Data envelopment analysis with embedded inputs and outputs

Abstract

Applications of data envelopment analysis (DEA) often include inputs and outputs that are embedded in some other inputs or outputs. For example, in a school assessment, the sets of students achieving good academic results or students with special needs are subsets of the set of all students. In a hospital application, the set of specific or successful treatments is a subset of all treatments. Similarly, in many applications, labour costs are a part of overall costs. Conventional variable and constant returns-to-scale DEA models cannot incorporate such information. Using such standard DEA models may potentially lead to a situation in which, in the resulting projection of an inefficient decision making unit, the value of an input or output representing the whole set is less than the value of an input or output representing its subset, which is physically impossible. In this paper, we demonstrate how the information about embedded inputs and outputs can be incorporated in the DEA models. We further identify common scenarios in which such information is redundant and makes no difference to the efficiency assessment and scenarios in which such information needs to be incorporated in order to keep the efficient projections consistent with the identified embeddings.

Appendix A: Proofs

Proof of Proposition 1

Denote $\mathcal {L}$ the finite set of DMUs $\left( \textbf{x}^l, \textbf{y}^l \right)$, $l=1,\dots ,L$. Let $\textrm{conv} \, \mathcal {L}$ denote the convex hull of the set $\mathcal {L}$. For the convex set $\mathcal {P}$ defined by (5), we have $\textrm{conv} \, \mathcal {P} = \mathcal {P}$. Because $\mathcal {L} \subset \mathcal {P}$, we have $\textrm{conv} \, \mathcal {L} \subseteq \textrm{conv} \, \mathcal {P} = \mathcal {P}$. $\square$

Lemma 1

Denote $\mathcal {K}$ the set on the right-hand side of (4). Then $\mathcal {K}$ satisfies Axioms IO, CT and BSD.

Proof of Lemma 1

Clearly, $\mathcal {K}$ satisfies Axiom IO. Because $\mathcal {K} = \mathcal {T}_{\textrm{VRS}} \cap \mathcal {P}$, where $\mathcal {P}$ is as defined by (5), and both sets $\mathcal {T}_{\textrm{VRS}}$ and $\mathcal {P}$ are convex, the set $\mathcal {K}$ is convex. Therefore, $\mathcal {K}$ satisfies Axiom CT. To prove that $\mathcal {K}$ satisfies Axiom BSD, let $\left( \textbf{x}, \textbf{y} \right) \in \mathcal {K}$. Then $\left( \textbf{x}, \textbf{y} \right)$ satisfies (4) with some vector $\bar{\varvec{\lambda }}$. The DMU $\left( \tilde{\textbf{x}}, \tilde{\textbf{y}} \right)$ in the statement of Axiom BSD satisfies (4) with the same vector $\bar{\varvec{\lambda }}$ and is in $\mathcal {K}$. Therefore, $\mathcal {K}$ satisfies Axiom BSD. $\square$

Proof of Theorem 1

Let $\mathcal {K}$ be the set on the right-hand side of (4). By Lemma 1, $\mathcal {K}$ satisfies Axioms IO, CT and BSD. By Definition 1, we have $\mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}} \subseteq \mathcal {K}$. Conversely, let $\left( \textbf{x}, \textbf{y} \right) \in \mathcal {K}$. Then, $\left( \textbf{x}, \textbf{y} \right)$ satisfies (4) with some vector $\bar{\varvec{\lambda }}$. We need to prove that $\left( \textbf{x}, \textbf{y} \right) \in \mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$. Because $\mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$ satisfies Axioms IO and CT, we have $\bigr ( \sum _{j \in \mathcal {J}} {\bar{\lambda }_{j} \textbf{x}_{j}}, \sum _{j \in \mathcal {J}} {\bar{\lambda }_{j} \textbf{y}_{j}} \bigr ) \in \mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$. Because $\mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$ also satisfies Axiom BSD, $\left( \textbf{x}, \textbf{y} \right) \in \mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$. Therefore, $\mathcal {K} \subseteq \mathcal {T}^{\textrm{BSD}}_{\textrm{VRS}}$. $\square$

Proof of Theorem 2

Consider program (8). The proof for program (9) is similar and is not given. Let $\varvec{\lambda }$ and $\theta \le 1$ be any feasible solution of program (8) from which constraint (8d) for $k=k'$ is removed. We need to prove that this constraint is also satisfied. Consider the three cases identified in the statement of Theorem 2.

(i) In this case, for all $\theta \le 1$, we have $\theta \textbf{a}^T_{k'} \textbf{x}_o \le \textbf{a}^T_{k'} \textbf{x}_o$. Because $(\textbf{x}_{o},\textbf{y}_{o})$ satisfies condition (2) for $k=k'$, we have

$$\begin{aligned} \theta \textbf{a}^T_{k'} \textbf{x}_o + \textbf{b}^T_{k'} \textbf{y}_o \le \textbf{a}^T_{k'} \textbf{x}_o + \textbf{b}^T_{k'} \textbf{y}_o \le 0. \end{aligned}$$

(ii) Because $\textbf{b}_{k'} \le \textbf{0}$ and $\textbf{y}_o \ge \textbf{0}$, we have $\textbf{b}^T_{k'} \textbf{y}_o \le 0$. By (8b), we have $\theta \ge 0$. If $\theta = 0$, condition $k'$ in (8d) becomes $\textbf{b}^T_{k'} \textbf{y}_o \le 0$ and, as shown, is satisfied. Let $\theta > 0$. Then, for all $\theta \le 1$, we have $(1 / \theta ) \textbf{b}^T_{k'} \textbf{y}_o \le \textbf{b}^T_{k'} \textbf{y}_o \le 0$. Therefore,

$$\begin{aligned} \begin{aligned} \theta \textbf{a}^T_{k'} \textbf{x}_o + \textbf{b}^T_{k'} \textbf{y}_o&= \theta \left( \textbf{a}^T_{k'} \textbf{x}_o + (1 / \theta ) \textbf{b}^T_{k'} \textbf{y}_o \right) \\&\le \theta \left( \textbf{a}^T_{k'} \textbf{x}_o + \textbf{b}^T_{k'} \textbf{y}_o \right) \le 0. \end{aligned} \end{aligned}$$

(iii) Because all observed DMUs satisfy (2) for $k=k'$, by Proposition 1, their convex combination

$$\begin{aligned} (\textbf{x}', \textbf{y}') = \biggl ( \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{x}_{j}, \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{x}_{j} \biggr ) \end{aligned}$$

also satisfies this condition. Therefore, we have

$$\begin{aligned} \textbf{a}^T_{k'} \textbf{x}' + \textbf{b}^T_{k'} \textbf{y}' \le 0. \end{aligned}$$

(22)

Because $\textbf{a}_{k'} \le \textbf{0}$ and $\textbf{b}_{k'} \ge \textbf{0}$, inequalities (8b) and (8c) imply (note the change of sign in the first inequality):

$$\begin{aligned} \begin{aligned} \textbf{a}^T_{k'} \textbf{x}'&= \textbf{a}^T_{k'} \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{x}_{j} \ge \theta \textbf{a}^T_{k'} \textbf{x}_{o}, \\ \textbf{b}^T_{k'} \textbf{y}'&= \textbf{b}^T_{k'} \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{y}_{j} \ge \textbf{b}^T_{k'} \textbf{y}_{o}. \end{aligned} \end{aligned}$$

(23)

Adding the two inequalities in (23) and noting (22), we have

$$\begin{aligned} \theta \textbf{a}^T_{k'} \textbf{x}_o + \textbf{b}^T_{k'} \textbf{y}_o \le \textbf{a}^T_{k'} \textbf{x}' + \textbf{b}^T_{k'} \textbf{y}' \le 0. \end{aligned}$$

$\square$

Proof of Corollary 3

Let $k' \in \left\{ 1,\dots ,K\right\}$ be such that the combined vector $(\textbf{a}_{k'},\textbf{b}_{k'})$ has exactly two nonzero components. If either $\textbf{a}_{k'}$ or $\textbf{b}_{k'}$ is a zero vector, the statement follows from Corollary 2. Otherwise, let ${a}_{k'i'} \ne 0$ and ${b}_{k'r'} \ne 0$, where $i' \in \mathcal {I}$ and $r' \in \mathcal {O}$. If ${a}_{k'i'} >0$ or ${b}_{k'r'} < 0$, the statement follows from Cases (i) and (ii) of Theorem 2. If not, the only possibility is ${a}_{k'i'} <0$ and ${b}_{k'r'} >0$, which is Case (iii) of Theorem 2. $\square$

Proof of Theorem 3

If $\varvec{\lambda }$, $\textbf{s}^{-}$ and $\textbf{s}^{+}$ is a feasible solution of program (13), then the same vectors and the vectors $\varvec{\zeta }^{-} = \textbf{0}$ and $\varvec{\zeta }^{+} = \textbf{0}$ are feasible in program (12). Therefore, $\sigma ^{\textrm{BSD}}_{o} \ge {\hat{\sigma }}^{\textrm{BSD}}_{o}$.

Conversely, consider any feasible solution $\tilde{\varvec{\lambda }}$, $\tilde{\textbf{s}}^{-}$, $\tilde{\textbf{s}}^{+}$, $\tilde{\varvec{\zeta }}^{-}$, $\tilde{\varvec{\zeta }}^{+}$ of program (12). Define $\hat{\varvec{\lambda }} = \tilde{\varvec{\lambda }}$, $\hat{\textbf{s}}^{-} = \tilde{\textbf{s}}^{-}+\tilde{\varvec{\zeta }}^{-}$ and $\hat{\textbf{s}}^{+} = \tilde{\textbf{s}}^{+}+\tilde{\varvec{\zeta }}^{+}$. Then, using conditions (12b) and (12c), we have

$$\begin{aligned} \left( \textbf{x}_o - \hat{\textbf{s}}^{-},\textbf{y}_{o} + \hat{\textbf{s}}^{+} \right) = \left( \textbf{x}_o - \tilde{\textbf{s}}^{-} - \tilde{\varvec{\zeta }}^{-}, \textbf{y}_{o} + \tilde{\textbf{s}}^{+}+\tilde{\varvec{\zeta }}^+ \right) =\biggl ( \sum \limits _{j \in \mathcal {J}} {\tilde{\lambda }_j} \textbf{x}_{j}, \sum \limits _{j \in \mathcal {J}} {\tilde{\lambda }_j} \textbf{y}_{j} \biggr ). \end{aligned}$$

(24)

Equality (24) shows that the vectors $\hat{\varvec{\lambda }}$, $\hat{\textbf{s}}^{-}$ and $\hat{\textbf{s}}^{+}$ satisfy conditions (13b) and (13c). By Proposition 1, equality (24) also implies that the vectors $\hat{\textbf{s}}^{-}$ and $\hat{\textbf{s}}^{+}$ satisfy condition (13d). Therefore, $\hat{\varvec{\lambda }}$, $\hat{\textbf{s}}^{-}$ and $\hat{\textbf{s}}^{+}$ is a feasible solution of program (13), and $\textbf{1}^T \tilde{\textbf{s}}^{-} + \textbf{1}^T \tilde{\textbf{s}}^{+} \le \textbf{1}^T \hat{\textbf{s}}^{-} + \textbf{1}^T \hat{\textbf{s}}^{+} \le {\hat{\sigma }}^{\textrm{BSD}}_{o}$. Because the feasible solution $\tilde{\varvec{\lambda }}$, $\tilde{\textbf{s}}^{-}$, $\tilde{\textbf{s}}^{+}$, $\tilde{\varvec{\zeta }}^{-}$, $\tilde{\varvec{\zeta }}^{+}$ of program (12) is arbitrary, we have $\sigma ^{\textrm{BSD}}_{o} \le {\hat{\sigma }}^{\textrm{BSD}}_{o}$. Taking into account the opposite inequality proved above, we have $\sigma ^{\textrm{BSD}}_{o} = {\hat{\sigma }}^{\textrm{BSD}}_{o}$. $\square$

Proof of Theorem 4

Consider any vectors $\varvec{\lambda }$, $\textbf{s}^{-}$ and $\textbf{s}^{+}$ that satisfy conditions (13b), (13c) and (13e). From the first two conditions,

$$\begin{aligned} \left( \textbf{x}_o - \textbf{s}^{-},\textbf{y}_{o} + \textbf{s}^{+} \right) = \biggl ( \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{x}_{j}, \sum \limits _{j \in \mathcal {J}} {\lambda _j} \textbf{y}_{j} \biggr ). \end{aligned}$$

By Proposition 1, vectors $\textbf{s}^{-}$ and $\textbf{s}^{+}$ satisfy constraints (13d). $\square$

Proof of Theorem 5

Under the assumed conditions, we have $\textbf{a}^{T}_{k'} \textbf{g}_{x} \ge 0$ and $\textbf{b}^{T}_{k'} \textbf{g}_{y} \le 0$. Because $(\textbf{x}_{o},\textbf{y}_{o}) \in {\mathcal {T}}^{\textrm{BSD}}_{\textrm{VRS}}$, we have $\textbf{a}^{T}_{k'} \textbf{x}_o + \textbf{b}^{T}_{k'} \textbf{y}_o \le 0$. Then, for all $\beta \ge 0$, we have

$$\begin{aligned} \textbf{a}^{T}_{k'} \left( \textbf{x}_o - \beta \textbf{g}_{x}\right) + \textbf{b}^{T}_{k'} \left( \textbf{y}_o + \beta \textbf{g}_{y} \right) \le \textbf{a}^{T}_{k'} \textbf{x}_o + \textbf{b}^{T}_{k'} \textbf{y}_o \le 0. \end{aligned}$$

Therefore, the constraint (14d) for $k=k'$ is satisfied and is redundant. $\square$

Proof of Theorem 6

Any DMU $\left( \textbf{x}, \textbf{y} \right) \in \mathcal {T}^\textrm{HD}_{\textrm{VRS}}$ satisfies (17) with some vectors $\bar{\varvec{\lambda }}$, $\bar{\varvec{\rho }}$ and $\bar{\varvec{\sigma }}$. Then it also satisfies (16) with the same vector $\bar{\varvec{\lambda }}$. Therefore, $\left( \textbf{x}, \textbf{y} \right) \in \tilde{\mathcal {T}}^{\textrm{BSD}}_{\textrm{VRS}}$. The second part of embedding (18) is established by (7). $\square$

Proof of Theorem 7

The proof follows from the fact that, if DMU $\left( \theta \textbf{x}_o, \textbf{y}_o \right)$ satisfies the constraints of program (8) with some vector $\varvec{\lambda }$, except the constraint $\textbf{1}^{T} \varvec{\lambda } =1$, then $\left( \textbf{x}_o, \eta \textbf{y}_o \right)$ satisfies the constraints of program (9) with $\eta =1 / \theta$ and vector $\varvec{\lambda }' = \varvec{\lambda } / \theta$ (also without the constraint $\textbf{1}^{T} \varvec{\lambda } =1$), and vice versa. $\square$