user (but maydiffer by user). The original problem is, in fact, equivalent to having the number of groupsequal to the number of subcarriers. The tractable subclass goes beyond the logarithmicrate function – the result holds as long as the single-user resource allocation is tractable.Moreover, we prove that recognizing this problem class is tractable as well.The remainder of the paper is organized as follows. The system model is given in Section 2.In Section 3, we provide and prove the base result of NP-hardness. Section 4 is devoted to3he problem’s tractability with restrictions on channel allocation. In Section 5, we consider theproblem class with structured channel gain and prove its tractability. In addition, we prove thatrecognizing this problem class is computable in polynomial time. Conclusions are given inSection 6.
Consider an OFDMA system with M users and N subcarriers. In this paper, the terms subcarrierand channel are used interchangeably. For convenience, we define sets M = { , . . . , M } and N = { , . . . , N } . Notation r mn ( P mn ) is reserved for the rate as a function of the transmissionpower P mn ≥ for the m th user and n th subcarrier. The inverse function r − mn , returning thepower for supporting a given rate, is denoted by f mn . The required rate of user m is denoted by R m .The forthcoming analysis focuses on the representative increasing, concave rate function r mn ( P mn ) = log (1 + g mn P mn ) , where g mn > represents the channel gain (normalized suchthat the noise variance is one). Hence, the corresponding signal-to-noise ratio (SNR) is g mn P mn .Some of the tractability results generalize to C inc ; we will explicitly mention the generalizationwhen it applies.The optimization problem is to minimize the total power by joint channel and rate allocation,such that the users’ required rate targets are met. The problem is formulated formally below.Throughout the rest of the paper, we refer to the problem as minimum-power channel allocation(MPCA). Input : User set M = { , . . . , M } and channel set N = { , . . . , N } with M ≤ N , positivechannel gain g mn , m ∈ M , n ∈ N , and positive rate targets R m , m ∈ M . Output : A channel partitioning represented by ( N , . . . , N M ) , where N m ⊂ N , m ∈ M , N m ∩ N m = ∅ , for all m , m ∈ M , m = m , and non-negative power P mn , m ∈ , n ∈ N m , such that P n ∈N m log (1 + P mn g mn ) ≥ R m , m ∈ M and the total power P m ∈M P n ∈N m P mn is minimized.By the rate function, there is a unique mapping between rate allocation and power expendi-ture. In addition, at optimum, the rate may be zero on some of the allocated channels of a user.Given the channel allocation, power optimization is determined by solving the single-user rateallocation problem by water-filling [11] in linear time (e.g., [7]). The combinatorial nature ofthe multi-user problem stems from the fact that the users can not share a subcarrier, and the coreof problem-solving is channel partitioning. To motivate the investigation of tractability in view of the current literature, two remarks arenoteworthy. First, an NP-hard problem may become tractable by imposing restrictions to thestructure of its input parameters (e.g., the shape of the objective function). An example is thetraveling salesman problem (TSP) having a cost function with the so called Klyaus-matrix struc-ture (meaning that inversed triangular distance inequality holds). In this case, TSP is solvablein polynomial time [12]. For MPCA, it is in fact polynomially solvable, if the power-rate func-tion would be linear (see the end of this section). As the second remark, an NP-hard problemmay become tractable by removing some of its constraints (and hence enlarging the solutionspace). For example, minimum spanning tree (MST) with constrained node degree is NP-hard,but becomes easily-solved if this constraint is removed.It has been widely accepted that MPCA and other related OFDMA resource allocation prob-lem are difficult. However, to the best of our knowledge, no formal analysis other than theresults in [9, 10] is available. The study in [10] formalizes the NP-hardness result with the as-sumption that the powers on all channels are given. This is equivalent to introducing constraintsfixing the power values. By the second remark above, the result does not answer the tractabilityif these constraints are removed (that is, the original problem with variable power). Indeed, for5ny linear rate function as well as the problem class in Section 5, the NP-hardness proof in [10]remains valid for fixed power, but these problem classes with variable power are solvable inpolynomial time. For the analysis in [9], the proof requires unbounded power growth for arbi-trarily small rate increase. Specifically, for the power function f , f ( n + k ) is k times higherthan f ( n ) , for arbitrary positive integers n and k . This assumption of ill-behaved power-ratefunction excludes not only f ( x ) = 2 x − (the inverse of the logarithmic rate function), but alsoall locally Lipschitz continuous functions. Recall that a (not necessarily continuous) function f is locally Lipschitz continuous, if for any x , there exists a small real number ǫ and an arbitrarilylarge but constant real number D , such that | f ( x + ǫ ) − f ( x ) | ≤ Dǫ , that is, the growth of thefunction is bounded when the change in the input diminishes. Clearly, f is not locally Lipschitzcontinuous, if f ( n + k ) increases by factor k over f ( n ) for any n and arbitrarily large k . Hence,by the previous remark of the impact of cost function on tractability, the tractability under morewell-behaved functions calls for investigation.We provide the tractability results that overcome the limitations of the currently availableanalysis in two aspects. First, we present a rigorous proof of the problem’s NP-hardness withthe representative logarithmic rate function. Second, in the next section, we extend the proofto reach a much stronger result, stating that the problem remains NP-hard even with two heavyrestrictions on channel allocation. Theorem 1.
MPCA, as defined in Section 2, is NP-hard.Proof.
As the proof is rather technical, we outline the basic idea and defer the details to Ap-pendix A. The proof uses a reduction from 3-satisfiability (3-SAT). Two groups of users aredefined. At optimum, each user in the first group either uses one channel of superior channelgain, or splits the rate on three inferior channels, but not both. This corresponds to the true/falsevalue assignment in 3-SAT. The optimal power for this group of users is a constant, while theoptimal power for the second user group gives the correct answer to 3-SAT.
Corollary 2.
MPCA remains NP-hard, even if the rate requirements of the users are uniform. roof. Follows immediately from the equal-rate values used in the proof of Theorem 1.Earlier in this section, it was claimed that MPCA with any linear rate function is tractable.Even if this case is not much of practical interest, it is instructive in showing the importance ofinput assumption on problem tractability.
Theorem 3.
MPCA with linear rate function r mn ( P ) = ℓ mn P , where ℓ mn ≥ , m ∈ M , n ∈N , is solvable in polynomial time.Proof. Since the rate (and hence power) function is linear, it follows that for any user m , itis optimal to allocate the entire rate R m to a single channel. Specifically, denoting by N m thechannel set allocated to m , the optimal selection is the channel giving min n ∈N m /ℓ mn . Becauseof this structure at optimum, MPCA reduces to pairing the M users with M out of the N channels. Hence the problem is equivalent to a minimum-weight matching problem (also knownas minimum-cost assignment [13]) in a bipartite graph with node sets M ∪ { M + 1 , . . . , N } and N ; the former represents the augmentation of M by M − N artificial users. For edge ( m, n ) ,with m ∈ M and n ∈ N , the cost is /ℓ mn . All edges adjacent to artificial users have zero cost.Because the assignment problem is polynomial-time solvable, the theorem follows. Remark
In light of Theorem 1, Corollary 2, and Theorem 3, we remark on the significance ofassumption of input on tractability, by revisiting the result provided in [10]. In this reference,the result is proven for fixed power, that is, the power of each channel is part of problem input.Under this condition, [10] provides an elegant hardness proof of a reduction from the numberpartitioning problem, by setting specific power values on the channels. As long as power isfixed, a line-by-line copy of the proof in [10] remains valid even if the underlying MPCA ratefunction is linear. A similar observation applies to the tractable problem class that will bedetailed in Section 5. In conclusion, for the original MPCA problem having power allocation aspart of the output (for which the key assumption of [10] does not apply), our analysis providesnew insights in the tractability rather than contradicting the previous results. (cid:3) Tractability with Restrictions on Channel Allocation
Consider imposing jointly two restrictions to channel allocation. First, the number of channelsto be allocated to each user is given. Tractability under this restriction is of significance to two-phase OFDMA resource allocation (see [14]) that determines the number of channels per userin phase one, followed by channel allocation in phase two. The second restriction is the use ofconsecutive channels, that is, the channels of every user must be consecutive in the sequence , . . . , N ; this channel-adjacency constraint has been considered in, for example, [15]. Weprove that MPCA remains NP-hard even with these two seemingly strong restrictions, althoughthere is a special case admitting polynomial-time tractability. Theorem 4.
MPCA remains NP-hard, even if the number of channels allocated to each user,i.e., the cardinality of N m , m = 1 , . . . , M , is given in the input, and N m , m = 1 , . . . , M mustcontain consecutive elements in the channel sequence , . . . , N .Proof. As we prove the result via an augmentation of reduction proof of in Appendix A, thedetails are deferred to Appendix B. In brief, the proof is built upon introducing a set of additionalchannels in the MPCA instance in Appendix A, in a way such that, at optimum, the channelsallocated to each user are consecutive and the corresponding cardinality is known.Consider further narrowing down the problem to the case where N consists of M uniform-sized subsets of consecutive channels, and N is a multiple of M . For this case, the problem istractable, as proven below. Theorem 5. If N is divisible by M , and it is restricted to allocate exactly NM consecutive chan-nels to each user, then MPCA is tractable with time complexity O (max { M , M N } ) .Proof. We prove the result by a polynomial-time transformation to the well-solved matchingproblem in a bipartite graph. For the problem setting in question, channel partition is unique- the M subsets created by the partition are { , . . . , N/M } , { N/M + 1 , . . . , N/M } , . . . , { N − N/M + 1 , . . . , N } , each containing N/M channels. The bipartite graph has M nodes8epresenting the users and the subsets of channels. For each pair of the two node groups, sayuser m and channel subset S , there is an edge of which the cost equals the power of meetingthe user rate using the channels in S . This cost is computed in O ( NM ) time (by single-user rateallocation). The complexity of computing all the edge costs is hence O ( M N ) . Maximum-weighted matching in a bipartite graph of V nodes and E edges is solved in O ( V log V + V E ) time [13]. In our case, the second term is dominating, giving a time complexity of O ( M ) ,which completes the proof.Theorem 5 provides a generalization of the trivial case of M = N . For M = N , solvingMPCA amounts to finding an optimal matching with complexity O ( M ) . Moreover, from theproof, it follows that the analysis generalizes to any rate function in C inc , except that the com-plexity of single-use rate allocation has to be accounted for accordingly. The observation yieldsthe following corollary. Corollary 6.
The time required for computing the optimum to the MPCA problem class inTheorem 5 is of O (max { M , M T ( NM ) } ) , where T ( NM ) denotes the time complexity of optimalrate allocation of a single user on NM channels. Denote by K a (possibly large) fixed positive integer independent of M or N . Let K = { , , · · · , K } . Consider MPCA in which the channels can be partitioned into (at most) K groups where every user in the same group has the same channel gain, but the channel gainsdiffer by user. Thus, we can write N = S k ∈K N k , such that for any channel n ∈ N k , thechannel gain depends only on the user index m , i.e., g mn = g m . Equivalently, the rate functionsbelong to the subclass that satisfy r mn = r m , n ∈ N k , k ∈ K , m ∈ M . Note that the channelgroups may vary in size, and the channel gain still differs by user within each group. In thesequel, we refer to the problem class as K -MPCA. The problem class is justified by scenarioswith K distinct bands and channel difference is overwhelmingly contributed by the separation9f the bands in the spectrum, whereas the subcarriers with each band are considered invariantfor each user.Consider -MPCA. The problem structure is significantly simpler than the general case.Namely, the optimization decision is no longer which, but how many channels each user shoulduse. It follows that, for any subset of users M ′ ⊂ M , the optimum allocation of h channels(with h ≥ |M ′ | ) among the users in M ′ is independent of channel allocation of the rest theusers. Thus -MPCA exhibits an optimal substructure, i.e., a part of the optimal solution is alsooptimal for that part of the problem. The observation leads to a dynamic programming line ofargument for problem-solving. As proven below, the solution strategy guarantees optimality inpolynomial time. Theorem 7.
Global optimum of -MPCA can be computed by dynamic programming in O ( M N ) time.Proof. Consider the partial problem of optimally allocating h channels to the users in { , . . . , m } with h ≥ m , and c m ( h ) the corresponding optimum power. Clearly, at the optimum of this sub-problem, the number of channels of user m is an integer in the set { , , . . . , h − m + 1 } . (Theupper bound h − m + 1 corresponds to having m − channels left for the other m − users.)Allocating k ∈ { , , . . . , h − m + 1 } channels to user m , the power equals p km + c m − ( h − k ) ,where p km denotes the power of user m with k channels. This gives the following recursiveformula for computing the optimal number of channels for user m . c m ( h ) = min k =1 ,...,h − m +1 { p km + c m − ( h − k ) } (1)We arrange the values c m ( h ) for m = 1 , . . . , M and h = 1 , . . . , N in an M × N matrix. Entriescorresponding to infeasible solutions are called invalid, and their values are denoted by ∞ . Inthe matrix, c m ( h ) = ∞ for all entries where h < m , or h > N − M + m . We compute the validentries as follows. For the first row, computing the entries c (1) , . . . , c ( N − M +1) in the givenorder are straightforward, and each entry requires O (1) computing time. Next, entries c m ( m ) ,10.e., one channel per user for the first m users, are calculated in O ( M ) time for m = 1 , . . . , M .The bulk of the computation calculates the remaining entries row by row, starting from rowtwo. For row m , the computations follow the order c m ( m + 1) , . . . , c m ( N − M + m ) . Each ofthese entries is calculated using formula (1). For the valid entries of a row, the total number ofcomparisons that they are used for computing the next row is · · · + N − M + 1 . Hence thecomplexity for computing row m, m = 2 , . . . , M , is of O ( N ) , and the overall time complexityis of O ( M N ) . The last entry computed, c M ( N ) , gives the optimal allocation of the N channelsto the M users and hence solves -MPCA. In parallel, the solution is stored in a second matrixof same size. Solution recording clearly has lower complexity than O ( M N ) , and the theoremfollows.In the following theorem, we generalize the dynamic programming concept to any positiveinteger K . The generalized algorithms are able to solve K -MPCA to global optimality. Theorem 8.
Global optimum of K -MPCA can be computed by dynamic programming in O ( M N K ) time.Proof. Let N j = |N j | , j = 1 , . . . , K . For the first m users, denote by c m ( h , . . . , h K ) theoptimum power of allocating h j channels of group, where h j ≤ N j , j = 1 , . . . , K . De-note by p ( k ,...,k K ) m the power for user m , if it is allocated k j channels of channel group j , j = 1 , . . . , K . We introduce the convention that p (0 ,..., m = ∞ for convenience. By enumeratinguser m ’s allocation of channels of the K groups, we obtain the following recursion formula for c m ( h , . . . , h K ) . c m ( h , . . . , h K ) = min k j ∈{ ,..., min { h j ,N − M +1 }} ,j =1 ,...,K { p ( k ,...,k K ) m + c m − ( h − k , . . . , h K − k K ) } (2)Extending the algorithm in the proof of Theorem 7, the corresponding matrix for K -MPCAhas dimension M Q j =1 ,...,K N j , which does not exceed O ( M N K ) . To compute an entry, there11re no more than O ( N K ) calculations (including addition and comparison) using (2). For eachcalculation, the time required to compute p ( k ,...,k K ) m is linear in K . Since K is a constant, thiscomputation does not add to the complexity. These observations lead to the overall complexityof O ( M N K ) . Finally, entry c M ( N , . . . , N K ) is clearly the optimum to K -MPCA. The proofis complete by observing that, similar to -MPCA, recording the channel allocation solutiondoes not form the computational bottleneck. Remark
The polynomial-time tractability of K -MPCA holds only if K is not dependent on M or N . In fact, the general setting of MPCA is equivalent to N -MPCA, i.e., N channel groupswith single channel each. The dynamic programming algorithm remains applicable for K = N .From Theorem 2, however, the algorithm corresponds to enumerating the solution space, andthe running time is exponential. (cid:3) Having concluded the tractability of K -MPCA, a natural question to ask next is whetheror not the identification of the problem class is tractable as well. Theorem 9 states that this isindeed the case. Theorem 9.
Recognizing K -MPCA can be performed in O ( M N ) time for K = 1 , and in O ( M N ) time for any K ≥ .Proof. For K = 1 , identifying the problem class simply amounts to verifying, for each user,whether or not all the N channels are of the same gain; this immediately leads to the O ( M N ) time complexity result. For K ≥ , we construct a graph G . The graph has N nodes, eachrepresenting a channel in N . Consider two arbitrary channels n , n ∈ N , n = n . If the twochannels have the same gain for each of the users, i.e., g mn = g mn , ∀ m ∈ M , we denote it by n ∼ = n . Checking whether or not this is the case runs obviously in O ( M ) time. If n ∼ = n ,we add edge ( n , n ) to G . Doing so for all unordered channel pairs has time complexity O ( M N ) . Next, note that the equivalence relation of channels is transitive, i.e., if n ∼ = n and This result follows by directly applying the single-user rate assignment to K channel groups, where the chan-nels in each group have uniform gain. ∼ = n , then n ∼ = n . Hence channels that are equivalent for all users form a clique in G ,whereas channels that differ in gain for at least one user are not connected in G . Consequentlythe number of strongly connected components in G equals the number of channel groups, eachof which contains channels being equivalent for any user. Identifying the number of stronglyconnected components requires no more than O ( N ) time for G . Therefore the bottleneck liesin the O ( M N ) complexity of obtaining the graph, and the theorem follows. Remark
The tractability results of this section are not restricted to the specific rate/powerfunction defined in the section of system model. The problem class remains tractable (al-though the overall complexity may grow) for any function in C inc , as long as the function admitspolynomial-time rate allocation of single user. (cid:3) We have considered the OFDMA resource allocation problem of minimizing the total powerof channel allocation, so as to satisfy some rate constraints. Although it has been known thatassuming the most general (and ill-behaved) increasing rate functions leads to NP-hard prob-lems, we have shown that the same conclusion holds even if we restrict the class to increasingand concave rate functions. Interestingly, the problem admits a polynomial-time solution ifthe rate function is an increasing linear function. Hence, progress in the fundamental under-standing on the tractability of the problem is made in the following sense: we have sharpenedthe boundary of tractability to between increasing concave and increasing linear rate functions.Finally, we have also identified specific cases when the problem remains NP-hard, or admitspolynomial-time solutions, under various restrictions.13
Proof of Theorem 1
There is no doubt that MPCA is in NP. The NP-hardness proof uses a polynomial-time reductionfrom the 3-satisfiability (3-SAT) problem that is NP-complete [16]. A 3-SAT instance consistsin a number of boolean variables, and a set of clauses each consisting of a disjunction of exactlythree literals. A literal is either a variable or its negation. The output is a yes/no answer towhether or not there is an assignment of boolean values to the variables, such that all the clausesbecome true. Denote by v and w the numbers of variables and clauses, respectively. For anybinary variable z , its negation is denoted by ˆ z . For the proof, we consider 3-SAT where eachvariable and its negation together appear at most 4 times in the clauses. Note that 3-SAT remainsNP-complete with this restriction [17]. Without loss of generality, we assume that each variable z appears in at least one clause, and the same holds for its negation ˆ z , because otherwise theoptimal value of the variable becomes known, and the variable can be discarded. Hence thetotal number of occurrences of each literal in the clauses is between one and three.We construct an MPCA instance with M = 2 v + w and N = 7 v + w . We categorize theusers and channels into groups, and, for convenience, name the groups based on their roles inthe proof. The users consist in v literal users and w clause users. The channels are composedby three groups: v super-channels, v literal channels, and w auxiliary channels. The rate target R m = 1 . , ∀ m ∈ M .Problem reduction is illustrated in Fig. 1. For each binary variable z , three identical literalchannels, denoted by z , z ′ and z ′′ , are defined. A similar construction is done for ˆ z . For thisgroup of six literal channels, one super-channel is defined. We introduce two literal users for theseven channels. One user has channel gain g l on the three literal channels z , z ′ , and z ′′ , and theother, complementary literal user has channel gain g l on the remaining three literal channels.Both users have channel gain g s on the super-channel. See Fig. 1(a). Next, recall that eachliteral appears at most three times in the clauses in the 3-SAT instance. In the proof, for anybinary variable z appearing t times in total in the clauses, with ≤ t ≤ , the occurrences arerepresented by any t elements in { z, z ′ , z ′′ } in any order. A similar representation is performed14 uper-channelLiteral channels Literal user z'z''z z'z''z^ Literal user g s g s g l g l g l g l g l g l ^^ (a) Users and channels for a variable. Auxiliary channelAn example of a clause z' g c v ( z z'' v ) Clause user g c g c g a ^ (b) User and channels for aclause. Figure 1: An illustration of problem reduction.for the negation ˆ z . For each clause, we introduce one clause user with gain g c on the channelscorresponding to the original literals in the clause. In addition, one auxiliary channel is definedper clause user with channel gain g a . See Fig. 1(b). We set g s = g c = 1 , g a = . w +0 . , g l = g a = · (0 . w +0 . . For the user-channel combinations other than those specified, thechannel gain is g ǫ = w . For each user, we refer to the four channels with gain higher than g ǫ as valid channels, and the other v + w − channels with gain g ǫ as invalid channels. From theconstruction, clearly the reduction is polynomial.We provide several lemmas characterizing the optimum to the MPCA instance. The firstthree lemmas use the following optimality conditions of single-user rate allocation (e.g., [7]).First, for any user, the derivatives of the power function, evaluated at the allocated rates, areequal on all channels with positive rates. Second, for channels not used, the function derivativesat zero rate are strictly higher than those of the used channels. For f ( x ) = x − g , where x is therate allocated and g is the channel gain, the derivative f ′ ( x ) = ln(2) x g . Lemma 1.
There is an optimum allocation in which no user is allocated any invalid channel.Proof.
Suppose that at optimum a clause user m is allocated at least one invalid channel.Assume m is also allocated any valid channel, then the invalid channels carry zero rate, becauseputting the entire rate of 1.0 on the auxiliary channel, the function derivative is at most ln(2) · g a = 2 ln(2)(0 . w + 0 . , whereas the function derivative for any invalid channel, at zero rate,15s ln(2) · w > . w + 0 . . Thus the invalid channels can be eliminated from theallocation of m . Assume now the auxiliary channel is allocated to another user, say m . Byconstruction, the auxiliary channel of m is invalid for m . Consider re-allocating any invalidchannel of m to m , and allocating the auxiliary channel to m . Clearly, the total power willnot increase. At this stage, the remaining invalid channels allocated to m carry zero flow.Repeating the argument, we obtain an optimal allocation in which no clause user is allocatedany invalid channel.Let m be any literal user and suppose it is allocated one or more invalid channels at opti-mum. If m is allocated any of its four valid channels, the function derivative at rate 1.0 is atmost ln(2) · . w + 0 . < ln(2) · w , and therefore the invalid channels carry zero rate andcan be removed from m ’s allocation. Assume therefore all four valid channels are allocatedto other users. Consider any literal channel of m , and suppose it is allocated to user m . For m , this literal channel is an invalid one. Swapping the allocation of the literal channel and anyinvalid channel currently allocated to m , the total power will not grow, and the remaining in-valid channels allocated to m can be released. The lemma follows from applying the procedurerepeatedly. Lemma 2.
If a literal user is allocated its super-channel in the optimal solution, then none ofthe three literal channels is allocated to the same user.Proof.
Putting the entire rate of 1.0 on the super-channel, the derivative value is . Forany literal channel, the derivative at zero rate is ln(2) g l . That g l = · (0 . w +0 . and w ≥ lead to ln(2) g l > , and the result follows. Lemma 3.
If a literal user is not allocated its super-channel in the optimal solution, then theuser is allocated all the three literal channelsProof.
By the assumption of the lemma and Lemma 1, the literal user in question is allocatedone, two, or all three of its literal channels, with total power f = g l , f = 2 · / − g l , and16 = 3 · / − g l , respectively. Note that, in the latter two cases, it is optimal to split the rateevenly because of the identical gain values. Clearly, f < f < f .We prove that cases one and two are not optimal. Suppose that, at optimum, the literal useris allocated two of the literal channels. Then the remaining literal channel is used to carry astrictly positive amount of flow of a clause user. Consider modifying the solution by allocatingall the three literal channels to the literal user, and letting the clause user use the auxiliarychannel only. The power saving for the literal user is exactly f − f > . g l > g a , whereas thepower increase for the clause user is less than g a . This contradicts the optimality assumption.Hence case two is not optimal. Since f − f > f − f , a similar argument applies to case one,and the result follows.By Lemmas 2–3, at optimum, a literal user will use either the super-channel only, or allthe three literal channels. Hence, for any literal in the 3-SAT instance, either none or all ofthe corresponding three literal channels become blocked for the clause users. Consequently,there is a unique mapping between a true/false variable assignment in the 3-SAT instance andthe availability of literal channels to the clause users in the MPCA instance. The total powerconsumption of all the literal users equals exactly v + 78 v · (2 / − . w + 0 . at optimum.In the remainder of the proof, we concentrate on the power consumption of the clause users. Lemma 4.
If every clause user is allocated at least one of the three literal channels correspond-ing to the literals in the clause in the 3-SAT instance, then the total power for all clause usersis at most w .Proof. Allocating the entire rate of 1.0 to one literal channel gives a power consumption of − g c = 1 for the clause user. As there are w clause users, the lemma follows. Lemma 5.
If at least one clause user is not allocated any of its three literal channels, the totalpower for all clause users is strictly higher than w .Proof. By the assumption, at least one clause user is allocated the auxiliary channel only, withpower g a . Each of the other w − clause users is allocated at most four channels. Assuming the17vailability of all four channels and setting g a = g c leads to an under-estimation of the powerconsumption. The under-estimation has a total power of w − / − g c + g a = 4( w − / −
1) + (0 . w + 0 . > . w − . . w + 0 . . w − . ≥ w .By Lemmas 4-5, the optimum power for the clause users is at most w if and only if theanswer is yes to the 3-SAT instance. Thus the recognition version of MPCA is NP-complete,and its optimization version is NP-hard. B Proof of Theorem 4
In the proof of Theorem 1, a 3-SAT instance of v variables and w clauses is reduced to an MPCAinstance with v + w users and v + w channels, such that no user will be allocated more thanthree channels at optimum. We make an augmentation by adding v channels, which we refer toas dummy channels. The channel gain of the dummy channels equals g ǫ (defined in the proof ofTheorem 1) for all users. After the augmentation, there is a total of v + w channels, organizedin three blocks. The sequence of channels is as follows. The first block has v channels,including the v super-channels and the v dummy channels, in a sequence of v chunks of channels each. Each chunk is composed by one super-channel and two dummy channels. Thenext block has the v literal channels, with v chunks having channels each. Every chunkcorresponds to a binary variable z in the 3-SAT instance, and the six literal channels appear inthe order z, z ′ , z ′′ , ˆ z, ˆ z ′ and ˆ z ′′ . The third block contains the w auxiliary channels. Consider theresulting MPCA instance with the restriction that, for the literal and clause users, respectively,the numbers of channels allocated per user are three and one. In addition, channel allocationmust be consecutive in the given sequence.To prove the hardness result, consider first a relaxation of the problem, in which the tworestrictions of channel allocation are ignored for the literal users. For this relaxation, it is clearthat Lemmas 1-3 remain valid. By Lemma 1 and the signal-channel restriction of the clauseusers, each of these users will be allocated one of the four valid channels. Obviously, the18ptimum is to allocate one literal channel, or the auxiliary channel if all the three literal channelsare allocated to literal users. Thus, as long as at least one literal channel is available to everyclause user, the result of Lemma 4 holds. In addition, the validity of Lemma 5 obviouslyremains. Therefore the optimum to the problem relaxation provides the correct answer to the3-SAT instance.By Lemma 3, in the optimum of the relaxed problem, each literal user is either allocated itsthree consecutive literal channels, or the super-channel. In the latter case, we modify the solu-tion by allocating the two dummy channels accompanying the super-channel, without changingthe total power. After the modification, the allocation satisfies the cardinality requirement andrestriction of using consecutive channels for all literal users, and the theorem follows. References [1] C. Y. Wong, R. S. Cheng, K. B. Letaief, and R. D. Murch, “Multiuser OFDM with adaptivesubcarrier, bit, and power allocation,”
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