Inverse DEA Model with Fuzzy Data for Output Estimation

Similar documents
A note on A New Approach for the Selection of Advanced Manufacturing Technologies: Data Envelopment Analysis with Double Frontiers

Exam-style practice: A Level

Proposing a Mixed Model Based on Stochastic Data Envelopment Analysis and Principal Component Analysis to Predict Efficiency

A L A BA M A L A W R E V IE W

Fun and Fascinating Bible Reference for Kids Ages 8 to 12. starts on page 3! starts on page 163!

Online Open Access publishing platform for Management Research. Copyright 2010 All rights reserved Integrated Publishing association

Cross Efficiency of Decision Making Units with the Negative Data in Data Envelopment Analysis

Scripture quotations marked cev are from the Contemporary English Version, Copyright 1991, 1992, 1995 by American Bible Society. Used by permission.

K E L LY T H O M P S O N

THIS PAGE DECLASSIFIED IAW E

ECEN474/704: (Analog) VLSI Circuit Design Spring 2018

T T V e g em D e j ) a S D } a o "m ek j g ed b m "d mq m [ d, )

c i l l is I D. a ^ ;5 3? i "0 > D N) NO o w

A B CDE F B FD D A C AF DC A F

Sensorless A.C. Drive with Vector Controlled Synchronous Motor

Concept Design Option C

Chapter 5. Long Waves

SMT 2018 Geometry Test Solutions February 17, 2018

-HYBRID LAPLACE TRANSFORM AND APPLICATIONS TO MULTIDIMENSIONAL HYBRID SYSTEMS. PART II: DETERMINING THE ORIGINAL

AGENDA REPORT. Payroll listing conforms to the approved budget except as noted and has been paid WILLIAM A HUSTON CITY MANAGER

Combinatorial proofs of a kind of binomial and q-binomial coefficient identities *

Answer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4.

Parts Manual. EPIC II Critical Care Bed REF 2031

D EFB B E B EAB ABC DEF C A F C D C DEF C AD C AEC D D E C D EF B ABC AB CD A EFD AD D E

Minimum Hyper-Wiener Index of Molecular Graph and Some Results on Szeged Related Index

Constructing an Evaluation Function for Playing Backgammon. ANDERS SJÖQVIST and ANDRÉ STENLUND

Today. The geometry of homogeneous and nonhomogeneous matrix equations. Solving nonhomogeneous equations. Method of undetermined coefficients

CLAIM No, HOLE No, FOOTAGE

I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o

I N A C O M P L E X W O R L D

- Prefixes 'mono', 'uni', 'bi' and 'du' - Why are there no asprins in the jungle? Because the parrots ate them all.

Chapter 3. Many-Electron Atoms

Chapter 3 Applications of resistive circuits

Instruction Sheet COOL SERIES DUCT COOL LISTED H NK O. PR D C FE - Re ove r fro e c sed rea. I Page 1 Rev A

Math K (24564) - Homework Solutions 02

OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9

A Fuzzy Mathematical Programming Approach to DEA Models

Dangote Flour Mills Plc

dm dt = 1 V The number of moles in any volume is M = CV, where C = concentration in M/L V = liters. dcv v

Standard Signs Manual

176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s

Software Process Models there are many process model s in th e li t e ra t u re, s om e a r e prescriptions and some are descriptions you need to mode

einpelledeetodeupport

Summary of Grade 1 and 2 Braille

T h e C S E T I P r o j e c t

EXAMPLE CFG. L = {a 2n : n 1 } L = {a 2n : n 0 } S asa aa. L = {a n b : n 0 } L = {a n b : n 1 } S asb ab S 1S00 S 1S00 100

M $ 4 65\ K;$ 5, 65\ M $ C! 4 /2 K;$ M $ /+5\ 8$ A5 =+0,7 ;* C! 4.4/ =! K;$,7 $,+7; ;J zy U;K z< mj ]!.,,+7;

0609ge. Geometry Regents Exam AB DE, A D, and B E.

c- : r - C ' ',. A a \ V

Chapter 2: Descriptive Statistics

Radiation Resistance of System G( Iron Torus is not used as we can see ) ( ) 2

Developing a Distributed Java-based Speech Recognition Engine

Chapter 2 Preliminaries

MATH 23 Exam 2 Review Solutions

Justification of Investment in IT systems

CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang

Consistency of Academic Performance in Higher Education: A Study of an Irish Business Degree Programme

College of Engineering Department of Electronics and Communication Engineering. Test 2 MODEL ANSWERS

Fuzzy Weights in Data Envelopment Analysis

Higher Order Linear Equations

THIS PAGE DECLASSIFIED IAW EO IRIS u blic Record. Key I fo mation. Ma n: AIR MATERIEL COMM ND. Adm ni trative Mar ings.

S U E K E AY S S H A R O N T IM B E R W IN D M A R T Z -PA U L L IN. Carlisle Franklin Springboro. Clearcreek TWP. Middletown. Turtlecreek TWP.

A RWA Performance Comparison for Hybrid Optical Networks combining Circuit and Multi-Wavelength Packet Switching

Module 09 From s-domain to time-domain From ODEs, TFs to State-Space Modern Control

NATIONAL SENIOR CERTIFICATE MEMORANDUM MATHEMATICS MEMORANDUM P2 SEPTEMBER 2016 GRADE 12

å Q d = 0 T dq =0 ò T reversible processes 1

Questions of Ethical Responsibility in the Research of Unaccompanied Minors

e*s EU[8tr r'glt il IIM IITTT D Cl : L25119AP1984PLiC XL719 New GIN: L27029APl 984PLC Bikrafi Keshari Prusty An ISO 9001 Company

St Peter's Anglican Primary School is offering HoLiday Care during JuLy 2016 school holidays.

χ be any function of X and Y then

Homework 1/Solutions. Graded Exercises

Ehrhart polynomial for lattice squares, cubes, and hypercubes

EECE 301 Signals & Systems Prof. Mark Fowler

Chapter 3. Differentiation 3.3 Differentiation Rules

P a g e 5 1 of R e p o r t P B 4 / 0 9

MODEL PAPER - I MATHEMATICS. Time allowed : 3 hours Maximum marks : 100

Copyright Birkin Cars (Pty) Ltd

SEMICLASSICAL STATES FOR THE NONLINEAR KLEIN-GORDON-MAXWELL SYSTEM

Architecture and development methodology for Location Based Services

EUROSTAT - OECD PPP PROGRAMME 1999 THE CALCULATION AND AGGREGATION OF PARITIES

4. Matrix inverses. left and right inverse. linear independence. nonsingular matrices. matrices with linearly independent columns

MATH 425, HOMEWORK 5, SOLUTIONS

_ J.. C C A 551NED. - n R ' ' t i :. t ; . b c c : : I I .., I AS IEC. r '2 5? 9

Nonhomogeneous Equations and Variation of Parameters

Proofs. by Bill Hanlon

An homomorphism to a Lie algebra of matrices is called a represetation. A representation is faithful if it is an isomorphism.

The Second Law implies:

Control Systems. Time response

rad / sec min rev 60sec. 2* rad / sec s

12781 Velp Avenue. West County B Rural Residential Development

7.5 Proportionality Relationships

Objective of curve fitting is to represent a set of discrete data by a function (curve). Consider a set of discrete data as given in table.

From our Guest Editor

Chair Susan Pilkington called the meeting to order.

A Rich History. AB OVE T H E F O L D: C O N S T R U C T E D by Dwight B. Heard between 1919 and HUMAN SPIDER ON TH E RADIO FI LM DE B UT

Lecture 4. Electrons and Holes in Semiconductors

8.6 Order-Recursive LS s[n]

WEYL MANIFOLDS WITH SEMI-SYMMETRIC CONNECTION. Füsun Ünal 1 and Aynur Uysal 2. Turkey.

Engine Control <2TR-FE> I

Informazioni per il lettore

Transcription:

Aaabe e at www.. Iaa Ja Optat 2200 388-4 Iaa Ja Optat Iee DEA Mde wt F Data Otpt Etat A Mad Rad a Rea Dea a Faad Heade Lt b a Depatet Mateatc Iac Aad Uet Maedea Bac Ia b Depatet Mateatc Iac Aad Uet Scece & Reeac Bac Tea Ia Abtact I t pape we w tat ee Data Eepet Aa DEA de ca be ed t etate tpt wt data a Dec Mak Ut DMU we e a pt ae ceaed ad decec ee te t ea caed. Kewd: Data Eepet Aa Mt-becte Pa Iee DEA Mde F Nbe Cepd at Ea addee: aadad@a.c

389. Itdct Data eepet aa DEA a -paaetc tecqe ea ad eaat te eate ecece a et ette caed dec ak t DMU wt te c pt ad tpt. Sce te t tdct te tecqe b Cae et a.[] kw a te CCR de. Sce te a pbcat DEA a bece a ppa etd aa te ecec a aat t [34]. Iteet Cae ad cpe ae a ad a cat pact te deepet tpe becte ea pa MOLP t te deepet a pa. A t Cae ad Cpe ae paed a cat e te deepet DEA ad MOLP eeace tee tw cap ae eea t pad c attet t eeac peed te te cap e deta ee[560]. Recet We et a. [9] pped ee DEA t awe te qet a w: a a p DMU we ceae ceta pt t a patca t ad ae tat te DMU ata t cet ecec ee wt epect t te DMU w c d te tpt te DMU ceae te tpt eed t be ceaed t a ceta ee ad te ecec te t ea caed w c e pt d be pded t te t? I ecet ea et te a bee pped a a wa t qat pece ad ae data DEA de []. Te DEA de wt data F DEA Mde ca e eatca epeet ea-wd pbe ta te ceta DEA de. F et te a aw tc data t be ed dect wt te DEA de. F DEA de take te ea pa de. Te qet dced ee DEA ca be cdeed wt data tat ae tat e data ae be ad we ceae e a pt ee a e DMU ad ae tat te DMU ata t cet ecec ee w c d te tpt te DMU cae? I t pape we

390 cde abta ca pt ee wt taa be ad we pped a MOLP de tpt etat. Te ede te pape aed a w: I te w ect we eew et ad be ea pa pbe. We cde DEA pbe ad t ee DEA pbe ect 3. I ect 4 we pped a MOLP tpt etate. We cde weak ecec cae I ect 5. I ect 6 we e a exape t tate cptat etd. Cc ae e ect 7. 2. Peae Sce te ke et be et te w be ed eqe; we a cde a ew ecea det. Det 2. I X a cect bect deted eeca b x te a et X a et deed pa: A x A x x X Wee A x caed te ebep ct wc acate wt eac x X a be [0] dcat t wat deee x a be. Det 2.2 Let A be a be.e. A cex aed bet te ea e te ee tat: ax R Ax. b A 0 0 x0 I a pecewe ct ct. Te - ee et A te A x Ax wee [0 ].

39 Det 2.3 A taa be deted b A A A A Wee A A A ad A A ad A ae ea be. I t pape we dete te et a taa be b F R. Let A = be. Dee: A A A ad B = B B B x 0 x R : xa xa xa xa x 0 x R : xa xa xa xa A B A A B A B B A A B B A A bt be taa B B T tdce a ea de be we t exteded peat ad ax ea be t cepd peat be MIN ad MAX. F a tw be A ad B we dee MIN A B MAX A B p x p. ax x A x B A x B

F a 392 R ee7. Te F R MIN MAX ca be expeed a te pa F R wee a pata de deed a: A B A B MIN A B A MAX A B B a a a Det 2.4 A be a 0 0 cpet a a ad a 0 0. ea a t Tee 2.. F a tw taa be A = A A A ad B = B B B we ae A B A B A B A B. Tee 2.2. F a tw taa be te ext k > c tat k. ad P. I ad k k k 3 : 2 : : k k k 3 2

393 I k = {k k 2 k 3 } te ae k. A ea pa pbe LPP deed a: M = cx S.t Ax=b 2. x 0 Wee c = c c b = b... b A a. I pbe 2. a te paaete ae cp[2]. Nw te LPP e cecet te pbe te becte ct tecca cecet te t-ad de cecet be te be te we a pbe a be ea pa pbe. Hee we cde te LPP wt be tecca ad t-ad de cecet. A be ea pa pbe FNLPP deed a w: ax. t x A x c 0 x B 2... 2.... 2.2 Wee A a a a F R ad B b b b FR.

394 B tee 2. pbe 2.2 ca be ewtte a Max c x S.t. a x b =2 a x b =2 2.3 a x b =2 Hwee ce a be ed pbe 2.3 ae cp te pbe 2.3 a ea pa pbe. I t pape we a pbe 2.3 eqaet t pbe 2.2. 3. F DEA pbe ad t ee pbe Cde DMU: DMU DMU wt pt ad tpt. Ipt ad tpt DMU ae X = x x 2... x ad Y... 2 = 2... epecte wee x 2... ad 2... ae Nbe. A X R ady R ad a X 0 ad Y 0 = 2. Cde te w tpt eted CCR de wee pt ad tpt ae taa be: Max

395 S.t X X 3. Y Y B tee 2. pbe 3. edce t te w: Max S.t x x =2 x x =2 x x =2 3.2 =2 =2 =2 0 =2. Te pbe 3.2 a tpt eted CCR de wee pt ad tpt ae ea be.

396 Sppe tat DMU pt DMU te pta ae 3. X t X X ad ae ceaed wee te ect X 0 ad X 0.e. at eat e cpet ceae. We eed t etate te cepd tpt ee we te ecec dex DMU ea caed wee T... ad F R =2. F ceece ppe DMU epeet DMU ate ca te pt ad tpt. Hece t eae te ecec te DMU we e te w de: Max S.t. X 3.3 Y 0 =2 +. Wee X X X 0 X 0. 4. Te eated MOLP

397 T d tpt ax c tat ecec dex ea caed we cde te w MOLP: Max 2... S.t.. X Y 4. Y 0 =2. Wee 3.. deed a bee ad e a te pta ae pbe Det 4. Let a eabe t pbe 4.. I tee eabe t 3.3 c tat te we ca weak Paet t pbe 4.. Tee 4.. Let be te pta ae pbe 3.. Te w pbe: be a weak Paet t pbe 4. ad a te pta ae te Max S.t. X

398 Y 4.2 0 =2. P. Becae pbe 4.2 a a pta ae ppe a pta t te pbe. A a weak Paet t 4. t ate te cdt:. X Y Y 0 =2. Becae a eabe t 4.2 we ae 0:. te I we wd ae Y Wee ad.

S 399 wd be a eabe t pbe 4. wc pbe becae a weak Paet t 4.. S we t ae.e. te pta ae 4.2. Tee 4.2. Let be eabe t pbe 4.. I te pta ae pbe 4.2 4.. te t be a weak Paet t P. I t tee wee t te te tee wd ext ate eabe t 4. c tat. A X Y 0 2.... F Eq. ad tee 2.2 k > c tat Y k k S te pta ae 4.2 at eat apt tat te pta ae 4.2. k wee k wc aat Cde te w pbe:

400 Max S.t X Y 4.3 0 2.... Tee 4.3. Y Ae. I te pta ae pbe 4.3 > te te pta ae 4.2 a ; cee te pta ae pbe 4.2 > te pta ae 4.3. P. Ft we ae tat te pta t 4.3 Cde te eqaet pbe 4.3: Max ad >. S.t x =2 x =2 x =2 4.4 =2

40 =2 =2 0 =2 We cde te da pbe 4.3: M S.t x x x 2... 0 4.5 0 2... 0... 2 0 Sppe......... Ad.........

402 Ae a pta t 4.4. S ad we ae = Accd t te cpeeta acke cdt LP pbe a pta t 4.3 te aabe λ wc cepd t te ctat te da pbe 4.5 t be λ = 0. Bt we λ = 0 pbe 4.3 t bece 4.2. S pbe 4.2 a ae pta ae. Cee te pta ae pbe 4.2 > cde te eqaet pbe 4.2: Max S.t x =2 x =2 x =2 4.6 =2 =2

403 =2 0 =2 Wte te da pbe 4.6 M S.t x x x 2... 0 4.7 2... 0... 2 0 A a a pta ae. Te deece betwee 4.5 ad 4.6 tat 4.5 cta e e ctat.e. ctat. Bt we ca w tat a pta t......... Ad......... pbe 4.5 t ctat t d a tct eqat. 0 I 0

404 Te te pta ae 4.5. Ad ece te pta ae 4.7 wd be wc ctadct te apt. S t addta ctat t be bd at a pta t ad ece pbe 4.5 eqaet t 4.7 wc pe tat a pta ae pbe 4.3 ad 4.5. Tee 4.4. Sppe tat te pta ae pbe 3. > ad te pt t DMU ae t ceae X t X X X 0 X 0. Let be a weak Paet t pbe 4. te te pta t 4.3 t. Cee Let be a eabe t pbe 4.. I te pta ae pbe 4.3 te t be a weak Paet t 4.. P. Ae a Paet t 4.. B tee 4. te pta ae 4.2. A > ad Y b tee 4.3 a te pta ae 4.3. Cee te pta ae 4.3 te b tee 4.3 a te pta ae 4.2. U tee 4.2 we kw tat a weak Paet t 4.. T det e Paet t 4. we cet t t a ebecte pa pbe b a p > 0 a te wet -t tpt =. Teee we w ae:

405 Max p p S.t. X Y 4.8 Y 0 =2. We kw a pta t 4.8 t be a weak Paet pbe 4.. Ca. Sppe te ecec dex DMU de de 3. > ad pt ae ceaed X t X X X 0 X 0. Let be a pta t pbe 4.8. Te we te tpt DMU ae ceaed t te ecec dex te DMU t. 5. Weak ecec cae We w t t te cae =. Cde te w LP pbe Max S.t. X 5.

406 Y Y 0 =2. Wee X X ad X 0 X 0. Let t pta ae be. I we dete te eabe e pbe 3. ad 5. b S 0 ad S epecte we ae: S 0 S. Teee =. Tee 5.. Sppe tat te pta ae pbe 3. DMU = ad te pt t DMU ae ceaed X t X X X 0 X 0. Te we te tpt DMU ae ceaed Y t Y wee te pta ae pbe 5. te pta ae pbe 5. wc cepd te ew DMU t. Y P. We te pt ad tpt DMU bece X X X 0 X 0 ad Y te ecec dex DMU eqa te pta ae te pbe bew: S.t. Max X 5.2 Y Y Y 0 =2.

407 Let te pta ae pbe 5.2 be ẑ. Wat we eed t pe tat ẑ =. I ẑ = te ẑ >. We wte te eqaet pbe 5.2: Max S.t x =2 x =2 x =2 5.3 =2 =2 =2 0 =2 We wte te da pbe 5.3 M S.t x x x 0 2...

408 0 2 2... 0... 2 0 Sppe......... ad......... ae t pta t. B dat et we kw tat 3 S te ctat 2 ad 3 we ae: 0 B te cpeeta acke tee at eac pta t 5.2 0. T ea ẑ a te pta ae te pbe Max S.t. X Y Y 0 =2. Ad teee te pta ae te pbe

409 Max S.t. X Y Y 5.4 0 =2. I. Bt te abe pbe wt epac te b t pbe 5. te pta ae pbe 5. wd be ad we ae ẑ > te ẑ > wc ctadct tat ax ae 5.. 6. Neca exape We cde tee DMU wt tee pt ad tw tpt. Te data pt ad tpt ae w te w tabe. DMU S I I 2 I 3 O O2 DMU 34 258 47 349 0820 DMU2 350 47 84 820 36 DMU3 8024 59 49 384 52 B eaat DMU 3 tpt-eted de 3. we ae O =.88. We w ceae te pt DMU3

40 X 3 = 8 0 24 5 8 4 9 t 5 34 8 6 28 7 6 29 te b te de 4.8 wt 3 p = p 2 = we w ae 3 8.23 0.8308 24.392 27.0770 48.7385 54.539. I t cae te ecec ad X 3 3 3 Y3 eqa. A b eaat DMU 2 de 3. we ae = we w ceae te pt DMU 2 X 2 = 3504784 t = 5 2 9 6 4 20 0 8 23 te b te de 5 2 we w ae =.4667 ad 2 2 Y.4667.7334 29.3334 4.4 6.334 23.4667 I t cae te ecec eqa ad bt =. ad X Y 2 2 2 2 7. Cc I t pape we dc tee pbe: I te peece data w d we ct te cae pt ee a e DMU c tat te ecec dex te DMU peeed. T e te pbe we pped a MOLP de. We pde te ecea ad cet cdt te pt cae de te ae ecec dex. Reeece [] Bake R. D. Cae A. Cpe W.W. Se de etat tecca ad cae ecece data eepet aa Maaeet cece 30 078-092 984.

4 [2] Baaaa M. S. Ja J. J. ad Sea H. D. Lea pa ad etwk w J We New Yk ecd edt 990. [3] Cae A. Cpe W.W. ad Rde E. Mea te ecec dec ak t Epea Ja Opeata Reeac 429-444 978. [4] Cae A. Cpe W.W. Lew A.Y. ad Sed L.M. Data eepet Aa- Te Metd ad Appcat Kwe Acadec Pbe Bt 994. [5] J R. Ke P. Wae J. Stcta cpa data eepet aa ad tpe becte ea pa Maaeet cece 447 926-970 999. [6] J R. Ke P. Zt S. A teacte appac t pe etate ae ecec data eepet aa Epea Ja peata eeac49 688-699 2003. [7] K G. J. Ya B. F et ad c: Te ad Appcat Petce- Ha Ida 200. [8] Maek H. R. Tata M. Mac M. Lea pa wt aabe F et ad te 09 2-33 2000. [9]We Q. Za L.J. ad Za X. A ee DEA de pt/tpt etate Epea Ja Opeata Reeac 2 5-63 2000. [0] W B. Y. H. Lqe M. Ya J. B. U teacte t becte etd t e DEA pbe wt ae det Cpte ad Opeat eeac 36 623-636 2009. [] Zea H. J. F et te ad t appcat Kwe Acadec pbe Ld 996.

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback