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Int. J.IndustrialMathematisVol. 3,No. 1(2011)41-53

Ranking EÆient Units by Regular Polygon Area

(RPA) in DEA

F.RezaiBalf

Department ofMathematis,QaemshahrBranh, IslamiAzadUniversity,Qaemshahr,Iran.

Reeived 29September2010;revised29January2011;aepted9February2011.

|||||||||||||||||||||||||||||||-Abstrat

In this paper, we address the problem of assessing the rank of the set of eÆient units

inData Envelopment Analysis (DEA). DEAmeasures theeÆienies of deision-making

units(DMUs)withintherangeoflessthanorequaltoone. TheorrespondingeÆienies

are referred to as the best relative eÆienies, whih measure the best performane of

DMUsand determinean eÆieny frontier. Thisresearh proposesa methodologyto use

anommonset ofweightsfortheperformaneindiesofonlyDEAeÆient DMUs. Then

DMUsareranked aordingtotheeÆienysoreweightedbytheommonsetofweights

intwo dimensionalspae.

Keywords: DataEnvelopmentAnalysis(DEA); WeightsAnalysis; EÆieny;Ranking.

||||||||||||||||||||||||||||||||{

1 Introdution

DataEnvelopmentAnalysis(DEA)isanonparametrimethodofmeasuringtheeÆieny

of a deision-making unit (DMU) suh as a rm, rst introdued by Charnes, Cooper,

and Rhodes (CCR) (1978). To measure the tehnial eÆieny of any observed

input-outputbundle,oneneeds toknowthemaximumquantityofoutputthatanbeprodued

from the relevant inputbundle. However, in DEAwe onstrut a benhmarktehnology

from theobserved input-output bundlesof thermsin thesample, suh asa prodution

frontier. Justifying eah unit on frontier is interpreted as eÆieny and any deviation

from this frontier is interpreted as ineÆieny. Firms that are found to be tehnially

ineÆientanberankedaspetoftheirmeasuredlevelsofeÆieny. Firmsthatarefound

to be eÆient are, also, all ranked equally bya riterion. Andersen and Petersen (1993)

suggested ariterion that permitsone to rank order rmsthat have allbeenfoundto be

at 100It is worthwhile that noted AP model an be infeasible, sometimes. A potential

(2)

problem of feasibility with these supper eÆieny models has been noted by Dula and

Hikman(1997), Seiford and Zhu (1999), Harker and Xue (2002), and Lovel and Rouse

(2003). Forsome eÆient observations,there maynotexist anyinput-orientedor

output-orientedprojetiononto a frontierthatisonstrutedfrom theremainingobservationsin

thedataset. Inthisstudy,weintrodueadierentapproahforrankingeÆientunits. In

ourapproah,we aim to obtainone ommonset of weights(CSW) to reate theidentity

ritial of projetion the eÆient units in two dimensional spae. Then, we alulate

the area of the regular polygon onstruter of all eÆient units. However, we rank the

projetedeÆientunits. Therankingthatadoptstheoneommonsetofweightsgenerated

from ourmethodology makessensebeause a deisionmakerobjetivelyhooses theone

ommon set of weights for the purpose of maximizing the group eÆieny. The seond

setion of thispaperrepresents disussionaboutregular polygonarea (RPA). In Setion

3, a method for nding CSW is briey disussed. Setion 4, present a brief disussion

aboutsupper-eÆienyranking tehniques. Setion5,givesa ompleterankingofDMUs

byRPA method. Numerial exampleand onlusionof themethodsare presentedinthe

lastsetions.

2 Regular polygon area (RPA)

In thissetion, we present a rulefor alulating of regular polygonarea (RPA). Forthis

purpose,werstonsideratriangleassimplestregularpolygonandthenintroduegeneral

ase for nding regular polygon area. Consider the Cartesian oordinates system intwo

dimensional spaes. Four possible ases exist for appointing origin, when we depit a

triangleinthissystem. Firstase, the originis one ofvertexes, seond ase theoriginlie

inside triangle,in thirdase theorigin lieoutsidetriangle and nallythe lastase origin

lieonone oftriangleedges. However, wewillshowtheareaofatriangleanbewrittenas

determinantform foreah ofabovequaternionases. Also,itan be shownthatifpoints

O;A and B be triangle vertexes in anti-lok wise sense respetively, then determinant

valueispositiveand itisnegative whenthese pointsarelokwise. It shouldbenotethe

valueof areais positive.

Theorem2.1. IfpointsO(0;0);A(x

1 ;y

1

)andB(x

2 ;y

2

)beoordinatesoftrianglevertexes

then the area of triangle 4OAB determine as follows:

S

4OAB =

1

2

x

1 y

1

x

2 y

2

Proof: Consider4OAB , aording to Fig. 1.

6

-o x

y

H

H

H

H

H

H

x

1

D y

2 x

2

y

1 A(x

1 ;y

1 ) B(x

2 ;y

2 )

x

1 x

2

y

2 y

(3)

Obviously,ndingthearea4OABequaltosubtratingtheareathreetriangles4OAD

,4OFB and4EAB ofthe retangulararea ODEF . Hene, itan beshown that

S 4OAB =S ODEF S 4OFB S 4OAD S 4EAB =x 1 y 2 x2y2 2 x1y1 2 (x 1 x 2 )(y 2 y 1 ) 2 =x 1 y 2 x 2 y 2 2 x 1 y 1 2 x 1 y 2 x 1 y 1 x 2 y 2 +x 2 y 1 2 = x 1 y 2 x 2 y 1 2 = 1 2 x 1 y 1 x 2 y 2

The proofis omplete.

Inontinueweextendthistopiwhenoriginisnottrianglevertex(SeeFig. 2andFig.

3).

Theorem 2.2. If points A(x

1 ;y

1 );B(x

2 ;y

2

) and C(x

3 ;y

3

) be arbitrarily oordinates of

triangle vertexes in anti-lok wise sense then the area of triangle 4ABC is determined

as follows: S 4ABC = 1 2 x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 + x 3 y 3 x 1 y 1

Proof: Forproof onsider twobelowases:

1. The originisoutside triangle(see Fig. 2)

2. The originisinside triangle(see Fig. 3)

In rstase aording to Fig. 2,we obtain:

S 4ABC =S 4OAB +S 4OBC S 4OAC

and withrespetto Theorem(2.1) wehave:

S 4ABC = 1 2 x 1 y 1 x 2 y 2 + 1 2 x 2 y 2 x 3 y 3 1 2 x 1 y 1 x 3 y 3

Aording to determinant property, Substitute;

1 2 x 1 y 1 x 3 y 3 = 1 2 x 3 y 3 x 1 y 1

then, we obtain

S 4ABC = 1 2 f x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 + x 3 y 3 x 1 y 1 g

The proofis omplete.

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6

-o x

y

H

H

H

H

H

H

H

H

H A(x

1 ;y

1 ) B(x

2 ;y

2 )

C(x

3 ;y

3 )

Fig. 2. Thegraphoftriangle4ABC thatisn'tontainorigin.

6

-o x

y

! !

! !

! !

! !

! !

! !

D

D

D

D

D

A(x

1 ;y

1 )

C(x

3 ;y

3 ) B(x

2 ;y

2 )

Fig. 3. Thegraphoftriangle4ABContainorigin.

At thispoint,weinterestto extendtheabove methodfornding theregularpolygon

area when the oordinate vertexes are available. It is mentionable that every regular

polygonwithnvertex an bepartitioned to n 2triangular.

Theorem2.3. Thearea of any regularpolygon withp

j (x

j ;y

j

);j =1;:::;nvertex in

anti-lok wise sense isas follows

S

p

1 p

2 :::p

n =

1

2 f

x

1 y

1

x

2 y

2

+

x

2 y

2

x

3 y

3

+

x

3 y

3

x

4 y

4

+:::+

x

n y

n

x

1 y

1

g

Proof: Theproof isbyindutionover: Considerthefollowingrelation,

S

p

1 p

2 :::p

n =S

4p1p2p3 +S

4p1p2p4

+:::+S

4p1p

k p

k +1

+:::+S

4p1pn 1pn

whereit an be writtenas

S

p1p2:::pn =

1

2

x

1 y

1

x y

+

x

2 y

2

x y

+

x

3 y

3

x y

+:::+

x

n y

n

x y

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This relationhasbeenproved for n=3 already. Supposethatit satisfyinase n=k as hypothesis indution: S p1p2:::p k = 1 2 x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 +:::+ x k y k x 1 y 1

We show theabove relationdueforn=k+1 (assertionindution)as

S p 1 p 2 :::p k +1 = 1 2 x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 +:::+ x k y k x k+1 y k+1 + x k+1 y k+1 x 1 y 1

Hene, for this purposeit is suÆient we add S

4p1p

k p

k +1

to the two parts of hypothesis

indution,weobtain

S p 1 p 2 :::p k +S 4p 1 p k p k +1 = 1 2 x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 +:::+ x k y k x 1 y 1 +S 4p 1 p k p k +1 Withsubstituting S 4p1p k p k +1 = 1 2 x 1 y 1 x k y k + x k y k x k+1 y k+1 + x k+1 y k+1 x 1 y 1

to above relation, theassertion issatised

S p 1 p 2 :::p n = 1 2 x 1 y 1 x 2 y 2 + x 2 y 2 x 3 y 3 + x 3 y 3 x 4 y 4 +:::+ x n y n x 1 y 1

Forfurtherdesriptiononsider thefollowingexample.

Example 2.1. Find the area of following 5- regular polygon, aording to Fig. 4.

6 -o x y H H H H H A A A A A A A p 1 (5; 1) p 2 (3;6) p 3 ( 3;4)

p

4

( 6; 2)

p

5

( 1; 4)

Fig.4. Thegraphof 5-regularpolygon.

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3 A method for nding ommon set of weights (CSW)

The exibilityin thehoie of weights isbotha weakness and strength. It isa weakness

beause the judiioushoie of weights by a unit possibly unrelated to the value of any

inputor output may allow a unit to appear eÆient but there may be onernthat this

is more to do with the hoie of weights than any inherent eÆieny. This exibility is

alsostrength,however,ifaunit turnsouttobeineÆientevenwhenthemostfavourable

weightshave beeninorporated inits eÆienymeasure thenthisisa strongstatement.

This setion presents a multipleobjetive programming proedure for nding a ommon

set of weights in DEA [9 ℄. An important outome of suh an analysis is a set of virtual

multipliersor weights aorded to eah fator taken into aount. These sets of weights

are, typially, dierent for eah of the partiipatingDMUs. In this setion, by means of

solvingonlyone problem,wean determine aommon setof weights.

In DEA for alulating the eÆieny of dierent DMUs, dierent set of weights are

ob-tained, whih seems to be unaeptableinreality. So the followingmodel is usedto nd

ommon set of weightswhih has some advantages that willbe disussed later on. This

idea is formulated as maximizing the ratio of outputs and inputs simultaneously for all

DMUs. Sowepresentsthefollowingmultipleobjetive funtionalprogrammingproblem.

max nP

s

r=1 ury

r1

P

m

i=1 v

i x

i1 ;:::;

P

s

r=1 uryrn

P

m

i=1 v

i x

in o

S:t: P

s

r=1 uryrj

P

m

i=1 vixij

1; j=1;:::;n

u

r

; r=1;:::;s

v

i

; i=1;:::;m

(3.1)

where U = (u

1 ;:::;u

s )

T

and V = (v

1 ;:::;v

m )

T

are the weights of outputs and inputs,

respetively, and is a positivenon-Arhimedean innitesimal.

For solving thisproblem, the following proedure is suggested. Here we onsider the

in-nite norm, so it tends the maximization of the objetive funtionpertaining the DMU

willminimumratioof outputsto inputs.

max n

min nP

s

r=1 uryr1

P

m

i=1 v

i x

i1 ;:::;

P

s

r=1 uryrn

P

m

i=1 v

i x

in oo

S:t: P

s

r=1 u

r y

rj

P

m

i=1 v

i x

ij

1; j=1;:::;n

u

r

; r=1;:::;s

v

i

; i=1;:::;m

(3.2)

By introduingnon-negative variablez , problem(3.2), anbeonverted to thefollowing

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max z

S:t: P

s

r=1 u

r y

rj z

P

m

i=1 v

i x

ij

0; j=1;:::;n

P

s

r=1 u

r y

rj P

m

i=1 v

i x

ij

0; j=1;:::;n

z0

u

r

; r=1;:::;s

v

i

; i=1;:::;m

(3.3)

Note that instead of solving n linear programming DEA models, only one non-linear

programmingproblem issolved and theeÆieny forall DMUsare obtained.

4 Supper-eÆieny ranking tehniques

Supposewehaveasetofn(produtiveunits),DMUs. EahDMU

j

; j=1;:::;nonsumes

m dierent inputsto produesdierent outputs. Two typesof orientationDEA models

are often used to evaluate DMUs' relative eÆieny: CRS models, suh as CCR model,

and VRS models, suh as BCC model. For example, CCR model in multiplier form is

denedas alinear programmingmodelasfollows:

max P

s

r=1 u

r y

ro

S:t: P

s

r=1 u

r y

rj P

m

i=1 v

i x

ij

0; j =1;:::;n

P

m

i=1 v

i x

io =1;

u

r

; r =1;:::;s

v

i

; i=1;:::;m

(4.4)

Whereisanon-Arhimedeanelementdenedtobesmallerthananypositiverealnumber.

The BCC (Bankeret al.,1984) model addsan additional onstant variable,u

o

, inorder

to permitvariable return-to-sale:

max P

s

r=1 u

r y

ro +u

o

S:t: P

s

r=1 u

r y

rj P

m

i=1 v

i x

ij +u

o

0; j =1;:::;n

P

m

i=1 v

i x

io =1;

u

r

; r =1;:::;s

v

i

; i=1;:::;m

(4.5)

In most models of Data Envelopment Analysis (DEA) (Charneset al., 1978 ; Banker et

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these eÆient units. SeeAndersenandPetersen(1993) [1 ℄,DoyleandGreen(1993, 1994)

[6 , 7℄, Stewart (1994) [13 ℄, Tofallis (1996)[10℄, Seiford and Zhu (1999)[11℄, Mehrabian

(1999)[10℄, Zhu (2001)[15℄ and Jahanshahloo (2005)[8℄, among others. It is mentioned

that only Jahanshahloo method is able to rank all kindof eÆient DMUs (extreme and

non-extreme eÆient DMUs), whilethe other methodsare notable to rank non-extreme

eÆient DMUs. Forexample, Andersen and Petersen (1993) developed a new proedure

forrankingeÆientunits. ThemethodologyenablesanextremeeÆientunit"o"toobtain

an eÆienysore greaterthanone byeliminatingtheo-thonstraint inthemodel (4.4),

asshown inmodel (4.6).

max P

s

r=1 u

r y

ro

S:t: P

s

r=1 u

r y

rj P

m

i=1 v

i x

ij

0; j=1;:::;n;j 6=o

P

m

i=1 v

i x

io =1;

u

r

r =1;:::;s

v

i

i=1;:::;m

(4.6)

The next setion develops the new method. Our goal is to translate the basi idea of

ombining RPA and DEAinorder to determiningranking ofeÆient units.

5 Ranking of DMUs by RPA method

ThissetiondealswithrankingofDMUsbyRPAmethod. SupposethatwehavenDMUs

eah with m inputsand s outputs. The vetorsv and u are theweight vetors for input

and output,respetively. We have to somehowsolve foreah DMUalinearprogramming

(LP) whih an lead to a dierent optimalsolution. The index "o" selets theDMU for

whih theoptimization shouldbe evaluated, asshowninmodels(4.4) or(4.5), aording

toonstant returnto saleorvariablereturnto sale. Throughoutthissetionweassume

thateÆieny isevaluatedbymeansof theCCRmodel(4.4) of tehnialeÆieny. This

measurewasintroduedbyCharnesetal. (1978)[3℄. We assumethatinevaluatingDMUs

withmodel(4.4),k DMUs(kn), areeÆient. ThesetE

o

=fj jDMU

j

is effiientg

and (v

;u

) be optimal ommon set of weights by model (3.3). Let us denef funtion

as:

f :R m+s

!R 2

f(x;y)=(v

x;u

y)

Dene thesetB asfollows:

B=fz

j j z

j =(v

x

j ;u

y

j

); j2E

o ; x

j 2R

m

; y

j 2R

s

g

It is obviously, z

j

0=(0;0); j2E

o

and B R 2

+

: Infat theset Bis projetionof A,

intwo dimension plane underommon set of weights. However, we rank themembers of

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5.1 The ranking method for members in the set B

In thissetion, we desribe theranking approah for themembers of theset B. Suppose

thatT beonvexhalloffz

j j z

j

2Bg. SetT =onvex(B). ItistrivialthatT isaregular

polygonalinR 2

+

. SupposethatSbeareaofT. Set,S =R PA(T). Forrankingz

p

inB,we

rstremoveitfrom thesetofT thenweobtain(another) onvexhallT

p

=T fz

p g. Let

S

p

=R PA(T

p

) . It isobvious T

p

T and S

p

S , (see Fig. 5), and set

p

=S S

p 0

. Thisvalue

p

isdened therank ofz

p .

Asalreadymentioned,z

j

isasapointofT =onvex(B). Ifz

j

beextremepointinT,then

p

>0 , meanwhileif z

j

be non-extreme point of T , then

p

=0 . Formore desription

dene B +

=f z

j j

j

>0; j 2Bg and B 0

=f z

j j

j

=0; j 2 Bg. Let Card(B 0

)=k

0

and Card(B +

) =k

1

, then it is obvious that, Card(B 0

)+Card(B +

) =Card(B) , that

is,k

0 +k

1

=k . Then, we present thebelowdenition:

Denition 5.1. Suppose that z

p ; z

q 2B

+

, then z

p

has higher rank of z

q

if and only if

p >

q .

6

-o vx

uy

A A A A A A z

1

z

2

z

3

z

4 z

5 z

6

z

7

r

r

r

r r r

r

Fig. 5. a. Convexhalloffz

j

g; =1;:::;7.

6

-o vx

uy

A A A A A A

B B B B B B z

1

z

2

z

3

z

4 z

5 z

6

z

7

1

S

1 r

r

r

r r r

r

Fig. 5. b. Convexhalloffz

j

g; =2;:::;7.

Fig. 5. a shows a onvex hall of 7 point together with its area S , meanwhile, Fig. 5. b

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Now ifwe verify points z

6 orz

7

we obtain

6 =

7

= 0 . In last ase we should be rank

theunitsz

6 and z

7

, intheonvex setT

0

=onvexfz

6 ;z

7 g.

Some notes areworthwhile:

Note 1. Supposethatz

p ; z

q 2B

+

,thentheprobability

p =

q

tendto vanish. It istrivial

thatpointsinB +

havehigherrankthanthepointsinB 0

. Nevertheless,weinterest

to rank pointsbelongto B 0

inseond order,aording to mentionedmethod based

on thenewonvexset T

0

=onvex(B B +

) .

Note 2. If T = onvex(B) be a segment line, that is,S = 0 , then we suppose that this

probabilityiszero. Otherwisethismethod enounter witha problem.

Note 3. Fornon-extremepointswhihhave thezero rankvalue,itis suggestedthe distane

of originbeasa riterion forranking.

6 Numerial example

Example6.1. Toillustrate theappliationofthis method, weonsider 19withtwoinputs

and two outputs (Table 1).

Table 1

The valueof inputsandoutputs

DMUs Input1 Input2 Output1 Output2

1 81 87.6 5191 205

2 85 12.8 3629 0

3 56.7 55.2 3302 0

4 91 78.8 3379 0

5 216 72 5368 639

6 58 25.6 1674 0

7 112.2 8.8 2350 0

8 293.2 52 6315 414

9 186.6 0 2865 0

10 143.4 105.2 7689 66

11 108.7 127 2165 266

12 105.7 134.4 3963 315

13 235 236.8 6643 236

14 146.3 124 4611 128

15 57 203 4869 540

16 118.7 48.2 3313 16

17 58 47.4 1853 230

18 146 50.8 4578 217

19 0 91.3 0 508

Table 2

The resultsof usingdierent modelsfor rankingof DMUs

DMUs 1 2 5 9 15 19

AP rank 4 1 3 - 2

-MAJ rank 5 3 2 6 4 1

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DMUs 1, 2, 5, 9, 15and 19 areCCR eÆient. Theresults of ranking using RPA method

are ompared with Thebyhenorm, AP andMAJmethods in Table2. Asshown in

Ta-ble 2, DMU19 and DMU9 havehighest and lowest rank in MAJand Thebyhe models,

respetively. Meanwhile, both of them (DMU19 and DMU9) are infeasible in AP model.

Also, notiethat, all DMUs areranked verylose to eah other in MAJ and Thebyhe

models,whilethis isnot inAPmodel. In modelAP,DMU2andDMU1haverstandlast

rankrespetively. TheoperationsforrankingeÆientunits1, 2,5, 9,15and19aording

to model (3.3), the ommon setof weights have obtained as follows:

u

1

=0:01000; u

2

=0:089560; v

1

=0:351901; v

2

=0:498316; z

=0:44

Therefore we aquire:

z

1

=(72:16;70:27); z

2

=(36:29;36:29); z

5

=(111:89;110:91); z

9

=(65:66;28:65);

z

15

=(121:22;97:05); z

19

=(45:5;45:5)

Now we rankthe units of the set B=fz

1 ;z

2 ;z

5 ;z

9 ;z

15 ;z

19 g

6

-o vx

uy

#

# #

# #H

H

H

` ` ` `

z

1

z

2

z

5

z

9

z

15

z

19

r

r

r

r

r

r

Fig.6. Convex hall ofmembersof B

For ranking units, rstwe ompute R PA(T)=S . Aording tothe Fig. 6, onsider that

z

1

don't eetivein onstrutive R PA(T)=S . Hene; we obtain:

S = 1

2

65:66 28:65

121:22 97:05

+

121:22 97:05

111:89 110:91

+

111:89 110:91

45:5 45:5

+

45:5 45:5

36:29 36:29

+

36:29 36:29

65:66 28:65

=4186:43

where

i

=S S

i

haveobtained follows:

1

=0;

2

=340:87;

5

=1029:17;

9

=2433:39;

15

=1408:23;

19

=9:03

7 Conlusion

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twodimensionspaebyaommonsetofweights. Thentheareafromtheregularpolygon

onstruter of all projeted eÆient units is onsidered alulatal. However, we ranked

the projeted eÆient units aording to the dierene between regular polygon areas

before and after removed. We also suggested that one an be work on thismethod over

impreisedata[5℄,andfousonStabilityregionsforkeepingeÆieny[15℄. Readershould

attend that thismethod hasa drawbak, beause the projetion funtionof f :R m+s

!

R 2

; f(x;y)=(vx;uy) isnotinjetivemap.

Referenes

[1℄ P. Andersen,N.C. Petersen, A proedure for rankingeÆient unitsin data

envelop-ment analysis,Management Siene 39 (1993)1261-1264.

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