Finding closest target for bank branches in the presence of weight restrictions using data envelopment analysis

Abstract

Data envelopment analysis technique, not only evaluates Decision making units (DMUs), but also introduces a benchmark for inefficient DMUs. By having the data related to the benchmark and using it appropriately, the DMU under the assessment determines how to eliminate its inefficiency and become efficient. The closer the evaluated DMU is to the presented target, the faster and easier it reaches the efficiency frontier. Furthermore, target setting in the presence of weight restrictions is an important issue. In this paper, a model for finding the closest target, in the presence of weight restrictions is presented. Thus, taking into account trade-offs, as well as weight restrictions, the closest target for each DMU is introduced. The proposed model is also compared with the previous models. Finally, administrating the proposed model for evaluating one of Iranian Banks, the least changes to inefficient branches is represented.

Appendices

Appendix A

1.1 Proof of Theorem (1)

Suppose $(\tilde{X},\tilde{Y}) \in D_{o}$ that is $(\tilde{X},\tilde{Y})$ is a Pareto-efficient point dominating DMU_o on the frontier T^′. $(\tilde{X},\tilde{Y})$ is efficient, thus the optimal solution of both of the following problems equals zero.

$$ \begin{aligned} & \begin{array}{*{20}c} {Max} & {1S^{ + } + 1S^{ - } } \\ \end{array} \\ & \begin{array}{*{20}l} {s.t.} \hfill & {\sum\limits_{j \in E}^{{}} {\lambda_{j} x_{j} + \sum\limits_{t = 1}^{L} {\mu_{t} P_{t} + S^{ - } = \tilde{X}} } } \hfill \\ {} \hfill & {\sum\limits_{j \in E}^{{}} {\lambda_{j} } y_{j} + \sum\limits_{t = 1}^{L} {\mu_{t} Q_{t} - S^{ + } = \tilde{Y}} } \hfill \\ {} \hfill & {\sum\limits_{j \in E}^{{}} {\lambda_{j} } x_{j} + \sum\limits_{t = 1}^{L} {\mu_{t} P_{t} \ge 0} } \hfill \\ {} \hfill & {\sum\limits_{j \in E}^{{}} {\lambda_{j} } y_{j} + \sum\limits_{t = 1}^{L} {\mu_{t} Q_{t} \ge 0} } \hfill \\ {} \hfill & {\lambda ,\mu ,S^{ + } ,S^{ - } \ge 0} \hfill \\ {} \hfill & {} \hfill \\ \end{array} \\ \end{aligned} $$

(8)

$$ \begin{aligned} & \begin{array}{*{20}c} {Min} & {V\tilde{X} - U\tilde{Y}} \\ \end{array} \\ & \begin{array}{*{20}c} {s.t.} & {VX_{j} - UY_{j} - W^{\prime \prime } X_{j} - W^{\prime } Y_{j} \ge 0} & {j \in E} \\ {} & {\sum\limits_{r = 1}^{s} {u_{r} Q_{rt} - \sum\limits_{i = 1}^{m} {v_{i} P_{it} + \sum\limits_{r = 1}^{s} {w_{r}^{\prime } P_{rt} } + } \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } Q_{it} } + d_{t}^{\prime } } } & {t = 1, \ldots ,L} \\ {} & {u_{r} \ge 1} & {r = 1, \ldots ,s} \\ {} & {v_{i} \ge 1} & {i = 1, \ldots ,m} \\ {} & {w^{\prime } ,w^{\prime \prime } \ge 0} & {} \\ \end{array} \\ \end{aligned} $$

(9)

As $(\tilde{X},\tilde{Y}) \ne 0$, and regarding the optimal solution of both of the proceeding problems, a non-negative scalar set $\tilde{\lambda }_{j} ;\;j \in E$ and $\tilde{\mu }_{t} ;\;(t = 1, \ldots ,L)$ can be found in which at least one of the $\tilde{\lambda }_{j} ,\tilde{\mu }_{t}$ is strictly positive, such that:

$$ (\tilde{X},\tilde{Y}) = \left( {\sum\limits_{j \in E} {\tilde{\lambda }_{j} x_{ij} + \sum\limits_{l} {\tilde{\mu }_{t}^{\prime \prime } b_{il} } ,\sum\limits_{j \in E} {\tilde{\lambda }_{j} y_{rj} + \sum\limits_{t} {\tilde{\mu }_{t}^{\prime } a_{rt} } } } } \right) $$

However, according to Complementary Slackness Theorem, for each $\tilde{\lambda }_{j} > 0$, its corresponding dual constraint is binding, that is:

$$ \exists \tilde{u}_{r} \ge 1,\tilde{v}_{i} \ge 1,\tilde{w}^{\prime } ,\tilde{w}^{\prime \prime } \ge 0\quad s.t.\quad \tilde{U}Y_{j} - \tilde{V}X_{j} + \tilde{W}^{\prime \prime } X_{j} + \tilde{W}^{\prime } Y_{j} = 0 $$

Then we define as follows:

$$ \tilde{\gamma }_{j} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\tilde{\lambda }_{j} > 0} \hfill \\ 1 \hfill & {o.w.} \hfill \\ \end{array} ,\quad \tilde{d}_{j} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\tilde{\lambda }_{j} > 0} \hfill \\ { - \tilde{U}Y_{j} + \tilde{V}X_{j} - \tilde{W}^{\prime \prime } X_{j} - \tilde{W}^{\prime } Y_{j} } \hfill & {o.w.} \hfill \\ \end{array} } \right.} \right. $$

Similarly, according to Complementary Slackness Theorem, for each $\tilde{\mu }_{t} > 0$, its corresponding dual constraint is binding, that is:

$$ \exists \tilde{v} \ge 1,\tilde{u} \ge 1,\tilde{w}^{\prime \prime } \ge 0,\tilde{w}^{\prime } \ge 0\quad s.t.\quad \sum\limits_{r = 1}^{s} {\tilde{u}_{r} Q_{rt} - \sum\limits_{i = 1}^{m} {\tilde{v}_{i} P_{it} + \sum\limits_{r = 1}^{s} {\tilde{w}_{r}^{\prime } P_{rt} } + } \sum\limits_{i = 1}^{m} {\tilde{w}_{i}^{\prime \prime } Q_{it} } = 0} $$

Then we define as follows:

$$ \tilde{\gamma }_{t}^{\prime } = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\tilde{\mu }_{t} > 0} \hfill \\ 1 \hfill & {o.w.} \hfill \\ \end{array} } \right.,\quad \tilde{d}_{t}^{\prime } = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\tilde{\mu }_{t} > 0} \hfill \\ { - \sum\limits_{r = 1}^{s} {\tilde{u}_{r} Q_{rt} + \sum\limits_{i = 1}^{m} {\tilde{v}_{i} P_{it} - \sum\limits_{r = 1}^{s} {\tilde{w}_{r}^{\prime } P_{rt} } - } \sum\limits_{i = 1}^{m} {\tilde{w}_{i}^{\prime \prime } Q_{it} } } } \hfill & {o.w.} \hfill \\ \end{array} } \right. $$

Finally, as we supposed $(\tilde{X},\tilde{Y})$ dominates DMU_o, we can define:

$$ \begin{aligned} \tilde{s}_{io}^{ - } = x_{io} - \tilde{x}_{i} \ge 0\quad i = 1, \ldots ,m. \hfill \\ \tilde{s}_{ro}^{ + } = \tilde{y}_{r} - y_{ro} \ge 0\quad r = 1, \ldots ,s. \hfill \\ \end{aligned} $$

So $\tilde{\lambda }_{j} ,\tilde{d}_{j} \ge 0,\,\tilde{\gamma }_{j} \in \left\{ {0,1} \right\};\;j \in E$, $\tilde{\mu }_{t} ,\tilde{d}_{t}^{\prime } \ge 0,\,\tilde{\gamma }_{t}^{\prime } \in \left\{ {0,1} \right\};\;t = 1, \ldots ,L$, $\tilde{S}^{ - } ,\tilde{S}^{ + } ,\tilde{w}^{\prime } ,\tilde{w}^{\prime \prime } \ge 0$ and $\tilde{u}_{r} \ge 1;\;r = 1, \ldots ,s,\;\tilde{v}_{i} \ge 1;\;i = 1, \ldots ,m,$ satisfy the theorem’s conditions.

Conversely suppose that the parameters $\tilde{\lambda }_{j} ,\tilde{d}_{j} \ge 0,\,\,\tilde{\gamma }_{j} \in \left\{ {0,1} \right\};\;j \in E$ ، $\tilde{\mu }_{t} ,\tilde{d}_{t}^{'} \ge 0,\,\,$ ، $\tilde{\gamma }_{t}^{'} \in \left\{ {0,1} \right\};\;t = 1, \ldots ,L$ and $\tilde{u}_{r} \ge 1;\;r = 1, \ldots ,s\,\,,\tilde{v}_{i} \ge 1\;\;i = 1, \ldots ,m,\tilde{S}^{ - } ,\tilde{S}^{ + } ,\tilde{w}^{\prime } ,\tilde{w}^{\prime \prime } \ge 0$ are true for the constraints (a.3)–(a.12).

We define:

$$ (\tilde{X},\tilde{Y}) = \left( {\sum\limits_{j \in E} {\tilde{\lambda }_{j} x_{ij} + \sum\limits_{t} {\tilde{\mu }_{t} P_{t} } ,\sum\limits_{j \in E} {\tilde{\lambda }_{j} y_{rj} + \sum\limits_{t} {\tilde{\mu }_{t} Q_{t} } } } } \right) $$

Now we prove that $(\tilde{X},\tilde{Y}) \in D_{o}$ According to (a.4) and (a.3) it is obvious that $(\tilde{X},\tilde{Y})$ dominates DMU_o = (X_o, Y_o)thus it is enough to prove that $(\tilde{X},\tilde{Y})$ is on the efficiency frontier.

We want to prove that $(\tilde{X},\tilde{Y})$ lies on the efficiency frontier that is to prove $U\tilde{Y} - V\tilde{X} + W^{\prime \prime } \tilde{X} + W^{\prime } \tilde{Y} = 0$. Based on what we already supposed, according to the (a.1) and (a.2), we know that $\tilde{X} = \sum\nolimits_{j \in E} {\tilde{\lambda }_{j} x_{j} } + \sum\limits_{t} {\tilde{\mu }_{t} P_{t} }$ and $\tilde{Y} = \sum\nolimits_{j \in E} {\tilde{\lambda }_{j} y_{j} } + \sum\limits_{t} {\tilde{\mu }_{t} Q_{t} }$.

by replacing, we have:

$U\tilde{Y} - V\tilde{X} + W^{\prime \prime } \tilde{X} + W^{\prime } \tilde{Y} = \underbrace {{(U + W^{\prime } )(\sum\limits_{j \in E} {\tilde{\lambda }_{j} y_{j} } ) - (V - W^{\prime \prime } )(\sum\limits_{j \in E} {\tilde{\lambda }_{j} x_{j} } )}}_{\prime \prime a\prime \prime } + \underbrace {{(U + W^{\prime } )(\sum\limits_{t} {\tilde{\mu }_{t} Q_{t} } ) - (V - W^{\prime \prime } )(\sum\limits_{t} {\tilde{\mu }_{t} P_{t} } )}}_{\prime \prime b\prime \prime }$so, to prove $U\tilde{Y} - V\tilde{X} + W^{\prime \prime } \tilde{X} + W^{\prime } \tilde{Y} = 0$, it is enough to prove that individual sentences “a” and “b” equals zero. To prove the sentence “a” is zero we have:

$$ \begin{aligned} \tilde{d}_{j} \ge 0 & \Rightarrow \sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} - \sum\limits_{i = 1}^{m} {v_{i} \tilde{x}_{ij} + \sum\limits_{r = 1}^{s} {w_{r}^{\prime } \tilde{y}_{rj} } + \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } \tilde{x}_{ij} } \le 0,\quad j \in E} } \\ & \Rightarrow \tilde{\lambda }_{j} \sum\limits_{r = 1}^{s} {u_{r} } \tilde{y}_{rj} - \tilde{\lambda }_{j} \sum\limits_{i = 1}^{m} {v_{i} \tilde{x}_{ij} + \tilde{\lambda }_{j} \sum\limits_{r = 1}^{s} {w_{r}^{\prime } \tilde{y}_{rj} } + \tilde{\lambda }_{j} \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } \tilde{x}_{ij} } \le 0,\quad j \in E} \\ & \Rightarrow U\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{y}_{j} - V\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{j} } } + w^{\prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} } \tilde{y}_{rj} + w^{\prime \prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{ij} } \le 0, \\ \end{aligned} $$

while

$$ \sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} - \sum\limits_{i = 1}^{m} {v_{i} \tilde{x}_{ij} + \sum\limits_{r = 1}^{s} {w_{r}^{\prime } \tilde{y}_{rj} } + \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } \tilde{x}_{ij} } } } = - d_{j} ,\,\,\,j \in E $$

therefore

$$ \,U\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{y}_{j} - V\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{j} } } + w^{\prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} } \tilde{y}_{rj} + w^{\prime \prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{ij} } = - \sum\limits_{j \in E} {\tilde{\lambda }_{j} d_{j} } $$

by (a.8) and (a.9), it is concluded that $\sum\nolimits_{j \in E} {\tilde{\lambda }_{j} d_{j} } = 0$ so

$$ (U + w^{\prime } )\sum\limits_{j \in E} {\tilde{\lambda }_{j} } \tilde{y}_{rj} - (V - w^{\prime \prime } )\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{ij} } = U\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{y}_{j} - V\sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{j} } } + w^{\prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} } \tilde{y}_{rj} + w^{\prime \prime } \sum\limits_{j \in E} {\tilde{\lambda }_{j} \tilde{x}_{ij} } = - \sum\limits_{j \in E} {\tilde{\lambda }_{j} d_{j} } = 0 $$

therefore, the sentence “a” is equal to zero.

To prove the sentence “b” is zero, we have:

$$ \begin{aligned} d_{t}^{'} \ge 0 & \Rightarrow \sum\limits_{r = 1}^{s} {u_{r} Q_{rt} - \sum\limits_{i = 1}^{m} {v_{i} P_{it} + \sum\limits_{r = 1}^{s} {w_{r}^{\prime } Q_{rt} } + } \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } P_{it} } \le 0\quad t = 1, \ldots ,L} \\ & \Rightarrow \tilde{\mu }_{t} \sum\limits_{r = 1}^{s} {u_{r} Q_{rt} - \tilde{\mu }_{t} \sum\limits_{i = 1}^{m} {v_{i} P_{it} + \tilde{\mu }_{t} \sum\limits_{r = 1}^{s} {w_{r}^{\prime } Q_{rt} } + } \tilde{\mu }_{t} \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } P_{it} } \le 0\quad t = 1, \ldots ,L} \\ & \Rightarrow U\sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} - V\sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} + w^{\prime } \sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} } + } w^{\prime \prime } \sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} } \le 0.} \\ \end{aligned} $$

while

$$ \,\sum\limits_{r = 1}^{s} {u_{r} Q_{rt} - \sum\limits_{i = 1}^{m} {v_{i} P_{it} + \sum\limits_{r = 1}^{s} {w_{r}^{\prime } Q_{rt} } + } \sum\limits_{i = 1}^{m} {w_{i}^{\prime \prime } P_{it} } = - d_{t}^{\prime } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = 1, \ldots ,L\,\,\,\,} $$

Therefore

$$ U\sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} - V\sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} + W^{\prime } \sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} } + } W^{\prime \prime } \sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} } } = - \sum\limits_{t} {\tilde{\mu }_{t} d_{t}^{\prime } } $$

By (a.11) and (a.12), it is concluded that $\sum\limits_{t} {\tilde{\mu }_{t}^{\prime } d_{t}^{\prime } } = 0$ so

$$ \left( {U + W^{\prime } } \right)\sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} Q_{it} - \left( {V - W^{\prime \prime } } \right)} \sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} P_{rt} = U\!\!\sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} } - V\!\!\sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} } +W^{\prime } \sum\limits_{r = 1}^{s} {\tilde{\mu }_{t} Q_{rt} + W^{\prime \prime } } \sum\limits_{i = 1}^{m} {\tilde{\mu }_{t} P_{it} = - \sum\limits_{t}^{{}} {\tilde{\mu }_{t} d_{t}^{\prime } = 0} } } $$

Therefore, the sentence “b” is equal to zero.

So $U\tilde{Y} - V\tilde{X} + W^{\prime \prime } \tilde{X} + W^{\prime } \tilde{Y} = 0$ and the desirable result is obtained.□

Appendix B

2.1 Proof of the Theorem (2)

With respect to Theorem (1), D_p is non-dominated strong efficient frontier ofDMU_p, and considering Definition (1) Least L₁-distance of DMU_pfrom D_pis equal to:

$$ \begin{aligned} & Min\left\{ {\left\| {\left( {\begin{array}{*{20}c} X \\ Y \\ \end{array} } \right) - \left( {\begin{array}{*{20}c} {X_{p} } \\ {Y_{p} } \\ \end{array} } \right)} \right\|_{1} ;\left( {\begin{array}{*{20}c} X \\ Y \\ \end{array} } \right) \in D_{p} } \right\} \\ & \quad = Min\left\{ {\left\| {\left( {\begin{array}{*{20}c} {\sum\limits_{j \in E} {\lambda_{j} x_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} P_{t} } } \\ {\sum\limits_{j \in E} {\lambda_{j} y_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} Q_{t} } } \\ \end{array} } \right) - \left( {\begin{array}{*{20}c} {X_{p} } \\ {Y_{p} } \\ \end{array} } \right)} \right\|_{1} ;\left( {\begin{array}{*{20}c} {\sum\limits_{j \in E} {\lambda_{j} x_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} P_{t} } } \\ {\sum\limits_{j \in E} {\lambda_{j} y_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} Q_{t} } } \\ \end{array} } \right) \in D_{p} } \right\} \\ & \quad = Min\left\{ {\left| {\sum\limits_{j \in E} {\lambda_{j} x_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} P_{t} } - X_{p} } \right| + \left| {\sum\limits_{j \in E} {\lambda_{j} y_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t} Q_{t} - Y_{p} } } \right|} \right\} \\ \quad s.t.\quad (7.1) - (7.15) \\ & \quad = Min\left\{ {\left| {S^{ - } } \right| + \left| {S^{ + } } \right|} \right\} \\ \quad s.t.\quad (7.1) - (7.15) \\ \end{aligned} $$

$$ \begin{aligned} & {\text{since }}\left( {S^{ - } , S^{ + } } \right) \ge 0; \\ & \quad = Min1S^{ - } + 1S^{ + } = Z^{*}_{p} \\ & \quad s.t.\begin{array}{*{20}c} {} & {(7.1) - (7.15)} \\ \end{array} \\ \end{aligned} $$

□

Appendix C

3.1 Proof of Theorem (3)

Let DMU_p is efficient and $(\lambda^{*} ,d^{*} ,d^{\prime *} ,\mu^{*} ,S^{ - *} ,S^{ + *} ,w^{\prime *} ,w^{\prime \prime *} )$ is an optimal solution of model (7), evaluating DMU_p. In contrast, suppose $Z_{p}^{*} \ne 0$, then at least one of the component of vector ($S ^{* - } , S^{* + }$) is positive. Hence

$$ \left( {\begin{array}{*{20}c} { - \left( {\sum\limits_{j \in E} {\lambda *_{j} x_{j} } + \sum\limits_{t = 1}^{L} {\mu *_{t} P_{t} } } \right)} \\ {\sum\limits_{j \in E} {\lambda *_{j} y_{j} } + \sum\limits_{t = 1}^{L} {\mu *_{t} Q_{t} } } \\ \end{array} } \right)\begin{array}{*{20}c} \ge \\ \ne \\ \end{array} \left( {\begin{array}{*{20}c} { - X_{p} } \\ {Y_{p} } \\ \end{array} } \right) $$

Which means DMU_p is dominated by $\left( {\begin{array}{*{20}c} { - \left( {\sum\limits_{j \in E} {\lambda_{j}^{*} x_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t}^{*} P_{t} } } \right)} \\ {\sum\limits_{j \in E} {\lambda_{j}^{*} y_{j} } + \sum\limits_{t = 1}^{L} {\mu_{t}^{*} Q_{t} } } \\ \end{array} } \right)$, and this is in contradiction with the assumption that DMU_p is efficient.

Conversely, let $Z_{p}^{*} = 0$. Regarding Theorem (2), $Z_{p}^{*}$ is the closest distance of DMU_p from D_P, and considering Theorem (1), D_P is the set of points on the frontier which dominate DMU_p. Accordingly, the distance between DMU_p and the set of efficient and dominated points is zero, which means DMU_p is efficient.□

Appendix D

4.1 Proof of Lemma (1)

Evaluating DMU_p, suppose that (λ^*, μ^*, $S^{ - *} , S^{ + *}$) is the optimal solution of model (8), and (V^*, U^*, $W^{\prime *}$, $W^{\prime \prime *}$) is its corresponding dual multipliers vector. According to the Complementary Slackness Theorem, for each $\lambda_{j}^{*} > 0$, the corresponding dual constraint is binding, that is [respecting model (9)]:

$$ \begin{array}{*{20}c} {\tilde{U}Y_{j} - \tilde{V}X_{j} + \tilde{W}^{\prime \prime } X_{j} + \tilde{W}^{\prime } Y_{j} = 0} & {} & {} \\ \end{array} $$

Then we define $\begin{array}{*{20}c} {\gamma_{j}^{*} } & {,d_{j}^{*} } \\ \end{array}$ as follows:

$$ \gamma^{*}_{j} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\tilde{\lambda }_{j} > 0} \hfill \\ 1 \hfill & {o.w.} \hfill \\ \end{array} ,\quad d_{j}^{*} = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\lambda_{j}^{*} > 0} \hfill \\ { - U^{*} Y_{j} + V^{*} X_{j} - W^{*\prime \prime } X_{j} - W^{*\prime } Y_{j} } \hfill & {o.w.} \hfill \\ \end{array} } \right.} \right. $$

Similarly, according to the Complementary Slackness Theorem for each $\mu_{t}^{*} > 0$, the corresponding dual constraint is binding, that is:

$$ \exists v_{i}^{*} \ge 1,u_{r}^{*} \ge 1,w^{*\prime } \ge 0,w^{*\prime \prime } \ge 0\quad s.t.\quad \sum\limits_{r = 1}^{s} {u_{r}^{*} Q_{rt} - \sum\limits_{i = 1}^{m} {v_{i}^{*} P_{it} + \sum\limits_{r = 1}^{s} {w_{r}^{*\prime } P_{rt} } + } \sum\limits_{i = 1}^{m} {w_{i}^{*\prime \prime } Q_{it} } = 0} $$

and we define:

$$ \gamma_{t}^{*\prime } = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\mu_{t}^{*} > 0} \hfill \\ 1 \hfill & {o.w.} \hfill \\ \end{array} } \right.,\quad d_{t}^{*\prime } = \left\{ {\begin{array}{*{20}l} 0 \hfill & {\mu_{t} > 0} \hfill \\ { - \sum\limits_{r = 1}^{s} {u_{r}^{*} Q_{rt} + \sum\limits_{i = 1}^{m} {v_{i}^{*} P_{it} - \sum\limits_{r = 1}^{s} {w_{r}^{*\prime } P_{rt} } - } \sum\limits_{i = 1}^{m} {w_{i}^{*\prime \prime } Q_{it} } } } \hfill & {o.w.} \hfill \\ \end{array} } \right. $$

Therefore,(λ^*, μ^*, S^−*, S^+*, V^*, U^*, W^′*, W^′′*, γ’^*, γ^*, d^*, d^′*) is a feasible solution for DMU_p in model(7).□

Appendix E

5.1 Proof of Lemma (2)

Regarding the proof of Lemma (1) and as (λ^*, μ^*, S^−*, S^+*, V^*, U^*, W^′*, W^′′*, γ^′*, γ^*, d^*, d^′*) is a feasible solution for model (7) and since the objective function of model (7) is $Min\left\{ {\sum\nolimits_{i = 1}^{m} {s_{i}^{ - } + \sum\nolimits_{r = 1}^{s} {s_{r}^{ + } } } } \right\}$, so $\gamma_{p}^{*} = \sum\nolimits_{i = 1}^{m} {s_{i}^{ - *} } + \sum\nolimits_{r = 1}^{s} {s_{r}^{ + *} }$ is an upper bound for model (7).□