Research Article Cost-Efficient Allocation of Additional ...Research Article Cost-Efficient...

Research ArticleCost-Efficient Allocation of Additional Resources forthe Service Placement Problem in Next-Generation Internet

Ding Ma12 M Onderwater34 F Wetzels3 G J Hoekstra3 R D van der Mei34

S Bhulai4 and Lei Zhuang1

1School of Information and Engineering Zhengzhou University Zhengzhou 450001 China2College of Information Science and Engineering Henan University of Technology Zhengzhou 450001 China3CWI 1098 XG Amsterdam Netherlands4VU University Amsterdam 1081 HV Amsterdam Netherlands

Correspondence should be addressed to Ding Ma dmagszzueducn

Received 24 March 2015 Revised 24 June 2015 Accepted 5 July 2015

Academic Editor Fabio Tramontana

Copyright copy 2015 Ding Ma et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

One of the major challenges in next-generation Internet is to allocate services to nodes in the network This problem known asthe service placement problem can be solved by layered graph approach However due to the existence of resource bottleneck therequests are rejected from the beginning in the resource constrained network In this paper we propose two iterative algorithmsfor efficient allocation of additional resources in order to improve the ratio of accepted service placement requests To this end we(1) introduce a new concept of sensitivity for each service node to locate the bottleneck node (2) state the problem of allocatingadditional resources and (3) use sensitivity to propose a simple iterative algorithm and an utilization-based iterative algorithmfor efficient resource allocation The performance of these two algorithms is evaluated by simulation experiments in a variety ofparameter settings The results show that the proposed algorithms increase request acceptance ratio significantly by allocatingadditional resources into the bottleneck node and links The utilization-based iterative algorithm also decreases the long-term costby making efficient use of additional resources

1 Introduction

A huge achievement has been made in the Internet technol-ogy over the last four decades in supporting a wide arrayof distributed applications and in providing fundamentalend-to-end communication connectivity However with theincrease of the Internetrsquos scale and scope of use someinherent deficiencies of the Internet architecture are graduallyexposed Innovations are needed in the following aspects ofthe current Internet architecture namely mobility supportreliability and availability and quality of service guarantees[1] To realize these innovations many promising networkarchitectures have been designed for the next-generationInternet

An important approach adopted by somenext-generationInternet architectures is to move the data processing func-tions from the end-systems to the routers inside the core ofnetwork In the context of such networks the data processing

function is considered as service (in-network service [2]) andthe transcoding encryption flow control multicast and soforth are examples of the service [2] This approach providesgreater flexibility with its ability to compose the servicesalong the data path to satisfy different communicationrequirements from the end-system applications [3 4] Insuch network architectures a major challenge is to determinewhich nodes the services are placed on along the data pathand determine the shortest path between these nodes Thisproblem is known as the service placement problem which isproven to be NP-hard when considering constraints of thenetwork resources [5]

In the service placement problem to establish an end-to-end connection the sequence of services that represents theapplication functionality and the required network resourcesneed to be specified in advance in the end userrsquos requestThe data from one end system to another needs to berouted to pass through the nodes where certain services

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 517409 15 pageshttpdxdoiorg1011552015517409

2 Mathematical Problems in Engineering

specified in the sequence of services are available while thenetwork resources are sufficient on these nodes Apparentlythe service placement problem is quite different from thetraditional routing problem in which the data always followthe shortest path Furthermore in the service-centric net-work architectures (egNetwork Service Architecture (NSA))the service controller performs the mapping algorithm todetermine where to place which services [2 3]

The layered graph algorithm with low computationalcomplexity is an efficient solution to the service placementproblem when the resource is unlimited Yet it shows lim-itations in finding valid end-to-end connections due to theexistence of resource bottleneck in the resource constrainednetwork [5] In this paper we focus on how to locate the bot-tleneck based on the layered graph algorithm and to increasethe resource capacity of the bottleneck to improve the per-formance of the resource constrained network To this endwe first introduce a new concept sensitivity to represent theimpact that a single service node canhave on the performanceof the entire network in terms of average ratio of acceptedservice placement requests To compute the sensitivity of aservice node we remove this service node from the servicenetwork while maintaining the network working to measurethe decrease rate in average request acceptance ratio Accord-ing to the value of sensitivities we locate the bottleneck nodewhich is corresponding to the service node with the greatestsensitivity value We then propose two sensitivity-based iter-ative algorithms called SI-AAR and UI-AAR for solving theproblemof allocating additional resources into the bottlenecknode and its adjacent links to increase the average requestacceptance ratio Both of them first compute the sensitivitiesfor each node in the network SI-AAR then provides asimple way to iteratively increase both the CPU capacity andbandwidth capacity by the same increase rate UI-AAR caniteratively supplement additional resources selectively basedon resource utilization ratio The results from our experi-ments show that our sensitivity-based iterative algorithmsincrease request acceptance ratio significantly by allocatingadditional resources into the bottleneck node and links UI-AAR outperforms SI-AAR in terms of the long-term averagecost and the amount of allocated additional resources bymaking more efficient use of additional resources

The rest of the paper is organized as follows In Section 2we first overview the related work briefly Then we statethe service placement problem in Section 3 In Section 4 wediscuss the concept of the sensitivity and state the problemof allocating additional resources Section 5 describes themethod of computing the sensitivity and two sensitivity-based iterative algorithms SI-AAR and UI-AAR Section 6presents experimental results in a variety of parameter set-tings In Section 7 we conclude this paper and outline futureresearch directions

2 Related Work

Some network architectures have been designed to supportin-network services and provide network functionalcomposition The Service Integration Control and Optimiza-tion (SILO) project considers building blocks of fine-grain

functionality as services and combines services to accomplishhighly configurable complex communication tasks [4 6]The NSA implements an abstraction for communicationbetween end-systems by providing packet processing serviceinside the network [2 3] Recent routing architectures such asprogrammable routers [7 8] and virtualized router platforms[9] have provided technical support for implementation ofthe above network architectures

Related service placement problems have been discussedextensively in recent work In programmable networks (1)Choi et al [5] presented a layered graph algorithm to solvethe service placement problem In this algorithm a multiple-layer graph is constructed and Dijkstrarsquos shortest path algo-rithm is applied on the layered graph to find the shortestpath Then they proposed a capacity tracking approach toprevent the overuse of resources when considering capacityconstraints But they did not consider the impact that thebottleneck has on the performance of entire network interms of request blocking rate (2) Huang et al [10 11]proposed a distributed solution to the service placementproblem in resource unconstrained network By introducinga service matrix this distributed algorithm can determinethe optimal or near-optimal routes for connection requestsThe advantage of this distributed algorithm is that it is moresuitable for large-scale networks

In service overlay network (1) Raman and Katz [12]also used the layered graph algorithm and focused on loadbalancing without considering the capacity constraints Byintroducing a least-inverse-available-capacity (LIAC) metricto reassign the link cost in the layered graph it is ensuredthat the links with lower load are preferred over the links withhigher load (2) Liang and Nahrstedt [13] presented a greedyheuristic algorithm to solve the service composition problemto find low-cost composition solutions (3) Qian et al [14]also used heuristic algorithm to establish composite servicesdelivery path with lowest cost In addition they consideredthe changes of data size along the data path when choosingservices hop-by-hop

In cloud environment Tran et al [15] used recurrencemethod to solve the service composition problem They dis-cussed three exact algorithms for three different topologiespath star and tree

To the best of our knowledge this paper is the firstproposal that studies the problem of bottleneck locatingand utilizing in the service-centric network architectureHere we introduce several related research works in othernetwork architectures In the Internet Hu et al [16] presenteda novel light-weight single-end active probing tool (Path-neck) which allows end users to efficiently and accuratelylocate bottleneck According to the location information ofbottlenecks they used multihoming and overlay routing toavoid bottlenecks In virtualized network (1) Butt et al [17]presented a topology-aware measure method to detect thebottleneck nodes and links in the substrate network Andthen they proposed a set of algorithms for reoptimizingand remapping initially rejected virtual network requests (2)Based on the analysis on resource utilization Fajjari et al[18] found that the existence of bottlenecked substrate linksis the main reason of most of the request rejections Given

Mathematical Problems in Engineering 3

s

d(s 0)

= 1

B(s 0

) =2Gbs

dd(s 1) =

3

B(s 1) =1Gbs

d(0 1)=3

B(0 1)=2

Gbp

s

d(0 3) = 2B(0 3) = 05Gs

d(0 2) =5

B(0 2) =2Gbps

d(1 2)

= 4

B(1 2)

= 05Gbs

d(3 5) = 4

B(3 5) = 1Gbs

d(2 3)=1

B(2 3)=1

Gbs

d(2 4) = 3

B(2 4) = 05Gbs

d(5 d) = 3B(5 d) = 1Gbs

d(4 5)=2

B(4 5)=2

Gbs

d(4d)

=6

B(4d)

=05

Gbs

P(0) = 80

3 4

S(0) = S1 S2S S

P(1) = 65

S(1) = S1 S3

P(3) = 95

S(3) = S2 S3

P(2) = 85S(2) = S1

S3 S4

P(4) = 60

P(5) = 75

S(5) = S1S2 S4

0

1

2

3

4

5

S(4) = S2S3 S4

Figure 1 Service Network

that they proposed a reactive and iterative algorithm forremapping the rejected request through migration of nodesand its bottlenecked attached links However none of themconsider adding additional resources to the bottleneck nodeand links to improve the performance of entire network

3 Service Placement Problem

In the network architecture where data processing functionscan be implemented inside the network we term this physicalnetwork as service network the node in this service networkas service node and the link in this service network as servicelinkThen we state the service placement problem formally asfollows

31 Service Network We model the service network as aweighted undirected graph and denote it by 119866 = (119873 119871)where119873 is the set of service nodes and 119871 is the set of servicelinks The number of service nodes and the number servicelinks are denoted by |119873| and |119871| respectively Each servicenode 119899

119894isin 119873 is associated with the CPU capacity weight value

119875(119899119894) available service set 119878(119899

119894) = 119878120591| service 119878

120591is available

on service node 119899119894 and service 119878

120591rsquos processing time weight

value 119889119878120591

(119899119894) on service node 119899

119894for service node resources

Each service link 119897(119894 119895) isin 119871 between service node 119899119894and 119899119895is

associated with the link bandwidth weight value 119861(119897) whichdenotes the total amount of bandwidth capacity and its linkdelay weight value 119889

119897equiv 119889(119894 119895) for service link resources

An example of the service network topology is shown inFigure 1

32 Request An end userrsquos request for end-to-end cus-tomized composite services can be represented as a setincluding six elements and denoted by 119877 = (119899

119904 119899119889 119905119886 119905119889 119887

119878119877) Here 119899119904is the source node 119899

119889is the destination node 119905

119886

is the arrival time 119905119889is the duration time 119887 is the required

bandwidth capacity and 119878119877 is the required service set whichis composed of service number sn and an ordered list ofservices sl Each service sl

120485isin sl where 120485 represents the index

of services in the ordered list is associated with the CPUcapacity requirement weight value 119901(sl

120485) For example a

request 119877 = (119904 119889 90 1050 200 (Mbs) 4 (1198783 rarr 1198781 rarr

1198784 rarr 1198782)) while 119901(sl1) = 5 119901(sl2) = 10 119901(sl3) = 10119901(sl4) = 5

33 Service Path Given the end userrsquos request 119877 and theservice network 119866 an end-to-end service path is such a pathfrom the source service node 119899

119904to the destination service

node 119899119889 and all required services in the service list sl should

be processed in sequence along this path

331 Service-to-Node Mapping Each service from theordered service list sl needs to be processed by a service nodein the end-to-end service path by a mapping MS sl rarr 119873

from services to service nodes such that for all sl120485isin sl

MS (sl120485) isin 119873 (1)

where (1) if MS(sl120485) = MS(sl1204851015840) 120485 = 1204851015840 then sl120485is not

necessarily equal to sl1204851015840 which means multiple services can

be performed on a single service node (2) if MS(sl120485) = that indicates the service sl

120485is not performed on any service

node

332 Service Path Given the service-to-node mapping theend-to-end service path is denoted by

P1198902119890

= (11989911989411198721198941) (11989911989421198721198942) (119899

119894119898

119872119894119898

) (2)

where 1198991198941is the source node 119899

1198941= 119899119904 119899119894119898

is the destinationnode 119899

119894119898

= 119899119889 1198991198942 1198991198943 119899

119894119898minus1

are the service nodes119897(119894119896 119894119896+1) is the service link the hops of P1198902119890 denoted by

hops(P1198902119890) are equal to 119898 minus 1 and119872119894119896

is a service-to-nodemapping set on service node 119899

119894119896

defined as

119872119894119896

= (sl120485997888rarr119899119894119896

) | M119878(sl120485) = 119899119894119896

(3)

where if119872119894119896

= that indicates no service is performed onservice node 119899

119894119896


Table 1 Service processing time on service nodes

Node Service1198781 1198782 1198783 1198784

0 1198891198781(0) = 3 119889

1198782(0) = 4 119889

1198783(0) = 4 119889

1198784(0) = 2

1 1198891198781(1) = 5 mdash 119889

1198783(1) = 3 mdash

2 1198891198781(2) = 2 mdash 119889

1198783(2) = 6 119889

1198784(2) = 1

3 mdash 1198891198782(3) = 1 119889

1198783(3) = 2 mdash

4 mdash 1198891198782(4) = 6 119889

1198783(4) = 8 119889

1198784(4) = 6

5 1198891198781(5) = 4 119889

1198782(5) = 3 mdash 119889

1198784(5) = 4

To guarantee the validity of the service path severalrequirements have to be met

(1) All service nodes have sufficient CPU capacity forperforming the mapped services such that for forall119899

119894119896

isin P1198902119890

119875 (119899119894119896

) ge sumforallsl120485rarr119899119894119896

isin119872119894119896

119901 (sl120485) (4)

where sl120485rarr 119899119894119896

isin 119872119894119896

indicates that service sl120485is performed

on service node 119899119894119896

(2) All service links have sufficient link bandwidth such

that for forall119897(119894119896 119894119896+1) isin P1198902119890

119861 (119897 (119894119896 119894119896+1)) ge 119887 lowast 119886119905 (5)

where the service link 119897(119894119896 119894119896+1) appears 119886119905 times inP1198902119890

Given the end userrsquos request 119877 outlined above servicenetwork (depicted in Figure 1) and the service processingtime on service nodes (depicted in Table 1) the service pathsP11989021198901 = (119904 ) (0 sl1 rarr 0) (2 sl2 rarr 2) (4 sl3 rarr

4) (5 sl4 rarr 5) (119889 ) and P11989021198902 = (119904 ) (1 ) (2sl1 rarr 2 sl2 rarr 2) (0 sl3 rarr 0) (3 sl4 rarr 3) (5 )(119889 ) are both valid for request 119877 In P11989021198901 each service isperformed on one service node inP11989021198902 services 1198783 (sl1) and1198781 (sl2) are performed on service node 2 and no service isperformed on service nodes 1 and 5

333 Objective The delay of an end-to-end service pathD(P1198902119890) is defined as the summation of service processingtime on service nodes and communication delay on servicelinks along the service path

D (P1198902119890) =

119898minus1sum119896=1

119889 (119894119896 119894119896+1) +

119898minus1sum119896=2

sumsl120485rarr119899119894119896

isin119872119894119896

119889sl120485

(119899119894119896

) (6)

where 119899119904= 1198991198941 119899119889= 119899119894119898

and119898 is the number of service nodesin P1198902119890 The objective of service placement problem in thispaper is to find a least delay service path from all validP1198902119890

Due to the finite nature of network resources capacityconstraints are the crucial considerations for solving theservice placement problemWhen an end-to-end connectionrequest arrives the service network has to determinewhetherto accept the request or not according to its specification Ifthe request is accepted the service network operator needsto place services on service nodes allocate the CPU capacityon the corresponding service nodes and link bandwidth on

service links to establish the least delay end-to-end servicepath Once the end user leaves the service path is destroyedand the allocated resources are released

In this paper we make several assumptions as follows

(1) We assume that requirements of resources and ser-vices specified in an end userrsquos connection request donot change over the duration time of the connection

(2) An end-to-end service path which is establishedaccording to an end userrsquos connection request is fixedduring the lifetime of this connection

4 Problem of Allocating Additional Resourcesinto Service Network Based on Sensitivity

The layered graph with capacity tracking algorithm is an effi-cient approach to solve the service placement problem How-ever the layered graph algorithm cannot perform well whenthe capacity of network resources is limitedThemain reasonsinclude the NP-hard nature of the problem and the existingresource bottleneck Therefore a valid service path cannotalways be found evenwhen a valid path exists and the end-to-end connection requests are blocked from the beginning [5]To solve the existed resource bottleneck problem we proposetwo iterative algorithms in this paper for efficient allocation ofadditional resources in order to improve the performance interms of average request acceptance ratio denoted by AR Tothis endwe (1) introduce a new concept of sensitivity for eachservice node to locate the bottleneck node (2) state the prob-lem of allocating additional resources into the service net-work based on sensitivity and (3) use sensitivity to proposea simple iterative algorithm and an utilization-based iterativealgorithm for efficient allocation of additional resources

41 Definition of Sensitivity In the service network a servicenode can perform complicated data processing functionsIn addition each service node has different resources forexample CPU capacity processing power available servicesstorage and memory The sensitivity of a service noderepresents the impact that this service node has on the perfor-mance of entire network (eg the impact on average requestacceptance ratio) When the most sensitive service node(bottleneck node) is located the owner of the service network(eg Infrastructure Provider (InP)) has an opportunity toimprove the performance of the entire network by simplysupplementing additional resource capacities into one node

To calculate the sensitivity of a service node we removeor shut down one different service node 119899

119894each time from

the service network and maintain the network working tomeasure the average request acceptance ratio without 119899

119894

denoted by AR(119894) If the AR(119894) drops significantly the servicenode 119899

119894plays a vital role in the service network and holds high

sensitivityThe set of sensitivities for all service nodes in the service

network (119866) is a vector defined as Sen = (Sen0 Sen1 Sen119894 Sen

|119873|minus1) where the element Sen119894representing the

sensitivity of service node 119899119894is defined as

Sen119894= ARminusAR (119894) forall119899

119894isin 119873 (7)


In the resource constrained network the average requestacceptance ratio is a significant performance metric whichdetermines how many end usersrsquo requests are accepted

After the calculation of every service nodersquos sensitivity weidentify the service nodewith the greatest sensitivity and termit as the most sensitive node We term the adjacent service linkof the most sensitive node as sensitive link Then we focus onincreasing the CPU capacity of the most sensitive node orthe bandwidth capacity of the sensitive links to improve theaverage request acceptance ratio of the entire network

42 Problem Statement Theproblem of allocating additionalresources based on sensitivity is stated as follows We firstdefine the total amount of additional resources added into theservice network as

120575 (Res) = 120572 sdot 119875 (119899120594) + 120573 sdot 119861 (119897120594)119879

(8)

where the most sensitive node is represented by 119899120594and 119897120594is a

vector representing 120582 adjacent sensitive links of 119899120594defined

as 119897120594= (1198971205941 1198971205942 119897120594120582

) 119861(119897120594) is also a vector representing

the bandwidth of each sensitive link defined as 119861(119897120594) =

(119861(1198971205941) 119861(1198971205942) 119861(119897

120594120582

)) 120572 is an integer indicating that theCPU capacity of the most sensitive node will be increased by120572 times 120573 is a vector defined as 120573 = (1205731 1205732 120573120484 120573120582)where the element 120573

120484is an integer indicating that the

bandwidth capacity of the sensitive link 119897120594120484

will be increasedby 120573120484times

Once the most sensitive node is located the main objec-tive is to devise the algorithms for efficient allocation ofadditional resources to improve the performance of entirenetwork

Similar to the previous work in [15 19 20] the revenue(ie economic profit) of accepting an end userrsquos request (119877)at time 119905 can be defined as the resources that 119877 requiresmultiplied by their prices

R (119877 119905) = sumsl120485isinsl119901 (sl120485) sdot 120583119901+ 119887 sdot (sn+ 1) sdot 120583119887 (9)

where 120583119901

represents the CPU capacity usage price perrequired resource unit (eg $instancesdothour) and 120583

119887repre-

sents the bandwidth usage price per required resource unit(eg $Gbsdothour) Given thatR(119877 119905) represents the total pricethat the end user needs to pay to the InP

The cost of building a service path for an end userrsquosrequest at time 119905 can be defined as the total amount ofresource capacity that the InP allocates to the service pathP1198902119890 multiplied by their costs

C (119877 119905) = sumsl120485isinsl 119899=M

119878(sl120485)

119901 (sl120485) sdot 119888 (119899) + hops (P1198902119890) sdot 119887

sdot 119888 (119897)

(10)

where 119888(119899) represents the CPU capacity usage cost per usedresource unit (eg $instancesdothour) and 119888(119897) represents thelink bandwidth usage cost per used resource unit (eg$Gbsdothour) The cost of serving an end userrsquos request mainlydepends on the hops of the chosen service path

Accordingly the cost per time unit caused by addingadditional resources to the service network is defined as thetotal amount of additional resourcesmultiplied by theirs costs

C (120575 (Res)) = 120572 sdot 119875 (119899120594) sdot 119888 (119899120594) + 120573 sdot 119861 (119897120594)119879

sdot 119888 (119897120594) (11)

After a service path is established the resources allocatedto it will be occupied in the whole lifetime of the correspond-ing requestThus the total revenue and cost of serving an enduserrsquos request are determined by its lifetime 119905

119889 denoted by

R(119877 119905) sdot 119905119889and C(119877 119905) sdot 119905

119889 respectively

In general the additional resources are allocated per-manently into the service network Hence the total cost isdetermined by the running time T of the service networkdenoted by C(120575(Res)) sdot T

From InPrsquos point of view an effective and efficientalgorithm of allocating additional resources would minimizethe amount of additional resources andmaximize the averagerequest acceptance ratio and the average revenue of the InPin the long run The long-term average revenue of the InPdenoted by R(119866) is defined as

R (119866) = limTrarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889

T (12)

The average request acceptance ratio (AR) of the servicenetwork is defined as

AR = limTrarrinfin

sumT119905=0

10038161003816100381610038161003816119877accepted

10038161003816100381610038161003816

|119877| (13)

where |119877accepted| is the number of requests successfullyaccepted by the service network and |119877| is the total numberof requests

Consider using the sensitivity-based iterative algorithmsfor allocating additional resources into the service networkthe long-term average cost of the InP which should takethe cost caused by taking additional resources into accountdenoted by C(119866) is defined as

C (119866) = limTrarrinfin

sumT119905=0 C (119877 119905) lowast 119905119889 + C (120575 (Res)) sdot T

T (14)

We measure the efficiency of allocating additionalresources in terms of the ratio of long-term average revenueto cost (RC) ratio which is defined as

R (119866)

C (119866)= lim

Trarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889


(15)

Our objective is to minimize the amount of additionalresources (120575(Res)) allocated into service network and acceptthe largest possible number of end userrsquos requests We alsowant to increase the long-term average revenue of InP (R(119866))and decrease the long-term average cost of InP (C(119866))Whenthe average request acceptance ratios of proposed algorithmsare nearly the same we prefer the one that supplements theleast amount of resources (120575(Res)) and offers highest long-term RC ratio


(1) Compute and record AR R(119866) C(119866) 119880119873 119880119871using LG-CT(119866)

(2) for all 119899119894isin 119873 do

(3) 119866119894larr Remove one different 119899

119894each time from 119866

(4) Compute AR(119894) using LG-CT(119866119894)

(5) Sen119894larr AR minus AR(119894)

(6) end for(7) Locate the most sensitive node 119899

120594which is the maximum of Sen

Algorithm 1 The sensitivity computing method

The problem of allocating additional resources statedabove is a multiobjective optimization problem with con-flicting objectives which is a combinatorial optimizationproblem known as NP-hard [21] As a matter of fact we canonly achieve balance among all above objectives by designingeffective and efficient algorithms For example we cannotsupplement additional resources unlimitedly although theaverage request acceptance ratio (AR) and long-term averagerevenue (R(119866)) increase sharply at the beginning With theincrease of 120575(Res) (1) the corresponding cost (C(120575(Res)))which is proportional to 120575(Res) increases (2) the increasein AR and R(119866) will converge eventually (3) the increasein C(119866) will significantly exceed the increase in R(119866) fromsome time Consequently the RC will eventually reach anunrealistic value (eg R(119866) lt 05) which is unacceptablefor an InP To achieve better performance we devise twoiterative algorithms for allocating additional resources basedon the computation of sensitivity denoted by SI-AAR andUI-AAR respectivelyWewill discuss these two algorithms in thefollowing Section 5 in detail

43 Measurement of Resources To allocate additional resour-ces efficiently some resource metrics need to be defined andcalculated in advance

431 Resources on Service Node The available capacity of aservice node denoted by 119860

119873(119899119894) is defined as the available

CPU capacity of the service node 119899119894isin 119873

119860119873(119899119894) = 119875 (119899

119894) minus sumforallsl120485rarr119899119894isin119872119894

119901 (sl120485) (16)

The capacity utilization ratio of a service node denotedby 119880119873(119899119894) is defined as the total amount of CPU capacity

allocated to different services performed on the service node119899119894isin 119873 divided by the CPU capacity of service node 119875(119899

119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)

The average utilization ratio of all service nodes is definedas the summation of utilization ratio of all service nodedivided by the number of service nodes

119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)

432 Resources on Service Link Similarly the availablecapacity of a service link denoted by 119860

119871(119897) is defined as the

total amount of bandwidth available on the service link 119897 isin 119871

119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)

The capacity utilization ratio of a service link denotedby 119880119873(119897) is defined as the total amount of link bandwidth

allocated to different links inP1198902119890 divided by the bandwidthof the service link 119861(119897)

119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)

The average utilization ratio of all service links is definedas

119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)

5 Sensitivity-Based Iterative Algorithms forAllocating Additional Resources

51 The Sensitivity Computing Method Themain task of thisalgorithm (Algorithm 1) is to set up the layered graph andrun the capacity tracking algorithm known as layered graphwith capacity tracking (LG-CT) to record the performancemetrics of the service network for example average requestacceptance ratio (AR) long-term average revenue (R(119866))long-termaverage cost (C(119866)) average node utilization (119880

119873)

and average link utilization (119880119871) We then remove one

different service node 119899119894each time from the service network

(119866) and run LG-CT again to compute corresponding Sen119894isin

SenThe most sensitive node 119899120594is the greatest element in the

vector SenTaking advantage of locating the most sensitive node

then we design two iterative algorithms for allocation ofadditional resources called SI-AAR andUI-AAR both takingthe service network as input We only consider the supple-ment of additional resources into the most sensitive nodeand sensitive links in these two algorithms in order to avoidthe rise of the average cost of the InP and the drop of theRC ratio of the InP in the long run The iteration methodis used for additional resources allocation since the exactvalues of 120572 and 120573 are impossible to predict directly On theone hand inadequate additional resources can not providesignificant improvement on performance On the other hand


(1) Locate the most sensitive node using Algorithm 1(2) 119905cpu = 0 119905bw = 0 120572 = 0 1205731015840 = (1 1 1

120582) 120573 = 0 sdot 1205731015840

(3) repeat(4) 119905cpu = 119905cpu + Δ119905cpu 120572 = 119905cpu(5) 119905bw = 119905bw + Δ119905bw 120573 = 119905bw sdot 120573

1015840

(6) Add 120575(Res) into 119866(7) Call LG-CT(119866)(8) untile (ΔAR lt 120598) or (RC lt 120590)

(9) 120572 = 119905cpu minus Δ119905cpu 120573 = (119905bw minus Δ119905bw) sdot 1205731015840

(10) Reset 119866 and supplement recalculated 120575(Res) into 119866

Algorithm 2 SI-AAR

(1) Locate the most sensitive node using Algorithm 1(2) if Δ119880 ge 120596 then(3) Add only CPU capacity into 119866 using Algorithm 4(4) else if Δ119880 le minus120596 then(5) Add only Bandwidth capacity into 119866 using Algorithm 5(6) else(7) Add both CPU and bandwidth capacity into 119866 using Algorithm 6(8) end if

Algorithm 3 UI-AAR

excessive additional resources can result in an overuse ofresources Iteration method provides an effective way to findproper values of 120572 and 120573 by gradually increasing the resourcecapacity Details of these two algorithms are given below

52 Simple Iterative Algorithm for Allocating AdditionalResources (SI-AAR) SI-AAR (Algorithm 2) describes a sim-ple way to iteratively increase both CPU capacity and band-width capacity simultaneouslyThe algorithmbegins by locat-ing the most sensitive node 119899

120594in service work (119866) SI-AAR

then supplements 120572 sdot119875(119899120594)CPU capacity into 119899

120594and 120573 sdot119861(119897

120594)

bandwidth capacity into 119897120594 Next it reruns LG-CT and calcu-

latesRC ratio and the increment inAR denoted byΔAR Foreach iteration SI-AAR compares ΔAR with a small positivefraction 120598 and RC with a threshold parameter 120590 which isalso a positive fraction The algorithm terminates under twoconditions (1) if ΔAR lt 120598 AR converges (2) if RC lt 120590the algorithm will become unacceptable from the point ofeconomic profit Otherwise SI-AAR enters the next iteration

To adjust the increment in additional resources addedinto the service network in each iteration we introducetwo increment parameters denoted by Δ119905cpu and Δ119905bw Bothof them are integers greater than zero and indicate theincrement in the amount of CPU capacity and bandwidthcapacity respectively in each iteration In realistic networkenvironment the capacities of network resources (eg CPUstorage bandwidth) can be supplemented by the number ofinstances which imply the times of resource capacity Forexample we can increase CPU capacity by adding one ormultiple CPU instances increase storage capacity by addingone or multiple hard disks and increase bandwidth capacityby adding one or multiple communication linksTherefore it

is reasonable to increase resource capacity by one or multipletimes in each iteration 1205731015840 is a constant vector with allelements having the same value 1 and its size is the same asthat of 120573 120573 = 119905bw sdot 120573

1015840 indicates that the bandwidth capacity ofall sensitive links will be increased by the same times

53 Utilization-Based Iterative Algorithm for Allocating Addi-tional Resources (UI-AAR) The average utilization ratio ofservice nodes and service links reflects how much capacityof the network resources is used and which resource is insuf-ficient The value of the average utilization ratio is decidedby several factors for example the arrival rate of requestsrequired resources and available capacities We observe therecorded average node utilization ratio 119880

119873and average link

utilization raio 119880119871and then compute the difference between

them denoted by Δ119880 defined as Δ119880 = 119880119873minus 119880119871 We

find that much higher increase in average request acceptanceratio can be achieved efficiently by only adding the resourcecapacity with higher average utilization ratio while reducingthe cost caused by adding additional resources For exampleif Δ119880 = 30 the CPU capacity is the scarce resource undermuch higher stress and we can improve average requestacceptance ratio significantly by only supplementing CPUcapacity into the most sensitive node In this case averagerequest acceptance ratio increases slightly if we only addbandwidth capacity into sensitive links

UI-AAR (Algorithm 3) introduces a mechanism to sup-plement additional resources into themost sensitive node andsensitive links separately or simultaneously as far as the incre-ment and threshold parameters allow To allocate additionalresources into the service networkUI-AARhas three choices


(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0)(2) repeat(3) 119905cpu = 119905cpu + Δ119905cpu 120572 = 119905cpu(4) Add 120575(Res) into 119866(5) Call LG-CT(119866)(6) until (ΔAR lt 120598) or (RC lt 120590)

(7) 120572 = 119905cpu minus Δ119905cpu(8) Reset 119866 and supplement recalculated 120575(Res) into 119866

Algorithm 4 AACC

(1) flag = true 120572 = 0 1205731015840 = (1 1 1120582) 120573 = 119905Vbw = 0 sdot 1205731015840

(2) while flag do(3) for all 119897

120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484

) bandwidth capacity into 119866(7) Call LG-CT(119866)(8) until (ΔAR lt 120598

119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484

minus Δ119905bw(10) Record AR(119897

120594120484

) which represents the AR after adding 1205731015840120484sdot 119861(119897120594120484

) bandwidth capacity into 119866(11) Reset 119866(12) end for(13) if all elements in 119905Vbw do not change then(14) flag = false(15) else(16) Locate the 119897

120594119896

which has the greatest AR(119897120594119896

)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)

(18) Supplement recalculated 120575(Res) into 119866(19) end if(20) end while

Algorithm 5 AABC

(1) adding CPU capacity to the most sensitive node(2) adding bandwidth capacity to sensitive links(3) adding both CPU capacity and bandwidth capacityUI-AAR makes a decision according to the value of

Δ119880 and the parameter 120596 which is a positive fraction andrepresents the threshold of Δ119880 We compare Δ119880 with 120596 todetermine which resource capacity should be added We willdiscuss the three scenarios as follows

(1) If Δ119880 ge 120596 UI-AAR only supplements the CPUcapacity into the most sensitive node using Algorithm 4Allocating Additional CPU Capacity (AACC)

The process is the same as that in SI-AAR if Δ119905bw = 0(2) IfΔ119880 le minus120596 UI-AARonly supplements the bandwidth

capacity into sensitive links using Algorithm 5 AllocatingAdditional Bandwidth Capacity (AABC)

Generally the most sensitive node has more than onesensitive links We cannot increase the bandwidth capacitiesof all sensitive links simultaneously by the same times 119905bwsince it is not cost-efficient (ie adding bandwidth capacityto some sensitive link makes no contribution to the perfor-mance in terms of acceptance ratio) AABC only selects one

sensitive link per iteration and adds bandwidth capacity intoit Like Algorithm 4 for each sensitive link AABC graduallysupplements its bandwidth capacity until AR convergesThenAABC records AR(119897

120594120484

) which represents the average requestacceptance ratio achieved by adding 119905Vbw

120484

times of bandwidthcapacity to the corresponding sensitive link After dealingwith the last sensitive link AABC identifies the sensitivelink which has the greatest AR(119897

120594120484

) increases its bandwidthcapacity by 119905Vbw

120484

times and writes it back to the servicenetwork topology This process will be repeated until allAR(119897120594120484

) convergeThat is to say adding additional bandwidthcapacity to any sensitive link can not increase ΔAR over 120598

119887

119905Vbw is a vector with the same size as that of 120573 Each element119905Vbw120484

isin 119905Vbw of which value represents the times of bandwidthcapacity added into sensitive links 119897

120594120484

can have different value(3) If minus120596 lt Δ119880 lt 120596 UI-AAR supplements the

CPU capacity into the most sensitive node and the band-width capacity into sensitive links simultaneously usingAlgorithm 6 Allocating Additional CPU and BandwidthCapacity (AACBC)


(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)

(2) repeat(3) 119905cpu = 119905cpu + Δ119905cpu 120572 = 119905cpu(4) Add 120572 sdot 119875(119899

120594) CPU capacity into 119866

(5) Computing 120573 using Algorithm 5 where in the first line(1) 120573 do not be reset to (0 0 0

120582) (2) set 119905Vbw = 120573

(6) Supplement 120575(Res) into 119866(7) Call LG-CT(119866)(8) until (ΔAR lt 120598) or (RC lt 120590)


Algorithm 6 AACBC

Table 2 Differences in three settings

Setting I Setting II Setting IIIRequired CPU capacity (0 50]a (0 20]a (0 50]a

Required bandwidth (0 20]a (0 50]a (0 50]aaThe values are real numbers uniformly distributed on the correspondingrange

AACBC combines the method in Algorithm 4 with themethod in Algorithm 5 Like the method in Algorithm 4 forthe most sensitive node AACBC gradually increases its CPUcapacity by 120572 times For each value of 120572 AACBC uses themethod in Algorithm 5 to compute 120573 and then supplementsthe corresponding resources 120575(Res) into service network Ifthe condition is true AACBC adds the value of Δ119905cpu to 120572and repeats the above process until AR converges

6 Performance Evaluation

In this section we first describe the evaluation environ-ment and then present our main experimental results toevaluate efficiency of the sensitivity-based iterative algo-rithms in terms of average request acceptance ratio allocatedadditional resources long-term average revenue long-termaverage cost andRC ratio Our evaluation primarily focuseson the performance comparison between proposed twoalgorithms and the advantages of sensitivity-based bottlenecklocating

61 Evaluation Environment We have implemented a dis-crete event simulator to evaluate our proposed algorithmsThree different settings are chosen for the following experi-ments Differences among the three settings are introducedin Section 613 and shown in Table 2

611 Service Network Topology In this paper we do notassume any specialized network topologies We first usethe GT-ITM Tool [22] to randomly generate service net-work topologies Each service network is a 20-node 27-linktopology with a choice of 4 different services a scale thatcorresponds to a small-sized ISP Specifically the number ofservices available on one single service node is an integer uni-formly distributed between 1 and 4 The bandwidth capacity

of service links and the CPU capacity of service nodes arereal numbers uniformly distributed between 50 and 100 Thecommunication delay of each service link is an integer whichis proportional to its Euclidean distance and normalizedbetween 1 and 10 Each servicersquos processing time is an integerwhich depends on its type and the CPU power of its corre-sponding service node and normalized between 1 and 10

612 Request We assume that end userrsquos requests arrive ina Poisson process with an average rate of 4 requests per100 time units Each end userrsquos request has an exponentiallydistributed duration time with an average of 1000 time unitsWe run our simulation for about 50000 time units whichcorresponds to about 2000 requests for an instance of thesimulation The required bandwidth for service links andthe required CPU capacity for each service are configuredaccording to different settings shown in Table 2 The numberof services in end userrsquos requests is fixed to 4The ordered ser-vices list consists of 4 different services randomly distributedamong 1198781 1198782 1198783 and 1198784

613 Differences in Three Settings To observe the perfor-mance of our algorithms under different resource utiliza-tion three different experimental settings are configuredfor our simulation The differences among them only existin required CPU capacity for each service and requiredbandwidth for service links Setting I represents the sce-nario that the service network is under more pressure fromrequested CPU capacity resources than it has from requestedbandwidth resourcesOn the contrary resource stress that theservice network encounters mainly stems from the requestedbandwidth in Setting II In Setting III with the increasingrequirements for resources the available capacities on bothservice nodes and service links become insufficient to acceptmore requests Details are shown in Table 2

62 Compared Algorithms and Parameter Settings In oursimulator we implement our sensitivity-based iteration algo-rithms for allocating additional resources alongside therelated strategies (1) SI-AAR and (2) UI-AAR We use twospecific cases of SI-AAR denoted by SI-AAR-Least and SI-AAR-RUB (referenced upper bound RUB) to provide alowest bound and an referenced upper bound respectively


on the amount of additional resources In SI-AAR-Least120572 is set to 1 and 120573 is set to (1 1 1) which representsthe situation of adding the least amount of resources to theservice network using SI-AAR In SI-AAR-RUB 120572 is set to 9and 120573 is set to (9 9 9) that is the amount of resources ofthe most sensitive node and sensitive links is increased to 10times which is not practical but provides a referenced upperbound on the performance of realistic algorithms Based onextensive simulations we adjust the parameters of proposedtwo algorithms We set 120583

119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and

Δ119905bw to 1 120598 to 1 120598119887to 05 120590 to 06 and 120596 to 10 to achieve

greatest increase in average request acceptance ratio and todecrease the amount of additional resources and the numberof iterations in the meantime

63 Performance Metrics In our experiments we use severalperformance metrics introduced in previous sections forthe purpose of evaluation for example average requestacceptance ratio (AR) the amount of allocated additionalresources (120575(Res)) long-term average revenue (R(119866)) long-term average cost (C(119866)) and long-term RC ratio Wemeasure the average request acceptance ratio (AR) andaverage request acceptance ratio without node 119894 (AR(119894)) tocompute the sensitivities and locate the most sensitive nodein the original service network (ie the service network with-out any additional resources being added) To evaluate theeffectiveness and the efficiency of our proposed algorithmswe alsomeasure the average request acceptance ratio averagerevenue average cost and RC ratio for end userrsquos requestsover time in the service network where the correspondingadditional resources have been added to In themeantime werecord the values of two essential parameters 120572 and 120573 to com-pute the amount of additional resources (120575(Res)) eventuallyallocated into the service network for evaluated algorithms

64 Evaluation Results We present the evaluation resultsto show the effectiveness and quantify the efficiency ofthe proposed algorithms under three different scenarios(depicted in Figures 2 3 and 4)

We first plot AR and AR(119894) against the index of servicenodes to show the computation of sensitivities (depicted inFigures 2(a) 3(a) and 4(a))Weuse a bar chart to compare theamount of allocated additional resources (120575(Res)) (depictedin Figures 2(b) 3(b) and 4(b)) Next we plot average requestacceptance ratio average revenue average cost andRC ratioagainst time to show how each of these algorithms actuallyperforms in the long run (depicted in Figures 2(c)ndash2(f) 3(c)ndash3(f) and 4(c)ndash4(f)) We summarize our key observations forthe three settings (Table 2) as follows

641 Sensitivity Computing We locate the most sensi-tive node through computation of sensitivities and choosethe strategy of allocating additional resources for UI-AARaccording to the results of resource utilization computing(shown in Table 3)

(1) Setting I as shown in Figure 2(a) and Table 3 node 7is the most sensitive node 119880

119873= 349 119880

119871= 235

Table 3 Resource utilization in three settings

Setting I Setting II Setting IIIAverage node utilization ratio 349 104 337Average link utilization ratio 235 325 308

Δ119880 = 114 gt 120596 and the CPU capacity of node 7 ischosen to be supplemented in UI-AAR

(2) Setting II from Figure 3(a) and Table 3 node 10 is themost sensitive node 119880

119873= 104 119880

119871= 325 Δ119880 =

minus221 lt minus120596 and the bandwidth of node 10rsquos sensi-tive links is chosen to be supplemented in UI-AAR

(3) Setting III the configurations of required CPUcapacity and bandwidth (Table 2) show that bothCPU capacity and bandwidth capacity are under hugepressure That is the reason why the average requestacceptance ratio is only 64 and 119880

119873is close to 119880

119871

(shown in Figure 4(a) and Table 3) Given that node7 is the most sensitive node minus120596 lt Δ119880 = 3 lt 120596 andthe CPU capacity of node 7 and the bandwidth ofnode 7rsquos sensitive links are chosen to be supplementedtogether in UI-AAR

642 Allocating Additional Resources into the Most SensitiveNodes and Sensitive Links Leads to Higher Average RequestAcceptance Ratio In Figures 2(a) 3(a) and 4(a) we also plotaverage request acceptance ratio against time to show howthe original LG-CT algorithm denoted byOriginal performswithout any additional resources being added to the servicenetwork It is important to note that we only present theeffectiveness of the proposed algorithms hereWewill discussthe efficiency of the proposed algorithms to see if and howthey can make efficient use of allocated additional resourcesin Section 643

From Figures 2(a) 3(a) and 4(a) it is evident thatthe proposed algorithms UI-AAR and SI-AAR and thetwo specific cases SI-AAR-Least and SI-AAR-RUB lead tohigher average request acceptance ratio than the Originalunder three different scenarios These three graphs showthat sensitivity-based bottleneck locating plays a vital rolein allocation of additional resources It is an effective wayto increase the average request acceptance ratio only bysupplementing additional resources into bottleneck node (themost sensitive node) and links (sensitive links)

In the three outlined settings after time unit of 20000the average request acceptance ratio of SI-AAR and UI-AARis nearly the same In addition the approximate incrementsin average request acceptance ratio (eg 134 and 136 inSetting I 156 and 154 in Setting II and 16 and 158in Setting III at the time unit of 50000) show that both SI-AAR and UI-AAR perform well In Setting I SI-AAR-RUBgains much higher acceptance ratio than SI-AAR and UI-AAR since it supplements ten times amount of additionalresources However in Setting II and Setting III the averagerequest acceptance ratio of SI-AAR-RUB is only 04 and22 higher than that of SI-AAR and 06 and 26 higher


0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio

Average request acceptance ratioAverage request acceptance ratiowithout node i

(a) Sensitivity computing

4000

3500

3000

2500

2000

1500

1000

500

0SI-AAR-least UI-AAR SI-AAR SI-AAR-RUB

120575(R

es)

(b) Additional resources allocated into the service network

2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095

Time (1000 time units)

Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)

(c) Comparisons of average request acceptance ratios over time

2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

(d) Comparisons of long-term average revenues

2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

(e) Comparisons of long-term average costs

2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

(f) Comparisons of long-term RC ratios

Figure 2 Comparisons between UI-AAR and SI-AAR in Setting I


0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)

(c) Comparisons of average request acceptance ratios overtime

2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


Figure 3 Comparisons between UI-AAR and SI-AAR in Setting II


0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


Figure 4 Comparisons between UI-AAR and SI-AAR in Setting III


than that of UI-AAR respectively On the contrary SI-AAR-Least produces the least increase in acceptance ratio amongthe four algorithms The reasons with respect to above twosituations will be analyzed in Section 643

643 UI-AAR Can Allocate the Additional Resources MoreEfficiently It is worth noting that the evaluated algorithmsthat lead to the higher acceptance ratio also produce higherlong-term average revenue (depicted in Figures 2(c) 2(d)3(c) 3(d) 4(c) and 4(d)) and higher long-term averagecost (excluding the cost produced by additional resources)(1) According to the definition of R(119866) more requests areaccepted while more revenue can be obtained and (2) allof them use the same LG-CT algorithm for solving theservice placement problem (ie same approaches of serviceplacement and resource allocation) However the amountof additional resources producing additional cost allocatedby the compared algorithms is different which implies that120575(Res) and RC ratio are two vital metrics to quantify theefficiency of the evaluated algorithms

FromFigures 2(b) 3(b) and 4(b) as the lowest bound andthe referenced upper bound SI-AAR-Least and SI-AAR-RUBuse the least and most amount of additional resources in SI-AAR respectively The additional resources used by UI-AARare close to that used by SI-AAR-Least in Setting I and SettingII and are twice greater in Setting III SI-AAR allocates moreadditional resources compared with UI-AAR for example 43 and 15 times greater in Setting I Setting II and SettingIII respectively The reason is that UI-AAR only adds CPUcapacity or bandwidth capacity in Setting I and Setting II andboth SI-AAR and UI-AAR add CPU capacity and bandwidthcapacity together in Setting III

Given that the evaluated algorithms that lead to thehigher acceptance ratio also produce higher additional cost(C(120575(Res)) and thus produce higher long-term average cost(C(119866)) (depicted in Figures 2(d) 3(d) and 4(d))

Based on the above illustration and analysis UI-AARper-forms considerably well in terms of acceptance ratio averagerevenue and cost and uses less additional resources Conse-quently UI-AAR produces the highest RC ratio among UI-AAR SI-AAR and SI-AAR-RUB (depicted in Figures 2(f)3(f) and 4(f)) The reasons why UI-AAR can make moreefficient use of additional resources are outlined as follow(1) UI-AAR selectively supplements the additional resourcesbased on resource utilization ratio For example in SettingI UI-AAR select only CPU capacity to supplement in eachiteration (see Figure 2(f) where120572 = 5 and120573 = (0 0 0 0)) thetotal amount additional resources is 4 times lower than thatof SI-AAR and even less than that of SI-AAR-Least (2) UI-AAR can find better combination of 120572 and 120573 For each valueof 120572 UI-AAR computes the optimal value of each element in120573 For example in Setting III for each value of 120572 UI-AARiteratively supplements only one sensitive linkrsquos bandwidthcapacity in each iteration and thus the bandwidth capacitieseventually added to sensitive links are different despite thevalue of 120572 being the same as that in SI-AAR (see Figure 3(f)where 120572 = 4 and 120573 = (2 4 3 3)) The average requestacceptance ratio of UI-AAR and SI-AAR is nearly the samebut the total amount of additional resources of UI-AAR is

15 times lower than that of SI-AAR Adding more additionalresources along with no improvement on acceptance ratioimplies that part of additional resources added by SI-AARdoes not make any contribution to the performance in termsof acceptance ratio Likewise SI-AAR-RUB is a suitable caseto illustrate the overuse of resources Although theRC ratioand 120575(Res) of SI-AAR-Least are close to those of UI-AAR(depicted in Figures 2(b) 2(f) 3(b) 3(f) 4(b) and 4(f)) UI-AAR significantly outperform SI-AAR-Least in terms of theaverage request acceptance ratio and the long-term averagerevenue The reasons are given in the following (1) Theadditional resources added by SI-AAR-Least are insufficientto accept more requests (2) Like SI-AAR SI-AAR-Least doesnot allocate the additional resource efficiently For examplecompared with UI-AAR SI-AAR-Least accepts less requestsusing more additional resources in Setting I

7 Conclusion and Future Work

The placement of services is an important problem inany network architecture that supports the implementationof data processing functions inside the network In thispaper we modeled and stated this problem To solve theresource bottleneck problem existing in LG-CT algorithmwe introduced a novel concept sensitivity We used thesensitivity to locate the most sensitive node and then allo-cated additional resources into the most sensitive nodeand sensitive links to improve the performance of entirenetwork After discussing and formulating the problem ofallocating additional resources we proposed two sensitivity-based iterative algorithms for efficient resource allocationThe first one SI-AAR provides a simple way to increase boththe CPU capacity and the bandwidth capacity by the sametimes The second one UI-AAR can supplement additionalresources selectively based on resource utilization ratioOur results from the experiments showed the effectivenessand efficiency of our proposed two algorithms under threespecific settings The increase in average request acceptanceratio was significant if we supplemented additional resourcesto themost sensitive node and sensitive linksThe utilization-based iterative algorithm (UI-AAR) can make more efficientuse of additional resources and thus outperforms SI-AAR interms of the amount of allocated additional resources long-term average cost and long-term RC ratio

In future work we will consider the number of thebottleneck nodes which we choose to supplement resourcesfocus on medium-size or large-scale network topology andinvestigate where the sensitivity can be further applied in theservice placement problem to improve the performance interms of optimizing capacity allocation and balancing load

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This research was partially supported by the National NaturalScience Foundation of China under Grant no 61379079 and


the National Basic Research Program of China (973) underGrant no 2012CB315900

References

[1] A Feldmann ldquoInternet clean-slate design what and whyrdquoACM SIGCOMMComputer Communication Review vol 37 no3 pp 59ndash64 2007

[2] T Wolf ldquoIn-network services for customization in next-generation networksrdquo IEEE Network vol 24 no 4 pp 6ndash122010

[3] S Ganapathy and T Wolf ldquoDesign of a network service archi-tecturerdquo inProceedings of the 16th IEEE International Conferenceon Computer Communications and Networks (ICCCN rsquo07) pp754ndash759 August 2007

[4] R Dutta G N Rouskas I Baldine A Bragg and D StevensonldquoThe silo architecture for services integration control andoptimization for the future internetrdquo in Proceedings of the IEEEInternational Conference on Communications (ICC rsquo07) pp1899ndash1904 June 2007

[5] S Choi J Turner and T Wolf ldquoConfiguring sessions inprogrammable networksrdquoComputer Networks vol 41 no 2 pp269ndash284 2003

[6] M Vellala A Wang G N Rouskas R Dutta I Baldineand D Stevenson ldquoA composition algorithm for the silocross-layer optimization service architecturerdquo in Proceedingsof the 1st International Conference on Advanced Network andTelecommunications Systems (ANTS rsquo07) 2007

[7] H H L Ruf K Farkas and B Plattner ldquoNetwork serviceson service extensible routersrdquo in Active and ProgrammableNetworks IFIP TC6 7th International Working ConferenceIWAN 2005 Sophia Antipolis France November 21-23 2005Revised Papers Lecture Notes in Computer Science pp 53ndash64Springer Berlin Germany 2005

[8] T Wolf ldquoChallenges and applications for network-processor-based programmable routersrdquo in Proceedings of the IEEE SarnoffSymposium pp 1ndash4 March 2006

[9] T Anderson L Peterson S Shenker and J Turner ldquoOvercom-ing the internet impasse through virtualizationrdquo Computer vol38 no 4 pp 34ndash41 2005

[10] X Huang S Ganapathy and T Wolf ldquoA distributed routingalgorithm for networks with data-path servicesrdquo in Proceedingsof 17th IEEE International Conference on Computer Communi-cations and Networks (ICCCN rsquo08) pp 1ndash7 2008

[11] H Xin S Ganapathy and T Wolf ldquoA scalable distributedrouting protocol for networks with data-path servicesrdquo in Pro-ceedings of the 16th IEEE International Conference on NetworkProtocols (ICNP rsquo08) pp 318ndash327 October 2008

[12] B Raman and R H Katz ldquoLoad balancing and stability issuesin algorithms for service compositionrdquo in Proceedings of the22nd Annual Joint Conference on the IEEE Computer andCommunications Societies (INFOCOM rsquo03) vol 2 pp 1477ndash1487 IEEE San Francisco Calif USA April 2003

[13] J Liang and K Nahrstedt ldquoService composition for advancedmultimedia applicationsrdquo in Multimedia Computing and Net-working vol 5680 of Proceedings of SPIE pp 228ndash240 Inter-national Society for Optics and Photonics San Jose Calif USAJanuary 2005

[14] Z Qian S Zhang K Yim and S Lu ldquoService orientedmultimedia delivery system in pervasive environmentsrdquo Journalof Universal Computer Science vol 17 no 6 pp 961ndash980 2011

[15] K-T Tran N Agoulmine and Y Iraqi ldquoCost-effective complexservice mapping in cloud infrastructuresrdquo in Proceedings of theIEEE Network Operations andManagement Symposium (NOMSrsquo12) pp 1ndash8 IEEE April 2012

[16] N Hu L E Li Z M Mao P Steenkiste and J Wang ldquoLocatinginternet bottlenecks algorithms measurements and implica-tionsrdquoACMSIGCOMMComputer Communication Review vol34 no 4 pp 41ndash54 2004

[17] N F Butt M Chowdhury and R Boutaba ldquoTopology-awareness and reoptimization mechanism for virtual networkembeddingrdquo inNETWORKING 2010 vol 6091 of Lecture Notesin Computer Science pp 27ndash39 Springer Berlin Germany 2010

[18] I Fajjari N Aitsaadi G Pujolle and H Zimmermann ldquoVnralgorithm a greedy approach for virtual networks reconfigu-rationsrdquo in Proceedings of the IEEE Global TelecommunicationsConference (GLOBECOM rsquo11) pp 1ndash6 IEEE 2011

[19] M Yu Y Yi J Rexford and M Chiang ldquoRethinking vir-tual network embedding substrate support for path splittingand migrationrdquo ACM SIGCOMM Computer CommunicationReview vol 38 no 2 pp 17ndash29 2008

[20] M Shen K Xu K Yang and H-H Chen ldquoTowards efficientvirtual network embedding across multiple network domainsrdquoin Proceedings of the 22nd IEEE International Symposium ofQuality of Service (IWQoS rsquo14) pp 61ndash70 IEEE May 2014

[21] M Ehrgott and X Gandibleux ldquoA survey and annotated bib-liography of multiobjective combinatorial optimizationrdquo OR-Spektrum vol 22 no 4 pp 425ndash460 2000

[22] E W Zegura K L Calvert and S Bhattacharjee ldquoHow tomodel an internetworkrdquo in Proceedings of the IEEE Conferenceon Computer Communications (INFOCOM rsquo06) 15th AnnualJoint Conference of the IEEE Computer Societies Networking theNext Generation pp 594ndash602 March 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


specified in the sequence of services are available while thenetwork resources are sufficient on these nodes Apparentlythe service placement problem is quite different from thetraditional routing problem in which the data always followthe shortest path Furthermore in the service-centric net-work architectures (egNetwork Service Architecture (NSA))the service controller performs the mapping algorithm todetermine where to place which services [2 3]

The layered graph algorithm with low computationalcomplexity is an efficient solution to the service placementproblem when the resource is unlimited Yet it shows lim-itations in finding valid end-to-end connections due to theexistence of resource bottleneck in the resource constrainednetwork [5] In this paper we focus on how to locate the bot-tleneck based on the layered graph algorithm and to increasethe resource capacity of the bottleneck to improve the per-formance of the resource constrained network To this endwe first introduce a new concept sensitivity to represent theimpact that a single service node canhave on the performanceof the entire network in terms of average ratio of acceptedservice placement requests To compute the sensitivity of aservice node we remove this service node from the servicenetwork while maintaining the network working to measurethe decrease rate in average request acceptance ratio Accord-ing to the value of sensitivities we locate the bottleneck nodewhich is corresponding to the service node with the greatestsensitivity value We then propose two sensitivity-based iter-ative algorithms called SI-AAR and UI-AAR for solving theproblemof allocating additional resources into the bottlenecknode and its adjacent links to increase the average requestacceptance ratio Both of them first compute the sensitivitiesfor each node in the network SI-AAR then provides asimple way to iteratively increase both the CPU capacity andbandwidth capacity by the same increase rate UI-AAR caniteratively supplement additional resources selectively basedon resource utilization ratio The results from our experi-ments show that our sensitivity-based iterative algorithmsincrease request acceptance ratio significantly by allocatingadditional resources into the bottleneck node and links UI-AAR outperforms SI-AAR in terms of the long-term averagecost and the amount of allocated additional resources bymaking more efficient use of additional resources

The rest of the paper is organized as follows In Section 2we first overview the related work briefly Then we statethe service placement problem in Section 3 In Section 4 wediscuss the concept of the sensitivity and state the problemof allocating additional resources Section 5 describes themethod of computing the sensitivity and two sensitivity-based iterative algorithms SI-AAR and UI-AAR Section 6presents experimental results in a variety of parameter set-tings In Section 7 we conclude this paper and outline futureresearch directions

2 Related Work

Some network architectures have been designed to supportin-network services and provide network functionalcomposition The Service Integration Control and Optimiza-tion (SILO) project considers building blocks of fine-grain

functionality as services and combines services to accomplishhighly configurable complex communication tasks [4 6]The NSA implements an abstraction for communicationbetween end-systems by providing packet processing serviceinside the network [2 3] Recent routing architectures such asprogrammable routers [7 8] and virtualized router platforms[9] have provided technical support for implementation ofthe above network architectures

Related service placement problems have been discussedextensively in recent work In programmable networks (1)Choi et al [5] presented a layered graph algorithm to solvethe service placement problem In this algorithm a multiple-layer graph is constructed and Dijkstrarsquos shortest path algo-rithm is applied on the layered graph to find the shortestpath Then they proposed a capacity tracking approach toprevent the overuse of resources when considering capacityconstraints But they did not consider the impact that thebottleneck has on the performance of entire network interms of request blocking rate (2) Huang et al [10 11]proposed a distributed solution to the service placementproblem in resource unconstrained network By introducinga service matrix this distributed algorithm can determinethe optimal or near-optimal routes for connection requestsThe advantage of this distributed algorithm is that it is moresuitable for large-scale networks

In service overlay network (1) Raman and Katz [12]also used the layered graph algorithm and focused on loadbalancing without considering the capacity constraints Byintroducing a least-inverse-available-capacity (LIAC) metricto reassign the link cost in the layered graph it is ensuredthat the links with lower load are preferred over the links withhigher load (2) Liang and Nahrstedt [13] presented a greedyheuristic algorithm to solve the service composition problemto find low-cost composition solutions (3) Qian et al [14]also used heuristic algorithm to establish composite servicesdelivery path with lowest cost In addition they consideredthe changes of data size along the data path when choosingservices hop-by-hop

In cloud environment Tran et al [15] used recurrencemethod to solve the service composition problem They dis-cussed three exact algorithms for three different topologiespath star and tree

To the best of our knowledge this paper is the firstproposal that studies the problem of bottleneck locatingand utilizing in the service-centric network architectureHere we introduce several related research works in othernetwork architectures In the Internet Hu et al [16] presenteda novel light-weight single-end active probing tool (Path-neck) which allows end users to efficiently and accuratelylocate bottleneck According to the location information ofbottlenecks they used multihoming and overlay routing toavoid bottlenecks In virtualized network (1) Butt et al [17]presented a topology-aware measure method to detect thebottleneck nodes and links in the substrate network Andthen they proposed a set of algorithms for reoptimizingand remapping initially rejected virtual network requests (2)Based on the analysis on resource utilization Fajjari et al[18] found that the existence of bottlenecked substrate linksis the main reason of most of the request rejections Given


s

d(s 0)

= 1

B(s 0

) =2Gbs

dd(s 1) =

3

B(s 1) =1Gbs

d(0 1)=3

B(0 1)=2

Gbp

s

d(0 3) = 2B(0 3) = 05Gs

d(0 2) =5

B(0 2) =2Gbps

d(1 2)

= 4

B(1 2)

= 05Gbs

d(3 5) = 4

B(3 5) = 1Gbs

d(2 3)=1

B(2 3)=1

Gbs

d(2 4) = 3

B(2 4) = 05Gbs

d(5 d) = 3B(5 d) = 1Gbs

d(4 5)=2

B(4 5)=2

Gbs

d(4d)

=6

B(4d)

=05

Gbs

P(0) = 80

3 4

S(0) = S1 S2S S

P(1) = 65

S(1) = S1 S3

P(3) = 95

S(3) = S2 S3

P(2) = 85S(2) = S1

S3 S4

P(4) = 60

P(5) = 75

S(5) = S1S2 S4

0

1

2

3

4

5

S(4) = S2S3 S4








119894) = 119878120591| service 119878

120591is available



value 119889119878120591

(119899119894) on service node 119899







119904 119899119889 119905119886 119905119889 119887



119886














MS (sl120485) isin 119873 (1)





node


P1198902119890

= (11989911989411198721198941) (11989911989421198721198942) (119899

119894119898

119872119894119898

) (2)


1198941= 119899119904 119899119894119898


119894119898

= 119899119889 1198991198942 1198991198943 119899

119894119898minus1




119894119896

defined as

119872119894119896

= (sl120485997888rarr119899119894119896

) | M119878(sl120485) = 119899119894119896

(3)

where if119872119894119896


119894119896



Node Service1198781 1198782 1198783 1198784

0 1198891198781(0) = 3 119889

1198782(0) = 4 119889

1198783(0) = 4 119889

1198784(0) = 2

1 1198891198781(1) = 5 mdash 119889

1198783(1) = 3 mdash

2 1198891198781(2) = 2 mdash 119889

1198783(2) = 6 119889

1198784(2) = 1

3 mdash 1198891198782(3) = 1 119889

1198783(3) = 2 mdash

4 mdash 1198891198782(4) = 6 119889

1198783(4) = 8 119889

1198784(4) = 6

5 1198891198781(5) = 4 119889

1198782(5) = 3 mdash 119889

1198784(5) = 4



119894119896

isin P1198902119890

119875 (119899119894119896


isin119872119894119896

119901 (sl120485) (4)

where sl120485rarr 119899119894119896

isin 119872119894119896


on service node 119899119894119896



119861 (119897 (119894119896 119894119896+1)) ge 119887 lowast 119886119905 (5)





D (P1198902119890) =

119898minus1sum119896=1

119889 (119894119896 119894119896+1) +

119898minus1sum119896=2

sumsl120485rarr119899119894119896

isin119872119894119896

119889sl120485

(119899119894119896

) (6)

where 119899119904= 1198991198941 119899119889= 119899119894119898













119894








119894isin 119873 (7)





120575 (Res) = 120572 sdot 119875 (119899120594) + 120573 sdot 119861 (119897120594)119879

(8)



as 119897120594= (1198971205941 1198971205942 119897120594120582



(119861(1198971205941) 119861(1198971205942) 119861(119897

120594120582








where 120583119901


119887repre-



C (119877 119905) = sumsl120485isinsl 119899=M

119878(sl120485)


sdot 119888 (119897)

(10)




sdot 119888 (119897120594) (11)


119889 denoted by

R(119877 119905) sdot 119905119889and C(119877 119905) sdot 119905

119889 respectively




sumT119905=0 R (119877 119905) lowast 119905119889

T (12)


AR = limTrarrinfin

sumT119905=0

10038161003816100381610038161003816119877accepted

10038161003816100381610038161003816

|119877| (13)





T (14)


R (119866)

C (119866)= lim

Trarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889


(15)




(2) for all 119899119894isin 119873 do













119860119873(119899119894) = 119875 (119899


119901 (sl120485) (16)



119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)


119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)




119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)



119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)


119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)



119873)









120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










s

d(s 0)

= 1

B(s 0

) =2Gbs

dd(s 1) =

3

B(s 1) =1Gbs

d(0 1)=3

B(0 1)=2

Gbp

s

d(0 3) = 2B(0 3) = 05Gs

d(0 2) =5

B(0 2) =2Gbps

d(1 2)

= 4

B(1 2)

= 05Gbs

d(3 5) = 4

B(3 5) = 1Gbs

d(2 3)=1

B(2 3)=1

Gbs

d(2 4) = 3

B(2 4) = 05Gbs

d(5 d) = 3B(5 d) = 1Gbs

d(4 5)=2

B(4 5)=2

Gbs

d(4d)

=6

B(4d)

=05

Gbs

P(0) = 80

3 4

S(0) = S1 S2S S

P(1) = 65

S(1) = S1 S3

P(3) = 95

S(3) = S2 S3

P(2) = 85S(2) = S1

S3 S4

P(4) = 60

P(5) = 75

S(5) = S1S2 S4

0

1

2

3

4

5

S(4) = S2S3 S4








119894) = 119878120591| service 119878

120591is available



value 119889119878120591

(119899119894) on service node 119899







119904 119899119889 119905119886 119905119889 119887



119886














MS (sl120485) isin 119873 (1)





node


P1198902119890

= (11989911989411198721198941) (11989911989421198721198942) (119899

119894119898

119872119894119898

) (2)


1198941= 119899119904 119899119894119898


119894119898

= 119899119889 1198991198942 1198991198943 119899

119894119898minus1




119894119896

defined as

119872119894119896

= (sl120485997888rarr119899119894119896

) | M119878(sl120485) = 119899119894119896

(3)

where if119872119894119896


119894119896



Node Service1198781 1198782 1198783 1198784

0 1198891198781(0) = 3 119889

1198782(0) = 4 119889

1198783(0) = 4 119889

1198784(0) = 2

1 1198891198781(1) = 5 mdash 119889

1198783(1) = 3 mdash

2 1198891198781(2) = 2 mdash 119889

1198783(2) = 6 119889

1198784(2) = 1

3 mdash 1198891198782(3) = 1 119889

1198783(3) = 2 mdash

4 mdash 1198891198782(4) = 6 119889

1198783(4) = 8 119889

1198784(4) = 6

5 1198891198781(5) = 4 119889

1198782(5) = 3 mdash 119889

1198784(5) = 4



119894119896

isin P1198902119890

119875 (119899119894119896


isin119872119894119896

119901 (sl120485) (4)

where sl120485rarr 119899119894119896

isin 119872119894119896


on service node 119899119894119896



119861 (119897 (119894119896 119894119896+1)) ge 119887 lowast 119886119905 (5)





D (P1198902119890) =

119898minus1sum119896=1

119889 (119894119896 119894119896+1) +

119898minus1sum119896=2

sumsl120485rarr119899119894119896

isin119872119894119896

119889sl120485

(119899119894119896

) (6)

where 119899119904= 1198991198941 119899119889= 119899119894119898













119894








119894isin 119873 (7)





120575 (Res) = 120572 sdot 119875 (119899120594) + 120573 sdot 119861 (119897120594)119879

(8)



as 119897120594= (1198971205941 1198971205942 119897120594120582



(119861(1198971205941) 119861(1198971205942) 119861(119897

120594120582








where 120583119901


119887repre-



C (119877 119905) = sumsl120485isinsl 119899=M

119878(sl120485)


sdot 119888 (119897)

(10)




sdot 119888 (119897120594) (11)


119889 denoted by

R(119877 119905) sdot 119905119889and C(119877 119905) sdot 119905

119889 respectively




sumT119905=0 R (119877 119905) lowast 119905119889

T (12)


AR = limTrarrinfin

sumT119905=0

10038161003816100381610038161003816119877accepted

10038161003816100381610038161003816

|119877| (13)





T (14)


R (119866)

C (119866)= lim

Trarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889


(15)




(2) for all 119899119894isin 119873 do













119860119873(119899119894) = 119875 (119899


119901 (sl120485) (16)



119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)


119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)




119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)



119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)


119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)



119873)









120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Node Service1198781 1198782 1198783 1198784

0 1198891198781(0) = 3 119889

1198782(0) = 4 119889

1198783(0) = 4 119889

1198784(0) = 2

1 1198891198781(1) = 5 mdash 119889

1198783(1) = 3 mdash

2 1198891198781(2) = 2 mdash 119889

1198783(2) = 6 119889

1198784(2) = 1

3 mdash 1198891198782(3) = 1 119889

1198783(3) = 2 mdash

4 mdash 1198891198782(4) = 6 119889

1198783(4) = 8 119889

1198784(4) = 6

5 1198891198781(5) = 4 119889

1198782(5) = 3 mdash 119889

1198784(5) = 4



119894119896

isin P1198902119890

119875 (119899119894119896


isin119872119894119896

119901 (sl120485) (4)

where sl120485rarr 119899119894119896

isin 119872119894119896


on service node 119899119894119896



119861 (119897 (119894119896 119894119896+1)) ge 119887 lowast 119886119905 (5)





D (P1198902119890) =

119898minus1sum119896=1

119889 (119894119896 119894119896+1) +

119898minus1sum119896=2

sumsl120485rarr119899119894119896

isin119872119894119896

119889sl120485

(119899119894119896

) (6)

where 119899119904= 1198991198941 119899119889= 119899119894119898













119894








119894isin 119873 (7)





120575 (Res) = 120572 sdot 119875 (119899120594) + 120573 sdot 119861 (119897120594)119879

(8)



as 119897120594= (1198971205941 1198971205942 119897120594120582



(119861(1198971205941) 119861(1198971205942) 119861(119897

120594120582








where 120583119901


119887repre-



C (119877 119905) = sumsl120485isinsl 119899=M

119878(sl120485)


sdot 119888 (119897)

(10)




sdot 119888 (119897120594) (11)


119889 denoted by

R(119877 119905) sdot 119905119889and C(119877 119905) sdot 119905

119889 respectively




sumT119905=0 R (119877 119905) lowast 119905119889

T (12)


AR = limTrarrinfin

sumT119905=0

10038161003816100381610038161003816119877accepted

10038161003816100381610038161003816

|119877| (13)





T (14)


R (119866)

C (119866)= lim

Trarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889


(15)




(2) for all 119899119894isin 119873 do













119860119873(119899119894) = 119875 (119899


119901 (sl120485) (16)



119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)


119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)




119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)



119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)


119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)



119873)









120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













120575 (Res) = 120572 sdot 119875 (119899120594) + 120573 sdot 119861 (119897120594)119879

(8)



as 119897120594= (1198971205941 1198971205942 119897120594120582



(119861(1198971205941) 119861(1198971205942) 119861(119897

120594120582








where 120583119901


119887repre-



C (119877 119905) = sumsl120485isinsl 119899=M

119878(sl120485)


sdot 119888 (119897)

(10)




sdot 119888 (119897120594) (11)


119889 denoted by

R(119877 119905) sdot 119905119889and C(119877 119905) sdot 119905

119889 respectively




sumT119905=0 R (119877 119905) lowast 119905119889

T (12)


AR = limTrarrinfin

sumT119905=0

10038161003816100381610038161003816119877accepted

10038161003816100381610038161003816

|119877| (13)





T (14)


R (119866)

C (119866)= lim

Trarrinfin

sumT119905=0 R (119877 119905) lowast 119905119889


(15)




(2) for all 119899119894isin 119873 do













119860119873(119899119894) = 119875 (119899


119901 (sl120485) (16)



119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)


119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)




119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)



119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)


119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)



119873)









120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(2) for all 119899119894isin 119873 do













119860119873(119899119894) = 119875 (119899


119901 (sl120485) (16)



119894)

119880119873(119899119894) =

119875 (119899119894) minus 119860119873(119899119894)

119875 (119899119894)

(17)


119880119873=sum|119873|minus1119894=0 119880

119873(119899119894)

|119873| (18)




119860119871 (119897) = 119861 (119897) minus 119887 lowast 119886119905 (19)



119880119871 (119897) =

119861 (119897) minus 119860119871 (119897)

119861 (119897) (20)


119880119871=sumforall119897isin119871

119880119871 (119897)

|119871| (21)



119873)









120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











120582) 120573 = 0 sdot 1205731015840


1015840




Algorithm 2 SI-AAR


Algorithm 3 UI-AAR





120594and 120573 sdot119861(119897

120594)















Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Algorithm 4 AACC



120594120484

isin 119897120594do

(4) repeat(5) 119905Vbw

120484

= 119905Vbw120484

+ Δ119905bw 1205731015840

120484= 119905Vbw

120484

(6) Add 1205731015840120484sdot 119861(119897120594120484


119887) or (RC lt 120590)

(9) 1205731015840

120484= 119905

Vbw120484

= 119905Vbw120484


120594120484



120594119896


)

(17) 120573119896= 1205731015840119896 119905Vbw = 120573 120573

1015840 = (1 1 1120582)


Algorithm 5 AABC








120594120484


120484


120594120484


120484



119887



120594120484




(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










(1) 119905cpu = 0 120572 = 0 120573 = (0 0 0120582)




120582) (2) set 119905Vbw = 120573



Algorithm 6 AACBC















119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119901 120583119887 119888(119899) and 119888(119897) to 1 Δ119905cpu and








119873= 349 119880

119871= 235





119873= 104 119880

119871= 325 Δ119880 =



119873is close to 119880

119871






0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 2 4 6 8 10 12 14 16 1805

06

07

08

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50065

07

075

08

085

09

095


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 503000

4000

5000

6000

7000

8000

9000

10000


Aver

age c

ost

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5004

05

06

07

08

09

1


Reve

nue

cost

ratio

UI-AAR 120572 = 5 120573 = (0 0 0 0)

SI-AAR 120572 = 5 120573 = (5 5 5 5)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)




0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 2 4 6 8 10 12 14 16 18045

05

055

06

065

07

075

Node index

Aver

age r

eque

st ac

cept

ance

ratio



5000

4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 50

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)

Original 120572 = 0 120573 = (0 0 0 0 0 0)


2 6 10 14 18 22 26 30 34 38 42 46 503000

3500

4000

4500

5000


Aver

age r

even

ue

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50

4000

6000

8000

10000


Aver

age c

ost

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 5002

03

04

05

06

07

08


Reve

nue

cost

ratio

UI-AAR 120572 = 0 120573 = (1 1 3 1 1 2)

SI-AAR 120572 = 3 120573 = (3 3 3 3 3 3)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9 9 9)




0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 2 4 6 8 10 12 14 16 1804

045

05

055

06

065

Node index

Aver

age r

eque

st ac

cept

ance

ratio



4500

4000

3500

3000

2500

2000

1500

1000

500


120575(R

es)


2 6 10 14 18 22 26 30 34 38 42 46 5005

06

07

08

09

1


Aver

age r

eque

st ac

cept

ance

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)



2 6 10 14 18 22 26 30 34 38 42 46 504000

5000

6000

7000

8000

9000


Aver

age r

even

ue

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 505000

7000

9000

11000

13000

15000


Aver

age c

ost

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)


2 6 10 14 18 22 26 30 34 38 42 46 50045

05

055

06

065

07

075


Reve

nue

cost

ratio

UI-AAR 120572 = 4 120573 = (2 4 3 3)

SI-AAR 120572 = 4 120573 = (4 4 4 4)

SI-AAR-least 120572 = 1 120573 = (1 1 1 1)

SI-AAR-RUB 120572 = 9 120573 = (9 9 9 9)















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





















Acknowledgment




References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Cost-Efficient Allocation of Additional ...Research Article Cost-Efficient...

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