CN104135743A - Resource allocation method based on cache control in LTE-A (Long Term Evolution-Advanced) cellular network - Google Patents

Abstract

The invention discloses a resource allocation method based on cache control in an LTE-A (Long Term Evolution-Advanced) cellular network. Through utilizing a timing-varying difference between return and access link channels, during each semi-static scheduling period, the number of resource blocks, which are pre-allocated by return sub-frames, is estimated, when the data transmission quantity of a return link is greater than the data transmission quantity of an access link, the excessive transmission data are stored in a relay-end cache so as to be read during the later semi-static scheduling period; the excessive transmission data are data, which are received on the return sub-frames by a relay and cannot be forwarded in access sub-frames during the scheduling period; throughput performance of a user and a system is improved, and meanwhile, fairness between a macro user and a relay user is improved in a self-adaptive manner for a history performance difference between the macro user and the relay user. According to the resource allocation method disclosed by the invention, fairness and throughput performance of different service stations are improved for application of a buffer queue. Meanwhile, the cost of cache equipment is relatively lower and is conveniently applied in an actual system.

Description

Resource allocation methods based on buffer control in a kind of LTE-A cellular network

Technical field

The invention belongs to wireless communication technology field, relate to the relaying technique of one of LTE key technology, studied the system resource allocation in junction network (Resource Allocation) method for designing.

Background technology

In recent years, 3G technology is quite ripe, the positive expanding day of 3G network coverage, but along with continuing to bring out and continuous rising that people require communication quality of new radio communication service, and frequency spectrum below 5GHz has almost been fully occupied, cannot configure new specific resource to radio communication, thus Radio Resource day aobvious phenomenon in short supply still day by day serious.This phenomenon has promoted the research of people to Next-Generation Wireless Communication Systems, at present, research for next generation network is also ripe gradually, some areas LTE technology has also been carried out relevant test job, relaying is as one of key technology of LTE, for the marginal user performance that improves community, improve the coverage of High Data Rate, realize green communications, expand cell coverage area, alleviate the operating pressure of base station, realize interim deployment etc. aspect and have outstanding contributions, simultaneously, it is also low many that the construction cost of relaying and power consumption are compared base station, and there is a lot of flexibilities.

Introduce after relaying, the interference scene in community is more complicated, and the resource allocation problem facing is more thorny, and whether the distribution of resource rationally can directly have influence on again the improvement of systematic function.Solve the resource allocation problem of relay system, reasonably allocation strategy must will be taken into account fairness and validity.Namely consider the compromise of suitable fairness and validity.Fairness in relay system comprises local fairness and global fair, local fairness has reflected the fairness between different user in same service station, global fair has reflected that the fairness between the user in different service stations is the serviced fairness of user, meanwhile, validity is intended to improve the accumulative total throughput of each sector.

Although have pertinent literature to distribute and launched research LTE relay system resource at present, all have different drawbacks.Wherein the simplest scheme is the Resource Allocation Formula of supposition Type I relay system, suppose that backhaul occupies identical frequency band from the access different service stations of subframe, and supposition backhaul number of subframes is fixed, and only decides the shared number of resource blocks in different service stations according to the number of excited users in different service stations on this basis.Obviously, the mode of this fixed backhaul subframe, lacks very large flexibility, and easily causes the waste of resource, and simultaneously resource allocation policy is failed fairness and the validity of the system of considering.On the basis of such scheme one, the subproblem existing is improved, we are referred to as scheme two, it has considered that the aggregated throughput of community is for the impact of resource distribution, but still fails to solve fixed backhaul subframe and do not consider the drawback that fairness is brought.Scheme three is for Utilizing Resource Pattern (FDM different in backhaul subframe, TDM), simultaneously respectively using fairness or validity as the target of distributing, article has provided the limit of reaching the standard grade that while taking into account fairness and validity, resource is distributed, article belongs to inspiration character, does not provide the compromise of effective fairness and validity.The backhaul subframe mode of scheme four based on TDM, maximizes the poorest user's throughput.For the algorithm of taking into account fairness and validity, we are designated as scheme five its typical scenario.In this scheme author can optimize by logarithmic utility function simultaneously more than two targets, but because the independence of each semi-persistent scheduling causes ensureing the long-term fairness of system, and herein with great majority research under the prerequisite without buffer memory, suppose that backhaul speed equals access rate, to prevent relay data congestion, but such equality constraint can make in back haul link or access link channel quality poor, make this link become the bottleneck that trunk subscriber throughput is improved.

Summary of the invention

For above-mentioned defect or deficiency, the invention provides the resource allocation methods based on buffer control in a kind of LTE-A cellular network.

For reaching above object, technical scheme of the present invention is:

Comprise the following steps:

1), obtain the user's of S set that total resources piece number in a radio frames in LTE-A system is Ω, service station s, service station s access set K _s, s service in service station in Cellular Networks user k throughput in access subframe

2), in the time carrying out transfer of data, during each semi-persistent scheduling, estimate the preallocated Resource Block number of backhaul subframe in the time that the volume of transmitted data of back haul link is greater than the transfer of data of access link, unnecessary transmission data are stored in the buffer memory of relay, with make after semi-persistent scheduling during be read; Described unnecessary transmission data are between this schedule periods, data that relaying receives in backhaul subframe and that cannot forward in access subframe;

3), according to the preallocated Resource Block number of backhaul subframe distribute remaining Resource Block $\mathrm{\Ω}-\underset{s}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{in},1}^s:$

3.1), basis obtain optimization aim, i.e. utility function U:

$U=\underset{k}{\mathrm{\Σ}}\log R_{a,s}^k---\left(1\right)$

$R_{a,s}^k\≈\frac{{\mathrm{\ω}}_a^s}{\left|K_s\right|}{\mathrm{\γ}}_{a,s}^k{G}_s---\left(2\right)$

Wherein, refer to that the user k that belongs to service station s in the time adopting PF scheduling is accessing the throughput in subframe; represent the resource block number that service station s distributes in access subframe; | K _s| represent the number of service station s institute service-user; represent service station s throughput on unit resource piece in access subframe; G _sthe additional gain that RR dispatching office brings is compared in expression PF scheduling;

3.2) the maximum utility function max (U), obtaining according to utility function U and formula (3), formula (4) and formula (5) are combined and are solved ω _{a, 2}, with

${\mathrm{\ω}}_{b,2}^s{\mathrm{\ρ}}_b^s+{\mathrm{\ω}}_{\mathrm{out},2}^s{\mathrm{\ρ}}_a^s-{\mathrm{\ω}}_{a,2}^s{\mathrm{\ρ}}_a^s=0---\left(3\right)$

${\mathrm{\ω}}_{a,2}^s\≤{\mathrm{\ω}}_{a,2}---\left(4\right)$

$\underset{s\≠0}{\mathrm{\Σ}}{\mathrm{\ω}}_{b,2}^s+{\mathrm{\ω}}_{a,2}=\mathrm{\Ω}-\underset{s\≠0}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{in},1}^s---\left(5\right)$

Wherein, ω _{a, 2}represent macro station shared resource block number in access subframe, the Resource Block consuming in subframe in access for relaying s, the Resource Block consuming in backhaul subframe for relaying s, represent the throughput of the each Resource Block of each user of relaying s backhaul subframe, represent the throughput of the each Resource Block of each user of service station s access subframe, the access subframe consuming at reading cache data for relaying s;

3.3), solve ω according to Lagrangian method multiplier method _{a, 2}, with :

${\mathrm{\ω}}_{a,2}=(\frac{1}{{\mathrm{\Σ}}_{s\∈L}{\mathrm{\η}}_s}\·\frac{{\mathrm{\Σ}}_{s\∈L}\left|K_s\right|}{\left|K\right|})(\mathrm{\Ω}-\underset{s}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{in},1}^s+\underset{s}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{out},2}^s),(s\∈L)---\left(6\right)$

${\mathrm{\ω}}_{a,2}^s=(\frac{1}{{\mathrm{\η}}_s}\·\frac{\left|K_s\right|}{\left|K\right|})(\mathrm{\Ω}-\underset{s}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{in},2}^s+\underset{s}{\mathrm{\Σ}}{\mathrm{\ω}}_{\mathrm{out},2}^s),(s\∉L)---\left(7\right)$

${\mathrm{\ω}}_{b,2}^s={\mathrm{\ω}}_{a,2}^s{\mathrm{\η}}_s-{\mathrm{\ω}}_{\mathrm{out},2}^s,(s\∉L)---\left(8\right)$

Wherein, | K| represents number of users total in this sector, for base station, η _svalue be 1, for relaying ${\mathrm{\η}}_s=\frac{{\mathrm{\ρ}}_a^s}{{\mathrm{\ρ}}_b^s}{G}_s,$ L is:

$L=\left\{s\right|{\mathrm{\ω}}_{a,2}^s={\mathrm{\ω}}_{a,2}\}---\left(9\right)$

3.4), according to the performance difference of the historical average throughput of different base station and relaying, the Resource Block ω to base station assigns adaptively _{a, 2}, relaying s distribute Resource Block revise;

4) basis ω _{a, 2}, and with final resource block assignments result is:

$\left\{\begin{array}{c}{\mathrm{\ω}}_a={\mathrm{\ω}}_{a,2}\\ {\mathrm{\ω}}_a^s={\mathrm{\ω}}_{a,2}^s+{\mathrm{\ω}}_{\mathrm{out},2}^s+{\mathrm{\ω}}_{\mathrm{out},3}^s\\ {\mathrm{\ω}}_b^s={\mathrm{\ω}}_{b,2}^s+{\mathrm{\ω}}_{\mathrm{in},1}^s\end{array}\right.---\left(10\right)$

Wherein, ω _afor the Resource Block that finally base station consumes in access subframe; for the Resource Block that finally relaying s consumes in access subframe; for the Resource Block that finally relaying s consumes in backhaul subframe.

Described step 2) obtain the shared Resource Block of data volume of the unnecessary transmission of back haul link detailed process is:

2.1), when the preallocated Resource Block number of backhaul subframe is time, obtain the data volume Δ I of the unnecessary transmission of back haul link ₊, and obtain the transmitted data amount Δ I of access link loss simultaneously _-; Then obtain Δ I ₊with Δ I _-ratio ξ _s:

${\mathrm{\ξ}}_s=\frac{\mathrm{\Δ}{I}_{+}}{\mathrm{\Δ}{I}_{-}}---\left(11\right)$

2.2), by formula (6), (7) and (8) substitution formulas (11), then define ξ _sequal the left expression formula of formula (11) and right expression formula, as follows:

${\mathrm{\ξ}}_s=\frac{\left|K\right|{\mathrm{\ρ}}_b^s}{\frac{{\mathrm{\Σ}}_{s\∈L}\left|K_s\right|}{{\mathrm{\Σ}}_{s\∈L}{\mathrm{\λ}}_s}{G}_0{\mathrm{\ρ}}_a^0+\underset{s}{\mathrm{\Σ}}\left|K_s\right|{\mathrm{\ρ}}_b^s}---\left(12\right)$

According to with ξ _sproportional relation, definition expression formula be:

${\mathrm{\&omega;}}_{\mathrm{in},1}^s=\left\{\begin{array}{c}0,({\mathrm{\&xi;}}_s<1)\\ B_{\max }^s\frac{{\mathrm{\&xi;}}_s}{{\mathrm{\&rho;}}_b^s{\mathrm{\&xi;}}_{\max }},(1\&le;{\mathrm{\&xi;}}_s<{\mathrm{\&xi;}}_{\max })\\ \frac{B_{\max }^s}{{\mathrm{\&rho;}}_b^s},({\mathrm{\&xi;}}_s\&GreaterEqual;{\mathrm{\&xi;}}_{\max })\end{array}\right.---\left(13\right)$

In above formula, be illustrated in ξ _swhile obtaining maximum, the maximum of memory buffers data volume; ξ _maxrepresent ξ _smaximum occurrences.

Described step 3.4 specifically comprises:

A), be less than relaying n when grand user's average throughput, when user's average throughput of n ∈ S, relaying n reading cache data, and consumption of natural resource access-in resource, in conjunction with equation (6) to upgrade ω _{a, 2};

B), be greater than relaying m when grand user's average throughput, when the user's of m ∈ S average throughput, relaying m reading cache data, basis on consumption of natural resource again access-in resource, make be modified to

In described step 3.4, a step specifically comprises:

In the time that grand user's average throughput is less than ID and is user's average throughput of relaying of n, reading cache data and consumption individual access-in resource piece, value is following formula:

${\mathrm{\&omega;}}_{\mathrm{out},2}^n=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^n,\frac{\left|K_n\right|}{\left|K_0\right|}(\mathrm{\&Omega;}-\underset{l\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^l))---\left(13\right)$

In formula, for reading the access-in resource piece number that relaying n is data cached consumed, | K ₀| be the user's of base station service number;

In the time that grand user's average throughput is greater than ID and is user's average throughput of relaying of n, value is 0.

In described step 3.4, b step specifically comprises:

In the time meeting grand user's average throughput and be greater than ID and be user's average throughput of relaying of m, reading cache data and consumption individual access-in resource piece, value be following formula:

${\mathrm{\&omega;}}_{\mathrm{out},3}^m=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^m,{\mathrm{\&omega;}}_{a,2}-{\mathrm{\&omega;}}_{a,2}^m)---\left(14\right)$

In the time meeting grand user's average throughput and be less than ID and be user's average throughput of relaying of m, value be 0.

Compared with the prior art, beneficial effect of the present invention is:

The invention provides the resource allocation methods based on buffer control in a kind of LTE-A cellular network, in the time that data are transmitted, unnecessary transmission data are carried out to caching process, the time variation that takes full advantage of backhaul and access link channel is different, user and throughput of system performance are promoted, while is for the historical performance difference of grand user and trunk subscriber, the adaptive fairness of improving between them.Therefore, this patent can be by having improved fairness and the throughput performance in different service stations for the application of buffer queue.Meanwhile, the price of buffer memory device is relatively low, is convenient to be applied in real system.

Brief description of the drawings

Fig. 1 is the flow chart of TSA algorithm of the present invention;

Fig. 2 is that the present invention disturbs scene illustraton of model, wherein, (a) is back haul link transmission, is (b) access link transmission;

Fig. 3 is resource allocator model figure of the present invention;

Fig. 4 is parameter ξ in the present invention _sthe scatter chart of excursion;

Fig. 5 is the flow chart of definite mode in L territory in algorithm steps 2 in the present invention;

Fig. 6 is the topological diagram of simulating scenes of the present invention;

Fig. 7 is the simulation curve figure of the average throughput ratio of grand user and trunk subscriber in any one sector of the present invention;

Fig. 8 is the distribution curve of the measurement factor of user's average throughput in different service stations in any one sector of the present invention;

Fig. 9 is the distribution curve of the aggregated throughput of the different sectors of the present invention;

Figure 10 is the distribution curve of all user throughputs in community of the present invention.

Embodiment

Below in conjunction with accompanying drawing, the present invention is described in detail.

As shown in Figure 1, the invention provides the resource allocation methods based on buffer control in a kind of LTE-A cellular network, comprise the following steps:

1), obtain the user's of S set that total resources piece number in a radio frames in LTE-A system is Ω, service station s, service station s access set K _s, s service in service station in Cellular Networks user k throughput in access subframe as shown in Figure 2, wherein, (a) for model of place is disturbed in back haul link transmission, (b) for model of place is disturbed in access link transmission.

2), in the time carrying out transfer of data, during each semi-persistent scheduling, estimate the preallocated Resource Block number of backhaul subframe in the time that the volume of transmitted data of back haul link is greater than the transfer of data of access link, unnecessary transmission data are stored in the buffer memory of relay, with make after semi-persistent scheduling during be read; Described unnecessary transmission data are between this schedule periods, data that relaying receives in backhaul subframe and that cannot forward in access subframe; As shown in Figure 3, Fig. 3 is resource allocator model.

Obtain the shared Resource Block of data volume of the unnecessary transmission of back haul link detailed process is:

2.1), when the preallocated Resource Block number of backhaul subframe is time, obtain the data volume Δ I of the unnecessary transmission of back haul link ₊, and obtain the transmitted data amount Δ I of access link loss simultaneously _-; Then obtain Δ I ₊with Δ I _-ratio ξ _s:

${\mathrm{\&xi;}}_s=\frac{\mathrm{\&Delta;}{I}_{+}}{\mathrm{\&Delta;}{I}_{-}}---\left(11\right)$

2.2), by formula (6), (7) and (8) substitution formulas (11), then define ξ _sequal the left expression formula of formula (11) and right expression formula, as follows:

${\mathrm{\&xi;}}_s=\frac{\left|K\right|{\mathrm{\&rho;}}_b^s}{\frac{{\mathrm{\&Sigma;}}_{s\&Element;L}\left|K_s\right|}{{\mathrm{\&Sigma;}}_{s\&Element;L}{\mathrm{\&lambda;}}_s}{G}_0{\mathrm{\&rho;}}_a^0+\underset{s}{\mathrm{\&Sigma;}}\left|K_s\right|{\mathrm{\&rho;}}_b^s}---\left(12\right)$

According to with ξ _sproportional relation, definition expression formula be:

${\mathrm{\&omega;}}_{\mathrm{in},1}^s=\left\{\begin{array}{c}0,({\mathrm{\&xi;}}_s<1)\\ B_{\max }^s\frac{{\mathrm{\&xi;}}_s}{{\mathrm{\&rho;}}_b^s{\mathrm{\&xi;}}_{\max }},(1\&le;{\mathrm{\&xi;}}_s<{\mathrm{\&xi;}}_{\max })\\ \frac{B_{\max }^s}{{\mathrm{\&rho;}}_b^s},({\mathrm{\&xi;}}_s\&GreaterEqual;{\mathrm{\&xi;}}_{\max })\end{array}\right.---\left(13\right)$

In above formula, be illustrated in ξ _swhile obtaining maximum, the maximum of memory buffers data volume; ξ _maxrepresent ξ _smaximum occurrences.In the present invention, as shown in Figure 5, by the measurement for system-level emulation platform, obtain ξ under the real network environment of simulation _smaximum be ξ _max=2.5.

3), according to the preallocated Resource Block number of backhaul subframe distribute remaining Resource Block $\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s:$

3.1), basis obtain optimization aim, i.e. utility function U:

$U=\underset{k}{\mathrm{\&Sigma;}}\log R_{a,s}^k---\left(1\right)$

$R_{a,s}^k\&ap;\frac{{\mathrm{\&omega;}}_a^s}{\left|K_s\right|}{\mathrm{\&gamma;}}_{a,s}^k{G}_s---\left(2\right)$

Wherein, refer to that the user k that belongs to service station s in the time adopting PF scheduling is accessing the throughput in subframe; represent the resource block number that service station s distributes in access subframe; | K _s| represent the number of service station s institute service-user; represent service station s throughput on unit resource piece in access subframe; G _sthe additional gain that RR dispatching office brings is compared in expression PF scheduling;

3.2) the maximum utility function max (U), obtaining according to utility function U and formula (3), formula (4) and formula (5) are combined and are solved ω _{a, 2}, with

${\mathrm{\&omega;}}_{b,2}^s{\mathrm{\&rho;}}_b^s+{\mathrm{\&omega;}}_{\mathrm{out},2}^s{\mathrm{\&rho;}}_a^s-{\mathrm{\&omega;}}_{a,2}^s{\mathrm{\&rho;}}_a^s=0---\left(3\right)$

${\mathrm{\&omega;}}_{a,2}^s\&le;{\mathrm{\&omega;}}_{a,2}---\left(4\right)$

$\underset{s\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{b,2}^s+{\mathrm{\&omega;}}_{a,2}=\mathrm{\&Omega;}-\underset{s\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s---\left(5\right)$

Wherein, ω _{a, 2}represent macro station shared resource block number in access subframe, the Resource Block consuming in subframe in access for relaying s, the Resource Block consuming in backhaul subframe for relaying s, represent the throughput of the each Resource Block of each user of relaying s backhaul subframe, represent the throughput of the each Resource Block of each user of service station s access subframe, the access subframe consuming at reading cache data for relaying s;

3.3), solve ω according to Lagrangian method multiplier method _{a, 2}, with :

${\mathrm{\&omega;}}_{a,2}=(\frac{1}{{\mathrm{\&Sigma;}}_{s\&Element;L}{\mathrm{\&eta;}}_s}\&CenterDot;\frac{{\mathrm{\&Sigma;}}_{s\&Element;L}\left|K_s\right|}{\left|K\right|})(\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s+\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{out},2}^s),(s\&Element;L)---\left(6\right)$

${\mathrm{\&omega;}}_{a,2}^s=(\frac{1}{{\mathrm{\&eta;}}_s}\&CenterDot;\frac{\left|K_s\right|}{\left|K\right|})(\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},2}^s+\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{out},2}^s),(s\&NotElement;L)---\left(7\right)$

${\mathrm{\&omega;}}_{b,2}^s={\mathrm{\&omega;}}_{a,2}^s{\mathrm{\&eta;}}_s-{\mathrm{\&omega;}}_{\mathrm{out},2}^s,(s\&NotElement;L)---\left(8\right)$

Wherein, | K| represents number of users total in this sector, for base station, η _svalue be 1, and for relaying territory L is as shown in Figure 4:

$L=\left\{s\right|{\mathrm{\&omega;}}_{a,2}^s={\mathrm{\&omega;}}_{a,2}\}---\left(9\right)$

3.4), according to the performance difference of the historical average throughput of different base station and relaying, the Resource Block ω to base station assigns adaptively _{a, 2}, relaying s distribute Resource Block revise, specifically comprise:

A), in the time that grand user's average throughput is less than user's average throughput of relaying n (n ∈ S), relaying n reading cache data, and consumption of natural resource access-in resource, in conjunction with equation (6) to upgrade ω _{a, 2};

Concrete, in the time that grand user's average throughput is less than ID and is user's average throughput of relaying of n, reading cache data and consumption individual access-in resource piece, value is following formula:

${\mathrm{\&omega;}}_{\mathrm{out},2}^n=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^n,\frac{\left|K_n\right|}{\left|K_0\right|}(\mathrm{\&Omega;}-\underset{l\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^l))---\left(13\right)$

In formula, for reading the access-in resource piece number that relaying n is data cached consumed, | K ₀| be the user's of base station service number;

In the time that grand user's average throughput is greater than ID and is user's average throughput of relaying of n, value is 0.

B), in the time that grand user's average throughput is greater than user's the average throughput of relaying m (m ∈ S), relaying m reading cache data, basis on consumption of natural resource again access-in resource, make be modified to

In the time meeting grand user's average throughput and be greater than ID and be user's average throughput of relaying of m, reading cache data and consumption individual access-in resource piece, value be following formula:

${\mathrm{\&omega;}}_{\mathrm{out},3}^m=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^m,{\mathrm{\&omega;}}_{a,2}-{\mathrm{\&omega;}}_{a,2}^m)---\left(14\right)$

In the time meeting grand user's average throughput and be less than ID and be user's average throughput of relaying of m, value be 0.

4) basis ω _{a, 2}, and with final resource block assignments result is:

$\left\{\begin{array}{c}{\mathrm{\&omega;}}_a={\mathrm{\&omega;}}_{a,2}\\ {\mathrm{\&omega;}}_a^s={\mathrm{\&omega;}}_{a,2}^s+{\mathrm{\&omega;}}_{\mathrm{out},2}^s+{\mathrm{\&omega;}}_{\mathrm{out},3}^s\\ {\mathrm{\&omega;}}_b^s={\mathrm{\&omega;}}_{b,2}^s+{\mathrm{\&omega;}}_{\mathrm{in},1}^s\end{array}\right.---\left(10\right)$

Wherein, ω _afor the Resource Block that finally base station consumes in access subframe; for the Resource Block that finally relaying s consumes in access subframe; for the Resource Block that finally relaying s consumes in backhaul subframe.

Finally provide simulated environment and simulation result and analysis:

Cell topology as shown in Figure 6, Fig. 7---Figure 10 provides respectively the comparison of system fairness and throughput performance under different resource allocation algorithm, comparison algorithm comprises distribution method (the Fixed Backhaul-subframes Algorithm of fixed backhaul number of subframes, FBA), take into account the scheme (Generalized Proportional Fairness, GPF) of fairness and validity.

Simulation analysis is as follows: the CDF curve (as Fig. 7) of first considering grand user and trunk subscriber throughput performance ratio in any one sector, simulation result show the TSA algorithm mentioned in this patent from 1 more close to, this shows to compare with FBA algorithm with GPF, TSA algorithm can be good at improving the fairness between grand user and trunk subscriber, in motion, to compare FBA algorithms to improve the most obvious for TSA algorithm, is because FBA algorithm is not considered the fairness between grand user and trunk subscriber;

Secondly, consider to weigh by the following factor for the fairness between different service stations:

$I=\frac{{\left({\mathrm{\&Sigma;}}_s{r}_s\right)}^2}{\left|S\right|{\mathrm{\&Sigma;}}_s{r}_s^2}$

It is more fair that this factor I more approaches 1 explanation, Fig. 8 has provided the comparison of the global fair of three kinds of schemes, the global fair that simulation result shows TSA algorithm is compared GPF algorithm has decline slightly, but known in conjunction with Fig. 9, the throughput of TSA algorithm has obvious lifting, this is because the scheme of this motion can significantly improve throughput, and from algorithm second and the 3rd step can find out that this motion is to improve as much as possible for global fair.In Fig. 9, TSA algorithm has obvious lifting than the sector accumulative total throughput of GPF algorithm and FBA algorithm.Figure 10 has described the distribution curve of user throughput under different schemes, and in the algorithm proposing, user's throughput can improve.Particularly, compare FBA algorithm, the more excellent user's proportion of the algorithm throughput that proposes be less than 10%, this is because FBA algorithm is not considered fairness, makes to exist certain customers can obtain more resource, this makes most of user throughput lower.

Claims (5)

1. the resource allocation methods based on buffer control in LTE-A cellular network, is characterized in that, comprises the following steps:

1), obtain the user's of S set that total resources piece number in a radio frames in LTE-A system is Ω, service station s, service station s access set K _s, s service in service station in Cellular Networks user k throughput in access subframe

2), in the time carrying out transfer of data, during each semi-persistent scheduling, estimate the preallocated Resource Block number of backhaul subframe in the time that the volume of transmitted data of back haul link is greater than the transfer of data of access link, unnecessary transmission data are stored in the buffer memory of relay, with make after semi-persistent scheduling during be read; Described unnecessary transmission data are between this schedule periods, data that relaying receives in backhaul subframe and that cannot forward in access subframe;

3), according to the preallocated Resource Block number of backhaul subframe distribute remaining Resource Block $\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s:$

3.1), basis obtain optimization aim, i.e. utility function U:

$U=\underset{k}{\mathrm{\&Sigma;}}\log R_{a,s}^k---\left(1\right)$

$R_{a,s}^k\&ap;\frac{{\mathrm{\&omega;}}_a^s}{\left|K_s\right|}{\mathrm{\&gamma;}}_{a,s}^k{G}_s---\left(2\right)$

Wherein, refer to that the user k that belongs to service station s in the time adopting PF scheduling is accessing the throughput in subframe; represent the resource block number that service station s distributes in access subframe; | K _s| represent the number of service station s institute service-user; represent service station s throughput on unit resource piece in access subframe; G _sthe additional gain that RR dispatching office brings is compared in expression PF scheduling;

3.2) the maximum utility function max (U), obtaining according to utility function U and formula (3), formula (4) and formula (5) are combined and are solved ω _{a, 2}, with

${\mathrm{\&omega;}}_{b,2}^s{\mathrm{\&rho;}}_b^s+{\mathrm{\&omega;}}_{\mathrm{out},2}^s{\mathrm{\&rho;}}_a^s-{\mathrm{\&omega;}}_{a,2}^s{\mathrm{\&rho;}}_a^s=0---\left(3\right)$

${\mathrm{\&omega;}}_{a,2}^s\&le;{\mathrm{\&omega;}}_{a,2}---\left(4\right)$

$\underset{s\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{b,2}^s+{\mathrm{\&omega;}}_{a,2}=\mathrm{\&Omega;}-\underset{s\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s---\left(5\right)$

Wherein, ω _{a, 2}represent macro station shared resource block number in access subframe, the Resource Block consuming in subframe in access for relaying s, the Resource Block consuming in backhaul subframe for relaying s, represent the throughput of the each Resource Block of each user of relaying s backhaul subframe, represent the throughput of the each Resource Block of each user of service station s access subframe, the access subframe consuming at reading cache data for relaying s;

3.3), solve ω according to Lagrangian method multiplier method _{a, 2}, with :

${\mathrm{\&omega;}}_{a,2}=(\frac{1}{{\mathrm{\&Sigma;}}_{s\&Element;L}{\mathrm{\&eta;}}_s}\&CenterDot;\frac{{\mathrm{\&Sigma;}}_{s\&Element;L}\left|K_s\right|}{\left|K\right|})(\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^s+\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{out},2}^s),(s\&Element;L)---\left(6\right)$

${\mathrm{\&omega;}}_{a,2}^s=(\frac{1}{{\mathrm{\&eta;}}_s}\&CenterDot;\frac{\left|K_s\right|}{\left|K\right|})(\mathrm{\&Omega;}-\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},2}^s+\underset{s}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{out},2}^s),(s\&NotElement;L)---\left(7\right)$

${\mathrm{\&omega;}}_{b,2}^s={\mathrm{\&omega;}}_{a,2}^s{\mathrm{\&eta;}}_s-{\mathrm{\&omega;}}_{\mathrm{out},2}^s,(s\&NotElement;L)---\left(8\right)$

Wherein, | K| represents number of users total in this sector, for base station, η _svalue be 1, for relaying ${\mathrm{\&eta;}}_s=\frac{{\mathrm{\&rho;}}_a^s}{{\mathrm{\&rho;}}_b^s}{G}_s,$ L is:

$L=\left\{s\right|{\mathrm{\&omega;}}_{a,2}^s={\mathrm{\&omega;}}_{a,2}\}---\left(9\right)$

3.4), according to the performance difference of the historical average throughput of different base station and relaying, the Resource Block ω to base station assigns adaptively _{a, 2}, relaying s distribute Resource Block revise;

4) basis ω _{a, 2}, and with final resource block assignments result is:

$\left\{\begin{array}{c}{\mathrm{\&omega;}}_a={\mathrm{\&omega;}}_{a,2}\\ {\mathrm{\&omega;}}_a^s={\mathrm{\&omega;}}_{a,2}^s+{\mathrm{\&omega;}}_{\mathrm{out},2}^s+{\mathrm{\&omega;}}_{\mathrm{out},3}^s\\ {\mathrm{\&omega;}}_b^s={\mathrm{\&omega;}}_{b,2}^s+{\mathrm{\&omega;}}_{\mathrm{in},1}^s\end{array}\right.---\left(10\right)$

Wherein, ω _afor the Resource Block that finally base station consumes in access subframe; for the Resource Block that finally relaying s consumes in access subframe; for the Resource Block that finally relaying s consumes in backhaul subframe.

2. the resource allocation methods based on buffer control in LTE-A cellular network according to claim 1, is characterized in that described step 2) obtain the shared Resource Block of data volume of the unnecessary transmission of back haul link detailed process is:

2.1), when the preallocated Resource Block number of backhaul subframe is time, obtain the data volume Δ I of the unnecessary transmission of back haul link ₊, and obtain the transmitted data amount Δ I of access link loss simultaneously _-; Then obtain Δ I ₊with Δ I _-ratio ξ _s:

${\mathrm{\&xi;}}_s=\frac{\mathrm{\&Delta;}{I}_{+}}{\mathrm{\&Delta;}{I}_{-}}---\left(11\right)$

2.2), by formula (6), (7) and (8) substitution formulas (11), then define ξ _sequal the left expression formula of formula (11) and right expression formula, as follows:

${\mathrm{\&xi;}}_s=\frac{\left|K\right|{\mathrm{\&rho;}}_b^s}{\frac{{\mathrm{\&Sigma;}}_{s\&Element;L}\left|K_s\right|}{{\mathrm{\&Sigma;}}_{s\&Element;L}{\mathrm{\&lambda;}}_s}{G}_0{\mathrm{\&rho;}}_a^0+\underset{s}{\mathrm{\&Sigma;}}\left|K_s\right|{\mathrm{\&rho;}}_b^s}---\left(12\right)$

According to with ξ _sproportional relation, definition expression formula be:

${\mathrm{\&omega;}}_{\mathrm{in},1}^s=\left\{\begin{array}{c}0,({\mathrm{\&xi;}}_s<1)\\ B_{\max }^s\frac{{\mathrm{\&xi;}}_s}{{\mathrm{\&rho;}}_b^s{\mathrm{\&xi;}}_{\max }},(1\&le;{\mathrm{\&xi;}}_s<{\mathrm{\&xi;}}_{\max })\\ \frac{B_{\max }^s}{{\mathrm{\&rho;}}_b^s},({\mathrm{\&xi;}}_s\&GreaterEqual;{\mathrm{\&xi;}}_{\max })\end{array}\right.---\left(13\right)$

In above formula, be illustrated in ξ _swhile obtaining maximum, the maximum of memory buffers data volume; ξ _maxrepresent ξ _smaximum occurrences.

3. the resource allocation methods based on buffer control in LTE-A cellular network according to claim 1, is characterized in that, described step 3.4 specifically comprises:

A), be less than relaying n when grand user's average throughput, when user's average throughput of n ∈ S, relaying n reading cache data, and consumption of natural resource access-in resource, in conjunction with equation (6) to upgrade ω _{a, 2};

B), be greater than relaying m when grand user's average throughput, when the user's of m ∈ S average throughput, relaying m reading cache data, basis on consumption of natural resource again access-in resource, make be modified to

4. the resource allocation methods based on buffer control in LTE-A cellular network according to claim 3, is characterized in that described step 3.4) in a) step specifically comprise:

In the time that grand user's average throughput is less than ID and is user's average throughput of relaying of n, reading cache data and consumption individual access-in resource piece, value is following formula:

${\mathrm{\&omega;}}_{\mathrm{out},2}^n=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^n,\frac{\left|K_n\right|}{\left|K_0\right|}(\mathrm{\&Omega;}-\underset{l\&NotEqual;0}{\mathrm{\&Sigma;}}{\mathrm{\&omega;}}_{\mathrm{in},1}^l))---\left(13\right)$

In formula, for reading the access-in resource piece number that relaying n is data cached consumed, | K ₀| be the user's of base station service number;

In the time that grand user's average throughput is greater than ID and is user's average throughput of relaying of n, value is 0.

5. the resource allocation methods based on buffer control in LTE-A cellular network according to claim 3, is characterized in that described step 3.4) in b) step specifically comprise:

In the time meeting grand user's average throughput and be greater than ID and be user's average throughput of relaying of m, reading cache data and consumption individual access-in resource piece, value be following formula:

${\mathrm{\&omega;}}_{\mathrm{out},3}^m=\min ({\mathrm{\&omega;}}_{\mathrm{outall}}^m,{\mathrm{\&omega;}}_{a,2}-{\mathrm{\&omega;}}_{a,2}^m)---\left(14\right)$

In the time meeting grand user's average throughput and be less than ID and be user's average throughput of relaying of m, value be 0.