Cost-efficiency measurement for two-stage DEA network using game approach: an application to electrical network in Iran

Abstract

This study proposes a two-stage game–data envelopment analysis (DEA) approach for cost-efficiency (CE) measurement using centralized and Stackelberg game models. Each decision-making unit (DMU) is proposed to make up two-stage network structures (or processes), where all the first-stage outputs are the only second-stage inputs. The main contribution of this study is the development of cooperative and non-cooperative models for CE measurement and implies a unique CE decomposition. We apply centralized and Stackelberg approaches for cooperative and non-cooperative assumption between two stages. We proposed a simplified version of the two-stage DEA network CE model. Applying the proposed model for CE calculation reduces both the number of constraints and variables, resulting in a sharp reduction in computational requirements. The applicability of the presented model is demonstrated in the context of studying the output of the electrical network in Iran. The results of the case study show that in a two-stage network, the proposed DEA model can provide accurate estimates of CE. This paper aids the two-stage network structures to control the costs by weak DMUs management. To the authors’ knowledge, this paper is the first research on the CE in a network by the DEA approach so that the relationships between the internal stages of the network are considered.

Appendix A

In this part, we prove that the area set of $S^{\prime}$ is convex.

Proof

Assume $\left( {v^{\prime}_{1} , \ldots ,v^{\prime}_{m} ,w^{\prime}_{1} , \ldots ,w^{\prime}_{d} ,u^{\prime}_{1} , \ldots ,u^{\prime}_{s} } \right) \in S$ and $\left( {v^{\prime\prime}_{1} , \ldots ,v^{\prime\prime}_{m} ,w^{\prime\prime}_{1} , \ldots ,w^{\prime\prime}_{d} ,u^{\prime\prime}_{1} , \ldots ,u^{\prime\prime}} \right) \in S$. Then for every $\lambda \in \left[ {0,1} \right]$ the following relationship is established:

$$ \begin{aligned} & \lambda v^{\prime}_{i} + (1 - \lambda )v^{\prime\prime}_{i} \ge 0,\quad i = 1, \ldots ,m \\ & \lambda w^{\prime}_{d} + (1 - \lambda )w^{\prime\prime}_{d} \ge 0,\quad d = 1, \ldots ,D \\ & \lambda u^{\prime}_{r} + (1 - \lambda )u^{\prime\prime}_{r} \ge 0,\quad r = 1, \ldots ,s. \\ \end{aligned} $$

Thus, we have

$$ \begin{aligned} \sum\limits_{i = 1}^{m} {[\lambda v^{\prime}_{i} + (1 - \lambda )v^{\prime\prime}_{i} ]x_{ij} } & = 1 \\ & = \lambda \sum\limits_{i = 1}^{m} {v^{\prime}_{i} x_{ij} } + (1 - \lambda )\sum\limits_{i = 1}^{m} {v^{\prime\prime}_{i} x_{ij} } \\ & = \lambda + (1 - \lambda ) = 1 \\ \end{aligned} $$

$$ \begin{aligned} \sum\limits_{d = 1}^{D} {\left[ {\lambda w^{\prime}_{d} + (1 - \lambda )w^{\prime\prime}_{d} } \right]} z_{dj} & = \lambda \sum\limits_{d = 1}^{D} {w^{\prime}_{d} z_{dj} + (1 - \lambda )\sum\limits_{d = 1}^{D} {w^{\prime\prime}_{d} z_{dj} } } \\ & \quad \le \lambda \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } + (1 - \lambda )\sum\limits_{i = 1}^{m} {v_{i} x_{ij} } \\ & \quad \le \sum\limits_{i = 1}^{m} {[\lambda v_{i} + (1 - \lambda )v_{i} ]x_{ij} } \\ \end{aligned} $$

$$ \begin{aligned} \sum\limits_{r = 1}^{s} {\left[ {\lambda u^{\prime}_{r} + (1 - \lambda )u^{\prime\prime}_{r} } \right]} y_{rj} & = \lambda \sum\limits_{r = 1}^{s} {u^{\prime}_{r} y_{rj} + (1 - \lambda )\sum\limits_{r = 1}^{s} {u^{\prime\prime}_{r} y_{rj} } } \\ & \quad \le \lambda \sum\limits_{d = 1}^{D} {w^{\prime}_{d} z_{dj} + (1 - \lambda )\sum\limits_{d = 1}^{D} {w^{\prime\prime}_{d} z_{dj} } } \\ & \quad \le \sum\limits_{d = 1}^{D} {\left[ {\lambda w^{\prime}_{d} + (1 - \lambda )w^{\prime\prime}_{d} } \right]} z_{dj} \\ \end{aligned} $$

$$ \begin{aligned} \lambda v^{\prime}_{{i^{a} }} + (1 - \lambda )v^{\prime\prime}_{{i^{a} }} - \frac{{p_{{i^{a} o}} }}{{p_{{i^{b} o}} }}(\lambda v^{\prime}_{{i^{b} }} + (1 - \lambda )v^{\prime\prime}_{{i^{b} }} ) & = \lambda (v^{\prime}_{{i^{a} }} - \frac{{p_{{i^{a} o}} }}{{p_{{i^{b} o}} }}v^{\prime}_{{i^{b} }} ) + (1 - \lambda )(v^{\prime\prime}_{{i^{a} }} - \frac{{p_{{i^{a} o}} }}{{p_{{i^{b} o}} }}v^{\prime\prime}_{{i^{b} }} ) \\ & = 0 \\ \end{aligned} $$

The last constraint is similarly convex.

Based on this:

$$ \begin{aligned} & \lambda w^{\prime}_{d} + (1 - \lambda )w^{\prime\prime}_{d} \in S,\quad d = 1, \ldots ,D \\ & \lambda u^{\prime}_{r} + (1 - \lambda )u^{\prime\prime}_{r} \in S,\quad r = 1, \ldots ,s \\ & \lambda v^{\prime}_{i} + (1 - \lambda )v^{\prime\prime}_{i} \in S,\quad i = 1, \ldots ,m. \\ \end{aligned} $$

Hence

$$ \lambda \left( {v^{\prime}_{1} , \ldots ,v^{\prime}_{m} ,w^{\prime}_{1} , \ldots ,w^{\prime}_{d} ,u^{\prime}_{1} , \ldots ,u^{\prime}_{s} } \right) + (1 - \lambda )\left( {v^{\prime\prime}_{1} , \ldots ,v^{\prime\prime}_{m} ,w^{\prime\prime}_{1} , \ldots ,w^{\prime\prime}_{d} ,u^{\prime\prime}_{1} , \ldots ,u^{\prime\prime}_{s} } \right) \in S. $$

Therefore, it can be concluded that the set $S$ is convex.