Identification of Congestion in DEA

Abstract

Productivity is a common descriptive measure for characterizing the resource-utilization performance of a production unit, or decision making unit (DMU). The challenge of improving productivity is closely related to a particular form of congestion, which reflects waste (overuse) of input resources at the production unit level. Specifically, the productivity of a production unit can be improved not only by reducing some of its inputs but also simultaneously by increasing some of its outputs, when such input congestion is present. There is thus a need first for identifying the presence of congestion, and then for developing congestion-treatment strategies to enhance productivity by reducing the input wastes and the output shortfalls associated with such congestion. Data envelopment analysis (DEA) has been considered a very effective method in evaluating input congestion. Because the assumption of strong input disposability precludes congestion, it should not be incorporated into the axiomatic modeling of the true technology involving congestion. Given this fact, we first develop a production technology in this contribution by imposing no input disposability assumption. Then we define both weak and strong forms of congestion based on this technology. Although our definitions are made initially for the output-efficient DMUs, they are well extended in the sequel for the output-inefficient DMUs. We also propose in this contribution a method for identifying congestion. The essential tool for devising this method is the technique of finding a maximal element of a non-negative polyhedral set. To our knowledge, our method is the only reliable method for precisely detecting both weak and strong forms of congestion. This method is computationally more efficient than the other congestion-identification methods developed in the literature. This is due to the fact that, unlike the others, our method involves solving a single linear program. Unlike the other methods, the proposed method also deals effectively with the presence of negative data, and with the occurrence of multiple projections for the output-inefficient DMUs. Based on our theoretical results, three computational algorithms are developed for testing the congestion of any finite-size sample of observed DMUs. The superiority of these algorithms over the other congestion-identification methods is demonstrated using four numerical examples, one of which is newly introduced in this contribution.

4.8 Appendix: Proofs

Proof of Theorem 1 Let $\left( \mathbf {u}^{*} , \mathbf {s}^{*} , \mathbf {t}^{*} , w^{*} \right)$ be an optimal solution to program (4.5). By dividing both sides of the linear constraints of program (4.5) by $w^{*}$ and assuming $\mathbf {u}^{\max } = \frac{1}{w^{*}} \left( \mathbf {s}^{*} + \mathbf {t}^{*} \right)$, we have $\mathbf {u}^{\max } \in \mathcal {P}$. If $\mathbf {u}^{\max } > \mathbf {0}_{k}$, there is nothing to prove. Otherwise, we assume without loss of generality that $\sigma \left( \mathbf {t}^{*} \right) = \left\{ 1,\ldots ,e \right\}$ ($e<k$). Because the linear constraints of program  (4.5) are homogeneous, it is straightforward to show that $t_{j}^{*}=1$ for all $j=1,\ldots ,e$. Therefore, $e = \mathbf {1}^{T}_{k} \mathbf {t}^{*}$.

By contradiction, we assume that $\mathbf {u}^{\max }$ is not a maximal element of $\mathcal {P}$. Then, by (4.3), there is a vector $\tilde{\mathbf {u}} \in \mathcal {P}$ that takes positive values in some zero components of $\mathbf {u}^{\max }$. This means that $\sum \limits _{j=e+1}^{k} {{\bar{u}}_{j}} > 0$. Let us define

$$\begin{aligned} {\tilde{s}}_{j} = \max \left\{ 0 , s_{j}^{*} + t_{j}^{*} + \tilde{u}_{j} - 1 \right\} ~\forall j, ~~{\tilde{t}}_{j} = \min \left\{ 1 , s_{j}^{*} + t_{j}^{*} + \tilde{u}_{j} \right\} ~ \forall j, ~~\tilde{\mathbf {v}} = \mathbf {v}^{*}, ~~ {\tilde{w}} = w^{*} + 1. \end{aligned}$$

Then, $\left( \tilde{\mathbf {s}} , \tilde{\mathbf {t}} , \tilde{\mathbf {v}} , \tilde{w} \right)$ is a feasible solution to program (4.5) that its corresponding objective value is greater than e. This contradicts the optimality of $\left( \mathbf {u}^{*} , \mathbf {s}^{*} , \mathbf {t}^{*} , w^{*} \right)$. Therefore, $\mathbf {u}^{\max }$ is a maximal element of $\mathcal {P}$. $\square$

Proof of Theorem 2 By contradiction, assume that technology $\mathcal {T}$ satisfying Axiom SID is congested. This means that there is an output-efficient DMU $\left( \bar{\mathbf {x}} ; \bar{\mathbf {y}} \right) \in \mathcal {T}$ that is weakly dominated by some $\left( \hat{\mathbf {x}};\hat{\mathbf {y}} \right) \in \mathsf {\mathcal {T}}$. Because technology $\mathcal {T}$ satisfies Axiom SID, we have $\left( \bar{\mathbf {x}} ; \hat{\mathbf {y}} \right) \in \mathcal {T}$, which contradicts the output efficiency of $\left( \bar{\mathbf {x}} ; \bar{\mathbf {y}} \right)$. Therefore, technology $\mathcal {T}$ is congestion-free. $\square$

Lemma 2

Denote $\mathcal {V}$ the set on the right-hand side of (4.6). Then $\mathcal {V}$ is a polyhedral set, which implies that it satisfies Axioms CT. Furthermore, $\mathcal {V}$ satisfies Axioms IO and SOD.

Proof of Lemma 2 From the projection lemma, it follows that $\mathcal {V}$ is a polyhedral set and, consequently, satisfies Axiom CT. Clearly, $\mathcal {V}$ also satisfies Axiom IO.

Let $\left( \mathbf {x};\mathbf {y} \right) \in \mathcal {V}$. Then $\left( \mathbf {x};\mathbf {y} \right)$ satisfies (4.6) with some vector $\bar{\varvec{\lambda }}$. To prove that $\mathcal {V}$ satisfies Axiom SOD, let $\mathbf {0}_{s} \le \hat{\mathbf {y}} \le \mathbf {y}$. Then $\left( \mathbf {x} ; \hat{\mathbf {y}} \right)$ satisfies (4.6) with the same vector $\bar{\varvec{\lambda }}$. Therefore, $\left( \mathbf {x} ; \hat{\mathbf {y}} \right) \in \mathcal {V}$, and $\mathcal {V}$ satisfies Axiom SOD. $\square$

Proof of Theorem 3 Let $\mathcal {V}$ be the set on the right-hand side of (4.6). By Lemma 2, the set $\mathcal {V}$ satisfies Axioms IO, CT, and SOD. Because $\mathcal {T}_\mathrm{CONG}$ is the smallest technology that satisfies Axioms IO, CT, and SOD, we have $\mathcal {T}_\mathrm{CONG} \subseteq \mathcal {V}$.

Conversely, let $\left( \mathbf {x};\mathbf {y} \right) \in \mathsf {\mathcal {V}}$. Then $\left( \mathbf {x};\mathbf {y} \right)$ satisfies (4.6) with some vector $\bar{\varvec{\lambda }}$. Because technology $\mathcal {T}_\mathrm{CONG}$ satisfies Axiom IO and CT, we have $\left( \mathbf {X} \bar{\varvec{\lambda }} ; \mathbf {Y} \bar{\varvec{\lambda }} \right) \in \mathcal {T}_\mathrm{CONG}$. Because this technology also satisfies Axiom SOD, it follows that $\left( \mathbf {x};\mathbf {y} \right) \in \mathcal {T}_\mathrm{CONG}$. Therefore, ${ \mathcal {V} \subseteq \mathcal {T}_\mathrm{CONG} }$. $\square$

Proof of Theorem 4 The fact that $\mathcal {T}_\mathrm{CONG}$ is a polyhedral set follows from the projection lemma. To prove that $\mathcal {T}_\mathrm{CONG}$ is also bounded, let $\left( \mathbf {x};\mathbf {y} \right) \in \mathcal {T}_\mathrm{CONG}$. Then $\left( \mathbf {x};\mathbf {y} \right)$ satisfies (4.6) with some vector $\bar{\varvec{\lambda }}$. By the normalizing equality $\mathbf {1}_{n}^{T} \bar{\varvec{\lambda }}=1$, it follows from the input and output constraints that $x_i \le \underset{j \in \mathcal {J}}{\max } \left\{ {x_{ij}} \right\}$ for all $i=1,\ldots ,m$, and $y_{r} \le \underset{j \in \mathcal {J}}{\max } \, \left\{ y_{rj} \right\}$ for all $r=1,\ldots ,s$. This implies that technology $\mathcal {T}_\mathrm{CONG}$ has no recession direction and is, therefore, bounded. Because any face of a polyhedral set is itself a polyhedral set, if follows that all faces of $\mathcal {T}_\mathrm{CONG}$ are polytopes. $\square$

Proof of Theorem 6 By Definition 2 and Corollary 1, it follows that $\partial _\mathrm{F} \mathcal {T}_\mathrm{VRS} \subseteq \partial _\mathrm{F} \mathcal {T}_\mathrm{CONG}$. Conversely, let $\left( \hat{\mathbf {x}} ; \hat{\mathbf {y}} \right) \in \partial _\mathrm{F} \mathcal {T}_\mathrm{CONG}$. By Corollary 1, $\left( \hat{\mathbf {x}} ; \hat{\mathbf {y}} \right)$ is in $\mathcal {T}_\mathrm{VRS}$. By contradiction, let there exist a DMU $\left( \mathbf {x}';\mathbf {y}' \right) \in \mathcal {T}_\mathrm{VRS}$ such that $\left( -\mathbf {x}' ; \mathbf {y}' \right) \gneqq \left( -\hat{\mathbf {x}} ; \hat{\mathbf {y}} \right)$. Without loss of generality, assume also that $\left( \mathbf {x}';\mathbf {y}' \right) \in \partial _\mathrm{F} \mathcal {T}_\mathrm{VRS}$. Denote $\mathcal {F}'$ as the smallest face of $\mathcal {T}_\mathrm{VRS}$ containing $\left( \mathbf {x}';\mathbf {y}' \right)$, i.e., the intersection of the collection of faces which contain $\left( \mathbf {x}';\mathbf {y}' \right)$. Then $\left( \mathbf {x}';\mathbf {y}' \right) \in {ri}\left( \mathcal {F}'\right)$ (see Proposition 1.19 in Tuy (1998)). Moreover, $\mathcal {F}'$ is a strong face of $\mathcal {T}_\mathrm{VRS}$ and is therefore a polytope (see Theorem 2 in Davtalab Olyaie et al. (2014)). Therefore, $\left( \mathbf {x}';\mathbf {y}' \right)$ can be expressed as a convex combination of extreme points of the face $\mathcal {F}'$. Because extreme points of face $\mathcal {F}'$ are extreme in $\mathcal {T}_\mathrm{VRS}$ itself (see Rockafellar (1970), page 163), $\left( \mathbf {x}';\mathbf {y}' \right)$ is a convex combination of extreme observed DMUs. By the convexity of technology $\mathcal {T}_\mathrm{CONG}$, it follows that $\left( \mathbf {x}';\mathbf {y}' \right) \in \mathcal {T}_\mathrm{CONG}$. This contradicts the full efficiency of $\left( \hat{\mathbf {x}} ; \hat{\mathbf {y}} \right)$ in technology $\mathcal {T}_\mathrm{CONG}$. Therefore, $\left( \hat{\mathbf {x}} ; \hat{\mathbf {y}} \right) \in \partial _\mathrm{F} \mathcal {T}_\mathrm{VRS}$, implying $\partial _\mathrm{F} \mathcal {T}_\mathrm{CONG} \subseteq \partial _\mathrm{F} \mathcal {T}_\mathrm{VRS}$. $\square$

Proof of Lemma 1 By contradiction, assume that $\varvec{\alpha } = \mathbf {0}_{m}$. Then we have $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} + \varvec{\beta } \right) \in \mathcal {T}_\mathrm{CONG}$. Because $\varvec{\beta } \ne \mathbf {0}_{s}$, it follows that DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is output-inefficient, which is a contradiction. Therefore $\varvec{\alpha } \ne \mathbf {0}_{m}$. $\square$

Proof of Theorem 7 Part (i) Let DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ be full-efficient in technology $\mathcal {T}_\mathrm{CONG}$. Then we have $\mathcal {S}_{o} = \left\{ \mathbf {0}_{m+s} \right\}$, because otherwise the full efficiency of DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is contradicted. Therefore, $\varvec{\alpha }^{\max }_{o} = \mathbf {0}_{m}$. Conversely, assume by contradiction that DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is not full-efficient. By definition, there is a DMU $\left( \mathbf {x}';\mathbf {y}' \right) \in \mathcal {T}$, such that $\left( -\mathbf {x}' ; \mathbf {y}' \right) \gneqq \left( -\mathbf {x} ; \mathbf {y} \right)$. Then the non-zero vector $\left( \varvec{\alpha }' ; \varvec{\beta }' \right) = \left( \mathbf {x}_{o} - \mathbf {x}' ; \mathbf {y}' - \mathbf {y}_{o} \right)$ satisfies all conditions in (4.9) with some $\varvec{\lambda }'$. Taking into account (4.3), it follows that $\left( \varvec{\alpha }^{\max }_{o} ; \varvec{\beta }^{\max }_{o} \right) \ne \mathbf {0}_{m+s}$. If $\varvec{\beta }^{\max }_{o} = \mathbf {0}_{s}$, then it is clear that $\varvec{\alpha }^{\max }_{o} \ne \mathbf {0}_{m}$. Otherwise, Lemma 1 implies that $\varvec{\alpha }^{\max }_{o} \ne \mathbf {0}_{m}$. In both cases, the assumption is contradicted. Therefore, DMU $(\mathbf {x}_{o} ; \mathbf {y}_{o} ;)$ is full-efficient.

Part (ii) Let DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ be input-inefficient in technology $\mathcal {T}_\mathrm{CONG}$. Then, by part (i) of the theorem, it follows that $\varvec{\alpha }^{\max }_{o} \ne \mathbf {0}_{m}$. Conversely, let $\varvec{\alpha }^{\max }_{o} \ne \mathbf {0}_{m}$. Because $\left( \mathbf {x}_{o} - \varvec{\alpha }^{\max }_{o} ; \mathbf {y}_{o} + \varvec{\beta }^{\max }_{o} \right) \in \mathcal {T}_\mathrm{CONG}$ we have $\left( \mathbf {x}_{o} - \varvec{\alpha }^{\max }_{o} ; \mathbf {y}_{o} \right) \in \mathcal {T}_\mathrm{CONG}$ Therefore, DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is input-inefficient.

Part (iii) Let DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ be weakly congested. By definition, there exists a DMU $\left( \mathbf {x}';\mathbf {y}' \right)$ in technology $\mathcal {T}_{\mathrm {CONG}}$, such that $-\mathbf {x}' \gneqq -\mathbf {x}$ and $\mathbf {y}' \gneqq \mathbf {y}$. The vector $\left( \varvec{\alpha }' ; \varvec{\beta }' \right) = \left( \mathbf {x}_{o} - \mathbf {x}' ; \mathbf {y}' - \mathbf {y}_{o} \right)$ satisfies all conditions in (4.9) with some $\varvec{\lambda }'$. Because $\varvec{\beta }' \ne \mathbf {0}_{s}$, (4.3) follows that $\varvec{\beta }^{\max }_{o} \ne \mathbf {0}_{s}$. Conversely, let $\varvec{\beta }^{\max }_{o} \ne \mathbf {0}_{s}$. Then Lemma 1 implies that $\varvec{\alpha }^{\max }_{o} \ne \mathbf {0}_{m}$. We also have $\left( \mathbf {x}_{o} - \varvec{\alpha }^{\max }_{o} ; \mathbf {y}_{o} + \varvec{\beta }^{\max }_{o} \right) \in \mathcal {T}_\mathrm{CONG}$. Therefore, DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is weakly congested.

Part (iv) Let DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ be strongly congested. By definition, there exists a DMU $\left( \mathbf {x}';\mathbf {y}' \right)$ in technology $\mathcal {T}_{\mathrm {CONG}}$, such that $\left( -\mathbf {x}' ; \mathbf {y}' \right) > \left( -\mathbf {x} ; \mathbf {y} \right)$. The vector $\left( \varvec{\alpha }' ; \varvec{\beta }' \right) = \left( \mathbf {x}_{o} - \mathbf {x}' ; \mathbf {y}' - \mathbf {y}_{o} \right)$ satisfies all conditions in (4.9) with some $\varvec{\lambda }'$. Because $\left( \varvec{\alpha }' ; \varvec{\beta }' \right) > \mathbf {0}_{m+s}$, (4.3) follows that $\left( \varvec{\alpha }^{\max }_{o} ; \varvec{\beta }^{\max }_{o} \right) > \mathbf {0}_{m+s}$. Conversely, let $\left( \varvec{\alpha }^{\max }_{o} ; \varvec{\beta }^{\max }_{o} \right) > \mathbf {0}_{m+s}$. Because $\left( \mathbf {x}_{o} - \varvec{\alpha }^{\max }_{o} ; \mathbf {y}_{o} + \varvec{\beta }^{\max }_{o} \right) \in \mathcal {T}_\mathrm{CONG}$, DMU $\left( \mathbf {x}_{o} ; \mathbf {y}_{o} \right)$ is strongly congested. $\square$

Proof of Theorem 8 Assume that $\left( \mathbf {x}^\mathrm{ri} ; \mathbf {y}^\mathrm{ri} \right)$ and $\varvec{\rho } \in \mathbb {R}^{l}_{++}$ are as in (4.14). For each $k=1,\ldots ,l$, let $\left( \varvec{\delta }_{k}^{*},\mathbf {s}_{k}^{-*},\mathbf {t}_{k}^{-*},\mathbf {s}_{k}^{+*},\mathbf {t}_{k}^{+*},w_{k}^{*} \right)$ be the optimal solution obtained from the evaluation of DMU $\left( \mathbf {x}_{j_k} ; \mathbf {y}_{j_k} \right)$ by model (4.10). Then, from the constraints of program (4.10) at optimality, we have the following equations:

$$\begin{aligned} \mathbf {X}\left( \frac{1}{1+w_{k}^{*}}\varvec{\delta }_{k}^{*} \right) {+}\varvec{\alpha }_{k}^{\max } {=} \mathbf {x}_{j_k}, \; \mathbf {Y}\left( \frac{1}{1+w_{k}^{*}}\varvec{\delta }_{k}^{*} \right) {-}\varvec{\beta }_{k}^{\max }{\ge } \mathbf {y}_{j_k}, \; \mathbf {1}_{n}^{T}\left( \frac{1}{1+w_{k}^{*}}\varvec{\delta }_{k}^{*} \right) {=}1, \; k=1,\ldots ,l, \end{aligned}$$

(4.30)

where the vector $\left( \varvec{\alpha }_{k}^{\max } ; \varvec{\beta }_{k}^{\max } \right)$ is as defined in (4.15). Multiplying both sides of the kth input, output, and normalizing equations in (4.30) by $\rho _{k}$, and then separately summing up the resulting input, output and normalizing equations over k lead to the following equations:

$$\begin{aligned} \mathbf {X} \varvec{\lambda }' = \mathbf {x}^\mathrm{ri} - \varvec{\alpha }', \quad \mathbf {Y} \varvec{\lambda }' \ge \mathbf {y}^\mathrm{ri} + \varvec{\beta }', \quad \mathbf {1}_{n}^{T} \varvec{\lambda }' = 1, \end{aligned}$$

(4.31)

where the vectors $\varvec{\lambda }'$, $\varvec{\alpha }'$, and $\varvec{\beta }'$ are as follows:

$$\begin{aligned} \varvec{\lambda }' = \sum \limits _{k=1}^{l}{\frac{\rho _{k}}{1+ w_k } \varvec{\delta }_{k}^{*}}, \quad \varvec{\alpha }' = \sum \limits _{k=1}^{l}{\rho _{k} \varvec{\alpha }_{k}^{\max }}, \quad \varvec{\beta }' = \sum \limits _{k=1}^{l} {\rho _{k} \varvec{\beta }_{k}^{\max }}. \end{aligned}$$

(4.32)

The equations obtained in (4.31) show that $\left( \mathbf {x}^\mathrm{ri} ; \mathbf {y}^\mathrm{ri} \right)$ satisfies all conditions in (4.10) with $\left( \varvec{\lambda }' , \varvec{\alpha }' , \varvec{\beta }' \right)$. This means that $\left( \varvec{\alpha }' , \varvec{\beta }' \right) \in \mathcal {S}_\mathrm{ri}$. Because $\varvec{\rho } \in \mathbb {R}_{++}^{l}$, it is clear from (4.16) and (4.32) that $\mathcal {I}_\mathcal {F} = \sigma \left( \varvec{\alpha }'\right)$ and $\mathcal {O}_\mathcal {F} = \sigma \left( \varvec{\beta }'\right)$. Consequently, taking into account (4.13), it follows that $\mathcal {I}_\mathcal {F} \subseteq \mathcal {I}_\mathrm {ri}$ and $\mathcal {O}_\mathcal {F} \subseteq \mathcal {O}_\mathrm {ri}$.

To prove that $\mathcal {I}_\mathrm {ri} \subseteq \mathcal {I}_\mathcal {F}$, let $\hat{i} \in \mathcal {I}_\mathrm{ri}$. Then, taking into account (4.12), it follows that some $\left( \hat{\varvec{\alpha }} ; \hat{\varvec{\beta }} \right) \in \mathcal {S}_\mathrm{ri}$ exists such that $\hat{\alpha }_{\hat{i}} > 0$. This means that $\left( \mathbf {x}^\mathrm{ri} - \hat{\varvec{\alpha }} ; \mathbf {y}^\mathrm{ri} + \hat{\varvec{\beta }}\right) \in \mathcal {T}_\mathrm{CONG}$ such that $x^\mathrm{ri}_{\hat{i}} - \hat{\alpha }_{\hat{i}} < x^\mathrm{ri}_{\hat{i}}$. From (4.14), we have $x^\mathrm{ri}_{\hat{i}} = \sum \limits _{k=1}^{l}{\rho _k x_{\hat{i} {j_k}}}$. Because $\sum \limits _{k=1}^{l}{\rho _k} = 1$, it follows that $x^\mathrm{ri}_{\hat{i}} - \hat{\alpha }_{\hat{i}} < x_{\hat{i} {j_{\hat{k}}}}$ for some $\hat{k} \in \left\{ 1,\ldots ,l\right\}$. Because $\left( \varvec{\alpha }_{\hat{k}}^{\max } ; \varvec{\beta }_{\hat{k}}^{\max } \right)$ is a maximal element of $\mathcal {S}_{j_{\hat{k}}}$, we therefore have $\hat{i} \in \sigma \left( \varvec{\alpha }_{\hat{k}}^{\max } \right)$, so $\hat{i} \in \mathcal {I}_{\mathcal {F}}$. Therefore, $\mathcal {I}_\mathrm {ri} \subseteq \mathcal {I}_\mathcal {F}$. Similarly, it can be proved that $\mathcal {O}_\mathrm {ri} \subseteq \mathcal {O}_\mathcal {F}$. $\square$

Proof of Theorem 9 Because $\mathcal {F}^{\min }_{o}$ is a face of the polyhedral set $\mathcal {T}_\mathrm{CONG}$, there is a supporting hyperplane of $\mathcal {T}_\mathrm{CONG}$, namely $\mathcal {H}^{\min }$, such that $\mathcal {F}^{\min }_{o} = \mathcal {H}^{\min } \cap \mathcal {T}_\mathrm{CONG}$.²¹ Because $\mathcal {F}^{\min }_{o}$ contains $\Pi _{o}$, the hyperplane $\mathcal {H}^{\min }$ is binding at all projections in $\Pi _{o}$, and therefore passes through all the reference DMUs in $\mathcal {G}_{o}$. By the convexity of $\mathcal {H}^{\min }$, it follows that $\mathrm {conv}\left( \mathcal {G}_{o}\right) \subseteq \mathcal {H}^{\min }$. Therefore, $\mathrm {conv}\left( \mathcal {G}_{o}\right) \subseteq \mathcal {F}^{\min }_{o}$.

As shown in Footnote 17, $\Pi _{o} \cap ri\left( \mathcal {F}^{\min }_{o} \right) \ne \emptyset$. This implies that all the observed DMUs on $\mathcal {F}^{\min }_{o}$, and therefore all the vertices of $\mathcal {F}^{\min }_{o}$, belong to $\mathcal {G}_{o}$. By Theorem 4, $\mathcal {F}^{\min }_{o}$ is a polytope. Therefore, $\mathcal {F}^{\min }_{o} \subseteq \mathrm {conv}\left( \mathcal {G}_{o}\right)$. $\square$

Proof of Theorem 10 Assume that $\left( \varphi ^{*} , \mathbf {q}^{*} , \varvec{\lambda }^{*} \right)$ is that optimal solution of program (4.7) by which $\Lambda _{o}$ has been stated as in (4.21). Let $\varvec{\lambda }^{\max }_{o}$ be a maximal element of $\Lambda _{o}$, and let $\varvec{\lambda }^{\max }_{o}$ satisfy (4.21) with some $\mathbf {q}^{\max }_{o}$. Then $\left( \varphi ^{*} , \varvec{\lambda }^{\max }_{o} , \mathbf {q}^{\max }_{o} \right)$ is an optimal solution of program (4.7). This proves that $\sigma \left( \varvec{\lambda }^{\max }_{o}\right) \subseteq \mathcal {J}_{o}$.

Conversely, let ${\hat{j}} \in \mathcal {J}_{o}$. Then there is exists an optimal solution $\left( \varphi ' , \varvec{\lambda }' , \mathbf {q}' \right)$ to program (4.7) such that ${\hat{j}} \in \sigma \left( \varvec{\lambda }'\right)$. Because $\varphi ' = \varphi ^{*}$ and $\mathbf {1}_{s}^{T} \mathbf {q}' = \mathbf {1}_{s}^{T} \mathbf {q}^{*}$, it follows that $\varvec{\lambda }' \in \Lambda _{o}$. By (4.3), we have $\sigma \left( \varvec{\lambda }'\right) \subseteq \sigma \left( \varvec{\lambda }^{\max }_{o}\right)$, and so ${\hat{j}} \in \sigma \left( \varvec{\lambda }^{\max }_{o}\right)$. Therefore, $\mathcal {J}_{o} \subseteq \sigma \left( \varvec{\lambda }^{\max }_{o}\right)$. $\square$