Properties of DEA-integrated balance and specialization measures

Abstract

A recently proposed approach measures balance and specialization degrees as opposite key performance indicators in addition to the efficiency and effectiveness scores well known from data envelopment analysis (DEA). It has been integrated into the DEA methodology by formulating output-oriented models of CCR- and BCC-type and has successfully been applied to two case studies of a European pharmacy business as well as German business schools’ research performance. Because the models are of (non-linear) minimax-type the calculation of the scores is not straightforward. Therefore this paper analyses the properties of the models in order to better understand them and to improve the solution process. The properties derived allow to propose an efficient heuristic solution procedure. While some properties hold for more general models others are true only for models with special convex polyhedric cones defining the balance set. A main result states that specialization of a decision-making unit is essentially measured by its angle distance to the balance cone.

Appendices

Appendix A

From the following proofs it becomes obvious that most of the properties of the DEA-integrated specialization model are true also for balance sets $\varvec{C}$ more general than the convex polyhedral balance cones defined in Sect. 3.2 (except for Theorem 5). These generalizations are based on particular characteristics of cones and formulated as remarks after the respective proofs. Before proving the theorems some basic characterization and implications relating to Assumptions 1 and 2 are given in the next definition and corollary.

Definition 1

• a) $\varvec{T}^\mathrm{env}$ is output bounded if and only if for all $(\varvec{x_0};\varvec{y_0}) \in \varvec{T}^\mathrm{env}$ there exists an (upper bound) $\varvec{y_0^u}$ fulfilling $\varvec{y_0^u} \ge \varvec{y}$ for all $(\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env}$ with $\varvec{x} \le \varvec{x_0}$.

• b) $\varvec{T}^\mathrm{env}$ fulfills free disposal if and only if with $(\varvec{x_0};\varvec{y_0}) \in \varvec{T}^\mathrm{env}$ also $(\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env}$ for all $\varvec{x} \ge \varvec{x_0}$ and $\varvec{0} \le \varvec{y} \le \varvec{y_0}$.

Corollary 2

Any balance cone $\varvec{C}$ satisfying Assumption 2 has the following attributes:

• a) $\varvec{C}$ is closed and convex.

• b) There is at least one balanced output mix $\varvec{y} \in \varvec{C}$ with $\varvec{y} > \varvec{0}$, i.e. $y_r > 0 \ \forall r$. Furthermore, if $\varvec{y} \in \varvec{C}$ with $y_{r} = 0$ for some output r then $\varvec{y} = \varvec{0}$.

• c) $\varvec{C}$ displays non-decreasing variability of scale, i.e. $\gamma \varvec{y} \in \varvec{C} \ \forall \varvec{y} \in \varvec{C} \ \forall \gamma \ge 1$.

• d) $\varvec{0} \in \varvec{C}$.

• e) $\varvec{C}$ displays non-increasing variability of scale, i.e. $\gamma \varvec{y}\in \varvec{C} \ \forall \ \varvec{y}\in \varvec{C}, 0 \le \gamma \le 1$.

Hence there is no balanced output combination with a zero output except $\varvec{y} = \varvec{0}$. The last attribute of the corollary holds for every convex balance set $\varvec{C}$ containing the origin. It is formally identical to the property of ‘weak disposability’ of outputs usually postulated in environmental economics with respect to bad outputs (e.g. Färe and Grosskopf 2005: 47). As regards content, however, the first property is concerned with performance or preferences ($\varvec{C}$) while the second considers the (technical) possibilities of production or disposability $(\varvec{T}^\mathrm{env})$.

Now we turn to the proofs of the theorems formulated in Sect. 4.

Proof of Theorem 1

• a) Note that $\varvec{T}^\mathrm{env}(\varvec{x_0};{\varvec{y^b})}:= \{ \left. (\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env} \right| \varvec{x} \le \varvec{x_0}, \varvec{y} \ge \varvec{y^b}\}$ is non-empty, bounded and closed, i.e. compact. Therefore, the minimum in (3) exists.

• b) Note that $\varvec{B(\varvec{x_0};\varvec{y_0})}:= \left\{ \left. (\varvec{x_0};\varvec{y}) \in \varvec{B} \right| \varvec{y} \le \varvec{y_0} \right\}$ is non-empty, bounded and closed, i.e. compact. Due to the continuity of $\eta (\varvec{x_0};\varvec{y^b})$ the minimum in (4) exists. Hence also (5) is well defined. Due to $\varvec{0} \in \varvec{C}$ and since $\varvec{T}^\mathrm{env}$ fulfills free disposal also $(\varvec{x_0};\varvec{0}) \in \varvec{B}$. Furthermore, $(\varvec{x_0};\varvec{0})$ is balanced and weakly dominated by $(\varvec{x_0};\varvec{y_0})$, i.e. $(\varvec{x_0};\varvec{0})$ fulfills (NC). This guarantees the existence of a corresponding balanced point and therefore the feasibility of (4) and (5).

• c)

  • (i) The proposition follows by (NC) and from the assumption that $\varvec{y} = \varvec{0}$ is the only balanced output mix with $y_{r} = 0$ for some r.

  • (ii) Due to the convexity of $\varvec{C}$ and the existence of at least one output mix $\varvec{y} \in \varvec{C}$ with $\varvec{y} > \varvec{0}$ there are $\epsilon >0$ and $\varvec{\tilde{y}} \in \varvec{C}$ with $\varvec{0}<\epsilon \cdot \varvec{y_0} \le \varvec{\tilde{y}} \le \varvec{y_0}$. $\varvec{\tilde{y}}$ fulfills (NC) and has a higher efficiency than $\varvec{0} \in \varvec{C}$. Hence $\varvec{y^b_0} \ge \varvec{\tilde{y}} > \varvec{0}$. $\square$

Remark

Theorem 1 also holds for more generalized balance sets only fulfilling attributes a, b and d of Corollary 2.

Proof of Theorem 2

Note that $\eta ^b \ge 1$. Corollary 2c implies $\eta ^b\varvec{y^{b}_0}\in \varvec{C}$. Hence $(\varvec{x_0};\eta ^b\varvec{y^{b}_0}) \in \varvec{B}$. $\square$

Remark

Theorem 2 also holds for more generalized balance sets only fulfilling attributes a, b and c of Corollary 2.

Proof of Theorem 3

• a) If $y_{r0} = 0$ for some r we are ready by Theorem 1ci. Therefore, let be $y_{r0} > 0$ for all outputs r. By Theorem 1cii we have $\varvec{y^b_0} > \varvec{0}$. We show Theorem 3 by contradiction. Assume $y_{r0}^b \ne y_{r0} \ \forall r$. Then by condition (NC) we have $\varvec{y^b_0} < \varvec{y_0}$. Now there exists $\epsilon > 0$ with $\varvec{\tilde{y}} := (1 + \epsilon ) \varvec{y^b_0} \le \varvec{y_0}$ and $\varvec{\tilde{y}} > \varvec{y^b_0}$. Note that due to $\varvec{C}$ fulfilling the condition of non-decreasing variability of scale we have $\varvec{\tilde{y}} \in \varvec{C}$. Now (3) implies $(*)$: $\eta (\varvec{x_0};\varvec{y^b_0}) > \eta (\varvec{x_0};\varvec{\tilde{y}})$ because of $\varvec{\tilde{y}} = (1 + \epsilon ) \varvec{y^b_0}$. On the other hand from (4) and $\varvec{\tilde{y}} < \varvec{y_0}$ we can conclude: $\eta (\varvec{x_0};\varvec{\tilde{y}}) \ge \mathrm{min}\{\eta =\eta (\varvec{x_0};\varvec{y})| \varvec{y} \le \varvec{y_0}; \ (\varvec{x_0};\varvec{y}) \in \varvec{B} \}= \eta (\varvec{x_0};\varvec{y^b_0})$ which contradicts $(*)$.

• b) If $\varvec{y^b_0}=\varvec{y_0}$ we are ready. Let $\varvec{y^b_0} \ne \varvec{y_0}$ and $\varvec{\tilde{y}}:= \gamma \varvec{y^b_0} + (1-\gamma )\varvec{y_0} \in \varvec{C}, 0 < \gamma < 1$. Furthermore let $\eta ^b = \eta (\varvec{x_0}; \varvec{y^b_0})$ be the optimal solution of (4), i.e. $\theta ^b = \frac{1}{\eta ^b}$ is maximal. By (4) follows $(*)$: $\tilde{\eta }:=\eta (\varvec{x_0}; \varvec{\tilde{y}}) \ge \eta ^b$, i.e. $(\varvec{x_0}; \varvec{\tilde{y}})$ has no higher efficiency score than $(\varvec{x_0}; \varvec{y^b_0})$. With (NC) and Theorem 3a we have $\varvec{y^b_0} \le \varvec{\tilde{y}} \le \varvec{y_0}$ and ${\tilde{y}_r} = y_{r0}$ for at least one r. For this reason $(\varvec{x_0};\varvec{y^b_0})$ is weakly dominated by $(\varvec{x_0};\varvec{\tilde{y}})$ and we have $\tilde{\eta } \le \eta ^b$. With $(*)$ follows: $\tilde{\eta } = \eta ^b$. Thus, $(\varvec{x_0};\varvec{\tilde{y}})$ and $(\varvec{x_0};\varvec{y^b_0})$ have the same efficiency score. Hence, $(\varvec{x_0};\varvec{\tilde{y}})$ has to be a corresponding balanced point for DMU $(\varvec{x_0};\varvec{y_0})$, too.

• c) This is an immediate implication of Theorem 3b.

• d) This is an immediate implication of Theorems 1b and 3b. $\square$

Remark

Theorem 3 also holds for more generalized balance sets only fulfilling attributes a, b and c of Corollary 2.

For Theorem 4 we need the following:

Lemma 1

Let be $\varvec{y_0} \notin \varvec{C}$. Let $(\varvec{x_0};\varvec{y^b_0})$ be a corresponding balanced point for DMU $(\varvec{x_0};\varvec{y_0})$ with balanced efficiency $\theta ^b = \frac{1}{\eta ^b}$. If the reference point $(\varvec{x_0};\varvec{y_{*}^b})$ with $\varvec{y_{*}^b}=\eta ^b \varvec{y^{b}_0}$ of $(\varvec{x_0};\varvec{y^b_0})$ referring to (3) is strongly (output-)efficient then $(\varvec{x_0};\varvec{y^b_0}) \in \partial \varvec{B}$ holds.

Proof

We show Lemma 1 by contradiction. Assume $(\varvec{x_0};\varvec{y^b_0}) \notin \partial \varvec{B}$. By Theorem 3c the boundary point $(\varvec{x_0};\varvec{\tilde{y}^b}) \in \partial \varvec{B}$ on the line segment between $(\varvec{x_0};\varvec{y^b_0})$ and $(\varvec{x_0};\varvec{y_0})$ is also a corresponding balanced point for DMU $(\varvec{x_0};\varvec{y_0})$ with same balanced efficiency $\theta ^b$ and the reference point $(\varvec{x_0};\eta ^b\varvec{\tilde{y}^b})$. Because of $\varvec{y^b_0} \le \varvec{\tilde{y}^b}$ (compare the proof of Theorem 3a) we have: $\eta ^b \varvec{y^b_0} \le \eta ^b\varvec{\tilde{y}^b}$. Thus, $(\varvec{x_0};\varvec{y_{*}^b})$ is weakly dominated by $(\varvec{x_0};\eta ^b\varvec{\tilde{y}^b})$ which contradicts that $(\varvec{x_0};\varvec{y_{*}^b})$ is strongly efficient. $\square$

Proof of Theorem 4

Follows immediately from Lemma 1. $\square$

Remark

Theorem 4 also holds for more generalized balance sets only fulfilling attributes a, b and c of Corollary 2.

Proof of Theorem 5

• a) This assertion summarizes some results of Theorems 1, 3 and 4. Furthermore, the characterization of the corresponding balanced outputs is an immediate consequence of the definition $\tau _k := \frac{\gamma _k}{\mu _k}$ with $\mu _k > 0$.

• b) Since $\varvec{y_0^b}$ is part of the boundary of the convex polyhedral cone $\varvec{C} \subset \mathbb R^s$ it can be constructed by the linear combination of at most $s-1$ (linearly independent) elements of $\varvec{\bar{Y}}$.

• c) Suppose $\sum _{k \in \varvec{\varOmega ^0}}\tau _k < 1$. Then $\varvec{y_0^b}=\sum _{k \in \varvec{\varOmega ^0}} \tau _k \varvec{\hat{y}_k} \le \varvec{y_0} \cdot \left( \sum _{k \in \varvec{\varOmega ^0}} \tau _k \right) < \varvec{y_0}$ because of $\varvec{y_0} > \varvec{0}$. This, however, is a contradiction to Theorems 3a and 5a that $y_{r0}^b = y_{r0}$ for at least one output r.

• d)

  • (i) Suppose that there exists $r \in {\varvec{\chi }}$ with $\hat{y}_{rk}=y_{r0} \ \forall k \in \varvec{\varOmega ^0}$. Then $y_{r0}^b = \sum _{k \in \varvec{\varOmega ^0}} \tau _k \hat{y}_{rk}$ $= y_{r0}\cdot \left( \sum _{k \in \varvec{\varOmega ^0}}\tau _k\right) = y_{r0}^b\cdot \left( \sum _{k \in \varvec{\varOmega ^0}}\tau _k\right)$, hence $\sum _{k \in \varvec{\varOmega ^0}} \tau _k = 1$. This implication is a logical equivalent to the first assertion.

  • (ii) Let be $r \in {\varvec{\chi }}$. From $\sum _{k \in \varvec{\varOmega ^0}} \tau _k = 1$ follows $0<y_{r0}=y_{r0}^b=\sum _{k \in \varvec{\varOmega ^0}} \tau _k \hat{y}_{rk} \le y_{r0}\cdot \left( \sum _{k \in \varvec{\varOmega ^0}}\tau _k\right) = y_{r0}$. Because of $\tau _k>0 \ \forall k \in \varvec{\varOmega ^0}$ we obtain the second assertion $\hat{y}_{rk}=y_{r0}^b=y_{r0} \ \forall k$. $\square$

Proof of Theorem 6

Let be $\varvec{C_1} \subset \varvec{C_2}$ and $\theta ^{i,b}$ together with $\varvec{y^{i,b}_0}$ optimal solutions of (5) for $i=1,2$. With $\varvec{C_1} \subset \varvec{C_2}$ we can conclude: $\varvec{B_1} \subset \varvec{B_2}$. Then with (4) we have:

$$\begin{aligned} \frac{1}{\theta ^{1,b}}=\eta ^{1,b}&= \mathrm{min} \left\{ \eta | \eta =\eta (\varvec{x_0};\varvec{y^{1,b}}); \varvec{y^{1,b}} \le \varvec{y_0}; (\varvec{x_0};\varvec{y^{1,b}}) \in \varvec{B_1} \right\} \\&\ge \mathrm{min} \left\{ \eta | \eta =\eta (\varvec{x_0};\varvec{y^{2,b}}); \varvec{y^{2,b}} \le \varvec{y_0}; (\varvec{x_0};\varvec{y^{2,b}}) \in \varvec{B_2} \right\} \\&= \eta ^{2,b} = \frac{1}{\theta ^{2,b}} \end{aligned}$$

which implies $\sigma _{1,0} \ge \sigma _{2,0}$. $\square$

Remark

Theorem 6 also holds for more generalized balance sets only fulfilling attributes a and b of Corollary 2.

Proof of Theorem 7

Let be $\theta ^{0}$ and $\tilde{\theta }^{0}$ the efficiency scores of $(\varvec{x_0};\varvec{y_0})$ and $(\varvec{x_0};\varvec{\tilde{y}_0})$. With $\varvec{y_0} \le \varvec{\tilde{y}_0}$ we can conclude: $\theta ^{0} \le \tilde{\theta }^{0}$. Then we have: $\frac{\theta ^b}{\theta ^{0}}\ge \frac{\theta ^b}{\tilde{\theta }^{0}}$ which implies $\sigma (\varvec{x_0;y_0}) \le \sigma (\varvec{x_0;\tilde{y}_0})$. $\square$

Remark

Theorem 7 also holds for more generalized balance sets only fulfilling attributes a and b of Corollary 2.

Proof of Theorem 8

Let be $(\varvec{x_0};\varvec{y_0}) \in \varvec{T}^\mathrm{env}$ and $(\varvec{x_0};\varvec{\tilde{y}_0}) \in \varvec{T}^\mathrm{env}$ such that $\varvec{\tilde{y}_0} = \alpha \varvec{y_0}$ with $\alpha >0$. We have to show that there exist two corresponding balanced points $(\varvec{x_0};\varvec{y_0^b})$ and $(\varvec{x_0};\varvec{\tilde{y}_0^b})$ with $\varvec{\tilde{y}_0^b} = \alpha \varvec{y_0^b}$ so that $\sigma (\varvec{x_0};\varvec{y_0}) = \sigma (\varvec{x_0};\varvec{\tilde{y}_0})$, i.e. the specialization degrees of all points on this ray in output space are identical.

The case $\varvec{y_0}= \varvec{0}$ is trivial. If $\varvec{y_0} \in \varvec{C}$ then, by Corollary 2c, $\varvec{\tilde{y}_0} \in \varvec{C}$, too, so that $\beta (\varvec{x_0};\varvec{y_0}) = 1 = \beta (\varvec{x_0};\varvec{\tilde{y}_0})$. Let be $\varvec{y_0} \notin \varvec{C}$. If $\varvec{y_0} \ne \varvec{0}$ and $y_{r0} = 0$ for at least one output r then $\varvec{y_0^b} = \varvec{0} = \varvec{\tilde{y}_0^b}$ and $\beta (\varvec{x_0};\varvec{y_0}) =0= \beta (\varvec{x_0};\varvec{\tilde{y}_0})$ by Theorem 1ci. So let be $\varvec{y_0} > \varvec{0}$ and hence $\varvec{\tilde{y}_0} = \alpha \varvec{y_0} > \varvec{0}$ for a given $\alpha > 0$. In view of Theorem 3 corresponding balanced points on the boundary of $\varvec{B}$ exist for both DMUs and hence the specialization degrees are well defined by (1), (3) and (4) through $\sigma = \sigma (\varvec{x_0};\varvec{y_0}) = 1- \frac{\eta ^0}{\eta ^b}$ and $\tilde{\sigma } = \sigma (\varvec{x_0};{\varvec{\tilde{y}_0}}) = 1- \frac{\tilde{\eta }^0}{\tilde{\eta }^b}$. As long as the maxima are defined the following general equalities hold for all optimization problems of type (1) and (3):

$$\begin{aligned} \eta (\varvec{\bar{x}};\varvec{\bar{y}}):&= \mathrm{max} \left\{ \eta | \varvec{x} \le \varvec{\bar{x}}, \varvec{y} \ge \eta \varvec{\bar{y}}, (\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env} \right\} \nonumber \\&= \mathrm{max} \left\{ \eta | \varvec{x} \le \varvec{\bar{x}}, \varvec{y} \ge \bar{\eta }(\alpha \varvec{\bar{y}}), \eta = \alpha \bar{\eta }, (\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env} \right\} \nonumber \\&= \alpha \cdot \mathrm{max} \left\{ \bar{\eta } | \varvec{x} \le \varvec{\bar{x}}, \varvec{y} \ge \bar{\eta }(\alpha \varvec{\bar{y}}), (\varvec{x};\varvec{y}) \in \varvec{T}^\mathrm{env} \right\} \nonumber \\&= \alpha \cdot \eta (\varvec{\bar{x}};\alpha \varvec{\bar{y}}) \end{aligned}$$

(7)

From (7) and the definitions $\varvec{\tilde{y}_0} := \alpha \varvec{y_0}$ and $\varvec{\tilde{y}_0^b} := \alpha \varvec{y_0^b}$ follows $\eta ^0 \equiv \eta (\varvec{x_0};\varvec{y_0}) = \alpha \cdot \eta (\varvec{x_0};\varvec{\tilde{y}_0}) \equiv \alpha \tilde{\eta }^0$ as well as $\eta ^b \equiv \eta (\varvec{x_0};\varvec{y_0^b}) = \alpha \cdot \eta (\varvec{x_0};\varvec{\tilde{y}_0^b}) \equiv \alpha \tilde{\eta }^b$ and hence $\tilde{\sigma } = 1-\frac{\tilde{\eta }^0}{\tilde{\eta }^b} = 1-\frac{{\eta }^0}{\eta ^b} = \sigma$. If $(\varvec{x_0};\varvec{y_0^b})$ is a corresponding balanced point for DMU $(\varvec{x_0};\varvec{y_0})$, to prove Theorem 8 it remains to show that then $(\varvec{x_0};\varvec{\tilde{y}_0^b})$ has to be a corresponding balanced point for DMU $(\varvec{x_0};\varvec{\tilde{y}_0})$, too, i.e. that $(\varvec{x_0};\varvec{\tilde{y}_0^b})$ is an optimal solution of the following minimization problem analogously to (4):

$$\begin{aligned} \tilde{\eta }^b \equiv \eta (\varvec{x_0};\varvec{\tilde{y}_0^b}) = \mathrm{min} \left\{ \tilde{\eta } | \tilde{\eta } = \eta (\varvec{x_0};\varvec{\tilde{y}^b}), \varvec{\tilde{y}^b} \le \varvec{\tilde{y}_0},(\varvec{x_0};\varvec{\tilde{y}^b)} \in \varvec{B} \right\} \end{aligned}$$

(8)

As supposed, $(\varvec{x_0};\varvec{y_0^b})$ is a corresponding balanced point of DMU $(\varvec{x_0};\varvec{y_0})$ so that it has to be an optimal solution of:

$$\begin{aligned} \eta ^b \equiv \eta (\varvec{x_0};\varvec{y_0^b}) = \mathrm{min} \left\{ \eta | \eta = \eta (\varvec{x_0};\varvec{y^b}), \varvec{y^b} \le \varvec{y_0},(\varvec{x_0};\varvec{y^b)} \in \varvec{B} \right\} \end{aligned}$$

(4)

Hence $\varvec{\tilde{y}_0^b} \equiv \alpha \varvec{y_0^b} \le \alpha \varvec{y_0} \equiv \varvec{\tilde{y}_0}$, i.e. $(\varvec{x_0};\varvec{\tilde{y}_0^b})$ is weakly dominated by DMU $(\varvec{x_0};\varvec{\tilde{y}_0})$ as requested by condition (NC). Moreover, this implies $(\varvec{x_0};\varvec{\tilde{y}_0^b}) \in \varvec{T}^\mathrm{env}$ in view of the free disposal property of $\varvec{T}^\mathrm{env}$. With $(\varvec{x_0};\varvec{y_0^b}) \in \varvec{B}$ and $\varvec{y_0^b} \in \varvec{C}$, in particular, by (4) we obtain $\varvec{\tilde{y}_0^b} \equiv \alpha \varvec{y_0^b} \in \varvec{C}$ and therefore $(\varvec{x_0};\varvec{\tilde{y}_0^b}) \in \varvec{B}$, too, by the Corollary 2c and e of $\varvec{C}$ displaying constant variability of scale. Thus, we can conclude that $(\varvec{x_0};\varvec{\tilde{y}_0^b})$ is a feasible solution of (8).

In order to show a contradiction let us assume that it is not an optimal solution of (8). I.e., there exists a point $(\varvec{x_0};\varvec{\bar{y}_0}) \in \varvec{B}$ with $\varvec{\bar{y}_0^b} \le \varvec{\tilde{y}_0}$ such that $\bar{\eta }^b := \eta (\varvec{x_0};\varvec{\bar{y}_0^b}) < \tilde{\eta }^b \equiv \eta (\varvec{x_0};\varvec{\tilde{y}_0^b})$. Applying (7) twice we get

$$\begin{aligned} \eta \left( \varvec{x_0};\frac{\varvec{\bar{y}_0^b}}{\alpha }\right) = \alpha \cdot \eta (\varvec{x_0};\varvec{\bar{y}_0^b}) < \alpha \cdot \eta (\varvec{x_0};\varvec{\tilde{y}_0^b}) = \eta (\varvec{x_0};\varvec{y_0^b}) \end{aligned}$$

which is the contradiction looked for because $(\varvec{x_0};\frac{\varvec{\bar{y}_0^b}}{\alpha })$ would be a feasible solution of (4) being smaller than the minimum. The feasibility follows analogously as before from Assumption 2 that $\varvec{C}$ is linear so that $\frac{\varvec{\bar{y}_0^b}}{\alpha } \in \varvec{C}$ and $(\varvec{x_0};\frac{\varvec{\bar{y}_0^b}}{\alpha }) \in \varvec{B}$ by .

$\square$

Remark

Theorem 8 also holds for more generalized cones only fulfilling attributes a, b, c, d and e of Corollary 2.

Appendix B

Using the properties shown in Sect. 4 Appendix B presents an exact solution procedure for the example of a multi-cone $\varvec{C}$ with three outputs, only, introduced in Sect. 5 and defined by Ahn et al. (2012). The eight points of Table 3 span the cone $\varvec{C}$ according to Assumption 2. Then the rays through C0 and C7 completely lie in the interior of $\varvec{C}$ and can be linearly combined by the other six points. Hence we can exclude them for the remaining examination because the whole cone $\varvec{C}$ is already spanned by C1–C6 (thus defining the new reduced set $\varvec{\bar{Y}})$.

With Assumption 1 we know from Theorem 3d that for each unbalanced DMU at least one of its corresponding balanced points belongs to the boundary of $\varvec{C}$. In our special case of the multi-cone (enveloping the cuboid) of Fig. 5 the boundary consists of six facets each composed by the linear combinations of only two adjacent corner points (C1/C2, C2/C3, C3/C4, C4/C5, C5/C6, C6/C1). For example, the facet C5/C6 can be expressed as follows:

$$\begin{aligned} \varvec{y} = \left( \begin{array}{l} 6{,}322 \\ 17{,}262 \\ 27{,}505 \end{array}\right) \gamma _5 + \left( \begin{array}{l} 10{,}353 \\ 17{,}262 \\ 27{,}505 \end{array}\right) \gamma _6 = \left( \begin{array}{l} 6{,}322\gamma _5 + 10{,}353\gamma _6 \\ 17{,}262(\gamma _5 + \gamma _6) \\ 27505(\gamma _5 + \gamma _6) \end{array}\right) , \gamma _5, \gamma _6 \ge 0 \end{aligned}$$

(9)

The example shows a special feature of all facets of the multi-cone used by Ahn et al. (2012). Each pair of adjacent corner points coincides in two of the three coordinates, e.g. the $y_2$- and $y_3$-coordinate for facet C5/C6 as we see in (9). This reduces the possibilities for the corresponding balanced points on the boundaries of $\varvec{C}$ and allows for an easier calculation.

As illustrative example let us consider (the unbalanced) DMU A. According to Theorem 5 we normalize the corner vectors so that they are maximum while fulfilling condition (NC) of weak dominance. Then a corresponding balanced point of DMU A lying on the facet C5/C6 must satisfy the following inequalities:

$$\begin{aligned} \varvec{y}&= \left( \begin{array}{l} 6{,}322 \\ 17{,}262 \\ 27{,}505 \end{array}\right) \cdot 0.5610 \tau _5 + \left( \begin{array}{l} 10{,}353 \\ 17{,}262 \\ 27{,}505 \end{array}\right) \cdot 0.3456 \tau _6 \nonumber \\&= \left( \begin{array}{l} 3{,}547 \\ 9{,}684 \\ 15{,}431 \end{array}\right) \tau _5 + \left( \begin{array}{l} 3{,}578 \\ 5{,}965 \\ 9{,}506 \end{array}\right) \tau _6 \le \left( \begin{array}{l} 3{,}578 \\ 12{,}329 \\ 15{,}431 \end{array}\right) = \varvec{y_A} , \ \tau _5, \tau _6 \ge 0 \end{aligned}$$

(10)

From Theorem 3 we know that there is at least one output r with $y_{rA}^b = y_{rA}$. In view of (9) and (10) this cannot be $r = 2$ (output OTC) because output $r = 3$ (PR) is proportional to $r = 2$ and always larger. Hence (10) reduces to

$$\begin{aligned} y_1&= 3{,}547\tau _5 + 3{,}578\tau _6 \le 3{,}578 = y_{1A} \nonumber \\ y_3&= 15{,}431\tau _5 + 9{,}506\tau _6 \le 15{,}431 = y_{3A} \end{aligned}$$

(11)

with at least one equality. In view of the efficiency condition (EF) a candidate for a corresponding balanced point for DMU A on facet C5/C6 has to have maximum outputs for the given input of DMU A. Since $y_2$ (OTC) is proportional to $y_3$ (PR) on this facet, according to (11) we distinguish two cases where (at least) one balanced output is equal to the original one and maximize the other output:

$$\begin{aligned} \mathrm{(a)}\quad \mathrm{max} \ y_3 \ \mathrm{s.t.} \ y_1 = y_\mathrm{1A} \nonumber \\ \mathrm{(b)} \quad \mathrm{max} \ y_1 \ \mathrm{s.t.} \ y_3 = y_\mathrm{3A} \end{aligned}$$

(12)

One of these two cases may not have a feasible solution. For (12), however, both solutions exist and are identical with $\tau _5=0.9862$ and $\tau _6 = 0.0224$ so that $\varvec{y^T} = (3{,}578, 9{,}684, 15{,}431)$ is the (best) candidate for the corresponding balanced point of DMU A on facet C5/C6. (Note that regarding Theorem 5 we have $\varvec{\varOmega ^0} = \{5, 6 \}$ with $\tau _5 + \tau _6 = 1.008 > 1$.)

By this procedure, for each of the six facets of the multi-cone at most two candidates for a corresponding balanced point of DMU A result. Table 5 shows these results in columns 2–4. As we can see, for four facets only one solution exists and in the other two cases both solutions are identical.

Table 5 Candidates for DMU A

A further look at the six candidates shows that five of them are dominated by the candidate of facet C5/C6. Because of the usual properties of DEA-models, in particular that $\varvec{T}^\mathrm{env}$ exhibits free disposal, this candidate has to be that one with the highest efficiency score and hence is the corresponding balanced point of DMU A on the boundary of $\varvec{C}$ looked for. For illustration, columns 5 and 6 display the efficiency scores of all candidates with respect to the BCC- (column 5: $\varvec{T}^\mathrm{env}$ convex) as well as to the CCR-model (column 6: $\varvec{T}^\mathrm{env}$ linear). Hence, we obtain the highest possible balanced efficiency for $\varvec{{y_A^b}^T} = (3{,}578, 9{,}684, 15{,}431)$ with $\theta _A^{b,\mathrm{BCC}} = 41.70 \, \%$ and $\theta _A^{b,\mathrm{CCR}} = 38.96 \, \%$. This procedure has to be applied for all specialized DMUs. The results are shown in Table 4 (Sect. 5).