Uplink Scheduling in Wireless Networks with Successive Interference Cancellation

1 Ulink Scheduling in Wireless Networks with Successive Interference Cancellation Majid Ghaderi, Member, IEEE, and Mohsen Mollanoori, Student Member, IEEE, Abstract In this aer, we study the roblem of ulink scheduling in wireless networks with successive interference cancellation SIC With SIC, concurrent transmissions, if roerly scheduled, can be successfully decoded at a receiver The scheduler decides: i in which time-slot to schedule, and ii in what order in a time-slot to decode each transmission in order to maximize the system utility and/or satisfy a system constraint These two scheduling decisions effectively determine the rates allocated to concurrent transmissions, which in turn determine the throughut and fairness of the system We consider several different scheduling roblems in this context The objective of the roblems is to either maximize the throughut of the system or to obtain some kind of fairness among the users We formulate and study each roblem from the ersective of comutational comlexity For each roblem, we either roose a olynomial time algorithm, if any exists, or show that the roblem is NP-hard Index Terms Successive Interference Cancellation, Scheduling in Wireless Networks, Throughut and Fairness Maximization 1 INTRODUCTION INTERFERENCE and noise are both limiting factors in wireless networks owever, their effects on the erformance of wireless networks are not identical In a multi-user wireless network, increasing the transmission ower can reduce the noise effect Nevertheless, an increase in the transmission ower, if not carefully controlled, may even worsen the interference effect In addition, interference is a structured signal since it is caused by other transmissions in the network, whereas noise is structureless Successive interference cancellation SIC is a multi-user detection technique that uses the structured nature of interference to decode multile concurrent transmissions Assume a comosite signal S = S 1 + + S n + Z of n overlaing signals S 1,, S n lus the noise signal Z is received at a receiver Emloying SIC, one of the signals, say S i, is decoded first while the rest of the signals are considered as noise After S i is decoded, the decoder reconstructs the corresonding ana signal and subtracts it from the original comosite signal S At this stage, the remaining signal is free from the interference of signal S i assuming error-free decoding and erfect removal of the signal S i The same technique is alied reeatedly to decode the remaining n 1 signals It is worth noting that if the decoder fails to decode one of the signals, it is unlikely that it can decode the subsequent signals Since at every stage, the remaining signals are treated as noise, the maximum rate achievable by a user deends not only on its corresonding received signal ower but M Ghaderi and M Mollanoori are equal contributing authors M Ghaderi is with the Deartment of Comuter Science, University of Calgary E-mail: mghaderi@ucalgaryca M Mollanoori is with the Deartment of Comuter Science, University of Calgary E-mail: mmollano@ucalgaryca also on the order at which its signal is decoded While SIC does not enforce a secific decoding order, because of ractical and theoretical reasons, it is common to decode the signals from the strongest to the weakest one we discuss the issue in Sections 6 and 7 Since the amount of interference is reduced in every stage because of subtraction of the decoded signal, late decoding of the weaker signals increases the corresonding SINR Signal to Interference lus Noise Ratio and consequently increases the robability of a successful decode ater, we show that while the above order of decoding does not decrease the sum caacity of the system, it actually results in a roortionally fair rate assignment among users With conventional decoders, transmission scheduling mechanisms are commonly used to avoid multile simultaneous transmissions because conventional receivers treated interference as random noise owever, to increase the throughut with SIC, the scheduling mechanism should allow multile concurrent transmissions in the network while controlling the rate and decoding order of the transmissions so that the comosite signal can be decoded at the receiver [1] SIC receivers are architecturally similar to traditional non-sic receivers in terms of hardware comlexity and cost [2] as they use the same decoder to decode the comosite received signal at different stages As a result, neither a comlicated decoder nor multile antennas are required to increase the throughut of the system [1], [3] It also makes SIC more ractical than other multi-user detection techniques such as joint detection [4] Furthermore, it is known that other multile access techniques such as CDMA and OFDMA are no more efficient than SIC [5, Ch 6] As such, SIC has been recently considered in commercial wireless systems as a way to increase system throughut [6] SIC can be emloyed for ulink [7] as well as down-

2 link [8] transmissions owever, SIC can achieve a higher throughut gain on the ulink of a wireless network since the total received ower is higher due to concurrent transmissions from multile users In Section 3, we further discuss the relation between the total received ower and caacity of the system In this aer, we consider several ulink scheduling roblems, assuming that SIC is imlemented at the hysical layer of the network To the best of our knowledge, none of the roblems considered in this aer and summarized as follows is studied by others in the context of SIC 1 Maximum Throughut Scheduling MTS: The objective of this roblem is to schedule a set of users at a given number of time slots such that the throughut of the system is maximized The throughut of the system is given by the summation of the throughut of individual users We show that MTS is NP-hard 2 Constrained Maximum Throughut Scheduling MTS C: This roblem is the same as MTS excet that at most l users are scheduled at each time slot We show that for l = 2, which is of ractical interest, the roblem is solvable in olynomial time, while for any l 3 the roblem is NP-hard 3 Proortional Fair Scheduling PFS: The objective of this roblem is to achieve roortional fairness among users [9] Proortional fairness among users is achieved by maximizing the summation of the arithm of individual users throughuts ence, the only difference between PFS and MTS scheduling roblems is the additional function in the objective function of PFS We show PFS can be solved in olynomial time 4 Scheduling with Minimum SINR MinSINR: This roblem requires that a minimum SINR for each user is rovided by the scheduling algorithm aving a minimum SINR for each user means a minimum rate guarantee for each user, which is a required roerty for some alications such as voice or video transmission We show this roblem is solvable in olynomial time In addition, we discuss variations of the above roblems as corollaries, wherever ossible The aroach we take in this aer is a combinatorial and comlexity theoretic one We aim to find the theoretical limits of comutation of the otimal solutions to the scheduling roblems with SIC To do so, for each of the roblems, we rove whether the roblem can be solved in olynomial time or it does not have any olynomial time solution, unless P = NP From the theoretical oint of view, NP-hardness of a roblem means that there is no olynomial time algorithm to comute the otimal schedule unless, P = NP From the ractical oint of view, it means that there is no otimal real-time scheduler that maximizes the system utility, excet for very small system sizes Note that in real world wireless systems, scheduling is erformed every few milliseconds This makes a brute-force search to find the otimal schedule imractical even for a few tens of users, the size of the search sace becomes too large The rest of the aer is organized as follows: In the next section, we describe the related work In Section 3 we describe the system model and assumtions Problems MTS, MTS C, PFS, and MinSINR are studied in Sections 4 to 7 Section 8 resents the simulation results Section 9 concludes the aer 2 REATED WORK Although scheduling is extensively studied in traditional wireless networks eg, [10] and [11], only a few works have exlicitly considered interference cancellation There have been a number of studies in the area of SIC-aware MAC rotocols It has been shown that SIC can achieve near Shannon caacity in cellular networks in which communications are closed-loo and accurate channel estimations exist see [2] and references therein It is also known that SIC can imrove the throughut of wireless ANs For instance, alerin et al [12] exerimentally studied the effect of SIC in unmanaged wireless networks with carrier sensing They concluded that SIC can effectively imrove the bandwidth utilization in wireless networks In contrast, Sen et al [13] reorted that, without a roer scheduling mechanism, SIC may not be a romising aroach to imrove the throughut of wireless networks In [14], Gollakota et al roosed a combination of interference alignment and cancellation as an effective technique to overcome the limitation of the number of antennas for imroving the throughut of MIMO wireless ANs They showed that the combined technique can be alied to scenarios where neither interference alignment nor cancellation alies alone In a recent work, Wu et al [15] utilized interference cancellation at the receiver to create a side channel that can be used to imlement a multi-user scheduling algorithm without requiring any out-of-band signalling The idea is to use the slack interference tolerance of the receiving hardware to allow simultaneous transmission of actual user data and control information We note that interference cancellation at the hysical layer is orthogonal to coding and transmission techniques that try to decode simultaneous transmissions at higher layers of the rotocol stack such as ZigZag decoding [16] and Rema [17] Yu and Giannakis roosed an algorithm called SICTA which uses SIC in a tree algorithm [18] SICTA retains the received signal vector resulting from a collision in a time slot The received signal vector of the subsequently decoded ackets is used to cancel the interference in the collided signal and decode it In [19], Mitran et al consider a multi-rate and multi-ower wireless multiho network with SIC caability They assume that the receiver is caable of decoding u to n signals

3 They formulate a joint routing and scheduling with SIC using a large linear rogram and comute the maximum achievable throughut of the network by solving the formulation numerically in a centralized manner There is also a large body of research on scheduling links in wireless multiho networks with a minimum frame length For instance, Goussevskaia et al [20] show that two roblems, namely scheduling and one-shot scheduling, under the geometric SINR model are NP-comlete In the first roblem given a set of links, it is asked to schedule the links in the minimum number of time slots ossible The second roblem, receives a weighted set of links and finds a subset of the links that can be scheduled in a single time slot so that the scheduled links have the maximum sum of weights See [20] and references therein for more details on this toic In the broader context of multi-acket recetion, there are several works on scheduling and MAC rotocols [21] [23], yet none of them is alicable to SICenabled networks As an examle, Zhao and Tong [21], [24] develoed a MAC rotocol with multi-acket recetion caability considering a recetion matrix as the underlying model A recetion matrix secifies the robability of k successful recetions when there are n concurrent transmissions in the network ence, the model does not differentiate between the concurrent transmissions eg, all the transmissions are suosed to have the same rate While the recetion matrix model is a clever abstraction of a general multi-acket recetion network, a more sohisticated model is required to accurately model the underlying dynamics of a SIC-based network such as the selection of transmission rates or the order of decoding see Section 3 Other works on multi-acket recetion, such as [22] and [23], consider a simler model in which a receiver is caable of receiving u to k simultaneous ackets, regardless of the transmission rates and channel conditions of users The closest work to ours is by Kumaran and Qian [25], in which the authors consider the ulink scheduling roblem with SIC Nevertheless, their work differs in that they only consider the case of scheduling multile transmissions in a single time-slot in order to maximize the network throughut In this work, we consider scheduling multile users in multile time-slots in order to otimize an objective function and/or satisfy a system constraint 3 SYSTEM MODE AND ASSUMPTIONS We consider a network consisting of a set of wireless users communicating with a single receiver eg, an access oint in a wireless AN or a base station in a cellular network Time is divided into scheduling frames, where each scheduling frame is divided into k time slots Scheduling is done once every scheduling frame, at the beginning of the frame this is similar to the scheduling structure of WiMAX networks [26] I 9 9 5 1 6 8 3 7 4 2 Fig 1 ogical reresentation of the scheduling frame with 3 time slots and 9 owers is the noise ower I i secifies the sum of interference owers from other users, eg, in the above figure I 9 = 1 + 5 Note that I i is a function of the location of u i s signal in the scheduling frame, ie, relocation may change I i Figure is drawn such that in each time slot i s are decoded from to to bottom Moreover, the height of the rectangles reresent the signal owers In every scheduling frame, a set of n users denoted by N = {u 1, u 2,, u n } is scheduled for ulink transmission We assume that the users reort their channel state information at the start of every scheduling frame and that channel fluctuations during a frame are negligible Without loss of generality, we ignore the ower control and assume that the users transmit at full ower so that the scheduler is able to estimate the received ower from each user et i denote the received ower of the user u i at the receiver and r i denote the throughut of the user u i Given the set of received owers 1, 2,, n, the aim of the scheduler is to schedule the set of users N in the k time slots so that a system utility function is maximized and/or a system constraint is satisfied et SINR i be the signal to noise lus interference ratio of user u i, ie, SINR i = ower that affects the signal of user u i To model the throughut of the users, we use the Shannon caacity formula as follows, i +I i, where I i is the interference r i = 1 + SINR i 1 Figure 1 shows the ical reresentation of a scheduling frame The figure is drawn such that signals are decoded from to to bottom For instance, in Figure 1, the signal of user u 9 is decoded just before the signal of user u 5 Therefore, the amount of interference affecting a user s signal is the sum of the received owers scheduled below that user in the same time slot The following well-known identity is crucial for many of the roofs throughout the aer and is formulated here for the ease of reference: The sum of the arithms is the arithm of the roduct of the terms More formally, let x i > 0 for i = 1,, n We have, x 1 x n = x 1 + + x n 2 We use the notation { a 1, a 2,, a k1, b 1, b 2,, b k2, } to show the ical reresentation of the frame That is, in one time slot a 1 is decoded first, a 2 is decoded

4 next, and so on until a k1 In the next time slot b 1 is decoded first, b 2 is decoded next, and so on For instance, the frame of Figure 1 is shown as: { 9, 5, 1, 6, 8, 3, 7, 4, 2 } Note that, { } shows an orderless list, while reresents an ordered list That is, the order of the time slots in a scheduling frame is not imortant while the order of decoding of the signals is imortant for us By this model, I i and consequently the throughut of each user deends on the time slot in which the user is scheduled as well as the order at which the signals are decoded owever, the sum caacity of the users scheduled at a time slot is indeendent of the order of decoding and deends solely on the sum of the received owers i i and the noise ower emma 1 states this roerty formally emma 1 et 1,, m be the received owers of the users scheduled in a single time slot The following equality holds, m 1 + m i = 1 + i 3 + I i Assume without loss of generality that the decoding order is m,, 1 Proof of emma 1 is simly by exanding the left hand side of 3 as, + 1 i + 2 i + + + 4 + 1 i + m 1 i + m i +, + m 2 i + m 1 i and using 2 to simlify the exression It is worth ointing out that emma 1 states two imortant roerties of the sum caacity: 1 Sum caacity is indeendent of the decoding order of the signals That is because m i is a constant regardless of the order of decoding 2 Increase in the sum caacity is roortional to the arithm of the sum of the received owers Theoretically, the second roerty means: if m goes to infinity, sum caacity goes to infinity as well Therefore, to maximize the system throughut, all the users have to be scheduled in every time slot if there is no constraint on the number of users that can be decoded using SIC [5], [25] owever, in ractice, because of a number of reasons, only a few users can be scheduled at each time slot First, since the increase in the sum caacity is arithmic in the sum of the received owers, having a few concurrent transmissions, significantly increases the sum caacity owever, adding more users to the concurrent transmissions would be less effective as the number of users is increased Second, due to the ractical limitations, only a few overlaing signals can be decoded successfully [1] One reason is that the decoding time in SIC increases linearly with the number of overlaing signals [5, Ch 6] aving many users, the linear decoding time causes ractical difficulties Another reason is that, in ractice, the decoded signals cannot be erfectly removed from the comosite signal Thus, some residual interference remains after each decoding stage, and roagates during the subsequent decoding stages [6] The accumulated residual interference results in a decoding error after a few users are decoded, limiting the number of simultaneous transmissions in a time slot To take into consideration the ractical limitations, we consider two variations of the MTS roblem namely, MTS C and MinSINR In MTS C, we assume that at most l users can be decoded in each time slot In MinSINR, we assume that the decoder requires a minimum SINR denoted by S min to successfully decode each signal We assume that the number of users is greater than the number of time slots, ie, N = n > k, since the solution for the case of n k, for all of the roblems, is trivial We assume that each node is scheduled exactly once in a scheduling frame, thus more than one node might be scheduled in a time slot recall that scheduling simultaneous transmissions is allowed with SIC While this restriction can be relaxed, its consideration here imoses some imlicit fairness among the users To relax the restriction, one may first schedule each of the nodes only once Then, based on some criteria, select a subset of the nodes and add them to the scheduling frame such that none of the nodes are scheduled more than once in a time slot In this manner, the exactly once constraint is relaxed to at least once constraint 4 MAXIMUM TROUGPUT SCEDUING The objective of the maximum throughut scheduling roblem is to maximize the sum of the throughut of the users in a scheduling frame In the following we formally state the roblem Problem 1 Maximum Throughut Scheduling MTS Given a multiset of received owers 1 2 n, and some noise ower, schedule the corresonding n users in k time slots so that the following objective function is maximized, n 1 + i, 5 + I i where I i is the amount of interference ower that affects the decoding of user u i s signal see Figure 1 Theorem 1 MTS is NP-hard Proof: The roof is by reduction from the artition roblem In the artition roblem, given a set S of integers, we are asked to find a subset S S such that,

5 x S x y S S y, 6 is minimized The Partition roblem is known to be NPcomlete [27] We show that the artition roblem can be reduced in olynomial time to a secial case of MTS, for k = 2 Since a secial case of the roblem is NP-hard, the general case is NP-hard as well et π 1 and π 2 denote the artition of the received owers for the two time slots, ie, the users corresonding to π 1 will be scheduled in time slot 1 and the users corresonding to π 2 will be scheduled in time slot 2 We can rewrite the objective function 5 as follows, π 1 + 1 1 + et X = know that, π 1 + 1 and Y = π 2 7 π 1 + 2 We X + Y = 2 + 1 + + n, 8 is constant regardless of how the set of received owers is artitioned Note that our objective is to maximize X Y Since, X+Y is constant, X Y is maximized when X Y is minimized This is similar to the objective function of the artition roblem Therefore, to solve an instance of the artition roblem for a given set S, we first sort the numbers in S in an increasing order Then, we solve the Maximum Throughut Scheduling roblem with k = 2 time slots for the sorted numbers as the set of owers and an arbitrary ositive number as the initial noise ower The reduction is obviously olynomial Therefore, MTS is NP-hard Since MTS is NP-hard, the otimal answer to the roblem cannot be found in olynomial time, unless P = NP Thus, we roose a greedy algorithm called GreedyMax which runs in olynomial time and gives a subotimal answer to the roblem Algorithm 1 shows the seudo code of the algorithm The algorithm first sorts the received owers in decreasing order Then, starting from the beginning of the list, in each ste, it assigns the ower to the time slot that has the minimum sum of received owers The algorithm actually tries to make the sum of received owers assigned to the time slots as equal as ossible The rational behind Algorithm 1 is to assign the smaller owers at the end, so that it can fine tune the summation of the received owers The same method is used in multirocessor scheduling algorithms [28], [29] 5 CONSTRAINED MAXIMUM TROUGPUT SCEDUING As described in Section 3, ractical limitations revent decoding of too many signals at a time slot In this section, we add a constraint to the MTS roblem that at Algorithm 1 GreedyMax: Subotimal Solution to MTS 1: P list of received owers 2: Sort P in descending order 3: S 4: for i 1 to lengthp do 5: Schedule P i at the time slot with minimum sum of owers 6: end for 7: return S most l signals can be scheduled in a time slot Considering this constraint, we show that the roblem remains NP-hard Problem 2 Constrained Maximum Throughut Scheduling MTS C Given a multiset of n = kl received owers 1 2 kl, and some noise ower, schedule the corresonding n users in k time slots so that exactly l users are assigned to each time slot and the following objective function is maximized, kl 1 + i, 9 + I i where I i is the amount of interference ower that affects the decoding of user u i s signal Theorem 2 For l 3, MTS C is NP-hard Proof: We show that for l = 3, the roblem is NPhard Therefore, for a general l 3, it is NP-hard as well Proof is by reduction from the 3-artition roblem The roblem is defined as follows [27] Definition 1 3-Partition Problem Given a set A of size 3m, a bound B Z +, and a size sa Z + for each a A, such that each sa satisfies B/4 < sa < B/2 and a A sa = mb The question is: Can A be artitioned into m disjoint sets S 1,, S m such that for 1 i m, a S i sa = B? Note that the constraint on the item sizes imlies that every S i must contain exactly 3 elements It is known that the 3-artition roblem is NP-comlete [27] et ϕ i = j denote that i is assigned to time slot j We have, n 1 + k i + P j =, 10 + I i N j=1 0 where P j is the sum of the owers assigned to time slot j, ie, P j = ϕ i =j i The identity can be roved using 2, emma 1 and some algebraic maniulation and is omitted for the sake of brevity We rewrite 9 using 10 as, k + ϕ j =i j 11 Note that, in the 3-artition roblem, a S i sa = B for every 1 i m if and only if m a S i sa = B m

6 Given an instance of 3-artition, we construct an instance of Problem 2 such that i = sa i for every a i A, k = m, l = 3, and an arbitrary value for is chosen Using 11, the answer to the instance of 3-artition is Yes if and only if the value of the objective function 9 equals k +B ie, the sum of the owers in each time slot equals B Note that for the sake of brevity, we do not roose a heuristic algorithm for MTS C for the case l 3 owever, it is worth ointing out that GreedyMax can handle MTS C after suitable modifications In addition, in Section 6, we roose the GreedyFair algorithm that results in a roortionally fair schedule GreedyFair can be used, out of the box, to obtain a subotimal solution to MTS C In Section 8, through simulations, it is shown that the throughut of GreedyFair is close to the otimal throughut Theorem 3 For l 2, MTS C can be solved in olynomial time Proof: For l = 1 Theorem 3 is obvious, since there is only one way to assign the owers to the time slots For l = 2, assume without loss of generality that the owers are sorted in non-decreasing order We show that the following assignment maximizes 9: 1 For 1 i k, assign i to time slot i 2 For k + 1 i 2k, assign i to time slot 2k i + 1 Why the above assignment is otimal? et i and j be two arbitrary time slots such that i < j et ˆ 1, ˆ 2, ˆ 3, and ˆ 4 be the four owers assigned to the time slots i and j Without loss of generality, assume ˆ 1 ˆ 2 ˆ 3 ˆ 4 Using 9, the throughut of any ower assignment of the four owers to the two time slots is one of the following cases, Case 1: 1 + ˆ 1 + ˆ 2 + 1 + ˆ 3 + ˆ 4 12 Case 2: Case 3: 1 + ˆ 1 + ˆ 3 + 1 + ˆ 2 + ˆ 4 13 1 + ˆ 1 + ˆ 4 + 1 + ˆ 2 + ˆ 3 14 owever, using algebraic maniulations, it can be shown that assuming 0 < ˆ 1 ˆ 2 ˆ 3 ˆ 4, Case 3 is always greater than or equal to the other two cases It is also simle to verify that the above assignment algorithm results in the throughut of Case 3 for all i and j Assume, the assignment of the above algorithm is not otimal and there is another assignment A which is otimal Therefore, there must be at least one air of time slots i and j in A such that i < j and the assignments do not follow Case 3 Since, we can imrove the throughut by swaing the ower assignments to time slots i and j to follow Case 3, A is not the otimal assignment Obviously, the assignment can be erformed in olynomial time Problem 3 Given a multiset of n kl received owers 1 2 n, and some noise ower, schedule the corresonding n users in k time slots so that at most l users are assigned to each time slot and the following objective function is maximized, n 1 + i, 15 + I i where I i is the amount of interference ower that affects the decoding of user u i s signal Corollary 1 Problem 3 is NP-hard This is obvious that if n = kl, Problem 3 is similar to MTS C In other words, Problem 2 is a secial case of MTS C As a result, since MTS C is NP-hard, Problem 3 is also NP-hard Corollary 2 Throughut of MTS in each time slot scales as On where n is the number of users, while the throughut of MTS C scales as O1 Based on the definitions of the roblems, one of k or l needs to be a constant for MTS, k is fixed while for MTS C, l is fixed Using emma 1, the throughut of the schedule in each time slot is of On/k Therefore, if k is a constant, this results in On while for a constant l, it results in Ol = O1 6 PROPORTIONA FAIR SCEDUING In addition to the system throughut, fairness is an imortant factor in wireless networks In this aer, we consider roortional fairness [9], which is widely imlemented in existing wireless systems [10] In general, the system throughut is affected by the fairness definition owever, while some fairness criteria, such as max-min fairness, may sacrifice the system throughut for fairness, roortional fairness achieves a reasonable trade-off between fairness and throughut [10] The objective of this section is to study short term fairness ie, fairness among the users scheduled in a scheduling frame in a wireless network with SIC at the hysical layer ong term fairness ie, fairness among the users during a longer time than the duration of a scheduling frame can be enforced by a higher level scheduler that selects the set of users {u 1,, u n } to be scheduled in each scheduling frame, which is not the focus of this aer Before defining the roortional fair scheduling roblem, we formally define roortional fairness first Definition 2 Proortional Fairness Assume that schedule Π gives the throughut vector r = r 1,, r n, where r i denotes the throughut of user u i A schedule Π with throughut vector r = r 1,, r n is roortionally fair if

7 Algorithm 2 GreedyFair: Proortional Fair Scheduling Require: 1 2 n 1: rocedure PF 1, 2,, n,, k 2: for i := 1 to n do 3: π m = argmin π π π 7 5 6 π 4: Schedule i on to of π m 5: end for 6: end rocedure 3 1 N 4 2 N and only if the following condition holds for any other schedule Π with throughut vector r, n r i r i r i 0 16 It has been shown that a roortionally fair schedule maximizes the sum of arithm of users utilities [9] In this case, the utility of a users is the rate achieved by the user Based on this roerty, we define the roortional fair scheduling roblem as follows Problem 4 Proortional Fair Scheduling PFS Given a multiset of received owers 1 2 n, and some noise ower, schedule the corresonding n users in k time slots so that the following objective function is maximized, n 1 + i, 17 I i + where I i is the amount of interference ower that affects the decoding of user u i s signal Theorem 4 PFS can be solved in olynomial time We rove that the following olynomial time algorithm gives the otimal solution to PFS: Assume owers are sorted in a non-decreasing order Starting from i = 1 u to n assign i to the time slot that minimizes I i see Algorithm 2 Since the assignment of the owers to the time slots takes linear time and sorting can be done in On n, the time comlexity of the algorithm is in On n Assume ϕ i shows the time slot that i is assigned to by the above algorithm ie, column number in Figure 2 and ρ i shows the index of i in the scheduled time slot That is, by the above algorithm, ϕ i = i 1 mod k+1 and if ρ i < ρ j then i j To comlete the roof, we need the results stated in emma 2 and emma 3 emma 2 et X Y > 0 and A B > 0 The following inequality holds for any C 0, 1 + X + 1 + Y A + C B 1 + X + 1 + Y 18 A B + C Fig 2 The roortional fair algorithm Algorithm 2 slits the owers into two iles π and π π contains 1 st, 3 rd, 5 th, smallest owers while π contains 2 nd, 4 th, 6 th, smallest owers In each ile the owers are sorted from bottom to to in a non-decreasing order Proof: Rewrite 18 as follows 1 + X A 1 + X A+C 19 1 + Y B 1 + Y B+C Define function fu as, 1 + X A+u fu =, 20 1 + Y B+u and show fu is an increasing function of u 0 We have, d du fu = Y 1+ A+u X B+uB+u+Y X 1+ B+u Y A+uA+u+X 2 1 + Y B+u 21 To show that fu is an increasing function of u, we need to show d dufu 0 Since the denominator is ositive, it suffices to show that the numerator is nonnegative More formally, we need to show, Y 1 + X A+u X 1 + Y B + ub + u + Y B+u A + ua + u + X, 22 or, equivalently, A + ua + u + X 1 + X A+u 23 X B + ub + u + Y 1 + Y B+u Y Next, define the function gs, t as follows ts + t gs, t = 1 + s 24 s t To establish the emma, we show that the function gs, t is non-decreasing in s and t It can be shown that the artial derivatives of the function are non-negative That is, t s t 1 + s s gs, t = t s 2 0, 25

8 which follows from the fact that z 1 + z Moreover, s + 2t 1 + s t gs, t = t s 0, 26 s which follows from the fact that 1 + z z z+2 emma 3 et π denote the number of owers in ile 1 π see Figure 2 and Sπ denote the sum of the owers in ile π ie, Sπ = i π i We always have π = π or π = π + 1 In addition, { Sπ Sπ, if π = π, 27 Sπ > Sπ, if π = π + 1 Proof: It is obvious that the way Algorithm 2 assigns the owers results in two iles similar to the ones shown in Figure 2 et ˆ i denote the i th smallest ower For n even, ˆ 1, ˆ 3,, ˆ n 1 are assigned to π while ˆ 2, ˆ 4,, ˆ n are assigned to π In this case the inequality Sπ Sπ is obtained by summing the airwise inequalities: ˆ 1 ˆ 2, ˆ 3 ˆ 4,, ˆ n 1 ˆ n For n odd, ˆ 1, ˆ 3,, ˆ n are assigned to π while ˆ 2, ˆ 4,, ˆ n 1 are assigned to π In this case the inequality Sπ > Sπ is obtained by summing the airwise inequalities: ˆ 3 ˆ 2, ˆ 5 ˆ 4,, ˆ n ˆ n 1 and ˆ 1 > 0 Proof of Theorem 4: The roof is by induction and is divided into 4 stes Given n received owers and k time slots, in Ste 1, we rove the theorem is correct for k = 1 and a general n In Ste 2, the theorem is roved for k = 2 and n = 2 In Ste 3, we establish the correctness of the theorem for k = 2 and a general n In Ste 4, we rove the theorem is correct for general k and n Ste 1 k = 1, general n: For k = 1, we show Algorithm 2 solves the Proortional Fair Scheduling roblem In other words, we show that for a single time slot, the otimal order of decoding is n, n 1,, 1, ie, decode n first, then n 1, and so on The roof is by contradiction Assume and are two subsequent owers in the otimal order of decoding so that < and is decoded just after We show that swaing the order of decoding of and will indeed increase the objective function in 17 Since and are decoded successively, swaing their decoding order will only affect their individual rates, while the rates of the remaining users will remain intact Therefore, we only need to show, 1 + N + 1 + + N + + 1 + < N 1 + N, 28 1 We use the term ile to refer to a ortion of the owers scheduled in a time slot where N denotes the noise ower lus the sum of the signal owers that are decoded after and Using arithm function roerties we can rewrite 28 as, 1 + N + 1 + < 29 N 1 + 1 + N + N It is clear that the sum of the two terms at each side of the inequality is constant, 1 + + 1 + = N + N 1 + + 1 + = N + N 1 + + 30 N In addition, it is known that the roduct of two terms with a constant sum is an increasing function of their difference Thus, the roduct of the terms is maximized when their difference is minimized It is then obtained that, 1 + N + 1 + N + which comletes the roof 1 + < 31 N 1 +, N Ste 2 k = 2, n = 2: We show for > 0 and N N > 0 we have, 1 + + 1 + N N 1 + + 1 + 32 N N That is, scheduling the greater ower on to of the higher noise and smaller ower on to of the lower noise will result in a greater value for 17 in comarison to the other way around This can simly be shown using emma 2 by substituting X =, Y =, A = N, B = N, and C = N N Ste 3 k = 2, general n: In this ste, we show the correctness of the theorem for k = 2 Figure 2 shows the solution of the roblem for n = 7 The algorithm slits the owers into two iles, π and π π contains 1 st, 3 rd, 5 th, smallest owers, while π contains 2 nd, 4 th, 6 th, smallest owers In each ile the owers are sorted from bottom to to in a non-decreasing order Additionally, π is always laced on to of N while π is laced on to of N Because of the recursive nature of the induction, we use a recursive version of Theorem 4 in this ste Algorithm 3 shows the stes of the recursive version of Algorithm 2 Given emma 3, it can be verified that Algorithm 3 roduces the same schedule as the one given by Algorithm 2 Clearly, the theorem is correct for n = 1 In addition, by ste 2, we know that the theorem is also correct for n = 2 We assume the algorithm works

9 11 12 13 q rs a By Theorem 4, the sequence of q s shown in the figure must be a non-decreasing sequence of owers 11 12 13 q rs b If q s are not sorted in a non-decreasing order, it cannot be an otimal schedule because it contradicts the result of Ste 1 and/or Ste 4 Fig 3 In the roortional fair schedule the received owers are sorted in a non-decreasing order Algorithm 3 Recursive Proortional Fair Scheduling for k = 2 Require: N + 0 N 1: rocedure RPF 0,, n, N, N 2: if n > 0 then 3: Schedule 0 on to of N 4: RPF 1,, n, N, N + 0 5: The requirement 1 + N 0 + N holds 6: end if 7: end rocedure for n = M owers and show the correctness of the algorithm for n = M + 1 owers Given two initial noise-lus-interference owers N, N, and M + 1 received owers, by the result of Ste 1, we know that the smallest ower is scheduled either on to of N or on to of N et 0 denote the smallest ower while 1, 2,, k denote the rest of the owers aving the induction assumtions, we only need to comare two cases, 1 0 is scheduled on to of N : In this case, since N + 0 N, it is obtained that 1, 3, are scheduled on to of N and 2, 4, are scheduled on to of N + 0 2 0 is scheduled on to of N : In this case, since N + 0 N using emma 3, it is obtained that 1, 3, are scheduled on to of N and 2, 4, are scheduled on to of N + 0 We show the second case is always the otimal one To that end, we construct an inequality that shows, value of 17 for case 1 value of 17 for case 2 33 et π i,<j denote the sum of owers in ile i that are laced below ower j ie, the amount of interference but not the noise that affects j For instance, in Figure 2, π,<2 = 0 and π,<5 = 3 + 1 et m = M if n is even, and M 1 otherwise For n even, the otimality of the second case is shown by slitting 33 into m 2 sub-inequalities see 34 and showing the correctness of each sub-inequality The correctness of 34a is established using the result of Ste 2 The correctness of 34b to 34c is verified by the result of emma 2 For instance, 34b is roven by substitution M = 3, Y = 2, A = π,<3 + N, B = π,<2 + N, and C = N N The correctness of 33 can be obtained by multilying left right hand side of 34a by the left right hand side of 34c In the case that n is odd, the following extra inequality is also required, 1 + M π,<m + N 1 + M, 35 π,<m + N which can be verified by noting that 1+ π,<m +u is a decreasing function of u, and N N This comletes the roof of Ste 3 Ste 4 general k, general n: In this ste, we show that the algorithm works correctly for a general k > 2 Note that a schedule is essentially a two-dimensional matrix of owers, where the columns of the matrix denote the time slots and the rows denote the decoding order of the received owers et q denote the ower i that is assigned to the row and column q of the schedule ie, ρ i = and ϕ i = q By the above algorithm, q s are sorted in a non-decreasing order see Figure 3 to That is, if < r or = r and q < s then q rs Assume for some q and rs that the condition does not hold ie, q > rs while < r or = r and q < s see Figure 3 bottom It cannot be the otimal scheduling since the fairness index in 17 can be imroved by swaing q and rs in the schedule For q = s, it contradicts the result of Ste 1 Otherwise, it contradicts the result of Ste 3 7 SCEDUING WIT MINIMUM SINR GUAR- ANTEE In some alications, such as voice calls [6], a minimum SINR guarantee is required for each user, whereas in many other alications this is a desired roerty A minimum SINR guarantee for each user means successful decoding of the signal with high robability lus a minimum rate guarantee In this section, we study the roblem of scheduling with minimum SINR guarantee A formal definition of the roblem follows M

10 1 + 1 + 2 π,<2 + N m 1 π,<m 1 + N 1 + 0 N 1 + 1 + 1 N 3 π,<3 + N 1 + 0 N 1 + 2 π,<2 + N 1 + 1 N 1 + 3 π,<3 + N 34a 34b m m 1 m 1 + 1 + 1 + 34c π,<m + N π,<m 1 + N π,<m + N Problem 5 Scheduling with Minimum SINR Guarantee MinSINR Given a multiset of received owers 1 2 n, and some noise ower, schedule the corresonding n users in k time slots so that the SINR of each user is at least S min ie, i +I i S min, i or outut IMPOSSIBE if the requirement cannot be satisfied As an examle, given = 1, S min = 2, and { 1 = 2, 2 = 3} as the set of received owers, it is IMPOSSIBE to schedule the users in k = 1 time slot The reason is that, for k = 1, { 2, 3 } is the best schedule ossible, which suorts a minimum SINR of 1 owever, for k = 2, the scheduling frame is { 2, 3 }, which suorts the minimum required SINR Proosition 1 et F be the frame that maximizes the minimum SINR of users Also, let S F be the minimum SINR rovided for the users by F A solution to MinSINR can be obtained by comaring S F and S min : if S F S min then F rovides the SINR requirement of the roblem, otherwise, it is imossible to find such a schedule As a result, to solve MinSINR, it is sufficient to find an algorithm that maximizes the minimum SINR Proosition 2 An instance of MinSINR might have more than one solution For instance, given = 1, S min = 01, k = 1, and {2, 3} as the set of received owers, both { 2, 3 } and { 3, 2 } are correct answers owever, the answer given by Proosition 1 is always correct Theorem 5 MinSINR can be solved in olynomial time Proof sketch: Algorithm 4 gives the solution to MinSINR and is based on Proosition 1 The algorithm uses a scheduling similar to Algorithm 2 Then, it comutes the SINR for every user If the SINR of any user is less than the required SINR, the algorithm oututs IMPOSSIBE The rest of the roof shows that the scheduling frame found by Algorithm 4, maximizes the minimum SINR The roof follows the four-ste rocedure we had for the roof of Theorem 4 with some changes in the details Ste 1 k = 1, general n: The roof is by contradiction Assume that, in the otimal decoding order, is decoded before and < It can be seen that swaing the order of decoding of and will either increase the minimum SINR or does not affect it That s because min, + I s + I s + min, + I s + I s +, 36 where I s is the amount of interference which affects the signal that is decoded last Ste 2 k = 2, n = 2: Given 1 2 3 4 as the set of received owers, the otimal schedule to maximize the minimum SINR is given by { 1, 3, 2, 4 } This can be shown by comaring all ossible scheduling frames for four users Ste 3 k = 2, general n: This ste is conducted by induction and uses the same technique, which is used in Section 6 owever, instead of 33 we need to show, minimum SINR for case 1 minimum SINR for case 2 37 Ste 4 general k, general n: This ste is roven by contradiction using an aroach similar to the one used in Section 6 The otimal schedule follows the one offered by Algorithm 4, otherwise, it will contradict with either Ste 1 or Ste 3 Corollary 3 The following roblem can be solved in olynomial time: Given a multiset of received owers 1 2 n, some noise ower, and a minimum SINR S min, find the minimum number of time slots k that suorts the required SINR S min It follows from the fact that Problem 5 can be solved in olynomial time The minimum k that gives a minimum SINR S min can be found using Algorithm 4 by searching over k in the range [1, n] That is the minimum k for which Algorithm 4 roduces a non IMPOSSIBE outut Since Algorithm 4 runs in On, a simle linear search over k results in the overall comlexity of On 2 Obviously, a binary search results in a faster running time Corollary 4 The following roblem is NP-hard: Given a multiset of received owers 1 2 n, and some noise ower, schedule the corresonding n users in k time slots so that the SINR of each user is at least S min ie, i +I i S min, i and sum caacity ie, 5 is maximized or outut IMPOSSIBE if the minimum SINR requirement cannot be satisfied It follows from the fact that if S min is small enough eg, S min min i i i, the minimum required SINR can i be easily ignored, because the requirement holds for any scheduling In this case, the roblem is reduced to Problem 1, which has already been shown to be NP-hard Note that, while the roblem of Corollary 4 is generally

11 Algorithm 4 Scheduling with Minimum SINR Requirement Require: 1 2 n 1: rocedure MINSNR 1, 2,, n,, k, S min 2: m = 3: for i := 1 to k do 4: S i = 5: N i = 6: end for 7: for i := 1 to n do 8: j = 1 + i 1 mod k 9: m = minm, i N j 10: S j = S j { i } 11: N j = N j + i 12: end for 13: if m S min then return {S 1,, S k } 14: else return IMPOSSIBE 15: end if 16: end rocedure NP-hard, some secial cases of the roblem can be solved in olynomial time For instance, if Algorithm 4 returns IMPOSSIBE for an instance of the roblem of Corollary 4, the answer is definitely IMPOSSIBE 8 NUMERICA RESUTS In this section, we rovide numerical results to show the utility and efficiency of the roosed scheduling algorithms in various simulated network scenarios 81 Simulation Setu Simulations are conducted using a custom built simulator written in Mathematica For simulations, we consider a disk-shae network with radius R max meters, where the access oint is laced at the center of the disk Node locations are drawn uniformly from inside the annulus bounded by concentric circles of radii R min and R max Throughout the simulations, we set R min = 1 m, R max = 1000 m, and = 174 dbm Consequently, we modify max to obtain different SNR levels Recall that our results and algorithms are insensitive to the secific roagation environment as we have assumed that the access oint has knowledge about users channel information as is the case in 3G/4G networks 82 Throughut Comarison by Varying n and k The simulations conducted in this section are aimed to analyze the effect of varying the number of users, n, and the number of time slots, k, on the throughut of the network Since the sum of received owers is roortional to n, according to emma 1, for a fixed k, we exect a arithmic increase in the throughut of the network by increasing n On the other hand, for a fixed n, by increasing k, throughut decreases because fewer users are scheduled in each time slot Figure 4a shows throughut er time slot comarison of different scheduling algorithms by varying n and k Throughut er time slot is defined as the sum throughut of the frame divided by the number of time slots k The lot asserts the exected behavior that for a constant number of users, decreasing k increases the throughut er time slot The lot also asserts the arithmic increase in the throughut by increase in sum of the received owers emma 1 For k = 1, GreedyMax and PFS work equally well The reason is that, in this case, all of the users are scheduled in one time slot by both of the algorithms and therefore they have the same throughut for any combination of inut owers For n = k No SIC scenario, we do not exect any increase or decrease in the average throughut with the increase of n For k = 2, k = 4 and k = 8, GreedyMax gives a higher throughut than PFS which is exected owever, the throughut of PFS is very close to the throughut of GreedyMax which is not aarent from the formulas Figure 4b shows throughut er time slot er user for the same set of inuts as Figure 4a Throughut er time slot er user is defined as throughut er time slot divided by the number of users n Recall that for a fixed k, throughut er time slot scales as O n with SIC while without SIC, it scales as O1 owever, since the number of users increases linearly, as n, throughut er time slot er user converges to zero although with SIC, convergence is slower This behavior is asserted by Figure 4b Figure 5 shows the throughut er time slot of the algorithms for different SNR levels and varying k As we exect, by increasing k, throughut decreases In addition, the lot shows that in the low SNR regime, the difference between the throughut of MTS and PFS is negligible owever, the difference increases as SNR increases Figure 5 also shows that the erformance of MTS and GreedyMax are very close in ractice Note that while in Figure 5 the erformance of MTS and GreedyMax may look indistinguishable, it is easy to find cases in which they differ For instance, consider received owers {9, 8, 6, 4, 3} and k = 2 GreedyMax slits the owers into {9, 4, 3} and {8, 6} while the otimal schedule given by MTS is {6, 9} and {4, 8, 3} 83 Fairness Comarison We use the Jain s fairness index, which is commonly used in the literature [30], as the measure of fairness for comarison For a given set of throughuts {x 1,, x n }, Jain s fairness index is defined as follows, J {x 1,, x n } = n x i 2 n n x2 i The index ranges between 1/n and 1 A higher value of the index means a fairer distribution of the throughuts

12 Throughut er Time Slot Throughut er Time Slot 8 k 1 k 2 6 k 4 k 8 4 k n 2 0 10 ProortionalFair GreedyMax NoSIC 10 15 20 25 30 a Throughut er time slot GreedyMax,k 1 GreedyMax,k 2 08 NoSIC,k n GreedyMax,k 4 GreedyMax,k 8 06 04 02 00 n 10 15 20 25 30 b Throughut er time slot er user Fig 4 Throughut of different SIC-enabled scheduling algorithms comared to no-sic scheduling Each lot shows the average of 500 runs for each case received owers are randomly generated so that the average of SNR is 10 db Figure 6 shows the result of Jain s fairness index for different algorithms and different levels of SNR by varying n The lots show that PFS erforms significantly betters in terms of fairness which is the exected behavior Furthermore, the figure shows that the difference between the erformance of the algorithms becomes more visible when SNR is high n Average Throughut 07 06 05 04 03 02 01 SNR 10 db SNR 0 db SNR 5dB 00 0 2 4 6 8 10 k ProortionalFair MaxThroughut GreedyMax Fig 5 Throughut er time slot er user for n = 10, different SNR levels, and varying k Plot shows the average of 100 runs 84 SIC Effectiveness The questions considered in this section are: ow better does SIC generally erform comared to no-sic scenario? and By what factors is the throughut increase affected if SIC is emloyed? By no-sic scenario, we refer to the case in which a single user is scheduled in each time slot To answer the question we define the ercentage increase in throughut as the ercentage of increase in the throughut er time slot of GreedyMax comared to the throughut er time slot for the no-sic scheduling eg, 150% increase means the throughut of GreedyMax is 25x the throughut of no-sic We consider the effect of two factors: SNR and coefficient of variation in the simulations of this section Figure 7 shows the ercentage of increase in throughut versus the coefficient of variation of the received owers for different SNR levels, n = 10 and k = 3 The figures also shows a linear model fitted to the oints along with the 68%, 95%, and 997% standard deviation bands The figures show that there is a ositive correlation between coefficient of variation and throughut increase Additionally, the increase in throughut is much higher for lower SNR levels Thus, we conclude, as the user channels become more diverse or the average SNR decreases, SIC becomes more effective in increasing the system throughut 85 Discussion In ractice, there are several issues with SIC that revent real systems from reaching the theoretical limits For examle, we assumed that signals are decoded in an error-free manner owever, in case an error occurs

13 Jain's Index 10 08 06 04 02 ProortionalFair MaxThroughut GreedyMax 00 2 4 6 8 10 n a SNR = 5 db Jain's Index 10 08 06 04 02 ProortionalFair MaxThroughut GreedyMax 00 2 4 6 8 10 n b SNR = 0 db Jain's Index 10 08 06 04 02 ProortionalFair MaxThroughut GreedyMax 00 2 4 6 8 10 n c SNR = +10 db Fig 6 Jain s fairness index for k = 2 and varying n with different SNR levels Plots show the average of 500 runs ThroughutIncrease 250 200 150 100 50 R 2 0953737 ThroughutIncrease 250 200 150 100 50 R 2 0933481 ThroughutIncrease 250 200 150 100 50 R 2 0864021 0 0 1 2 3 4 5 6 7 CoefficientofVariationCV a SNR = 5 db 0 0 1 2 3 4 5 6 7 CoefficientofVariationCV b SNR = 0 db 0 0 1 2 3 4 5 6 7 CoefficientofVariationCV c SNR = +10 db Fig 7 Percentage increase in throughut using GreedyMax vs Coefficient of Variation of received owers for n = 10 and k = 3 Each lot shows 250 runs while decoding a signal, it is unlikely that the signal is correctly removed from the comosite signal Thus, at later stages the signals cannot be decoded correctly [5] Also, due to imerfect channel estimation and anato-digital quantization errors, the ana version of a decoded signal cannot be erfectly reconstructed Thus, the interference removal might be imerfect, which causes lower throughut gains at later stages [31] 9 CONCUSION In this aer, we considered ulink scheduling in wireless networks suorting SIC at the hysical layer We considered four different scheduling roblems with different objectives and constraints In the first roblem MTS we considered maximizing the throughut of the system and showed that the roblem is NP-hard We roosed a greedy On n algorithm for the maximum throughut scheduling roblem, which gives a subotimal answer to the roblem The second roblem MTS C is similar to the first roblem excet it uts a constraint on the number of decoded signals in each time slot We showed that while the general form of the roblem is NP-hard, a secial case l 2 of the roblem is solvable in olynomial time In the third roblem PFS we considered roortional fair scheduling We also roosed an On n algorithm for PFS roblem and roved its correctness analytically The last roblem considered is scheduling with some minimum SINR guarantee We roosed an algorithm to find the otimal schedule for this roblem in olynomial time Simulations verify the arithmic increase in the system throughut by increase in the number of users using SIC In addition, it is shown that while PFS erforms significantly better in terms of fairness, the overall system throughut is not sacrificed by the scheduler Since, PFS runs in olynomial time, the results suggest that PFS is

14 a good scheduler for the real world scenarios Finally, simulations show that in the case that users have diverse channels or low average SNR, SIC rovides a higher throughut gain comared to the non-sic scenarios REFERENCES [1] D alerin, J Ammer, T Anderson, and D Wetherall, Interference cancellation: better receivers for a new wireless MAC, in ACM otnets, Atlanta, USA, Nov 2007 [2] J Andrews, Interference cancellation for cellular systems: a contemorary overview, IEEE Wireless Communications Magazine, vol 12, no 2, 2005 [3] J ou, J Smee, Pfister, and S Tomasin, Imlementing interference cancellation to increase the EV-DO Rev A reverse link caacity, IEEE Communications Magazine, vol 44, no 2, 2006 [4] J Blomer and N Jindal, Transmission caacity of wireless ad hoc networks: successive interference cancellation vs joint detection, in IEEE ICC, Dresden, Germany, Jun 2009 [5] D Tse and P Viswanath, Fundamentals of Wireless Communication Cambridge University Press, 2005 [6] S Sambhwani, W 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Comuter Science Deartment at the University of Calgary Dr Ghaderi received BSc and MSc degrees from Sharif University of Technoy, and a PhD degree from the University of Waterloo, all in Comuter Science is research interests include wireless networking and mobile comuting with emhasis on modeling and erformance analysis of wireless networks Mohsen Mollanoori is a PhD candidate in the Deartment of Comuter Science at University of Calgary and a member of EISA network grou since 2009 e received his MSc in Software Engineering in 2007 from the Deartment of Electrical & Comuter Engineering, Tarbiat Modares University with a first class honors Mohsen also received his BSc in Software Engineering from Tarbiat Moallem University of Tehran with a first class honors in 2005