# On Critical Point for Functions with Bounded Parameters

Priyanka Roy‡ and Geetanjali Panda∗
Department of Mathematics, Indian Institute of Technology Kharagpur,
Kharagpur, West Bengal-721302, India.
‡<EMAIL_ADDRESS>
∗<EMAIL_ADDRESS>
###### Abstract

Selection of descent direction at a point plays an important role in numerical
optimization for minimizing a real valued function. In this article, a descent
sequence is generated for the functions with bounded parameters to obtain a
critical point. First, sufficient condition for the existence of descent
direction is studied for this function and then a set of descent directions at
a point is determined using linear expansion. Using these results a descent
sequence of intervals is generated and critical point is characterized. This
theoretical development is justified with numerical example.

keywords Line search technique, Markov difference, Interval ordering, Descent
direction, Critical point.

##  1 Introduction

In most of the mathematical models, the parameters vary in some bounds which
can be estimated from historical data. These uncertain parameters can be
considered as intervals. In that case, the functions involved in the model
here known as functions with bounded parameters as intervals. Interval
analysis plays an important role to handle these functions. This article has
studied some properties of these type functions, $\breve{F}$ from
$\mathbf{R}^{n}$ to the set of closed intervals, whose parameters are
intervals. An example of such type function is
$\breve{F}(x,y)=e^{[-3,3]x^{3}}\oplus[4,5]xy^{5}$. In the literature of
interval analysis, Markov is the pioneer who has studied different areas of
modern mathematics using interval analysis ( see [5], [6], [10], [11], [12],
[13] etc. ). Markov has introduced nonstandard substraction between two closed
intervals( known as Markov difference, $\ominus_{M}$), which has explored
calculus of functions with bounded parameters and its application in several
areas in recent times (see [1], [2], [9], [15], [16] etc.). Another
nonstandard difference, known as gH-difference, in the set of intervals is
introduced in [18] and further developed in [4], [7],[17] etc. . It is
justified in [3] that gH difference coincides with Markov difference in case
of compact intervals. Markov difference is more comfortable for use in
numerical computations. Hence in this article we have accepted Markov
difference and proceed to develop an iterative process for generating a
descent sequence of intervals. This concept may be extended further to develop
a descent sequence of points which may converge to a local minimum point of a
function with interval parameters under reasonable conditions. This can
explore a new area of numerical optimization, which may be considered as the
possible scope of the present contribution. At this present stage we focus on
characterizing descent direction, generating descent sequence of intervals for
a function with interval parameters, which provides the critical point of the
function.

Some prerequisites on interval analysis are discussed in Section 2. In Section
3, Markov difference is used to derive the linear expansion of $\breve{F}$ and
existence of descent direction of $\breve{F}$ at a point $x$ is studied.
Section 4 is devoted for generating descent sequence of intervals which
determines the critical point of $\breve{F}$. Section 5 provides concluding
remarks with future scope.

##  2 Prerequisites

$K(\textbf{R})$ denotes the set of all compact intervals on R throughout this
article . $\breve{\alpha}\in K(\mathbf{R})$ is the closed interval of the form
$[\underline{\alpha},\overline{\alpha}]$ where
$\underline{\alpha}\leq\overline{\alpha}$. For two points $\alpha_{1}$ and
$\alpha_{2}$,(not necessarily $\alpha_{1}\leq\alpha_{2}$), $\breve{\alpha}$
can be written as $\breve{\alpha}=[\alpha_{1}\vee\alpha_{2}]$. A real number
$r$ can be represented as a degenerate interval denoted by $\breve{r}$ as
$\breve{r}=[r,r]$ or $r.\breve{I}$, where $\breve{I}=[1,1]$. The null interval
is $\breve{0}=[0,0]=0$.
In $K(\mathbf{R})$, the norm $(\|.\|)$ of an interval $\breve{\alpha}$ is
defined as
$\|\breve{\alpha}\|=\max\left\\{|\underline{\alpha}|,|\overline{\alpha}|\right\\}$
([8] ) which is associated with the metric
$d(\breve{\alpha},0)=\|\breve{\alpha}\|$ and
$d(\breve{\alpha},\breve{\beta})=\max\left\\{|\underline{\alpha}-\underline{\beta}|,\leavevmode\nobreak\
|\overline{\alpha}-\overline{\beta}|\right\\}$.
$K(\mathbf{R})$ is not a complete ordered set. Following interval ordering is
used throughout the article..
For $\breve{\alpha}$, $\breve{\beta}\in K(\mathbf{R})$,
$\breve{\alpha}\preceq\breve{\beta}\Leftrightarrow\underline{\alpha}\leq\underline{\beta}$,
$\overline{\alpha}\leq\overline{\beta}$ and $\breve{\alpha}\neq\breve{\beta}$;
$\breve{\alpha}\prec\breve{\beta}\Leftrightarrow\underline{\alpha}<\underline{\beta}$
and $\overline{\alpha}<\overline{\beta}$.

The other interval order relations $`\succeq^{\prime}$ and
${}^{\prime}\succ^{\prime}$ can be defined in a similar way.

Additive inverse in $\left\langle K(\mathbf{R}),\oplus,\odot\right\rangle$ may
not exist, that is, $\breve{\alpha}\ominus\breve{\alpha}$ is not necessarily
$\breve{0}$ according to this approach. The non-standard subtraction due to
[9], denoted by $\ominus_{M}$, provides additive inverse, which is

$\displaystyle\breve{\alpha}\ominus_{M}\breve{\beta}=\left[\min\left\\{\underline{\alpha}-\underline{\beta},\overline{\alpha}-\overline{\beta}\right\\},\max\left\\{\underline{\alpha}-\underline{\beta},\overline{\alpha}-\overline{\beta}\right\\}\right]$
(1)

Following properties of $\ominus_{M}$ due to [5] and [9] are used throughout
the article.
(i)  $\breve{\alpha}\ominus_{M}\breve{\alpha}=\breve{0}$ ; (ii)
$\ominus_{M}\breve{\alpha}=(-1)\breve{\alpha}=[-\overline{\alpha},-\underline{\alpha}]$

Limit and continuity of $\breve{F}$ are understood in the sense of
$\parallel.\parallel$ due to [9]. Following results due to [9] are summarized
for $\breve{F}:\mathbf{R}\rightarrow K(\mathbf{R})$,
$\breve{F}(x)=[\underline{F}(x),\overline{F}(x)]$, where
$\underline{F},\overline{F}:\mathbf{R}\rightarrow\mathbf{R}$,
$\underline{F}(x)\leq\overline{F}(x)$ $\forall\leavevmode\nobreak\
\leavevmode\nobreak\ x\in\mathbf{R}$.

######  Definition 1 (Definition 2, [9]).

$\breve{F}:\mathbf{R}\rightarrow K(\mathbf{R})$ is differentiable at $x_{0}$
if $\displaystyle\lim_{h\rightarrow
0}\frac{\breve{F}(x_{0}+h)\ominus_{M}\breve{F}(x_{0})}{h}$ exists. The
limiting value is the derivative of $\breve{F}$ at $x_{0}$, denoted by
$\breve{F}^{\prime}(x_{0})$.
Alternatively if $\exists\breve{F}^{\prime}(x_{0})\in K(\mathbf{R})$ and an
error function
$\breve{E}_{x_{0}}:\mathbf{R}\rightarrow K(\mathbf{R})$ at $x$ such that

$\displaystyle\breve{F}(x_{0}+h)\ominus_{M}\breve{F}(x_{0})=h\odot\breve{F}^{\prime}(x_{0})\oplus\breve{E}_{x_{0}}(h)$
(2) $\displaystyle\text{ where }\lim_{h\rightarrow
0}\breve{E}_{x_{0}}(h)=\breve{0}$

######  Theorem 2.1 (Theorem 9, [9]).

If $\breve{F}:\mathbf{R}\rightarrow I(\mathbf{R})$ is
continuous in $\Delta$, where $\Delta=\left[p,q\right]$ and differentiable in
$(p,q)$,then
$\breve{F}(q)\ominus_{M}\breve{F}(p)\subset\breve{F}^{\prime}(\Delta)(q-p)$,
where
$\displaystyle\breve{F}^{\prime}(\Delta)=\cup_{\xi\in\Delta}\breve{F}^{\prime}(\xi)$.

Above results discuss the calculus of interval function on $\mathbf{R}$. In
next section some of these results are extended to develop calculus of
interval functions on $\mathbf{R}^{n}$.

##  3 Descent direction for interval function over $\mathbf{R}^{n}$

######  Definition 2.

For $\breve{F}:\mathbf{R}^{n}\rightarrow K(\mathbf{R})$, the partial
derivative
of $\breve{F}$ with respect to $x_{i}$ at ‘$x$ ’ exists if
$\underset{h_{i}\rightarrow
0}{\lim}\dfrac{\breve{F}(x_{1},\cdots,x_{i-1},x_{i}+h_{i},x_{i+1},\cdots
x_{n})\ominus_{M}\breve{F}(x)}{h_{i}}$ exists. The limiting value is denoted
by $\frac{\partial\breve{F}(x)}{\partial x_{i}}$.

In the light of concept of differentiability in (2) of Definition 1, the
following can be stated as follows.

######  Definition 3.

$\breve{F}:\mathbf{R}^{n}\rightarrow K(\mathbf{R})$ is called differentiable
at $x_{0}$ if $\frac{\partial\breve{F}(x_{0})}{\partial x_{i}}$ exists
$\forall\leavevmode\nobreak\ i$ and an interval function
$\breve{E}_{x_{0}}(h):\mathbf{R}^{n}\rightarrow K(\mathbf{R})$ such that

$\breve{F}(x_{0}+h)\ominus_{M}\breve{F}(x_{0})=\sum\limits_{i=1}^{n}\left(\oplus
h_{i}\odot\frac{\partial\breve{F}(x_{0})}{\partial
x_{i}}\right)\oplus\|h\|\odot\breve{E}_{x_{0}}(h)$

for $\|h\|<\delta$ for some $\delta>0$, where $\lim_{\|h\|\rightarrow
0}\breve{E}_{x_{0}}(h)=\breve{0}$.

Gradient of $\breve{F}$ at $x$ is $(\frac{\partial\breve{F}(x)}{\partial
x_{1}},\frac{\partial\breve{F}(x)}{\partial
x_{2}},\cdots,\frac{\partial\breve{F}(x)}{\partial x_{n}})^{\prime}$ and
denoted by $\nabla\breve{F}(x)$.

Following result is about linear expansion of $\breve{F}$ in inclusion form
which will be used further to derive descent direction.

######  Theorem 3.1.

Let $\Omega$ be an open convex subset of $\mathbf{R}^{n}$ and
$\breve{F}:\Omega\rightarrow K(\mathbf{R})$ be differentiable on $\Omega$.
Then for any $u,\leavevmode\nobreak\ v\in\Omega$,

$\displaystyle\breve{F}(v)\ominus_{M}\breve{F}(u)\subset\underset{c\in
L.S\\{u,v\\}}{\cup}\sum_{i=1}^{n}(v_{i}-u_{i})\odot\frac{\partial\breve{F}(c)}{\partial\gamma_{i}}$
(3)

where $\gamma(t)=u+t(v-u)$, $t\in[0,1]$ and $L.S\\{u,v\\}$ denotes the line
segment joining $u$ and $v$.

###### Proof.

Since $\Omega$ is convex subset of $\mathbf{R}^{n}$, for $u,v\in\Omega$,
$u+t(v-u)$ with $t\in[0,1]$ must belongs to $\Omega$.
Let $\breve{\phi}:[0,1]\rightarrow K(\mathbf{R})$ is defined by
$\breve{\phi}(t)=\breve{F}(\gamma_{1}(t),\gamma_{2}(t),\cdots\gamma_{n}(t))$.
Since $\breve{F}(\gamma)$ and $\gamma(t)$ are differentiable, using Definition
3 and first order Taylor expansion of $\gamma$ it can be shown that their
composite function $\breve{\phi}$ is differentiable and
$\breve{\phi}^{\prime}(t)=\displaystyle\sum_{i=1}^{n}\gamma_{i}^{\prime}(t)\odot\frac{\partial\breve{F}(\gamma(t))}{\partial\gamma_{i}}=\displaystyle\sum_{i=1}^{n}(v_{i}-u_{i})\odot\frac{\partial\breve{F}(\gamma(t))}{\partial\gamma_{i}}$.

From Theorem 2.1,
$\breve{\phi}(1)\ominus_{M}\breve{\phi}(0)\subset\underset{\theta\in[0,1]}{\cup}\breve{\phi}^{\prime}(\theta)$.
Here $\breve{\phi}(1)=\breve{F}(v)$ and $\breve{\phi}(0)=\breve{F}(u)$. Hence
(3) follows, where $c=a+\theta(b-a)$ for some $\theta$ and
$\breve{\phi}^{\prime}(\theta)=\displaystyle\sum_{i=1}^{n}(v_{i}-u_{i})\odot\frac{\partial\breve{F}(c)}{\partial\gamma_{i}}$.
∎

######  Proposition 1.

Let $\breve{\alpha},\breve{\beta}\in K(\mathbf{R})$. Then
$\breve{\alpha}\prec\breve{\beta}$ holds if and only if
$\breve{\alpha}\ominus_{M}\breve{\beta}\prec\breve{0}$.

Proof of this result is straight forward from the definition of $\ominus_{M}$.

######  Definition 4.

$d\in\mathbf{R}^{n}$ is called a descent direction of $\breve{F}$ at $x$ if
$\exists$ some $\delta>0$ so that $\breve{F}(x+\alpha
d)\prec\breve{F}(x)\leavevmode\nobreak\ \leavevmode\nobreak\
\forall\alpha\in(0,\delta)$.

######  Notation 1.

Denote $\nabla\breve{F}(x)$ by $\breve{g}(x)$ where
$\breve{g}(x)=\begin{pmatrix}\breve{g}_{1}(x)&\breve{g}_{2}(x)&\cdots&\breve{g}_{n}(x)\end{pmatrix}^{\prime}$,
with
$\breve{g}_{i}(x)=[\underline{g}_{i}(x),\overline{g}_{i}(x)]\leavevmode\nobreak\
\forall i=1,2,\cdots,n$ where
$\underline{g}_{i}(x)=\min\\{\frac{\partial\underline{F}(x)}{\partial
x_{i}},\frac{\partial\overline{F}(x)}{\partial x_{i}}\\},\leavevmode\nobreak\
{\overline{g}_{i}(x)=\max\\{\frac{\partial\overline{F}(x)}{\partial
x_{i}},\frac{\partial\underline{F}(x)}{\partial x_{i}}\\}}$.
Denote
$\underline{g}(x)\triangleq\begin{pmatrix}\underline{g}_{1}(x)&\underline{g}_{2}(x),&\cdots,&\underline{g}_{n}(x)\end{pmatrix}^{\prime}$
and
$\overline{g}(x)\triangleq\begin{pmatrix}\overline{g}_{1}(x)&\overline{g}_{2}(x),&\cdots,&\overline{g}_{n}(x)\end{pmatrix}^{\prime}$.

######  Theorem 3.2.

Let $\breve{F}:\Omega\subseteq\mathbf{R}^{n}\rightarrow K(\mathbf{R})$ be
continuously differentiable. If
$\displaystyle\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x)\prec\breve{0}$, then
$d$ is a descent direction of $\breve{F}$ at $x$.

###### Proof.

$\displaystyle\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x)$ is continuous since
$\breve{g}$ is continuous. Since
$\displaystyle\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x)\prec\breve{0}$, so
$\displaystyle 0\notin\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x)$. Therefore
$\exists\leavevmode\nobreak\ \leavevmode\nobreak\ \delta>0$ such that

$\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x+\alpha
d)\prec\breve{0}\leavevmode\nobreak\ \leavevmode\nobreak\
\forall\alpha\in(0,\delta).$ (4)

Using Theorem 3.1,

$\displaystyle\breve{F}(x+\alpha
d)\ominus_{M}\breve{F}(x)\subset\underset{c\in\leavevmode\nobreak\
L.S\\{x,x+\alpha d\\}}{\cup}\alpha\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(c)$
(5)

From (4),
$\displaystyle\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(c)\prec\breve{0}$, for
each $c\in L.S\left\\{x,x+\alpha d\right\\}$. Hence from (5),
$\breve{F}(x+\alpha d)\ominus_{M}\breve{F}(x)\prec\breve{0}$.
$\breve{F}(x+\alpha d)\prec\breve{F}(x)$ follows from Proposition 1. Therefore
$d$ is the descent direction at $x$. ∎

Figure 1: Shaded region indicates the set of descent directions

From the Theorem 3.2, one may conclude that the descent direction of
$\breve{F}(x,y)=[\underline{F}(x,y),\overline{F}(x,y)]$ at point
$(x_{0},y_{0})$ can be determined by solving
$\sum_{i=1}^{n}d_{i}\odot\breve{g}_{i}(x_{0},y_{0})\prec\breve{0}$ i.e.
$\breve{g}(x_{0},y_{0})^{\prime}\odot d\prec\breve{0}$. The set of descent
directions is

$\left\\{d\in\mathbf{R}^{n}:\underline{g}(x_{0},y_{0})^{\prime}d<0;\overline{g}(x_{0},y_{0})^{\prime}d<0\right\\}.$

In Figure 1, $C\underline{F}$ and $C\overline{F}$ are the contours of lower
and upper bound functions at levels $\underline{F}(x_{0},y_{0})$ and
$\overline{F}(x_{0},y_{0})$ respectively. $T\underline{F}$ and $T\overline{F}$
are the tangent lines to the contours $C\underline{F}$ and $C\overline{F}$ at
$(x_{0},y_{0})$ respectively. The set of descent directions at $(x_{0},y_{0})$
is the set of vectors in $R^{2}$, which make obtuse angle with both
$\underline{g}(x_{0},y_{0})$ and $\overline{g}(x_{0},y_{0})$. This is the
shaded region in Figure 1.

##  4 Descent direction and generating descent sequence

Selection of suitable descent direction plays an important role while
developing an efficient numerical algorithm for minimizing a real valued
function $f(x)$. Objective of this section is to develop a descent sequence
$\breve{F}(x^{k})$ with respect to the interval ordering $\preceq$, staring at
an initial point $x^{0}$. For the descent direction $d^{k}\in\mathbf{R}^{n}$
at $x^{k}$ of $\breve{F}(x)$, $\breve{F}(x^{k+1})\preceq\breve{F}(x^{k})$
holds causing reduction on lower and upper function simultaneously. Here
$x^{k=1}=x^{k}+\alpha_{k}d^{k}$, where $\alpha_{k}>0$ is the step length at
$x^{k}$ in the descent direction $d^{k}$, selected in such a way that
$\breve{F}(x^{k+1})\preceq\breve{F}(x^{k})$ holds. This process terminates at
a point $x^{*}$, when either $\alpha_{k}=0$ or no such descent direction
exists. We say a point where no such descent direction exists as a critical
point of $\breve{F}(x)$, which is defined below.

######  Definition 5.

A point $x^{*}$ is called a critical point of $\breve{F}$ if
$\nexists\leavevmode\nobreak\ d\in\mathbf{R}^{n}\mbox{ such
that}\leavevmode\nobreak\ \leavevmode\nobreak\ p^{T}d<0\leavevmode\nobreak\
\leavevmode\nobreak\ \forall\leavevmode\nobreak\ p\in\breve{g}(x^{*})$ .

######  Remark 1.

Weak efficient solution of the optimization problem $\min_{x\in
R^{n}}\breve{F}(x)$ is a critical point of $\breve{F}(x)$. For a detailed
study of weak efficient solution, the readers may see [2] and [14].

Following results are studied in this direction, which may be considered as a
stepping stone for generating descent sequence.

######  Theorem 4.1.

Let $\breve{F}:\Omega\rightarrow K(\mathbf{R})$ be differentiable such that
$0\notin int(\breve{g}_{i}(x))\leavevmode\nobreak\ \forall i=1,2,\cdots,n$.
Then $\overline{d}$ is a descent direction at $x$ where
$\overline{d}_{i}\in\ominus_{M}\breve{g}_{i}(x)$ for each $i=1,2,\cdots,n$ .

###### Proof.

Since $0\notin int(\breve{g}_{i}(x))\forall i=1,2,\cdots,n$, either
$\frac{\partial\breve{F}}{\partial x_{i}}\succeq\breve{0}$ or
$\frac{\partial\breve{F}}{\partial x_{i}}\preceq\breve{0}\forall
i=1,2,\cdots,n$.
Since $\overline{d}_{i}\in\ominus_{M}\breve{g}_{i}(x)$ for each
$i=1,2,\cdots,n$, so $d_{i}\geq 0$ if $\frac{\partial\breve{F}}{\partial
x_{i}}\preceq\breve{0}$ and vice versa.
So, $\sum_{i=1}^{n}\overline{d}_{i}\odot\frac{\partial\breve{F}(x)}{\partial
x_{i}}\prec\breve{0}$.
Therefore using Theorem 3.2, $\overline{d}$ is descent direction at $x$. ∎

######  Remark 2.

From the above theorem the following results can be concluded.

  * $\bullet$

For a descent direction $\overline{d}$, since
$\overline{d}_{i}\in\ominus_{M}\breve{g}_{i}(x)=[-\overline{g}_{i}(x),-\underline{g}_{i}(x)]$
for each $i=1,2,\cdots,n$ , each $d_{i}$ can be written as
$d_{i}=-\overline{g}_{i}(x)+t_{i}(\overline{g}_{i}(x)-\underline{g}_{i}(x))$
for each $t_{i}\in[0,1]$.

  * $\bullet$

If $\exists$ some $i$, for which $0\in int(\breve{g}_{i}(x))$, then it may not
guarantee that $\overline{d}$ is a descent direction at $x$. In that case for
preserving descent property one may choose $\overline{d}_{i}=0$ for those $i$.

### Generating a descent sequence of interval function

For generating descent sequence of intervals $\\{\breve{F}(x^{k})\\}$
satisfying $\breve{F}(x^{k+1})\preceq\breve{F}(x^{k})$ and
$x^{k+1}=x^{k}+\alpha_{k}d^{k}$, selection of $\alpha_{k}$ and suitable
stopping condition play important role. The iterative process will be stopped
at critical point. Therefore a tolerance level $\epsilon>0$ can be fixed for
$\parallel d^{k}\parallel$ as a breaking condition. Here exact line search
technique for real valued functions is used for step-length selection in this
iterative process, which is explained below.
Suppose $\underline{\alpha}_{k}=\arg\leavevmode\nobreak\
\underset{\alpha>0}{\min}\underline{F}(x^{k}+\alpha d^{k})$ and
$\overline{\alpha}_{k}=\arg\leavevmode\nobreak\
\underset{\alpha>0}{\min}\overline{F}(x^{k}+\alpha d^{k})$ . Choose

$\alpha_{k}=\min\\{\underline{\alpha}_{k},\overline{\alpha}_{k}\\}$ (6)

along the descent direction $\\{d^{k}\\}$. In particular if
$\underline{F}(x^{k}+\alpha d^{k})$ and $\overline{F}(x^{k}+\alpha d^{k})$ are
convex functions in $\alpha$ then
$\underline{\alpha}_{k}=\min\\{\alpha>0:\underline{g}(x^{k}+\alpha
d^{k})^{\prime}d^{k}=0\\}$ and
$\overline{\alpha}_{(k)}=\min\\{\alpha>0:\overline{g}(x^{k}+\alpha
d^{k})^{\prime}d^{k}=0\\}$
The above theoretical development is summarized in following steps.

  1. 1.

Fix tolerance level $\epsilon$, initialize $k=0$, $x^{0}$, $\alpha_{0}$, fix
$t_{i}\in$ rand $[0,1]$.

  2. 2.

Choose $\overline{d}_{i}(t)$ according to Remark 2.

  3. 3.

Select suitable step length $\alpha_{k}$ according to (6).

  4. 4.

$x^{k+1}=x^{k}+\alpha_{k}\overline{d}^{k}$.

  5. 5.

Check if $\parallel\overline{d}^{k+1}\parallel<\epsilon$, stop. Otherwise
$k:=k+1$.

The above steps can be verified with the following example. In this example a
descent sequence is generated and critical point is verified.

######  Example 1.

Consider
$\breve{F}(x_{1},x_{2})=[2,4]x_{1}^{2}\oplus[2,3]x_{1}x_{2}\oplus[1,2]x_{2}^{2}\oplus[1,2]x_{1}\ominus_{M}[1,3]x_{2}$.

$\breve{F}(x_{1},x_{2})=\begin{cases}\left[2x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}+x_{1}-x_{2}\right.\vee\\\
\left.4x_{1}^{2}+3x_{1}x_{2}+2x_{2}^{2}+2x_{1}-3x_{2}\right];\\\ \text{ if
}x_{1}\geq 0,x_{2}\geq 0\\\
\left[2x_{1}^{2}+3x_{1}x_{2}+x_{2}^{2}+2x_{1}-x_{2}\right.\vee\\\
\left.4x_{1}^{2}+2x_{1}x_{2}+2x_{2}^{2}+x_{1}-3x_{2}\right];\\\ \text{ if
}x_{1}\leq 0,x_{2}\geq 0\\\
\left[2x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}+2x_{1}-3x_{2}\right.\vee\\\
\left.4x_{1}^{2}+3x_{1}x_{2}+2x_{2}^{2}+x_{1}-x_{2}\right];\\\ \text{ if
}x_{1}\leq 0,x_{2}\leq 0\\\
\left[2x_{1}^{2}+3x_{1}x_{2}+x_{2}^{2}+x_{1}-3x_{2}\right.\vee\\\
\left.4x_{1}^{2}+2x_{1}x_{2}+2x_{2}^{2}+2x_{1}-x_{2}\right];\\\ \text{ if
}x_{1}\geq 0,x_{2}\geq 0\end{cases}$

$\nabla\breve{F}(x_{1},x_{2})=\begin{cases}\begin{pmatrix}[4x_{1}+2x_{2}+1\vee
8x_{1}+3x_{2}+2]\\\ [2x_{1}+2x_{2}-1\vee 3x_{1}+4x_{2}-3]\end{pmatrix};\\\
\text{ if }x_{1}\geq 0,x_{2}\geq 0\\\ \begin{pmatrix}[4x_{1}+3x_{2}+2\vee
8x_{1}+2x_{2}+1]\\\ [3x_{1}+2x_{2}-1\vee 2x_{1}+4x_{2}-3]\end{pmatrix};\\\
\text{ if }x_{1}\leq 0,x_{2}\geq 0\\\ \begin{pmatrix}[4x_{1}+2x_{2}+2\vee
8x_{1}+3x_{2}+1]\\\ [2x_{1}+2x_{2}-3\vee 3x_{1}+4x_{2}-1]\end{pmatrix};\\\
\text{ if }x_{1}\leq 0,x_{2}\leq 0\\\ \begin{pmatrix}[4x_{1}+3x_{2}+1\vee
8x_{1}+2x_{2}+2]\\\ [3x_{1}+2x_{2}-3\vee 2x_{1}+4x_{2}-1]\end{pmatrix};\\\
\text{ if }x_{1}\geq 0,x_{2}\leq 0\\\ \end{cases}$

Selection of descent direction at $(1,-1)$.
Here $\nabla\breve{F}(1,-1)=\begin{pmatrix}[2,8]&[-3,-2]\end{pmatrix}$.
$\overline{d_{1}}=-8+6t_{1}$, $\overline{d_{2}}=2+t_{2}$, with
$t_{1},t_{2}\in[0,1]$. Choose $t_{1}=\dfrac{5}{6}$ and $t_{2}=0$. Therefore
$\overline{d}=\begin{pmatrix}-3\\\ 2\end{pmatrix}$. Then
$\sum_{i=1}^{2}\overline{d}_{i}\odot\breve{g}_{i}(1.-1)=[-30,-10]\prec\breve{0}$.
From Theorem 3.2, $\overline{d}$ is a descent direction at $(1,-1)$.

In this example a descent sequence of intervals $\\{\breve{F}(x^{k})\\}$ is
generated starting with randomly chosen initial point. At every iteration,
$\alpha_{k}$ is selected according to (6). $\overline{d}^{k}$ is determined
using Remark 2. All the iterations are summarized in the following table.

Table 1: Iterations with $x^{0}=(1,1)^{\prime}$ $k$ | $x^{k}$ | $\breve{F}(x^{k})$ | $\overline{d}^{k}$ | $\alpha_{k}$
---|---|---|---|---
$0$ | $(1.00000,1.00000)$ | $[5.00000,8.00000]$ | $(-13.00000,-4.00000)$ | $0.10703$
$1$ | $(-0.39141,0.57188)$ | $[-1.90307,-0.77751]$ | $(-0.58438,1.88672)$ | $0.37676$
$2$ | $(-0.61157,1.28271)$ | $[-2.63791,-1.06984]$ | $(-1.11913,-0.34227)$ | $0.09160$
$3$ | $(-0.71409,1.25136)$ | $[-2.69151,-1.16687]$ | $(-0.64637,-0.00000)$ | $0.00801$
$4$ | $(-0.71926,1.25136)$ | $[-2.69162,-1.17016]$ | $(-0.00000,-0.00000)$ | $-$

From the Table 1, one may observe that
$\breve{F}(x^{4})\preceq\breve{F}(x^{3})\preceq\breve{F}(x^{2})\preceq\breve{F}(x^{1})\preceq\breve{F}(x^{0})$.
The iterative process terminates at $x^{4}$ since
$\parallel\overline{d}^{k}\parallel<\epsilon$ where $\epsilon=10^{-5}$.

Justification of critical point:
To justify $x^{4}=(-0.71926,1.25136)$, generated in Table 1 as a critical
point, from Definition 5, it is enough to show that
$\sum_{i=1}^{2}d_{i}\odot\breve{g}_{i}(x^{4})\prec\breve{0}$ has no solution
for any non zero $d\in\mathbf{R}^{2}$.
$\breve{g}(x^{4})=\begin{pmatrix}[-2.25136,2.87704]&[-0.65506,0.56692]\end{pmatrix}$.
Therefore the interval inequality will be

$d_{1}\odot[-2.25136,2.87704]\oplus
d_{2}\odot[-0.65506,0.56692]\prec\breve{0}.$

The above inequality can be reduced to two different system of inequalities
(in real form) for different signs of $d_{1}$ and $d_{2}$.
Case 1: $d_{1}$ and $d_{2}$ are of same sign.
Therefore either $d_{1}\geq 0$, $d_{2}\geq 0$ or $d_{1}\leq 0$, $d_{2}\leq 0$
with $(d_{1},d_{2})\neq(0,0)$. In that case the above interval inequality
reduces to the following system of inequalities.

$\displaystyle-2.25136d_{1}-0.65506d_{2}<0$ $\displaystyle
2.87704d_{1}+0.56692d_{2}<0$

For $d_{1}\geq 0$, $d_{2}\geq 0$, the first inequality holds but the second
one doesn’t hold.
The second inequality satisfies the condition $d_{1}\leq 0$, $d_{2}\leq 0$ but
the first one does not. Thus this system has no solution.

Case 2: $d_{1}$ and $d_{2}$ are of opposite sign.
Therefore either $d_{1}\geq 0$, $d_{2}\leq 0$ or $d_{1}\leq 0$, $d_{2}\geq 0$
with $(d_{1},d_{2})\neq(0,0)$. In that case the above interval inequality
reduces to the following system of inequalities.

$\displaystyle 2.87704d_{1}-0.65506d_{2}<0$
$\displaystyle-2.25136d_{1}+0.56692d_{2}<0$

Here the first inequality holds for $d_{1}\leq 0$, $d_{2}\geq 0$ but the
second inequality doesn’t hold for the same condition. Similarly in case of
$d_{1}\geq 0$, $d_{2}\leq 0$, second inequality holds but the second
inequality cannot hold. Hence the above system has no solution.

From the above cases we can conclude that $\nexists$ non zero
$d\in\mathbf{R}^{2}$ such that
$\sum_{i=1}^{2}d_{i}\odot\breve{g}_{i}(x^{4})\prec\breve{0}$. Hence $x^{4}$ is
a critical point of $\breve{F}$.

Since $\breve{F}(x)$ is a set valued mapping so critical point of
$\breve{F}(x)$ is not unique. Starting with different initial points one may
have different critical points.

##  5 Conclusion and future scope

The set of intervals is partially ordered. So it is not always easy to create
a descent sequence for any general interval valued function. Objective of this
article is to develop an iterative process to construct a descent sequence of
intervals which provides a critical point of a function with bounded
parameters as intervals. Some natural questions raise after the theoretical
discussions of this article. Though $\\{\breve{F}(x^{k})\\}$ is a descent
sequence of intervals, convergence of $\\{x^{k}\\}$ to the critical point can
be guaranteed only with several assumptions on $\breve{F}$. This part is not
studied here and kept for future research. The article focuses only on the
iterative process. Exact line search technique is used to decide the step
length. Determination of suitable step length satisfying conditions (6) is
cumbersome for complex functions. There are several inexact line search
methods for selection of step length, which may be used to justify the
convergence, which remains the scope of the present contribution.

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