CN103731922B - Channel allocation method based on efficiency - Google Patents

Abstract

The invention discloses a kind of channel allocation method based on efficiency, belong to the resource allocation techniques field in mobile radio networks, specifically include following steps: determine original subchannel allocative decision；On the basis of original subchannel allocative decision, the home subscriber of change libron channel is the user that in this subchannel, speed is maximum；A non-free subchannel can be chosen from big to small by room for promotion according to speed, it is carried out substitute operation.Subchannel, on the premise of ensureing each user rate requirements, is reasonably distributed to community user by the method for the present invention so that community total energy effect maximizes, thus improves energy efficiency.

Description

Channel allocation method based on efficiency

Technical field

The invention belongs to the resource allocation techniques field in mobile radio networks, be specifically related to wireless mobile communication network Channel allocation method in network.

Background technology

Along with the raising of mobile radio networks scale He its data rate requirement, the energy consumed in communication process Notable rising, greenhouse gas emission is continuously increased, and the environmental pollution caused is on the rise, and its power saving has caused full society The concern of meeting, the concept of green communications is arisen at the historic moment.It is not only one, communications field core topic, is also to affect sustainable development One major obstacles of exhibition.In the face of the severe challenge of radio communication energy consumption problem, focus on resource utilization, save network operation Required energy consumption, reduces network operation cost and has become operator's development focus, around green communications academic research the most increasingly Extensively.

In the design of existing Radio Resource allocation algorithm based on efficiency, in order to reduce the computation complexity of algorithm, greatly The power studied under the subchannel distribution being broken down under fix power allocation distributes with stator channel controls two sons more Problem is inquired into.It should be noted that stator channel emission power distributes, to maximize the system energy efficiency letter as target Road assignment problem is equivalent to the subchannel assignment problem of the total speed of maximization system, and typically one 0-1 linear integer rule The problem of drawing, belongs to NP-difficulty problem, and the typical method solving optimal solution is branch and bound method and cutting plane algorithm, but both approaches Worst case time complexity the highest, therefore be difficult to meet for the practical communication system that there is big quantum channel and user The system time requirement to resource allocation algorithm.Therefore, it is possible to try to achieve the algorithm of sub-channel assignment result in polynomial time more There is practical significance.

About subchannel assignment problem, document Energy efficient resource allocation in non- Cooperative multi-cell OFDMA systems proposes DSA algorithm, and its core concept is greedy strategy, but this calculation Method can not ensure the minimum-rate requirement of user, and after entering, in DSA algorithm, the security mechanism that user's minimum-rate requires, greedy Greedy strategy no longer can reach optimal solution.Document Energy-Efficient Resource Allocation in OFDMA Networks proposes MDSA algorithm, obtains the suboptimal solution of allocative decision by maximizing minimum user's efficiency, but emulate Show that under some scene, the performance of this suboptimal solution is the best.

Summary of the invention

The invention aims to solve the problem of existing mobile radio networks subchannel allocated phase, propose A kind of channel allocation method based on efficiency.

Understand the present invention for the ease of those of ordinary skill in the art, first technical term made and being defined below:

Subchannel: being that mobile radio networks downlink distributes to the least unit of user at frequency domain, base station is permissible In communication every time, dynamically distribute subchannel, and control the transmitting power on every sub-channels.

Efficiency: transfer rate and the ratio of power consumption, wherein, power consumption includes circuit power and launches power.

Subchannel achievable rate: subchannel is distributed to the user in a certain user set, can obtain in this subchannel The maximum rate obtained.

Home subscriber: if a certain subchannel is distributed to a certain user, then this user is called the home subscriber of this channel.

User's minimum-rate retrains: for ensureing the minimum-rate that customer service must reach.

Libron channel: if a certain subchannel to be distributed to arbitrarily other users, the minimum of the former home subscriber of this subchannel Rate requirement still meets, then be called libron channel；On the contrary, if the minimum-rate of the former home subscriber of this subchannel requires not Meet, the most non-free subchannel.

Substitute: for non-free subchannel i, by a, its home subscriber being changed to b, there is a sub-channels j simultaneously, j is Libron channel or distribute to the non-free subchannel of user b, the home subscriber of change j is a, if the speed of the most all users Minimum-rate requires to still meet, and after changing i, j and speed rise, the most referred to as j substitutes i.

Sub-channel rates can room for promotion: the maximum achievable rate of subchannel and the difference of actual speed rate are this subchannel Speed can room for promotion.

The technical scheme is that: a kind of channel allocation method based on efficiency, specifically include following steps:

S1. original subchannel allocative decision is determined；

S2., on the basis of step S1 original subchannel allocative decision, the home subscriber of change libron channel is this son The user that on channel, speed is maximum；

S3. can choose a non-free subchannel from big to small by room for promotion according to speed, it is carried out substitute operation.

Further, step S1 specifically includes the most step by step:

S11. initialize, according to formula $r_{s,i,n}=W{\log }_2(1+P_{s,n}{\left|h_{s,i,n}\right|}^2/(\underset{\underset{s^{\′}\≠s}{s^{\′}=1}}{\overset{S}{\mathrm{\Σ}}}{P}_{s^{\′},n}{\left|h_{s,i,n}^{s^{\′}}\right|}^2+{\mathrm{WN}}_0))$ Seek each use Family achievable rate on each of the sub-channels, wherein, r_s,i,nFor obtainable speed in the user i the n-th subchannel in the s of community Rate, W is the bandwidth of every sub-channels；h_s,i,nFor the channel gain in the n-th subchannel of the user i in the s of community；For little The user i in district s channel gain in community s ' the n-th subchannel；P_s,nFor the base station of community s in the n-th subchannel Launch power, P_s′,nFor the base station of the community s ' transmitting power in the n-th subchannel, S is number of cells；N₀For white noise Average power spectral density；Defining total user in optional user Ji Wei community, residue channel set is all channels in community；

If S12. residue channel set is not empty, chooses the subchannel that maximum achievable rate is minimum, assign them at this letter The user that on road, speed is maximum, leaves out this channel in residue channel set；Update this user obtains speed, if greater than this use The minimum-rate constraint at family, then concentrate from optional user and leave out this user；If optional user collection is sky, then it represents that met institute There is user rate requirements, enter step S13, otherwise repeat step S12；

S13. the user in residue channel set is distributed to the user that speed is maximum.

Further, step S2 specifically includes the most step by step:

S21. according to formula $r_{s,i,n}=W{\log }_2(1+P_{s,n}{\left|h_{s,i,n}\right|}^2/(\underset{\underset{s^{\′}\≠s}{s^{\′}=1}}{\overset{S}{\mathrm{\Σ}}}{P}_{s^{\′},n}{\left|h_{s,i,n}^{s^{\′}}\right|}^2+{\mathrm{WN}}_0))$ Ask each user each Achievable rate in subchannel；

S22. according to achievable rate and existing distribution, ask each user's and speed, and the reality on every sub-channels Speed；

S23. for every sub-channels and its home subscriber, if this user's deducts the reality in this subchannel with speed Speed is more than the minimum-rate constraint of this user, then this subchannel is libron channel.Maximum achievable rate and reality in subchannel The difference of border speed is that the speed of this subchannel can room for promotion；Finding out speed can the maximum r of room for promotion^incmaxAnd it is right Subchannel i answered；

If S24. r^incmaxLess than or equal to zero, step S2 terminates, and otherwise, the home subscriber of change i is speed on this channel Maximum user, forwards step S22 to.

Further, step S3 specifically includes the most step by step:

S31. for every sub-channels n each user i ', according to formulaAsk and the home subscriber of n is become The speed of more i ' changes matrixWherein,The speed that user i ' can reach is distributed to for subchannel n,For The actual speed rate of existing allocation result sub-channels n；

S32. find outMiddle maximum r_changeAnd subchannel n of correspondence₁, userSubchannel n₁Former ownership is used Family is $i_1^{\mathrm{src}};$

If S33. r_changeLess than or equal to zero, step S3 terminates, and otherwise, initializes subchannel n₁Substituted set C₁For Empty；For n₁Outside anyon channel n₂, it may be judged whether meeting following three condition simultaneously, if meeting, being included in C₁；

(1) .n₂Former home subscriber be not

(2) .n₂Change home subscriber isThe speed change caused and n₁Speed can room for promotion sum more than zero；

(3) if. change n₁Home subscriber beChange n₂Home subscriber beSoAnd n₂'s Former home subscriber still meets respective minimum speed limit requirement；

If S34. C₁For sky, thenFor-1, i.e. do not consider n before performing substitute next time and operating₁Home subscriber become MoreForward step S32 to；Otherwise: choose C₁In subchannel n₂Meet following condition, i.e. if change n₂Home subscriber isMake n₂Speed promotes maximum,For subchannel n₂Former home subscriber；Perform substitute operation, i.e. change n₁Home subscriber It is changed toChange n₂Home subscriber isForward step S31 to.

Beneficial effects of the present invention: The present invention gives one and be effectively and flexibly subchannel allocation algorithm, especially by Step S2 achieves the maximization of free channel efficiency；Constantly promoted the energy of non-free channel by substitute operation by step S3 Effect；Simulation result shows, the system total energy effect of the complete subchannel allocation algorithm comprising three steps can reach optimal solution More than 95%.On the other hand, its motility is embodied in, and step S1 and step S3 can be saved in some scenarios, step S1, S2 Algorithms T-cbmplexity in the worst cases is all o (N²I), during for being strict with convergence in polynomial time, Step S3 can be saved.Subchannel, on the premise of ensureing each user rate requirements, is reasonably distributed by the method for the present invention To community user so that community total energy effect maximizes, thus improves energy efficiency.

Accompanying drawing explanation

Fig. 1 is the schematic flow sheet of the channel allocation method based on efficiency of the present invention.

Detailed description of the invention

For making the object, technical solutions and advantages of the present invention clearer, below in conjunction with accompanying drawing and concrete example to this Invention is described in further detail.

Assume that certain mobile radio networks comprises S neighbor cell, the N number of available subchannels of each little Division, often In individual community, total I is individual is random equally distributed user in spatial domain, and the community ownership of each user determines,For in the s of community User collection, I_sIt it is the user's number in the s of community.Definition β_s,i,n{ 0,1} is that resource distributes designator, β to ∈_s,i,n=1 represents community The n-th subchannel in s distributes to user i, on the contrary then β_s,i,n=0.In order to avoid the cochannel in community disturbs, it is stipulated that each Subchannel at most can only distribute to a user, i.e.

${\mathrm{\Σ}}_{i=1}^{I_s}{\mathrm{\β}}_{s,i,n}=1,\∀s,n---\left(1\right)$

Definition r_s,i,nAchievable rate in the user i the n-th subchannel in the s of community:

$r_{s,i,n}=W{\log }_2(1+P_{s,n}{\left|h_{s,i,n}\right|}^2/(\underset{\underset{s^{\′}\≠s}{s^{\′}=1}}{\overset{S}{\mathrm{\Σ}}}{P}_{s^{\′},n}{\left|h_{s,i,n}^{s^{\′}}\right|}^2+{\mathrm{WN}}_0))---\left(2\right)$

Wherein, W is the bandwidth of every sub-channels；h_s,i,nIncrease for the channel in the n-th subchannel of the user i in the s of community Benefit；For the channel gain in community s ' the n-th subchannel of the user i in the s of community；P_s,nFor the base station of community s n-th Transmitting power on sub-channels；N₀Average power spectral density for white noise.

DefinitionEfficiency for community s:

${\mathrm{\η}}_{{\mathrm{EE}}_s}=\frac{{\mathrm{\Σ}}_{i=1}^{I_s}{\mathrm{\Σ}}_{n=1}^N{\mathrm{\β}}_{s,i,n}{r}_{s,i,n}}{{\mathrm{\Σ}}_{n=1}^N{P}_{s,n}+{P_s}^C}---\left(3\right)$

Wherein,Circuit power consumption for the base station of community s.

In sum, during stator channel power, the maximum efficiency resource optimization problem of multi-cell system can mathematical modeling For:

Wherein,Minimum speed limit requirement for user i.

Based on foregoing description, embodiments provide a kind of channel allocation method based on efficiency, idiographic flow Figure is as it is shown in figure 1, comprise the following steps:

S1. original subchannel allocative decision is determined；

S2., on the basis of step S1 original subchannel allocative decision, the home subscriber of change libron channel is this son The user that on channel, speed is maximum；

S3. can choose a non-free subchannel from big to small by room for promotion according to speed, it is carried out substitute operation.

In the present embodiment, step S1 specifically uses sorting consistence greedy algorithm, and this algorithm is the most effective, it is possible to take into account user On the premise of minimum-rate constraint, obtain bigger efficiency the most on sub-channels.

The operation of step S1 divides following two stages: the first stage, in the user of not met rate requirement gathers, greedy Subchannel is distributed to have the user of maximum achievable rate.Second stage is carried out after all users meet rate requirement, The user that residue subchannel is distributed to have maximum achievable rate i.e. thrown the reins to.Owing to the subchannel of second stage is complete Distribute to optimal user freely, therefore, first the subchannel that maximum achievable rate is less is allocated, is conducive to acquisition bigger Community efficiency, this is introduced into the purpose of sorting consistence.

The implementation process of step S1 is as follows in this example:

S11. initialize, seek each user achievable rate on each of the sub-channels according to formula (2), define optional user collection For total user in community, residue channel set is all channels in community；

If S12. residue channel set is not empty, chooses the subchannel that maximum achievable rate is minimum, assign them at this letter The user that on road, speed is maximum, leaves out this channel in residue channel set；Update this user obtains speed, if greater than this use The minimum-rate constraint at family, then concentrate from optional user and leave out this user；If optional user collection is sky, then it represents that met institute There is user rate requirements, enter step S13, otherwise repeat step S12；

S13. the user in residue channel set is distributed to the user that speed is maximum.

Step S2 is in existing allocation resultOn the basis of, promote order from big to small according to speed, libron Channel distributes to moving of the user that achievable rate is maximum, referred to as libron channel.This distribution change ensure that single little District's total energy effect does not declines.Additionally, major part subchannel already at optimum or can not freely be divided in existing allocation result Timing, the execution time of algorithm is the shortest.In this example, the implementation process of step S2 is as follows:

S21. each user achievable rate on each of the sub-channels is sought according to formula (2)；

S22. according to achievable rate and existing distribution, ask each user's and speed, and the reality on every sub-channels Speed；

S23. for every sub-channels and its home subscriber, if this user's deducts the reality in this subchannel with speed Speed is more than the minimum-rate constraint of this user, then this subchannel is libron channel.Maximum achievable rate and reality in subchannel The difference of border speed is that the speed of this subchannel can room for promotion；Finding out speed can the maximum r of room for promotion^incmaxAnd it is right Subchannel i answered；

If S24. r^incmaxLess than or equal to zero, step S2 terminates, and otherwise, the home subscriber of change i is speed on this channel Maximum user, forwards step S22 to.

After step S2, the speed of all libron channels can room for promotion be all zero.

Step S3 is aiming at speed can maximum non-free subchannel n of room for promotion₁, choose the n meeting following condition₂ As substitute, whereinIt is n₁Former home subscriber,It is n₁The user reaching maximum rate.

(1) .n₂Former home subscriber be not

(2) .n₂Change home subscriber isThe speed change caused and n₁Speed can room for promotion sum more than zero；

(3) if. change n₁Home subscriber beChange n₂Home subscriber beSoAnd n₂ Former home subscriber still meet respective minimum speed limit requirement.

Clearly for the n meeting above three point condition₂, changing its home subscriber isChange n₁Home subscriber beCommunity total energy effect will be made to increase.And according to efficiency can room for promotion from big to small choose n₁Condition and energy is met with selection Effect increases maximum or reduces minimum subchannel as n₂So that step S3 decreases during continuous improving energy efficiency need not The substitute operation wanted, reduces its execution time.In this example, the implementation process of step S3 is as follows:

S31. for every sub-channels n each user i ', according to formulaAsk and the home subscriber of n is become The speed of more i ' changes matrixWherein,The speed that user i ' can reach is distributed to for subchannel n,For The actual speed rate of existing allocation result sub-channels n；

S32. find outMiddle maximum r_changeAnd subchannel n of correspondence₁, userSubchannel n₁Former ownership is used Family is $i_1^{\mathrm{src}};$

If S33. r_changeLess than or equal to zero, step S3 terminates, and otherwise, initializes subchannel n₁Substituted set C₁For Empty；For n₁Outside anyon channel n₂, it may be judged whether meet above three condition (1), (2) and (3) simultaneously, if meeting, Included in C₁。

If S34. C₁For sky, thenFor-1, i.e. do not consider n before performing substitute next time and operating₁Home subscriber become MoreForward step S32 to；Otherwise: find out C₁In a certain subchannel n₂Meet following condition, i.e. if change n₂Ownership is used Family isMake n₂Speed promotes maximum,For subchannel n₂Former home subscriber；Perform substitute operation, i.e. change n₁Ownership User is changed toChange n₂Home subscriber isForward step S31 to.

Simulation result shows, the only system total energy effect containing step S1, the sub-channel allocation scheme of S2 can reach optimal solution More than 60%.Control, for comprising subchannel distribution and power, the resource allocation algorithm that two stage iteration are carried out, can only exist Using step S1 first during iteration, every time sub-channel assignment result is that the subchannel of this iteration is initial before iteration afterwards Distribution, i.e. saves step S1, and in practical problem, operates few owing to meeting the substitute of condition, and therefore step S3 is time-consuming Less.

Those of ordinary skill in the art it will be appreciated that embodiment described here be to aid in reader understanding this Bright principle, it should be understood that protection scope of the present invention is not limited to such special statement and embodiment.This area It is each that those of ordinary skill can make various other without departing from essence of the present invention according to these technology disclosed by the invention enlightenment Planting concrete deformation and combination, these deform and combine the most within the scope of the present invention.

Claims (2)

1. a channel allocation method based on efficiency, it is characterised in that specifically include following steps:

S1. original subchannel allocative decision is determined；

Specifically include the most step by step:

S11. initialize, according to formulaAsk each user often Achievable rate on sub-channels, wherein, r_s,i,nFor obtainable speed in the user i the n-th subchannel in the s of community, W is The bandwidth of every sub-channels；h_s,i,nFor the channel gain in the n-th subchannel of the user i in the s of community；For in the s of community User i channel gain in community s ' the n-th subchannel；P_s,nFor the transmitting in the n-th subchannel of the base station of community s Power, P_s′,nFor the base station of the community s ' transmitting power in the n-th subchannel, S is number of cells；N₀Average for white noise Power spectral density；Defining total user in optional user Ji Wei community, residue channel set is all channels in community；

If S12. residue channel set is not empty, chooses the subchannel that maximum achievable rate is minimum, assign them on this channel The user that speed is maximum, leaves out this channel in residue channel set；Update this user obtains speed, if greater than this user's Minimum-rate retrains, then concentrate from optional user and leave out this user；If optional user collection is sky, then it represents that met institute useful Family rate requirement, enters step S13, otherwise repeats step S12；

S13. the user in residue channel set is distributed to the user that speed is maximum；

S2., on the basis of step S1 original subchannel allocative decision, the home subscriber of change libron channel is this subchannel The user that upper speed is maximum；

Specifically include the most step by step:

S21. according to formulaEach user is asked to believe at every height Achievable rate on road；

S22. according to achievable rate and existing distribution, ask each user's and speed, and the actual speed rate on every sub-channels；

S23. for every sub-channels and its home subscriber, if this user's deducts the actual speed rate in this subchannel with speed Retrain more than the minimum-rate of this user, then this subchannel is libron channel, maximum achievable rate and reality speed in subchannel The difference of rate is that the speed of this subchannel can room for promotion；Finding out speed can the maximum r of room for promotion^incmaxAnd correspondence Subchannel i；

If S24. r^incmaxLess than or equal to zero, step S2 terminates, and otherwise, the home subscriber of change i is that speed is maximum on this channel User, forward step S22 to；

S3. can choose a non-free subchannel from big to small by room for promotion according to speed, it is carried out substitute operation.

Channel allocation method the most according to claim 1, it is characterised in that step S3 specifically includes the most step by step:

S31. for every sub-channels n each user i ', according to formulaAsk and the home subscriber of n is changed to i ' Speed change matrixWherein,The speed that user i ' can reach is distributed to for subchannel n,For existing distribution The actual speed rate of result sub-channels n；

S32. find outMiddle maximum r_changeAnd subchannel n of correspondence₁, userSubchannel n₁Former home subscriber is

If S33. r_changeLess than or equal to zero, step S3 terminates, and otherwise, initializes subchannel n₁Substituted set C₁For sky；Right In n₁Outside anyon channel n₂, it may be judged whether meeting following three condition simultaneously, if meeting, being included in C₁；

(1).n₂Former home subscriber be not

(2).n₂Change home subscriber isThe speed change caused and n₁Speed can room for promotion sum more than zero；

(3) if. change n₁Home subscriber beChange n₂Home subscriber beSoAnd n₂Former return Belong to user and still meet respective minimum speed limit requirement；

If S34. C₁For sky, thenFor-1, i.e. do not consider n before performing substitute next time and operating₁Home subscriber be changed toForward step S32 to；Otherwise: choose C₁In subchannel n₂Meet following condition, i.e. if change n₂Home subscriber isMake Obtain n₂Speed promotes maximum,For subchannel n₂Former home subscriber；Perform substitute operation, i.e. change n₁Home subscriber be changed toChange n₂Home subscriber isForward step S31 to.