Data Envelopment Analysis of Missing Data in Crisp and Interval Cases

Transcription

1 Int Journal of Math Analysis, Vol 3, 2009, no 20, Data Envelopment Analysis of Missing Data in Crisp and Interval Cases Tamaddon a 1, G R Jahanshahloo b, F Hosseinzadeh otfi, c M R Mozaffari c, K Gholami a a Department of Mathematics, Science and Research Branch,Islamic Azad niversity, Fars, Iran b Department of Mathematics, Tarbiat Moallem niversity, Tehran, Iran c Department of Mathematics, Science and Research Branch, Islamic Azad niversity, Tehran, Iran Abstract In this paper, we propose a method for finding the missing data in the case that the data are crisp and interval Firstly, we determine the missing amounts via the sum of other DM s inpus and outputs in the crisp case; then, in the case that the data are interval, we obtain the upper and lower bounds of the missing data via crisp processes And by using convex combination of the interval beginings and endings, we can obtain a linear function of an analogous variable with each one of the inputs and output s components; so that we can obtain a function for the missing data via crisp processes Finally, we represent an algorithm for improving the missing interval Keywords: Data envelopment analysis, missing data, interval DEA 1 Introduction Data envelopment analysis (DEA), is considered as a useful tool for management and decision, and from the time that was developed by Charnes [1], it has been expanded wonderfully in problems, technology and the existed application Data envelopement analysis is a planning technique of mathematics which recognizes the ideal efficiency of DMs In fact, for measuring the relative efficiencies of a set DMs which produce multiple outputs by consuming multiple inputs 1 Corresponding author, address: tamaddon64@yahoocom,

2 956 Tamaddon et al In classic models of DEA such as CCR and BCC models and etc our supposition is that all the input and output data have been defined exactly; but, in fact, these hypothesises are not always true in the universal condition and we will deal with all kinds of defined data and also non-defined data such as missing data, comparative, ordinal ones and data with having limits, and etc The case of missing values in DEA models have been examined in the literature in different ways Some DEA application [2] propose the exclusion from the analysis of the units that have missing values, a common practice followed in statistical applications this approach is not suitable for DEA as it affects the efficiency of other units due to the comparative evaluation and may possibly disturb the statistical properties of the efficiency estimators [3] Other approaches use imputation techniques to estimate exact approximation of the missing values (for example average value of the other units) Such an arrangement may lead to misleading efficiency results due to the stability problems according to which a unit accepting an infinitesimal perturbation may change its classification from an efficient to an inefficient status or vice-versa [4] In the same line, Kuosmanen [5] proposes the use of dummy entries(zero for the outputs and sufficiency large number for inputs)and particular weight restriction to reduce the impact of the units with the missing values to the efficiency evaluation of the other unitskao and iu [6]propose the extention of DEA based on fuzzy theory (fuzzy DEA) They replace the missing values with intervals and use the observed data to estimate membership function of fuzzy efficiency scores The current article proceeds as follows: In Section 2, we review DEA models and interval data In the section 3 we will find the missing data when the inputs and outputs are crisp In section 4 we will argue the envelopment analysis of missing data in interval case and also we will represent a method for finding the missing data and, consequently, an algorithm is represented for improving missing interval inwhich the evaluated DM becomes efficient In section 5 we will represent some examples in crisp and interval cases and we execate algorithm for them Sectin 6 provides some concluding remarks DEA 2 Preliminary Notes 21 DEA and interval DEA Consider ndmswith m inputs and s outputs The input and output vectors of DM j (j =1,,n) are X j =(x 1j,,x mj ) t,y j =(y 1j,,y sj ) t, respectively, where X j 0, X j 0,Y j 0, Y j 0 By using the constant returns to scale, convexity, and possibility postulates, the non-empty production possibility set (PPS) is defined as follows:

3 Missing data 957 T c = {(X, Y ):X λ j X j,y λ j Y j,λ j 0,j =1,n} j=1 j=1 By using the above definition, the CCR model is defined as the following: min θ st λ j x ij θx ik, i =1,,m j=1 j=1 λ j 0, λ j y rj y rk, j =1,,n (1) r =1,,s If the optimal answer of the above model is θ<1, DM k is not efficient and if θ = 1, it means DM k is efficient In model (1), if we suppose that the inputs and outputs are located between interval with limits; it means that: x ij [x ij,x ij] and y rj [y rj,y rj] Then, we have model (1) as the following: min st θ λ j [x ij,x ij ] θ[x ik,x ik ], j=1 j=1 λ j 0, λ j [y rj,y rj ] [y rk,y rk ], j =1,,n i =1,m r =1,,s (2) Model (2) is not a linear model; since, it s parameters which are x ij and y rj are interval Two models of (3) and (4) calculate the limits of the relative efficiency of DM k : θ = min θ st λ j x ij + λ kx ik θx ik, j=1,j k λ j 0, λ j y rj + λ k y rk y rk, j =1,,n i =1,,m r =1,,s (3) Model (3) is a DEA model with exact data in which the levels of inputs and outputs are adjusted unfavourably of unit k (inputs are set to the upper bound and outputs to the lower bound) and in favour of other units

4 958 Tamaddon et al θ = min θ st λ j x ij + λ k x ik θx ik, j=1,j k λ j 0, λ j y rj + λ ky rk y rk, j =1,,n i =1,,m r =1,,s (4) Say θ, is the efficiency score for that unit that derives from its most favorable position(inputs are set to the lower bound and outputs to the upper bound)while all the rest units are set to their least favorable position(inputs are set to the upper bound and outputs to the lower bound) Models (3) and (4) provide, for all the evaluated units,bounded intervals of efficiency scores [θ,θ ],j =1,,nwhich can be used to further discriminate them in three classes of efficiency as follows: E ++ = {j J θj =1}, E + = {j J θj < 1 and θj =1} and E = {j J θj < 1} The set E ++ consists of units that are efficient in any case(any combination of input/output levels) The set E + consists of units that are efficient in a maximal sens, but there are input/output adjustments under which they cannot maintain their efficiency Finally the set E consists of the definitely inefficient units model(fdea) 3 The method for finding the missing data in crisp case In the data envelopment analysis, may be one of the inputs or outputs among them from a DM is missing; so, for finding the missing data, the following method is suggested: suppose that there are n DMs with m inputs and s outputs which the data of inputs and outputs are crisp but the ith input from DM k ; that is x ik, is missing For obtaining the x ik, consider the data of table (1) DM j I 1 I 2 I i I m O 1 O 2 O s DM 1 x 11 x 21 x i1 x m1 y 11 y 21 y s1 DM 2 x 12 x 22 x i2 x m2 y 12 y 22 y s2 DM k x 1k x 2k? x mk y 1k y 2k y sk DM n x 1n x 2n x in x mn y 1n y 2n y sn Table 1 The missing data among the crisp data

5 Missing data 959 For each one of the inputs, of DM k, the first input to m is divided on the sum of other DM inputs of the first to m columns and we call them ρ 1,,ρ m (be careful that ρ i is not obvious) We do this process for outputs, that is, we divide the first to sth outputs from DM k on the sum of the other DM outputs of the first to sth columns and we call them ρ 1,,ρ m ρ 1 = ρ 1 = x 1k y 1k x 1j,ρ 2 = y 1j,ρ 2 = x 2k y2k x 2j,,ρ i =?,,ρ m = y 2j,,ρ m = ymk y mj x mk x mj Then, we obtain their average and call it ρ, at the end, we obtain the x ik by multiplying ρ in the sum of the ith inputs from all the DMs expect DM k therefore, we have: m s ρ = ρ i + ρ r i=1 r=1 x ik x ij = ρ = x ik = ρ x ij measuring interval efficiencies 4 Missing data among interval data 41 Finding the missing data Suppose that ndmswith m inputs and s outputs which are the data of interval inputs and outputs are existed, but the ith input of DM k ; that is [x ik,x ik ] is missing For obtaining the missing data, consider the data of table (2) DM j I 1 I 2 I i O 1 O 2 O s DM 1 [x 11,x 11 ] [x 21,x 21 ] [x i1,x i1 ] [y 11,y 11 ] [y 21,y 21 ] [y s1,y s1 ] DM 2 [x 12,x 12 ] [x 22,x 22 ] [x i2,x i2 ] [y 12,y 12 ] [y 22,y 22 ] [y s2,y s2 ] DM k [x 1k,x 1k ] [x 2k,x 2k ]? [y 1k,y 1k ] [y 2k,y 2k ] [y sk,y sk ] DM n [x 1n,x 1n] [x 2n,x 2n] [x in,x in ] [y 1n,y1n] [y2n,y 2n] [ysn,y sn] Table 2 The missing data among the interval data

6 960 Tamaddon et al The way that was suggested for obtaining the missing data among crisp data was used separately for obtaining the upper and lower bounds of the missing interval; therefore, the lower bound of the missing interval is calculated as the following: ρ 1 = ρ 1 = x 1k x 1j y 1k y1j Therefore: m s ρ i + ρ i=1 r=1 = F inally : x ik x ij,ρ 2 = x 2k,ρ 2 = ρ r x 2j y 2k y2j = ρ = x ik = ρ,,ρ i =?,,ρ m = x mk,,ρ s = n x ij y sk ysj x mj Also, the upper bound of the missing data is calculated similarly: ρ 1 = x 1k ρ 1 = x 1j y 1k y1j Then: m s ρ i + ρ i=1 r=1 = F inally : x ik x ij,ρ 2 = x 2k,ρ 2 = ρ r x 2j y 2k y2j = ρ = x ik = ρ,,ρ i =?,,ρ m = x mk,,ρ s = n x ij ysk ysj x mj 42 The combination of the interval data and finding a function of a variable for the missing interval ntil now, the missing interval which is [x ik,x ik ] is obtained through the followings via a suggested solution of the missing data:

7 Missing data 961 [x n ik,x ik ]=[ ρ x ij, ρ which is equal to: n x ij ] [ x x ( 1k y ++ mk y )+( 1k ++ sk ) x 1j x mj y1j ysj ( x 1k ++ x mk )+( y 1k ++ y sk ) x ij, (5) x 1j x mj y1j ysj x ij] Now, we should show that every points considered within the intervals of defined inputs and outputs (It means that we considre the inputs and outputs as the crisp data)and we use the suggested way for finding the missing data among crisp data, the obtained data is placed in the interval (5) So, we consider the convex combination of all inputs and outputs which are interval Therefore, the data of table (2) change through the followings: DM j I 1 I i I m O 1 O s DM 1 µx 11 +(1 µ)x 11 µx i1 +(1 µ)x i1 µy11 +(1 µ)y 11 DM k µx 1k +(1 µ)x 1k? µy1k +(1 µ)y 1k DM n µx 1n +(1 µ)x 1n µx in +(1 µ)x in µy1n +(1 µ)y1n Table 3 The combination of the interval data Which 0 μ 1 In order to obtain missing data x ik in table (3) the solution for finding the missing data among crisp data should be executed: ρ 1 = ρ 1 = Therefore: µx 1k +(1 µ)x 1k (μx 1j +(1 μ)x 1j ),,ρ i =?,,ρ m = µy 1k +(1 µ)y 1k (μy 1j +(1 μ)y 1j ),,ρ s = µx mk +(1 µ)x mk µy sk +(1 µ)y sk (μysj +(1 μ)y sj ) (μx mj +(1 μ)x mj )

8 962 Tamaddon et al m s ρ i + ρ r i=1 r=1 ρ = And: x ik (μx ij +(1 μ)x ij ) = ρ Finally: x ik = ρ (μx ij +(1 μ)x ij) (6) In fact, if we substitute ρ in(6), the x ik will be obtained through the following: ( µx 1k +(1 µ)x 1k ++ (μx 1j +(1 μ)x 1j) µx mk +(1 µ)x mk (μx mj +(1 μ)x mj) + µy 1k +(1 µ)y 1k ++ (μy 1j +(1 μ)y 1j) µy sk +(1 µ)y sk (μy sj +(1 μ)y sj) ) (7) (μx ij +(1 μ)x ij ) Therefore, by considering the convex combination of all obvious inputs and outputs, the missing data x ik is obtained through a function on the basis of μ such as f(μ) Theorem 41 The interval [f(0),f(1)] is that interval which is related to the missing data Proof If we call the relation (7) f(μ); then, f(0) is obtained through the following: f(0) = x x ( 1k y ++ mk y )+( 1k ++ sk ) x 1j x mj y1j ysj x ij f(0) is that lower bound in reality which was obtained by using a process followed for interval data So (7) is the smallest lower bounded that is obtained

9 Missing data 963 from this process f(1) is also obtained through the following: f(1) = x x ( 1k y ++ mk y )+( 1k ++ sk ) x 1j x mj y1j ysj x ij Similarly, f(1) is that upper bound which was obtained via the process followed for interval data So, because for the highest amount of μ which is equal to one, it is obtained; then, the largest upper bound which can be obtained from the previous process is like that amount Now, we should show that for each amount: x ik <x ik <x ik Since, when a<b,for 0 <λ<1, the relation a<λa+(1 λ)b <bwill be existedthen: (a) μx 1k +(1 μ)x 1k >x 1k, μx mk +(1 μ)x mk >x mk, μy1k +(1 μ)y1k >y1k, (b) (μx 1j +(1 μ)x 1j) > x 1j (μx mj +(1 μ)x mj ) > n x mj (μy1j +(1 μ)y1j) > y1j μy sk +(1 μ)y sk >y 1k, (μy sj +(1 μ)y sj) > ysj By dividing the sides of relations (a) on relations(b), the following results are obtained: µx 1k +(1 µ)x 1k x > 1k (μx 1j +(1 μ)x 1j ) x 1j µx mk +(1 µ)x mk (μx mj +(1 μ)x mj ) > x mk x mj

10 964 Tamaddon et al µy 1k +(1 µ)y 1k (μy 1j +(1 μ)y 1j ) > y1k y1j µy sk +(1 µ)y sk (μy sj +(1 μ)y sj ) > ysk ysj Therefore, we have the following relations through the sum of the non-equal sides: µx 1k +(1 µ)x 1k + + (μx 1j +(1 μ)x 1j) µy1k +(1 µ)y 1k (μy 1j +(1 μ)y 1j ) + + x + mk x mj + y1k y1j + + ysk ysj µx mk +(1 µ)x mk + (μx mj +(1 μ)x mj) µy sk +(1 µ)y sk (μy sj +(1 μ)y sj ) > x 1k x 1j We have the following relation through dividing the above non-equal sides on and multiplying the sides of that relation on (μx ij +(1 μ)x ij ) > x ij we have: x ik >x ik Similarly, it can be proved that: x ik <x ik + 43 Improving the interval of efficiency Now, we should know that where DM k which has the missing data be placed; which one of the sets E + or E ++? Therefore, we chang the models of (3) and (4) through the followings:

11 Missing data 965 θ = min st θ j=1,j k j=1,j k λ j 0, λ j x ij + λ kx ik θx ik, i =1,,m i t λ j x tj + λ kf(μ) =θf(μ), λ j y rj + λ ky rk y rk, j =1,,n r =1,,s (8) And also: θ = min st θ j=1,j k j=1,j k λ j 0, λ j x ij + λ kx ik θx ik, i =1,,m i t λ j x tj + λ k f(μ) =θf(μ), λ j y rj + λ ky rk y rk, j =1,,n r =1,,s (9) Regarding that in model (8), the evaluated DM is in the worst conditons and in model (9), the evaluated DM is the idealest conditions, we can substitute f(1) and f(0) in models (8) and (9) instead of f(μ) respectively Also, in model (9), the constraint θ f(0) θ f(1) is added to the model in order to establish the condition x tk x tk Now, if we represent an algorithm in order to improve the missing interval;so that the evaluated DM will be placed in one of the E + or E ++ sets 44 algorithm Step 1 We solve models (8) and (9), -ifdm k is placed in E ++ go to step (2) -ifdm k is placed in E + go to step (3) -ifdm k is placed in E go to step (4) Step2 The missing values that we substituted them in models (8) and (9) and the algorithm is stop Step3 The missing values that we substituted them in models (8) and (9) but for improving the interval of efficiency the values of θ f(0), θ f(1) are

12 966 Tamaddon et al considered equal to f( μ) and f(ˆμ) respectively; then, we obtain μ and ˆμ and we substitute the new values of f( μ) and f(ˆμ) instead of f(1) and f(0) in the models, if models (8) and (9) are feasible go to step (1) otherwise the algorithm is stop Step4 The improved condition of DM k is obtained regarding models (8) and (9) then, we determine the missing interval regarding the improve condition and the algorithm is stop efficienciese 5 Numerical example In this part, we bring some examples about crisp and interval cases and then, we bring an example by using convex combination and we perform the algorithm on it Finding the missing data in crisp case consider table (4), In this table 16 decision making units have been shown with 3 inputs and 10 outputs from which the second input and outputs 3, 4 and 5 are missing in DM 6 DM I 1 I 2 I 3 O 1 O 2 O 3 O 4 O 5 O 6 O 7 O 8 O 9 O ? ??? Table 4 The missing data among crisp inputs and outputs For obtaining the missing data of table (4), we use the process which was suggested in section (30 for crisp data; therefore, the missing data are obtained through the followings: x 26 = , y 36 =12123, y 46 =13133, y 56 =41420

13 Missing data 967 Finding the missing data in interval case Consider the following table In this table, there are 5, DMs which have two inputs and two outputs which the data of inputs and outputs are interval and the second input which is related to DM 3, which itself is an interval, is missing input input output output DM j x 1j x 2j y 1j y 2j ?? Table 5 The missing interval among interval inputs and outputs For finding the missing interval [x 23,x 23 ], we use the process expressed in section (41) So the amount of x 23 and the amount of x 23 become equal to and respectivelytherefore, the missing interval is obtained through [00899, 03497] Convex combination of interval data and performance the algorithm on it Consider table (6), the data of this table are the convex combination of the intervals of table (5) which in this table, the second of DM 3 is missing input input output output DM j x 1j x 2j y 1j y 2j 1 15µ + 12(1 µ) 048µ +021(1 µ) 144µ + 138(1 µ) 22µ + 21(1 µ) 2 17µ + 10(1 µ) 07µ +01(1 µ) 159µ + 143(1 µ) 35µ + 28(1 µ) 3 12µ + 4(1 µ)? 198µ + 157(1 µ) 29µ + 21(1 µ) 4 22µ + 19(1 µ) 019µ +012(1 µ) 181µ + 158(1 µ) 25µ + 21(1 µ) 5 15µ + 14(1 µ) 009µ +006(1 µ) 161µ + 157(1 µ) 40µ + 28(1 µ) Table 6 The convex combination of interval data Now, for obtaining the missing data, we use the process which was expressed in section (3) The missing data x 23 will be obtained through a linear function of a variable on the basis of μ; so that this function is as the following: x 23 = ( 12µ+4(1 µ) 69µ+55(1 µ) + 198µ+157(1 µ) 645µ+596(1 µ) + 29µ+21(1 µ) 122µ+98(1 µ) ) (146μ + 049(1 μ)) 3 If μ = 1, the amount of x 23 is equal to which is that x 23 related to example (52) and if μ =0,x 23 is equal to which is that x 23 related to example (52) If μ =1/2, the amount of x 23 is equal to which this amount is located in the interval [x 23,x 23 ] which has been obtained from example (52) Also, if we make the μ related to inputs be equal to 1 and μ related to outputs be equal to 0 and vice versa, the amounts of and

14 968 Tamaddon et al are respectively obtained that these amounts are located within interval [x 23,x 23 ] Now, we execute algorithm 44 for data of table (5): We perform the first process of algorithm 44 the optimal answer of models (8) and (9) are obtained through θ =02176 and θ = 1 Therefore, DM 3 is located in E + and the missing data is that amount of [00899, 03497] But for improving the efficient interval, the process 2 of algorithm can be perform for model (8) So, θ =04783 and the upper bound of the missing interval decreases from to and the DM 3 is located in E + for a further time 6 Conclusion In the data envelopment analysis, for evaluating the operation of the decision making units, we solve a P analogous with each decision making unit and; so, the efficiency of DM is distinguished, But, when some of the data are missing, finding the missing data is very important In this paper, a method for finding the missing data in crisp and interval cases is represented for improving the missing interval in the cases is represented and finally, an algorithm will be represented for improving the missing interval in the case that DM which has interval is located in E + or E ++ sing the suggested algorithm evaluates DM which has the missing interval; but, when the number of inputs and outputs are high, calculating f(μ) has so many problems In this paper, the methods were in crisp and interval conditions A method for finding the missing data in fuzzy condition can be represented, or in the interval condition a method is represented which has more than one missing input(output) and or one DM which doesn t have any of it s input and output data References [1] Yannis GSmirlis, Elias KMaragos, Dimitris KDespotis, Data envelopment analysis with missing values: An interval DEA approach, Eur J Operat Res 177(2006) 1-10 [2] PV O Neal, YA Ozcan, MYanqiang, Benchmarking mechanical ventilation services in teaching hospitals, J Medi Sys 26(3) (2002) [3] Simar, PWilson, Statistical inference in non-parametric frontier models: The state of the art, J Product Anal 13(1)(2000)49-78

15 Missing data 969 [4] WCooper, Seiford, Extending the frontiers of data envelopment analysis A comprehensive text with models, applications, refrences and DEA Solver software Kluwer A cademic Publishers, 1999, p253 [5] TKuosmanen, Modelling blank data entries in Data Envelopment Analysis, Econometrics Working Paper Archive at WST, available from: abs, 2001 [6] CKao, STiu, Data envelopment analysis with missing data: An application to niversity ibraries in Taiwan, Eur J Operat Res 51(8)(2000) Received: September 9, 2008