CN101136713B - Time-frequency resource allocation method of OFDM access system - Google Patents

Abstract

This invention relates to a distribution method for time frequency resources of OFDMA access system, which applies optimization to residual space structure after one-time of distribution to get a distribution result of high usability and gets good performance at the same time when reducing count volume so as to realize real time system, in which, this invention gives a free degree of residual space area and residual space edges and a definition of second moment sub-block area of residual space to make up of a recursion search method with a fixed depth to improve the effect.

Description

A kind of time-frequency resource allocation method of OFDM access system

Technical field

The invention belongs to communication technical field, be specifically related to a kind of time-frequency resource allocation method of OFDM access system.

Background technology

OFDMA (OFDM access) transmission technology is as the IEEE802.16 standard ^{[2] [3]}An important component part obtaining increasing concern.As the multiple access technique that comes from OFDM (OFDM), OFDMA has inherited the advantage of OFDM: have than strong robustness for intersymbol interference and multipath interference.OFDMA has adopted TDMA (time division multiplexing) and these two kinds of access waies of FDMA (frequency division multiplex) simultaneously, can be subdivided into the OFDM symbol on time-domain, is subdivided into subcarrier in frequency domain.Subcarrier can consist of subchannel on the logical concept by certain mapping principle.Each user can take different running time-frequency resources, and thinner resource division granularity and time-frequency resource allocating mode more flexibly are provided.

IEEE802.16OFDMA descending sub frame structure as shown in Figure 1, the OFDMA descending sub frame can be divided in time time-multiplexed zone (Zone), there is subcarrier mapping mode separately in different types of zone, wherein " part is used subcarrier zone (PUSC Zone) " is the zone that descending sub frame must comprise, and other zones all are optional ^{[2] [3]}The application scenarios that the present invention considers is the descending PUSC Zone that the pressure of IEEE802.16 standard comprises.PUSC Zone has used the subcarrier mapping that disperses ^{[2] [3]}Therefore can resist the transmission hydraulic performance decline that frequency selective fading causes, and can think that the average channel conditions of each user on each sub-channels is similar to identical [4], thereby can under the condition that does not affect user's transmission rate, be abstracted into a simple two dimensional surface segmentation problem to the channel resource allocation problem.

Time/frequency source block is described with the rectangle in the two dimensional surface.The unit of the rectangular block longitudinal axis is subchannel, and the every lattice on the longitudinal axis represent 1 sub-channels; The unit of rectangular block transverse axis is two OFDM symbols, and the every lattice on the transverse axis represent two continuous OFDM symbols.It is a groove (slot) that resource is distributed the smallest particles degree, and single groove is made of two OFDM symbols and a sub-channels, shown in the lattice among Fig. 2.In the method for back was described, all resources divided that to be equipped with slot be least unit.Distribute to the continuum that certain user transmits on the time/frequency source block and be called burst transfer (burst).The burst that the user is distributed in the IEEE802.16 standard code must be rectangle, so burst has taken a rectangle sub-block in the time/frequency source block.Each user proposes by this method the channel time-frequency resource allocating to be distributed to each user with the form of burst after the slot request of some, and specifies subchannel that each user's transmission takies and the position at the whole story of OFDM symbol.

The channel resource allocation problem belongs to the NP-complete problem, traditional method ^[1]Because computation complexity is too high, is not suitable for real-time calculating.The present invention proposes a kind of computation complexity is O (n ²) quick heuristic method resource allocation problem is carried out approximate solution.

Summary of the invention

The object of the present invention is to provide the method for the distribution of low, the fireballing orthogonal frequency division multiple access system running time-frequency resource of a kind of computation complexity.

The DFDMA method for distributing system resource that the present invention proposes is to adopt small freedom of motion weighting (Weighted LessFlexibility First is hereinafter to be referred as WLFF) method that the time-frequency resource allocating problem is found the solution.

The assignment problem of running time-frequency resource has rectangle constraints, and its step of finding the solution is as follows:

(1) the first big after small order that at first occupies resource according to each user transmission sorts to user's assignment order; (2) secondly when being some user resource allocations according to the suitable allocative decision of Information Selection of the remaining space limit degree of freedom, remaining space sub-block area second moment; Improve distribution effects by the greedy distribution method of the recurrence of a constant depth at last.

Wherein, the suitable allocative decision of so-called selection lists all allocative decisions when the steps include: as some user resource allocations; For each allocative decision of enumerating out, at first this scheme is carried out " virtual assigned "; Then calculate the remaining space limit degree of freedom; Pick out the allocative decision with limit, least residue space degree of freedom and form a set, called after " set of the minimum edge degree of freedom "; For each allocative decision in " set of the minimum edge degree of freedom ", at first this scheme is carried out " virtual assigned ", then calculate remaining space sub-block area second moment, select the scheme that wherein has least residue space sub-block second moment, as final allocative decision and carry out actual allocated; Repeat said process, until finish the distribution of all user's transmission.

The greedy distribution method of so-called last recurrence the steps include: at first to cancel two users' transmission of last distribution and the state of replacement problem input parameter; Then constantly utilize the recurrence greedy function that user transmission is distributed, until the user transmits that set becomes empty set or without unallocated resource; Assigning process finishes.

The below is described the implementation of WLFF method.The input parameter of WLFF method is a set, and the slot size of being asked by each user consists of; The Output rusults of WLFF method is each user's the positional information of burst on time/frequency source block and total resource utilization.The solution procedure of WLFF method is divided into two stages:

Phase I, consisted of by a series of separate " first kind distribution "." first kind distribution " distributes the needed resource of burst transfer (burst) for some users in time/frequency source block." first kind distribution " result be made of two elements: length and width configuration and the burst residing position in time/frequency source block of burst.The allocative decision of mentioning later in this specification refers to by these two examples that element consists of.This method adopts a series of a priori criteria that " virtual assigned " result in " first kind distribution " process is assessed, and here " virtual assigned " refers to such process: according to allocative decision the part resource is distributed to the user and this sub-distribution is labeled as virtual assigned.Corresponding inverse process is " virtual cancelling ": will distribute accordingly according to the virtual assigned information of mark and cancel, " virtual cancelling " is implicit the execution, describes in the process being not described in detail in method.The a priori criteria that assessment is used comprise: the geometric shape of remaining area should be oversimplified after distributing; Remaining area after distributing should be able to provide optional allocative decision as much as possible for the subsequent allocations process.

After the method for phase I finishes, in time/frequency source block, often there are some remaining blank Resource Block to distribute for the user of remainder.This is that any one allocative decision all might " be blocked " current blank Resource Block, causes subsequent allocations to carry out because less at the final stage remaining space area that distributes.In order to address this problem, after the distribution of phase I finished, the distribution at latter two burst of time/frequency source block cancellation phase I joined these two users' requests in the unallocated sequence, obtains unappropriated running time-frequency resource.Unappropriated user request is sorted according to first big after small order, be taken out to many 6 users requests from the end of sequence, form request set to be allocated.The second stage method as input, adopts the searching method of recurrence to carry out the resource distribution with request set to be allocated, running time-frequency resource to be allocated.

The below describes describing term and the symbol used in the WLFF procedure:

Slot: the least unit of time-frequency resource allocating is made of continuous two the OFDM symbols of time domain and frequency domain one sub-channels, as shown in Figure 2.

Burst: burst transfer, distribute to the continuum that certain user is transmitted on the time/frequency source block, as shown in fig. 1.

L: the time length of field (L=Resource Block time domain OFDM symbolic number/2) of initial time/frequency source block.

W: the frequency domain length (number of W=subchannel) of initial time/frequency source block.

L * W: by L, the definition of W and slot can be known: the slot number that the initial time/frequency source block of L * W=comprises.

r _i: i user's request size (i user transmitted the slot number that needs), stipulate r here _iIt must be integer.

REQ: by the set that each user's request consists of, REQ={r ₁, r ₂..., r _i..., r _n| r _i∈ Z}

l _i: the burst that distributes to i user is long.

w _i: the burst that distributes to i user is wide.

Factor (r _i): by r _iThe set that consists of of all factor pairs, r _iEach factor pair by r _iTwo factors consist of, these two factors must satisfy condition: product equals r _i

MAP: the data structure of describing time/frequency source block.MAP is that a transverse axis length is L+2, and longitudinal extent is the two-dimensional array of W+2.Method with (transverse axis, the longitudinal axis) is carried out addressing (initial coordinate is 1), (1 to the point in the coordinate; 1)～(1, W+2), (L+2; 1)～(L+2, W+2), (1; 1)～(L+2; 1), (1, W+2)～(L+2; W+2) point in these 4 zones is the boundary belts outside the time/frequency source block, and the numerical value of these points is-1.The resource points of MAP (point within the boundary belt) identifies the allocation situation of this slot, is not assigned with if this point is 0 this slot of expression; If this point is i then represents that this slot distributes to the user and asks size to be the burst of i.

FRM (free_space_map): unappropriated resource in the time/frequency source block.

Surround the limit: all and FRM is tangent and length equals the line segment of tangent length, as among Fig. 3 with shown in the line segment of double-head arrow.

FRM_Edge: the set of the encirclement side information of record FRM.

Allocation_Scheme: the set that is consisted of by allocative decision.

The below is elaborated to describing the definition of using in the WLFF procedure:

1 allocative decision

Allocative decision is by the configuration of the length and width of burst and the burst example that the present position consists of in time/frequency source block, described the physical dimension of the burst that a user may be assigned to and residing positional information in Resource Block.

The length and width configuration definition of burst is as follows: the length and width configuration of burst depends on that user corresponding to this burst asks size.For example i user asks size to be ri, is a slot because resource is distributed least unit, and the burst of distribution occupies an integer unit length at time shaft and frequency axis, the length and width configuration w of burst _i, l _iCan only be from r _iThe factor in select and w _i, l _iSatisfy condition: r _i=w _i* l _iR for example _i=8 factor pair is: factor (r _i)={ 1 * 8,2 * 4}.

Defined function Factorization (r _i), the burst length and width of finding the solution i user configure:

Input parameter: r _i

Output rusults: factor (r _i)

Factorization (r _i) algorithm steps as follows:

1) r _iThe numerical value fractions omitted of evolution partly obtains integer sqrt_r _i

2) definition integer variable j, j is at interval [1, sqrt_r _i] in value one by one from small to large, for each value of j, carry out following calculating:

2.1) if Integer, so

Be a factor pair, it is joined factor (r _i) among the set;

Otherwise current j is Components pair not.

Algorithm finishes.

The feasible location of burst in time/frequency source block is defined as follows: FRM is that burst only allows to be placed on the position of being close to the two ends that surround the limit by all " encirclement limits " zones of surrounding, here be close to the w that refers to burst _iThe limit is parallel overlapping with an end points that surrounds the limit, as shown in Figure 4, and the e among the figure _kFor the encirclement limit of considering in the example.In order to list the feasible location of burst in time/frequency source block, we only need to find the encirclement limit of FRM, two feasible location that drift angle is exactly burst on every limit.

Defined function GetEdge (MAP) obtains FRM_Edge information:

Input parameter: MAP

Output rusults: FRM_Edge

The algorithm steps of GetEdge (MAP) is as follows:

1) definition integer variable i rises to W+1 successively from 2, increases by 1 at every turn.

1.1) for each value of i, at first set variable edge_start and equal 0, edge_start and equal the current point of 1 expression and be positioned at and surround on the limit, otherwise 0; Then setting variable j initial value is 1.

1.2) if j surpasses L+2, jump to 1), calculate (1-value (j, i)) * (value (j, i) XORvalue (j, i-1)), wherein value (j, i) function is judged for the numerical value of (j, i) position, if this numerical nonzero then value (j, i)=1; Otherwise value (j, i)=0 then.XOR is xor operation.If:

1.2.1) value be 0 and edge_start be 0, j from adding 1, jump to 1.2)

1.2.2) value be 0 and edge_start be 1, finish on current encirclement limit, with this information recording/that surrounds the limit in FRM_Edge.Edge_start becomes 0, j from adding 1, jumps to 1.2)

1.2.3) value be 1 and edge_start be 1, j from adding 1, jump to 1.2)

1.2.4) value be 1 and edge_start be 0, edge_start becomes 1, found a new initial end of surrounding the limit, j jumps to 1.2 from adding 1)

2) definition integer variable i rises to L+1 successively from 2, increases by 1 at every turn.

2.1) for each value of i, at first set variable edge_start and equal 0; Then setting variable j initial value is 1.

2.2) if j surpasses W+2, jump to 2), calculate (1-value (i, j)) * (value (i, j) XOR value (i-1, j)), if:

2.2.1) value be 0 and edge_start be 0, j from adding 1, jump to 2.2)

2.2.2) value be 0 and edge_start be 1, finish on current encirclement limit, and this is surrounded information note on limit

Record is in FRM_Edge.Edge_start becomes 0, j from adding 1, jumps to 2.2)

2.2.3) value be 1 and edge_start be 1, j from adding 1, jump to 2.2)

2.2.4) value be 1 and edge_start be 0, edge_start becomes 1, found a new initial end of surrounding the limit, j jumps to 2.2 from adding 1)

3) definition integer variable i is decremented to 2 successively from W+1, reduces 1 at every turn.

3.1) for each value of i, at first set variable edge_start and equal 0; Then setting variable j initial value is 1.

3.2) if j surpasses L+2, jump to 3), calculate (1-value (j, i)) * (value (j, i) XOR value (j, i+1)), if:

3.2.1) value be 0 and edge_start be 0, j from adding 1, jump to 3.2)

3.2.2) value be 0 and edge_start be 1, finish on current encirclement limit, with this information recording/that surrounds the limit in FRM_Edge.Edge_start becomes 0, j from adding 1, jumps to 3.2)

3.2.3) value is 1 for l and edge_start, j jumps to 3.2 from adding 1)

3.2.4) value be 1 and edge_start be 0, edge_start becomes 1, found a new initial end of surrounding the limit, j jumps to 3.2 from adding 1)

4) definition integer variable i is decremented to 2 successively from L+1, reduces 1 at every turn.

4.1) for each value of i, at first set variable edge_start and equal 0; Then setting variable j initial value is 1.

4.2) if j surpasses W+2, jump to 4), calculate (1-value (i, j)) * (value (i, j) XOR value (i+1, j)), if:

4.2.1) value be 0 and edge_start be 0, j from adding 1, jump to 4.2)

4.2.2) value be 0 and edge_start be 1, finish on current encirclement limit, with this information recording/that surrounds the limit in FRM_Edge.Edge_start becomes 0, j from adding 1, jumps to 4.2)

4.2.3) value be 1 and edge_start be 1, j from adding 1, jump to 4.2)

4.2.4) value be 1 and edge_start be 0, edge_start becomes 1, found a new initial end of surrounding the limit, j jumps to 4.2 from adding 1)

Algorithm finishes.

Obtained factor (r _i) and FRM_Edge after, list factor (r _i) in element (r _iFactor pair) and FRM_Edge in all permutation and combination situations of element (the encirclement limit of unallocated resource in the time/frequency source block), obtain distributing draft.For example:

factor(r _i)＝{1×8，2×4}，FRM_Edge＝{Edge1，Edge2}

Can obtain following distribution draft:

Distribute draft=(1 * 8, Edge1, head), (1 * 8, Edge1, tail),

(1×8，Edge2，head)，(1×8，Edge2，tail)，

(2×4，Edge1，head)，(2×4，Edge1，tail)，

(2×4，Edge2，head)，(2×4，Edge2，tail)

The head here, tail represent that respectively same surrounds two end points zones on limit.Weed out in these drafts can not meeting geometric constraint (can not be overlapping between burst and the burst) situation, just obtained all allocative decisions.

Defined function AllocScheme (factor (r _i), FRM_Edge), find the solution all allocative decisions:

Input parameter: factor (r _i), FRM_Edge

Output rusults: Allocation_Scheme

AllocScheme (factor (r _i), algorithm steps FRM_Edge) is as follows:

1) for factor (r _i) in each factor pair, carry out following operation:

1.1) suppose that current factor pair is w _i* l _i, each the encirclement side information among the FRM_Edge proceeds as follows:

1.1.1) suppose that current encirclement side information is e _k, with { w _i* l _i, e _k, head}, { w _i* l _i, e _k, tail} information is added among the Allocation_Scheme.

2) check and reject the draft that distributes generation to conflict with MAP among the Allocation_Scheme.

Algorithm finishes.

The definition of 2DOF

For each a priori criteria are carried out numerical metric, defined at this: describe the remaining space limit degree of freedom of unallocated resource geometric shape complexity, describe the remaining space sub-block area second moment of subsequent allocations scheme variation degree.Below provide definition and the related algorithm step of these degrees of freedom.The remaining space that relates in the definition in the back just refers to unallocated Resource Block.

2.1 the remaining space limit degree of freedom (FSED)

The limit degree of freedom of remaining space refers to current allocative decision is carried out " virtual assigned " afterwards encirclement limit number of unallocated Resource Block.Represent the allocative decision of current investigation is carried out " virtual assigned " running time-frequency resource block data structure afterwards with MAP.

Definition CalFSED (MAP), calculate FSED:

Input parameter: MAP

Output rusults: FSED

The algorithm steps of CalFSED (MAP) is as follows:

1) carries out GetEdge (MAP) function calculation and obtain the FRM_Edge set.

2) number of element is exactly the value of FSED in the FRM_Edge set.

Algorithm finishes

2.2 remaining space sub-block area second moment (FSSAV)

At first the remaining space sub-block is defined: the remaining space sub-block refers to occupy a certain encirclement limit of unallocated Resource Block (FRM), along the rectangular block of the area maximum that can reach perpendicular to this Directional Extension that surrounds the limit, the rectangular block that identifies out with thick dashed line in Fig. 5 is exactly the remaining space sub-block in FRM inside.Before the definition of considering FSSAV, suppose that first FRM has the K bar to surround the limit, that is to say in the FRM_Edge set has K element, supposes that these elements are: { e ₁, e ₂..., e _k..., e _K.For surrounding limit e _k, definition by the remaining space sub-block area that this encirclement limit consists of is:

${\mathrm{ss}}_k=(e_k\×e_k^{\⊥})$

Wherein

Expression remaining space sub-block is perpendicular to e _kThe limit.Defined function OrthEdge (info_e _k, MAP) right

Calculate.Info_e in input parameter _kThe information on limit is surrounded in expression: two apex coordinate (x ₁, y ₁), (x ₂, y ₂), this surrounds the directional information on limit and propagation direction information in FRM.Be parallel to time shaft if surround the limit, directional information is 0 so; Otherwise be 1.If propagation direction is along the coordinate figure growing direction, vertical direction information is 1; Otherwise be-1.MAP is previously defined running time-frequency resource block data structure.

OrthEdge(info_e _k，MAP)：

Input parameter: info_e _k, MAP

Output rusults:

Length

OrthEdge (e _k, algorithm steps MAP) is as follows:

1) initialization

Length be 0, defined variable i initial value is 0.Defined variable Δ and determine as follows the value of Δ: if directional information be 0 and vertical direction information be 1, Δ=(0,1) so; If directional information be 0 and vertical direction information be-1, Δ=(0 ,-1) so; If directional information be 1 and vertical direction information be 1, Δ=(1,0) so; If directional information be 1 and vertical direction information be-1, Δ=(1,0) so.

2) obtain new line segment for surrounding after the coordinate figure of having a few on the limit is offset according to the size of i * Δ, if all point value on this line segment all are 0, so will

Length add 1, i variable from adding 1, re-execute 2); Otherwise algorithm finishes.

Algorithm finishes.

Obtain after all remaining space sub-block areas that surround the limit these areas being asked its second moment, just obtained FSSAV.Definition CalFSSAV (MAP), calculate FSSAV:

Input parameter: MAP

Output rusults: FSSAV

The algorithm steps of CalFSSAV (MAP) is as follows:

1) carries out GetEdge (MAP) function calculation and obtain the FRM_Edge set.

2) for each element among the set FRM_Edge, by carrying out OrthEdge (info_e _k, MAP) function is tried to achieve the length of perpendicular edge, thereby obtains the area of remaining space sub-block.

3) calculate the variance of these remaining space sub-block areas: $\mathrm{Var}=\frac{1}{K}\underset{k=1}{\overset{K}{\mathrm{\Σ}}}{({\mathrm{ss}}_k-\mathrm{ss}\_\mathrm{average})}^2,$ Wherein ss_average is the average of these sub-block areas.

Algorithm finishes.

3 recurrence greedy functions

The form of recurrence greedy function is: RecursiveGreedyAlloc (REQ, MAP).This function adopts the method for recurrence greediness to distribute user's request.The recurrence greediness here refers to: at first the possible situation of various distribution is carried out enumerating of recurrence, then from these distribution condition, take out have the highest " virtual utilization rate " allocative decision as a result of.

Definition RecursiveGreedyAlloc (REQ, MAP), distribute user's request:

Input parameter: REQ, MAP

Output rusults a: user's burst allocative decision

The algorithm steps of RecursiveGreedyAlloc (REQ, MAP) is as follows:

1) for each the user's request among the REQ: at first calculate FRM_Edge by GetEdge (MAP); Then by Factorization (r _i) obtain factor (r _i); Then by AllocScheme (factor (r _i), FRM_Edge) obtain the allocative decision that this user asks.After in allocative decision, increasing the information of user ID, this user's allocative decision is joined among the S set CHT.

2) with all allocative decisions among the S set CHT according to first big after small arranged sequentially of wi, then have identical w _iThe allocative decision inside of value is according to l _iFirst big after small these allocative decisions arranged sequentially.

3) for all allocative decisions among the S set CHT, carry out following steps:

3.1) suppose that current allocative decision is i among the SCHT.Copy S set CHT, called after SCHT_G copies MAP, called after MAP_G.Implement to distribute at MAP_G with current allocative decision, then from SCHT_G, delete all and ask relevant allocative decision with this user.

3.2) if SCHT_G is empty set, jump to 3.3).Otherwise take out first allocative decision among the SCHT_G, if this allocative decision can access distribution in MAP_G, then implement this and distribute, then from SCHT_G, delete all and ask relevant allocative decision and re-execute 3.2 with this user); Otherwise delete this allocative decision and re-execute 3.2).

3.3) calculate the resource utilization of MAP_G and be recorded as: Utility (i).

4) from all allocative decisions of SCHT, pick out have maximum Utility value scheme as a result of.

Algorithm finishes.

The below is illustrated the process of WLFF method:

1 phase I distribution method

The problem that the phase I method solves is: each user who describes according to REQ asks size, and time/frequency source block MAP is distributed to each user.At first the element among the REQ is rearranged according to first big after small order, the user who then is followed successively by among the REQ according to sequencing is transmitted distribution burst.In the time of for user assignment burst, at first for existing unallocated resource, list all allocative decisions; Then carry out " virtual assigned " and the dispensed remaining space limit degree of freedom (FSED) afterwards for all allocative decisions; Then count minimum FSED value and extract the FSED value and enter the next round screening with the identical allocative decision of minimum FSED value; Carry out " virtual assigned " and dispensed remaining space sub-block area second moment (FSSAV) afterwards for the allocative decision that screens afterwards, select to have the final allocative decision of allocative decision conduct of minimum FSSAV value and carry out actual distribution.If certain user's request has obtained distribution, then request corresponding to deletion in set REQ; Otherwise there is not allocative decision in this user's request, adds among the backup set Alloc_Temp after this user's request is deleted from set REQ.User assignment process in phase I constantly repeats, until do not have unappropriated user's request or do not have unappropriated running time-frequency resource in REQ.

2 second stage modification methods

Preparatory stage: latter two burst that cancels at first that the phase I distributes, the cancellation here refers to: latter two user's request that obtains distributing is added in the REQ set, and running time-frequency resource value corresponding to burst about latter two distribution was set to for 0 (representing that these resources are unallocated) among the MAP.Then all user's requests among the backup set Alloc_Temp are added among the REQ set.To gathering element among the REQ this moment according to first big after small arranged sequentially, if element number surpasses 6 among the REQ, keeps 6 last elements, and upgrade REQ and gather.Through these two steps, REQ and MAP become the input parameter of second stage.

Utilize greedy recursive function RecursiveGreedyAlloc (REQ, MAP) dispensed scheme, obtain after the result at every turn, carry out actual distribution and the user that will obtain distributing asks deletion with it from the REQ set.This process constantly repeats until greedy recursive function provides the result for empty, perhaps on time/frequency source block without remaining space.

Description of drawings

Fig. 1: IEEE802.16OFDMA descending sub frame structure.

Fig. 2: the description of time/frequency source block and the definition of slot.

Fig. 3: the definition of surrounding the limit.

Fig. 4: burst is in the feasible location of FRM inside.

Fig. 5: remaining space sub-block example.

Embodiment

The specific embodiment of the present invention is as follows:

Input:

N burst: $\{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DEST\_PATH\_S07144922220071130E000011.png,S07144922220070910E000021.png"\; state="\{\{state\}\}">r</figure-callout>}}_1,{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="S07144922220070910E000021.png"\; state="\{\{state\}\}">r</figure-callout>}}_2,...,r_i,...,r_n|r_i\&Element;Z,\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{r}_i\&GreaterEqual;L*W\}$

Output:

The actual allocated scheme of each burst, running time-frequency resource area utilization method flow:

Phase I:

1) all burst is obtained REQ after according to first big after small ordering, definition intermediate variable set A lloc_Temp.

2) if the REQ set for empty set, jumps to 4); Otherwise: ask for first user among the REQ: at first calculate FRM_Edge by GetEdge (MAP); Then by Factorization (r _i) obtain factor (r _i); Then by AllocScheme (factor (r _i), FRM_Edge) obtain the allocative decision that this user asks.

3) if the set that allocative decision consists of is empty set, first user's request among the set REQ is moved among the Alloc_Temp, jump to 2); Otherwise:

3.1) all allocative decisions are carried out one by one " virtual assigned " and utilized CalFSED (MAP) to calculate corresponding FSED, then count minimum FSED value and extract FSED value and enter next round with the identical allocative decision of minimum FSED value and screen.

3.2) carry out " virtual assigned " and utilize CalFSSAV (MAP) dispensed remaining space sub-block area second moment (FSSAV) afterwards for the allocative decision that screens, select to have the final allocative decision of allocative decision conduct of minimum FSSAV value and carry out actual distribution.In the REQ set, delete distributing user request, jump to 2).

Second stage:

4) latter two user's request that obtains distributing is added in the REQ set, and running time-frequency resource value corresponding to burst about latter two distribution among the MAP is set to 0, then all users' requests among the backup set Alloc_Temp added among the REQ set.

5) to gathering element among the REQ this moment according to first big after small arranged sequentially, if element number surpasses 6 among the REQ, keeps 6 last elements, and upgrade REQ and gather.

6) utilize function RecursiveGreedyAlloc (REQ, MAP) dispensed scheme, utilize its result to carry out actual distribution, and the user that will obtain distributing from the REQ set asks deletion.If REQ gathers non-NULL, re-execute 6); Otherwise jump to 7).

7) calculate final area utilization, method finishes.

Allocation example

The below's process that Demo Asset distributes that cites a plain example:

The problem parameter:

1) length and width of time/frequency source block all are 5 units

The set that the size of 2) being asked by the user consists of: REQ={3,5,4,6,5,4}

The method solution procedure:

1) will gather REQ obtains after arranged sequentially according to first big after small: REQ={6,5,5,4,4,3}

2) phase I: distribute r=6: at first the configuration of the length and width of r can be decomposed into 6=(1 * 6,2 * 3); The MAP of this moment is:

If with the information that (summit 1_x coordinate, summit 1_y coordinate, increment _ x coordinate, increment _ y coordinate, encirclement edge lengths) represents to surround the limit, then 4 current encirclement side informations are: (2,2,1,0,5); (6,2,0,1,5); (6,6 ,-1,0,5); (2,6,0 ,-1,5).Wherein " increment " expression summit 2 is with respect to the symbol of the changes in coordinates direction on summit 1.

Obviously the allocative decision one that can see this moment has 8:

{2×3，(2，2，1，0，5)，head}，{2×3，(2，2，1，0，5)，tail}，{2×3，(6，2，0，1，5)，head}，

{2×3，(6，2，0，1，5)，tail}，{2×3，(6，6，-1，0，5)，head}，{2×3，(6，6，-1，0，5)，tail}，

{2×3，(2，6，0，-1，5)，head}，{2×3，(2，6，0，-1，5)，tail}

The FSED of these allocative decisions is 6, so all allocative decisions enter the screening stage of FSSAV.Calculate us by FSSAV and know that all allocative decisions have identical FSSAV, therefore get first allocative decision as the actual allocated scheme, MAP is updated to after distributing:

-1 -1 -1 -1 -1 -1 -1

-1 0 0 0 0 0 -1

-1 0 0 0 0 0 -1

-1 6 6 0 0 0 -1

-1 6 6 0 0 0 -1

-1 6 6 0 0 0 -1

-1 -1 -1 -1 -1 -1 -1

The REQ set is updated to: REQ={5,5,4,4,3}

Next distribute r=5: at first the configuration of the length and width of r can be decomposed into 5=(1 * 5);

Surround limit one and have 6: (4,2,1,0,3); (6,2,0,1,5); (6,6 ,-1,0,5); (2,6,0 ,-1,2); (2,5,1,0,2); (4,4,0 ,-1,3).

The allocative decision of this moment has 4:

{1×5，(6，2，0，1，5)，head}，{1×5，(6，2，0，1，5)，tail}，{1×5，(6，6，-1，0，5)，head}，{1×5，(6，6，-1，0，5)，tail}

The FSED of these allocative decisions is 6, so all allocative decisions enter the screening stage of FSSAV.

Calculate by FSSAV, these allocative decisions FSSAV separately is respectively: 4.5556,4.5556,14.25,14.25.Select first allocative decision as the actual allocated scheme, MAP is updated to after distributing:

-1 -1 -1 -1 -1 -1 -1

-1 0 0 0 0 5 -1

-1 0 0 0 0 5 -1

-1 6 6 0 0 5 -1

-1 6 6 0 0 5 -1

-1 6 6 0 0 5 -1

-1 -1 -1 -1 -1 -1 -1

The REQ set is updated to: REQ={5,4,4,3}

Next distribute r=5: at first the configuration of the length and width of r can be decomposed into 5=(1 * 5);

Surround limit one and have 6: (4,2,1,0,2); (5,2,0,1,5); (5,6 ,-1,0,4); (2,6,0 ,-1,2); (2,5,1,0,2); (4,4,0 ,-1,3).

The allocative decision of this moment has 2:

{1×5，(5，2，0，1，5)，head}，{1×5，(5，2，0，1，5)，tail}

The FSED of these allocative decisions is 6, so all allocative decisions enter the screening stage of FSSAV.

Calculate by FSSAV, these allocative decisions FSSAV separately is respectively: 1.1389,1.1389.Select first allocative decision as the actual allocated scheme, MAP is updated to after distributing:

-1 -1 -1 -1 -1 -1 -1

-1 0 0 0 5 5 -1

-1 0 0 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 -1 -1 -1 -1 -1 -1

The REQ set is updated to: REQ={4,4,3}

Next distribute r=4: at first the configuration of the length and width of r can be decomposed into 4=(1 * 4,2 * 2);

Surround limit one and have 6: (4,2,1,0,1); (4,2,0,1,5); (4,6 ,-1,0,3); (2,6,0 ,-1,2); (2,5,1,0,2); (4,4,0 ,-1,3).

The allocative decision of this moment has 8:

{1×4，(4，2，0，1，5)，head}，{1×4，(4，2，0，1，5)，tail}，{2×2，(4，6，-1，0，3)，head}，

{2×2，(4，6，-1，0，3)，tail}，{2×2，(2，6，0，-1，2)，head}，{2×2，(2，6，0，-1，2)，tail}，

{2×2，(2，5，1，0，2)，head}，{2×2，(2，5，1，0，2)，tail}

Their FSED separately are respectively: 6,8,8,4,4,4,4,4; 5 allocative decisions of back enter the screening stage of FSSAV.Calculate by FSSAV, these allocative decisions FSSAV separately is respectively: 0,0,0,0,0.Select first allocative decision as the actual allocated scheme, MAP is updated to after distributing:

-1 -1 -1 -1 -1 -1 -1

-1 4 4 0 5 5 -1

-1 4 4 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 -1 -1 -1 -1 -1 -1

The REQ set is updated to: REQ={4,3}

Next distribute r=4: at first the configuration of the length and width of r can be decomposed into 4=(1 * 4,2 * 2);

Surround limit one and have 4: (4,2,1,0,1); (4,2,0,1,5); (4,6 ,-1,0,1); (4,6,0 ,-1,5).

The allocative decision of this moment has 4:

{1×4，(4，2，0，1，5)，head}，{1×4，(4，2，0，1，5)，tail}，{1×4，(4，6，0，-1，5)，head}，{1×4，(4，6，0，-1，5)，tail}

Their FSED separately are respectively: 4,4,4,4; All enter screening stage.Calculate by FSSAV, these allocative decisions FSSAV separately is respectively: 0,0,0,0.Select first allocative decision as the actual allocated scheme, MAP is updated to after distributing:

-1 -1 -1 -1 -1 -1 -1

-1 4 4 0 5 5 -1

-1 4 4 4 5 5 -1

-1 6 6 4 5 5 -1

-1 6 6 4 5 5 -1

-1 6 6 4 5 5 -1

-1 -1 -1 -1 1 -1 -1

The REQ set is updated to: REQ={3}

By calculating as can be known, there is not allocative decision in r=3, and the phase I method finishes.

3) second stage: after the allocation result of cancelling last two-wheeled, obtain following problem:

The REQ set is: REQ={4,4,3};

MAP is:

-1 -1 -1 -1 -1 -1 -1

-1 0 0 0 5 5 -1

-1 0 0 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 -1 -1 -1 -1 -1 -1

List all allocative decisions according to form { l * w, (surrounding the surrounding edge letter), head/tail, the sequence number in the REQ set }, obtain after the ordering:

{2×2，(4，6，-1，0，3)，head，1}，{2×2，(4，6，-1，0，3)}，

{2×2，(2，6，0，-1，2)，head，1}，{2×2，(2，6，0，-1，2)，tail，1}，

{2×2，(2，5，1，0，2)，head，1}，{2×2，(2，5，1，0，2)，tail，1}，

{2×2，(4，6，-1，0，3)，head，2}，{2×2，(4，6，-1，0，3)，tail，2}，

{2×2，(2，6，0，-1，2)，head，2}，{2×2，(2，6，0，-1，2)，tail，2}，

{2×2，(2，5，1，0，2)，head，2}，{2×2，(2，5，1，0，2)，tail，2}，{1×4，(4，2，0，1，5)，head，1}，

{1×4，(4，2，0，1，5)，tail，1}，{1×4，(4，2，0，1，5)，head，2}，{1×4，(4，2，0，1，5)，tail，2}，

{1×3，(4，2，0，1，5)，head，3}，{1×3，(4，2，0，1，5)，tail，3}，{1×3，(4，6，-1，0，3)，head，3}，

{1×3，(4，6，-1，0，3)，tail，3}，{1×3，(4，4，0，-1，3)，head，3}，{1×3，(4，4，0，-1，3)，tail，3}

After above each allocative decision carried out that " virtual " recurrence is greedy and distribute, its utilance was respectively: 23/25,24/25,24/25,24/25,24/25,24/25,23/25,24/25,24/25,24/25,24/25,24/25,24/25,24/25,24/25,24/25,23/25,23/25,23/25,23/25,23/25,23/25.Choose the wherein scheme of utilance maximum: 2 * 2, (4,6 ,-1,0,3), tail, 1}; After the distribution, MAP is updated to:

-1 -1 -1 -1 -1 -1 -1

-1 4 4 0 5 5 -1

-1 4 4 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 6 6 0 5 5 -1

-1 -1 -1 -1 -1 -1 -1

The REQ set is: REQ={4,3};

List all allocative decisions, obtain after the ordering:

{1×4，(4，2，0，1，5)，head，1}，{1×4，(4，2，0，1，5)，tail，1}，

{1×4，(4，6，0，-1，5)，head，1}，{1×4，(4，6，0，-1，5)，tail，1}，

{1×3，(4，2，0，1，5)，head，2}，{1×3，(4，2，0，1，5)，tail，2}，{1×3，(4，6，0，-1，5)，head，2}，

{1×3，(4，6，0，-1，5)，tail，2}

After above each allocative decision carried out that " virtual " recurrence is greedy and distribute, its utilance was respectively: 24/25,24/25,24/25,24/25,23/25,23/25,23/25,23/25.Choose the wherein scheme of utilance maximum: 1 * 4, (4,2,0,1,5), head, 1}; After the distribution, MAP is updated to:

-1 -1 -1 -1 -1 -1 -1

-1 4 4 0 5 5 -1

-1 4 4 4 5 5 -1

-1 6 6 4 5 5 -1

-1 6 6 4 5 5 -1

-1 6 6 4 5 5 -1

-1 -1 -1 -1 -1 -1 -1

Second stage finishes, and final utilance is 24/25, and the actual allocated scheme is as shown in process.

The standard sequence test result

Cycle tests in the document [5] has been carried out emulation, and the allocation result in the document [1] compares, can see that the WLFF method that we propose can obtain higher resource utilization.

Cycle tests	Thing piece number	The resource gross area	The unallocated ratio of LFF (%)	The unallocated ratio of WLFF (%)
					prob1.dat	16	400	2.00	0.00
prob2.dat	17	400	2.00	0.00
					prob3.dat	16	400	2.50	0.00
prob4.dat	25	600	0.67	0.33
					prob5.dat	25	600	0.00	0.00
prob6.dat	25	600	0.00	0.50
					prob7.dat	28	1800	0.67	0.67
prob8.dat	29	1800	0.83	0.78
					prob9.dat	28	1800	0.78	0.56
prob10.dat	49	3600	0.97	0.39
					prob11.dat	49	3600	0.22	0.00
prob12.dat	49	3600	N/A	0.17
					prob13.dat	73	5400	0.30	0.06
prob14.dat	73	5400	0.04	0.00
					prob15.dat	73	5400	0.83	0.06
prob16.dat	97	9600	0.25	0.17
					prob17.dat	97	9600	3.74	0.00
prob18.dat	97	9600	0.54	0.14
					Average unallocated ratio (%)			0.92	0.21

The LFF algorithm computation complexity that document [1] proposes is O (n ⁵Logn), this method is at O (n ²) can obtain to approach even surpass the performance of LFF arithmetic result in the situation of complexity.The time-frequency resource allocation method of OFDM access system that the present invention proposes is a kind of fast and effectively distribution method.

List of references (References)

[1]Yu-Liang Wu，Wenqi Huang，et al.An effective quasi-human based heuristic for solving therectangle packing problem[J].European Journal of Operational Research，2002，141(2)：341-358.

[2]IEEE802.16，802.16-2004：IEEE Standard for Local and Metropolitan Area Networks Part16：Air Interface for Fixed Broadband Wireless Access Systems，IEEE802.16-2004，October2004.

[3]IEEE802.16e，802.16e：IEEE Standard for Local and Metropolitan Area Networks Part16：Air Interface for Fixed and Mobile Broadband Wireless Access Systems Amendment forPhysical and Media Access Control Layers for Combined Fixed and Mobile Operation inLicensed Bands，IEEE802.16e，September2005.

[4]Yehuda Ben-Shimol，Itzik Kitroser，et al.Two-dimensional mapping for wireless OFDMAsystems[J].IEEE Transactions on Broadcasting，2006，52(3)：388-396.

[5]E Hopper，B C H Turton.An empirical investigation of meta-heuristic and heuristicalgorithms for a2d packing problem[J].European Journal of Operational Research，2001，128(1)：34-57.

Claims (1)

1. time-frequency resource allocation method of OFDM access system is characterized in that adopting following steps to find the solution to the resource allocation problem under the rectangle constraints: the first big after small order that at first occupies resource according to each user's transmission sorts to user's assignment order; Secondly when being some user resource allocations according to the suitable allocative decision of Information Selection of the remaining space limit degree of freedom, remaining space sub-block area second moment; Improve distribution effects by the greedy distribution method of the recurrence of a constant depth at last; Wherein,

The limit degree of freedom of described remaining space refers to current allocative decision is carried out " virtual assigned " afterwards encirclement limit number of unallocated Resource Block;

Described remaining space sub-block area second moment refers to occupy a certain of unallocated Resource Block and surrounds the limit, in unallocated Resource Block inside along the second moment of the rectangular block of the area maximum that can reach perpendicular to this Directional Extension that surrounds the limit;

The allocative decision that described selection is suitable lists all allocative decisions when the steps include: as some user resource allocations;

For each allocative decision of enumerating out, at first this scheme is carried out " virtual assigned ", then calculate the remaining space limit degree of freedom; Pick out the allocative decision with limit, least residue space degree of freedom and form a set, called after " set of the minimum edge degree of freedom ";

For each allocative decision in " set of the minimum edge degree of freedom ": at first this scheme is carried out " virtual assigned ", then calculate remaining space sub-block area second moment; Select the final allocative decision of scheme conduct that wherein has least residue space sub-block area second moment and carry out actual allocated;

Repeat said process, until finish the distribution of all user's transmission;

The greedy distribution method of described recurrence, the state that the steps include: at first to cancel two user's transmission of last distribution and upgrade the resource allocation problem input parameter; Then constantly utilize the recurrence greedy function that user transmission is distributed, until the user transmits that set becomes empty set or without unallocated resource; Assigning process finishes.

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