Efficiency measurement for general network systems: a slacks-based measure model

Abstract

Traditional radial DEA models treat DMUs as black boxes, whose internal structures are ignored, and measure input and output changes proportionally. On the other hand, network DEA models consider the internal structure of a DMU as a network system where sub-processes are connected by intermediate products aside from consuming main inputs and producing final outputs. The interest in network DEA has increased over the years producing different types of formulations including the network slacks-based measure DEA (NSBM). NSBM is a non-radial network approach suitable for measuring efficiencies when inputs and outputs may change non-proportionally. However, NSBM models and some of its recent developments failed to adhere to three important efficiency measurement issues: (a) able to properly define overall and divisional efficiencies, (b) able to identify Pareto-Koopmans efficiency status, and (c) able to show equivalence of primal (multiplier) and dual (envelopment) forms of the network DEA model. This study addressed these issues based on the NSBM approach and compared the results of the proposed network SBM model with other existing models in the literature. A numerical study on 10 electric power companies further illustrated the use and significance of the proposed network SBM model in efficiency measurement.

Appendix A: Proofs of Theorems

Theorem 1. To prove Theorem 1, we show that every feasible solution in NSBM model should also be feasible in the proposed model. Of the three sets of constraints in both models, the first and second sets of constraints are the same. The difference in the feasible region of both models lies on the third set of constraints. Model (1–5) uses equality sign in the intermediate products constraints while model (8–12) uses inequality sign (≥). Clearly, the intermediate product constraints (11) are a relaxation of intermediate product constraints (4). Therefore, any feasible solution in model (1–5) should also be feasible in model (8–12). ■

Theorem 2. If DMU₀ is sub-vector efficient, then $\rho _0^ \ast = 1$, i.e., $\lambda _0^{k \ast } = 1,s_{i0}^{k - \ast } = 0$ and $s_{r0}^{k + \ast } = 0$ ∀k, i, r. To prove by contradiction, suppose DMU_a completely dominates DMU₀ where both DMUs are in {1, …, n}. It implies ${\boldsymbol{x}}_a^k \le {\boldsymbol{x}}_0^k$ and ${\boldsymbol{y}}_a^k \ge {\boldsymbol{y}}_0^k$. Consequently, the vector (λ^k*, s^k−*, s^k+*) is an optimal solution to model (8–12) where $\lambda _a^{k \ast } = 1$ and $\lambda _j^{k \ast } = 0$ for j ≠ a as well as $s_{ia}^{k - \ast } = s_{ra}^{k + \ast } = 0$ ∀k, i, r. By constraints (9) and (10), we have ${\boldsymbol{x}}_a^k = {\boldsymbol{x}}_0^k$ and ${\boldsymbol{y}}_a^k = {\boldsymbol{y}}_0^k$. This indicates that DMU₀ is also sub-vector efficient. Therefore, there does not exist any DMU that sub-vector dominates DMU₀ which is sub-vector efficient. ■

Theorem 3. We build the proof for Theorem 3 from Theorem 2. Given that the difference between sub-vector efficiency (Theorem 2) and Pareto-Koopmans efficiency (Theorem 3) is the latter’s consideration of intermediate products in the evaluation of efficiency, then the proof of Theorem 2 holds true and can be extended. It follows from equations (4.14) and (4.15) that ${\sum \nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L}} {\boldsymbol{z}}_a^{\left( {k,g} \right)} = {\sum \nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L}} {\boldsymbol{z}}_0^{\left( {k,g} \right)}$ (∀g) for the above optimal solution of model (8–12). Moreover, we have

$${\sum \limits_{h = 1}^K} {{\sum \limits_{g:\left( {k,g} \right){\it{\epsilon }}L}} {{\boldsymbol{z}}_a^{\left( {k,g} \right)}} } = {\sum \limits_{h = 1}^K} {{\sum \limits_{g:\left( {k,g} \right){\it{\epsilon }}L}} {{\boldsymbol{z}}_0^{\left( {k,g} \right)}} }$$

(A1)

Since DMU_a fully dominates DMU₀, then we have $\mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_a^{\left( {g,k} \right)} \le \mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_0^{\left( {g,k} \right)}$ (∀k). Provided that ${\boldsymbol{x}}_a^k = {\boldsymbol{x}}_0^k$ and ${\boldsymbol{y}}_a^k = {\boldsymbol{y}}_0^k$ and $\mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_a^{\left( {k,g} \right)} \le \mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_0^{\left( {k,g} \right)}$ for at least a pair of indices in d and k, we have a strict inequality constraint, i.e., there exist indices d and k such that $\mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_a^{\left( {g,k} \right)} < \mathop {\sum }\nolimits_{g:\left( {k,g} \right){\it{\epsilon }}L} {\boldsymbol{z}}_0^{\left( {g,k} \right)}$. It implies that

$$\mathop {\sum}\limits_{h = 1}^K {\mathop {\sum}\limits_{g:\left( {k,g} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_a^{\left( {g,k} \right)}} } < \mathop {\sum}\limits_{k = 1}^K {\mathop {\sum}\limits_{g:\left( {g,k} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_0^{\left( {g,k} \right)}} } .$$

(A2)

On the other hand, we have

$$\mathop {\sum}\limits_{k = 1}^K {\mathop {\sum}\limits_{g:\left( {g,k} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_a^{\left( {g,k} \right)}} } = \mathop {\sum}\limits_{k = 1}^K {\mathop {\sum}\limits_{g:\left( {k,g} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_a^{\left( {k,g} \right)}} } \,{\mathrm{and}}$$

(A3)

$$\mathop {\sum}\limits_{k = 1}^K {\mathop {\sum}\limits_{g:\left( {g,k} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_0^{\left( {g,k} \right)}} } = \mathop {\sum}\limits_{k = 1}^K {\mathop {\sum}\limits_{g:\left( {k,g} \right){\it{\epsilon }}L} {{\boldsymbol{z}}_0^{\left( {k,g} \right)}} } ,$$

(A4)

which are the sums of all intermediate products linked among all divisions of DMU_a and DMU₀, respectively. Then, we arrived with the desired contradiction. Therefore, there does not exist any DMU that fully dominates DMU₀ that is Pareto-Koopmans efficient. ■

Theorem 4. The proof for Theorem 4 is similar to that of Theorem 3 except that the point of comparison is at the division level and hence we omit it. ■