research article on modeling and analysis of mimo wireless...

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Research Article On Modeling and Analysis of MIMO Wireless Mesh Networks with Triangular Overlay Topology Zhanmao Cao, 1,2 Chase Q. Wu, 2 Yuanping Zhang, 3 Sajjan G. Shiva, 2 and Yi Gu 4 1 Department of Computer Science, South China Normal University, Guangzhou, Guangdong 510631, China 2 Department of Computer Science, University of Memphis, Memphis, TN 38152, USA 3 School of Computer Science & Education Soſtware, Guangzhou University, Guangzhou, Guangdong 510006, China 4 Department of Computer Science, Middle Tennessee State University, Murfreesboro, TN 37132, USA Correspondence should be addressed to Chase Q. Wu; [email protected] Received 27 September 2014; Revised 17 January 2015; Accepted 27 January 2015 Academic Editor: Hsuan-Ling Kao Copyright © 2015 Zhanmao Cao et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Multiple input multiple output (MIMO) wireless mesh networks (WMNs) aim to provide the last-mile broadband wireless access to the Internet. Along with the algorithmic development for WMNs, some fundamental mathematical problems also emerge in various aspects such as routing, scheduling, and channel assignment, all of which require an effective mathematical model and rigorous analysis of network properties. In this paper, we propose to employ Cartesian product of graphs (CPG) as a multichannel modeling approach and explore a set of unique properties of triangular WMNs. In each layer of CPG with a single channel, we design a node coordinate scheme that retains the symmetric property of triangular meshes and develop a function for the assignment of node identity numbers based on their coordinates. We also derive a necessary-sufficient condition for interference-free links and combinatorial formulas to determine the number of the shortest paths for channel realization in triangular WMNs. 1. Introduction e WiMax group advocated the last-mile broadband servi- ces, IEEE 802.16 Standard [1], which defines broadband back- bones as wireless mesh networks (WMNs). Such networks typically consist of two types of nodes, that is, base station (BS) and subscriber station (SS). BS is a wireless gateway con- nected to the Internet, while SS is a node that acts as a relay station. In multiple input multiple output (MIMO) WMNs, all nodes are equipped with multiple interfaces and support both multicast and mesh modes. Particularly, in a mesh mode, nodes can communicate with neighbors without the help of BS and the relay strategy provides an economical way to expand the mesh covering area. MIMO WMNs (in the rest of the paper, we use the term WMNs for conciseness) are well recognized as an efficient extension to the Internet backhaul [2]. WMNs possess some inherent characteristics that are different from ad hoc or wireless sensor networks. Since the nodes in WMNs are almost fixed and typically powered by electrical wires, the links or routing paths in WMNs generally last longer than those in mobile ad hoc networks. Also, every node in WMNs typically has nonzero traffic requests because it needs to route aggregated traffic from the terminal devices in its region for either upload or download. e topology of WMNs may be determined based on the predicted traffic requests or geographical environments. Since both BS and SS can be considered static, it is reasonable to view the mesh topology as a fixed graph. e rapidly growing demand for ubiquitous Internet access requires an effective mathematical model for WMNs as it may simplify the tasks of routing, scheduling, and channel assignment. To achieve a maximum fair usage of multiple channels in WMNs, it is important to employ an efficient channel allocation scheme and an appropriate overlay graph topology for a given area [3]. In [4], the virtual topology is viewed as CPG to simplify the channel assignment problem through a graph. In addition, the CPG model also brings convenience for the analysis of routing and scheduling in WMNs. As nodes are static in WMNs, they can be identified Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 185262, 11 pages http://dx.doi.org/10.1155/2015/185262

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Page 1: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Research ArticleOn Modeling and Analysis of MIMO Wireless Mesh Networkswith Triangular Overlay Topology

Zhanmao Cao12 Chase Q Wu2 Yuanping Zhang3 Sajjan G Shiva2 and Yi Gu4

1Department of Computer Science South China Normal University Guangzhou Guangdong 510631 China2Department of Computer Science University of Memphis Memphis TN 38152 USA3School of Computer Science amp Education Software Guangzhou University Guangzhou Guangdong 510006 China4Department of Computer Science Middle Tennessee State University Murfreesboro TN 37132 USA

Correspondence should be addressed to Chase Q Wu chasewumemphisedu

Received 27 September 2014 Revised 17 January 2015 Accepted 27 January 2015

Academic Editor Hsuan-Ling Kao

Copyright copy 2015 Zhanmao Cao et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Multiple input multiple output (MIMO) wireless mesh networks (WMNs) aim to provide the last-mile broadband wireless accessto the Internet Along with the algorithmic development for WMNs some fundamental mathematical problems also emerge invarious aspects such as routing scheduling and channel assignment all of which require an effective mathematical model andrigorous analysis of network properties In this paper we propose to employ Cartesian product of graphs (CPG) as a multichannelmodeling approach and explore a set of unique properties of triangularWMNs In each layer ofCPGwith a single channel we designa node coordinate scheme that retains the symmetric property of triangular meshes and develop a function for the assignment ofnode identity numbers based on their coordinates We also derive a necessary-sufficient condition for interference-free links andcombinatorial formulas to determine the number of the shortest paths for channel realization in triangular WMNs

1 Introduction

The WiMax group advocated the last-mile broadband servi-ces IEEE 80216 Standard [1] which defines broadband back-bones as wireless mesh networks (WMNs) Such networkstypically consist of two types of nodes that is base station(BS) and subscriber station (SS) BS is a wireless gateway con-nected to the Internet while SS is a node that acts as a relaystation In multiple input multiple output (MIMO) WMNsall nodes are equipped with multiple interfaces and supportboth multicast and mesh modes Particularly in a meshmode nodes can communicate with neighbors without thehelp of BS and the relay strategy provides an economical wayto expand the mesh covering area MIMOWMNs (in the restof the paper we use the termWMNs for conciseness) are wellrecognized as an efficient extension to the Internet backhaul[2]

WMNs possess some inherent characteristics that aredifferent from ad hoc or wireless sensor networks Since thenodes in WMNs are almost fixed and typically powered by

electrical wires the links or routing paths inWMNs generallylast longer than those in mobile ad hoc networks Also everynode inWMNs typically has nonzero traffic requests becauseit needs to route aggregated traffic from the terminal devicesin its region for either upload or download The topologyof WMNs may be determined based on the predicted trafficrequests or geographical environments Since both BS and SScan be considered static it is reasonable to view the meshtopology as a fixed graph

The rapidly growing demand for ubiquitous Internetaccess requires an effectivemathematicalmodel forWMNs asit may simplify the tasks of routing scheduling and channelassignment To achieve a maximum fair usage of multiplechannels in WMNs it is important to employ an efficientchannel allocation scheme and an appropriate overlay graphtopology for a given area [3] In [4] the virtual topology isviewed as CPG to simplify the channel assignment problemthrough a graph In addition the CPG model also bringsconvenience for the analysis of routing and scheduling inWMNs As nodes are static in WMNs they can be identified

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 185262 11 pageshttpdxdoiorg1011552015185262

2 Mathematical Problems in Engineering

geographically through their coordinates [5] Therefore therouting and scheduling problems can be analyzed using thenode coordinates Interference is another fundamental issuein either scheduling or routing and the properties of interfer-ence under a given coordinate scheme if defined properlyand described rigorously may bring benefits to resource utili-zation and interference avoidance

In this work we use CPG as a modeling approach andexplore a set of unique properties of WMNs with a triangu-lar topology In each layer of CPG with a single channel thenetwork topology is a planar mesh Our work makes severaltheoretical contributions to the analysis of WMN properties(i) we design a coordinate scheme that retains the symmetricproperty of triangular meshes and develop a function toassign a unique identity number to a specific wireless nodebased on its coordinates (ii) We derive a necessary-sufficientcondition for interference-free links under the proposednode coordinate scheme (iii) We derive combinatorial for-mulas in terms of the number of transceivers and channelsto determine the number of the shortest paths for channelrealization in triangular WMNs

The rest of the paper is organized as follows Section 2surveys related work Section 3 presents a channel-layeredCPG model with emphasis on interference detection Ineach layer of CPG we propose a coordinate scheme namedparallel cluster coordinate derive a necessary-sufficient con-dition for interference avoidance and develop a function fornode identity number assignment to support efficient WMNmaintenance and administration Section 4 derives formulasfor determining the number of the shortest paths in WMNs

2 Related Work

We conduct a brief survey of work directly related to mathe-matical models for WMNs

In the past decade most efforts in WMN topology werefocused on interference and performance in planar meshes[6ndash9] With geographical information from satellite or con-trol channel communications it is relatively easy to acquire aplanar topology since the nodes in WMNs are almost staticFor example in IEEE Standard 80221 formedia-independenthandover Media Independent Information Service (MIIS)stores the geographical information of all access networkoperators available in a particular region [10]

Square and hexagonal meshes have been proposed to actas wireless broadband backbones [11] However they are lesscompetitive than triangular meshes as the latter outperformsthe former and other random meshes in terms of variousperformance metrics such as coverage area link qualityper-user fair rate and node density [12] Hong and Huaconducted a comparative evaluation of the throughput per-formance between square hexagonal and triangular meshesTheir experiments showed that triangular meshes achievehigher throughput than others in several cases and theirtotal throughput does not vary significantly in response totopology changes in large wireless networks with a constantdensity [13] Therefore we also adopt a triangular meshtopology in our model

A unified network model based on super graph mayfurther facilitate the analysis of various aspects of WMNssuch as interference scheduling routing and channel assign-ment However research efforts along this line are still quitelimited Several researchers considered some of these aspectssimultaneously [3 11 14] which motivates us to design aunified model for WMNs

In a given network topology a properly designed coor-dinate scheme may facilitate link interference detection andpath finding In hexagonal meshes Chin et al proposed anode coordinate scheme with three parallel line clusters [15]where a node is represented by a 3-tuple In triangularmeshesconsisting of BSnodeswith a node degree of six Cao et al pro-posed two BS-centered coordinate schemes [5] and exploredinterference and link groups in each of these schemesFurthermore in addition to coordinates a router should alsobe assigned a unique identity number to support convenientsimulation administration and maintenance In [4 5] Caoet al also represented the coordinates of a node by a 3-tuplebut did not tackle the identity number assignment problem

The performance of WMNs is largely affected by linkinterference Most research efforts on this subject have beenmade through generic methods or experimental studiesinstead of conclusive results in the form of necessary-sufficient conditions [3 16 17] In our work we attempt todesign a suitable node coordinate scheme and thenmodel theinterference in WMNs as a specific checking list based on settheory

Routing in WMNs is a 2-step procedure that is pathfinding followed by channel assignment One basic approachto find an alternative interference-free path is to count thenumber of shortest paths and the number of all possiblechannel assignment schemes A tree-like path finding schemeis proposed in [6] without any node coordinate Cao andXiaoproposed path counting formulas for a source-destinationpair in square grids [18] while the path counting problem intriangular meshes is still left unexplored In our work basedon the proposed CPG model and coordinate scheme wetackle this problem in triangular meshes with multiple chan-nels and interfaces Channel assignment is another importantproblem involving several network layers inWMNs which isessentially an NP-complete edge coloring problem [14 19]

3 A Channel-Layered Graph Model

As massive MIMO is on its way from theory to realisticdeployment one of the key problems is the interchannelcooperation which calls for the development of sophisticatedanalytical channel models Larsson et al provide an overviewof massive MIMO and motivate researchers to developchannel models capturing the essential channel behaviorsdespite their limitations [20] For example the Kroneckermodel which is widely used to model channel correlationis not an exact representation of reality but provides a usefulmodel for certain types of analysis

WMNs are conventionally modeled by a directed graphwhere a directed edge between twoneighbor nodes representsa communication link over a specific channel Since a node

Mathematical Problems in Engineering 3

(a) Mesh and virtual nodes (b) Layered mesh

(c) Links in a channel (d) Maximal links

AA

A

A

BB

B

B

CC

C

C

DD

D

D

EE

E

E

c1

c2

Figure 1 The CPG-modeled virtual topology of a physical network with five physical nodes and two channels

equipped with 120585 transceivers may have (at most) 120585 simul-taneous links over 120585 orthogonal channels (assuming thatmore than 120585 channels are available) we can split a physicalnode into 120585 fully connected virtual nodes each of which isequipped with a single transceiver This way we are able torepresent the originalWMN as 120585 identical layers of networkseach of which operates over a different channel

31 Cartesian Product of Graphs The topology-basedmodel-ing approach has been commonly used in wireless networksfor various purposes but often in a planar view [8 12 14]and most of the discussions on scheduling routing andchannel assignment are also based on a planar topology [611 13 17] The recent development of MIMO WMNs callsfor a suitable model to describe MIMO-specific propertiesand understand the cooperative activities across differentinterfaces over multiple channels [20] The planar topologycan be used to determine the internode interference [8] but isinsufficient to provide a visual representation for analyzingthe cooperation between links or channels On the otherhand modelingMIMOWMNs as a super graph still remainslargely unexploited except the work in [4] In this paper ourgoal is to develop an effective model to facilitate the analysisof MIMO channel cooperation

We propose to employ the CPG to model WMNs bycombining a triangularmesh of physical nodes and a graph of

fully connected virtual nodes Together with the coordinatesof triangular overlay nodes the CPG model provides aconvenient way to analyze the properties of interferenceavoidance channel assignment and routing path countingThis model retains the independence between orthogonalchannelswhile providing a general approach to analyzing linkbehaviors over multiple channels

Cartesian Product of Graphs Given two graphs 119866 and119867 theCartesian product 119866 times 119867 is a graph such that

(i) the graph 119866times119867 has a vertex set119881(119866) times119881(119867) that isa vertex in119866times119867 is denoted by a pair (V V1015840) V isin 119881(119866)and V1015840 isin 119881(119867)

(ii) any two nodes (119906 1199061015840) and (V V1015840) and 119906 V isin 119881(119866) and1199061015840 V1015840 isin 119881(119867) are adjacent in 119866times119867 if and only if oneof the following holds (a) 119906 = V and 1199061015840 is adjacent toV1015840 in119867 or (b) 1199061015840 = V1015840 and 119906 is adjacent to V in 119866

For illustration Figure 1(a) shows a mesh network of fivephysical nodes (left side) and a graph of two connected virtualnodes (right side) corresponding to a physical node equippedwith two transceivers each operating on a different channel1198881or 1198882 Figure 1(b) shows a channel-layered virtual topology

of the original mesh network modeled by CPGIn this example the CPG of two graphs in Figure 1(a)

results in a two-layered graph in Figure 1(b) A solid directed

4 Mathematical Problems in Engineering

edge in the top or bottom planar meshes in Figures 1(b)1(c) and 1(d) represents a communication link 119897 = (119878 119877)

119888

transmitting data from node 119878 to node 119877 over a channel 119888while a dashed edge has a conflict with some active links

The maximum possible number of concurrent activelinks on a given channel in a mesh is largely affected bythe selection of the senders As shown in Figure 1(b) whennode 119860 in the top layer is sending data on channel 119888

1 the

three neighbors of 119860 that is nodes 119861 119862 and 119863 in thesame layer (over channel 119888

1) cannot send data Furthermore

node 119864 can be selected as a sender but neither of itsneighbors that is nodes 119862 and 119863 can receive data from 119864without interference illustrated as the dashed directed edgesin Figure 1(c) In this case there is no other active link except(119860 119861)

1198881

However if we choose the initial senders properlyas shown in Figure 1(d) there could exist two concurrentinterference-free links on channel 119888

119894 119894 = 1 2 that is (119861 119860)

119888119894

and (119864119863)119888119894

Since all active links must be interference-free on the

same channel at the same time links are generally sparselydistributed in a planar mesh with respect to a certainchannel Note that more links mean better service to trafficrequests The CPG model allows us to consider concurrentpaths in multiple layers over different channels throughradio cooperation among channel layers for a given traffictask hence providing more capacity and higher throughputcompared to the situation with a single channel and radioThis is consistent with the experimental results presentedby Draves et al [16] The proposed CPG model is alignedwell with existing research in terms of interference relationlink activity and network throughput as well as routing andscheduling [6 7 14 19 21] and enables us to conduct deepertheoretical analysis of WMNs

One advantage that CPG brings is to simplify the expres-sion of multichannel links Since different layers (operatingon different channels) are of the same topology the schedul-ing strategy derived in one layer is readily applicable toanother layer For example there exists a certain link distri-bution pattern among concurrent links In CPG it is obviousthat such an interference-free link distribution pattern on onechannel also exists on others

32 Coordinate Scheme A well-designed coordinate schememay facilitate the analysis of WMNs Since all channellayers are of the same topology we only need to design thecoordinate scheme for one layer or channel The channelinformation can be added to the node coordinates to uniquelyidentify a specific layer

Chin et al proposed a coordinate scheme in hexagonalcellular networks where each node has a degree of three[15] Inspired by their work we propose a parallel clustercoordinate scheme in a local triangular mesh with one BSnode and a number of SS nodes with a degree of six Thisscheme can be readily extended to larger networks withmultiple BSs by inserting the BS information to the nodecoordinates

321 Parallel Cluster Coordinate Scheme In a triangularmesh we first define three clusters of parallel lines along three

A

BC

D

E

j

F

BS

i minus 2 i minus 1 i + 1

j + 2

j + 1

j minus 1

j minus 2

k minus 2 k minus 1 k + 1k

i

Figure 2 The parallel cluster coordinate scheme

different directions that is north-east east and south-eastSince each node is a point intersected by three lines eachfrom one of the three clusters we propose a parallel clustercoordinate scheme (PCCS) which uses a 3-tuple (119894 119895 119896) torepresent the coordinates of a node intersected by the 119894th 119895thand 119896th line in the corresponding clusters [5] As illustrated inFigure 2 the BS node with the coordinates (0 0 0) is locatedat the center and the SS nodes 119860 119861 119862 119863 119864 and 119865 whichare one hop away from the BS node have the coordinates119860 = (1 0 1) 119861 = (0 1 1) 119862 = (minus1 minus1 0) 119863 = (minus1 0 minus1)119864 = (0 minus1 minus1) and 119865 = (1 1 0)

Note that not every combination of three integers canrepresent a node in a triangular mesh because some linesdefined by combinatorial 3-tuples do not intersect at acommon point We summarize such lines as follows

Ω = (119894 minus 1 119895 minus 1 119896) (119894 + 1 119895 + 1 119896) (119894 119895 minus 1 119896 + 1)

(119894 119895 + 1 119896 minus 1) (119894 + 1 119895 119896 minus 1) (119894 minus 1 119895 119896 + 1) (1)

Under the proposed PCCS scheme for a given sender(119894 119895 119896) there are six possible receivers which form itsneighbor set119873

119887(119894 119895 119896)

119873119887(119894 119895 119896) = (119894 minus 1 119895 + 1 119896) (119894 + 1 119895 minus 1 119896)

(119894 119895 + 1 119896 + 1) (119894 119895 minus 1 119896 minus 1)

(119894 + 1 119895 119896 + 1) (119894 minus 1 119895 119896 minus 1)

(2)

PCCS retains the symmetric nature of a triangular meshand facilitates the calculation of the distance between a pairof nodes

322 Symmetric Property of PCCS In PCCS the coordinatesof any two nodes that are symmetric with respect to thecentral BS node located at (0 0 0) are negated For examplethe pairs of nodes 119860 and119863 119861 and 119864 and 119862 and 119865 in Figure 2are symmetric and their coordinates are negated from theircounterparts

The symmetric properties of CPG and Cayley graphhave been well studied [22] According to the vertexedge

Mathematical Problems in Engineering 5

transitive properties a link group can be transited to generateanother one in any layer

Similarly due to the symmetric property we are able totransform a link group to another one through rotation Forexample link group of ((0 2 2) (0 1 1))

119888and ((2 minus2 0)

(1 minus1 0))119888is interference-free After a clockwise rotation of

1205873 a new group of interference-free links is of ((2 0 2)(1 0 1))

119888and ((0 minus2 minus2) (0 minus1 minus1))

119888 respectively With

another clockwise rotation of 1205873 the derived interference-free links are ((2 minus2 0) (1 minus1 0))

119888and ((minus2 0 minus2) (minus1 0

minus1))119888 respectively Note that the rotation operation is edge

transitive and it generates a new link group because of thesymmetric property

323 Distance between Two Nodes Asmost traffic is uploaddownload (tofromBS) we need to count the number of hopsfrom a router node to BS in PCCS We have the followingproperties

Property 1 In triangular WMNs with PCCS the minimumnumber ℏ of hops from 119860 = (119894 119895 119896) to BS = (0 0 0) is

ℏ =|119894| +10038161003816100381610038161198951003816100381610038161003816 + |119896|

2= max |119894| 1003816100381610038161003816119895

1003816100381610038161003816 |119896|(3)

which is consistent with the one in [4]

Property 2 Suppose that BS is positioned at an arbitrarylocation (119894

0 1198950 1198960) instead of (0 0 0) Node (119894 119895 119896) can be

translated to (119894 119895 119896) minus (1198940 1198950 1198960) by a translation function

120575 (119894 119895 119896) 997888rarr ((119894 119895 119896) minus (1198940 1198950 1198960)) (4)

In general suppose that (1198940 1198950 1198960) is the destination of a

traffic path we may virtually view (1198940 1198950 1198960) as the BS node

after applying the translation of (4) Then the formula in (3)may facilitate further analysis

Given two nodes 119860 = (1198941 1198951 1198961) and 119861 = (119894

2 1198952 1198962) in

PCCS the distance between 119860 and 119861 denoted by 119889(119860 119861)is the minimum number of hops between them which iscalculated as 119889(119860 119861) = (|119894

1minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2

324 Mapping to 2D Points In order to draw a planar meshwe need to map 3-tuple coordinates (119894 119895 119896) to 2-dimensional(2D) points (119909 119910) To do this we first overlap a rectangularplane coordinate system to the PCCS triangular mesh Letthe 119909-axis overlap the axis 119895 = 0 while keeping the positiverightward direction Meanwhile the 119910-axis passes the BSnode and is vertical to the line 119895 = 0 with a positive upwarddirectionThen we can determine (119909 119910) by projecting (119894 119895 119896)onto axes 119909 and 119910 The PCCS supports a function 119891mapping(119894 119895 119896) to (119909 119910) as follows

Property 3 A one-to-one mapping function 119891 maps nodecoordinates (119894 119895 119896) in PCCS to 2D coordinates (119909 119910)

(119909 119910) = 119891 (119894 119895 119896) =

119909 =119894 + 119896

2119910 = 119895

(5)

For example node (0 1 1) is mapped to (12 1) and(minus1 3 2) is mapped to (12 3) If a node is on the 119910-axis 119894 =minus119896 always holds For example (minus1 2 1) is mapped to (0 2) Ifnode is located on 119896 = 0 in PCCS 119909 = 1198942 For example(minus2 2 0) is mapped to (minus1 2) This property facilitates theplotting of a triangular mesh in PCCS

33 Interference-Free Conditions To analyze the interferencebetween links we need to consider node interference rela-tions which are critical to scheduling links in WMNs Asinterference is an inherent nature for radio media the wire-less communication performance may be severely degradedif radios operate without a proper scheduling scheme [17]

Minimizing interference has been extensively investi-gated in the literature [9 17] Subramanian et al discussedchannel assignment in a multiradio situation [9] Tan et aldesigned algorithms to set up a skeleton ofminimum interfer-ence for a single channel [17] Scheduling links in a coopera-tive way will improve the energy efficiency and reduce colli-sion These discussions assume variable transmitting poweror interface channel switching However the variation oftransmitting power may lead to the variation of networktopology which may cause changes in the interferencerelation Xu et al use a sensing scheme to achieve power effi-ciency for convergence communication [23] Their methodcan help set the initial power in an almost-static WMNtopology while promising interference-free cognitive accesswith link status as busy or idle

When sender 119878 is sending over channel 1198880 neither can

119878 receive data over channel 1198880 nor can its neighbors in its

effective radio coverage send data If a valid neighbor node119877 receives data over channel 119888

0 we have a link (119878 119877)

1198880

Furthermore the neighbors of receiver 119877 cannot send dataover channel 119888

0at the same time Therefore to be link

interference-free in multiradio multichannel environmentsusing PCCS we need to consider three classes of node inter-ferences sender-to-sender receiver-to-receiver and sender-to-receiver

331 Sender-to-Sender Given a certain channel any twosenders must be at least two hops away in a triangular meshto avoid mutual interference which could help construct aninterference-free candidate set for possible senders

Definition 1 A router set 119878119862is called a sender candidate set if

every two nodes in 119878119862are at least two hops away

Given two nodes 119878119894= (119886 119887 119888) and 119878

119895= (119889 119890 119891) in 119878

119862 the

distance between them must satisfy 119889(119878119894 119878119895) ge 2

For a given node 119878119895notin 119878119862 we can add 119878

119895to 119878119862only if it

is at least two hops away to any node 119878119894in 119878119862 We have the

following necessary condition on interference-free links

119889 (119878119894 119878119895) =|119886 minus 119889| + |119887 minus 119890| +

1003816100381610038161003816119888 minus 1198911003816100381610038161003816

2ge 2 (6)

332 Receiver-to-Receiver Two nodes 119877119894and 119877

119895can each act

as a receiver simultaneously if they have a distance larger than1 Similar to the sender candidate set we use 119877

119862to denote

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 2: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

2 Mathematical Problems in Engineering

geographically through their coordinates [5] Therefore therouting and scheduling problems can be analyzed using thenode coordinates Interference is another fundamental issuein either scheduling or routing and the properties of interfer-ence under a given coordinate scheme if defined properlyand described rigorously may bring benefits to resource utili-zation and interference avoidance

In this work we use CPG as a modeling approach andexplore a set of unique properties of WMNs with a triangu-lar topology In each layer of CPG with a single channel thenetwork topology is a planar mesh Our work makes severaltheoretical contributions to the analysis of WMN properties(i) we design a coordinate scheme that retains the symmetricproperty of triangular meshes and develop a function toassign a unique identity number to a specific wireless nodebased on its coordinates (ii) We derive a necessary-sufficientcondition for interference-free links under the proposednode coordinate scheme (iii) We derive combinatorial for-mulas in terms of the number of transceivers and channelsto determine the number of the shortest paths for channelrealization in triangular WMNs

The rest of the paper is organized as follows Section 2surveys related work Section 3 presents a channel-layeredCPG model with emphasis on interference detection Ineach layer of CPG we propose a coordinate scheme namedparallel cluster coordinate derive a necessary-sufficient con-dition for interference avoidance and develop a function fornode identity number assignment to support efficient WMNmaintenance and administration Section 4 derives formulasfor determining the number of the shortest paths in WMNs

2 Related Work

We conduct a brief survey of work directly related to mathe-matical models for WMNs

In the past decade most efforts in WMN topology werefocused on interference and performance in planar meshes[6ndash9] With geographical information from satellite or con-trol channel communications it is relatively easy to acquire aplanar topology since the nodes in WMNs are almost staticFor example in IEEE Standard 80221 formedia-independenthandover Media Independent Information Service (MIIS)stores the geographical information of all access networkoperators available in a particular region [10]

Square and hexagonal meshes have been proposed to actas wireless broadband backbones [11] However they are lesscompetitive than triangular meshes as the latter outperformsthe former and other random meshes in terms of variousperformance metrics such as coverage area link qualityper-user fair rate and node density [12] Hong and Huaconducted a comparative evaluation of the throughput per-formance between square hexagonal and triangular meshesTheir experiments showed that triangular meshes achievehigher throughput than others in several cases and theirtotal throughput does not vary significantly in response totopology changes in large wireless networks with a constantdensity [13] Therefore we also adopt a triangular meshtopology in our model

A unified network model based on super graph mayfurther facilitate the analysis of various aspects of WMNssuch as interference scheduling routing and channel assign-ment However research efforts along this line are still quitelimited Several researchers considered some of these aspectssimultaneously [3 11 14] which motivates us to design aunified model for WMNs

In a given network topology a properly designed coor-dinate scheme may facilitate link interference detection andpath finding In hexagonal meshes Chin et al proposed anode coordinate scheme with three parallel line clusters [15]where a node is represented by a 3-tuple In triangularmeshesconsisting of BSnodeswith a node degree of six Cao et al pro-posed two BS-centered coordinate schemes [5] and exploredinterference and link groups in each of these schemesFurthermore in addition to coordinates a router should alsobe assigned a unique identity number to support convenientsimulation administration and maintenance In [4 5] Caoet al also represented the coordinates of a node by a 3-tuplebut did not tackle the identity number assignment problem

The performance of WMNs is largely affected by linkinterference Most research efforts on this subject have beenmade through generic methods or experimental studiesinstead of conclusive results in the form of necessary-sufficient conditions [3 16 17] In our work we attempt todesign a suitable node coordinate scheme and thenmodel theinterference in WMNs as a specific checking list based on settheory

Routing in WMNs is a 2-step procedure that is pathfinding followed by channel assignment One basic approachto find an alternative interference-free path is to count thenumber of shortest paths and the number of all possiblechannel assignment schemes A tree-like path finding schemeis proposed in [6] without any node coordinate Cao andXiaoproposed path counting formulas for a source-destinationpair in square grids [18] while the path counting problem intriangular meshes is still left unexplored In our work basedon the proposed CPG model and coordinate scheme wetackle this problem in triangular meshes with multiple chan-nels and interfaces Channel assignment is another importantproblem involving several network layers inWMNs which isessentially an NP-complete edge coloring problem [14 19]

3 A Channel-Layered Graph Model

As massive MIMO is on its way from theory to realisticdeployment one of the key problems is the interchannelcooperation which calls for the development of sophisticatedanalytical channel models Larsson et al provide an overviewof massive MIMO and motivate researchers to developchannel models capturing the essential channel behaviorsdespite their limitations [20] For example the Kroneckermodel which is widely used to model channel correlationis not an exact representation of reality but provides a usefulmodel for certain types of analysis

WMNs are conventionally modeled by a directed graphwhere a directed edge between twoneighbor nodes representsa communication link over a specific channel Since a node

Mathematical Problems in Engineering 3

(a) Mesh and virtual nodes (b) Layered mesh

(c) Links in a channel (d) Maximal links

AA

A

A

BB

B

B

CC

C

C

DD

D

D

EE

E

E

c1

c2

Figure 1 The CPG-modeled virtual topology of a physical network with five physical nodes and two channels

equipped with 120585 transceivers may have (at most) 120585 simul-taneous links over 120585 orthogonal channels (assuming thatmore than 120585 channels are available) we can split a physicalnode into 120585 fully connected virtual nodes each of which isequipped with a single transceiver This way we are able torepresent the originalWMN as 120585 identical layers of networkseach of which operates over a different channel

31 Cartesian Product of Graphs The topology-basedmodel-ing approach has been commonly used in wireless networksfor various purposes but often in a planar view [8 12 14]and most of the discussions on scheduling routing andchannel assignment are also based on a planar topology [611 13 17] The recent development of MIMO WMNs callsfor a suitable model to describe MIMO-specific propertiesand understand the cooperative activities across differentinterfaces over multiple channels [20] The planar topologycan be used to determine the internode interference [8] but isinsufficient to provide a visual representation for analyzingthe cooperation between links or channels On the otherhand modelingMIMOWMNs as a super graph still remainslargely unexploited except the work in [4] In this paper ourgoal is to develop an effective model to facilitate the analysisof MIMO channel cooperation

We propose to employ the CPG to model WMNs bycombining a triangularmesh of physical nodes and a graph of

fully connected virtual nodes Together with the coordinatesof triangular overlay nodes the CPG model provides aconvenient way to analyze the properties of interferenceavoidance channel assignment and routing path countingThis model retains the independence between orthogonalchannelswhile providing a general approach to analyzing linkbehaviors over multiple channels

Cartesian Product of Graphs Given two graphs 119866 and119867 theCartesian product 119866 times 119867 is a graph such that

(i) the graph 119866times119867 has a vertex set119881(119866) times119881(119867) that isa vertex in119866times119867 is denoted by a pair (V V1015840) V isin 119881(119866)and V1015840 isin 119881(119867)

(ii) any two nodes (119906 1199061015840) and (V V1015840) and 119906 V isin 119881(119866) and1199061015840 V1015840 isin 119881(119867) are adjacent in 119866times119867 if and only if oneof the following holds (a) 119906 = V and 1199061015840 is adjacent toV1015840 in119867 or (b) 1199061015840 = V1015840 and 119906 is adjacent to V in 119866

For illustration Figure 1(a) shows a mesh network of fivephysical nodes (left side) and a graph of two connected virtualnodes (right side) corresponding to a physical node equippedwith two transceivers each operating on a different channel1198881or 1198882 Figure 1(b) shows a channel-layered virtual topology

of the original mesh network modeled by CPGIn this example the CPG of two graphs in Figure 1(a)

results in a two-layered graph in Figure 1(b) A solid directed

4 Mathematical Problems in Engineering

edge in the top or bottom planar meshes in Figures 1(b)1(c) and 1(d) represents a communication link 119897 = (119878 119877)

119888

transmitting data from node 119878 to node 119877 over a channel 119888while a dashed edge has a conflict with some active links

The maximum possible number of concurrent activelinks on a given channel in a mesh is largely affected bythe selection of the senders As shown in Figure 1(b) whennode 119860 in the top layer is sending data on channel 119888

1 the

three neighbors of 119860 that is nodes 119861 119862 and 119863 in thesame layer (over channel 119888

1) cannot send data Furthermore

node 119864 can be selected as a sender but neither of itsneighbors that is nodes 119862 and 119863 can receive data from 119864without interference illustrated as the dashed directed edgesin Figure 1(c) In this case there is no other active link except(119860 119861)

1198881

However if we choose the initial senders properlyas shown in Figure 1(d) there could exist two concurrentinterference-free links on channel 119888

119894 119894 = 1 2 that is (119861 119860)

119888119894

and (119864119863)119888119894

Since all active links must be interference-free on the

same channel at the same time links are generally sparselydistributed in a planar mesh with respect to a certainchannel Note that more links mean better service to trafficrequests The CPG model allows us to consider concurrentpaths in multiple layers over different channels throughradio cooperation among channel layers for a given traffictask hence providing more capacity and higher throughputcompared to the situation with a single channel and radioThis is consistent with the experimental results presentedby Draves et al [16] The proposed CPG model is alignedwell with existing research in terms of interference relationlink activity and network throughput as well as routing andscheduling [6 7 14 19 21] and enables us to conduct deepertheoretical analysis of WMNs

One advantage that CPG brings is to simplify the expres-sion of multichannel links Since different layers (operatingon different channels) are of the same topology the schedul-ing strategy derived in one layer is readily applicable toanother layer For example there exists a certain link distri-bution pattern among concurrent links In CPG it is obviousthat such an interference-free link distribution pattern on onechannel also exists on others

32 Coordinate Scheme A well-designed coordinate schememay facilitate the analysis of WMNs Since all channellayers are of the same topology we only need to design thecoordinate scheme for one layer or channel The channelinformation can be added to the node coordinates to uniquelyidentify a specific layer

Chin et al proposed a coordinate scheme in hexagonalcellular networks where each node has a degree of three[15] Inspired by their work we propose a parallel clustercoordinate scheme in a local triangular mesh with one BSnode and a number of SS nodes with a degree of six Thisscheme can be readily extended to larger networks withmultiple BSs by inserting the BS information to the nodecoordinates

321 Parallel Cluster Coordinate Scheme In a triangularmesh we first define three clusters of parallel lines along three

A

BC

D

E

j

F

BS

i minus 2 i minus 1 i + 1

j + 2

j + 1

j minus 1

j minus 2

k minus 2 k minus 1 k + 1k

i

Figure 2 The parallel cluster coordinate scheme

different directions that is north-east east and south-eastSince each node is a point intersected by three lines eachfrom one of the three clusters we propose a parallel clustercoordinate scheme (PCCS) which uses a 3-tuple (119894 119895 119896) torepresent the coordinates of a node intersected by the 119894th 119895thand 119896th line in the corresponding clusters [5] As illustrated inFigure 2 the BS node with the coordinates (0 0 0) is locatedat the center and the SS nodes 119860 119861 119862 119863 119864 and 119865 whichare one hop away from the BS node have the coordinates119860 = (1 0 1) 119861 = (0 1 1) 119862 = (minus1 minus1 0) 119863 = (minus1 0 minus1)119864 = (0 minus1 minus1) and 119865 = (1 1 0)

Note that not every combination of three integers canrepresent a node in a triangular mesh because some linesdefined by combinatorial 3-tuples do not intersect at acommon point We summarize such lines as follows

Ω = (119894 minus 1 119895 minus 1 119896) (119894 + 1 119895 + 1 119896) (119894 119895 minus 1 119896 + 1)

(119894 119895 + 1 119896 minus 1) (119894 + 1 119895 119896 minus 1) (119894 minus 1 119895 119896 + 1) (1)

Under the proposed PCCS scheme for a given sender(119894 119895 119896) there are six possible receivers which form itsneighbor set119873

119887(119894 119895 119896)

119873119887(119894 119895 119896) = (119894 minus 1 119895 + 1 119896) (119894 + 1 119895 minus 1 119896)

(119894 119895 + 1 119896 + 1) (119894 119895 minus 1 119896 minus 1)

(119894 + 1 119895 119896 + 1) (119894 minus 1 119895 119896 minus 1)

(2)

PCCS retains the symmetric nature of a triangular meshand facilitates the calculation of the distance between a pairof nodes

322 Symmetric Property of PCCS In PCCS the coordinatesof any two nodes that are symmetric with respect to thecentral BS node located at (0 0 0) are negated For examplethe pairs of nodes 119860 and119863 119861 and 119864 and 119862 and 119865 in Figure 2are symmetric and their coordinates are negated from theircounterparts

The symmetric properties of CPG and Cayley graphhave been well studied [22] According to the vertexedge

Mathematical Problems in Engineering 5

transitive properties a link group can be transited to generateanother one in any layer

Similarly due to the symmetric property we are able totransform a link group to another one through rotation Forexample link group of ((0 2 2) (0 1 1))

119888and ((2 minus2 0)

(1 minus1 0))119888is interference-free After a clockwise rotation of

1205873 a new group of interference-free links is of ((2 0 2)(1 0 1))

119888and ((0 minus2 minus2) (0 minus1 minus1))

119888 respectively With

another clockwise rotation of 1205873 the derived interference-free links are ((2 minus2 0) (1 minus1 0))

119888and ((minus2 0 minus2) (minus1 0

minus1))119888 respectively Note that the rotation operation is edge

transitive and it generates a new link group because of thesymmetric property

323 Distance between Two Nodes Asmost traffic is uploaddownload (tofromBS) we need to count the number of hopsfrom a router node to BS in PCCS We have the followingproperties

Property 1 In triangular WMNs with PCCS the minimumnumber ℏ of hops from 119860 = (119894 119895 119896) to BS = (0 0 0) is

ℏ =|119894| +10038161003816100381610038161198951003816100381610038161003816 + |119896|

2= max |119894| 1003816100381610038161003816119895

1003816100381610038161003816 |119896|(3)

which is consistent with the one in [4]

Property 2 Suppose that BS is positioned at an arbitrarylocation (119894

0 1198950 1198960) instead of (0 0 0) Node (119894 119895 119896) can be

translated to (119894 119895 119896) minus (1198940 1198950 1198960) by a translation function

120575 (119894 119895 119896) 997888rarr ((119894 119895 119896) minus (1198940 1198950 1198960)) (4)

In general suppose that (1198940 1198950 1198960) is the destination of a

traffic path we may virtually view (1198940 1198950 1198960) as the BS node

after applying the translation of (4) Then the formula in (3)may facilitate further analysis

Given two nodes 119860 = (1198941 1198951 1198961) and 119861 = (119894

2 1198952 1198962) in

PCCS the distance between 119860 and 119861 denoted by 119889(119860 119861)is the minimum number of hops between them which iscalculated as 119889(119860 119861) = (|119894

1minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2

324 Mapping to 2D Points In order to draw a planar meshwe need to map 3-tuple coordinates (119894 119895 119896) to 2-dimensional(2D) points (119909 119910) To do this we first overlap a rectangularplane coordinate system to the PCCS triangular mesh Letthe 119909-axis overlap the axis 119895 = 0 while keeping the positiverightward direction Meanwhile the 119910-axis passes the BSnode and is vertical to the line 119895 = 0 with a positive upwarddirectionThen we can determine (119909 119910) by projecting (119894 119895 119896)onto axes 119909 and 119910 The PCCS supports a function 119891mapping(119894 119895 119896) to (119909 119910) as follows

Property 3 A one-to-one mapping function 119891 maps nodecoordinates (119894 119895 119896) in PCCS to 2D coordinates (119909 119910)

(119909 119910) = 119891 (119894 119895 119896) =

119909 =119894 + 119896

2119910 = 119895

(5)

For example node (0 1 1) is mapped to (12 1) and(minus1 3 2) is mapped to (12 3) If a node is on the 119910-axis 119894 =minus119896 always holds For example (minus1 2 1) is mapped to (0 2) Ifnode is located on 119896 = 0 in PCCS 119909 = 1198942 For example(minus2 2 0) is mapped to (minus1 2) This property facilitates theplotting of a triangular mesh in PCCS

33 Interference-Free Conditions To analyze the interferencebetween links we need to consider node interference rela-tions which are critical to scheduling links in WMNs Asinterference is an inherent nature for radio media the wire-less communication performance may be severely degradedif radios operate without a proper scheduling scheme [17]

Minimizing interference has been extensively investi-gated in the literature [9 17] Subramanian et al discussedchannel assignment in a multiradio situation [9] Tan et aldesigned algorithms to set up a skeleton ofminimum interfer-ence for a single channel [17] Scheduling links in a coopera-tive way will improve the energy efficiency and reduce colli-sion These discussions assume variable transmitting poweror interface channel switching However the variation oftransmitting power may lead to the variation of networktopology which may cause changes in the interferencerelation Xu et al use a sensing scheme to achieve power effi-ciency for convergence communication [23] Their methodcan help set the initial power in an almost-static WMNtopology while promising interference-free cognitive accesswith link status as busy or idle

When sender 119878 is sending over channel 1198880 neither can

119878 receive data over channel 1198880 nor can its neighbors in its

effective radio coverage send data If a valid neighbor node119877 receives data over channel 119888

0 we have a link (119878 119877)

1198880

Furthermore the neighbors of receiver 119877 cannot send dataover channel 119888

0at the same time Therefore to be link

interference-free in multiradio multichannel environmentsusing PCCS we need to consider three classes of node inter-ferences sender-to-sender receiver-to-receiver and sender-to-receiver

331 Sender-to-Sender Given a certain channel any twosenders must be at least two hops away in a triangular meshto avoid mutual interference which could help construct aninterference-free candidate set for possible senders

Definition 1 A router set 119878119862is called a sender candidate set if

every two nodes in 119878119862are at least two hops away

Given two nodes 119878119894= (119886 119887 119888) and 119878

119895= (119889 119890 119891) in 119878

119862 the

distance between them must satisfy 119889(119878119894 119878119895) ge 2

For a given node 119878119895notin 119878119862 we can add 119878

119895to 119878119862only if it

is at least two hops away to any node 119878119894in 119878119862 We have the

following necessary condition on interference-free links

119889 (119878119894 119878119895) =|119886 minus 119889| + |119887 minus 119890| +

1003816100381610038161003816119888 minus 1198911003816100381610038161003816

2ge 2 (6)

332 Receiver-to-Receiver Two nodes 119877119894and 119877

119895can each act

as a receiver simultaneously if they have a distance larger than1 Similar to the sender candidate set we use 119877

119862to denote

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Stochastic AnalysisInternational Journal of

Page 3: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Mathematical Problems in Engineering 3

(a) Mesh and virtual nodes (b) Layered mesh

(c) Links in a channel (d) Maximal links

AA

A

A

BB

B

B

CC

C

C

DD

D

D

EE

E

E

c1

c2

Figure 1 The CPG-modeled virtual topology of a physical network with five physical nodes and two channels

equipped with 120585 transceivers may have (at most) 120585 simul-taneous links over 120585 orthogonal channels (assuming thatmore than 120585 channels are available) we can split a physicalnode into 120585 fully connected virtual nodes each of which isequipped with a single transceiver This way we are able torepresent the originalWMN as 120585 identical layers of networkseach of which operates over a different channel

31 Cartesian Product of Graphs The topology-basedmodel-ing approach has been commonly used in wireless networksfor various purposes but often in a planar view [8 12 14]and most of the discussions on scheduling routing andchannel assignment are also based on a planar topology [611 13 17] The recent development of MIMO WMNs callsfor a suitable model to describe MIMO-specific propertiesand understand the cooperative activities across differentinterfaces over multiple channels [20] The planar topologycan be used to determine the internode interference [8] but isinsufficient to provide a visual representation for analyzingthe cooperation between links or channels On the otherhand modelingMIMOWMNs as a super graph still remainslargely unexploited except the work in [4] In this paper ourgoal is to develop an effective model to facilitate the analysisof MIMO channel cooperation

We propose to employ the CPG to model WMNs bycombining a triangularmesh of physical nodes and a graph of

fully connected virtual nodes Together with the coordinatesof triangular overlay nodes the CPG model provides aconvenient way to analyze the properties of interferenceavoidance channel assignment and routing path countingThis model retains the independence between orthogonalchannelswhile providing a general approach to analyzing linkbehaviors over multiple channels

Cartesian Product of Graphs Given two graphs 119866 and119867 theCartesian product 119866 times 119867 is a graph such that

(i) the graph 119866times119867 has a vertex set119881(119866) times119881(119867) that isa vertex in119866times119867 is denoted by a pair (V V1015840) V isin 119881(119866)and V1015840 isin 119881(119867)

(ii) any two nodes (119906 1199061015840) and (V V1015840) and 119906 V isin 119881(119866) and1199061015840 V1015840 isin 119881(119867) are adjacent in 119866times119867 if and only if oneof the following holds (a) 119906 = V and 1199061015840 is adjacent toV1015840 in119867 or (b) 1199061015840 = V1015840 and 119906 is adjacent to V in 119866

For illustration Figure 1(a) shows a mesh network of fivephysical nodes (left side) and a graph of two connected virtualnodes (right side) corresponding to a physical node equippedwith two transceivers each operating on a different channel1198881or 1198882 Figure 1(b) shows a channel-layered virtual topology

of the original mesh network modeled by CPGIn this example the CPG of two graphs in Figure 1(a)

results in a two-layered graph in Figure 1(b) A solid directed

4 Mathematical Problems in Engineering

edge in the top or bottom planar meshes in Figures 1(b)1(c) and 1(d) represents a communication link 119897 = (119878 119877)

119888

transmitting data from node 119878 to node 119877 over a channel 119888while a dashed edge has a conflict with some active links

The maximum possible number of concurrent activelinks on a given channel in a mesh is largely affected bythe selection of the senders As shown in Figure 1(b) whennode 119860 in the top layer is sending data on channel 119888

1 the

three neighbors of 119860 that is nodes 119861 119862 and 119863 in thesame layer (over channel 119888

1) cannot send data Furthermore

node 119864 can be selected as a sender but neither of itsneighbors that is nodes 119862 and 119863 can receive data from 119864without interference illustrated as the dashed directed edgesin Figure 1(c) In this case there is no other active link except(119860 119861)

1198881

However if we choose the initial senders properlyas shown in Figure 1(d) there could exist two concurrentinterference-free links on channel 119888

119894 119894 = 1 2 that is (119861 119860)

119888119894

and (119864119863)119888119894

Since all active links must be interference-free on the

same channel at the same time links are generally sparselydistributed in a planar mesh with respect to a certainchannel Note that more links mean better service to trafficrequests The CPG model allows us to consider concurrentpaths in multiple layers over different channels throughradio cooperation among channel layers for a given traffictask hence providing more capacity and higher throughputcompared to the situation with a single channel and radioThis is consistent with the experimental results presentedby Draves et al [16] The proposed CPG model is alignedwell with existing research in terms of interference relationlink activity and network throughput as well as routing andscheduling [6 7 14 19 21] and enables us to conduct deepertheoretical analysis of WMNs

One advantage that CPG brings is to simplify the expres-sion of multichannel links Since different layers (operatingon different channels) are of the same topology the schedul-ing strategy derived in one layer is readily applicable toanother layer For example there exists a certain link distri-bution pattern among concurrent links In CPG it is obviousthat such an interference-free link distribution pattern on onechannel also exists on others

32 Coordinate Scheme A well-designed coordinate schememay facilitate the analysis of WMNs Since all channellayers are of the same topology we only need to design thecoordinate scheme for one layer or channel The channelinformation can be added to the node coordinates to uniquelyidentify a specific layer

Chin et al proposed a coordinate scheme in hexagonalcellular networks where each node has a degree of three[15] Inspired by their work we propose a parallel clustercoordinate scheme in a local triangular mesh with one BSnode and a number of SS nodes with a degree of six Thisscheme can be readily extended to larger networks withmultiple BSs by inserting the BS information to the nodecoordinates

321 Parallel Cluster Coordinate Scheme In a triangularmesh we first define three clusters of parallel lines along three

A

BC

D

E

j

F

BS

i minus 2 i minus 1 i + 1

j + 2

j + 1

j minus 1

j minus 2

k minus 2 k minus 1 k + 1k

i

Figure 2 The parallel cluster coordinate scheme

different directions that is north-east east and south-eastSince each node is a point intersected by three lines eachfrom one of the three clusters we propose a parallel clustercoordinate scheme (PCCS) which uses a 3-tuple (119894 119895 119896) torepresent the coordinates of a node intersected by the 119894th 119895thand 119896th line in the corresponding clusters [5] As illustrated inFigure 2 the BS node with the coordinates (0 0 0) is locatedat the center and the SS nodes 119860 119861 119862 119863 119864 and 119865 whichare one hop away from the BS node have the coordinates119860 = (1 0 1) 119861 = (0 1 1) 119862 = (minus1 minus1 0) 119863 = (minus1 0 minus1)119864 = (0 minus1 minus1) and 119865 = (1 1 0)

Note that not every combination of three integers canrepresent a node in a triangular mesh because some linesdefined by combinatorial 3-tuples do not intersect at acommon point We summarize such lines as follows

Ω = (119894 minus 1 119895 minus 1 119896) (119894 + 1 119895 + 1 119896) (119894 119895 minus 1 119896 + 1)

(119894 119895 + 1 119896 minus 1) (119894 + 1 119895 119896 minus 1) (119894 minus 1 119895 119896 + 1) (1)

Under the proposed PCCS scheme for a given sender(119894 119895 119896) there are six possible receivers which form itsneighbor set119873

119887(119894 119895 119896)

119873119887(119894 119895 119896) = (119894 minus 1 119895 + 1 119896) (119894 + 1 119895 minus 1 119896)

(119894 119895 + 1 119896 + 1) (119894 119895 minus 1 119896 minus 1)

(119894 + 1 119895 119896 + 1) (119894 minus 1 119895 119896 minus 1)

(2)

PCCS retains the symmetric nature of a triangular meshand facilitates the calculation of the distance between a pairof nodes

322 Symmetric Property of PCCS In PCCS the coordinatesof any two nodes that are symmetric with respect to thecentral BS node located at (0 0 0) are negated For examplethe pairs of nodes 119860 and119863 119861 and 119864 and 119862 and 119865 in Figure 2are symmetric and their coordinates are negated from theircounterparts

The symmetric properties of CPG and Cayley graphhave been well studied [22] According to the vertexedge

Mathematical Problems in Engineering 5

transitive properties a link group can be transited to generateanother one in any layer

Similarly due to the symmetric property we are able totransform a link group to another one through rotation Forexample link group of ((0 2 2) (0 1 1))

119888and ((2 minus2 0)

(1 minus1 0))119888is interference-free After a clockwise rotation of

1205873 a new group of interference-free links is of ((2 0 2)(1 0 1))

119888and ((0 minus2 minus2) (0 minus1 minus1))

119888 respectively With

another clockwise rotation of 1205873 the derived interference-free links are ((2 minus2 0) (1 minus1 0))

119888and ((minus2 0 minus2) (minus1 0

minus1))119888 respectively Note that the rotation operation is edge

transitive and it generates a new link group because of thesymmetric property

323 Distance between Two Nodes Asmost traffic is uploaddownload (tofromBS) we need to count the number of hopsfrom a router node to BS in PCCS We have the followingproperties

Property 1 In triangular WMNs with PCCS the minimumnumber ℏ of hops from 119860 = (119894 119895 119896) to BS = (0 0 0) is

ℏ =|119894| +10038161003816100381610038161198951003816100381610038161003816 + |119896|

2= max |119894| 1003816100381610038161003816119895

1003816100381610038161003816 |119896|(3)

which is consistent with the one in [4]

Property 2 Suppose that BS is positioned at an arbitrarylocation (119894

0 1198950 1198960) instead of (0 0 0) Node (119894 119895 119896) can be

translated to (119894 119895 119896) minus (1198940 1198950 1198960) by a translation function

120575 (119894 119895 119896) 997888rarr ((119894 119895 119896) minus (1198940 1198950 1198960)) (4)

In general suppose that (1198940 1198950 1198960) is the destination of a

traffic path we may virtually view (1198940 1198950 1198960) as the BS node

after applying the translation of (4) Then the formula in (3)may facilitate further analysis

Given two nodes 119860 = (1198941 1198951 1198961) and 119861 = (119894

2 1198952 1198962) in

PCCS the distance between 119860 and 119861 denoted by 119889(119860 119861)is the minimum number of hops between them which iscalculated as 119889(119860 119861) = (|119894

1minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2

324 Mapping to 2D Points In order to draw a planar meshwe need to map 3-tuple coordinates (119894 119895 119896) to 2-dimensional(2D) points (119909 119910) To do this we first overlap a rectangularplane coordinate system to the PCCS triangular mesh Letthe 119909-axis overlap the axis 119895 = 0 while keeping the positiverightward direction Meanwhile the 119910-axis passes the BSnode and is vertical to the line 119895 = 0 with a positive upwarddirectionThen we can determine (119909 119910) by projecting (119894 119895 119896)onto axes 119909 and 119910 The PCCS supports a function 119891mapping(119894 119895 119896) to (119909 119910) as follows

Property 3 A one-to-one mapping function 119891 maps nodecoordinates (119894 119895 119896) in PCCS to 2D coordinates (119909 119910)

(119909 119910) = 119891 (119894 119895 119896) =

119909 =119894 + 119896

2119910 = 119895

(5)

For example node (0 1 1) is mapped to (12 1) and(minus1 3 2) is mapped to (12 3) If a node is on the 119910-axis 119894 =minus119896 always holds For example (minus1 2 1) is mapped to (0 2) Ifnode is located on 119896 = 0 in PCCS 119909 = 1198942 For example(minus2 2 0) is mapped to (minus1 2) This property facilitates theplotting of a triangular mesh in PCCS

33 Interference-Free Conditions To analyze the interferencebetween links we need to consider node interference rela-tions which are critical to scheduling links in WMNs Asinterference is an inherent nature for radio media the wire-less communication performance may be severely degradedif radios operate without a proper scheduling scheme [17]

Minimizing interference has been extensively investi-gated in the literature [9 17] Subramanian et al discussedchannel assignment in a multiradio situation [9] Tan et aldesigned algorithms to set up a skeleton ofminimum interfer-ence for a single channel [17] Scheduling links in a coopera-tive way will improve the energy efficiency and reduce colli-sion These discussions assume variable transmitting poweror interface channel switching However the variation oftransmitting power may lead to the variation of networktopology which may cause changes in the interferencerelation Xu et al use a sensing scheme to achieve power effi-ciency for convergence communication [23] Their methodcan help set the initial power in an almost-static WMNtopology while promising interference-free cognitive accesswith link status as busy or idle

When sender 119878 is sending over channel 1198880 neither can

119878 receive data over channel 1198880 nor can its neighbors in its

effective radio coverage send data If a valid neighbor node119877 receives data over channel 119888

0 we have a link (119878 119877)

1198880

Furthermore the neighbors of receiver 119877 cannot send dataover channel 119888

0at the same time Therefore to be link

interference-free in multiradio multichannel environmentsusing PCCS we need to consider three classes of node inter-ferences sender-to-sender receiver-to-receiver and sender-to-receiver

331 Sender-to-Sender Given a certain channel any twosenders must be at least two hops away in a triangular meshto avoid mutual interference which could help construct aninterference-free candidate set for possible senders

Definition 1 A router set 119878119862is called a sender candidate set if

every two nodes in 119878119862are at least two hops away

Given two nodes 119878119894= (119886 119887 119888) and 119878

119895= (119889 119890 119891) in 119878

119862 the

distance between them must satisfy 119889(119878119894 119878119895) ge 2

For a given node 119878119895notin 119878119862 we can add 119878

119895to 119878119862only if it

is at least two hops away to any node 119878119894in 119878119862 We have the

following necessary condition on interference-free links

119889 (119878119894 119878119895) =|119886 minus 119889| + |119887 minus 119890| +

1003816100381610038161003816119888 minus 1198911003816100381610038161003816

2ge 2 (6)

332 Receiver-to-Receiver Two nodes 119877119894and 119877

119895can each act

as a receiver simultaneously if they have a distance larger than1 Similar to the sender candidate set we use 119877

119862to denote

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

4 Mathematical Problems in Engineering

edge in the top or bottom planar meshes in Figures 1(b)1(c) and 1(d) represents a communication link 119897 = (119878 119877)

119888

transmitting data from node 119878 to node 119877 over a channel 119888while a dashed edge has a conflict with some active links

The maximum possible number of concurrent activelinks on a given channel in a mesh is largely affected bythe selection of the senders As shown in Figure 1(b) whennode 119860 in the top layer is sending data on channel 119888

1 the

three neighbors of 119860 that is nodes 119861 119862 and 119863 in thesame layer (over channel 119888

1) cannot send data Furthermore

node 119864 can be selected as a sender but neither of itsneighbors that is nodes 119862 and 119863 can receive data from 119864without interference illustrated as the dashed directed edgesin Figure 1(c) In this case there is no other active link except(119860 119861)

1198881

However if we choose the initial senders properlyas shown in Figure 1(d) there could exist two concurrentinterference-free links on channel 119888

119894 119894 = 1 2 that is (119861 119860)

119888119894

and (119864119863)119888119894

Since all active links must be interference-free on the

same channel at the same time links are generally sparselydistributed in a planar mesh with respect to a certainchannel Note that more links mean better service to trafficrequests The CPG model allows us to consider concurrentpaths in multiple layers over different channels throughradio cooperation among channel layers for a given traffictask hence providing more capacity and higher throughputcompared to the situation with a single channel and radioThis is consistent with the experimental results presentedby Draves et al [16] The proposed CPG model is alignedwell with existing research in terms of interference relationlink activity and network throughput as well as routing andscheduling [6 7 14 19 21] and enables us to conduct deepertheoretical analysis of WMNs

One advantage that CPG brings is to simplify the expres-sion of multichannel links Since different layers (operatingon different channels) are of the same topology the schedul-ing strategy derived in one layer is readily applicable toanother layer For example there exists a certain link distri-bution pattern among concurrent links In CPG it is obviousthat such an interference-free link distribution pattern on onechannel also exists on others

32 Coordinate Scheme A well-designed coordinate schememay facilitate the analysis of WMNs Since all channellayers are of the same topology we only need to design thecoordinate scheme for one layer or channel The channelinformation can be added to the node coordinates to uniquelyidentify a specific layer

Chin et al proposed a coordinate scheme in hexagonalcellular networks where each node has a degree of three[15] Inspired by their work we propose a parallel clustercoordinate scheme in a local triangular mesh with one BSnode and a number of SS nodes with a degree of six Thisscheme can be readily extended to larger networks withmultiple BSs by inserting the BS information to the nodecoordinates

321 Parallel Cluster Coordinate Scheme In a triangularmesh we first define three clusters of parallel lines along three

A

BC

D

E

j

F

BS

i minus 2 i minus 1 i + 1

j + 2

j + 1

j minus 1

j minus 2

k minus 2 k minus 1 k + 1k

i

Figure 2 The parallel cluster coordinate scheme

different directions that is north-east east and south-eastSince each node is a point intersected by three lines eachfrom one of the three clusters we propose a parallel clustercoordinate scheme (PCCS) which uses a 3-tuple (119894 119895 119896) torepresent the coordinates of a node intersected by the 119894th 119895thand 119896th line in the corresponding clusters [5] As illustrated inFigure 2 the BS node with the coordinates (0 0 0) is locatedat the center and the SS nodes 119860 119861 119862 119863 119864 and 119865 whichare one hop away from the BS node have the coordinates119860 = (1 0 1) 119861 = (0 1 1) 119862 = (minus1 minus1 0) 119863 = (minus1 0 minus1)119864 = (0 minus1 minus1) and 119865 = (1 1 0)

Note that not every combination of three integers canrepresent a node in a triangular mesh because some linesdefined by combinatorial 3-tuples do not intersect at acommon point We summarize such lines as follows

Ω = (119894 minus 1 119895 minus 1 119896) (119894 + 1 119895 + 1 119896) (119894 119895 minus 1 119896 + 1)

(119894 119895 + 1 119896 minus 1) (119894 + 1 119895 119896 minus 1) (119894 minus 1 119895 119896 + 1) (1)

Under the proposed PCCS scheme for a given sender(119894 119895 119896) there are six possible receivers which form itsneighbor set119873

119887(119894 119895 119896)

119873119887(119894 119895 119896) = (119894 minus 1 119895 + 1 119896) (119894 + 1 119895 minus 1 119896)

(119894 119895 + 1 119896 + 1) (119894 119895 minus 1 119896 minus 1)

(119894 + 1 119895 119896 + 1) (119894 minus 1 119895 119896 minus 1)

(2)

PCCS retains the symmetric nature of a triangular meshand facilitates the calculation of the distance between a pairof nodes

322 Symmetric Property of PCCS In PCCS the coordinatesof any two nodes that are symmetric with respect to thecentral BS node located at (0 0 0) are negated For examplethe pairs of nodes 119860 and119863 119861 and 119864 and 119862 and 119865 in Figure 2are symmetric and their coordinates are negated from theircounterparts

The symmetric properties of CPG and Cayley graphhave been well studied [22] According to the vertexedge

Mathematical Problems in Engineering 5

transitive properties a link group can be transited to generateanother one in any layer

Similarly due to the symmetric property we are able totransform a link group to another one through rotation Forexample link group of ((0 2 2) (0 1 1))

119888and ((2 minus2 0)

(1 minus1 0))119888is interference-free After a clockwise rotation of

1205873 a new group of interference-free links is of ((2 0 2)(1 0 1))

119888and ((0 minus2 minus2) (0 minus1 minus1))

119888 respectively With

another clockwise rotation of 1205873 the derived interference-free links are ((2 minus2 0) (1 minus1 0))

119888and ((minus2 0 minus2) (minus1 0

minus1))119888 respectively Note that the rotation operation is edge

transitive and it generates a new link group because of thesymmetric property

323 Distance between Two Nodes Asmost traffic is uploaddownload (tofromBS) we need to count the number of hopsfrom a router node to BS in PCCS We have the followingproperties

Property 1 In triangular WMNs with PCCS the minimumnumber ℏ of hops from 119860 = (119894 119895 119896) to BS = (0 0 0) is

ℏ =|119894| +10038161003816100381610038161198951003816100381610038161003816 + |119896|

2= max |119894| 1003816100381610038161003816119895

1003816100381610038161003816 |119896|(3)

which is consistent with the one in [4]

Property 2 Suppose that BS is positioned at an arbitrarylocation (119894

0 1198950 1198960) instead of (0 0 0) Node (119894 119895 119896) can be

translated to (119894 119895 119896) minus (1198940 1198950 1198960) by a translation function

120575 (119894 119895 119896) 997888rarr ((119894 119895 119896) minus (1198940 1198950 1198960)) (4)

In general suppose that (1198940 1198950 1198960) is the destination of a

traffic path we may virtually view (1198940 1198950 1198960) as the BS node

after applying the translation of (4) Then the formula in (3)may facilitate further analysis

Given two nodes 119860 = (1198941 1198951 1198961) and 119861 = (119894

2 1198952 1198962) in

PCCS the distance between 119860 and 119861 denoted by 119889(119860 119861)is the minimum number of hops between them which iscalculated as 119889(119860 119861) = (|119894

1minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2

324 Mapping to 2D Points In order to draw a planar meshwe need to map 3-tuple coordinates (119894 119895 119896) to 2-dimensional(2D) points (119909 119910) To do this we first overlap a rectangularplane coordinate system to the PCCS triangular mesh Letthe 119909-axis overlap the axis 119895 = 0 while keeping the positiverightward direction Meanwhile the 119910-axis passes the BSnode and is vertical to the line 119895 = 0 with a positive upwarddirectionThen we can determine (119909 119910) by projecting (119894 119895 119896)onto axes 119909 and 119910 The PCCS supports a function 119891mapping(119894 119895 119896) to (119909 119910) as follows

Property 3 A one-to-one mapping function 119891 maps nodecoordinates (119894 119895 119896) in PCCS to 2D coordinates (119909 119910)

(119909 119910) = 119891 (119894 119895 119896) =

119909 =119894 + 119896

2119910 = 119895

(5)

For example node (0 1 1) is mapped to (12 1) and(minus1 3 2) is mapped to (12 3) If a node is on the 119910-axis 119894 =minus119896 always holds For example (minus1 2 1) is mapped to (0 2) Ifnode is located on 119896 = 0 in PCCS 119909 = 1198942 For example(minus2 2 0) is mapped to (minus1 2) This property facilitates theplotting of a triangular mesh in PCCS

33 Interference-Free Conditions To analyze the interferencebetween links we need to consider node interference rela-tions which are critical to scheduling links in WMNs Asinterference is an inherent nature for radio media the wire-less communication performance may be severely degradedif radios operate without a proper scheduling scheme [17]

Minimizing interference has been extensively investi-gated in the literature [9 17] Subramanian et al discussedchannel assignment in a multiradio situation [9] Tan et aldesigned algorithms to set up a skeleton ofminimum interfer-ence for a single channel [17] Scheduling links in a coopera-tive way will improve the energy efficiency and reduce colli-sion These discussions assume variable transmitting poweror interface channel switching However the variation oftransmitting power may lead to the variation of networktopology which may cause changes in the interferencerelation Xu et al use a sensing scheme to achieve power effi-ciency for convergence communication [23] Their methodcan help set the initial power in an almost-static WMNtopology while promising interference-free cognitive accesswith link status as busy or idle

When sender 119878 is sending over channel 1198880 neither can

119878 receive data over channel 1198880 nor can its neighbors in its

effective radio coverage send data If a valid neighbor node119877 receives data over channel 119888

0 we have a link (119878 119877)

1198880

Furthermore the neighbors of receiver 119877 cannot send dataover channel 119888

0at the same time Therefore to be link

interference-free in multiradio multichannel environmentsusing PCCS we need to consider three classes of node inter-ferences sender-to-sender receiver-to-receiver and sender-to-receiver

331 Sender-to-Sender Given a certain channel any twosenders must be at least two hops away in a triangular meshto avoid mutual interference which could help construct aninterference-free candidate set for possible senders

Definition 1 A router set 119878119862is called a sender candidate set if

every two nodes in 119878119862are at least two hops away

Given two nodes 119878119894= (119886 119887 119888) and 119878

119895= (119889 119890 119891) in 119878

119862 the

distance between them must satisfy 119889(119878119894 119878119895) ge 2

For a given node 119878119895notin 119878119862 we can add 119878

119895to 119878119862only if it

is at least two hops away to any node 119878119894in 119878119862 We have the

following necessary condition on interference-free links

119889 (119878119894 119878119895) =|119886 minus 119889| + |119887 minus 119890| +

1003816100381610038161003816119888 minus 1198911003816100381610038161003816

2ge 2 (6)

332 Receiver-to-Receiver Two nodes 119877119894and 119877

119895can each act

as a receiver simultaneously if they have a distance larger than1 Similar to the sender candidate set we use 119877

119862to denote

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Mathematical Problems in Engineering 5

transitive properties a link group can be transited to generateanother one in any layer

Similarly due to the symmetric property we are able totransform a link group to another one through rotation Forexample link group of ((0 2 2) (0 1 1))

119888and ((2 minus2 0)

(1 minus1 0))119888is interference-free After a clockwise rotation of

1205873 a new group of interference-free links is of ((2 0 2)(1 0 1))

119888and ((0 minus2 minus2) (0 minus1 minus1))

119888 respectively With

another clockwise rotation of 1205873 the derived interference-free links are ((2 minus2 0) (1 minus1 0))

119888and ((minus2 0 minus2) (minus1 0

minus1))119888 respectively Note that the rotation operation is edge

transitive and it generates a new link group because of thesymmetric property

323 Distance between Two Nodes Asmost traffic is uploaddownload (tofromBS) we need to count the number of hopsfrom a router node to BS in PCCS We have the followingproperties

Property 1 In triangular WMNs with PCCS the minimumnumber ℏ of hops from 119860 = (119894 119895 119896) to BS = (0 0 0) is

ℏ =|119894| +10038161003816100381610038161198951003816100381610038161003816 + |119896|

2= max |119894| 1003816100381610038161003816119895

1003816100381610038161003816 |119896|(3)

which is consistent with the one in [4]

Property 2 Suppose that BS is positioned at an arbitrarylocation (119894

0 1198950 1198960) instead of (0 0 0) Node (119894 119895 119896) can be

translated to (119894 119895 119896) minus (1198940 1198950 1198960) by a translation function

120575 (119894 119895 119896) 997888rarr ((119894 119895 119896) minus (1198940 1198950 1198960)) (4)

In general suppose that (1198940 1198950 1198960) is the destination of a

traffic path we may virtually view (1198940 1198950 1198960) as the BS node

after applying the translation of (4) Then the formula in (3)may facilitate further analysis

Given two nodes 119860 = (1198941 1198951 1198961) and 119861 = (119894

2 1198952 1198962) in

PCCS the distance between 119860 and 119861 denoted by 119889(119860 119861)is the minimum number of hops between them which iscalculated as 119889(119860 119861) = (|119894

1minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2

324 Mapping to 2D Points In order to draw a planar meshwe need to map 3-tuple coordinates (119894 119895 119896) to 2-dimensional(2D) points (119909 119910) To do this we first overlap a rectangularplane coordinate system to the PCCS triangular mesh Letthe 119909-axis overlap the axis 119895 = 0 while keeping the positiverightward direction Meanwhile the 119910-axis passes the BSnode and is vertical to the line 119895 = 0 with a positive upwarddirectionThen we can determine (119909 119910) by projecting (119894 119895 119896)onto axes 119909 and 119910 The PCCS supports a function 119891mapping(119894 119895 119896) to (119909 119910) as follows

Property 3 A one-to-one mapping function 119891 maps nodecoordinates (119894 119895 119896) in PCCS to 2D coordinates (119909 119910)

(119909 119910) = 119891 (119894 119895 119896) =

119909 =119894 + 119896

2119910 = 119895

(5)

For example node (0 1 1) is mapped to (12 1) and(minus1 3 2) is mapped to (12 3) If a node is on the 119910-axis 119894 =minus119896 always holds For example (minus1 2 1) is mapped to (0 2) Ifnode is located on 119896 = 0 in PCCS 119909 = 1198942 For example(minus2 2 0) is mapped to (minus1 2) This property facilitates theplotting of a triangular mesh in PCCS

33 Interference-Free Conditions To analyze the interferencebetween links we need to consider node interference rela-tions which are critical to scheduling links in WMNs Asinterference is an inherent nature for radio media the wire-less communication performance may be severely degradedif radios operate without a proper scheduling scheme [17]

Minimizing interference has been extensively investi-gated in the literature [9 17] Subramanian et al discussedchannel assignment in a multiradio situation [9] Tan et aldesigned algorithms to set up a skeleton ofminimum interfer-ence for a single channel [17] Scheduling links in a coopera-tive way will improve the energy efficiency and reduce colli-sion These discussions assume variable transmitting poweror interface channel switching However the variation oftransmitting power may lead to the variation of networktopology which may cause changes in the interferencerelation Xu et al use a sensing scheme to achieve power effi-ciency for convergence communication [23] Their methodcan help set the initial power in an almost-static WMNtopology while promising interference-free cognitive accesswith link status as busy or idle

When sender 119878 is sending over channel 1198880 neither can

119878 receive data over channel 1198880 nor can its neighbors in its

effective radio coverage send data If a valid neighbor node119877 receives data over channel 119888

0 we have a link (119878 119877)

1198880

Furthermore the neighbors of receiver 119877 cannot send dataover channel 119888

0at the same time Therefore to be link

interference-free in multiradio multichannel environmentsusing PCCS we need to consider three classes of node inter-ferences sender-to-sender receiver-to-receiver and sender-to-receiver

331 Sender-to-Sender Given a certain channel any twosenders must be at least two hops away in a triangular meshto avoid mutual interference which could help construct aninterference-free candidate set for possible senders

Definition 1 A router set 119878119862is called a sender candidate set if

every two nodes in 119878119862are at least two hops away

Given two nodes 119878119894= (119886 119887 119888) and 119878

119895= (119889 119890 119891) in 119878

119862 the

distance between them must satisfy 119889(119878119894 119878119895) ge 2

For a given node 119878119895notin 119878119862 we can add 119878

119895to 119878119862only if it

is at least two hops away to any node 119878119894in 119878119862 We have the

following necessary condition on interference-free links

119889 (119878119894 119878119895) =|119886 minus 119889| + |119887 minus 119890| +

1003816100381610038161003816119888 minus 1198911003816100381610038161003816

2ge 2 (6)

332 Receiver-to-Receiver Two nodes 119877119894and 119877

119895can each act

as a receiver simultaneously if they have a distance larger than1 Similar to the sender candidate set we use 119877

119862to denote

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

6 Mathematical Problems in Engineering

Table 1 Conditions for simultaneous links over one channel

Three neighbor pairs Link interference-free conditionsNodes Nodes Necessary Sufficientin 119878119862

in 119877119862

condition condition119878119894 119878119895

forall119894 = 119895 119889(119878119894 119878119895) ge 2 (1)

119877119894 119877119895

forall119894 = 119895 119889(119877119894 119877119895) ge 1 (2) (1) and (2) and (3)

119878119894

119877119895

forall119894 = 119895 119889(119878119894 119877119895) gt 1 (3)

a receiver candidate set in which any two nodes 119877119894= (119897 119898 119899)

and 119877119895= (119906 119901 119902) satisfy the following condition

119889 (119877119894 119877119895) =|119897 minus 119906| +

1003816100381610038161003816119898 minus 1199011003816100381610038161003816 +1003816100381610038161003816119899 minus 119902

10038161003816100381610038162

ge 1 (7)

Note that both 119878119862and 119877

119862are candidate sets and the

actual sender and receiver sets are a subset of 119878119862and 119877

119862

respectively The relations between nodes largely depend onthe previous selected links

Property 4 If links (1198781 1198771)119888and (119878

2 1198772)119888can be scheduled at

the same time then 119889(1198781 1198782) must satisfy condition (6) and

119889(1198771 1198772)must satisfy condition (7)

Note that Property 4 is only a necessary condition forinterference-free links

333 Sender-to-Receiver Given a certain channel a nodecannot receive data if it is in the effective radio range ofthe sender of any other active link otherwise interferenceoccurs If two links 119897

119894and 119897119895(119894 = 119895) coexist the nodes involved

in these two links must satisfy the following condition

119889 (119878119894 119877119895) =|119886 minus 119906| +

1003816100381610038161003816119887 minus 1199011003816100381610038161003816 +1003816100381610038161003816119888 minus 119902

10038161003816100381610038162

gt 1 119894 = 119895 (8)

where 119878119894is the sender of link 119897

119894and 119877

119895is the receiver of link

119897119895For example in Figure 1(d) since119860rsquos neighbor119863 does not

conflict with link (119861 119860)1198881

119863 can be a receiver of another link(119864119863)

1198881

We summarize three necessary conditions for link coex-

istence in the aforementioned three classes in Table 1We have the following theorem

Theorem 2 Two links 119897119894and 119897119895can be simultaneously sched-

uled (coexist) on the same channel if and only if they satisfy allof the three necessary conditions in Table 1

Theorem 2 is based on the PCCS scheme and the settheory As long as the PCCS node coordinates are givenwe are able to determine the interference between linksThese known conditions are helpful to find as many links aspossible while contributing to concurrent central scheduling

34 Transformations of Link Groups We attempt to findas many coexisting links as possible in a given local areaSince a link can be established only between a valid sendercandidate and a valid receiver candidate the actual (final)

k

j

(i minus 2 j + 3 k + 1) (i j + 3 k + 3)

(i minus 1 j + 3 k + 2)

(i minus 1 j + 2 k + 1) (i j + 2 k + 2)

(i minus 1 j + 1 k) (i + 1 j + 1 k + 2)

(i j + 1 k + 1)

(i j k) (i + 1 j k + 1)

i

Figure 3 Coexisting links in a local area

scheduled link set is a subset of 119878119862times119877119862 Generally coexisting

links are sparsely distributed in the network Based on aknown group of coexisting links around one triangle we wishto obtain a new group of coexisting links through certaintransformations

Starting from a link 1198971= (1198781 1198771)1198881

where 1198781= (119894 119895 119896) and

1198771= (119894 119895 + 1 119896 + 1) we want to set up a dense link group in

a local area The set of 1198781rsquos neighbor nodes is 119873

119887(1198781) = V |

119889(V (119894 119895 119896)) le 1 V isin 119881 The set of 1198771rsquos neighbor nodes is

119873119887(1198771) = V | 119889(V (119894 119895 + 1 119896 + 1)) le 1 forallV isin 119881 As any

neighbor node of the sender or the receiver of an active linkshould remain silent if it does not receive data from the sender1198781 to avoid interference with 119897

1 nodes in119873

119887(119894 119895 119896) cup119873

119887(1198771)

cannot send data on 1198881when 119897

1is active

To expand the active link group containing 1198971= (1198781 1198771)1198881

two nodes 119878

2= (119894 119895 + 3 119896 + 3) and 119877

2= (119894 119895 + 2 119896 + 2)

which are not in119873119887(1198781) and119873

119887(1198771) may negotiate for a new

link If successful (1198782 1198772)1198881

is added to the active link groupAccording to the conditions inTheorem 2 link 119897

2= (1198782 1198772)1198881

can coexist with 1198971

Any node in 119873119887(1198771) cup 119873

119887(1198772) cannot be a new sender

except 1198781and 119878

2 However it is completely different on the

receiver side where one node may be the receiver of a newlink even if it is in119873

119887(1198771)cap119873119887(1198772) For example in Figure 3

1198783= (119894minus2 119895+3 119896+1) and 119877

3= (119894minus1 119895+2 119896+1) form a new

link 1198973= (1198783 1198773)1198881

which can coexist with 1198971and 1198972 Around

the central triangle in Figure 3 we find three interference-freelinks coexisting over one channel

Although the three nodes of the central triangle areinvolved in three active links as shown in Figure 3 neither ofthese three nodes can act as a sender while the other two arereceiving data from their respective senders To obtain thesame number of new coexisting links we need to keep thereceivers unchanged while considering a certain switchingto the three senders of active links 119897

1 1198972 and 119897

3

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Mathematical Problems in Engineering 7

By passing the sending token from the current senderto its neighbor node clockwise while keeping the originalreceiver we are able to establish three possible new links

1198971015840

1= ((119894 minus 1 119895 + 1 119896) (119894 119895 + 1 119896 + 1))

1198881

1198971015840

2= ((119894 + 1 119895 + 2 119896 + 3) (119894 119895 + 2 119896 + 2))

1198881

1198971015840

3= ((119894 minus 1 119895 + 3 119896 + 2) (119894 minus 1 119895 + 2 119896 + 1))

1198881

(9)

Since the sender (119894 minus 1 119895 + 3 119896 + 2) of 11989710158403is one hop

away from the receiver (119894 119895 + 2 119896 + 2) of 11989710158402 which violates

the third condition inTheorem 2 11989710158402and 11989710158403cannot coexist on

channel 1198881 which means that 1198971015840

1 11989710158402 and 1198971015840

3cannot coexist In

other words switching the senders of interference-free linksmay incur new interferences Therefore we must performinterference check after switching the sender of an active link

Sender switching may generate a new dense link group ina local mesh For example we obtain a new link 119897lowast

1= ((119894 +

1 119895 119896 + 1) (119894 119895 + 1 119896 + 1))1198881

by switching the sender of link 1198971

to its neighbor node anticlockwise The three links 119897lowast1 1198972 and

1198973are interference-free Similarly link 119897lowast

3= ((119894minus2 119895+2 119896) (119894minus

1 119895 + 2 119896 + 1))1198881

is also interference-free with links 1198971and 1198972

Since triangular meshes possess symmetric propertiestransformations such as rotation and translation can retainthe interference-free features which may save computingtime in finding new link groups [24 25]

35 Node Identity Number Assignment The nodes in tri-angular meshes can be viewed not only as wireless routernodes but also as resources or data sets Assigning a uniqueidentification (ID) number to each node brings severalbenefits For example such IDs can help to locate or identifynodes quickly for various administration or maintenancepurposes

Given the coordinates (119894 119895 119896) of a node in PCCS thereexists a general function 119892 that maps (119894 119895 119896) to a uniqueinteger For example in Figure 2 we canmap BS = (0 0 0) to0 while 119861 = (1 0 minus1) to 1 and 119862 = (minus1 1 0) to 2

Definition 3 In PCCS119872-circle in a triangular mesh is a setof nodes that have exact distance of119872 hops to the node BS

Suppose that (119894 119895 119896) is in119872-circle where119872 = max|119894||119895| |119896| according to (3) To construct a mapping function119892(119894 119895 119896) we classify nodes according to their coordinates Forany node 119860 = (119894 119895 119896) we consider the following seven casesin Figure 4 which is a logical route extracted from Figure 2The first and special case119872 = 0 119892(0 0 0) = 0 The node IDsin119872-circle increase along an anticlockwise direction Thesecases are applicable to any 119872-circle The parallel solid linesegments are corresponding to identity counting piecewisefunctions for example the 2nd case includes three parallelsegments of different119872-circle in Figure 4 in which nodes onthese segments get their ID number following the 2nd one in(11) The dashed line means that the inner node is countedalready and the outer node is the new start of the next circle

Case 1

Case 2

Case 3

Case 4

Case 5

Case 6

Case 7

Figure 4 The cases for node identity number assignment

To assign an identity to each node on119872-circle we needto know the total number 120581 of nodes inside this circle

120581 = 1 + 6 (1 + 2 + sdot sdot sdot + 119872 minus 1) = 1 + 3119872 (119872 minus 1) (10)

LetΔ = 3119872(119872minus1) we have 120581 = Δ+1 For example thereis one node inside circle119872 = 1 there are Δ + 1 = 7 nodesinside circle119872 = 2 and there are Δ + 1 = 18 + 1 = 19 nodesinside circle119872 = 3

The number of nodes on119872-circle depends on the valueof119872 According to (11) for any119872-circle with the six solidline segments in the order as shown by arrows in Figure 4we assign each node a unique integer on 119872-circle with anID number in Δ + 1 Δ + 2 Δ + 119895 + 1 Δ + 6119872After overlapping Figure 4 on Figure 2 a node is assignedwith its identity number in one of the seven cases in (11)corresponding to the line segment case in Figure 4

119892 (119894 119895 119896) =

0 119872 = 0 1stΔ + 119895 + 1 119872 = 0 and119872 = 119896 2ndΔ +119872 minus 119894 + 1 119872 = 0 and119872 = 119895 3rdΔ + 2119872 minus 119896 + 1 119872 = 0 and119872 = minus119894 4thΔ + 3119872 minus 119895 + 1 119872 = 0 and119872 = minus119896 5thΔ + 4119872 + 119894 + 1 119872 = 0 and119872 = minus119895 6thΔ + 5119872 + 119896 + 1 119872 = 119894 and 119894 = 119896 0 7th

(11)

For example in the 6th case node (1 minus4 minus3) is assignedwith an ID number of 3 times 4 times (4 minus 1) + 4 times 4 + 1 + 1 = 54

Some nodes may satisfy two cases in (11) while getting thesame ID For example node (0 3 3) which is on the 3-circleand the line segment in parallel with 119896-axis is assigned withan ID number of 3 times119872times (119872minus1) + 119895 + 1 = 3 times 3 times 2 + 3 + 1 =22 in the 2nd case in (11) Meanwhile node (0 3 3) as it ison one line segment of Case 3 in Figure 4 can get its ID of3 times 119872 times (119872 minus 1) + 119872 minus 119894 + 1 = 3 times 3 times 2 + 3 + 1 = 22 byfollowing the 3rd case in (11)

We shall provide more explanations for the 2nd and7th cases to facilitate a better understanding of (11) In the2nd case where 119872 = 0 and 119872 = 119896 the coordinate 119896 ofnode (119894 119895 119896) must be positive Node (119894 119895 119896) on this119872-circle(119872 = 119896) segment should be assigned with an ID number in

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

8 Mathematical Problems in Engineering

Δ+1 Δ+2 Δ+119895+1 Δ+119872 In this case node (119894 119895 119896)is located above the lines 119895 = 0 and 119896 gt 0 as shown in Case 2in Figure 4 Hence with the increasing 119895 each node (119894 119895 119896) isassigned with an ID number 119892(119894 119895 119896) = Δ + 119895 + 1 Generallyon each119872-circle (119872 gt 0) from the 2nd case to the 7th caseeach with119872 nodes the total number of nodes is 6119872

In the 7th case we need to avoid repeatedly counting thenode on line 119895 = 0 as shown in Figure 2The number of hopsto the BS is 119872 = 119894 The last node on this circle is on a linesegment of the 119894-axis parallel cluster with 119895 = 0 according toFigure 4 The condition 119872 = 119894 and 119894 = 119896 excludes the nodesalready counted in the previous Case 2 in this circle while119872 = 119894 and 119894 = 0 prevents the 1st case from being reconsideredMeanwhile the condition119872 = 119896 ensures that we count nodeson the next circle following the 2nd case again where119872 = 119896In the 7th case a node can be assigned with an ID number ina finite set Δ+5119872+1 Δ+5119872+119896+1 Δ+6119872 Withthe increasing 119896 where 119896 = 0 1 119872 minus 1 node (119894 119895 119896) onthis segment is assigned with an ID number ofΔ+5119872+119896+1

Note that the mapping function 119892(119894 119895 119896) is a segmentedlinear function which is invertible for any finite set

4 Path Counting

Routing is one fundamental problem in WMNs To developa good routing scheme one needs to know the number ofalternative paths and the number of channel assignmentsfor a given pair of source and destination We discuss pathfinding and realization based on the proposed CPG modeland coordinate schemeThe total number of shortest paths ingrid meshes was discussed in [18] In this section we tacklethe path counting problem in triangular meshes

We use 119889(119878 119863) to denote the distance between source 119878 =(1198941 1198951 1198961) and destination 119863 = (119894

2 1198952 1198962) Firstly 119889(119878 119863) =

(|1198941minus 1198942| + |1198951minus 1198952| + |1198961minus 1198962|)2 In order to transmit data

from 119878 to 119863 we need to select one path from 119875(119878119863) whichis the set of all shortest paths from 119878 to119863

41 PathAlternatives Every step along the shortest path from119878 to 119863 is one hop forward in one of the directions 119894 119895 and 119896The two smaller numbers of |119894

1minus1198942| |1198951minus1198952| |1198961minus1198962| indicate

the lines of parallel clusters that form a grid mesh for pathselection as illustrated in the grid of dashed lines in Figure 5We refer to the two corresponding directions from 119878 to 119863 inthe grid mesh as the correct directions

The correct directions ensure that the data is transmittedthrough one of the shortest path where every hop selectionmakes one hop closer to the destination Through the useof correct directions we are able to reduce the problem ofcounting all shortest paths from 119878 to 119863 in triangular meshesto a problem in grid meshes

We provide an example in Figure 5 to count the numberof paths where |119894

1minus1198942| = 2 |119895

1minus1198952| = 3 and |119896

1minus1198962| = 5The

grid with dashed lines contains all the shortest paths from 119878to119863

In a simple situation where one of |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| is 0 the dashed grid degrades to a line Therefore there

is only one shortest path available

D

XW

U

V

S

Figure 5 The shortest path alternatives in triangular meshes

For convenience let weierp = |1198941minus 1198942| |1198951minus 1198952| |1198961minus

1198962| The correct directions are consistent with min(weierp) (the

minimum of three elements in weierp) and mid(weierp) (the middleof three elements in weierp) To select a shortest path it isnecessary to remove the direction corresponding to max(weierp)(the maximum of three elements in weierp) otherwise it wouldlead to a longer path

If the smallest in |1198941minus 1198942| |1198951minus 1198952| |1198961minus 1198962| is 1 or 2

in the corresponding direction the sender and receiver mustbe on two neighbor parallel lines or two parallel lines withone line between themWe can use the corresponding grid tocalculate the number 119901

(119878119863)of paths from 119878 to119863 as follows

119901(119878119863)=

1 if min (weierp) = 0mid (weierp) + 1 if min (weierp) = 1(mid (weierp) + 1) (mid (weierp) + 2)

2if min (weierp) = 2

(12)

For example in Figure 5 the number of paths from 119878 to119863is determined by two directions 119894 and 119895 as |119894

1minus1198942| = 2 and |119895

1minus

1198952| = 3 are smaller than |119896

1minus1198962| = 5The total number of path

alternatives from 119878 to 119863 is 10 However one step along thedirection of 119896 (it becomes either 119896+1 or 119896minus1) obviously leadsto a longer path We present two more lemmas on direction-related properties as follows

Lemma 4 On a shortest path the coordinate displacementsbetween 119878 and 119863 along the three directions satisfy

min (weierp) +mid (weierp) = max (weierp) (13)

Proof The number of hops on a path defined by the correctdirection of min(weierp) and mid(weierp) is min(weierp) +mid(weierp) and thepath traverses exactlymin(weierp)+mid(weierp)+1 different points Aseach node is intersected by three lines each from one clusterone of the three lines must belong to the max(weierp) cluster Itfollows that the path traverses min(weierp) + mid(weierp) + 1 linesin the max(weierp) cluster Hence the displacement of the twoline numbers (ie max(weierp)) traversing 119878 and 119863 is equal tothe height of the tree (min(weierp) + mid(weierp) + 1) minus 1 We havemax(weierp) = (min(weierp)+mid(weierp)+1)minus1 = min(weierp)+mid(weierp)

Lemma 5 The distance from source 119878 to destination119863 is

119889 (119878 119863) = min (weierp) +mid (weierp) (14)

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Mathematical Problems in Engineering 9

The proof of Lemma 5 simply follows the definition of119889(119878 119863) and Lemma 4

In triangular WMNs given a source-destination pair 119878and 119863 the number of shortest paths satisfies the followingtheorem

Theorem 6 When min(weierp) gt 0 the number 119901(119878119863)

of paths isdetermined in a grid of the two correct directions correspondingtomin(weierp) andmid(weierp) as follows

119901(119878119863)= (119889 (119878 119863)min (weierp)) (15)

which is the number of min(weierp) combinations chosen from119889(119878 119863) objects

Proof Without loss of generality let 119894 and 119895 be the correctdirections Note that 119894 is determined by min(weierp) while 119895 isdetermined by mid(weierp) The coordinates of the next receivernode would lead to one hop closer to the destination alongthe direction 119894 or 119895

The number 119901(119878119863)

of paths is equal to the number ofstrings of 119894rsquos and 119895rsquos 119894 is repeated min(weierp) times and 119895 isrepeated mid(weierp) times in a permutation of min(weierp) +mid(weierp)elements that is 119889(119878 119863) in Lemma 5 The total number ofpermutations is 119889(119878 119863) = (min(weierp) + mid(weierp)) but withrepetitions Note that the same permutated strings can onlybe counted once The number of duplicated permutations ismin(weierp) and mid(weierp) in the direction of 119894 and 119895 respectivelyThen the total number of different paths is obtained bydividing the total number of permutations by the number ofduplications in both directions

119889 (119878 119863)

(min (weierp) timesmid (weierp))= (119889 (119878 119863)min (weierp)) = (

119889 (119878 119863)mid (weierp)) (16)

For example in Figure 5 weierp = 2 3 5 with 119894 and 119895 beingthe correct directions where 119894 appears min(weierp) = 2 times and119895 appears mid(weierp) = 3 times in every permutation A validrouting path is determined by the number of 119894rsquos and 119895rsquos aswell as their relative positions in the string For example 119894119895119895119894119895119894119894119895119895119895 and 119894119895119894119895119895 are all valid paths with two 119894rsquos and three 119895rsquosThepath 119901

1= 119878 rarr 119880 rarr 119881 rarr 119882 rarr 119883 rarr 119863 which is in the

set 119875(119878119863) of all shortest paths from 119878 to119863 can be expressedas a constrained permutation of the two correct directionsthat is 119894119895119895119894119895

Given the node coordinates data packets are deliveredhop by hop along the correct directions taking the receiver ofthe current hop as the sender of the next hop until max(weierp) =0 This can be done recursively and may help avoid theoverhead of maintaining a routing table

42 Path Counting with Channel Assignment For a givenpath from 119878 to119863with a constant number of available orthogo-nal channels we need to decide the number of feasiblechannel assignment schemes for implementing this path byusing three channels

D

XW

U

V

Sc0c1c2c3c4

Figure 6 Channel assignment to realize a path from 119878 to119863

To illustrate this problem we show a path with a channelassigned to each hop in Figure 6 which is derived fromFigure 5 119878 rarr

1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863 where thelabel 119888

119894between two neighbor nodes is the channel assigned

to the corresponding linkThe directed edges distributed in different layers form a

feasible orthogonal channel assignment to a path from 119878 to119863A valid path realization allows simultaneous transmission ofall the component links on the path For example in Figure 6links (119878 119880)

1198882

(119882119883)1198882

(119880 119881)1198881

(119883119863)1198881

and (119881119882)1198883

cancoexist at the same time

Let the correct directions be 119894 and 119895 Given 120603 orthogonalchannels the first hop has ( 120603

1) channel choices the second

hop has ( 120603minus11) channel choices and the third hop has ( 120603minus2

1)

channel choices Then the three selected channels can beused repeatedly without interference that is the fourth hopselects the same channel as the first hop and the fifth hopselects the same channel as the second hop and so onChannel 119888

119894can be selected at most lceil119889(119878 119863)3rceil times

In realizing a path using three channels at time 119905 the firstthree hops determine the channel assignment order in everythree downstream hops For example if the first three hopsare arranged in a channel order of 119888

2 1198881 and 119888

3 then the

second three hops should be assigned channels in the sameorder Otherwise a realization of the path from 119878 to119863 wouldrequire more channels For example with 119889(119878 119863) = 4 if thefirst three hops use channels in the order of 119888

2 1198881 and 119888

3 the

fourth hop cannot use channels 1198881and 1198883 but only 119888

2or a new

channelNote that using 3 channels the channel assignment

permutations of the first three hops (ie 3) are all possibleschemes for downstream three-hop groups

Lemma7 Whenmin(weierp) = 0 with120603 orthogonal channels thenumber of valid channel assignment schemes for a given pathfrom source 119878 to destination119863 using three channels at the sametime is

120582 = (1206033) times 3 (17)

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

10 Mathematical Problems in Engineering

Based onTheorem 6 and Lemma 7 we have the followingtheorem

Theorem 8 When using three channels to realize a path thenumber of channel assignments for all shortest paths from 119878 to119863 is

(119889 (119878119863)min (weierp)) sdot 120582 (18)

Proof Whenmin(weierp) = 0 there is only one shortest path andthe number of channel assignments for this path is obtainedby Lemma 7

We focus on a general case where min(weierp) gt 0 Thenumber of channel assignments can be counted in twoindependent steps

The first step is to count all shortest paths from 119878 to119863 in aplane mesh This step does not assign channels For exampleFigure 5 shows one such path 119878 rarr 119880 rarr 119881 rarr 119882 rarr119883 rarr 119863

The second step assigns channels to the selected pathusing three channels without interference For exampleFigure 6 shows the channel assignment for 119901

1from 119878 to 119863

in Figure 5 119878 rarr1198882

119880 rarr1198881

119881 rarr1198883

119882 rarr1198882

119883 rarr1198881

119863Since the above two steps are independent by the mul-

tiplication principle the total number of path realizationschemes satisfies (18) This result could be extended todifferent numbers of channels assigned to one path

5 Conclusion

We conducted a theoretical exploration on mathematicalmodels and combinatorial characteristics of MIMO WMNsFor a single-channeled mesh we designed a coordinatescheme and a node identity assignment scheme and derivedthe interference-free conditions FormultiradiomultichannelWMNs we derived rigorous formulas to count the number ofshortest paths from source to destination

It is of our future interest to find some transformationsto generate new link groups from the known ones Alongthis direction we plan to investigate the CPG vertexedgetransitive properties for performance improvement

Conflict of Interests

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

Acknowledgments

The research is funded in part by Nature Science project ofGuangdong Province under Grant no S2012040010974 andChina Scholarship Council no 201306755013This research isalso partly sponsored by US Department of Energyrsquos Officeof Science under Grant no DE-SC0010641 with University ofMemphis

References

[1] IEEE Standards Association ldquoIEEE std 80216-2004 for localand metropolitan area networks part 16 air interface for fixedandmobilewireless access systemsrdquoOctober 2004 httpstand-ardsieeeorggetieee802download80216-2004pdf

[2] I F Akyildiz XWang andWWang ldquoWirelessmesh networksa surveyrdquo Computer Networks vol 47 no 4 pp 445ndash487 2005

[3] H Huang X Cao X Jia and X Wang ldquoChannel assignmentusing block design in wireless mesh networksrdquo Computer Com-munications vol 32 no 7-10 pp 1148ndash1153 2009

[4] Z Cao W Xiao and L Peng ldquoA MeshtimesChain graph model forMIMO scheduling in IEEE80216 WMNrdquo in Proceedings of the2nd IEEE International Conference on Computer Modeling andSimulation (ICCMS rsquo10) vol 2 pp 547ndash551 IEEE Sanya ChinaJanuary 2010

[5] Z M Cao Y P Zhang Z L Shan and Y C Jiang ldquoA schemeto address routers in MIMO triangular wireless overlay meshrdquoApplied Mechanics and Materials vol 427ndash429 pp 2584ndash25872013

[6] B Han W Jia and L Lin ldquoPerformance evaluation of schedul-ing in IEEE 80216 based wireless mesh networksrdquo ComputerCommunications vol 30 no 4 pp 782ndash792 2007

[7] F Jin A Arora J Hwang and H-A Choi ldquoRouting and packetscheduling in WiMAX mesh networksrdquo in Proceedings of the4th International Conference on Broadband CommunicationsNetworks Systems (BROADNETS rsquo07) pp 574ndash582 RaleighNC USA September 2007

[8] Q Liu X Jia and Y Zhou ldquoTopology control for multi-channel multi-radio wireless mesh networks using directionalantennasrdquoWireless Networks vol 17 no 1 pp 41ndash51 2011

[9] A P Subramanian H Gupta S R Das and J Cao ldquoMinimuminterference channel assignment in multiradio wireless meshnetworksrdquo IEEE Transactions on Mobile Computing vol 7 no12 pp 1459ndash1473 2008

[10] IEEE Standards Association IEEE Standard for Local andMetropolitan Area NetworksmdashPart 21 Media Independent Han-dover Services IEEE Standards Association 2008 httpstandardsieeeorggetieee802download80221-2008pdf

[11] H Viswanathan and SMukherjee ldquoThroughput-range tradeoffof wireless mesh backhaul networksrdquo IEEE Journal on SelectedAreas in Communications vol 24 no 3 pp 593ndash602 2006

[12] J Robinson and E W Knightly ldquoA performance study ofdeployment factors in wireless mesh networksrdquo in Proceedingsof the IEEE 26th IEEE International Conference on ComputerCommunications (INFOCOM rsquo07) pp 2054ndash2062 AnchorageAlaska USA May 2007

[13] K Hong and Y Hua ldquoThroughput of large wireless networks onsquare hexagonal and triangular gridsrdquo in Proceedings of the 4thIEEE Workshop on Sensor Array and Multichannel Processingpp 461ndash465 2006

[14] M K Marina S R Das and A P Subramanian ldquoA topologycontrol approach for utilizing multiple channels in multi-radiowireless mesh networksrdquo Computer Networks vol 54 no 2 pp241ndash256 2010

[15] F Y L Chin Y Zhang and H Zhu ldquoA 1-local 139-competitivealgorithm for multicoloring hexagonal graphsrdquo in Computingand Combinatorics vol 4598 of Lecture Notes in ComputerScience pp 526ndash536 Springer Berlin Germany 2007

[16] R Draves J Padhye and B Zill ldquoRouting in multi-radiomulti-hop wireless mesh networksrdquo in Proceedings of the 10th

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Mathematical Problems in Engineering 11

ACM Annual International Conference on Mobile Computingand Networking (MobiCom rsquo04) pp 114ndash128 October 2004

[17] H Tan T Lou Y Wang Q-S Hua and F C Lau ldquoExact algo-rithms to minimize interference in wireless sensor networksrdquoTheoretical Computer Science vol 412 no 50 pp 6913ndash69252011

[18] Z Cao and W Xiao ldquoAn algorithm to generate regularmesh topology for wireless networksrdquo International Journal ofAdvancements in Computing Technology vol 3 no 3 pp 123ndash133 2011

[19] L Iannone R Khalili K Salamatian and S Fdida ldquoCross-layerrouting in wireless mesh networksrdquo in Proceedings of the 1stInternational Symposium on Wireless Communication Systems(ISWCS rsquo04) pp 319ndash323 September 2004

[20] E G Larsson O Edfors F Tufvesson and T L MarzettaldquoMassive MIMO for next generation wireless systemsrdquo IEEECommunications Magazine vol 52 no 2 pp 186ndash195 2014

[21] H Shetiya and V Sharma ldquoAlgorithms for routing and central-ized scheduling in IEEE 80216 mesh networksrdquo in Proceedingsof the IEEE Wireless Communications and Networking Confer-ence (WCNC rsquo06) vol 1 pp 147ndash152 Las Vegas Nev USA April2006

[22] C Godsil and G Royle Algebraic Graph Theory vol 207Springer New York NY USA 2001

[23] Z Xu W Qin Q Tang and D Jiang ldquoEnergy-efficient cogni-tive access approach to convergence communicationsrdquo ScienceChina Information Sciences vol 57 no 4 pp 1ndash12 2014

[24] Z Cao and L Peng ldquoDestination-oriented routing and max-imum capacity scheduling algorithms in cayley graph modelfor wireless mesh networkrdquo Journal of Convergence InformationTechnology vol 5 no 10 pp 82ndash90 2010

[25] Z M Cao and J C Tang ldquoRouting methods and schedulingpatterns inMIMOWMNvirtualmodelrdquoAppliedMechanics andMaterials vol 519-520 pp 216ndash221 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article On Modeling and Analysis of MIMO Wireless ...downloads.hindawi.com/journals/mpe/2015/185262.pdf · it may simplify the tasks of routing, scheduling, and channel assignment

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of