Finding the strong efficient frontier and strong defining hyperplanes of production possibility set using multiple objective linear programming

Abstract

Data envelopment analysis (DEA) models use the frontier of the production possibility set (PPS) to evaluate decision making units (DMUs). However, the explicit-form equations of the frontier cannot be obtained using the traditional DEA models. To fill this gap, the current paper proposes an algorithm to generate all strong-efficient DMUs and the explicit-form equations of the strong-efficient frontier and the strong defining hyperplanes for the PPS with the variable returns to scale (VRS) technology. The algorithm is based on a multiple objective linear programming (MOLP) problem in the DEA methodology, which is solved through the multicriteria simplex method. Also, Isermann’s test is employed to specify strong-efficient nonbasic variables in each strong-efficient multicriteria simplex table. Before presenting the algorithm, a theoretical framework is introduced to characterize the relationships between the feasible region in the decision space of the MOLP problem and the PPS with the VRS technology. It is shown that the algorithm which has four phases is finitely convergent and has less computational complexity than other algorithms in the related literature. Finally, two examples are used to illustrate the algorithm.

Appendices

Appendix 1

1.1 Proof of Theorem 2

Suppose $\chi ^*$ is a Pareto optimal solution for the problem (9). It is proven that $(x^*,y^*)$ is an efficient unit in $T_V$. By contradiction, suppose $(x^*,y^*)$ is not efficient. Hence, there exists $({\widehat{x}},{\widehat{y}})\in T_V$ such that $(-{\widehat{x}},{\widehat{y}})\ge (-{x}^*,{y}^*)$. Because $({\widehat{x}},{\widehat{y}})\in T_V$, there exist ${\widehat{\lambda }}\geqslant 0$ and slack variables ${\widehat{s}}^-\geqslant 0$ and ${\widehat{s}}^+\geqslant 0$ such that

$$\begin{aligned} \begin{array}{rlll} &&\displaystyle \sum _{j=1}^n{\widehat{\lambda }}_{j}{\overline{x}}_{i j}+{\widehat{s}}_{i}^-={\widehat{x}}_{i},&i=1,\ldots ,m,\\ &&\displaystyle \sum _{j=1}^n{\widehat{\lambda }}_{j}{\overline{y}}_{r j}-{\widehat{s}}_{r}^+={\widehat{y}}_{r},&r=1,\ldots ,s,\\ &&\displaystyle \sum _{j=1}^n{\widehat{\lambda }}_{j}=1. \end{array} \end{aligned}$$

(26)

Constraints (26) imply that ${\widehat{\chi }}=({\widehat{\lambda }},{\widehat{s}}^-,{\widehat{s}}^+,{\widehat{x}},{\widehat{y}})$ is a feasible solution for (9) with $(-{\widehat{x}},{\widehat{y}})\ge (-{x}^*,{y}^*)$, which is a contradiction.

Conversely, suppose $(x^*,y^*)$ is an efficient unit in $T_V$. It is proven that $\chi ^*$ is a Pareto optimal solution for (9). Similarly, as $(x^*,y^*)\in T_V$, there are $\lambda ^*\geqslant 0$ and slack variables $s^{-*}\geqslant 0$ and $s^{+*}\geqslant 0$ with

$$\begin{aligned} \begin{array}{rlll} &&\displaystyle \sum _{j=1}^n\lambda _{j}^*{\overline{x}}_{i j}+s_{i}^{-*}=x_{i}^*,&i=1,\ldots ,m,\\ &&\displaystyle \sum _{j=1}^n\lambda _{j}^*{\overline{y}}_{r j}-s_{r}^{+*}=y_{r}^*,&r=1,\ldots ,s,\\ &&\displaystyle \sum _{j=1}^n\lambda _{j}^*=1. \end{array} \end{aligned}$$

(27)

Constraints (27) imply that ${\chi }^*$ is a feasible solution for (9). With $(x^*,y^*)$ being an efficient unit in $T_V$, there does not exist any $({\widehat{x}},{\widehat{y}})\in T_V$ where $(-{\widehat{x}},{\widehat{y}})\ge (-{x}^*,{y}^*)$. Therefore, $\chi ^*$ is a Pareto optimal solution for (9). $\square$

1.2 Proof of Theorem 3

Suppose ${\chi }^\imath$ is an extreme point in $S^\prime$. It is proven that $({\lambda }^\imath ,{x}^\imath ,{y}^\imath )$ is an extreme point in ${\overline{S}}$. With ${\chi }^\imath$ being an extreme point, for each $(\lambda ^1,s^{-1},s^{+1},x^1,y^1)$ and $(\lambda ^2,s^{-2},s^{+2},x^2,y^2)$ belonging to $S^\prime$ and $\mu \in (0,1)$, where

$$\begin{aligned} ({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath },{x}^\imath ,{y}^\imath )=\mu (\lambda ^1,s^{-1},s^{+1},x^1,y^1)+(1-\mu )(\lambda ^2,s^{-2},s^{+2},x^2,y^2), \end{aligned}$$

(28)

we have

$$\begin{aligned} ({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath },{x}^\imath ,{y}^\imath )=(\lambda ^1,s^{-1},s^{+1},x^1,y^1)=(\lambda ^2,s^{-2},s^{+2},x^2,y^2). \end{aligned}$$

(29)

Since $(\lambda ^1,s^{-1},s^{+1},x^1,y^1)$ and $(\lambda ^2,s^{-2},s^{+2},x^2,y^2)$ belong to $S^\prime$, points $(\lambda ^1,x^1,y^1)$ and $(\lambda ^2,x^2,y^2)$ are in ${\overline{S}}$. By contradiction, suppose $({\lambda }^\imath ,{x}^\imath ,{y}^\imath )$ is not an extreme point. Hence, there exist two points $(\lambda ^1,x^1,y^1)$ and $(\lambda ^2,x^2,y^2)$ belonging to ${\overline{S}}$ and $\mu \in (0,1)$ such that

$$\begin{aligned} ({\lambda }^\imath ,{x}^\imath ,{y}^\imath )=\mu (\lambda ^1,x^1,y^1)+(1-\mu )(\lambda ^2,x^2,y^2). \end{aligned}$$

(30)

(29) and (30) result in

$$\begin{aligned} ({\lambda }^\imath ,{x}^\imath ,{y}^\imath )=(\lambda ^1,x^1,y^1)=(\lambda ^2,x^2,y^2), \end{aligned}$$

(31)

which is a contradiction.

Conversely, suppose $({\lambda }^\imath ,{x}^\imath ,{y}^\imath )$ is an extreme point in ${\overline{S}}$. It is proven that ${\chi }^\imath$ is an extreme point in $S^\prime$. By contradiction, suppose ${\chi }^\imath$ is not an extreme point. Therefore, there exist distinct points $(\lambda ^1,s^{-1},s^{+1},x^1,y^1)$ and $(\lambda ^2,s^{-2},s^{+2},x^2,y^2)$ belonging to $S^\prime$ and $\mu \in (0,1)$ with

$$\begin{aligned} ({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath },{x}^\imath ,{y}^\imath )=\mu (\lambda ^1,s^{-1},s^{+1},x^1,y^1)+(1-\mu )(\lambda ^2,s^{-2},s^{+2},x^2,y^2). \end{aligned}$$

(32)

(32) implies that $({\lambda }^\imath ,{x}^\imath ,{y}^\imath )=\mu (\lambda ^1,x^1,y^1)+(1-\mu )(\lambda ^2,x^2,y^2)$ where $(\lambda ^1,x^1,y^1)$ and $(\lambda ^2,x^2,y^2)$ are two distinct points belonging to ${\overline{S}}$ and $\mu \in (0,1)$, which is a contradiction. $\Box$

1.3 Proof of Theorem 4

In each basic feasible solution $\chi ^\imath$, there are two cases for $(x^\imath ,y^\imath )\in \mathfrak {R}^{m+s}$:

• 1. Consider $(x^\imath ,y^\imath )>0$. To construct the basis B with the rank $m+s+1$, we should select only one column from the constraint coefficients matrix A associated with the variable $\lambda _{j}$ $(j\in \{1,\ldots ,n\})$. Therefore, $\displaystyle \sum \nolimits _{j=1}^n \lambda _j^\imath =1$ and $\lambda _j^\imath \geqslant 0$ $(j=1,\ldots ,n)$ result in $\lambda ^\imath =e_j$ $(j \in \{1,\ldots ,n\})$.

• 2. Consider $(x^\imath ,y^\imath )\ge 0$, in which there exists at least one zero component. Suppose $x_{i}$ and $s_i$ are the nonbasic variables. Hence, the ith constraint of the problem (9) is converted to:

  $$\begin{aligned} \lambda _1^\imath {\overline{x}}_{i1}+\lambda _2^\imath {\overline{x}}_{i2}+\ldots +\lambda _n^\imath {\overline{x}}_{i n}+0-0=0. \end{aligned}$$

  (33)

  Given this assumption, two columns of the constraint coefficient matrix A associated with the variables $\lambda _{j^\prime }$ and $\lambda _{j^{\prime \prime }}\,(j^\prime ,j^{\prime \prime }\in \{1,\ldots ,n\},\,j^\prime \ne j^{\prime \prime })$ should be selected to construct the basis B. However, in (33), $\lambda _{j^\prime }^\imath$ and $\lambda _{j^{\prime \prime }}^\imath$ cannot be simultaneously positive because they result in a contradiction for constructing the $i\hbox {th}$ row of the basis B (a zero row is appeared). Therefore, $\chi ^\imath$ is a degenerate basic feasible solution in $S^\prime$. Finally, $\displaystyle \sum \nolimits _{j=1}^n \lambda _j^\imath =1$ and $\lambda _j^\imath \geqslant 0$ $(j=1,\ldots ,n)$ lead to $\lambda ^\imath =e_j$ $(j \in \{1,\ldots ,n\})$. Similarly, this contradiction occurs for $y_{r}^\imath =0$. $\square$

1.4 Proof of Theorem 5

Theorem 4 implies that in each basic feasible solution $\chi ^\imath$ belonging to $S^\prime$, ${\lambda }^\imath$ is a unit vector $e_{j} \in \mathfrak {R}^n$ $(j \in \{1,\ldots ,n\})$ whose $j\hbox {th}$ element is 1 and the others are 0. Without loss of generality, suppose ${\lambda }^\imath =e_1 \in \mathfrak {R}^n$. Regarding the basic feasible solution $({\lambda }^\imath =e_1,{s}^{-\imath },{s}^{+\imath },{x}^\imath ,{y}^\imath )$, the constraints of the problem (9) are converted to:

$$\begin{aligned} \begin{array}{rlll} &&{\overline{x}}_{i 1}+{s}_{i}^{-\imath } - {x}_{i}^\imath =0,&i=1,\ldots ,m,\\ &&{\overline{y}}_{r 1}-{s}_{r}^{+\imath } - {y}_{r}^\imath =0,&r=1,\ldots ,s,\\ &&1=1. \end{array} \end{aligned}$$

(34)

System (34) implies that there exist two cases for ${s}_{i}^{-\imath }$ $(i=1,\ldots ,m)$:

• 1. All of the input slack variables are zero i.e. ${s}^{-\imath }=0$, and the proof is complete in this case.

• 2. There exists at least one nonzero input slack variable. Without loss of generality, suppose ${s}_1^{-\imath }> 0$, which means that ${s}_1^{-}$ is a basic variable. Hence, ${x}_1$ cannot be a basic variable, and thus, ${x}_1^\imath =0$. Then, the first constraint in (34) is converted to:

  $$\begin{aligned} \begin{array}{rlll}&\,&{\overline{x}}_{1 1}+{s}_1^{-\imath } - 0=0. \end{array} \end{aligned}$$

  (35)

  In view of the fact that ${\overline{x}}_{1 1}\geqslant 0$ and ${s}_1^{-\imath }> 0$, inconsistent is the constraint (35).

As a consequence, all the input slack variables for each basic feasible solution in $S^\prime$ are zero. $\square$

1.5 Proof of Theorem 6

Suppose $({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath }=0,{x}^\imath ,{y}^\imath )$ is a basic feasible solution in $S^\prime$. It is proven that $({x}^\imath ,{y}^\imath )$ is a DMU in $T_V$. With $({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath }=0,{x}^\imath ,{y}^\imath )$ being a basic feasible solution, Theorems 4 and 5 result in ${\lambda }^\imath =e_{j} \in \mathfrak {R}^n\,(j\in \{1,\ldots ,n\})$ and ${s}^{-\imath }=0 \in \mathfrak {R}^m$. As regards the basic feasible solution $({\lambda }^\imath =e_{j},{s}^{-\imath }=0,{s}^{+\imath }=0,{x}^\imath ,{y}^\imath )$, the constraints of the problem (9) are converted to:

$$\begin{aligned} \begin{array}{rlll} &&\displaystyle \sum _{j=1}^ne_{j}{\overline{x}}_{i j}+0={x}_{i}^\imath ,&i=1,\ldots ,m,\\ &&\displaystyle \sum _{j=1}^ne_{j}{\overline{y}}_{r j}-0={y}_{r}^\imath ,&r=1,\ldots ,s,\\ &&\displaystyle \sum _{j=1}^ne_{j}=1. \end{array} \end{aligned}$$

(36)

System (36) implies that $({x}^\imath ,{y}^\imath )$ is one of the n observed DMUs.

Conversely, suppose $({x}^\imath ,{y}^\imath )$ is a DMU in $T_V$. It is proven that $({\lambda }^\imath ,{s}^{-\imath },{s}^{+\imath }=0,{x}^\imath ,{y}^\imath )$ is a basic feasible solution in $S^\prime$. Without loss of generality, suppose $({x}^\imath ,{y}^\imath )=({\overline{x}}_1,{\overline{y}}_1)$. Therefore, there exists a feasible solution $({\lambda }^\imath =e_1,{s}^{-\imath }=0,{s}^{+\imath }=0,{x}^\imath ={\overline{x}}_1,{y}^\imath ={\overline{y}}_1)$ in $S^\prime$. Now, it is shown to be a basic solution. There exist two cases for $({\overline{x}}_1,{\overline{y}}_1)$:

• 1. All of the ${\overline{x}}_{i 1}\,(i=1,\ldots ,m)$ and ${\overline{y}}_{r 1}\,(r=1,\ldots ,s)$ are positive. Hence, the column vectors corresponding to the basic variables ${\lambda }_1 ,{x}_1 ,\ldots ,{x}_m ,{y}_1 ,\ldots ,{y}_s$ in the constraint coefficient matrix A are as follows:

  $$\begin{aligned} {B}= \begin{bmatrix} {{\overline{x}}}_{11}&-1&\ldots&0&0&\ldots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots \\ {{\overline{x}}}_{m1}&0&\ldots&-1&0&\ldots&0\\ {{\overline{y}}}_{11}&0&\ldots&0&-1&\ldots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots \\ {{\overline{y}}}_{s1}&0&\ldots&0&0&\ldots&-1\\ 1&0&\ldots&0&0&\ldots&0 \\ \end{bmatrix}. \end{aligned}$$

  (37)

  It is obvious that the set of the column vectors for the constructed basis B is linearly independent in $\mathfrak {R}^{m+s+1}$; it completes the proof of this case.

• 2. Some of the ${\overline{x}}_{i 1}\,(i \in \{1,\ldots ,m\})$ and/or ${\overline{y}}_{r 1}\,(r \in \{1,\ldots ,s\})$ are zero. Hence, with respect to the case 1, the set of the column vectors

  $$\begin{aligned} \left\{ \begin{bmatrix} {\overline{x}}_{i1}\\ {\overline{y}}_{r1}\\ 1\\ \end{bmatrix} :\, {\lambda }_1>0 \right\} \bigcup \left\{ \begin{bmatrix} -I_{.i}\\ 0\\ 0\\ \end{bmatrix} :\, {x}_{i}>0 \right\} \bigcup \left\{ \begin{bmatrix} 0\\ -I_{.r}\\ 0\\ \end{bmatrix} :\, {y}_{r}>0 \right\} \end{aligned}$$

  (38)

  is linearly independent in $\mathfrak {R}^{m+s+1}$ (Murty 1983, p. 118), and the proof is complete in this case. $\square$

1.6 Proof of Theorem 8

There exists at least one feasible solution for the DEA weighted-sum problem (13) as $\lambda _o=1$ $( o\in \{1,\ldots ,n\} )$, $\lambda _{j}=0$ $(j=1,\ldots ,n,\,j\ne o)$, $s_{i}^-=0\,(i=1,\ldots ,m)$, $s_{r}^+=0,\,(r=1,\ldots ,s)$, $x_{i}={\overline{x}}_{i o}\,(i=1,\ldots ,m)$, and $y_{r}={{\overline{y}}_{r o}\,(r=1,\ldots ,s)}$; therefore, this problem is always feasible. It is clear that there is no extreme direction $d=(d^\lambda ,d^x,d^y)$ with $d^y\ne 0$ in ${\overline{S}}$, which implies that there is no extreme direction in $S^\prime$ with $d^y\ne 0$. Consequently, given the objective function of the DEA weighted-sum problem (13), by considering any positive weighting vector $\vartheta \in \mathfrak {R}^{m+s}$, the optimal objective value never increases indefinitely. $\Box$

1.7 Proof of Theorem 9

Let $({\overline{\delta }}^*,{\overline{u}}^*)$ be the optimal solution for the DEA Isermann subproblem (14). Its dual

$$\begin{aligned} \begin{array}{rlll} &\hbox {Min}&\displaystyle \sum _{l=1}^{m+s}\vartheta _l\\ &\hbox {s.t.}&\displaystyle \sum _{l=1}^{m+s}\vartheta _l{\overline{c}}_{lt}^\imath \geqslant 0,&t\in {\widetilde{N}}^\imath /{\widetilde{P}}^\imath ,\\ &&\displaystyle \sum _{l=1}^{m+s}\vartheta _l{\overline{c}}_{l{\overline{r}}}^\imath =0,&{\overline{r}}\in {\widetilde{P}}^\imath ,\\ &&\vartheta _l\geqslant 1,&l=1,\ldots ,m+s, \end{array} \end{aligned}$$

(39)

has an optimal solution $\vartheta ^*=(\vartheta ^*_1,\ldots ,\vartheta ^*_m,\vartheta ^*_{m+1},\ldots ,\vartheta ^*_{m+s})$. By considering the obtained weighting vector $\vartheta ^*$ for the objective function of the DEA weighted-sum problem (13), the optimal solution $\chi ^\imath$ is obtained. On the other hand, in each nondegenerate pivot step for this problem, the theorems provided in Sect. 3.1 make sure that the nonbasic entering variable and the basic leaving variable are corresponding to $\lambda _{{\overline{r}}}$ and $\lambda _\tau$ $({\overline{r}},\tau \in \{1,\ldots ,n\},\,{\overline{r}}\ne \tau )$, i.e the pivot element is $a_{\tau {\overline{r}}}=1$ which lies in the constraint corresponding to $\displaystyle \sum \nolimits _{j=1}^n\lambda^\imath _{j}=1$. Then, each basic feasible solution $\chi ^\jmath$ which can be evaluated by pivoting in the ${\overline{r}}$th column $({\overline{r}}\in {\widetilde{P}}^\imath )$ is an optimal solution for the DEA weighted-sum problem (13) with $\vartheta =\vartheta ^*$. Hence, $\chi ^\jmath$ is a Pareto optimal basic feasible solution for the problem (9) which is dual feasible and adjacent to $\chi ^\imath$. $\square$

1.8 Proof of Theorem 10

With $\lambda _{{\overline{r}}}\,({\overline{r}}\in {\widetilde{P}}^\imath )$ entering the basis $\chi ^\imath$ and after performing the efficient pivot operation, the next Pareto optimal basic feasible solution $\chi ^\jmath$ is obtained. For a nondegenerate pivot, the minimum ratio test is equal to 1 $(\theta =1)$. Hence, $Z(\chi ^\jmath )=Z(\chi ^\imath )-\theta ({\overline{c}}_{:{\overline{r}}})^\imath$ leads to $Z(\chi ^\imath )-Z(\chi ^\jmath )=({\overline{c}}_{:{\overline{r}}})^\imath$ where $({\overline{c}}_{:{\overline{r}}})^\imath$ is an $m+s$ column vector in the criteria row of the efficient multicriteria simplex table for $\chi ^\imath$. From a geometrical point of view, $({\overline{c}}_{:{\overline{r}}})^\imath$ is described as the transfer vector of the efficient points $Z(\chi ^\imath )$ and $Z(\chi ^\jmath )$ in the feasible region of the criterion space for the problem (9). $\square$

1.9 Proof of Theorem 11

The efficient multicriteria simplex table corresponding to $\chi ^\imath$ is considered. For each maximal index set $( {\widetilde{P}}^\imath )$ taken by the DEA Isermann subproblem (14), the dual of the subproblem specifies the optimal positive vector $\vartheta ^*\in \mathfrak {R}^{m+s}$ which is perpendicular to all the transfer vectors $(w_{:{\overline{r}}})^\imath \,({\overline{r}}\in {\widetilde{P}}^\imath )$. Thus, regarding Theorem 10, $\vartheta ^*$ is the gradient vector of the efficient face in the feasible region of the criterion space for the problem (9) which contains $Z(\chi ^\imath )$ and all $Z(\chi ^\jmath )$ $({\overline{r}}\in {\widetilde{P}}^\imath )$. $\square$

1.10 Proof of Theorem 12

Constructing ${\widetilde{U}}^\varepsilon$ ensures the existence of some ${\widetilde{Q}}^\imath$ $(\imath \in {\widetilde{I}})$ so that ${\widetilde{U}}^\varepsilon ={\widetilde{Q}}^\imath$. This implies that the DEA Isermann subproblem (14) with ${\widetilde{P}}^\imath ={\widetilde{Q}}^\imath /{\widetilde{D}}^\imath$ has an optimal solution, and thus, the respective dual problem possesses an optimal solution $\vartheta ^\varepsilon$. Each ${\widehat{\chi }}=({\widehat{\lambda }},{\widehat{s}}^-,{\widehat{s}}^+,{\widehat{x}},{\widehat{y}}) \in F^\varepsilon$ is an optimal solution for the DEA weighted-sum problem (13) with the positive weighting vector $\vartheta ^\varepsilon$. In light of Theorem 1, ${\widehat{\chi }}$ is a Pareto optimal solution for the problem (9). Thus, for each $\varepsilon \in \{1,\dots ,{\overline{\varepsilon }}\}$, $F^\varepsilon$ is a subset of Pareto optimal solutions for the DEA MOLP problem. Theorem 2 implies that there is an efficient unit $({\widehat{x}},{\widehat{y}})$ in $T_V$ corresponding to the Pareto optimal solution ${\widehat{\chi }}$. Consequently, $T^\varepsilon$ is a subset of efficient units in the frontier of $T_V$ for each $\varepsilon \in \{1,\ldots ,{\overline{\varepsilon }}\}$. $\square$

1.11 Proof of Theorem 13

Let $\chi ^o=(\lambda ^o,s^{-o},s^{+o},x^o,y^o)$ be a Pareto optimal solution for the problem (9). Due to Theorem 1, there exists a positive weighting vector $\vartheta ^o$ such that $\chi ^o$ is an optimal solution for the DEA weighted-sum problem (13). Let ${\widetilde{I}}^o$ denote the index set of all the Pareto optimal basic feasible solutions while being dual feasible for (9) are optimal in the DEA weighted-sum problem. Then, $\chi ^o$ can be represented as

$$\begin{aligned} \chi ^o=\displaystyle \sum _{\imath \in {\widetilde{I}}^o}\alpha _\imath (\lambda ^\imath ,s^{-\imath },s^{+\imath },x^\imath ,y^\imath ),\,\,\,\,\displaystyle \sum _{\imath \in {\widetilde{I}}^o}\alpha _\imath =1,\,\,\,\,\alpha _\imath \geqslant 0,\,\,\,\,\imath \in {\widetilde{I}}^o. \end{aligned}$$

(40)

The DEA Isermann subproblem (14) has an optimal solution for each $\imath \in {\widetilde{I}}^o$ and ${\widetilde{P}}^o={\widetilde{Q}}^o /{\widetilde{D}}^\imath$ where ${\widetilde{Q}}^o=\bigcup _{\jmath \in {\widetilde{I}}^o}{\widetilde{D}}^\jmath$. The search for all maximal index sets in Phase 2 implies that some maximal index sets ${\widetilde{P}}^\imath$ have been provided in a way ${\widetilde{P}}^o\subseteq {\widetilde{P}}^\imath$. From ${\widetilde{P}}^\imath \cup {\widetilde{D}}^\imath \subseteq {\widetilde{U}}^\varepsilon$, for some $\varepsilon$, the assertion follows. Moreover, (40) results in

$$\begin{aligned} (x^o,y^o)=\displaystyle \sum _{\imath \in {\widetilde{I}}^o}\alpha _\imath (x^\imath ,y^\imath ),\,\,\,\,\displaystyle \sum _{\imath \in {\widetilde{I}}^o}\alpha _\imath =1,\,\,\,\,\alpha _\imath \geqslant 0,\,\,\,\,\imath \in {\widetilde{I}}^o. \end{aligned}$$

(41)

The above-mentioned notes and (41) imply that there exists at least one $\varepsilon \in \{1,\ldots ,{\overline{\varepsilon }}\}$ so that $(x^o,y^o)\in T^\varepsilon$. $\square$

Appendix 2