Data envelopment analysis in hierarchical category structure with fuzzy boundaries

Abstract

Data envelopment analysis (DEA) is used for the performance evaluation of a set of decision making units (DMUs). Such performance scores are necessary for taking managerial decisions like allocation of resources, improvement plans for the poor performers, and maintaining high efficiency of the leaders. In classical DEA, it is assumed that the DMUs are operating in a similar environment. But in practice, this assumption is normally broken as DMUs operate in a varied environment due to several uncontrollable factors like socio-economic differences, competitiveness in the region and location. In order to address this issue, categorical DEA was proposed for the construction of peer groups by creating crisp categories based on the level of competitiveness. However, such categorizations suffer from indeterminate factors, for example, human judgment and biases, linguistic ambiguity and vagueness. In this paper, we propose a more realistic DEA approach which is capable of handling categories defined in natural languages or with vague boundaries and generates efficiency as triangular fuzzy number. The analysis indicates that if a higher degree of fuzziness is allowed while defining the boundaries of the reference set, it results in (1) a compromise with the accuracy, signified by the spread of the fuzzy efficiency, (2) degradation of the quality, signified by the centre of the fuzzy efficiency, of the decision. Finally, the applicability of this approach has been demonstrated using public library data for different regions in Tokyo city. The sensitivity of the optimal decisions to the changes in fuzzy parameters has also been investigated.

Appendix

1.1 Appendix A: Impact of possibility levels or belief degrees of the inequality constraints on the solution space

In the de-fuzzification technique of model (7), the degree of confidence or belief degree of each constraint is denoted by $h_{i}, i=1, \dots ,m$. It marks the minimum possibility that the constraint is satisfied. Therefore, in the proposed model, if the possibility of any binding constraints is modified then the optimal solution also changes. It’s worthy of note that, the aspiration level and corresponding possibility measure must be reconsidered rationally in light of the new belief degree; otherwise it may lead to infeasibility of the model.

Inevitably, the total spread, the objective function, decreases if the solution space reduces. It is shown by Tanaka and Asai (1981) that the solution space reduces with higher belief degree. Threfore, a higher belief degree leads to smaller spread and also a smaller Y-intercept, ceteris paribus.

Fig. 7

Distortion of shape of optimal fuzzy variable at different levels of possibility

It is clear from Fig. 7, the centre needs to shift towards the origin for a higher belief degree if the spread has reduced. Intuitively, a higher belief degree h manifests itself in a lower uncertainty in the decision variable as well. As a result, the possibility distribution of decision variables contracts.

A higher $h_l$ implies lower Y-intercept. It can be easily determined from: $\mu _{\tilde{Y}_j}(0) \le 1-h_l$. One can also refer to Fig. 7. For different values of $h_l, l=1, \, 2$, and 3 such that $h_1> h_2 > h_3$, the corresponding functions of Y-intercepts are denoted by $Y_1, Y_2$, and $Y_3$, respectively. Reduction of spread, together with lower Y-intercept jointly make the centre shift left towards the origin. As a result the centre also reduces e.g., $C_3> C_2 > C_1$. This is to say that, lower return, as symbolised by centre value, is accompanied with lower amount of risk, where risk is conveyed by the spread of the possibility distribution.

1.2 Appendix B: Feasibility of the proposed fully fuzzy DEA model (7)

Recollect that corollary of Farka’s lemma states that for $A \in R^{m \times n}$ and $b \in R^{m \times 1}$ exactly one of the Eqs. (15.1, 15.2) holds:

$$\begin{aligned} \exists x \in R^{ n \times 1} \; such \; that \; Ax \le b, x \ge 0 \end{aligned}$$

(15.1)

$$\begin{aligned} \exists y \in R^{ 1 \times m} \; such \; that \; A^T y \ge 0, \; y^Tb < 0 \end{aligned}$$

(15.2)

To improve the readability, let’s declare the following vectors for the decision variables $\mathbf {U^{g}}=[u^g_1,u^g_2, \dots , u^g_s]$ and $\mathbf {V^{g}}=[v^g_1,v^g_2, \dots , v^g_m]$ where $g=c \, or \, s \,$ for centre or spread respectively. The inequality constraints of model (7), can easily be expressed in the form of $Ax \le b$ as:

$$\begin{aligned} \begin{pmatrix} 0&0&Y_o&-hY_o&-1&-h \end{pmatrix} \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ u^s_0 \end{bmatrix} \ge (A^c_o-h A^s_o)=\bar{Z} \end{aligned}$$

equivalently,

$$\begin{aligned} \begin{pmatrix} 0&0&-Y_o&hY_o&1&h \end{pmatrix} \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ u^s_0 \end{bmatrix} \le -\bar{Z}. \end{aligned}$$

Similarly, the $j^{th}$ constraint from the third set of constraints of model (7) can be abstracted as:

$$\begin{aligned} \begin{pmatrix} -X_j&h_jX_j&Y_j&-h_jY_j&-1&h_j \end{pmatrix} \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ u^s_0 \end{bmatrix} \le 0. \end{aligned}$$

It is a well-received idea that an equality constraint, e.g., $A_2x_2=b_2$, can be substituted by two inequalities e.g., $A_2x_2 \le b_2$ and $A_2x_2 \ge b_2$. Hence, the inequalities for the second constraint in model (7), can be written in matrix multiplication form as:

$$\begin{aligned} \begin{pmatrix} X_o&(h-1)X_o&0&0&0&0 \end{pmatrix} \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ u^s_0 \end{bmatrix} \le (1^c+(1-h)1^s)=\bar{1} \end{aligned}$$

and,

$$\begin{aligned} \begin{pmatrix} -X_o&(1-h)X_o&0&0&0&0 \end{pmatrix} \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ u^s_0 \end{bmatrix} \le -(1^c+(1-h)1^s)=-\bar{1}. \end{aligned}$$

The Eq. (15.3) arises after combining all the constraints of model (7).

$$\begin{aligned} \left[ \begin{array}{cccccc} 0 &{}\quad 0 &{}\quad Y_o &{}\quad -hY_o &{}\quad -1 &{}\quad -h \\ X_o &{}\quad (h-1)X_o &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ -X_o &{}\quad (1-h)X_o &{}\quad 0 &{} \quad 0 &{}\quad 0 &{}\quad 0 \\ -X_1 &{}\quad h_1X_1 &{}\quad Y_1 &{}\quad -h_1Y_1 &{}\quad -1 &{}\quad h_1 \\ \vdots \\ -X_n &{}\quad h_nX_n &{} \quad Y_n &{}\quad -h_nY_n &{} \quad -1 &{} \quad h_n \\ \end{array} \right] \begin{bmatrix} V^c \\ V^s \\ U^c \\ U^s \\ u^c_0 \\ V^s_0 \end{bmatrix} \le \begin{bmatrix} -\bar{Z} \\ \bar{1} \\ -\bar{1} \\ 0 \\ \vdots \\ 0 \end{bmatrix} \end{aligned}$$

(15.3)

Let $\mathbf {A} \in R^{(n+3) \times 6 }$, $\mathbf {x} \in R^{6 \times 1}$ and $\mathbf {b} \in R^{(n+3) \times 1}$ denote the coefficient matrix, decision variable matrix and constant matrix, respectively. As per Farka’s lemma, for a $\mathbf {y} \in R^{(n+3) \times 1}_+$ and $\mathbf {A}^T \mathbf {y}=0$, we have to prove $\mathbf {b}^T \mathbf {y} \nless 0$ i.e., Eq. (15.2) doesn’t hold. Only then it can be asserted that Eq. (15.1) holds and $\mathbf {A}\mathbf {x} \le \mathbf {b}$ has a feasible solution. The following set of equations is generated when $\mathbf {A}^Ty=0$ is expanded:

$$\begin{aligned} \left[ \begin{array}{cccccc} 0 &{}\quad X_o &{}\quad -X_o &{}\quad -X_1 &{}\quad \dots &{}\quad -X_n \\ 0 &{}\quad (h-1)X_o &{}\quad (1-h)X_o &{} \quad h_1X_1 &{}\quad \dots &{}\quad h_nX_n \\ -Y_o &{}\quad 0 &{}\quad 0 &{}\quad Y_1 &{}\quad \dots &{}\quad Y_n \\ hY_o &{} \quad 0 &{}\quad 0 &{}\quad Y_1 &{}\quad \dots &{}\quad Y_n \\ +1 &{}\quad 0 &{}\quad 0 &{}\quad -1 &{}\quad \dots &{} \quad -1 \\ h &{}\quad 0 &{}\quad 0 &{}\quad h_1 &{}\quad \dots &{}\quad h_n \\ \end{array} \right] \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ \vdots \\ y_{n+3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{aligned}$$

Equation (15.4) can be derived from the fifth equation of the above set of equations.

$$\begin{aligned} y_1=y_4+y_5+ \dots +y_{n+3}. \end{aligned}$$

(15.4)

Similarly, Eq. (15.5) represents the last equation of that set.

$$\begin{aligned} hy_1+h_1y_4+h_2y_5+ \dots +h_ny_{n+3}=0. \end{aligned}$$

(15.5)

substituting the value of $y_1$ from Eq. (15.4), Eq. (15.5) can be converted to

$$\begin{aligned} y_4(h_1+h)+y_5(h_2+h)+ \dots +y_{n+3}(h_n+h)=0. \end{aligned}$$

Recall, every component of $\mathbf {y}$ i.e., $y_i, \, i=1 \dots (n+3)$ is non-negative. Since, all the bracketed terms $(h+h_i), \, \forall i=1\dots n$, denoting summation of possibility levels, are inevitably positive it can be ascertained that,

$$\begin{aligned} y_4=y_5= \dots =y_{n+3}=0. \end{aligned}$$

(15.6)

The value of $y_1$, as followed from Eqs. (15.4) and (15.6), turns out to be 0. Now, consider the first constraint of $\mathbf {A}^Ty=0$.

$$\begin{aligned} X_oy_2-X_oy_3-X_1y_4+ \dots +X_ny_{n+3}=0. \end{aligned}$$

(15.7)

Substituting the values of $y_i \, , \, where \, i=4 \dots n+3$, from Eq. (15.6) into Eq. (15.7) leads to:

$$\begin{aligned} y_2=y_3. \end{aligned}$$

(15.8)

For Eq. (15.2) to have a solution, $\mathbf {b}^Ty<0$ must hold. Expanding matrix $\mathbf {b}$, we can write:

$$\begin{aligned} -\bar{Z}y_1+\bar{1}y_2-\bar{1}y_3 < 0. \end{aligned}$$

(15.9)

Using Eq. (15.8), we derive $\bar{Z}y_1>0$. Recall, we have also derived the value of $y_1$ to be 0. Hence, $\bar{Z}y_1 \nless 0$. This completes the proof. $\square$

Table 6 Fuzzy membership of DMUs to fuzzy categories based on their population for scenario-B

Table 7 Fuzzy membership of DMUs to fuzzy categories based on their population for scenario-C

1.3 Appendix C: At the optimality of model (7), the aspiration level constraint becomes binding

Proof

We prove this lemma using the logic of the following reductio ad absurdum. Assume that the optimal solution of model (7) is denoted by $(u^{c*}_r,u^{s*}_r,v^{c*}_i,v^{s*}_i,u^{c*}_0,u^{s*}_0)$. The convexity constraint is binding, and suppose rest of the constraints are non-binding. Then it follows that

$$\begin{aligned} \sum \limits _{r=1}^{s} u^{c*}_ry_{ro}-u^{c*}_0-A^{c}_{o} > h\left( \sum \limits _{r=1}^{s} u^{s*}_ry_{ro}+u^{s*}_0+A^{s}_{o}\right) \end{aligned}$$

(16.1)

$$\begin{aligned} \sum \limits _{i=1}^{m} v^{c*}_ix_{ij} + u^{c*}_0 - \sum \limits _{r=1}^{s}u^{c*}_ry_{rj} > h_{j}\left( \sum \limits _{r=1}^{s} u^{s*}_ry_{rj}+u^{s*}_0+\sum \limits _{i=1}^{m} v^{s*}_ix_{io}\right) \,\, \, \forall j=1 \dots n. \end{aligned}$$

(16.2)

Let $k_0$ and $k_j \, \forall j=1 \dots n$ be non-negative variables such that,

$$\begin{aligned}&\sum \limits _{r=1}^{s} u^{c*}_ry_{ro}-u^{c*}_0-A^{c}_{o} = k_0 + h\left( \sum \limits _{r=1}^{s} u^{s*}_ry_{ro}+u^{s*}_0+A^{s}_{o}\right) \\&\sum \limits _{i=1}^{m} v^{c*}_ix_{ij} + u^{c*}_0 - \sum \limits _{r=1}^{s}u^{c*}_ry_{rj} = k_j + h_{j}\left( \sum \limits _{r=1}^{s} u^{s*}_ry_{rj}+u^{s*}_0+\sum \limits _{i=1}^{m} v^{s*}_ix_{io}\right) \,\, \, \forall j=1 \dots n. \\ \end{aligned}$$

Hence, $(u^{c*}_r,u^{s*}_r,v^{c*}_i,v^{s*}_i,u^{c*}_0, u^{s*}_0+min(k_0,k_1,\dots ,k_n))$ is a feasible solution of model (7). This contradicts the optimal solution $(u^{c*}_r,u^{s*}_r,v^{c*}_i,v^{s*}_i,u^{c*}_0,u^{s*}_0)$. Therefore, the assumption that all inequality constraints remain non-binding in the optimal solution is incorrect.

Let’s denote the dual variables for the constraints of model (7) by vector $Y \in R^{(n+2) \times 1}$ with components as $y_1, \; y_2 \; and, \; y_{3j}$ where j varies from 1 to n. Recall, a variant of Farka’s lemma states that the system $Ax \le b$, where $x \ge 0$ and the system $A^Ty \ge 0$ and $b^Ty<0$, where $y \ge 0$ can not hold simultaneously. As a corollary, it is correct to state that the system $Ax \le b$, where $x \ge 0$ and the system $A^Ty \ge 0$ and $b^Ty \ge 0$, where $y \ge 0$ shall hold simultaneously. The constraints of model (7) of the main paper can be framed as $Ax \le b$, where $x \ge 0$ as shown in Eq. (15.3). Similarly, $A^Ty \ge 0$ can be represented as:

$$\begin{aligned} \left[ \begin{array}{cccccc} 0 &{}\quad X_o &{}\quad -X_o &{}\quad -X_1 &{}\quad \dots &{} \quad -X_n \\ 0 &{}\quad (h-1)X_o &{}\quad (1-h)X_o &{}\quad h_1X_1 &{}\quad \dots &{}\quad h_nX_n \\ -Y_o &{}\quad 0 &{}\quad 0 &{} \quad Y_1 &{}\quad \dots &{}\quad Y_n \\ hY_o &{}\quad 0 &{}\quad 0 &{}\quad Y_1 &{} \quad \dots &{}\quad Y_n \\ +1 &{}\quad 0 &{}\quad 0 &{}\quad -1 &{}\quad \dots &{}\quad -1 \\ h &{}\quad 0 &{}\quad 0 &{}\quad h_1 &{}\quad \dots &{}\quad h_n \\ \end{array} \right] \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ \vdots \\ y_{n+3} \end{bmatrix} \ge \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}. \end{aligned}$$

(16.3)

The fifth constraint of the system sets the minimum value of $y_1$ as:

$$\begin{aligned} y_1 \ge y_4 + y_5+ \dots +y_{n+3}. \end{aligned}$$

(16.4)

Moreover, in Eq. (16.3) $y_k, \; k=1,4,\dots ,(n+3)$ values signify the dual variables of the inequality constraints of model (7). Dual variables for the aspiration level constraint is $y_1$. The dual variables for each constraints in the third set of constraints in model (7) is signified by $y_4, y_5, \dots , y_{n+3}$ respectively. As we have already proved one of these inequalities stand binding at optimality, the following equation can be formed from duality theorem:

$$\begin{aligned} y_1 + y_4 + y_5+ \dots +y_{n+3}>0. \end{aligned}$$

(16.5)

It is not hard to understand from Eqs. (16.4, 16.5) that $y_1 > 0$. This is sufficient condition for the aspiration constraint to be binding at optimality. $\square$