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\newtheorem{th}{Theorem}[section]
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\title{\bf Some Optimization Problems on Solubility Sets of
Separable Max-Min Equations and Inequalities}
\author{{\bf Asem Tharwat} \and {\bf Karel Zimmermann}\thanks{Department
of Operation Research, Faculty of Mathematics and Physics,
Charles University, Malostransk\'{e} n\'{a}m.~25, 118~00~Praha~1,
Czech Republic.
 Supported by GA €R 201/95/1484}}

\date{}



\begin{document}

\maketitle

\section{Introduction}

The aim of this paper is to suggest a direct Parametric method
for solving some optimization problems on attainable sets of so
called max-separable operators. Such problems in a less general
form connected with the fuzzy set theory were considered e.g. in
\cite{1}, \cite{4}. The problem considered in this paper is
presented independently of the fuzzy sets context as a non-linear
nonconvex optimization problem. Parametic approach to its
solution suggested is flexible enough to allow further extension
and generalization, which are briefly discussed in the concluding
sections.

\section{Notations and Formulation of the Basic Problem}

In this paper, we shall consider the following system of
equations and inequalities

\beqn
& & R_i(x) \equiv \max_{j \in N} (a_{ij} \wedge r_{ij} (x_j)) = b_i,
\hspace{0.5cm} \forall i:i\in S \nonumber \\
& & h_j \leq x_j \leq H_j, \hspace{0.5cm} \forall j: j \in N
\label{1}
\eeqn

\noindent where $N \equiv \{ 1, 2, \dots n \}$, $S \equiv \{ 1, 2, \dots
m \}$,  $x \equiv ( x_1, \dots x_n )^T \in R^n$,  $b \equiv (b_1,
\dots b_m)^T\!\!\!~\in~\!\!\!R^m$,  $h \equiv (h_1, \dots h_n)^T \in R^n$,
$H \equiv (H_1, \dots H_n)^T \in R^n$, $a_{ij} \wedge r_{ij}
(x_j) \equiv \min (a_{ij}, r_{ij} (x_j))$, $R(x) = (R_1(x) ,
\dots, R_m (x))^T$, let us assume further that $r_{ij}: R
\rightarrow R$ are given strictly increasing continuous functions
$\forall i : i \in S$, $\forall j : j \in N$. Using the above
vector notation we can reformulate the system(\ref{1}) as
follows:

\be
R(x) = b , \hspace{1cm} h \leq x \leq H
\label{2}
\ee

Denote the set of all solutions of the system (\ref{1}) (or
(\ref{2})) by $M(b)$.
Each component of $R: E^n \rightarrow
E^m$ is a function depending on $n$ variables; this
function is expressed as the maximum
$n$ nondecreasing functions of one variable of the form
$a_{ij} \wedge r_{ij} (x_j)$, so that these functions are
separated by a max-operation. By similarity with the additive
separability, we call this property of the functions $R_i(x)$
max-separablity and $R(x)$ is called a max-separable operator.

The vector $b$ in the system (\ref{2}) can be understood as a
vector, which can be attained by an appropriate choice of $x \in
M(b)$. Therefore those $b$'s, for which $M(b) \neq \emptyset$,
are called attainable elements and the set

\be
A \equiv \{ b / M(b) \neq \emptyset \}
\label{3}
\ee

\noindent is called the attainable set.

If an element $\hat{b} \in A$, then there exists a solution of the
system (\ref{2}) with $b = \hat{b}$ which can be obtained using
some of the methods described in the literature (see e.g.
\cite{2} \cite{3}). If $\hat{b} \not\in A$, we want to find an
approximate solution of the system (\ref{2}) with the right hand
side $\hat{b}$. For this purpose, we look for an element $b^{opt}
\in A$, which has in some sense the minimal distance from
$\hat{b}$ and accept the elements of $M(b^{opt})$ as appropriate
approximate solutions.

In this article, we shall use the Tshebyshev distance, i.e. the
following distance:

\be
\parallel b - \hat{b} \parallel \equiv \max_{i \in S} | b_i -
\hat{b}_i |
\label{4}
\ee

\noindent The problem, we are going to solve here is thus in the
following form:

\be
\parallel b - \hat{b} \parallel \rightarrow \min \hspace{0.5cm}
{\rm subject \; to} \hspace{0.5cm} b \in A
\label{5}
\ee

\noindent Since if $b \in A$, it means that there exists $x$ such
that $b = R(x)$ so that we can reformulate the problem (\ref{5})
as follows:

\be
\parallel R(x) - \hat{b} \parallel \equiv \max_{i \in S} |R_i(x)
- \hat{b}_i| \rightarrow \min \hspace{0.5cm} {\rm subject \; to}
\hspace{0.5cm} h \leq x \leq H
\label{6}
\ee

\noindent The reformulation (\ref{6}) shows that we minimize a
continuous function of $x$ on a compact set, so that there exists
always at least one optimal solution $x^{opt}$ of the system
(\ref{6}); thus if we set $b^{opt} \equiv R (x^{opt})$, we will
obtain an optimal solution of the problem (\ref{5}).

Let us define  the set $M(t)$ for any $t$: $t \in [0, \infty)$ as
follows:

\be
M(t) \equiv \{ x / h \leq x \leq H \; \& \; \parallel R(x) -
\hat{b} \parallel \leq t \}.
\label{7}
\ee

\noindent The set $M(t)$ is nonempty if and only if the following
system of inequalities:

\be
R_i(x) \leq \hat{b}_i + t, \; \; i \in S \; \; \& \; \; R_i(x)
\geq \hat{b}_i - t, \; \; i \in S \; \; \& \; \; h \leq x \leq H,
\label{8}
\ee

\noindent is soluble with respect to $x$; note that the set
$M(t)$ is the set of all solutions $x$ of (\ref{8}). We can
replace our original problems (\ref{5}), (\ref{6}) by the
following problem:

\be
t \rightarrow \min \hspace{0.5cm} {\rm subject \; to}
\hspace{0.5cm} M(t) \neq \emptyset
\label{9}
\ee

We shall show in the sequel that there  exists always the optimal
solution $t^{opt} \geq 0$ of the problem(\ref{9}) and also we
will derive a direct numerical procedure for determining
$t^{opt}$.
If $x^{opt}$ is an any element of $M(t^{opt})$, then $\parallel
R(x^{opt}) - \hat{b} \parallel \leq t^{opt}$; since the strict
inequality can't hold, the equality must occur i.e. $b^{opt}
\equiv R(x^{opt}) \in A$ is the optimal solution of the problem
(\ref{5}), and the vector $x^{opt}$ can be accepted as an
approximate solution of the system (\ref{2}) in the case that
$\hat{b} \not\in A$, since for any solution $x$ of the system
(\ref{2}), we have $\parallel R(x) - \hat{b} \parallel \geq
t^{opt}$.

In the next section we investigate some properties of the set
$M(t)$ where $t \in [0, \infty)$. This will enable us to derive
the direct solution method for the system (\ref{5}).

\section{Properties of $M(t)$}

We shall introduce the following notations $\forall i : i \in S$,
$\forall j : i \in N$, $t \in [0, \infty)$:

{\small
\beq
V_{ij}(t) & = & \{ x_j / h_j \leq x_j \leq H_j \hspace{0.5cm} {\rm
and} \hspace{0.5cm} a_{ij} \wedge r_{ij} (x_j) \leq \hat{b}_i + t
\} \\
V_j (t) & = & \bigcap_{i \in S} V_{ij} (t) \; \; \& \; \;
W_{ij}(t) = \{ x_j / h_j \leq x_j \leq H_j \hspace{0.5cm} {\rm
and} \hspace{0.5cm} a_{ij} \wedge r_{ij} (x_j) \geq \hat{b}_i - t
\}
\eeq
}

\noindent For the illustration of these sets see the Appendix.
The following theorem gives the necessary and sufficient
conditions for $M(t) \neq \emptyset$.

\begin{th}
$$M(t) \neq \emptyset  \Leftrightarrow \left\{ \begin{array}{l}
1) \; V_j (t) \neq
\emptyset, \; \; \forall j \in N \\
2) \forall i \in S \; \exists j(i) \in N \; \; {\rm such \;
that} \; \; W_{ij(i)} (t) \cap V_{j(i)}(t) \neq \emptyset.
\end{array} \right\} $$
\label{t31}
\end{th}

\noindent {\bf Proof:} Define the interval $I_j = [h_j, H_j]$ and
introduce the following notations:

\beq
& & a1_{ij} = a_{ij} - (\hat{b}_i + t), \; \; a2_{ij} = a_{ij} -
(\hat{b}_i - t) \; \; \& \; \; r1_{ij} (x_j) = r_{ij} (x_j) -
(\hat{b}_i + t), \\
& &  r2_{ij} (x_j) = r_{ij} (x_j) -
(\hat{b}_i - t)
\eeq

\noindent {\sl Sufficiency:} \\
Assume that we have a point $x \equiv (x_1, \dots , x_n)$
which satisfies the R.H.S. of the $\Leftrightarrow$-relation.

\beq
x_j \in V_j(t), \; \; \forall j \in N & \rightarrow & x_j \in
\bigcap_{i \in S} V_{ij} (t), \; \; \forall j  \in N \\
& \rightarrow & (a1_{ij} \wedge r1_{ij} (x_j) \leq 0,
\; \; \forall i  \in S; x_j \in I_j), \; \; \forall j  \in N \\
& \rightarrow & \max_{j \in N} (a1_{ij} \wedge r1_{ij} (x_j) \leq
0), \; \; \forall i  \in S;  \; \; \forall x_j  \in I_j, \\
x_{j(i)} \in W_{ij(i)}(t) & \rightarrow & a2_{ij(i)} \wedge
r2_{ij(i)} (x_j) \geq 0; \; \;  x_{j(i)}  \in I_{j(i)} \\
& \rightarrow & \max_{j \in N} (a2_{ij} \wedge
r2_{ij} (x_j) \geq 0; \; \;  x_{j(i)}  \in I_{j(i)}.
\eeq

\noindent Then we can deduce that $x \in M(t)$.

\vspace{0.5cm}

\noindent {\sl Necessity:}
{\small
\beq
x \in M(t) & \rightarrow & x_j \in I_j, \; \; \forall j \in N \;
\; \& \; \; (\max_{j \in N} a1_{ij} \wedge r1_{ij} (x_j) \leq 0
\; \; \forall i  \in S) \\
& & \hspace{3.8cm} (\max_{j \in N} a2_{ij} \wedge r2_{ij} (x_j)
\geq 0 \; \; \forall i  \in S) \\
& \rightarrow & (x_j \in I_j, \; \; {\rm and} \; \; a1_{ij}
\wedge r1_{ij} (x_j) \leq 0 \; \; \forall i  \in S), \forall j
\in N \\
& & \& \forall i  \in S \;\; \exists j(i) \in N \; \; {\rm such \;
that} \; \;  a2_{ij(i)} \wedge r2_{ij(i)} (x_j) \geq 0; \; \;
x_{j(i)} \in I_{j(i)} \\
& & \\
& \rightarrow & 1) \; V_j(t) \neq \emptyset, \; \; \forall j \in
N \\
& & 2) \;  \forall i  \in S \;\; \exists j(i) \in N \; \; {\rm such \;
that} \; \;  W_{ij(i)} (t) \cap V_{j(i)}(t) \neq \emptyset.
\eeq
}

\noindent Thus the proof of the theorem is complete. \hfill\
$\Box$

\section{Properties of $V(t)$  \& $W(t)$}

We shall investigate here the conditions for $V_{ij}(t) \neq
\emptyset$, $V_j(t) \neq \emptyset$, $W_{ij}(t) \neq \emptyset$.
Define the following variables:

\beqn
\eta^{ij} & \equiv & \min (\max \{a_{ij} - \hat{b}_i, 0\}, \;
\max \{r_{ij} (h_j) - \hat{b}_i, 0\}), \hspace{0.5cm} \eta^j
\equiv \max_{i \in S} \eta^{ij}, \nonumber \\
\eta & \equiv & \max_{j \in N} \eta^j \hspace{0.5cm} {\rm and}
\hspace{0.5cm} \tau^{ij} \equiv \max (0, \hat{b}_i - a_{ij}, \hat{b}_i
- r_{ij} (H_j)).
\label{10}
\eeqn


\noindent For the illustration of these variables see the
appendix.

\begin{th}
For each $j \in N$, $\exists \eta^{ij} \geq 0$ such that
$V_{ij}(t) \neq \emptyset \Leftrightarrow t \geq \eta^{ij}$.
\label{t41}
\end{th}

\newpage

\noindent {\bf Proof:}
{\small
\beq
V_{ij}(t) = \emptyset & \Leftrightarrow & a_{ij} \wedge r_{ij}
(x_j) > \hat{b}_i + t; \hspace{0.5cm} \forall x_j \in I_j \\
& \Leftrightarrow & \hat{b}_i + t < r_{ij} (x_j) < a_{ij} \; \;
{\rm or} \; \; \hat{b}_i + t < a_{ij}  < r_{ij} (x_j); \; \;
{\rm it \; is \; further} \; \; t \geq 0 \\
& \Leftrightarrow & t < \min (\max \{ r_{ij} (h_j) - \hat{b}_i, 0
\}, \max \{ a_{ij} - \hat{b}_i, 0 \}) = \eta^{ij}
\eeq
}

\noindent where $\eta^{ij}$ is given by (\ref{10}); this
completes the proof of the theorem. \hfill\ $\Box$

\begin{th}
For each $j \in N$, $V_j (t) \neq \emptyset \Leftrightarrow
t \geq \eta^j$; where $\eta^j$ is given by (\ref{10}).
\label{t42}
\end{th}

\noindent {\bf Proof:}
$V_j(t) = \emptyset$ equivalent to the fact $V_{i_0j}(t) = \emptyset$
for some $i_0 \in S$; since $V_{ij_0}$ are nested\footnote{Since
for each $j$, $1 \leq j \leq n$, there exists a permutation $\{i,
\dots , i_m\}$ such that \\
$V_{i_1j} \subset V_{i_2j} \subset \dots
\subset V_{i_mj}$ because of the fact that $a_{ij}\wedge r_{ij}
(x_j)$ are nondecreasing in $x_j$.} sets  {\em for
fixed} $j_0 \in N$; which means that $t < \eta^{i_0j} \leq
\eta^j$ which is the maximum of $\eta^{ij}$ on $S$. \hfill\
$\Box$

\begin{cl}
For each $j \in N$, $V_j (t) \neq \emptyset \Leftrightarrow
t \geq \eta$; where $\eta$ is given by (\ref{10}).
\label{c41}
\end{cl}

\noindent {\bf Proof:} The proof is obviously derived from
theorem~\ref{t42}.

\begin{cl}
$M(t) \neq \emptyset \Leftrightarrow
t \geq \eta$
\label{c42}
\end{cl}

\noindent {\bf Proof:} The proof is obviously derived from
theorem~\ref{t31}, theorem~\ref{t42} and corollary~\ref{c41};
where $\eta$ {\em is given by} (\ref{10}).


\begin{th}
For each $i \in S$, $j \in N$; $\exists \tau^{ij}$ such that
$W_{ij} (t) \neq \emptyset \Leftrightarrow
t \geq \tau^{ij}$.
\label{t43}
\end{th}

\noindent {\bf Proof:}
{\small
\beq
W_{ij}(t) = \emptyset & \Leftrightarrow & a_{ij} \wedge r_{ij}
(x_j) < \hat{b}_i - t; \hspace{0.5cm} \forall x_j \in I_j \\
& \Leftrightarrow & a_{ij} < r_{ij} (x_j) < \hat{b}_i - t  \; \;
{\rm or} \; \; r_{ij}(x_j) < a_{ij} < \hat{b}_i - t; \; \;
{\rm it \; is \; further} \; \; t \geq 0 \\
& \Leftrightarrow & t < -\min (0, a_{ij} - \hat{b}_i, r_{ij}
(H_j) - \hat{b}_i) \\
& \Leftrightarrow & t < \max (0, \hat{b}_i - a_{ij},
\hat{b}_i) r_{ij} (H_j)) = \tau^{ij}
\eeq
}

\noindent where $\tau^{ij}$ is given by (\ref{10}); this
completes the proof of the theorem. \hfill\ $\Box$

\begin{cl}
For each $j \in N$, $i \in S$; $\exists \tau^{ij} \geq 0$
such that \\
$\forall t : t < \max\limits_{i \in S} \min\limits_{j \in N}
\max\limits_{i
\in S} (\eta, \tau^{ij}) \Rightarrow M(t) \neq \emptyset$
\label{c43}
\end{cl}

\noindent {\bf Proof:} It is clear from
corollary~\ref{c41},~\ref{c42} and theorem~\ref{t31},
where $\tau^{ij}$ is given by~(\ref{10})

\vspace{0.3cm}

Let us define the following sets: $P_{ikj} (t) = W_{ij}(t) \cap
V_{kj} (t)$; $\forall i,k \in S$, $j \in N$. To investigate the
necessary and sufficient conditions for $P_{ikj}(t) \neq
\emptyset$, assume that the variable $\eta_{ikj}$ satisfies the
following equation $r_{kj}^{-1} (\hat{b}_k + \eta_{ikj}) =
r_{ij}^{-1} (\hat{b}_i - \eta_{ikj})$ for some $i,k \in S$, $j
\in N$ and define the variables $\xi_{ikj}$, $\zeta_{ikj}$ and
$\gamma_{ikj}$ for some $i,k \in S$ and $j \in N$ as follows:

{\small
\beqn
\xi_{ikj} & = & \left\{ \begin{array}{lll}
\eta_{ikj} & {\rm if} & \max (\tau^{ij}, \eta^{ij}) < \eta_{ikj}
< \min (a_{kj} - \hat{b}_k, \hat{b}_i - r_{ij} (h_j)) \\
a_{kj} - \hat{b}_k & {\rm if} & r_{kj}^{-1} (\hat{b}_k +t) \leq
r_{ij}^{-1} (\hat{b}_i - t) < H_j \\
\max (\tau^{ij}, \eta^{ij}) & & {\rm otherwise}
\end{array} \right.
\label{11a} \\
\zeta_{ikj} & = & \left\{ \begin{array}{lll}
\eta_{ikj} & {\rm if} & \max (\tau^{ij}, \eta^{ij}) < \eta_{ikj}
< \min (r_{kj} (H_j) - \hat{b}_k, \hat{b}_i - r_{ij} (h_j)) \\
\max (\tau^{ij}, \eta^{ij}) & & {\rm otherwise}
\end{array} \right. \nonumber \\
& & \label{11b} \\
\gamma_{ikj} & = & \max (\tau^{ij}, \eta^{ij})
\label{11c}
\eeqn
}

\noindent For the illustration of these variables see also the
appendix.


Concerning the definition of the sets $V_{kj}(t)$ \& $W_{ij}
(t)$, if we assume that

$$r_{ij} (h_j) \leq a_{ij} \leq r_{ij}(H_j) \hspace{0.5cm} {\rm
for \; all} \hspace{0.5cm} i \in S, \; \; j \in N$$

\noindent then it is easy to recognize the following remarks:

\vspace{0.5cm}

\noindent {\bf Remark 1:}

If the two sets $[r_{kj}(h_j) - \hat{b}_k, a_{kj} -
\hat{b}_k]~\&~[\hat{b}_i - a_{ij}, \hat{b}_i - r_{ij}(h_j)]$ have
an empty intersection, then we can deduce the following:
\begin{itemize}
\item if $a_{kj} - \hat{b}_k < \hat{b}_i - a_{ij} = \tau^{ij} =
\max (\tau^{ij}, \eta^{ij})$, for some $k \in S$, then \\
$V_{kj}(t)
\cap W_{ij}(t) = \emptyset$ if $t < \hat{b}_i - a_{ij}$; and if
$t \geq \hat{b}_i - a_{ij}$ the intersection equals $W_{ij}(t)$.

\item if $\hat{b}_i - r_{ij}(h_j) < r_{kj}(h_j)- \hat{b}_k  =
\eta^{ij} =
\max (\tau^{ij}, \eta^{ij})$, for some $k \in S$, then $V_{kj}(t)
\cap W_{ij}(t) = \emptyset$ if $t < r_{kj}(h_j) - \hat{b}_k$; and if
$t \geq r_{kj}(h_j) - \hat{b}_k$ the intersection equals $V_{kj}(t)$.
\end{itemize}


\vspace{0.5cm}

\noindent {\bf Remark 2:}

If the two sets $[r_{kj}(h_j) - \hat{b}_k, a_{kj} -
\hat{b}_k]~\&~[\hat{b}_i - a_{ij}, \hat{b}_i - r_{ij}(h_j)]$ have
a single point in their intersection, then we can deduce the following:
\begin{itemize}
\item if the point of intersection is $x= \hat{b}_i - r_{ij}(h_j)
= r_{kj}(h_j) - \hat{b}_k = \eta^{ij} =
\max (\tau^{ij}, \eta^{ij})$, for some $k \in S$, then $V_{kj}(t)
\cap W_{ij}(t) = \emptyset$ if $t < x$; and if
$t \geq x$ the intersection equals $V_{kj}(t)$.

\item if the point of intersection is $x= a_{kj} - \hat{b}_k
=  \hat{b}_i - a_{ij}$, for some $k \in S$, then we have the
following two cases: \\
$V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t < x$, given that
$r^{-1}_{ij} (\hat{b}_i - t) \leq r_{kj}^{-1} (\hat{b}_k+t) < H_j$; and if
$t \geq x$ the intersection equals $W_{ij}(t)$; $x = a_{kj} -
\hat{b}_k$. \\
$V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t < x$, given that
$r^{-1}_{kj} (\hat{b}_k + t) \leq r_{ij}^{-1} (\hat{b}_i-t) < H_j$; and if
$t \geq x$ the intersection equals $V_{kj}(t)$; $x =
\hat{b}_i -a_{ij} = \tau^{ij} = \max (\tau^{ij}, \eta^{ij})$.
\end{itemize}



\vspace{0.5cm}

\noindent {\bf Remark 3:}

Let
\beq
z_1 = r_{kj} (h_j) - \hat{b}_k &, & y_1 = a_{kj} - \hat{b}_k \\
z_2 =  \hat{b}_i - a_{ij} &, & y_2 = \hat{b}_i - r_{ij}(h_j).
\eeq

Assuming that $[z,y] = [z_1,y_1] \cap [z_2,y_2]$ one can found
the following cases:
\begin{itemize}
\item if $[z,y] = [z_1,y_1]$, then there exits some $t_0$ such
that

$$\tau^{ij} < \eta^{ij} = z_1 < t_0 < y_1 < y_2 \hspace{0.5cm}
{\rm and} \hspace{0.5cm} r_{kj}^{-1} (\hat{b}_k +t_0) =
r_{ij}^{-1} (\hat{b}_i - t_0);$$

\noindent i.e. $\max (\tau^{ij}, \eta^{ij}) < t_0 < \min (y_1,
y_2)$, \\
then $V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t \geq t_0$ the
intersection is nonempty, (similarly the case $[z,y] =
[z_2,y_2])$.

\item if $[z,y] = [z_0,y_1]$, $z_0 = z_1 = z_2$ then there exits
some $t_0$ such that

$$\tau^{ij} = \eta^{ij} = z_0 < t_0 < y_1 < y_2 \hspace{0.5cm}
{\rm and} \hspace{0.5cm} r_{kj}^{-1} (\hat{b}_k +t_0) =
r_{ij}^{-1} (\hat{b}_i - t_0);$$

\noindent i.e. $\max (\tau^{ij}, \eta^{ij}) < t_0 <\min (y_1,
y_2)$, \\
then $V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t < t_0$; and if
$t \geq t_0$ the
intersection is nonempty, (similarly the case $[z,y] =
[z_0,y_2])$.

\item if $[z,y] = [z_1,y_0]$, $y_0 = y_1 = y_2$ then there exits
some $t_0$ such that

$$\tau^{ij} < \eta^{ij} = z_1 < t_0 < y_0 \hspace{0.5cm}
{\rm and} \hspace{0.5cm} r_{kj}^{-1} (\hat{b}_k +t_0) =
r_{ij}^{-1} (\hat{b}_i - t_0);$$

\noindent i.e. $\max (\tau^{ij}, \eta^{ij}) < t_0 <\min (y_1,
y_2)$, \\
then $V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t < t_0$; and if
$t \geq t_0$ the
intersection is nonempty, (similarly the case $[z,y] =
[z_2,y_0])$.

\item if $[z,y] = [z_1,y_2]$, then there exits
some $t_0$ such that

$$\tau^{ij} < \eta^{ij} = z_1 < t_0 < y_2 < y_1 \hspace{0.5cm}
{\rm and} \hspace{0.5cm} r_{kj}^{-1} (\hat{b}_k +t_0) =
r_{ij}^{-1} (\hat{b}_i - t_0);$$

\noindent i.e. $\max (\tau^{ij}, \eta^{ij}) < t_0 <\min (y_1,
y_2)$, \\
then $V_{kj}(t) \cap W_{ij}(t) = \emptyset$ if $t < t_0$; and if
$t \geq t_0$ the
intersection is nonempty, (similarly the case $[z,y] =
[z_2,y_1])$.
\end{itemize}



\begin{th}
Let $i,k \in S$, $j \in N$; $r_{ij}(h_j) \leq a_{ij} \leq
r_{ij}(H_j)$ for all $i \in S$, $j \in N$; then $\exists \xi_{ikj}
\geq 0$ such that $P_{ikj}(t) \neq \emptyset \Leftrightarrow t
\geq \xi_{ikj}$.
\label{t44}
\end{th}

\noindent {\bf Proof:}
The proof is obviously derived from the above remarks and from the
definition of $\xi_{ikj}$ which is given in (\ref{11a}).


\begin{th}
Let $i,k \in S$, $j \in N$; $r_{ij}(h_j) \leq r_{ij} (H_j) \leq
a_{ij}$ for all $i \in S$, $j \in N$; then $\exists \zeta_{ikj}
\geq 0$ such that $P_{ikj}(t) \neq \emptyset \Leftrightarrow t
\geq \zeta_{ikj}$.
\label{t45}
\end{th}

\noindent {\bf Proof:}
The proof is obviously derived from the above remarks and from the
definition of $\zeta_{ikj}$ which is given in (\ref{11b}).

\begin{th}
Let $i,k \in S$, $j \in N$; $a_{ij} \leq r_{ij}(h_j) \leq r_{ij}
(H_j)$ for all $i \in S$, $j \in N$; then $\exists \gamma_{ikj}
\geq 0$ such that $P_{ikj}(t) \neq \emptyset \Leftrightarrow t
\geq \gamma_{ikj}$.
\label{t46}
\end{th}

\noindent {\bf Proof:}
The proof is obviously derived from the above remarks and from the
definition of $\gamma_{ikj}$ which is given in (\ref{11c}).

\vspace{0.3cm}

To generalize the above three theorems we introduce the following
lemmas and remarks.

\begin{lm}
Let $i,k \in S$, $j \in N$;  then $\exists \delta^{ij}
\geq 0$ such that $V_j(t) \cap W_{ij} (t) \neq \emptyset \Rightarrow t
\geq \delta^{ij}$.
\label{l41}
\end{lm}

\noindent {\bf Proof:} Let $\delta^{ij} = \max (\tau^{ij},
\eta^j)$; assume that for some fixed $i \in S$, $j \in N$ (say
$i_0$, $j_0$); $\eta^{j_0} \geq \tau^{i_0j_0}$; then $t <
\delta^{i_0j_0} \Rightarrow t < \eta^{j_0} \Rightarrow V_{j_0}
(t) \cap W_{i_0j_0} (t) = \emptyset$ (th.~\ref{t42}). Similarly we
can treat the other case, and then the proof is complete.
\hfill\ $\Box$

\begin{lm}
Let $i,k \in S$, $j \in N$;  then $\exists \beta \geq 0$ such that
$V_j(t) \cap W_{ij} (t) \neq \emptyset \Rightarrow t \geq \beta$.
\label{l42}
\end{lm}

\noindent {\bf Proof:} Let $\beta = \max_{i \in S} \min_{j \in N}
\delta^{ij}$. \\
Assume that $t < \beta \Rightarrow $ for some fixed $i \in S$,
$j \in N$ (say $i_0$, $j_0$) we have $t < \delta^{i_0j_0}$ \\
\hspace*{3.4cm} $\rightarrow t < \max (\tau^{i_0j_0}, \eta^{j_0})$
\\
\hspace*{3.4cm} $\Rightarrow V_{j_0} (t) \cap W_{i_0j_0} (t) =
\emptyset$ (th.~\ref{t41}), \\
and then the proof is complete.
\hfill\ $\Box$


\vspace{0.5cm}

\noindent {\bf Remarks:}

\begin{itemize}
\item From lemma~\ref{l41} and lemma~\ref{l42} we have \\
$V_j(t) \neq \emptyset$ and $W_{ij}(t) \neq \emptyset$ if $t \geq
\beta$ i.e. $t \in [\beta, \infty)$.
\item If $t \geq \max (\tau^{ij}, \eta^j)$, then $V_j(t) \neq
\emptyset$ and $W_{ij}(t) \neq \emptyset$ .
\item From th.~\ref{t41} and th.~\ref{t42} we have: \\
$V_j(t) \neq \emptyset$ and $W_{ij}(t) \neq \emptyset$ if $t <
\hat{b}_i - r_{ij}(h_j)$ and $ t \leq \max (\tau^{ij}, \eta^j)$
i.e. $t \in [\max (\tau^{ij}, \eta^j), \hat{b}_i - r_{ij} (h_j))$.
\end{itemize}

If we redefine $\xi_{ikj}$, $\zeta_{ikj}$  and $\gamma_{ikj}$ by
replacing $\eta^{ij}$ by $\eta^j$ and then by $\eta$ in
(\ref{11a}), (\ref{11b}) (\ref{11c}), then from
theorems~\ref{t41}~\&~\ref{t42}, lemmas~\ref{l41}~\&~\ref{l42}
and also from the above remarks, we can prove again the
generalized form of theorems~\ref{t44}~\&~\ref{t45}~\&~\ref{t46}
which obtained by the new formulas of $\xi_{ikj}$, $\zeta_{ikj}$
and $\gamma_{ikj}$.

\vspace{0.5cm}

Now let us define the following maximum variables:

\be
\xi^{ij} = \max_{k \in S} \xi_{ikj}, \hspace{0.5cm}
\zeta^{ij} = \max_{k \in S} \zeta_{ikj}, \hspace{0.5cm} {\rm and}
\hspace{0.5cm} \gamma^{ij} = \max_{k \in S} \gamma_{ikj}.
\label{12}
\ee

The following three theorems give another sufficient and
necessary conditions for

$$V_j (t) \cap W_{ij}(t) \neq \emptyset.$$

\begin{th}
Let $i \in S$, $j \in N$; $r_{ij}(h_j) \leq a_{ij} \leq r_{ij}
(H_j)$ for all $i \in S$, $j \in N$; then $\exists \xi^{ij}
\geq 0$ such that $V_j (t)\cap W_{ij}(t) \neq \emptyset \Leftrightarrow t
\geq \xi^{ij}$; $\xi^{ij}$ is given by (\ref{12}).
\label{t47}
\end{th}

\noindent {\bf Proof:}
The assertion follows immediately from theorem~\ref{t44} and the
definition of $V_j(t)$.

\begin{th}
Let $i \in S$, $j \in N$; $r_{ij}(h_j) \leq r_{ij} (H_j) \leq
a_{ij}$ for all $i \in S$, $j \in N$; then $\exists \zeta^{ij}
\geq 0$ such that $V_j (t)\cap W_{ij}(t) \neq \emptyset \Leftrightarrow t
\geq \zeta^{ij}$; $\zeta^{ij}$ is given by (\ref{12}).
\label{t48}
\end{th}

\noindent {\bf Proof:}
The assertion follows immediately from theorem~\ref{t45} and the
definition of $V_j(t)$.


\begin{th}
Let $i \in S$, $j \in N$; $a_{ij} \leq r_{ij} (h_j) \leq
r_{ij} (H_j)$ for all $i \in S$, $j \in N$; then $\exists \gamma^{ij}
\geq 0$ such that $V_j (t)\cap W_{ij}(t) \neq \emptyset \Leftrightarrow t
\geq \gamma^{ij}$; $\gamma^{ij}$ is given by (\ref{12}).
\label{t49}
\end{th}

\noindent {\bf Proof:}
The assertion follows immediately from theorem~\ref{t46} and the
definition of $V_j(t)$.

\vspace{0.3cm}

>From the above results we conclude that there exist some values,
say $t^j$~\&~$T^{ij}$ for which the relations $V_j (t) \neq
\emptyset \Leftrightarrow t \geq t^j$ and  $W_{ij} (t) \cap
V_j (t) \neq \emptyset \Leftrightarrow t \geq T^{ij}$ hold
$\forall i : i \in S$, $\forall j : j \in N$; where $T^{ij}$ is
equal to one of the values $\xi^{ij}$, $\zeta^{ij}$ or
$\gamma^{ij}$ according to which of the conditions from
Theorems~4.7, 4.8 and 4.9 are satisfied
and $t^j$ is the same as $\eta^j$ which
defined in (\ref{10}). Then the optimal value of $t(t^{opt})$
calculated according to the following formula:

\be
t^{opt} = \max (\max_{j \in N} t^j, \max_{i \in S} \min_{j \in N}
T^{ij}).
\label{13}
\ee

\noindent Consequently we can deduce that, the optimal value of
$t(t^{opt})$ is calculated according to the following theorem:

\begin{th}
If $t$ is the solution of problem (\ref{9}), then $t$ holds one
of the following relations:

\beqn
{\rm If} \; r_{ij} (h_j) & \leq & a_{ij} \leq r_{ij}(H_j) \; \;
{\rm for \; all} \; \; i \in S, \; j \in N; \; {\rm then}
\nonumber \\
t & \geq & t^{opt} = \max_{i \in S} \min_{j \in N} \xi^{ij}.
\label{14a} \\
{\rm If} \; r_{ij} (h_j) & \leq & r_{ij} (H_j) \leq a_{ij} \; \;
{\rm for \; all} \; \; i \in S, \; j \in N; \; {\rm then}
\nonumber \\
t & \geq & t^{opt} = \max_{i \in S} \min_{j \in N} \zeta^{ij}.
\label{14b} \\
{\rm If} \; a_{ij} & \leq & r_{ij} (h_j) \leq r_{ij} (H_j) \; \;
{\rm for \; all} \; \; i \in S, \; j \in N; \; {\rm then}
\nonumber \\
t & \geq & t^{opt} = \max_{i \in S} \min_{j \in N} \gamma^{ij}.
\label{14c}
\eeqn

\noindent Where $\xi^{ij}$, $\zeta^{ij}$ and $\gamma^{ij}$ are
given in (\ref{12}).
\label{t410}
\end{th}

\noindent {\bf Proof:} In our proof we will concentrate on the
first case. Let $\xi^{i_0j_0} = \max_{i \in S} \min_{j \in N}
\xi^{ij}$, and assume that $t < \xi^{i_0j_0}$, then from
theorem~\ref{t47} we can deduce that:

$$W_{i_0j}(t) \cap V_j(t) = \emptyset; \hspace{0.5cm} \forall j
\in N,$$

\noindent hence, according to theorem~\ref{t31}, $M(t) =
\emptyset$; this complete the proof of the theorem. \hfill\
$\Box$

\section{Algorithm for Calculating $t^{opt}$}

\noindent {\sl Step 1}: \\
Find $\eta^{ij}$, $\eta^j$, $\eta$ and $\tau^{ij}$ from relations
(\ref{10}), for each $i \in S$ and each $j \in N$.

\vspace{0.3cm}

\noindent {\sl Step 2}: \\
Calculate $\eta_{ikj}$ from the equation

$$r_{kj}^{-1} (\hat{b}_k + \eta_{ikj}) = r_{kj}^{-1} (\hat{b}_i -
\eta_{ikj})$$

\noindent for each $i,k \in S$ and each $j \in N$.


\vspace{0.3cm}

\noindent {\sl Step 3}: \\
Find $\xi_{ikj}$, $\zeta_{ikj}$, or $\gamma_{ikj}$ from relations
(\ref{11a}), (\ref{11b}), (\ref{11c}) for each $i,k \in S$ and
each $j \in N$.


\vspace{0.3cm}

\noindent {\sl Step 4}: \\
Find $\xi^{ij}$, $\zeta^{ij}$ or $\gamma^{ij}$ from relations
(\ref{12}) for each $i \in S$ and $j \in N$.

\vspace{0.3cm}

\noindent {\sl Step 5}: \\
Find $t^{opt}$ from relations (\ref{14a}), (\ref{14b}),
(\ref{14c}).

\vspace{0.5cm}

%\newpage

\noindent {\bf Example}

Here we want to solve the following problem

$$R_i \equiv \max_{j \in N} (a_{ij} \wedge r_{ij} (x_j)) = b_i
\hspace{0.5cm} \forall i \in S$$

\noindent and

$$h_j \leq x_j \leq H_j \hspace{0.5cm} \forall j \in N$$

\noindent where

$$N = \{ 1,2,3,4 \}; \; \; \; S = \{ 1,2,3 \}; \; \; \; x = [
x_1 x_2 x_3 x_4 ]^T; $$

$$b = [5 \;  7 \;  3]^T; \; \; \; h = [ 1 \; 2 \; 0 \; 1]^T; \; \; \;
H = [ 5 \; 6 \; 4 \; 3]^T;$$

$$\left[ \begin{array}{llll}
r_{11}(x_1) & r_{12}(x_2) & r_{13}(x_3) &  r_{14}(x_4) \\
r_{21}(x_1) & r_{22}(x_2) & r_{23}(x_3) &  r_{24}(x_4) \\
r_{31}(x_1) & r_{32}(x_2) & r_{33}(x_3) &  r_{34}(x_4)
\end{array} \right] \equiv
\left[ \begin{array}{cccc}
4 x_1 & 7 x_2 &  6 x_3 &  x_4 +1 \\
2 x_1 &  x_2 &  6 x_3 +1 &  2 x_4 \\
x_4 + 1 &  x_2 -1 &  2 x_3 -1 &   x_4 - 3
\end{array} \right]$$

\noindent and

$$\left[ \begin{array}{llll}
a_{11} & a_{12} & a_{13} &  a_{14} \\
a_{21} & a_{22} & a_{23} &  a_{24} \\
a_{31} & a_{32} & a_{33} &  a_{34}
\end{array} \right] =
\left[ \begin{array}{lllr}
7 & 15 & 4 &  2 \\
3 & 4 & 1 &  4 \\
3 & 3 & 2 &  -1
\end{array} \right]$$

\noindent Note that this problem has no solution in general. \\
It is clear that $r_{ij}(h_j) \leq a_{ij} \leq r_{ij} (H_j)$
$\forall i \in S, j \in N$. Using the relations (\ref{10}) we can
deduce that

$$\left[ \begin{array}{llll}
\eta^{11} & \eta^{12} & \eta^{13} &  \eta^{14} \\
\eta^{21} & \eta^{22} & \eta^{23} &  \eta^{24} \\
\eta^{31} & \eta^{32} & \eta^{33} &  \eta^{34}
\end{array} \right] =
\left[ \begin{array}{llll}
0 & 9 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right]$$

\noindent this gives that

$$\left[ \begin{array}{llll}
\eta^1 & \eta^2 & \eta^3 & \eta^4 \end{array} \right] =
\left[  \begin{array}{llll}
0 & 9 & 0 & 0 \end{array} \right]$$

\noindent which implies that $\eta \leq 9$.

Also we get from (\ref{10}), the following

$$\left[ \begin{array}{llll}
\tau^{11} & \tau^{12} & \tau^{13} &  \tau^{14} \\
\tau^{21} & \tau^{22} & \tau^{23} &  \tau^{24} \\
\tau^{31} & \tau^{32} & \tau^{33} &  \tau^{34}
\end{array} \right] =
\left[ \begin{array}{llll}
0 & 0 & 1 & 3 \\
4 & 3 & 6 & 3 \\
0 & 0 & 1 & 4
\end{array} \right]$$

\noindent From the equality

$$r_{kj}^{-1} (b_k + \eta_{ikj}) = r_{ij}^{-1} (b_i -
\eta_{ikj})$$

\noindent We can calculate $\eta_{ikj}$ for each $k \in S$; $i
\in S$ and $j \in N$, then the application of the above relation
will give us the following three matrices:

\noindent at $k = 1$

$$\left[ \begin{array}{llll}
\eta_{111} & \eta_{112} & \eta_{113} &  \eta_{114} \\
\eta_{211} & \eta_{212} & \eta_{213} &  \eta_{214} \\
\eta_{311} & \eta_{312} & \eta_{313} &  \eta_{314}
\end{array} \right] =
\left[ \begin{array}{llll}
0 & 0 & 0 & 0 \\
3 & \frac{11}{2} & \frac{31}{7} & \frac{-1}{3} \\
\frac{3}{5} & \frac{23}{8} & \frac{7}{4} & 1
\end{array} \right]$$

\noindent at $k = 2$

$$\left[ \begin{array}{llll}
\eta_{121} & \eta_{122} & \eta_{123} &  \eta_{124} \\
\eta_{221} & \eta_{222} & \eta_{223} &  \eta_{224} \\
\eta_{321} & \eta_{322} & \eta_{323} &  \eta_{324}
\end{array} \right] =
\left[ \begin{array}{rrrl}
-3 & \frac{-11}{2} & \frac{-31}{7} & \frac{1}{3} \\
0 & 0 & 0 & 0 \\
-1 & \frac{-3}{2} & \frac{-8}{3} & \frac{5}{3}
\end{array} \right]$$

\noindent at $k = 3$

$$\left[ \begin{array}{llll}
\eta_{131} & \eta_{132} & \eta_{133} &  \eta_{134} \\
\eta_{231} & \eta_{232} & \eta_{233} &  \eta_{234} \\
\eta_{331} & \eta_{332} & \eta_{333} &  \eta_{334}
\end{array} \right] =
\left[ \begin{array}{rrrr}
\frac{-3}{5} & \frac{-23}{8} & \frac{-7}{4} & -1 \\
1 & \frac{3}{2} & \frac{8}{3} & \frac{5}{3} \\
0 & 0 & 0 & 0
\end{array} \right]$$


\noindent Now from equation (\ref{11a}) and the values of the
above parameters, we can obtain the following matrices

\noindent at $k=1$
\beq
& & \left[ \begin{array}{llll}
\xi_{111} & \xi_{112} & \xi_{113} &  \xi_{114} \\
\xi_{211} & \xi_{212} & \xi_{213} &  \xi_{214} \\
\xi_{311} & \xi_{312} & \xi_{313} &  \xi_{314}
\end{array} \right] \equiv    \\
& & \equiv
\left[ \begin{array}{llll}
0 & 9 & 1 &  3 \\
2 \; {\rm if} \; 0 \leq t <3 & \left\{ \begin{array}{ll}
10 & {\rm if} \; \; 1 < t \leq \frac{11}{2} \\
3 & {\rm otherwise} \end{array} \right. & \left\{ \begin{array}{rl}
-1 & {\rm if} \; \; 2 < t \leq \frac{31}{7} \\
6 & {\rm otherwise} \end{array} \right. & 3 \\
3/5 & \left\{ \begin{array}{ll}
10 & {\rm if} \; \; 0 \leq t \leq \frac{23}{8} \\
0 & {\rm otherwise} \end{array} \right. & \left\{ \begin{array}{rl}
-1 & {\rm if} \; \; 0 < t \leq \frac{7}{4} \\
1 & {\rm otherwise} \end{array} \right. & 4
\end{array} \right]
\eeq

\vspace{0.6cm}

\noindent at $k=2$
$$\left[ \begin{array}{llll}
\xi_{121} & \xi_{122} & \xi_{123} &  \xi_{124} \\
\xi_{221} & \xi_{222} & \xi_{223} &  \xi_{224} \\
\xi_{321} & \xi_{322} & \xi_{323} &  \xi_{324}
\end{array} \right] =
\left[ \begin{array}{llll}
0 & 9 & 1 &  3 \\
4 & 3 & 6 &  3 \\
0 & 0 & 1 &  4
\end{array} \right]$$

\newpage

\noindent at $k=3$
\beq
& & \left[ \begin{array}{llll}
\xi_{131} & \xi_{132} & \xi_{133} &  \xi_{134} \\
\xi_{231} & \xi_{232} & \xi_{233} &  \xi_{234} \\
\xi_{331} & \xi_{332} & \xi_{333} &  \xi_{334}
\end{array} \right] \equiv \\
& & \equiv
\left[ \begin{array}{llll}
0 & 9 & 1 &  3 \\
0 \; {\rm if} \; 0 \leq t <1 & \left\{ \begin{array}{ll}
0 & {\rm if} \; \; 1 < t \leq \frac{3}{2} \\
3 & {\rm otherwise} \end{array} \right. & \left\{ \begin{array}{rl}
-1 & {\rm if} \; \; 2 < t \leq \frac{10}{3} \\
6 & {\rm otherwise} \end{array} \right. & 3 \\
0 & 0 & 1 & 4
\end{array} \right]
\eeq

\noindent From the definition of $\xi^{ij}$ given in (\ref{12}),
it is easy to obtain the following


$$\left[ \begin{array}{llll}
\xi^{11} & \xi^{12} & \xi^{13} &  \xi^{14} \\
\xi^{21} & \xi^{22} & \xi^{23} &  \xi^{24} \\
\xi^{31} & \xi^{32} & \xi^{33} &  \xi^{34}
\end{array} \right] \equiv
\left[ \begin{array}{llll}
0 & 9 & 1 &  3 \\
4  & \left\{ \begin{array}{ll}
10 & {\rm if} \; \; 1 < t \leq \frac{11}{2} \\
3 & {\rm otherwise} \end{array} \right. & 6 & 3 \\
3/5  & \left\{ \begin{array}{ll}
10 & {\rm if} \; \; 0 < t \leq \frac{23}{8} \\
0 & {\rm otherwise} \end{array} \right. & 1 & 4
\end{array} \right]$$


\noindent Hence $t^{opt}$ which given by equation (\ref{13}) will be

$$t^{opt} = \max (9,3) = 9$$

\noindent Take any point (say $x^*$) from the set

$$M(b) = \{x: h \leq x \leq H, \hspace{0.5cm} \parallel R(x) - b
\parallel \leq t, \hspace{0.5cm} t \geq 9\},$$

\noindent then $x^*$ will be accepted as an approximate solution
of our problem.

\vspace{0.5cm}

In the original case $V_{12} = \emptyset$ $\{$since $15 \wedge 7x_2
>5$, $x_2 \in [2,6]\}$,

$$V_{12} = \emptyset \rightarrow V_2 = \emptyset \longrightarrow
M = \emptyset$$

\noindent i.e. there is no solution for the original problem.

In the modified case, if we take $t = 9$, then we try to solve
the following problem

\beq
& & -4 \leq \max 7 \wedge 4 x_1 , 15 \wedge 7x_2 , 4 \wedge 6x_3,
2 \wedge (x_4+1) \leq 14 \\
& & -2 \leq \max 3 \wedge 2 x_1 , 4 \wedge x_2 , 1 \wedge (x_3
+1),
4 \wedge 2x_4 \leq 16 \\
& & -6 \leq \max 3 \wedge (x_1+1) , 3 \wedge (x_2 -1) , 2 \wedge (2x_3
-1) , -1 \wedge (x_4 -3) \leq 12
\eeq

$$1 \leq x_1 \leq 5, \hspace{0.5cm} 2 \leq x_2 \leq 6, \hspace{0.5cm}
0 \leq x_3 \leq 4, \hspace{0.5cm}  1 \leq x_4 \leq 3.$$

\noindent Then

$$V_1 = [1,5] \hspace{0.5cm} V_2 = \{ 2\} \hspace{0.5cm}
V_3 = [0,4] \hspace{0.5cm} V_4 = [1,3],$$

\noindent and

$$V_1 \cap W_{11} \neq \emptyset, \hspace{0.5cm} V_2 \cap W_{22}
\neq \emptyset, \hspace{0.5cm} V_3 \cap W_{33} \neq \emptyset,$$

\noindent where

$$W_{11} = [1,5] \hspace{0.5cm} W_{22} = [2,6] \hspace{0.5cm}
W_{33} = [0,4]$$

\noindent then choose any $x$ such that

$$x = (x_1, 2, x_3, x_4)$$

\noindent where

$$x_1 \in [1,5] \hspace{0.5cm} x_3 \in [0,4] \hspace{0.5cm}
x_4 \in [1,3]$$

\noindent will be an approximate solution  for the original
problem.

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\end{document}



