A minimax approach for selecting the overall and stage-level most efficient unit in two stage production processes

Abstract

Data envelopment analysis (DEA) is a very effective management tool in assessing the performance of a set of decision making units (DMUs). In the efficiency evaluation using classic single stage DEA models, the internal processes of the DMUs are often neglected. In most real-world problems, it may be more realistic to evaluate the efficiency evaluation in two-stage production systems. In some cases: however, the decision-maker must need to identify the most efficient single unit. Numerous methods have been introduced to find the most efficient unit in single stage systems whereas no methods have been proposed for this aim in two stage production systems. Therefore, a new model based on mixed-integer programming was proposed to determine the most efficient DMU in two-stage systems and sub-stages in this study. The most important innovation of the suggested approach is that the most efficient DMUs of both stages can be found separately using only one model. Numerical examples for real world problems and a simulation study were provided for the validity of the proposed model.

Appendix

1.1 Appendix A. Proofs of theorems

Theorem 1

The MILP model (4) is feasible.

Proof

To ease of notation, we let $I_{r}^{u} = \frac{1}{{\left( {s + q + m} \right)\hbox{max} \left( {y_{rj} } \right)}};\; r = 1, \ldots ,s$, $I_{p}^{w} = \frac{1}{{\left( {s + q + m} \right)\hbox{max} \left( {z_{pj} } \right)}}; \;p = 1, \ldots ,q$, and $I_{i}^{v} = \frac{1}{{\left( {s + q + m} \right)\hbox{max} \left( {x_{ij} } \right)}}; \;i = 1, \ldots ,m$.

For $I_{r}^{u}$ and $I_{p}^{w}$, in the third set of constraints of model (4), there are two situations.

One is $\sum\nolimits_{r = 1}^{s} {I_{r}^{u} y_{rj} } - \sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } \le I_{2j}$ for $\forall j \in \left\{ {1, \ldots ,n} \right\}$ and the other is $\sum\nolimits_{r = 1}^{s} {I_{r}^{u} y_{rj} } - \sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } \ge I_{2j}$ for $\exists j \in \left\{ {1, \ldots ,n} \right\}$.

Let’s start with $\sum\nolimits_{r = 1}^{s} {I_{r}^{u} y_{rj} } - \sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } \le I_{2j}$.

If $\sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} } \le I_{1j}$, the model has a possible solution. Conversely, if $\sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} } \ge I_{1j}$ $\exists j \in \left\{ {1, \ldots ,n} \right\}$, we consider the set of deviation variable $d_{1j} > 0 \left( {j = 1, \ldots ,n} \right)$ meet the conditions $\sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} - d_{1j} } \le I_{1j}$. Let $\delta_{1} = \mathop {\hbox{max} }\limits_{j} \left\{ {{{d_{1j} } \mathord{\left/ {\vphantom {{d_{1j} } {\sum\nolimits_{{{\text{i}} = 1}}^{m} {x_{ij} } }}} \right. \kern-0pt} {\sum\nolimits_{{{\text{i}} = 1}}^{m} {x_{ij} } }}} \right\}$ and $v_{i}^{a} = I_{i}^{v} + \delta_{1}$ for $\forall {\text{i}} \in \left\{ {1, \ldots ,m} \right\}$. Then we can write $\delta_{1} > 0,v_{i}^{a} > I_{i}^{v}$ for $\forall i \in \left\{ {1, \ldots ,m} \right\}$, and using the ${{\delta_{1} \ge d_{1j} } \mathord{\left/ {\vphantom {{\delta_{1} \ge d_{1j} } {\sum\nolimits_{p = 1}^{q} {x_{ij} } }}} \right. \kern-0pt} {\sum\nolimits_{p = 1}^{q} {x_{ij} } }}$, we obtain

$$ \begin{aligned} & \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - \mathop \sum \limits_{i = 1}^{m} v_{i}^{a} x_{ij} \\ & = \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - \mathop \sum \limits_{i = 1}^{m} (I_{i}^{v} + \delta_{1} )x_{ij} \\ & = \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - \mathop \sum \limits_{i = 1}^{m} I_{i}^{v} x_{ij} - \mathop \sum \limits_{i = 1}^{m} \delta_{1} x_{ij} \\ & \le \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - \mathop \sum \limits_{i = 1}^{m} I_{i}^{v} x_{ij} - d_{1j} \\ & \le I_{1j} \\ \end{aligned} $$

for $\forall j \in \left\{ {1, \ldots ,n} \right\}$. Thus, $I_{p}^{w} \left( {p = 1, \ldots ,q} \right),v_{i}^{a} \left( {i = 1, \ldots ,m} \right)$, $I_{r}^{u} \;\left( {r = 1, \ldots ,s} \right)$ are in feasible region.

Consider the $\sum\nolimits_{r = 1}^{s} {I_{r}^{u} y_{rj} } - \sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } \ge I_{2j}$ $\exists j \in \left\{ {1, \ldots ,n} \right\}$. Let a set of deviation variable $d_{2j} > 0 \left( {j = 1, \ldots ,n} \right)$ meet the conditions $\sum\nolimits_{r = 1}^{s} {I_{r}^{u} y_{rj} } - \sum\nolimits_{p = 1}^{q} {I_{p}^{w} z_{pj} } - d_{2j} \le I_{2j}$. Let $\delta_{2} = \mathop {\hbox{max} }\limits_{j} \left\{ {{{d_{2j} } \mathord{\left/ {\vphantom {{d_{2j} } {\sum\nolimits_{p = 1}^{q} {z_{pj} } }}} \right. \kern-0pt} {\sum\nolimits_{p = 1}^{q} {z_{pj} } }}} \right\}$ and $w_{p}^{a} = I_{p}^{w} + \delta_{2}$ for $\forall p \in \left\{ {1, \ldots ,q} \right\}$ for this case. Then we can write $\delta_{2} > 0, w_{p}^{a} > I_{p}^{w}$ for $\forall p \in \left\{ {1, \ldots ,q} \right\}$, and using ${{\delta_{2} \ge d_{2j} } \mathord{\left/ {\vphantom {{\delta_{2} \ge d_{2j} } {\sum\nolimits_{p = 1}^{q} {z_{pj} } }}} \right. \kern-0pt} {\sum\nolimits_{p = 1}^{q} {z_{pj} } }}$, i.e., $\delta_{2} \sum\nolimits_{p = 1}^{q} {z_{pj} } \ge d_{2j}$ we have

$$ \begin{aligned} & \mathop \sum \limits_{r = 1}^{s} I_{r}^{u} y_{rj} - \mathop \sum \limits_{p = 1}^{q} w_{p}^{a} z_{pj} \\ & = \mathop \sum \limits_{r = 1}^{s} I_{r}^{u} y_{rj} - \mathop \sum \limits_{p = 1}^{q} \left( {I_{p + }^{w} \delta_{2} } \right)z_{pj} \\ & = \mathop \sum \limits_{r = 1}^{s} I_{r}^{u} y_{rj} - \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - \mathop \sum \limits_{p = 1}^{q} \delta_{2} z_{pj} \\ & \le \mathop \sum \limits_{r = 1}^{s} I_{r}^{u} y_{rj} - \mathop \sum \limits_{p = 1}^{q} I_{p}^{w} z_{pj} - d_{2j} \\ & \le I_{2j} \\ \end{aligned} $$

for $\forall j \in \left\{ {1, \ldots ,n} \right\}$.

If $\sum\nolimits_{p = 1}^{q} {w_{p}^{a} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} } \le I_{1j}$ $\forall j \in \left\{ {1, \ldots ,n} \right\}$, the model has a feasible region.

If $\sum\nolimits_{p = 1}^{q} {w_{p}^{a} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} } \ge I_{1j}$ $\exists j \in \left\{ {1, \ldots ,n} \right\}$, we consider a set of deviation variable $\tilde{d}_{1j} > 0 \left( {j = 1, \ldots ,n} \right)$ meet the conditions $\sum\nolimits_{p = 1}^{q} {w_{p}^{a} z_{pj} } - \sum\nolimits_{i = 1}^{m} {I_{i}^{v} x_{ij} } - \tilde{d}_{1j} \le I_{1j}$. Let $\tilde{\delta }_{1} = \mathop {\hbox{max} }\limits_{j} \left\{ {{{\tilde{d}_{1j} } \mathord{\left/ {\vphantom {{\tilde{d}_{1j} } {\sum\nolimits_{{{\text{i}} = 1}}^{m} {x_{ij} } }}} \right. \kern-0pt} {\sum\nolimits_{{{\text{i}} = 1}}^{m} {x_{ij} } }}} \right\}$ and $v_{i}^{b} = I_{i}^{v} + \tilde{\delta }_{1}$ for $\forall i \in \left\{ {1, \ldots ,m} \right\}$. Then we can write $\tilde{\delta }_{1} > 0,v_{i}^{a} > I_{i}^{v}$ for $\forall i \in \left\{ {1, \ldots ,m} \right\}$, and using the $\tilde{\delta }_{1} \ge {{\tilde{d}_{1j} } \mathord{\left/ {\vphantom {{\tilde{d}_{1j} } {\sum\nolimits_{p = 1}^{q} {x_{ij} } }}} \right. \kern-0pt} {\sum\nolimits_{p = 1}^{q} {x_{ij} } }}$, we obtain

$$ \begin{aligned} & \mathop \sum \limits_{p = 1}^{q} w_{p}^{a} z_{pj} - \mathop \sum \limits_{i = 1}^{m} v_{i}^{b} x_{ij} \\ & = \mathop \sum \limits_{p = 1}^{q} w_{p}^{a} z_{pj} - \mathop \sum \limits_{i = 1}^{m} (I_{i}^{v} + \tilde{\delta }_{1} )x_{ij} \\ & = \mathop \sum \limits_{p = 1}^{q} w_{p}^{a} z_{pj} - \mathop \sum \limits_{i = 1}^{m} I_{i}^{v} x_{ij} - \mathop \sum \limits_{i = 1}^{m} \tilde{\delta }_{1} x_{ij} \\ & \le \mathop \sum \limits_{p = 1}^{q} w_{p}^{a} z_{pj} - \mathop \sum \limits_{i = 1}^{m} I_{i}^{v} x_{ij} - \tilde{d}_{1j} \\ & \le I_{1j} \\ \end{aligned} $$

for $\forall j \in \left\{ {1, \ldots ,n} \right\}$. So, $w_{p}^{a} \left( {p = 1, \ldots ,q} \right),$ $v_{i}^{b}$ $\left( {i = 1, \ldots ,m} \right)$, $I_{r}^{u}$ $\left( {r = 1, \ldots ,s} \right)$ are in feasible region. That is, the MILP model for two stages is always feasible. If lower bound solutions are infeasible then, these can be made feasible by increasing the values of input weight ($v_{i}$) or of the intermediate measures weights ($w_{p}$).

Summing the constraints $\sum\nolimits_{p = 1}^{q} {w_{p} z_{pj} } - \sum\nolimits_{i = 1}^{m} {v_{i} x_{ij} } \le I_{1j}$ and $\sum\nolimits_{r = 1}^{s} {u_{r} y_{rj} } - \sum\nolimits_{p = 1}^{q} {w_{p} z_{pj} } \le I_{2j}$ over $j$ from $j = 1$ to $n$ seperately, we have

$$ \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) \ge - 1 $$

and

$$ \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) - \mathop \sum \limits_{r = 1}^{s} u_{r} \left( {\mathop \sum \limits_{j = 1}^{n} y_{rj} } \right) \ge - 1. $$

Summing up the last two inequalities, we can write

$$ \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{r = 1}^{s} u_{r} \left( {\mathop \sum \limits_{j = 1}^{n} y_{rj} } \right) \ge - 2. $$

Therefore, the objective function of the model (4) has a lower bound. As the MILP model is a minimization problem, model (4) is a feasible solution. This completes the proof.

Theorem 2

Model (4) and model (5) are equivalent.

Proof

From the $\sum\nolimits_{p = 1}^{q} {w_{p} z_{pj} } - \sum\nolimits_{i = 1}^{m} {v_{i} x_{ij} + S_{1j} } = I_{1j}$ constraint, we can write

$$ S_{1j} = \mathop \sum \limits_{i = 1}^{m} v_{i} x_{ij} - \mathop \sum \limits_{p = 1}^{q} w_{p} z_{pj} + I_{1j} . $$

Summing up these constraints over $j$ from $j = 1$ to $j = n$ and considering $\sum\nolimits_{j = 1}^{n} {I_{1j} } = 1,$ we can write

$$ \mathop \sum \limits_{j = 1}^{n} S_{1j} = \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) + 1. $$

Similarly, from the third type of constraints in the model (5), summing up these constraints over j from j = 1 to j = n, and considering $\sum\nolimits_{j = 1}^{n} {I_{2j} } = 1,$ we can write

$$ \mathop \sum \limits_{j = 1}^{n} S_{2j} = \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) - \mathop \sum \limits_{r = 1}^{s} u_{r} \left( {\mathop \sum \limits_{j = 1}^{n} y_{rj} } \right) + 1. $$

By summing up the last two inequalities, we obtain

$$ \mathop \sum \limits_{j = 1}^{n} S_{1j} + \mathop \sum \limits_{j = 1}^{n} S_{2j} = \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) + 2. $$

If the constant 2 is ignored, the objective function of the model (5) is written as

$$ \mathop \sum \limits_{j = 1}^{n} S_{1j} + \mathop \sum \limits_{j = 1}^{n} S_{2j} = \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{r = 1}^{s} u_{r} \left( {\mathop \sum \limits_{j = 1}^{n} y_{rj} } \right) $$

which is equal to the objective function of model (4). This completes the proof.

Theorem 3

The following model (6) and model (4) are equivalents.

Proof

In model (6), let $d_{1j} = 1 - I_{1j}$ and $\beta_{1j} = 1 - S_{1j}$ for $j = 1, \ldots , n$. Summing up these equations over $j$ from $j = 1$ to $j = n$, this results in

$$ \mathop \sum \limits_{j = 1}^{n} d_{1j} = n - \mathop \sum \limits_{j = 1}^{n} I_{1j} = n - 1 $$

and

$$ \mathop \sum \limits_{j = 1}^{n} \beta_{1j} = - \left( {\mathop \sum \limits_{j = 1}^{n} S_{1j} - n} \right) = \left( {n - 1} \right) - \left( {\mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right)} \right). $$

Similarly, if $d_{2j} = 1 - I_{2j}$ and $\beta_{2j} = 1 - S_{2j}$ are taken for $j = 1, \ldots , n$, it is obtained as

$$ \mathop \sum \limits_{j = 1}^{n} \beta_{2j} = \left( {n - 1} \right) - \left( {\mathop \sum \limits_{p = 1}^{q} w_{p} \left( {\mathop \sum \limits_{j = 1}^{n} z_{pj} } \right) - \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right)} \right). $$

Hence, $\sum\nolimits_{j = 1}^{n} {\left( {\beta_{1j} + \beta_{2j} } \right)}$ is obtained as $\left( {2n - 2} \right) - \sum\nolimits_{i = 1}^{m} {v_{i} } \left( {\sum\nolimits_{j = 1}^{n} {x_{ij} } } \right) - \sum\nolimits_{r = 1}^{s} {u_{r} } \left( {\sum\nolimits_{j = 1}^{n} {y_{rj} } } \right).$

Since $2n - 2$ is a constant, the fact that $\sum\nolimits_{j = 1}^{n} {\left( {\beta_{1j} + \beta_{2j} } \right)}$ is equal to minimum of

$$ \mathop \sum \limits_{i = 1}^{m} v_{i} \left( {\mathop \sum \limits_{j = 1}^{n} x_{ij} } \right) - \mathop \sum \limits_{r = 1}^{s} u_{r} \left( {\mathop \sum \limits_{j = 1}^{n} y_{rj} } \right) $$

completes the proof.

Theorem 4

The proposed minimax model (8) is always feasible and bounded.

Proof

Let $\left( {u^{0} ,w^{0} ,v^{0} ,I_{1}^{0} ,I_{2}^{0} } \right)$ be a feasible solution to model (7) and we proved that such a solution exists in Theorem 1. Let $d_{1j}^{0} = 1 - I_{1j}^{0}$, $\beta_{1j}^{0} = 1 - I_{1j}^{0} + w^{0} z_{j} - v^{0} x_{j}$ for $j = 1$ to $j = n$ and $d_{max1}^{0} = max\left\{ {d_{1j}^{0} - \beta_{1j}^{0} :j = 1, \ldots , n } \right\}$ for the first stage. Summing up $d_{1j}^{0} = 1 - I_{1j}^{0}$ over $j = 1$ to $j = n$ results in $\sum\nolimits_{j = 1}^{n} {d_{1j}^{0} } = n - 1$. For $\sum\nolimits_{p = 1}^{q} {w_{p} z_{pj} } - \sum\nolimits_{i = 1}^{m} {v_{i} x_{ij} + d_{1j} - \beta_{1j} } = 0$ constraints, $w^{0} z_{j} - v^{0} x_{j} + 1 - I_{1j}^{0} - 1 + I_{1j}^{0} - w^{0} z_{j} + v^{0} x_{j} = 0$ could be proved. In the first stage, while for the most efficient unit, $w^{0} z_{j} - v^{0} x_{j} \le 1$ satisfies, for the other units, $w^{0} z_{j} - v^{0} x_{j} \le 0$ satisfies. Therefore, for $d_{{max_{1} }} - d_{1j} + \beta_{1j} \ge 0$ constraints, $d_{max1}^{0} \ge v^{0} x_{j} - w^{0} z_{j}$ could be obtained. Hence, when considering the most efficient unit and other units for the first stage, $d_{max1}^{0} \ge - 1$ is obtained. Similarly, if $d_{2j}^{0} = 1 - I_{2j}^{0}$ and $\beta_{2j}^{0} = 1 - I_{2j}^{0} + u^{0} y_{j} - w^{0} z_{j}$ for $j = 1$ to $j = n$ and $d_{max2}^{0} = max\left\{ {d_{2j}^{0} - \beta_{2j}^{0} :j = 1, \ldots , n } \right\}$ are taken for the second stage, the same results are obtained. It is proved that $\left( {u^{0} ,w^{0} ,v^{0} ,I_{1}^{0} ,I_{2}^{0} } \right)$ is a feasible solution to model (8).

To show that the proposed model has the bounded, $\left( {u^{0} ,w^{0} ,v^{0} ,d_{1}^{0} ,\beta_{1}^{0} ,d_{2}^{0} ,\beta_{2}^{0} ,d_{max1}^{0} ,d_{max2}^{0} } \right)$ is taken any arbitrary feasible solution to model (8). From the constraints of this model,

$$ d_{max1}^{0} \ge d_{1j}^{0} - \beta_{1j}^{0} \ge - \beta_{1j}^{0} \ge - 1 $$

and similarly

$$ d_{max2}^{0} \ge d_{2j}^{0} - \beta_{2j}^{0} \ge - \beta_{2j}^{0} \ge - 1 $$

is obtained. Summing up the last two equations,

$$ d_{max1}^{0} + d_{max2}^{0} \ge d_{1j}^{0} - \beta_{1j}^{0} + d_{2j}^{0} - \beta_{2j}^{0} \ge - \beta_{1j}^{0} - \beta_{2j}^{0} \ge - 2 $$

means the objective function is bounded from below. Since the proposed model (8) is a minimization problem, the proof is completed.

1.2 Appendix B. Detailed graphs on cases in simulation

See Figs. 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 and 21.

Fig. 4

Graph of correlation results in 100 repeats for Case 1

Fig. 5

Graph of correlation results in 100 repeats for Case 2

Fig. 6

Graph of correlation results in 100 repeats for Case 3

Fig. 7

Graph of correlation results in 100 repeats for Case 4

Fig. 8

Graph of correlation results in 100 repeats for Case 5

Fig. 9

Graph of correlation results in 100 repeats for Case 6

Fig. 10

Graph of correlation results in 100 repeats for Case 7

Fig. 11

Graph of correlation results in 100 repeats for Case 8

Fig. 12

Graph of correlation results in 100 repeats for Case 9

Fig. 13

Graph of correlation results in 100 repeats for Case 10

Fig. 14

Graph of correlation results in 100 repeats for Case 11

Fig. 15

Graph of correlation results in 100 repeats for Case 12

Fig. 16

Graph of correlation results in 100 repeats for Case 13

Fig. 17

Graph of correlation results in 100 repeats for Case 14

Fig. 18

Graph of correlation results in 100 repeats for Case 15

Fig. 19

Graph of correlation results in 100 repeats for Case 16

Fig. 20

Graph of correlation results in 100 repeats for Case 17

Fig. 21

Graph of correlation results in 100 repeats for Case 18