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.
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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.
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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
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
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.
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
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.
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
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
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
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
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
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
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.
[2℄ R.D. Banker, A. Charnes, W.W. Cooper, Some model for estimating tehnial and
saleineÆieniesindataenvelopmentanalysis,ManegementSiene30(1984)
1078-1092.
[3℄ A. Charnes, W.W. Cooper, E.Rhodes, Measuring the eÆieny of deision making
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