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MMRU-ALLOC: An Optimal Resource Allocation Framework for OFDMA in IEEE

802.11ax

Conference Paper · November 2020

DOI: 10.1109/PIMRC48278.2020.9217154

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MMRU-ALLOC: An Optimal Resource AllocationFramework for OFDMA in IEEE 802.11axAvik Dutta*

Department of CSEIIIT Delhi

Delhi, Indiaavikd@iiitd.ac.in

Naman Gupta*

Department of CSEIIIT Delhi

Delhi, Indianamang@iiitd.ac.in

Syamantak DasDepartment of CSE

IIIT DelhiDelhi, India

syamantak@iiitd.ac.in

Mukulika MaityDepartment of CSE

IIIT DelhiDelhi, India

mukulika@iiitd.ac.in

Abstract—IEEE 802.11ax introduces OFDMA (OrthogonalFrequency Division Multiple Access) that allows multiple users totransmit or receive frames concurrently. The OFDMA transmis-sion duration is decided by the client with maximum transmissionduration. We focus on minimizing the maximum transmissionduration. The standard restricts assignment of at max one RUto a client and provides a specific way of splitting the channelinto smaller RUs. In this paper, we come up with a genericframework, MMRU-ALLOC (Min Max Resource Unit Allocation)to allocate RUs to clients under the given constraints for a singleOFDMA transmission. The framework allows one to define anynon-negative cost function such as data transmission time orthe padding length required for synchronized end time of allclients for a client-RU pair and we believe this can capturea wide variety of scenarios. We design provably efficient andoptimal algorithms for this general problem. We demonstratethe applicability of our framework for two specific problemspertaining to transmission of one OFDMA frame. (1) Minimizingthe transmission duration of one OFDMA frame for a given set ofclients. (2) Minimizing the maximum padding length required byany client to ensure synchronized end time. We implemented (1)in NS-3 and evaluated its performance. We compare it with twopopular scheduling and resource allocation algorithms: Max Rate(MR) and Proportional Fairness (PF). We find that it outperformsboth MR and PF by upto 91.3% in terms of frame transmissiontime and upto 91.1% terms of throughput achieved.

I. INTRODUCTION

The density of WiFi connected devices is very high insettings such as stadiums, airports, classrooms, other crowdedplaces and even at homes. However, the WiFi performance isstill poor in such settings due to the huge contention faced by alarge number of devices. IEEE 802.11ax focuses on improvingthe WLAN (Wireless Local Area Network) performance indense network scenarios by reducing contention/collision withMU (Multiple Users) support. It introduces OFDMA (Orthog-onal Frequency Division Multiple Access) where the AccessPoint (AP) divides the entire frequency band into multiplesubsets of orthogonal sub-carriers, termed as Resource Units(RU). These RUs are assigned to different users/clients totransmit in parallel with the constraint that a maximum of oneRU is assigned to one user. For Multi-User Multiple-Input-Multiple-Output (MU-MIMO), an RU can be allocated to more

*These two authors contributed equally to the work.

than one user (user group). However, in this paper, we onlyfocus on the OFDMA transmission as the MU transmissiontechnique. Moreover, there is a specific way as to how theRUs can be split.

To avoid synchronization issues and interference from Over-lapping Basic Service Set (OBSS), IEEE 802.11ax mandatesthat all the transmissions inside one OFDMA frame must startand end simultaneously [1]. This OFDMA frame transmissionduration is called scheduling duration. In a WLAN, thereis a mix of different types of traffic generated by differentapplications running on the clients. Therefore, there is a varietyof sizes of the frames [2]. Moreover, the clients will havevariety in their bit-rates based on their channel conditions.There are various scheduling algorithms already adapted forIEEE 802.11ax, such as Round Robin, Max Rate, ProportionalFair [3], [4] etc. Most prior work assume that concurrenttransmission in an OFDMA transmissions are of equal lengthand can realize the maximum gain possible. However, alarge fraction of Internet packets are of very short length(< 200B) [5]. This co-exists with other applications such asstreaming video or file download that are MTU (MaximumTransfer Unit) sized packets. In this paper, we address thefollowing question - once a scheduler decides the set ofusers to participate in one OFDMA transmission, what is theoptimal way to allocate the resources to the users withoutviolating any of the constraints? Let us take, for example, theproblem of minimizing scheduling duration. The schedulingduration of one OFDMA transmission is clearly decided by theclient with maximum transmission duration. Hence, a naturalobjective is to minimize this quantity. On the other hand, allparticipating clients, whose transmission time is less than thescheduling duration, might have to transmit null data bits,also called padding, in order to align their end time. Thisleads to a wastage of energy since the power consumptionis an increasing function of the padding length. One possiblesolution is to make the transmissions as “balanced” as possible,that is, the transmission times of the clients should not be toodifferent from each other. Note that the above two objectivescan indeed be conflicting - minimizing the maximum durationdoes not necessarily mean small padding duration for allclients.

In this paper, we provide a generic framework, MMRU-978-1-7281-4490-0/20/$31.00 © 2020 IEEE

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978-1-7281-4490-0/20/$31.00 ©2020 IEEE

ALLOC that aims to solve the problem of allocating RUsto clients in a single OFDMA transmission. The frameworkallows one to define a general non-negative cost function forevery pair of client and RU. The objective is to determinea feasible allocation that minimizes the maximum allocationcost of any client. We give a polynomial time algorithm forthis problem and argue its optimality for the defined objectivefunction. We showcase how this framework can be utilizedfor both the objectives mentioned above: (1) minimize themaximum transmission time of any client in one OFDMAtransmission, (2) minimize the maximum padding durationof any client in one OFDMA transmission. We subsequentlyshow that (1) can, in fact, be solved optimally using a fasterand simpler algorithm.

We implemented (1) in NS-3 and evaluated its performancefor different frame lengths generated by clients. We usedthree different frame length distributions: two uniform and oneskew. To study the impact of MCS (Modulation and CodingScheme), we evaluated for two MCS settings: same maximalMCS and random MCS. We also study the effect of numberof clients in one OFDMA frame by varying the same. We thencompare MMRU-ALLOC with 802.11ax adaptation of popularscheduling and resource allocation algorithms such as Propor-tional Fair (PF) and Max Rate (MR) [4]. It outperforms both,PF and MR by upto 91.3% in terms of frame transmissiontime and upto 91.1% in terms of throughput achieved.

II. RELATED WORK

Prior work has looked at the resource allocation problemin IEEE 802.11ax. The authors in [3] look at the resourcemanagement problem in IEEE 802.11ax with an objective tominimize the padding. The authors propose two models todetermine the scheduling duration. (1) Fixed PPDU model:the scheduling duration is fixed for all OFDMA transmissions.(2) Dynamic-PPDU model: here, the scheduling durationis decided dynamically such that the padding overhead isminimized while considering airtime fairness and energy con-sumption of the users. The authors show case that fixed-PPDUmodel causes under utilization, in our paper as well we followD-PPDU to optimize the scheduling duration dynamically.This paper does not consider the specific way of splitting theRUs as IEEE 802.11ax mandates.

The authors in [6] consider the problem of assignment ofRUs to users and user groups with the goal of maximizingthe sum rate. However, the optimal algorithm that the authorsproposed, violate the IEEE 802.11ax specification by allowingmultiple RU allocations to users and user groups. Then theyprovide greedy and recursive algorithms. However, the focusis on maximizing the sum rate. Here, we focus on minimizingthe scheduling duration.

In [4], [7] the authors consider the problem of schedulingthe uplink users in IEEE 802.11ax. The authors formulateda generalized utility function of maximizing the sum anddemonstrated how other well-known scheduling algorithmsin LTE like Max Rate (MR), Proportional Fair (PF) andShortest Remaining Processing Time (SRPT) could be adapted

TABLE ITHE MAXIMUM NUMBER OF RUS FOR EACH BANDWIDTH [8]. *+n

MEANS ”PLUS N 26-TONE RUS”

RU type 20 MHz 40 MHz 80 MHz 160(80 + 80) MHz26-tone 9 18 37 7452-tone 4+1 8+2 16+5 32+10

106-tone 2+1 4+2 8+5 16+10

242-tone 1 2 4+1 8+2

484-tone NA 1 2+1 4+2

996-tone NA NA 1 2

for 802.11ax OFDMA. Given a fixed RU set and MCS, theauthors use Hungarian algorithm for the assignment of RUs tousers such that the utility function is maximized. The authorsshow, given a specific splitting of RUs, how to optimallyallocate RUs to users. However, they do not propose a wayof efficiently splitting the RUs themselves. In our paper, wefocus on efficiently splitting the RUs, subject to minimizingthe maximum cost function.

III. MOTIVATION

IEEE 802.11ax proposes OFDMA where orthogonal subsetsof subcarriers, i.e., RUs (Resource Units), are assigned toconcurrent users. The standard proposes that the AP announcesa scheduling duration to the users. All the users participating inone OFDMA transmission must obey this scheduling durationand must start and finish their transmissions concurrently. Thedata transmission time of a user in the OFDMA frame isinversely proportional to the RU width allocated to it. The802.11ax standard follows a specific way of splitting the RUs.An RU can consist of 26, 52, 106, 242, 484, 996 tones. Theset of available RUs depend on the channel bandwidth. Eachwide RU can be split into two narrower RUs that, in turn, canbe split again independently. The only exception is for 242tone and 996 tone RUs where they can be split into two 106tone RU & one 26 tone and two 484 tone RU & one 26 tonerespectively. The maximum number of RUs of different tonesper channel bandwidth is shown in Table I.

Such splitting restricts the maximum number of concurrentclients per OFDMA frame. We assume that a schedulerprovides us the list of n users to participate in one OFDMAtransmission. Given this, we investigate ways to minimizethe scheduling duration. If the clients participating in anOFDMA transmission have frames with different sizes and usedifferent MCS (Modulation and Coding Scheme), the clientwith maximum transmission duration would determine theOFDMA frame transmission time. All of the other clientsmight stuff padding bits for the extra duration. We depict thiswith an example in Fig. 1. In this case, 4 users participate inan uplink OFDMA transmission. The third user decides thescheduling duration, the other three users pad dummy bits forthe extra duration. Padding causes wastage of resources. Asdescribed in [3], the overall throughput of the network ismaximized when this scheduling duration is minimized.

IV. PROBLEM STATEMENT

As mentioned in Section I, we shall define our problem ina very general fashion. Let [n] denote the set of first n natural

2020 IEEE 31st Annual International Symposium on Personal, Indoor and Mobile Radio Communications: Track 2: Networking and MAC

Fig. 1. An example of Uplink OFDMA frame exchanges

numbers. We are given a set of clients C = {c1, c2, · · · cn} anda set of RUs R = {r1, r2, · · · rm}. We shall abuse notationsslightly and use rj to denote both the RU and its size. Further,we are given a cost function λ(ci, rj) that denotes the costof assigning client ci to the resource unit rj . For each RUrj ∈ R, there is an upper bound bj ∈ [n] on the number ofRU rj that can be allocated. Let b(R′) = suprj∈R′ bj for anysubset R′ ⊆ R.

Finally, there is an upper bound F on the total size of RUunits that can be allocated in the given bandwidth. Let S be afeasible allocation satisfying all the constraints and let rS(i)denote the RU allocated to client i in S. We consider thefollowing problems in this paper

1) GEN-MMRU-ALLOC : The task is to determine thefeasible allocation S that minimizes the maximum al-location cost of any client, that is find S to minimizemaxi∈C λ(ci, r

S(i))2) MMRU-ALLOC-TX : Given a set Q = {q1, q2, · · · qn}

of client frames and MCS rates P = {p1, p2, · · · pn},the transmission duration of client i, allocated RU rjis qi

pi·rj . The task is to find a feasible allocation S thatminimizes the maximum transmission duration.

3) MMRU-ALLOC-PAD : Assume that T ? is the maximumtransmission duration and ti is the transmission durationof client i under a feasible allocation S. Then paddinglength of i is defined as `i = T ? − ti. The task is tominimize maxi∈[n] `i.

V. ALGORITHMS FOR GEN-MMRU-ALLOC

Theorem 1 There exists a deterministic algorithm that solvesthe GEN-MMRU-ALLOC problem optimally and runs in timeO((mn) 3

2 log(max{m,n}W ) · logmn).

Our algorithm for has two ingredients. The first one is to“guess” the correct value of the optimal solution - let us callit T ?. We observe that the value of T ? must be λ(ci, rj), forsome client i ∈ [n] and RU j ∈ [m]. Hence, there are only mnpossible choices for T ?. We shall maintain a sorted list of allsuch values and then use a a binary search framework to findthe minimum value of T ? that admits a feasible allocation.In order to guide our binary search, we shall appeal to thesecond ingredient of our algorithm which is summarized inthe following lemma.

Algorithm ALG-GEN-MMRU-ALLOC (C,R)L : Array of all possible values of T ?, sorted in

non-decreasing orderL′ ← L, start← 1, end← |L|

If |L′| > 1 \\ Binary search on values of T ?

mid← b start+end2 c

T ? ← L(mid)If FeasibleAlloc(T ?) = feasible ThenL′ ← L′[start : mid− 1], end← mid− 1ElseL′ → L′[mid : end], start← mid

ElseIf FeasibleAlloc(T ?) = feasibleReturn the matching from FeasibleAlloc(T ?)

ElseReturn ”Not feasible”

FeasibleAlloc(T ?)Construct threshold graph GT?

.Solve Minimum Cost Bipartite b-matching on GT?

- let C be the cost of the matchingIf (C ≤ F ) Then

Return feasible ElseReturn infeasible

Definition 2 Bipartite b-matching. Given a bipartite graphG(A∪B,E) a non-negative integer parameter b(v),∀v ∈ B,a perfect b-matching of A is a subset of edges that containsevery vertex of A exactly once and each vertex v ∈ B at themost b(v) times.

Suppose in addition to the above definition, there is a weightfunction w(e), e ∈ E defined on the edges of G. Then findingthe b-matching of minimum total weight is a classical problemin combinatorial optimization and there are known efficientalgorithms to solve this in time O(|E| 32 log(nW )) [9] underthe condition that supv∈B b(v) = O(|E|), where W is themaximum weight on any edge in the graph. Equipped withthese definitions, we now describe the allocation algorithm.

In the algorithm, we require a structure which we call thethreshold graph. For a fixed value of the parameter T ?, weconstruct a bipartite graph GT?

= (A∪B,E). The left partiteset A consists of a node for each client ci ∈ C. The right partiteset consists of one vertex for each resource unit rj ∈ R. Forany client ci ∈ C, we introduce an edge between the node forci and the node for RU rj , if and only if, λ(ci, rj) ≤ T ?.Finally, we define the cost function w : E ← R≥0 - for anye ∈ E incident on a node rj , w(e) = rj and b(rj) = bj .

Lemma 3 There exists a perfect b-matching for the setA in the threshold graph GT?

of cost at the most F ifand only if there exists a feasible allocation S such thatmaxi∈[n] λ(ci, r

S(i)) ≤ T ?.

Proof: We first prove the sufficiency condition. Assume thereexists such a solution S. We consider the following set of

2020 IEEE 31st Annual International Symposium on Personal, Indoor and Mobile Radio Communications: Track 2: Networking and MAC

edges M ⊆ E - for each client, take the edge (ci, rS(i)).

We first argue that such an edge always exists for anyclient ci. Assume not, but this implies by our constructionλ(ci, r

S(i)) > T ?, which is a contradiction to the conditionsfor S. Now we argue that the set of selected edges forms ab-matching with cost at the most F . Since S is a feasiblesolution, the total number of RU units of type rj allocatedcannot exceed bj and hence the same property holds for theselected edges. Finally, it is straightforward to observe thatsupe∈M w(e) = supci∈C r

S(i) ≤ F , again by feasibility of S.Now we prove the necessity condition. Suppose there exists

a perfect b-matching M of A of cost at the most F in GT?

and suppose node ci is matched to the node rM(i) inM. Wecreate an allocation where we assign client ci the RU rM(i).We claim that this allocation never uses more than bj units ofRU rj . This follows directly from the b-matching property ofvertices in B. Further, the sum of the allocated RUs is equalto the total weight of the matching which is at the most F .

We use the above lemma to prove Theorem 1. The algorithmALG-GEN-MMRU-ALLOC uses a binary search frameworkon a sorted array L of size mn - the list contains all possiblevalues of the maximum transmission time. For a specificchoice of T ?, if the algorithm FeasibleAlloc(T ?) does notreturn a feasible allocation, then by Lemma 3, there does notexist a feasible allocation with maxi∈[n] λ(ci, r

S(i)) ≤ T ?.Hence, we restrict the main algorithm to the right half of thearray L. Conversely, if the FeasibleAlloc(T ?) indeed returnsa feasible allocation, then we restrict to the left half of thelist L. By standard correctness of binary search, the algorithmreturns the correct value of T ? upon termination.

The total number of edges in the threshold graph is mn.The maximum weight W = maxj∈R rj . Hence the run-time for a given T ? is O((mn) 3

2 log(max{m,n}W ). Com-bined with the binary search, the runtime is dominated byO((mn) 3

2 log(max{m,n}W ) · logmn).

A. Solutions for MMRU-ALLOC-TX , MMRU-ALLOC-PAD

The MMRU-ALLOC-TX problem reduces to the GEN-MMRU-ALLOC problem where λ(ci,rj) = qi

pi·rj . In order toreduce MMRU-ALLOC-PAD to GEN-MMRU-ALLOC , weagain guess the maximum transmission duration T ? by uti-lizing binary search. Now, the padding duration of ci, i ∈ [n],if allocated RU rj would be T ?− qi

pi·rj . Note that this is onlya function of ci, rj and hence we can define λ(ci, rj) as thisquantity and our framework is immediately applicable.

B. Faster Algorithm for MMRU-ALLOC-TX

The basic framework of this algorithm is the same as thatfor GEN-MMRU-ALLOC . However, the FeasibleAlloc(T ?)subroutine now uses a simpler greedy strategy which is muchfaster than solving the b-matching problem that was requiredin the GEN-MMRU-ALLOC algorithm.

FeasibleAlloc(T ?) \\ for MMRU-ALLOC-TXSort client set C in non-decreasing order of

qi/pi, i ∈ [n], R in non-decreasing order of rjFor i=1 to n do

Allocate the least indexed RU, say rj in the sortedlist R such qi

pirj≤ T ?

Remove rj from RIf ci not allocated Then

Return InfeasibleReturn Feasible

Lemma 4 (Monotonicity Lemma) There exists a feasibleallocation S with the maximum transmission time T ? suchthat for two client i, i′ ∈ C, if qi

pi≤ qi′

pi′, then rS(i) ≤ rS(i′)

Lemma 5 The above algorithm finds a feasible allocation Swith maximum transmission time T ? if and only if one exists.

To summarize, our greedy algorithm constructs a feasiblesolution with maximum transmission time of any client T ?,provided such a solution exists. It is easy to observe that asuitable implementation of the above algorithm along withbinary search runs in time O(mn logmn) and we skip thedetails in the interest of space.

VI. EVALUATIONIn this section, we first discuss our simulation setup and

then present our evaluation results.

A. Simulation Setup

We implemented MMRU-ALLOC-TX algorithm in NS-3on top of OFDMA support developed by the NS-3 commu-nity [10]. The parameters for our simulation is mentioned inTable II. We consider a WLAN with clients running differenttypes of applications such as browsing (HTTP), interactive(voice/video call), file download (FTP), etc.. To simulatethe variety in frame lengths generated by these applications,we generated a set of virtual traffic in the downlink usingthree different distributions two uniform and one skew. Thepayload size of each frame is randomly uniformly pickedfrom (a) [100, 1440] bytes interval to simulate video streaming[11], (b) [200, 11454] bytes interval [5]. (c) The payload sizeis randomly picked either from [200, 400] or [8000, 10000]bytes interval to simulate HTTP and FTP applications [5]respectively. We considered two MCS settings: (a) all theclients operate at same maximal MCS, i.e., 1024 QAM, 5/6with only one exception clients with RUs<242 tones operateat 256 QAM, 5/6 MCS, (b) clients operate at different MCS(randomly uniformly distributed between 1 to 12). Note thatMMRU-ALLOC-TX works on a given set of users participat-ing in one OFDMA. Therefore, we limit the number of clientsto the maximum concurrent ones supported in one OFDMA(e.g., 18 for 40 MHz). Further, to understand the effect ofnumber of clients, we vary this in each OFDMA transmission.

To validate the optimality, we implemented and comparedour results with exhaustive search. We also compare it with

2020 IEEE 31st Annual International Symposium on Personal, Indoor and Mobile Radio Communications: Track 2: Networking and MAC

TABLE IIEXPERIMENTAL SETUP

Parameter ValueChannel Bandwidth 40 MHzMCS [1, 12]

No. of STAs [2, 18]

Propagation loss model Friis Loss Model

two popular SRA algorithms: Max Rate (MR) and Propor-tional Fair (PF). We implemented 11ax adaptation [4] of MRand PF. We run each simulation for 20 seconds. In each simu-lation, around 6000 to 26, 000 OFDMA frames are transmitteddepending upon the number of STAs and their payload size.We then look at mean (with 95% confidence intervals) ofthe following metrics: OFDMA frame transmission time andaggregate throughput to gain some insights. Note that thoughin this paper we present the results for downlink traffic only,it is easy to run similar experiments for uplink traffic as ouralgorithm remains the same.B. Results

Fig. 2. Mean transmission time with payload size following uniform distri-bution [100, 1440] at random MCS.

Fig. 3. Mean Aggregate downlink Throughput with payload size followinguniform distribution [100, 1440] at random MCS.

First, we consider when the frame size is generated usinguniform distribution [100, 1400]. Fig. 2 shows that MMRU-ALLOC-TX yields the same transmission time, as, exhaustivesearch, thus validating the optimality of MMRU-ALLOC-TX .We observe that MMRU-ALLOC-TX always outperforms thePF and MR algorithms by achieving upto 77.6% and 80.4%lesser transmission time respectively when all clients operateat different MCS and upto 58.2% and 73.5% less transmission

time respectively when all clients operate at maximal MCS(graph omitted due to space constraints). The reason for thissuperior performance of MMRU-ALLOC-TX over MR and PF(in terms of transmission time) is that MMRU-ALLOC-TXalways gives an optimal allocation to minimize the maximumtransmission time whereas MR simply allocates RUs to STAswhich operates at maximum rate regardless of the payloadsize of each STA, and PF allocates RUs to STAs by takinginto account the current rate and average service rate of eachSTA. Also, the results show that as the number of clients inone OFDMA frame increases, so does the frame transmissionduration. This is expected, as the number of clients increases,the RU size per client decreases to accommodate all clients.It is noteworthy to mention that all four algorithms yieldoptimal allocation, when either only 2 STAs or all 18 STAsare scheduled. This is because a 40 MHz bandwidth can onlybe split into 18 26 tones RUs to accommodate all 18 STAs andinto 2 242 tones RUs for 2 STAs. In Fig. 3, we observe thatgain in throughput of MMRU-ALLOC-TX over PF and MRis upto 67.8% and 76.1%, respectively Moreover, we noticesimilar behaviour when all clients operate at the maximal MCS(57.4% and 74.6%). Additionally, we observe that PF performscomparatively better than MR as PF allocates resources basedon the current rate and history of the throughput achieved byeach user, consequently increasing the aggregate throughput,whereas MR only considers the current state of an STA. Wealso notice that as the number of clients in one OFDMA frameincreases, the throughput degrades (122 Mbps to 43 Mbps).

Fig. 4. Mean Aggregate downlink Throughput with payload size followinguniform distribution [200, 11454] at maximal MCS.

Next, we consider when the frame size is generated usinguniform distribution [200, 11454] and skew distribution. Here,again we observe that average time taken by MMRU-ALLOC-TX (graphs omitted due to space constraints) to transmit anOFDMA is always less as compared to PF and MR forboth MCS settings: maximal and random. In the former case,where all clients operate at maximal MCS, MMRU-ALLOC-TX takes upto 49.2% and 52.1% less transmission time re-spectively when payload size follows uniform distribution in[200, 11454], and upto 91.3% (both PF and MR) for skewdistribution. Whereas in the latter case (each client operatingat different MCS), MMRU-ALLOC-TX takes upto 70.8%

2020 IEEE 31st Annual International Symposium on Personal, Indoor and Mobile Radio Communications: Track 2: Networking and MAC

Fig. 5. Mean Aggregate downlink Throughput with payload size followingskew distribution [200, 400] ∪ [8000, 10000] at maximal MCS.

TABLE IIIPOSSIBLE OPTIMAL RU ALLOCATION VECTOR FOR 40MHZ CHANNEL

# of clients RU split13 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 52, 52, 106}14 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 52, 106}15 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 106}16 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 52, 52}17 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 52}18 {26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26}

and 75.9% less transmission time respectively for uniformdistribution in [200, 11454], and upto 75.5% (both PF andMR) for skew distribution. It is interesting to note that theframe transmission time beyond 12 clients is approximatelythe same (≈1330 µsec for uniform and ≈1200 µsec for skew).When we looked into the RU allocation for this, we noticedthat majority of the clients were allocated with an RU widthof 26 tones, while only few clients were allocated with 52tones or 106 tones RU. The RUs in 5 GHz 40 MHz channelare of 26, 52, 106, 242 and 484 tones (as shown in Table. I).Hence, for 13 clients the only possible optimal allocationwould be 106, 52, 52, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26. Notethat 52, 52, 52, 52, 52, 26, 26, 26, 26, 26, 26, 26, 26 is not anoptimal allocation as this would not minimize the maximumtransmission time.

Fig. 4 shows that the throughput achieved for uniformdistribution [200, 11454]. It is roughly similar when numberof clients are 13 and above. In this case, we would generallyget 2− 3 large-sized frame payload (10, 000 Bytes to 11, 400Bytes) followed by 2 small-sized frame payload (500 Bytesto 1000 Bytes) and rest medium-sized frame payload (2500Bytes to 7500 Bytes). Table. III shows that there is only onepossible RU allocation vector for 13 clients and above. Hence,after assigning an RU of 106 tones to a client with largestpayload size, we are left with only 26 tones RUs. These arethen assigned to clients with both small and medium-sizedpayloads. As a result, the clients with medium-sized payloadscontribute to the OFDMA frame transmission time rather thanthe clients with small ones. On the other hand, in case of skewdistribution as shown in Fig. 5, there are either very large orvery small payload sizes i.e., the number of frames with 8000Bytes to 10, 000 Bytes payload, and 200 Bytes to 400 Bytespayload are roughly the same. Initially, throughput increasesupto 5 clients as 106 tones RU can be allocated to largepayload sized clients (hence small frame transmission time

and high throughput). Now, after 12 clients, only one clientwith large-sized payload frame is assigned a higher width RU(106/52) tones among all other clients with large-sized frames.In this case, the transmission time is determined by the clientswith large-sized frames that were allocated with 26 tones. Thethroughput achieved by clients (14 to 18) is roughly similarsince most of them were assigned the same RU width. It isevident from above figures that MMRU-ALLOC-TX alwaysyields higher throughput than PF and MR irrespective of thepayload size. The aforementioned trend holds true even wheneach client is operating at a random MCS, where maximumgain in throughput of MMRU-ALLOC-TX against PF and MRis 55.6% & 73.7% for uniform distribution in [200, 11454], and70.9% (both PF and MR) for skew distribution respectively.

VII. CONCLUSION

In this paper, we have studied the resource allocationproblem for IEEE 802.11ax. We provided a generic frameworkfor minimizing the maximum cost of resource allocation to anyclient and demonstrated the applicability of this frameworkwith two important objectives, viz., transmission duration andpadding length. We implemented the first one in NS-3 andevaluated its performance. We noticed that beyond a certainnumber of clients, the OFDMA frame’s transmission time andthroughput suffer significantly. This suggests that there is alimit on the maximum number of concurrent clients beyondwhich OFDMA performance will degrade.

ACKNOWLEDGEMENTThis work was funded in part by IMPRINT II and ECR

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2020 IEEE 31st Annual International Symposium on Personal, Indoor and Mobile Radio Communications: Track 2: Networking and MAC

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