An efficiency measure satisfying the Dmitruk–Koshevoy criteria on DEA technologies

Abstract

The purpose of this paper is to develop an efficiency measurement model by enhancing a CCR (Charnes–Cooper–Rhodes) model and then to prove that the enhanced model satisfies five desirable properties: indication, strict monotonicity, homogeneity, continuity and units unvariance. In order for our model to be empirically tractable, we also provide an algorithm aimed at estimating efficiency scores.

Appendix

Lemma 1

${\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) \Leftrightarrow$ there exists a pair (v, w) such that ${\mathbf{v}} > {\mathbf{0}},\;{\mathbf{w}} \ge{\mathbf{0}},\;{\mathbf{vx}} = {\mathbf{w}\hat{\mathbf{y}}}$ and vx _j  ≥ wy _j for all j = 1,2, …, J.

Proof

Note that ${\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) \Leftrightarrow ``{\mathbf{x}}^{\prime } \le {\mathbf{x}}\,{\text{and}}\,{\mathbf{x}}^{\prime } \ne {\mathbf{x}}\quad {\text{imply}}\,{\text{that}}\,{\mathbf{x}}^{\prime } \notin T\left( {{\hat{\mathbf{y}}}} \right)''$. Therefore, there is no solution of the following linear inequalities system:

$$ \begin{gathered} - \sum\limits_{j = 1}^{J} {{\mathbf{x}}_{j} \lambda_{j} \le \mu {\mathbf{x}}} , - \sum\limits_{j = 1}^{J} {{\mathbf{x}}_{j} \lambda_{j} \ne \mu {\mathbf{x}}} \hfill \\ \sum\limits_{j = 1}^{J} {{\mathbf{y}}_{j} \lambda_{j} \ge \mu {\hat{\mathbf{y}}}} , \hfill \\ \lambda_{j} \ge 0,\quad \forall j = 1, \ldots ,J \hfill \\ \mu > 0. \hfill \\ \end{gathered} $$

(28)

From Slater’s theorem of the alternative, a linear inequalities system

$$ \begin{gathered} - {\mathbf{vx}}_{j} + {\mathbf{wy}}_{j} \le 0,\quad \forall j = 1, \ldots ,J \hfill \\ {\mathbf{vx}} - {\mathbf{wy}} + \alpha = 0, \hfill \\ \end{gathered} $$

(29)

has a solution v > 0, w ≥ 0, α = 0 or v ≥ 0, w ≥ 0, α > 0. Since ${\mathbf{x}} \in T\left( {{\hat{\mathbf{y}}}} \right)$, it follows that vx – wy = 0 and α = 0. Therefore, the above system (26) has a solution v > 0, w ≥ 0, α = 0 but v ≥ 0, w ≥ 0, α > 0.

Conversely, we assume that v > 0, w ≥ 0, α = 0 satisfies the system (26). It follows from Slater’s theorem of the alternative that the system (27) has no solution. Therefore, we see form ${\mathbf{x}} \in T\left( {{\hat{\mathbf{y}}}} \right)$ that ${\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right).$ $\square$

Lemma 2

$\left( {{\mathbf{x}},{\mathbf{y}}} \right) \in Eff\left( T \right) \Leftrightarrow$ there exists a pair ( v , w ) such that v > 0 , w > 0 , vx = wy and vx _j  ≥ wy _j for all j = 1,2, …, J.

Proof

Similar to the proof of Lemma 1, we can prove Lemma 2. $\square$

Lemma 3

For any face F of T, the set $\left\{ {{\mathbf{x}}\left| {\left( {{\mathbf{x}},{\hat{\mathbf{y}}}} \right) \in F,{\mathbf{x}} \in T\left( {{\hat{\mathbf{y}}}} \right)} \right.} \right\}$ is a face of $T\left( {{\hat{\mathbf{y}}}} \right)$ . For any face G of $T\left( {{\hat{\mathbf{y}}}} \right)$ , there exists a face F of T such that $\left\{ {\left( {{\mathbf{x}},{\hat{\mathbf{y}}}} \right)\left| {{\mathbf{x}} \in G} \right.} \right\} \subseteq F$ .

Proof

It follows from two definitions of face, (6) and (7). $\square$

Lemma 8

If 1 ≥ ε₁ ≥ ε₂ ≥ 0, then we have

$$ {\mathbf{I}}^{N} (\varepsilon_{2} ){\mathbf{s}}_{{}}^{x} \ge {\mathbf{0}}\, \Rightarrow \,{\mathbf{I}}^{N} (\varepsilon_{1} ){\mathbf{s}}^{x} \ge {\mathbf{0}}. $$

(30)

Proof

For an arbitrary positive number ε ≤ 1, I ^N(ε) is a square matrix, its maximum eigenvalue is 1 + (N − 1)ε and e ^N, an N-dimensional vector of ones, is a corresponding eigenvector. Denoting an N by N dimensional square matrix of ones by E, we have

$$ {\mathbf{EI}}^{N} (\varepsilon ) = (1 + (N - 1)\varepsilon ){\mathbf{E}}. $$

(31)

Consider the case where 0 < ε₂ ≤ ε₁ = 1. We multiply both sides of (31) by s ^x from the right and utilize I ^N(ε₁) = I ^N(1) = E to obtain

$$ {\mathbf{EI}}^{N} (\varepsilon_{2} ){\mathbf{s}}^{x} = (1 + (N - 1)\varepsilon_{2} ){\mathbf{Es}}^{x} = (1 + (N - 1)\varepsilon_{2} ){\mathbf{I}}^{N} (\varepsilon_{1} ){\mathbf{s}}^{x} . $$

If I ^N(ε₂) = s ^x ≥ 0, then EI ^N(ε₂)s ^x ≥ 0. This result along with (1 + (N − 1)ε₂) > 0 yields I ^N(ε₁) = s ^x ≥ 0.

Now consider the case where 1 > ε₁ ≥ ε₂ ≥ 0. Let I be an N by N identity matrix. For any ε ∈ (0,1), we have

$$ \frac{1}{1 - \varepsilon }\left( {{\mathbf{E}} - {\mathbf{I}}^{N} (\varepsilon )} \right) = {\mathbf{E}} - {\mathbf{I}}. $$

(32)

Applying (10) twice to I ^N(ε₂) = I + ε₂(E − I), we obtain

$$ {\mathbf{I}}^{N} (\varepsilon_{2} ) = E - \frac{1}{{1 - \varepsilon_{1} }}\left( {{\mathbf{E}} - {\mathbf{I}}^{N} \left( {\varepsilon_{1} } \right)} \right) + \frac{{\varepsilon_{2} }}{{1 - \varepsilon_{1} }}\left( {{\mathbf{E}} - {\mathbf{I}}^{N} \left( {\varepsilon_{1} } \right)} \right), $$

which yields

$$ {\mathbf{I}}^{N} (\varepsilon_{2} ) = \frac{{\varepsilon_{2} - \varepsilon_{1} }}{{1 - \varepsilon_{1} }}{\mathbf{E}} + \frac{{1 - \varepsilon_{2} }}{{1 - \varepsilon_{1} }}{\mathbf{I}}^{N} \left( {\varepsilon_{1} } \right). $$

A simple manipulation along with the use of (31) yields

$$ {\mathbf{I}}^{N} (\varepsilon_{2} ) = \frac{1}{{1 - \varepsilon_{1} }}\left( {\left( {1 - \varepsilon_{2} } \right){\mathbf{I}} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}} \right){\mathbf{I}}^{N} (\varepsilon_{1} ) . $$

(33)

Since the maximum eigenvalue of matrix E is N, we have

$$ 1 - \varepsilon_{2} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}N \ge 1 - \varepsilon_{2} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{\varepsilon_{1} }} = \frac{{\varepsilon_{2} }}{{\varepsilon_{1} }} - \varepsilon_{2} = \varepsilon_{2} \left( {\frac{1}{{\varepsilon_{1} }} - 1} \right) > 0. $$

The maximum eigenvalue of matrix $\frac{{\varepsilon_{1} - \varepsilon_{2} }}{{\varepsilon_{1} + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}$ is less than $1 - \varepsilon_{2}$. This means that the inverse matrix of $\left( {\left( {1 - \varepsilon_{2} } \right){\mathbf{I}} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}} \right)$ in Eq. (33) is a positive matrix. Multiplying both sides of Eq. (33) by s ^x from the right and by $\left( {\left( {1 - \varepsilon_{2} } \right){\mathbf{I}} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}} \right)^{ - 1}$ from the left, we obtain

$$ \left[ {\left( {1 - \varepsilon_{2} } \right){\mathbf{I}} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}} \right]^{ - 1} {\mathbf{I}}^{N} \left( {\varepsilon_{2} } \right){\mathbf{s}}^{x} = \frac{1}{{1 - \varepsilon_{1} }}{\mathbf{I}}^{N} \left( {\varepsilon_{1} } \right){\mathbf{s}}^{x} . $$

(34)

If ${\mathbf{I}}^{N} (\varepsilon_{2} ){\mathbf{s}}^{x} \ge {\mathbf{0}}$, then ${\mathbf{I}}^{N} (\varepsilon_{1} ){\mathbf{s}}^{x} \ge {\mathbf{0}}$ due to the fact that $\left( {\left( {1 - \varepsilon_{2} } \right){\mathbf{I}} - \frac{{\varepsilon_{1} - \varepsilon_{2} }}{{1 + \left( {N - 1} \right)\varepsilon_{1} }}{\mathbf{E}}} \right)^{ - 1}$ is a positive matrix and 1 – ε₁ > 0. □

Lemma 9

For any ε ∈ [0,1], we have

$$ T\left( {{\hat{\mathbf{y}}}} \right) \subseteq \left\{ {{\mathbf{x}} + \, {\mathbf{s}}^{x} \left| {{\mathbf{x}} \in {\text{Eff}}\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) , { }I^{N} \left( \varepsilon \right){\mathbf{s}}^{x} \ge {\mathbf{0}}} \right.} \right\} $$

(35)

where the equality (=) holds if and only if ε = 0.

Proof

For any ${\mathbf{x}} \in T\left( {{\hat{\mathbf{y}}}} \right)$ there exists ${\mathbf{x}}^{*} \in {\text{Eff}}\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right)$ such that x = x* + s ^x for some s ^x ≥ 0. Therefore, we have $T\left( {{\hat{\mathbf{y}}}} \right) = \left\{ {{\mathbf{x}} + \, {\mathbf{s}}^{x} \left| {{\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) , { }I^{N} \left( 0 \right){\mathbf{s}}^{x} \ge {\mathbf{0}}} \right.} \right\}$. Lemma 8 implies that

$$ \begin{aligned} T\left( {{\hat{\mathbf{y}}}} \right) & = \left\{ {{\mathbf{x}} + \, {\mathbf{s}}^{x} \left| {{\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) ,\, \, I^{N} \left( 0 \right){\mathbf{s}}^{x} \ge {\mathbf{0}}} \right.} \right\} \\ & \subseteq \left\{ {{\mathbf{x}} + \, {\mathbf{s}}^{x} \left| {{\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) , { }I^{N} \left( \varepsilon \right){\mathbf{s}}^{x} \ge {\mathbf{0}}} \right.} \right\} \\ \end{aligned} $$

for all ε ∈ [0,1]. □

Lemma 10

We have

$$ 0 < \mathop {\min }\limits_{{f = 1, \ldots ,f_{2} }} \,\mathop {\max }\limits_{{\left( {{\mathbf{v}},{\mathbf{w}}} \right) \in VW\left( {\mathbb{F}\left( {{\mathbf{v}}^{f} ,{\mathbf{w}}^{f} } \right)} \right)}} \quad \min \left\{ {v_{n} \left| {n = 1, \ldots ,N} \right.} \right\}. $$

Proof

From the choice of $\left( {{\mathbf{v}}^{f} ,{\mathbf{w}}^{f} } \right) \in \Re_{ + + }^{N} \times \Re_{ + }^{M}$ for all f = 1, …, f ₂ we have $v_{n}^{f} > 0\quad \forall n = 1, \ldots ,N\;{\text{and}}\quad \forall f = 1, \ldots ,f_{2}$. This means that for all f = 1, …, f ₂

$$ 0 < \min \left\{ {v_{1}^{f} , \ldots ,v_{N}^{f} } \right\} \le \mathop {\max }\limits_{{\left( {{\mathbf{v}},{\mathbf{w}}} \right) \in VW\left( {\mathbb{F}\left( {{\mathbf{v}}^{f} ,{\mathbf{w}}^{f} } \right)} \right)}} \,\min \left\{ {v_{n} \left| {n = 1, \ldots ,N} \right.} \right\}. $$

□

Lemma 11

Let η* be the optimal value of (19) and let $\bar{\varepsilon } = 1/\left( {\tfrac{1}{{\eta^{*} }} - N + 1} \right)$, then the set $Q\left( \varepsilon \right) \equiv \left\{ {{\mathbf{x}} + {\mathbf{s}}^{x} \left| {{\mathbf{x}} \in Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) , { }{\mathbf{I}}^{N} \left( \varepsilon \right){\mathbf{s}}^{x} \ge {\mathbf{0}}} \right.} \right\}$ satisfies [Q1]–[Q5] for any $\varepsilon \in (0,\bar{\varepsilon }]$.

Proof

For any $\varepsilon \in (0,\bar{\varepsilon }]$ we need to prove the following properties:

$$ \begin{array}{*{20}c} {[Q1]} \hfill & {Q\left( \varepsilon \right)\;{\text{is closed}}.} \hfill \\ {[Q2]} \hfill & {Q\left( \varepsilon \right) + R_{ + }^{N} \subseteq Q\left( \varepsilon \right)} \hfill \\ {[Q3]} \hfill & {Isoq\left( {Q\left( \varepsilon \right)} \right) = Eff\left( {Q\left( \varepsilon \right)} \right)} \hfill \\ {[Q4]} \hfill & {T\left( {{\hat{\mathbf{y}}}} \right) \subseteq \,Q\left( \varepsilon \right)} \hfill \\ {[Q5]} \hfill & {Eff\left( {T\left( {{\hat{\mathbf{y}}}} \right)} \right) \subseteq \,Eff\left( {\,Q\left( \varepsilon \right)} \right)} \hfill \\ \end{array} $$

For any ε ∈ (0,1] Property [Q1] trivially holds. For any ε ∈ (0,1], Properties [Q2] and [Q4] follow directly from Lemma 8.

By using the alternative theorem of Slater, we have the following equivalent condition to Property [Q3] : the optimal value of (20) for ${\hat{\mathbf{x}}} \in Q\left( \varepsilon \right)$ is 1 if and only if there exists a pair (v, w) such that ${\mathbf{v}} > {\mathbf{0}},\,{\mathbf{w}} \ge {\mathbf{0}},\,{\mathbf{u}} \ge {\mathbf{0}},\,{\mathbf{v}\hat{\mathbf{x}}} = {\mathbf{w}\hat{{y}}},\,{\mathbf{v}} = {\mathbf{u}}I^{N} \left( \varepsilon \right)$ and vx _j  ≥ wy _j for all j = 1,2, …, J.

Choose any ε ∈ (0,1] arbitrarily. Assume that ${\hat{\mathbf{x}}} \in Isoq\left( {Q\left( \varepsilon \right)} \right)$, equivalently, ${\hat{\mathbf{x}}}$ satisfies $1 = \min \left\{ {\theta \left| {\theta {\hat{\mathbf{x}}} \in Q\left( \varepsilon \right)} \right.} \right\}$ The dual problem of $\min \left\{ {\theta \left| {\theta {\hat{\mathbf{x}}} \in Q\left( \varepsilon \right)} \right.} \right\}$is

$$ \max \left\{ {{\mathbf{w}\hat{\mathbf{y}}}}\left|{{\mathbf{vx}}_{j} - {\mathbf{wy}}_{j} \ge 0\quad \forall j = 1,\ldots ,J,\,{\mathbf{v}\hat{\mathbf{x}}}} = 1,\,{\mathbf{v}} ={\mathbf{u}}I^{N} \left( \varepsilon\right),\,{\mathbf{v}},{\mathbf{w}},{\mathbf{u}} \ge {\mathbf{0}}\right. \right\}. $$

(36)

By the duality theorem of LP, the dual problem (36) has an optimal solution (v ^*, w ^*, u ^*) and its optimal value is $1 = {\mathbf{w}}^{*} {\hat{\mathbf{y}}}$. Suppose that u ^* = 0, then v ^* = u ^* I ^N(ε) = 0. This contradicts the supposition ${\mathbf{v}}^{*} {\hat{\mathbf{x}}} = 1$. We have u ^* ≥ 0 and u ^* ≠ 0. It follows from ε ∈ (0,1] that v ^* = u ^* I ^N(ε) > 0.

Conversely, assume that there exists a pair (v, w) such that v > 0, W ≥ 0, u ≥ 0, ${\mathbf{v}\hat{\mathbf{x}}} = {\mathbf{w}\hat{\mathbf{y}}},\,{\mathbf{v}} = {\mathbf{u}}I^{N} \left( \varepsilon \right)$ and vx _j  ≥ wy _j for all j = 1,2, …, J, then, the dual problem (36) has an optimal solution and its optimal value is 1. By the duality theorem of LP, we have $1 = \min \left\{ {\theta \left| {\theta {\hat{\mathbf{x}}} \in Q\left( \varepsilon \right)} \right.} \right\}$ and hence, ${\hat{\mathbf{x}}} \in Isoq\left( {Q\left( \varepsilon \right)} \right)$.

Let $\bar{\varepsilon } = 1/\left( {\tfrac{1}{{\eta^{*} }} - N + 1} \right)$and choose any $\varepsilon \in (0,\bar{\varepsilon }]$ arbitrarily, then we will show Property [Q5] by the following assertions:

Assertion A

Let E be an N × N matrix whose components are all one, then $\left( {{\mathbf{I}}^{N} \left( \varepsilon \right)} \right)^{ - 1} = \frac{1}{1 - \varepsilon }\left( {{\mathbf{I}}^{N} \left( 0 \right) - \frac{\varepsilon }{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}}} \right)$ where $0 < \varepsilon \le \bar{\varepsilon } < 1$ .

Proof

For any ε ∈ (0,1), we have

$$ \begin{aligned} \frac{1}{1 - \varepsilon }\left( {{\mathbf{I}}^{N} \left( 0 \right) - \frac{\varepsilon }{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}}} \right){\mathbf{I}}^{N} \left( \varepsilon \right) & = \frac{1}{1 - \varepsilon }\left( {{\mathbf{I}}^{N} \left( 0 \right) - \frac{\varepsilon }{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}}} \right)\left( {\left( {1 - \varepsilon } \right){\mathbf{I}}^{N} \left( 0 \right) + \varepsilon {\mathbf{E}}} \right) \\ & = \frac{1}{1 - \varepsilon }\left( {\left( {1 - \varepsilon } \right){\mathbf{I}}^{N} \left( 0 \right) - \frac{{\varepsilon \left( {1 - \varepsilon } \right)}}{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}} + \varepsilon {\mathbf{E}} - \frac{{N\varepsilon^{2} }}{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}}} \right) \\ & = \frac{1}{1 - \varepsilon }\left( {\left( {1 - \varepsilon } \right){\mathbf{I}}^{N} \left( 0 \right) - \frac{{\varepsilon \left( {1 + \left( {N - 1} \right)\varepsilon } \right)}}{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}} + \varepsilon {\mathbf{E}}} \right) \\ & = \frac{1}{1 - \varepsilon }\left( {\left( {1 - \varepsilon } \right){\mathbf{I}}^{N} \left( 0 \right) - \varepsilon {\mathbf{E}} + \varepsilon {\mathbf{E}}} \right) = \frac{1}{1 - \varepsilon }\left( {1 - \varepsilon } \right){\mathbf{I}}^{N} \left( 0 \right) = {\mathbf{I}}^{N} \left( 0 \right). \\ \end{aligned} $$

□

Assertion B

For every f = 1, …, f ₂ , each problem ( 36 ) for any ${\hat{\mathbf{x}}} \in \hat{\mathbb{F}}\left( {{\mathbf{v}}^{f} ,{\mathbf{w}}^{f} } \right)$ has an optimal solution attaining the optimal objective function value 1.

Proof

Choose f ∈ {1, …f ₂} arbitrarily and let $\left( {{\bar{\mathbf{v}}}^{f} ,{\bar{\mathbf{w}}}^{f} } \right)$ be an optimal solution of $\mathop {\max }\limits_{{\left( {{\mathbf{v}},{\mathbf{w}}} \right) \in VW\left( {\mathbb{F}\left( {{\mathbf{v}}^{f} ,{\mathbf{w}}^{f} } \right)} \right)}} \;\;\min \left\{ {v_{n} \left| {n = 1, \ldots ,N} \right.} \right\}$, then it follows from Lemma 10 that

$$ \bar{v}_{n}^{f} \ge \eta^{*} > 0\quad {\text{for}}\,{\text{all}}\,n = 1, \ldots ,N $$

(37)

$$ \sum\limits_{n = 1}^{N} {\bar{v}_{n}^{f} } = 1 $$

(38)

$$ {\bar{\mathbf{v}}}^{f} {\mathbf{x}}_{j} - {\bar{\mathbf{w}}}^{f} {\mathbf{y}}_{j} \ge 0\quad \forall j = 1, \ldots ,J $$

(39)

$$ {\bar{\mathbf{v}}}^{f} {\hat{\mathbf{x}}} = {\bar{\mathbf{w}}}^{f} {\hat{\mathbf{y}}} $$

(40)

$$ {\bar{\mathbf{w}}}^{f} \ge {\mathbf{0}}. $$

(41)

Let $\hat{v}_{n}^{f} \equiv {{\bar{v}_{n}^{f} } \mathord{\left/ {\vphantom {{\bar{v}_{n}^{f} } {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }}$ for all n = 1, …, N and let $\hat{w}_{m}^{f} \equiv {{\bar{w}_{m}^{f} } \mathord{\left/ {\vphantom {{\bar{w}_{m}^{f} } {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }}$ for all m = 1, …, M, then

$$ \hat{v}_{n}^{f} \ge 0\quad {\text{for}}\,{\text{all}}\,n = 1, \ldots ,N $$

(42)

$$ \sum\limits_{n = 1}^{N} {\hat{v}_{n}^{f} } \hat{x}_{n} = 1 $$

(43)

$$ {\hat{\mathbf{v}}}^{f} {\mathbf{x}}_{j} - {\hat{\mathbf{w}}}^{f} {\mathbf{y}}_{j} \ge 0\quad \forall j = 1, \ldots ,J $$

(44)

$$ {\hat{\mathbf{v}}}^{f} {\hat{\mathbf{x}}} = 1,\quad {\hat{\mathbf{w}}}^{f} {\hat{\mathbf{y}}} = 1 $$

(45)

$$ {\hat{\mathbf{w}}}^{f} \ge {\mathbf{0}} . $$

(46)

By the definition of $\bar{\varepsilon }$ we have $\bar{\varepsilon } > 0$. If we find a vector ${\hat{\mathbf{u}}}^{f} \ge {\mathbf{0}}\;{\text{such}}\,{\text{that}}\;{\hat{\mathbf{v}}}^{f} = {\hat{\mathbf{u}}}^{f} I^{N} \left( \varepsilon \right)$, then we will complete the proof of Assertion B. Hereafter, we discuss the existence of such a ${\hat{\mathbf{u}}}^{f}$.

Firstly, we consider the case of $\eta^{*} < {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 N}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$N$}}$, equivalently $\bar{\varepsilon } < 1$, then we have 0 < ε < 1. Let e be an N-dimensional vector whose component is all one, then it follows from Assertion A, ${1 \mathord{\left/ {\vphantom {1 \varepsilon }} \right. \kern-\nulldelimiterspace} \varepsilon } \ge {1 \mathord{\left/ {\vphantom {1 {\bar{\varepsilon }}}} \right. \kern-\nulldelimiterspace} {\bar{\varepsilon }}},\,\eta^{*} = 1/\left( {\tfrac{1}{{\bar{\varepsilon }}} + N - 1} \right)$ and (37) that

$$ \begin{aligned} {\hat{\mathbf{v}}}^{f} \left( {I^{N} \left( \varepsilon \right)} \right)^{ - 1} & = \frac{1}{1 - \varepsilon }{\hat{\mathbf{v}}}^{f} \left( {{\mathbf{I}}^{N} \left( 0 \right) - \frac{\varepsilon }{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{E}}} \right) = \frac{1}{{\left( {1 - \varepsilon } \right)\sum\nolimits_{n = 1}^{N} {\bar{v}_{n}^{f} } }}\left( {{\bar{\mathbf{v}}}^{f} - \frac{\varepsilon }{{1 + \left( {N - 1} \right)\varepsilon }}{\mathbf{e}}} \right) \\ & = \frac{1}{{\left( {1 - \varepsilon } \right)\sum\nolimits_{n = 1}^{N} {\bar{v}_{n}^{f} } }}\left( {{\bar{\mathbf{v}}}^{f} - \frac{1}{{{1 \mathord{\left/ {\vphantom {1 \varepsilon }} \right. \kern-\nulldelimiterspace} \varepsilon } + \left( {N - 1} \right)}}{\mathbf{e}}} \right) \\ & \ge \frac{1}{{\left( {1 - \varepsilon } \right)\sum\nolimits_{n = 1}^{N} {\bar{v}_{n}^{f} } }}\left( {{\bar{\mathbf{v}}}^{f} - \frac{1}{{{1 \mathord{\left/ {\vphantom {1 {\bar{\varepsilon }}}} \right. \kern-\nulldelimiterspace} {\bar{\varepsilon }}} + \left( {N - 1} \right)}}{\mathbf{e}}} \right) = \frac{1}{{\left( {1 - \varepsilon } \right)\sum\nolimits_{n = 1}^{N} {\bar{v}_{n}^{f} } }}\left( {{\bar{\mathbf{v}}}^{f} - \eta^{*} {\mathbf{e}}} \right) \ge {\mathbf{0}}. \\ \end{aligned} $$

In the case of $\eta^{*} < {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 N}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$N$}}$ we have ${\hat{\mathbf{u}}}^{f} = {\hat{\mathbf{v}}}^{f} \left( {I^{N} \left( \varepsilon \right)} \right)^{ - 1} \ge {\mathbf{0}}$ and hence, the dual problem (36) has an optimal solution $\left( {{\hat{\mathbf{v}}}^{f} ,{\hat{\mathbf{w}}}^{f} ,{\hat{\mathbf{u}}}^{f} } \right)$ and its optimal value is 1.

In the case of $\eta^{*} = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 N}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$N$}}$ we have $\bar{\varepsilon } = 1$. When $\varepsilon < \bar{\varepsilon }$, as stated above, the dual problem (36) has an optimal solution $\left( {{\hat{\mathbf{v}}}^{f} ,{\hat{\mathbf{w}}}^{f} ,{\hat{\mathbf{u}}}^{f} } \right)$ and the optimal value 1. Otherwise, $\varepsilon = \bar{\varepsilon } = 1$, then it follows from (37) and (38) that $\bar{v}_{n}^{f} = \eta^{*} = 1/N$ for all n = 1, …N. Hence, we have $\hat{v}_{n}^{f} = {{\bar{v}_{n}^{f} } \mathord{\left/ {\vphantom {{\bar{v}_{n}^{f} } {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{l = 1}^{N} {\bar{v}_{l}^{f} \hat{x}_{l} } }} = {1 \mathord{\left/ {\vphantom {1 {\sum\nolimits_{l = 1}^{N} {\hat{x}_{l} } }}} \right. \kern-\nulldelimiterspace} {\sum\nolimits_{l = 1}^{N} {\hat{x}_{l} } }}$ for all n = 1, …, N. Let $\hat{u}_{n}^{f} = {1 \mathord{\left/ {\vphantom {1 {\left( {N\sum\nolimits_{l = 1}^{N} {\hat{x}_{l} } } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {N\sum\nolimits_{l = 1}^{N} {\hat{x}_{l} } } \right)}}$ for all n = 1, …, N, then we have ${\hat{\mathbf{u}}}^{f} \ge {\mathbf{0}}$ and

$$ {\hat{\mathbf{u}}}^{f} I^{N} \left( {\bar{\varepsilon }} \right) = {\hat{\mathbf{u}}}^{f} I^{N} \left( 1 \right) = {\hat{\mathbf{u}}}^{f} E = \frac{1}{{\sum\nolimits_{n = 1}^{N} {\hat{x}_{n}^{{}} } }}{\mathbf{e}} = {\hat{\mathbf{v}}}^{f} . $$

This means that the dual problem (36) for $\varepsilon = \bar{\varepsilon } = 1$ has an optimal solution $({\hat{\mathbf{v}}}^{f} ,{\hat{\mathbf{w}}}^{f} ,{\hat{\mathbf{u}}}^{f} )$ and its optimal value is 1. □

Since ${\hat{\mathbf{x}}} \in Eff (Q(\varepsilon ) )$is equivalent to the existence of an optimal solution of (36) attaining the optimal value 1, it follows from Assertion B that a face $\hat{\mathbb{F}}({\mathbf{v}}^{f} ,{\mathbf{w}}^{f} )$ of $T({\hat{\mathbf{y}}})$ is included in $Eff (Q(\varepsilon ) )$ for all f = 1, …f ₂, and it follows from (11) that Property [Q5] is valid for the fixed $\varepsilon \in (0,\bar{\varepsilon }]$. □

Theorem 12

Assume $1 = \mathop {\min }\limits_{j = 1, \ldots ,J} x_{nj} \;\left( {n = 1, \ldots ,N} \right)$ and choose ε satisfying the assumptions of Lemma 11, then eCCR ( 22 ) model satisfies [I], [M], [H], [C] and [U].

Proof

This assertion follows from Lemma 11 and the input transformation of D _x . □