A Radial Framework for Estimating the Efficiency and Returns to Scale of a Multi-component Production System in DEA

Abstract

This chapter provides radial measurements of efficiency for the production process possessing multi-components under different production technologies. Our approach is based on the construction of various empirical production possibility sets. Then we propose a procedure that is unaffected affected by multiple optima for estimating returns to scale. The theoretical connections between the traditional black box and the proposed multi-component approach are established, which ascertains consistency in estimating the efficiency and returns to scale. Moreover, we introduce two homogeneity conditions, which clarify the difference between our approach and the existing one, and are important for evaluating performance in multi-component setting. Finally, an empirical study of the pollution treatment processes in China is presented, and compared to the results from black-box approach. Many insightful findings related to the operations of the pollution treatment processes in China are secured.

Appendix

Proof of Theorem 1

Before we prove theorem 1, we establish Lemma 1.

Lemma A1

Define ${\widehat{T}}_b^{VRS},{\widehat{T}}^{VRS}$ as follows:

$$ {\widehat{T}}_b^{VRS}=\left\{\left(X,Y\right)\Big|{\displaystyle \sum_{j=1}^{n^2}{\lambda}_j}{x}_{ij}={x}_i,i=1,\dots, m,\right.\left.{\displaystyle \sum_{j=1}^{n^2}{\lambda}_j}{y}_{rj}={y}_r,r=1,\dots, s,{\displaystyle \sum_{j=1}^{n^2}{\lambda}_j}=1,{\lambda}_j\ge 0\right\} $$

and

${\widehat{T}}^{VRS}=\left\{\left(X,Y\right)\Big|{\displaystyle \sum_{k=1}^2{\displaystyle \sum_{j=1}^n{\lambda}_j^k{x}_{ij}^k}}={x}_i,i=1,\dots, m,{\displaystyle \sum_{k=1}^2{\displaystyle \sum_{j=1}^n{\lambda}_j^k}}{y}_{rj}^k={y}_r,r=1,\dots, s,{\displaystyle \sum_{j=1}^n{\lambda}_j^k}=1,{\lambda}_j^k\ge 0\right\}$. Then ${\widehat{T}}_b^{VRS}={\widehat{T}}^{VRS}$.

Proof

(1) ${\widehat{T}}_b^{VRS}\subseteq {\widehat{T}}^{VRS}$;

Let DMU_j be some DMU in EDS, and (x _1j , …, x _mj , y _1j , …, y _rj ) be its input–output bundle. Suppose it is made of SDMU_1k, and SDMU_2m, where $k,m\in \left\{1,\dots, n\right\}$. Obviously, $\left({x}_{1j},\dots, {x}_{mj},{y}_{1j},\dots, {y}_{rj}\right)\in {\widehat{T}}_b^{VRS}$, since it can be decomposed into input–output bundle of SDMU_1k, and that of SDMU_2m. To put it another way, if we set a multiplier corresponding to SDMU_1k and SDMU_2m equal to 1 and other multipliers equal to zero, we can see that (x _1j , …, x _mj , y _1j , …, y _rj ) satisfies the condition to be an element of ${\widehat{T}}^{VRS}$. Therefore ${\widehat{T}}_b^{VRS}\subseteq {\widehat{T}}^{VRS}$ holds.

(2) ${\widehat{T}}_b^{VRS}\supseteq {\widehat{T}}^{VRS}$;

For any $\left(X,Y\right)\in {\widehat{T}}^{VRS}$, there exist two sets of convex multipliers (λ ¹₁ , …, λ ¹ _n ) and $\left({\lambda}_1^2,\dots, {\lambda}_n^2\right)\left({\lambda}_j^1,{\lambda}_j^2\ge 0,{\displaystyle \sum_{j=1}^n{\lambda}_j^1}=1,{\displaystyle \sum_{j=1}^n{\lambda}_j^2}=1\right)$ such that

$$ \begin{array}{l}{x}_i={\displaystyle \sum_{j=1}^n{\lambda}_j^1}{x}_{ij}^1+{\displaystyle \sum_{j=1}^n{\lambda}_j^2}{x}_{ij}^2\left(i=1,\dots, m\right),\\ {}{y}_r={\displaystyle \sum_{j=1}^n{\lambda}_j^1}{y}_{rj}^1+{\displaystyle \sum_{j=1}^n{\lambda}_j^2}{y}_{rj}^2\left(r=1,\dots, s\right).\end{array} $$

(14.19)

We need to show that there always exists a convex multiplier ${\displaystyle {\sum}_{j=1}^{n^2}{\lambda}_j}=1,{\lambda}_j\ge 0$, such that ${x}_i={\displaystyle {\sum}_{j=1}^{n^2}{\lambda}_j{x}_{ij}},{y}_r={\displaystyle {\sum}_{j=1}^{n^2}{\lambda}_j{y}_{rj}}$, where (x _1j , …, x _mj , y _1j , …, y _rj ) is the input–output bundle of DMU_j in EDS. In other words, there is a convex multiplier such that the following equations hold:

$$ \begin{array}{l}{x}_i={\displaystyle \sum_{j=1}^n{\lambda}_j}\left({x}_{i1}^1+{x}_{ij}^2\right)+{\displaystyle \sum_{j=n+1}^{2n}{\lambda}_j}\left({x}_{i2}^1+{x}_{i\left(j-n\right)}^2\right)+,\dots, +{\displaystyle \sum_{j={n}^2-n+1}^{n^2}{\lambda}_j}\left({x}_{in}^1+{x}_{i\left(j-{n}^2-n\right)}^2\right)\\ {}{y}_r={\displaystyle \sum_{j=1}^n{\lambda}_j}\left({y}_{r1}^1+{y}_{rj}^2\right)+{\displaystyle \sum_{j=n+1}^{2n}{\lambda}_j}\left({y}_{r2}^1+{y}_{r\left(j-n\right)}^2\right)+,\dots, +{\displaystyle \sum_{j={n}^2-n+1}^{n^2}{\lambda}_j}\left({y}_{rn}^1+{y}_{r\left(j-{n}^2-n\right)}^2\right)\end{array} $$

(14.20)

where (x ¹_1j , …, x ¹ _mj , y ¹_1j , …, y ¹ _sj ) and (x ²_1j , …, x ² _mj , y ²_1j , …, y ² _sj ), $j=1,\dots, n$, are the respective input bundle and output bundle of SDMU_1j, and SDMU_2j. That is to say, ${\displaystyle \sum_{j=1}^{n^2}{\lambda}_j}=1,{\lambda}_j\ge 0$ must satisfy the following conditions:

$$ {\lambda}_j^1={\displaystyle \sum_{k=\left(j-1\right)n+1}^{\left(j-1\right)n+n}{\lambda}_k},\ {\lambda}_j^2={\displaystyle \sum_{k=1}^n{\lambda}_{n\left(j-1\right)+k}},\ j=1,\dots, n $$

(14.21)

To facilitate understanding, we organize the conditions as matrix products.

$$ \left[\begin{array}{l}{\lambda}_1\kern0.45em {\lambda}_{n+1}\kern0.45em \dots \kern0.45em {\lambda}_{n^2-n+1}\\ {}{\lambda}_2\kern0.45em {\lambda}_{n+2}\kern0.45em \dots \kern0.45em {\lambda}_{n^2-n}\\ {} \dots \dots \dots \kern0.45em \dots \\ {}{\lambda}_n\kern0.45em {\lambda}_{n+n}\kern0.45em \dots \kern0.45em {\lambda}_{n^2}\end{array}\right]\left[\begin{array}{l}1\\ {}1\\ {}\dots \\ {}1\end{array}\right]=\left[\begin{array}{l}{\lambda}_1^2\\ {}{\lambda}_2^2\\ {}\dots \\ {}{\lambda}_n^2\end{array}\right] $$

(14.22)

$$ {\left[\begin{array}{l}{\lambda}_1\kern0.75em {\lambda}_{n+1}\kern0.75em \dots \kern0.75em {\lambda}_{n^2-n+1}\\ {}{\lambda}_2\kern0.75em {\lambda}_{n+2}\kern0.75em \dots \kern0.75em {\lambda}_{n^2-n}\\ {} \dots \dots \kern1em \dots \kern1em \dots \\ {}{\lambda}_n\kern0.75em {\lambda}_{n+n}\kern0.75em \dots \kern0.75em {\lambda}_{n^2}\end{array}\right]}^T\left[\begin{array}{l}1\\ {}1\\ {}\dots \\ {}1\end{array}\right]=\left[\begin{array}{l}{\lambda}_1^1\\ {}{\lambda}_2^1\\ {}\dots \\ {}{\lambda}_n^1\end{array}\right] $$

(14.23)

The above illustration indicates that the row j of the matrix is summed to λ ² _j , and the column j the matrix is summed to λ ¹ _j . Let us now combine (14.22) and (14.23) into the following equations where A is 2n by n ².

$$ \mathbf{A}\lambda =\left[\begin{array}{l}\overset{n}{\overbrace{11,\dots, 1}}\ \overset{n}{\overbrace{00,\dots, 0}}\ \overset{n}{\overbrace{00,\dots, 0}} \dots \overset{n}{\overbrace{00,\dots, 0}}\\ {}00,\dots, 0\ 11,\dots, 1\kern0.75em 00,\dots, 0 \dots 00,\dots, 0\\ {}\kern1.5em \dots \kern2em \dots \kern2.5em \dots \kern1.75em \dots \kern2em \dots \\ {}00,\dots, 0\ 00,\dots, 0\kern0.75em 00,\dots, 0 \dots 11,\dots, 1\\ {}10,\dots, 0\ 10,\dots, 0\kern0.5em 10,\dots, 0 \dots 10,\dots, 0\\ {}01,\dots, 0\ 01,\dots, 0\kern0.5em 01,\dots, 0\kern0.5em \dots \kern0.5em 01,\dots, 0\\ {}\kern1.25em \dots \kern2em \dots \kern2.5em \dots \kern1.75em \dots \kern2em \dots \\ {}00,\dots, 1\ 00,\dots, 1\kern0.5em 00,\dots, 1\kern0.75em \dots \kern0.5em 00,\dots, 1\end{array}\right]\left[\begin{array}{l}{\lambda}_1\\ {}{\lambda}_2\\ {}\dots \\ {}{\lambda}_{n^2}\end{array}\right]=\left[\begin{array}{l}{\lambda}_1^1\\ {}{\lambda}_2^1\\ {}\dots \\ {}{\lambda}_n^1\\ {}{\lambda}_1^2\\ {}{\lambda}_2^2\\ {}\dots \\ {}{\lambda}_n^2\end{array}\right]=\boldsymbol{\Gamma} $$

(14.24)

We are going to prove (14.24) always has a nonnegative solution ${\lambda}_1^{*},\dots, {\lambda}_{n^2}^{*}$. Note that ${\displaystyle \sum_{j=1}^{n^2}{\lambda}_j^{*}}=1$ automatically holds provided ${\displaystyle \sum_{j=1}^n{\lambda}_j^1}=1$ and ${\displaystyle \sum_{j=1}^n{\lambda}_j^2}=1$. Our problem reduces to the existence of nonnegative solution to (14.24). We claim the nonnegative solution always exists, by way of contradiction. Before moving on, we reduce (14.24) to (14.25).

$$ \overline{\mathbf{A}}\lambda =\left[\begin{array}{l}\overset{n}{\overbrace{00,\dots, 0}}\ \overset{n}{\overbrace{11,\dots, 1}}\ \overset{n}{\overbrace{00,\dots, 0}} \dots \overset{n}{\overbrace{00,\dots, 0}}\\ {}00,\dots, 0\ 00,\dots, 0\kern0.5em 11,\dots, 1 \dots 00,\dots, 0\\ {}\kern1em \dots \kern2em \dots \kern2.5em \dots \kern1.75em \dots \kern2em \dots \\ {}00,\dots, 0\ 00,\dots, 0\kern0.5em 00,\dots, 0 \dots 11,\dots, 1\\ {}10,\dots, 0\ 10,\dots, 0\kern0.5em 10,\dots, 0 \dots 10,\dots, 0\\ {}01,\dots, 0\ 01,\dots, 0\kern0.5em 01,\dots, 0\kern0.5em \dots \kern0.5em 01,\dots, 0\\ {}\kern1.25em \dots \kern2em \dots \kern2.5em \dots \kern1.75em \dots \kern2em \dots \\ {}00,\dots, 1\ 00,\dots, 1\kern0.5em 00,\dots, 1\kern0.75em \dots \kern0.5em 00,\dots, 1\end{array}\right]\left[\begin{array}{l}{\lambda}_1\\ {}{\lambda}_2\\ {}\dots \\ {}{\lambda}_{n^2}\end{array}\right]=\left[\begin{array}{l}{\lambda}_2^1\\ {}{\lambda}_3^1\\ {}\dots \\ {}{\lambda}_n^1\\ {}{\lambda}_1^2\\ {}{\lambda}_2^2\\ {}\dots \\ {}{\lambda}_n^2\end{array}\right]=\overline{\boldsymbol{\Gamma}} $$

(14.25)

Note that we have eliminated the first row of A and the first element of Γ by elementary row operation. Assume, now, that $\overline{A}\lambda =\overline{\Gamma}$ doesn’t have a nonnegative solution, i.e., $\overline{\boldsymbol{\Gamma}}$ doesn’t belong to the conic hull constructed by the column vectors of Ā. By Farkas lemma, there exists $\mathbf{x}\in {R}^{2n-1}$, such that

• (1) ${\mathbf{x}}^{\mathbf{T}}\overline{\Gamma}>0$;

• (2) ${\mathbf{x}}^{\mathbf{T}}\overline{\mathbf{A}}(i)\le 0,\overline{\mathbf{A}}(i)$ denotes the i th column of Ā, $i=1,\dots, {n}^2$.

By (2), it follows that

• (1) $\mathbf{x}(i)\le 0,\ i=n,\dots, 2n-1,\ \Big(\mathbf{x}(i)$ denotes the ith component of vector x);

• (2) For any $k=1,\dots, n-1$, we have $x(k)+x(i)\le 0,\ i=n,\dots, 2n-1$, i.e., $x(k)\le \underset{j=n,\ldots 2n-1}{ \min }-x(j).$

Combining the previous two conditions, we obtain

$$ \begin{array}{l}{x}^T\overline{\Gamma}={\displaystyle \sum_{k=1}^{n-1}x}(k){\lambda}_{k+1}^1+{\displaystyle \sum_{j=n}^{2n-1}x}(j){\lambda}_j^2\le \left(\underset{j=n,\dots, 2n-1}{ \min }-x(j)\right){\displaystyle \sum_{k=1}^{n-1}{\lambda}_{k+1}^1}+{\displaystyle \sum_{j=n}^{2n-1}x}(j){\lambda}_j^2\\ {}=\left(-\underset{j=n,\dots, 2n-1}{ \max }x(j)\right){\displaystyle \sum_{k=1}^{n-1}{\lambda}_{k+1}^1}+{\displaystyle \sum_{j=n}^{2n-1}x}(j){\lambda}_j^2\\ {}\le \left(-\underset{j=n,\dots, 2n-1}{ \max }x(j)\right){\displaystyle \sum_{k=1}^{n-1}{\lambda}_{k+1}^1}+\underset{j=n,\dots, 2n-1}{ \max }x(j)\\ {}=\left(\underset{j=n,\dots, 2n-1}{ \max }x(j)\right)\left(1-{\displaystyle \sum_{k=1}^{n-1}{\lambda}_{k+1}^1}\right)\le 0\end{array} $$

(14.26)

To see why the last relation holds, note that ${\displaystyle {\sum}_{j=1}^n{\lambda}_j^1}=1$ and $\mathbf{x}(i)\le 0,\ i=n,\dots, 2n-1$. So it follows that $1-{\displaystyle {\sum}_{k=1}^{n-1}{\lambda}_{k+1}^1}={\lambda}_1^1\ge 0$, and $\underset{j=n,\dots, 2n-1}{ \max }x(j)\le 0$. Therefore, the product of the two parts is less than or equal to zero.

This contradicts ${\mathbf{x}}^T\overline{\boldsymbol{\Gamma}}>0$. Therefore, $\overline{\boldsymbol{\Gamma}}$ belongs to the conic hull constructed by the column vectors of Ā, i.e., there is $\lambda =\left({\lambda}_1,{\lambda}_2,\dots, {\lambda}_{n^2}\right)\ge 0$ such that $\overline{\mathbf{A}}\lambda =\overline{\boldsymbol{\Gamma}}$, which also means that $\mathbf{A}\lambda =\boldsymbol{\Gamma}$. By our construction, we know that there exists $\lambda =\left({\lambda}_1,{\lambda}_2,\dots, {\lambda}_{n^2}\right)\ge 0$ such that (14.22) and (14.23) hold. In turn, this establishes that $\left(X,Y\right)\in {\widehat{T}}_b^{VRS}$. □

Proof of Theorem 1

Let (x _1j , …, x _mj , y _1j , …, y _rj ) be an arbitrary point in T ^VRS _b . We first prove that ${T}_b^{VRS}\subseteq {T}^{VRS}$. By definition, there exists one point $\left({\overline{x}}_{1j},\dots, {\overline{x}}_{mj},{\overline{y}}_{1j},\dots, {\overline{y}}_{rj}\right)$ in ${\widehat{T}}_b^{VRS}$ such that ${x}_{ij}\ge {\overline{x}}_{ij}$ and ${y}_{rj}\le {\overline{y}}_{rj}$. In light of Lemma 1, $\left({\overline{x}}_{1j},\dots, {\overline{x}}_{mj},{\overline{y}}_{1j},\dots, {\overline{y}}_{rj}\right)$ also belongs to ${\widehat{T}}^{VRS}$. Therefore $\left({x}_{1j},\dots, {x}_{mj},{y}_{1j},\dots, {y}_{rj}\right)\in {T}^{VRS}$, since there is a point in T ^VRS such that ${x}_{ij}\ge {\overline{x}}_{ij}$ and ${y}_{rj}\le {\overline{y}}_{rj}$ hold. By analogy, we can prove ${T}_b^{VRS}\supseteq {T}^{VRS}$. Therefore, ${T}_b^{VRS}={T}^{VRS}$ holds.

By substituting the convex condition in the definition of T ^VRS and T ^VRS _b for ${\displaystyle {\sum}_{j=1}^n{\lambda}_j^k}=t\ \left(k=1,2\right)$ and ${\displaystyle {\sum}_{j=1}^{n^2}{\lambda}_j}=t\ \left(t\ge 0\right)$ respectively, it follows that ${T}^{VRS}(t)={T}_b^{VRS}(t)$, since they are obtained by scaling up or down T ^VRS and T ^VRS _b by the same factor t. Given the fact that ${T}_b^{CRS}={\displaystyle \underset{t\in \left[0,\infty \right)}{\cup }{T}_b^{VRS}(t)}$, ${T}_b^{NIRS}={\displaystyle \underset{t\in \left[0,1\right]}{\cup }{T}_b^{VRS}(t)}$, and ${T}^{CRS}={\displaystyle \underset{t\in \left[0,\infty \right)}{\cup }{T}^{VRS}(t)},\ {T}^{NIRS}={\displaystyle \underset{t\in \left[0,1\right]}{\cup }{T}^{VRS}(t)}$, it follows ${T}_b^{CRS}={T}^{CRS}$ and ${T}_b^{NIRS}={T}^{NIRS}$. □