Ranking units in Data Envelopment Analysis with fuzzy data

Suppose that the exact level of x _ij lies in [ x ij 1 , L , x ij 1 , U ] or [ x ij 2 , L , x ij 2 , U ], such that [ x ij 1 , L , x ij 1 , U ] ⊆ [ x ij 2 , L , x ij 2 , U ] , i = 1 , . . . , m , j = 1 , . . . , n and similarly, the exact level of y _rj lies in [ y rj 1 , L , y rj 1 , U ] or [ y rj 2 , L , y rj 2 , U ], such that [ y rj 1 , L , y rj 1 , U ] ⊆ [ y rj 2 , L , y rj 2 , U ] , r = 1 , . . . , s , j = 1 , . . . , n .

In this case, the efficiency value interval of DMU _o can be attained by two intervals [ h o 1 , L , h o 1 , U ] and [ h o 2 , L , h o 2 , U ], where h o k , L and h o k , U are the optimal solutions of (1) and (2), respectively, when x _ij and y _rj are in [ x ij k , L , x ij k , U ] and [ y rj k , L , y rj k , U ], respectively, (k=1,2). It can be easily shown that the efficiency value intervals do not enlarge when data interval shrink.

Theorem 1. With the above assumptions, [ h o 1 , L , h o 1 , U ] ⊆ [ h o 2 , L , h o 2 , U ] .

Proof. It is sufficient to show h o 2 , L ≤ h o 1 , L and h o 1 , U ≤ h o 2 , U. Suppose that (λ₁, . . . , λ _n , θ) is an optimal solution for (2) according to [ x ij 1 , L , x ij 1 , U ] and [ y rj 1 , L , y rj 1 , U ] interval data. As mentioned previously θ - λ o ≥ 0 , 1 - λ o ≥ 0 .

Since we have x ij 2 , L ≤ x ij 1 , L , x ij 1 , U ≤ x ij 2 , U , y rj 2 , L ≤ y rj 1 , L , y rj 1 , U ≤ y rj 2 , U for j = 1, . . . , n, i = 1, . . . , m and r = 1, . . . , s, we can obtain ∑ j ≠ o λ j x ij 2 , L ≤ ∑ j ≠ o λ j x ij 1 , L ≤ ( θ - λ o ) x io 1 , U ≤ ( θ - λ o ) x io 2 , U and ∑ j ≠ o λ j y rj 2 , U ≥ ∑ j ≠ o λ j y rj 1 , U ≥ ( 1 - λ o ) y ro 1 , L ≥ ( 1 - λ o ) y ro 2 , L . Hence (λ₁, . . . , λ _n , θ) is a feasible solution for (2) according to [ x ij 2 , L , x ij 2 , U ] and [ y rj 2 , L , y rj 2 , U ] interval data, therefore h o 2 , L ≤ h o 1 , L. Similarly, it can be shown that h o 1 , U ≤ h o 2 , U and the proof is completed.□

Hence when all interval data of inputs and outputs are shrinking for all DMUs, then the efficiency value interval of DMU _o is not enlarging. Therefore

- If a DMU is in E⁺⁺ for an interval data, then it remains in E⁺⁺ after the shrinking of the data intervals of all DMUs.

- If a DMU is in E^- for an interval data, then it remains in E^- after the shrinking of the data intervals of all DMUs.

3.1 DEA with fuzzy data

Assume we have n DMUs where DMU _j  (j = 1, . . . , n) consumes input levels x ˜ ij ( i = 1 , . . . , m ) to produce output levels y ˜ rj ( r = 1 , . . . , s ), where all x ˜ ij and y ˜ rs are convex bounded fuzzy numbers.

As the input and the output levels of DMUs are fuzzy numbers, we expect the relative efficiency score of DMU _o  (o ∈ {1, . . . , n}) also to be a fuzzy number.

One can evaluate the relative efficiency score of DMU _o by the CCR model in the output-oriented form, as follows: θ ~ 0 * = M i n θ ~ s . t . ∑ j ≠ o λ j x ~ i j ≤ θ ~ x ~ i o , i = 1 , ... , m ∑ j ≠ o λ j y ~ r j ≥ y ~ r o , r = 1 , ... , s λ j ≥ 0 , j = 1 , ... , n ,(4) where we assume that λ _j is a real variable. In the above model, we use the algebraic sum and product as defined in [12]. Essentially, the inequality concept used in Model (4) must be in the form of fuzzy relations. Therefore, Model (4) is not a linear programming problem, so we cannot solve it with linear techniques. In the following section, a method for solving such models with a review of the efficiency concept is presented.

A target of DEA models is the relative efficiency score evaluation for each DMU that is obtained by computing a numerical level θ. When the i-th input level of the j-th decision making unit is a fuzzy number such as a triangular fuzzy number x ˜ ij = ( m , δ , β ), it means DMU _j has, in an uncertain manner, used an imprecise level of the i-th input and m is one of them with a complete degree of membership 1. Any number that is less than m - δ and greater than m + β cannot be the value for x _ij . All numbers in interval [m - δ, m + β] can be a value for x _ij with some degree of membership.

With the assumption of fuzzy numbers x ˜ ij and y ˜ rj being convex and bounded, from [12] it can be derived that each α-level of x ˜ ij and y ˜ rj is a bounded interval as follows: [ x ˜ ij ] α = [ x ij α , L , x ij α , U ] , [ y ˜ rj ] α = [ y rj α , L , y rj α , U ] , α ∈ ( 0 , 1 ] , j = 1 , . . . , n , i = 1 , . . . , m , r = 1 , . . . , s .

By any choice of α, a set is obtained that consists of n DMUs with interval data, and we can compute its efficiency interval by using (1), (2). Therefore, for any α, an efficiency interval is obtained. As previously mentioned, the efficiency interval of DMU _o for an α-level is denoted by [ h o α , L , h o α , U ], where h o α , L = M i n θ s . t . ∑ j ≠ o λ j x i j α , L + λ o x i o α , U ≤ θ x i o α , U , i = 1 , ... , m ∑ j ≠ o λ j y r j α , U + λ o y r o α , L ≥ y r o α , L , r = 1 , ... , s λ j ≥ 0 , j = 1 , ... , n(5) and h o α , U = M i n θ s . t . ∑ j ≠ o λ j x i j α , U + λ o x i o α , L ≤ θ x i o L , i = 1 , ... , m ∑ j ≠ o λ j y r j α , L + λ o y r o α , U ≥ y r o α , U , r = 1 , ... , s λ j ≥ 0 , j = 1 , ... , n(6) with h o α , L and h o α , U sometimes being equal.

Theorem 3.If 0 < α₁ < α₂ ≤ 1, then [ h o α 2 , L , h o α 2 , U ] ⊆ [ h o α 1 , L , h o α 1 , U ] .

Proof. From α-level definition, it can be inferred that for any i, j and r, we have:

[ x ij α 2 , L , x ij α 2 , U ] ⊆ [ x ij α 1 , L , x ij α 1 , U ]

[ y rj α 2 , L , y rj α 2 , U ] ⊆ [ y rj α 1 , L , y rj α 1 , U ]

and from Theorem 1, the proof is completed.□

From the above theorem, the efficiency value interval shrinks as α increases from 0 to 1. From properties of nested intervals the efficiency score can be introduced by a fuzzy set.

Definition 5. With the above assumptions, the fuzzy set θ ˜ o on interval (0, 1] with the membership function is the relative efficiency score of DMU _o , θ ˜ o ( x ) = sup { t ∣ 0 < t ≤ 1 , x ∈ [ h o t , L , h o t , U ] } if there is some t such that x ∈ [ h o t , L , h o t , U ], otherwise θ ˜ o ( x ) = 0.

Figure. 1. represents an example for fuzzy efficiency score θ ˜ o. In the following part, we show that fuzzy set θ ˜ o satisfies fuzzy number conditions [12]. But, first, we present some important results. The following theorem states that the efficiency interval obtained from a special α-level data is equal to the α-level θ ˜ o, that is denoted by [ θ ˜ o ] α.

Theorem 4.If 0 < α ≤ 1, then[ θ ˜ o ] α = [ h o α , L , h o α , U ].

Proof. From Definition 5, we have, [ θ ˜ o ] α = { x ∈ ( 0 , 1 ] ∣ , θ ˜ o ( x ) ≥ α } .

Let x ∈ [ h o α , L , h o α , U ]; therefore from Definition 5 we concluded θ ˜ o ( x ) ≥ α and so x ∈ [ θ ˜ o ] α. Conversely, let x ∈ [ θ ˜ o ] α, then θ ˜ o ( x ) ≥ α. By considering

θ ˜ o ( x ) = sup { t ∣ x ∈ [ h o t , L , h o t , U ] } = α ¯

we get α ¯ ≥ α, and from Theorem 3 we have, [ h o α ¯ , L , h o α ¯ , U ] ⊆ [ h o α , L , h o α , U ] hence x ∈ [ h o α , L , h o α , U ] and so proof is completed.□

Conclusion 1.θ ˜ o is a convex bounded fuzzy set.

Unimodality is another property of function θ ˜ o ( x ) which is necessary for a fuzzy number.

Theorem 5. The function θ ˜ o ( x ) is unimodal.

Proof. Since x ˜ ij and y ˜ rj are fuzzy numbers, they are unimodal. By choosing α = 1, Models (5) and (6) are converted to a linear programming problem (with crisp data), which is exactly the CCR model and h o 1 , L = h o 1 , U. Therefore [ θ ˜ o ] 1 = { x ¯ } where 0 < x ¯ = h o 1 , L = h o 1 , U ≤ 1.□

Hereafter, we assume [ θ ˜ o ] 1 = { x ¯ o }.

Conclusion 2. If [ θ ˜ o ] α = { x } for any 0 < α < 1, then x = x ¯ o.

From the above theorems, the function θ ˜ o ( x ) is maximized at point x ¯ o. In the next theorem, we show that this function is increasing and decreasing before and after x ¯ o, respectively.

Theorem 6.The functionθ ˜ o ( x )is increasing on interval( 0 , x ¯ o ]and ifx ¯ o < 1, then it is decreasing on interval[ x ¯ o , 1 ].

Proof. Suppose that 0 < x 1 < x 2 ≤ x ¯ o and θ ˜ o ( x 1 ) = α 1 , θ ˜ o ( x 2 ) = α 2 . To complete the proof, we need to show that α₁ ≤ α₂. By contradiction, suppose α₂ < α₁. Therefore by theorems 3,4 we have [ θ ˜ o ] α 1 ⊆ [ θ ˜ o ] α 2.

From the above assumption and Definition 5, we have x 2 ∉ [ θ ˜ o ] α 1, therefore, 0 < x 2 < x 1 ≤ x ¯ o, which contracts the initial assumption. In a similar manner, it can be shown that the function θ ˜ o ( x ) is decreasing on interval [ x ¯ o , 1 ] if x ¯ o < 1.□

The following theorem shows that θ ˜ o ( x ) is a piecewise continuous function on the interval (0, 1].

Theorem 7.a) The functionθ ˜ o ( x )is either continuous or right continuous on the interval( 0 , x ¯ o ).

b) Ifx ¯ o < 1, thenθ ˜ o ( x )is either continuous or left continuous on the interval( x ¯ o , 1 ].

Proof.(a) Suppose x ˆ ∈ ( 0 , x ¯ o ) and θ ˜ o ( x ˆ ) = α ˆ, therefore, 0 < α ˆ < 1. Consider an arbitrary number ɛ > 0 and first let α ˆ > 0. If α ˆ - ɛ ⊁ 0, then without loss of generality it can be supposed that ɛ < α ˆ 2 and similarly, if α ˆ + ɛ ⊀ 1 , then it can be supposed that ɛ < 1 - α ˆ 2, hence, 0 < α ˆ - ɛ < α ˆ < α ˆ + ɛ < 1. From Theorems 3,4 we have, [ θ ˜ o ] α ˆ + ɛ ⊆ [ θ ˜ o ] α ˆ ⊆ [ θ ˜ o ] α ˆ - ɛ

and from Theorem 4, we also have [ θ ˜ o ] α ˆ - ɛ = [ h o α ˆ - ɛ , L , h o α ˆ - ɛ , U ] [ θ ˜ o ] α ˆ = [ h o α ˆ , L , h o α ˆ , U ] [ θ ˜ o ] α ˆ + ɛ = [ h o α ˆ + ɛ , L , h o α ˆ + ɛ , U ] .

We know that x ¯ o belongs to all above sets and 0 < h o α ˆ - ɛ , L ≤ h o α ˆ , L ≤ h o α ˆ + ɛ , L ≤ x ¯ o .

Since x ˆ ∉ [ θ ˜ o ] α ˆ + ɛ, it results in h o α ˆ , L < h o α ˆ + ɛ , L.

There are two cases for the relationship between h o α ˆ , L and h o α ˆ - ɛ , L, as follows:

h o α ˆ - ɛ , L < h o α ˆ , L

In case i, we show that the function θ ˜ o ( x ) is continuous at x ˆ, because by choosing 0 < δ < min { ( h o α ˆ + ɛ , L - h o α ˆ , L ) , ( h o α ˆ , L - h o α ˆ - ɛ , L ) }

for any x, x ˆ - δ < x < x ˆ, and also from Theorem 6 and Definition 5, we have: θ ˜ o ( x ˆ ) - ɛ = α ˆ - ɛ < θ ˜ o ( x ) < θ ˜ o ( x ˆ ) = α ˆ and for any x, x ˆ < x < x ˆ + δ, we have α ˆ = θ ˜ o ( x ˆ ) < θ ˜ o ( x ) < θ ˜ o ( x ˆ ) + ɛ = α ˆ + ɛ ; therefore, the result is obtained. In case ii, we show that the function θ o ˜ ( x ) is right continuous at x ˆ. By choosing 0 < δ < h o α ˆ + ɛ , l - h o α ˆ , l, for any x, x ˆ < x < x ˆ + δ, we have α ˆ = θ ˜ o ( x ˆ ) < θ ˜ o ( x ) < θ ˜ o ( x ˆ ) + ɛ = α ˆ + ɛ and the result is obtained.

Similar to case (ii), the result also follows for α = 0 and the proof of (b) is similar to (a) and is left to the readers.□

From the above theorems, we conclude that if the data for the DMUs are fuzzy numbers, then the relative efficiency scores of the DMUs are also fuzzy numbers. Below, we define the concept of an efficient DMU in fuzzy DEA.

Definition 6.DMU _o is efficient if and only if [ θ o ˜ ] α = { 1 } for any 0 < α ≤ 1.

The above definition has a very strong condition for a DMU to be efficient (see Fig. 2). Even it may happen that none of DMUs is efficient. Because it is possible that E⁺⁺ is empty when all DMUs have interval data. In the next section, we present a numerical method for the ranking of DMUs with fuzzy data.

Definitions 5 and 6 are fundamentally theoretical criteria with respect to the efficiency score θ ˜ o and the efficient DMUs, but in practice, it is not possible that θ ˜ o or efficient DMUs are exactly determined by these definitions.

Here we present a numerical ranking method to determine the rank order of DMUs approximately. Consider the following k real numbers

0 < α 1 < ... < α k = 1 Hence, these numbers form k nested interval of inputs and outputs as follows [ x ˜ i ] α l + 1 ⊆ [ x ˜ i ] α l

[ y ˜ r ] α l + 1 ⊆ [ y ˜ r ] α l

i = 1 , . . . , m , r = 1 , . . . , s , l = 1 , . . . , k - 1 .

So from Theorems 3, 4, we have [ θ ˜ o ] α l + 1 ⊆ [ θ ˜ o ] α l , l = 1 , . . . , k - 1 .

Since each α-level set of θ ˜ o is a closed interval, as mentioned previously, the following statements can be considered. If for some l′ (l′ < k):

–DMU _o  ∈ E⁺⁺, then DMU _o  ∈ E⁺⁺ for all l where l > l′.

–DMU _o  ∈ E^-, then DMU _o  ∈ E^- for all l where l > l′.

Moreover, for α _k  = 1, we know the inputs and outputs are crisp data and [ θ ˜ o ] 1 = θ o = { x ¯ o }. At this level, if DMU _o is efficient (θ _o  = 1), then DMU _o is in E⁺⁺ or E⁺ for all α _l  < 1, and otherwise (θ _o  < 1) then DMU _o is in E⁺ or E^- for all α₁< 1.

It must be considered that the nearer α₁ is to zero, the more exactly the efficient units are assessed. This does not mean that at least one unit will be certainly efficient. Also, after studying the ranking method below, it will be clear that the higher the number of , with the distances between them selected equally, the more fairly and precisely the ranking of the DMUs is carried out.

Therefore, suppose α₁> 0 is sufficiently small and the other α₁ are selected as follows: α l = α 1 + ( 1 − α 1 ) ( l − 1 ) k − 1 , l = 2 , ... , k hence α _k  = 1.

Definition 7. The set of DMUs are in E⁺⁺ for -level data are denoted by ; that is E α l + + = { D M U o h o α l , L = 1 } , l = 1 , ... , k and also the set of DMUs are in E⁺ for -level data are denoted by ; that is E α l + + = { D M U o h o α l , L < 1 & h o α l , U = 1 } , l = 1 , ... , k Obviously E + 1 ≠ φ and the DMUs in E α l ++ have the highest efficiency score. If a DMU is efficient as defined in Definition 6, then it must be in , but the reverse does not necessarily hold true.

Suppose p = 1 and J = {DMU _l , . . . , DMU _n }. There are two cases for E α l ++ as:

E α l ++ ≠ φ

If E α l ++ ≠ φ, then one can ask which DMUs are more efficient, the DMUs in E α l + or in ? We claim the DMUs in E α 2 ++ are more efficient because these DMUs are the DMUs in E α l ++ φ that join E α 2 ++ for φ₂-level data. Therefore, when E α l ++ = φ, we turn to E α 2 ++ = φ.

With the E α l ++ φ assumption, there are also two cases for E α 2 ++. If E α 2 ++ = φ, similary, we turn to E α k ++ and this continues until E α 2 ++ ≠ φ for some l′ (l′ = 1, . . . , k). There is certainly such an l′, because E α k ++ = E l ++ ≠ φ the end.

In this case, let E p = E α l ' + + J = J - E p p = p + 1 .

Now again we form E α l ++ and E α l + (l=1,...,k) for the reduced set J and compute E _p similarly. This process continues until J = φ. It can be at most repeated for n iterations. In the end, we will obtain E₁, . . . , E _p (p ≤ n) from the above process. Clearly the DMUs in E₁ have higher efficiency scores than the ones in E₂, . . . , E _p and also the DMUs in E₂ have the same relation with other DMUs in E₃, . . . , E _p , and so on. Therefore, by arranging DMUs on the basis of these sets we can rank them. Figure. 3 shows a DMU with fuzzy efficiency score θ ˜ 1 is more efficient than the DMU with fuzzy efficiency score θ ˜ 2.

It is possible that E₁, . . . , E _p have more than one member, so some DMUs may have the same rank. This problem can be removed by the following change to determine E₁, . . . , E _p . In each iteration, if E _p has more than one member, then by going back to one level before , i.e., , we choose a DMU in E _p which has the highest h j L , α k - 1 as the member of E _p and omit the other members. Even, if E _p still has more than one member, we can go back to α_l-2,...,α₁ and choose E _p as a singleton. In the end, if E _p is not a singleton yet, then we can conclude that all members of E _p have the same rank order. In the following figure we can see that DMU with fuzzy efficiency score θ ˜ 1 is more efficient than the DMU with fuzzy efficiency score θ ˜ 2 when E _p is not singleton.

Definition 8.q denotes the rank order of a DMU when the DMU ∈E _q , q = 1, . . . , p, after using the ranking method above.

3.3 Algorithm

Main Algorithm

Step 1- Let J = {DMU₁, . . . , DMU _n }, p = 1. Choose the first level α₁ > 0 and choose k as the number of levels and let α l = α 1 + ( 1 - α 1 ) ( l - 1 ) k - 1 , l = 2 , . . . , k . Step 2- Let t = 1 and form E α l + + by Models(5),(6) and Definition 7 for any α _l (l = 1,...,k).

Step 3- If E α t + + = φ, then go to step 4, otherwise go to step 5.

Step 4- Let t = t + 1 and go to step 3.

Step 5- Let E p = E α t + + and go to subalgorithm E _p .

Step 6- Let J = J - E _p . If J = φ then go to step 7, otherwise let p = p + 1 and go to step 2.

Step 7- set q as the rank order of each DMU in E _q and stop.

Subalgorithm E _p

Step 1- Let c = 0 and go to step 2.

Step 2- If E _p is a singleton then return to the main algorithm (step 6), otherwise let c = c + 1, F = E _p and go to step 3.

Step 3- If c > k - 1 then return to the main algorithm (step 6), otherwise let E p = { DMU j ′ ∣ h j ′ α k - c , L = max j ∈ F h j α k - c , L } and go to step 2.

3.4 Application in a practical example

The performance of teachers in an Iranian educational department is evaluated based on the indices introduced in the evaluation questionnaires which are filled out by students.

Assume that the fuzzy triangular number x ˜ ij = ( x ij , δ ij x , β ij x ) is the i-th input value, i = 1, . . . , m, and the fuzzy triangular number y ˜ rj = ( y rj , δ rj y , β rj y ) is the r-th out value, r = 1, . . . , s, of the j-th DMU, j = 1, . . . , n, where δ ij x, δ rj y, β ij x, β rj y are all non-negative numbers.

If δ ij x, β ij x are positive numbers, then x ˜ ij has the following membership function in ℝ: x ˜ ij ( t ) = { t - x ij + δ ij x δ ij x x ij - δ ij x ≤ t < x ij x ij + β ij x - t β ij x x ij ≤ t < x ij + β ij x 0 otherwise , and if δ ij x = 0 and β ij x is positive, then x ˜ ij = ( x ij , 0 , β ij x ) has the following membership function in ℝ: x ˜ ij ( t ) = { x ij + β ij x - t β ij x x ij ≤ t < x ij + β ij x 0 otherwise , and if β ij x = 0 and δ ij x is positive, then x ˜ ij = ( x ij , δ ij x , 0 ) has the following membership function in ℝ:

x ˜ ij ( t ) = { t - x ij + δ ij x δ ij x x ij - δ ij x ≤ t < x ij 0 otherwise .

One can discuss the membership function of y ˜ rj in a similar manner.

By previously mentioned, for α ∈ (0, 1], the α-level of x ˜ ij and y ˜ rj will be as follows ( x ˜ ij ) α = ( x ij , δ ij x , β ij x ) α = [ x ij - δ ij x ( 1 - α ) , x ij + β ij x ( 1 - α ) ](7)( y ˜ rj ) α = ( y rj , δ rj y , β rj y ) α = [ y rj - δ rj y ( 1 - α ) , y rj + β rj y ( 1 - α ) ] .(8) The important advantage of using fuzzy triangular numbers is that for α = 0, as well, one can define an α-level to form a convex bounded interval, since the 0-level of fuzzy triangular number N = (m, δ, β) will be as fillows N 0 = ( m - δ , m + β ) = supp ( N ) .

So, for the fuzzy triangular numbers x ˜ ij and y ˜ rj, we have ( x ˜ ij ) 0 = ( x ij - δ ij x , x ij + β ij x ) ( y ˜ rj ) 0 = ( y rj - δ rj y , y rj + β rj y ) , which can also be obtained by setting α = 0 in (7-8).

Now, for α ∈ [0, 1], the efficiency bounds of DMU _o resulting from an α-level, [ h o α , L , h o α , U ], can be obtained as follows, regarding models (5-6)

h o α , L = M a x ∑ r = 1 s u r y r o − ( 1 − α ) ∑ r = 1 s u r δ r o y s . t . ∑ i = 1 m v i x i o + ( 1 − α ) ∑ i = 1 m v i β i o x = 1 , ∑ r = 1 s u r y r j − ∑ i = 1 m v i x i j + ( 1 − α ) ( ∑ r = 1 s u r β r j y + ∑ i = 1 m v i δ i j x ) ≤ 0 , ( j = 1 , … , n , j ≠ o ) ∑ r = 1 s u r y r o − ∑ i = 1 m v i x i o − ( 1 − α ) ( ∑ r = 1 s u r δ r o y + ∑ i = 1 m v i β i o x ) ≤ 0 , ( v i , u r ≥ ε , ∀ i , r . )(9) and h o α , U = M a x ∑ r = 1 s u r y r o + ( 1 − α ) ∑ r = 1 s u r β r o y s . t . ∑ i = 1 m v i x i o − ( 1 − α ) ∑ i = 1 m v i δ i o x = 1 , ∑ r = 1 s u r y r j − ∑ i = 1 m v i x i j − ( 1 − α ) ( ∑ r = 1 s u r δ r j y + ∑ i = 1 m v i β i j x ) ≤ 0 , ( j = 1 , … , n , j ≠ o ) ∑ r = 1 s u r y r o − ∑ i = 1 m v i x i o + ( 1 − α ) ( ∑ r = 1 s u r β r o y + ∑ i = 1 m v i δ i o x ) ≤ 0 , ( v i , u r ≤ ε , ∀ i , r . )(10) The indices by which the students were asked to evaluate teachers are presented in Table 1.

For each teacher, one of the qualitative expressions: Excellent, Good, Average, Poor, and Very Poor is assigned to each index by the students.

A suitable choice of fuzzy numbers for the students’ responses is to use fuzzy triangular numbers. Table 2 shows five typical triangular numbers in the interval [0, 1], corresponding to the responses.

The above numbers are depicted in Fig. 5.

To calculate the score of a certain teacher for each index, we perform the fuzzy addition on all the responses to that certain index and then divide the sum by the number of respondents. This operation, called normalizing the data, causes the resulting scores to lie in the interval [0, 1]. Normalizing data are independent of the unit of measurement and the number of respondents. Therefore, the effect of the number of the respondents not being equal from one teacher to the other is removed.

The indices for the evaluation of units are always divided into two categories, inputs and outputs,which are represented in Table 1.

Therefore, in the model employed for evaluating the teachers, they are considered as DMUs, such that have 11 outputs (s = 11) but no inputs (m = 0). The outputs of the j-th teacher are denoted by y_1j to y_11j in order of appearance in Table 1. The CCR model for this situation is as follows.

θ = M a x ∑ r = 1 11 u r y r o s . t . ∑ r = 1 11 u r y r j ≤ 1 , j = 1 , ... , n u r ≥ ε , r = 1 , ... , 11(11)

where n is the number of the teachers who are compared together. Thus, models (9-10) to determine the efficiency bounds of DMU _o (o ∈ {1, . . . , 11}) for a certain α-level will be as follows.

h o α , U = M a x ∑ r = 1 11 u r y r o + ( 1 − α ) ∑ r = 1 11 u r β r o y s . t . ∑ r = 1 11 u r y r j − ( 1 − α ) ∑ r = 1 11 u r δ r j y ≤ 1 , ( j = 1 , … , n , j ≠ o ) ∑ r = 1 11 u r y r o + ( 1 − α ) ∑ r = 1 11 u r β r o y ≤ 1 , ( u r ≥ ε , r = 1 , ... , 11. )(12)

and h o α , L = M a x ∑ r = 1 11 u r y r o − ( 1 − α ) ∑ r = 1 11 u r δ r o y s . t . ∑ r = 1 11 u r y r j + ( 1 − α ) ∑ r = 1 11 u r β r j y ≤ 1 , ( j = 1 , … , n , j ≠ o ) ∑ r = 1 11 u r y r o − ( 1 − α ) ∑ r = 1 11 u r δ r o y ≤ 1 , ( u r ≥ ε , r = 1 , ... , 11 )(13)

Replacing model (5-6) by models (12-13) in the algorithm in the previous section makes it possible to rank the teachers.

Now, consider a ranking to be carried out for five teachers, the overall data of which is presented in Tables 3–7, according to the 11 questions in Table 1. In the last column, the score for each index is obtain for each teacher as a fuzzy triangular number, using the fuzzy valuing method in Table 2. Note that the number of respondents varies from one teacher to another, a problem which has been solved by normalizing the data.

By employing the ranking algorithm, all the teachers belong to the class E⁺ for α < 0.4, and teacher number 4 belongs to the class E⁺⁺ for α = 0.4, therefore, teacher number 4 has the first rank. By omitting and employing the ranking algorithm again, teachers number 1 and 5 belong to the class E⁺⁺ for α = 0.2, by using the algorithm E _p , teacher number 5 has the second rank. Now, we also omit teacher 5, and employ algorithm again, teacher number 1 has the third rank for α = 0.2, and after omitting teacher number 1, teacher number 2 has the fourth rank for α = 0.1, and in the end, we will obtain that teacher number 3 has the fifth rank.