Data envelopment analysis with embedded inputs and outputs

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.

1. The origins of the CRS and VRS models can also be found in the earlier economics literature—see Farrell (1957), Afriat (1972), Shephard (1974), Färe and Lovell (1978) and Färe et al. (1983).

2. For different rules of thumb concerning the acceptable number of inputs and outputs, see, for example, Cooper et al. (2007) and Peyrache et al. (2020).

3. The exact meaning of the term “redundant” is described in the mathematical results obtained in Sect. 5. In simple words, redundant information about the embedded inputs and outputs does not affect the evaluation of efficiency of the DMUs. However, such information may still affect the feasible region of the corresponding linear programs.

4. For the assessment of efficiency of DMU B, condition (8d) becomes \(-15\theta +20 -5 \le 0\), which is rearranged as \(\theta \ge 1\), and condition (9d) becomes \(-15+20\eta -5\eta \le 0\), which is rearranged as \(\eta \le 1\).

5. It may still be interesting to compare the input and output-oriented models based on the BSD technology incorporating conditions (20) with the corresponding HD models. As noted, for conditions (20), the BSD model is equivalent to the standard VRS model. Mehdiloo and Podinovski (2019) show that, in line with the theoretical embedding (18), the latter model and, therefore, its equivalent BSD analogue, provide better discrimination on efficiency than the HD model.

6. We can also note that measuring the efficiency of schools with respect to output \(y_2\) is equivalent to the nonradial evaluation of efficiency in the direction of combined vector \(\left( \textbf{g}_{x},\textbf{g}_{y} \right) \) whose only nonzero component is the second component of vector \(\textbf{g}_{y}\) corresponding to output \(y_2\). By Theorem 5, the constraint (20a) is redundant in such model.

7. In our computations, efficient schools (having efficiency equal to 1) in the VRS and BSD models are the same. Theoretically, a DMU may be efficient in the BSD model but inefficient in the VRS model, which is based on the larger technology. For example, DMU L in Fig. 5 is efficient in the direction \(y_2\) in the BSD technology but is inefficient in the VRS technology. Note that DMU L satisfies inequality (20b) as equality, which places L on the boundary of the BSD technology. A similar situation does not occur in our experiments because, according to the DGP employed, output \(y_2\) of every observed school does not exceed \(90\%\) of its output \(y_1\).

Authors and Affiliations

1. Department of Mathematics and Applications, University of Mohaghegh Ardabili, Ardabil, Iran

   Mahmood Mehdiloo

2. Loughborough Business School, Loughborough University, Loughborough, Leicestershire, LE11 3TU, UK

   Victor V. Podinovski

Corresponding author

Correspondence to Mahmood Mehdiloo.

Conflict of interest

The authors have no conflicting interests.

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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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

(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,

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

also satisfies this condition. Therefore, we have

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):

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

\(\square \)

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

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,

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

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. \(\square \)

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