1 Introduction

One of the most encountered problems in our life are optimization problem (mathematical programming problem). Translating this problem into mathematical language, give us an objective function. Optimization problems are called according to objective functions. For example, an interval programming problem (interval-valued optimization problem) arises when the objective function is the interval-valued function. Interval programming problems are a special form of the set-valued programming problems because the interval-valued function is a special case of the set-valued map. Also, interval programming problems are a generalization of scalar programming problems.

The purpose of a mathematical programming problem is to find the best among the options. Naturally, it has been attracted the attention of scientists, who have been working in mathematics, engineering, economics, and many other disciplines, for many years. There are many methods to solve programming problems as scalarization [5, 14, 25], vectorization [8, 12, 16, 17], derivative [13], subdifferential [6, 10, 11], duality [21, 24] etc.

Scalarization is an effective tool for solving mathematical programming problems by using scalar-valued functions. This method replaces a mathematical programming problem with a scalar programming problem, which can be solved by using known methods. Solutions of scalar programming problems obtained via these methods are also solution of mathematical programming problems. Xu and Li [25], Karaman et al. [14], and Hernández and Rodríguez-Marín [5] have obtained optimality criteria for set-valued programming problems with the aid of scalarization.

Vectorization is frequently used to get solutions and optimality conditions of set and vector programming problems. Vectorization reduces a programming problem to a vector programming problem as well as scalarization. In the vectorization, mathematical programming problems are characterized by vector programming problems by using vector-valued functions. The first vectorization is given by Küçük et al. [16, 17]. Later, Jahn [8] and Karaman et al. [12] obtained vectorizations for set-valued programming problems with respect to (shortly, wrt) the set approach.

Order relation is required in order to obtain the solution of set-valued programming and interval programming problems. In the interval programming problems, Ishibuchi and Tanaka [7], Costa et al. [4], Moore [18] and references therein are used some order relations via ordinary order on \(\mathbb R\). In the interval analysis, many authors have presented valuable studies to get solutions and optimality criteria for interval programming problems [1,2,3,4, 7, 9, 15, 18,19,20, 23]. To maximize the interval objective function, the order relations which represent the decision maker’s preference between interval profits are defined in [7]. Interval-valued function is defined in the parametric form and its properties are studied in [1]. Costa et al. [4] present a method for endowing the generalized interval space with some different structures, such as vector spaces, order relations and an algebraic calculus. Karush–Kuhn–Tucker type optimality conditions for optimization problems with interval-valued objective function are given by Chalco-Cano et al. in [3]. Interval-valued optimization problems are examined in which both objective and constraint functions are non-differentiable in [2].

The aim of this paper is to obtain the optimality criteria for interval programming problems. Scalarization, vectorization and some basic calculus rule are used to achieve this idea.

The layout of this manuscript is as tracks: In Sect. 2, some basic notations, definitions, and solution concepts are recalled in the interval and vector programming problems. In Sect. 3, a real-life example, which includes investment for an investor, for interval programming problem is given. Some necessary and sufficient optimality criteria are obtained in the last section.

2 Preliminaries

In this section, primal notations, definitions, and solution concepts are overviewed in the interval and vector programming problems.

2.1 Interval Numbers and Arithmetics

Throughout this work, a bounded and closed interval A on \(\mathbb R\) is defined as:

$$\begin{aligned} A=[a_1,a_2]:=\{x\in \mathbb R\ | \ a_1\le x\le a_2\}=\left\{ x\in \mathbb R\ | \ \left| x-\frac{a_1+a_2}{2}\right| \le \frac{a_2-a_1}{2}\right\} \end{aligned}$$

where \(a_1\) and \(a_2\) are called left and right end points of the interval A, respectively. Each real number can be defined by using interval as \(a=[a,a]\) for \(a\in \mathbb R\). If \(a_1=a_2\), then the interval A is called degenerate and it equals to single point \(a_1\) or \(a_2\). In this work, the family of closed and bounded intervals in \(\mathbb R\) are denoted by \(\mathcal {I}\).

Let \(A=[a_1,a_2]\), \(B=[b_1,b_2]\in \mathcal {I}\) and \(\lambda \in \mathbb R\). The sum of two intervals A and B is defined by:

$$\begin{aligned} A+B:=\{x+y\in \mathbb R\ | \ x\in A \text { and } \ y\in B\}=[a_1+b_1,a_2+b_2], \end{aligned}$$

the difference of two interval A and B is given as:

$$\begin{aligned} A-B:=\{x-y\in \mathbb R\ | \ x\in A \text { and } \ y\in B\}=[a_1-b_2,a_2-b_1] \end{aligned}$$

and the scalar multiplication is given by:

$$\begin{aligned} \lambda A:=\left\{ \begin{array}{ll} {[}\lambda a_1,\lambda a_2] &{} ;\lambda \ge 0 \\ {[}\lambda a_2, \lambda a_1], &{} \lambda <0. \end{array} \right. \end{aligned}$$

Now we give some terms used in the interval analysis. Let \(A=[a_1,a_2]\in \mathcal {I}\). Then, the radio and the midpoint of the interval A are defined by \(w(A):=\frac{1}{2}(a_2-a_1)\) and \(m(A):=\frac{1}{2}(a_1+a_2)\), respectively. Therefore, using these notations an interval can be described by using its midpoint-radio representation as

$$\begin{aligned} \begin{array}{ll} A=[a_1,a_2] &{} =[m(A)-w(A),m(A)+w(A)] \\ &{} =m(A)+[-w(A),w(A)] \\ &{} =m(A)+w(A)\left[ -1,1\right] \\ &{} =\{x\in \mathbb R\ | \ \left| x-m(A)\right| \le w(A)\}. \end{array} \end{aligned}$$

Let \(A,B,C\in \mathcal {I}\). Then, intervals have the following properties: \(A+B=B+A\), \(A+(B+C)=(A+B)+C\), \(0+A=A+0=A\) where \(0=[0,0]\), \(1A=A\), \(A-A=A+(-A)\) and \(C+B=A+B\Longrightarrow C=A\) (Cancellation law). \(A+(-A)=0\) if and only if \(a_1=a_2\). Because there is not the additive inverse of an interval, the set \(\mathcal {I}\) isn’t a linear vector space.

Let \(A=[a_1,a_2]\), \(B=[b_1,b_2]\in \mathcal {I}\). A equals to B if and only if \(a_1=b_1\) and \(a_2=b_2\). If \(a_1=-a_2\) then A is called symmetric. Midpoints of symmetric intervals are 0. If \(a_1\le 0\le a_2\), then positive and negative parts of A are defined as \(A^+:=[0,a_2]\) and \(A^-:=[a_1,0]\), respectively. Then \(A=A^++A^-\).

2.2 Interval Programming Problems

Let \(X\ne \emptyset \) and \(F:X\rightarrow \mathcal {I}\) be an interval-valued function, that is \(F(x)=[f_L(x),f_U(x)]\) for all \(x\in X\), where the scalar-valued functions \(f_L, f_U:X\rightarrow \mathbb R\) are such that \(f_L(x)\le f_U(x)\) for all \(x\in X\). \(f_L\) and \(f_U\) are called the left and right end points functions of F, respectively. Also, if \(F(x)=[f_L(x),f_U(x)]\) is a symmetric interval for all \(x\in X\), then the interval-valued function is called symmetric valued on X.

Primal interval programming problem is defined as:

$$\begin{aligned} ({\hbox {IVOP}}) \left\{ \begin{array}{ll} \max F(x) &{} \\ {\hbox {s.t.}} \ x\in X &{} \end{array} \right. \end{aligned}$$

where the nonempty set X is called the constraint set.

In order to solve this interval programming problem, we need an order relation, which is defined on interval numbers. There are some order relations in the literature. In this study, we will use the following order relation defined as: For \(A=[a_1,a_2],B=[b_1,b_2]\in \mathcal {I}\)

$$\begin{aligned} A\le _{mw}B :\Longleftrightarrow m(A)\le m(B) \text { and }w(A)\ge w(B). \end{aligned}$$

This partial order relation is defined by Ishibuchi and Tanaka [7] and used for determine the maximization problems. They say that “Since the center and the radio of the interval can be considered as the expected value and the uncertainty of an interval respectively, this order relation represents the decision maker’s preference for the alternative with the higher expected value and less uncertainty”.

Note that \(\le _{mw}\) is compatible with addition and nonnegative scalar multiplication. \(A\le _{mw}B\) implies \(A-B\le _{mw}0\) for all \(A,B\in \mathcal {I}\). But the other direction may not be true always. For example, \(A-B\le _{mw}0\) and \(A\not \le _{mw}B\) for \(A=[-1,1]\) and \(B=[0,3]\).

Now, we define the strict version of \(\le _{mw}\) to obtain the weak solution of (IVOP) as follow: For \(A=[a_1,a_2],B=[b_1,b_2]\in \mathcal {I}\)

$$\begin{aligned} A<_{mw}B :\Longleftrightarrow m(A)<m(B) \text { and }w(A)>w(B). \end{aligned}$$

Because the order relation \(\le _{mw}\) is a partial (reflexive, transitive and antisymmetric) order, we use the following definition to determine the efficient elements.

Definition 1

Let \(S\subset \mathcal {I}\) and \(A\in S\). Then, we say that

  1. (i)

    A is a maximal interval of S if there is not any \(B\in S\) such that \(A\le _{mw}B\) and \(A\ne B\),

  2. (ii)

    A is a weak maximal interval of S if there is not any \(B\in S\) such that \(A<_{mw}B\),

  3. (iii)

    A is a strongly maximal interval of S if \(B\le _{mw}A\) for all \(B\in S\).

Let \(\mathcal {F}(X)\) be the image family of the interval-valued function F, that is \(\mathcal {F}(X):=\{F(x) \ | \ x\in X\}\). We say that \(x_0\) is a solution of (IVOP) if \(F(x_0)\) is a maximal interval of \(\mathcal {F}(X)\). Similarly, we say that \(x_0\) is a weak solution of (IVOP) if \(F(x_0)\) is a weak maximal interval of \(\mathcal {F}(X)\). Also, \(x_0\) is a strongly solution of (IVOP) if \(F(x_0)\) is a strongly maximal interval of \(\mathcal {F}(X)\).

Note that every solution of a (IVOP) is also a weak solution of the problem and every strongly solution of (IVOP) is also a solution of the problem. Moreover, when we consider \(-F\) instead of F and calculate (IVOP), we obtain minimal solution of (IVOP). Namely, all minimal interval programming problems are expressed by a maximum interval programming problem.

2.3 Vector Programming Problems on \(\mathbf {\mathbb R^2}\)

Now, we recall some basic definitions and notations on vector programming.

Nonnegative and positive orthants in \(\mathbb R^2\) are denoted and defined by \(\mathbb R^2_+:=\{(x_1,x_2)^\mathrm{T} \ | \ x_1\ge 0\text { and }x_2\ge 0\}\) and \(\mathbb R^2_{++}:=\{(x_1,x_2)^\mathrm{T} \ | \ x_1>0\text { and }x_2>0\}\), respectively. It is known that \(\mathbb R^2_+\) induces the following ordering relations on \(\mathbb R^2\): For \(a,b\in \mathbb R^2\)

$$\begin{aligned} \begin{array}{ll} a\preceq b :\Longleftrightarrow &{} b-a\in \mathbb R^2_+\\ a\prec b :\Longleftrightarrow &{} b-a\in \mathbb R^2_{++}. \end{array} \end{aligned}$$

Because \(\mathbb R^2_+\) is convex and pointed cone, the order relation \(\preceq \) is a partial order relation.

Let \(A\subset \mathbb R^2\) and \(a_0\in A\). \(a_0\) is a maximal vector of A wrt \(\preceq \) if \(A\cap (a_0+\mathbb R^2_+)=\{a_0\}\). Similarly, \(a_0\) is a weak maximal vector of A wrt \(\prec \) if \(A\cap (a_0+\mathbb R^2_{++})=\emptyset \).

Basic vector programming problem in \(\mathbb R^2\) is given by

$$\begin{aligned} (VOP) \left\{ \begin{array}{l} \max f(x) \\ {\hbox {s.t.}} \ x\in X \end{array} \right. \end{aligned}$$

where \(f:X\rightarrow \mathbb R^2\) is a vector-valued function and X is the constraint set.

Definition 2

An element \(x_0\in X\) is called

  1. (i)

    the solution of (VOP) wrt \(\preceq \) iff there isn’t any \(x\in X{\setminus }\{x_0\}\) such that \(f(x_0)\preceq f(x)\),

  2. (ii)

    the weak solution of (VOP) wrt \(\prec \) iff there isn’t any \(x\in X\) such that \(f(x_0)\prec f(x)\),

  3. (iii)

    the strongly solution of (VOP) wrt \(\preceq \) iff \(f(x)\preceq f(x_0) \text { for all } x\in X\).

Note that every strongly solution of (VOP) is also a solution of the problem. Besides, every solution is also a weak solution of the problem.

3 A Real-Life Example

An investor intends to invest 500,000 TL Turkish Liras ( ) in foreign currencies, the potential foreign currencies are USD Dollar ($), EUR Euro (€) and GBP Pound (£). The investor has to use only one investment instrument. Which currency exchange rate should the investor choose to get maximum profit?

In Table 1, buying (buy.) and selling (sel.) prices of USD, EUR, and GBP on the first working days of the months between 01.01.2017 and 01.12.2019 are given. The data was taken from the Central Bank of the Republic of Turkey—TCMB’s website in [22]. When the first day of the month coincides with a holiday, the first working day of the corresponding month is considered.

Table 1 The value of the USD, EUR and GBP in TL between 01.01.2017 and 01.12.2019 in [22]
Full size table

One year later, the interval of currencies are found by using Microsoft Excel Forecast program. The Forecast Function in the Excel is defined as: The equation of forecast is \(a+bx\), where \(a=\bar{y}-b\bar{x}\) and \(b=\dfrac{\sum (x-\bar{x})(y-\bar{y})}{\sum (x-\bar{x})^2}\), where y and x are the sample means average(known y’s) and average (known x’s).

The value of USD, EUR, and GBP in TL in the past and future are shown in Figs. 1, 2 and 3, respectively. In Table 2, Estimation, Lower and Upper Reliability Orders of USD, EUR, and GBP in TL are offered at 01.12.2020.

Let \(F:X\rightarrow \mathcal {I}\) be defined as \(F(x)=\frac{T}{V_x}Z=\frac{T}{V_x}[z_1,z_2]\), where x is the investment tool, T is the amount of invested money, \(V_x\) represents current TL value of an investment at the time for buying, for example \(V_{\$}=3.526\) for dollar at 01.01.2017, interval-valued number \(Z=[z_1,z_2]\), which is obtained by using the excel forecast program, is the interval of the investment instrument in TL at the end of period, \(z_1\) is the Lower Reliability Order of the investment at the end of the period and \(z_2\) is the Upper Reliability Order of the investment at the end of the period. Take into account the following interval programming problem:

(1)

Then, \(F(\$)=\dfrac{500{,}000}{3.526}[4.596,8.896]\cong [651 \ 730, 1 \ 261 \ 486]\), and \(F(\text {\pounds }){=}\dfrac{500{,}000}{4{.}341}[3{.}038{,}13{.}294]\cong [349 \ 919, 1 \ 531 \ 214]\).

Maximal of the set is \(F(\$)\) since . So, $ is the solution of the interval programming problem (1). That is, if the investor chooses the dollar, then he/she gets the most profit.

4 Optimality Criteria for Interval Programming Problems

In this section, necessary and sufficient optimality criteria are obtained via basic calculus rule, scalarization and vectorization.

Theorem 1

If \(x_0\in X\) is a solution of (IVOP), then \(f_L(x)<f_U(x_0)\) or \(f_L(x)<f_U(x)\) for all \(x\in X{\setminus }\{x_0\}\).

Fig. 1
figure 1

The value of USD in TL in the past and forecast interval

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Fig. 2
figure 2

The value of EUR in TL in the past and forecast interval

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Fig. 3
figure 3

The value of GBP in TL in the past and forecast interval

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Table 2 The estimation, lower and upper reliability orders of USD, EUR and GBP in TL on 01.12.2020
Full size table

Proof

Because \(x_0\) is a solution of (IVOP), we have \(F(x_0)\not \le _{mw}F(x)\) for all \(x\in X{\setminus }\{x_0\}\). Then,

$$\begin{aligned} \frac{1}{2}(f_L(x_0)+f_U(x_0))\not \le \frac{1}{2}(f_L(x)+f_U(x))\text { or }\frac{1}{2}(f_U(x)-f_L(x))\not \le \frac{1}{2}(f_U(x_0)-f_L(x_0)), \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) This gives

$$\begin{aligned} f_L(x_0)+f_U(x_0)-f_L(x)-f_U(x)\not \le 0 \text { or }f_U(x)-f_L(x)-f_U(x_0)+f_L(x_0)\not \le 0, \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) That is,

$$\begin{aligned} f_L(x_0)+f_U(x_0)-f_L(x)-f_U(x)>0 \text { or }f_U(x)-f_L(x)-f_U(x_0)+f_L(x_0)>0, \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) Since \(f_L(x)\le f_U(x)\) for all \(x\in X\), we get

$$\begin{aligned} f_U(x_0)-f_L(x)>0 \text { or }f_U(x)-f_L(x)>0 \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) Hence, \(f_L(x)<f_U(x_0)\) or \(f_L(x)<f_U(x)\) for all \(x\in X{\setminus }\{x_0\}\).

Corollary 1

Let \(x_0\in X\). Then, we have

  1. (i)

    if \(x_0\in X\) is a weak solution of (IVOP), then \(f_L(x)\le f_U(x_0)\) for all \(x\in X\),

  2. (ii)

    if \(x_0\in X\) is a strongly solution of (IVOP), then \(f_L(x)\le f_U(x_0)\) for all \(x\in X\).

Proof

Similar to the proof of Theorem 1.

Theorem 2

Let the interval-valued function F be symmetric valued on X. Then, \(x_0\in X\) is a solution of (IVOP) if and only if \(f_U(x)+f_L(x_0)>0\) for all \(x\in X{\setminus }\{x_0\}\).

Proof

Assume that \(x_0\) is a solution of (IVOP). \(F(x_0)\not \le _{mw}F(x)\) is obtained for all \(x\in X{\setminus }\{x_0\}\). Then, we have

$$\begin{aligned}&\frac{1}{2}(f_L(x_0)+f_U(x_0))\not \le \frac{1}{2}(f_L(x)\nonumber \\&\quad +f_U(x))\text { or }\frac{1}{2}(f_U(x)-f_L(x))\not \le \frac{1}{2}(f_U(x_0)-f_L(x_0)), \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) Hence, we get

$$\begin{aligned} f_L(x_0)+f_U(x_0)-f_L(x)-f_U(x)\not \le 0 \text { or }f_U(x)-f_L(x)-f_U(x_0)+f_L(x_0)\not \le 0, \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) That is,

$$\begin{aligned} f_L(x_0)+f_U(x_0)-f_L(x)-f_U(x)>0 \text { or }f_U(x)-f_L(x)-f_U(x_0)+f_L(x_0)>0, \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) Since F is symmetric valued on X, we obtain

$$\begin{aligned} f_U(x)+f_L(x)<0 \end{aligned}$$
(2)

or

$$\begin{aligned} f_U(x)+f_L(x_0)>0 \end{aligned}$$
(3)

for all \(x\in X{\setminus }\{x_0\}.\) Since F is symmetric valued on X, Eq. (2) is not satisfy for all \(x\in X\). Hence, we have \(f_U(x)+f_L(x_0)>0\) for all \(x\in X{\setminus }\{x_0\}\).

Conversely, let

$$\begin{aligned} f_U(x)+f_L(x_0)>0 \end{aligned}$$
(4)

be satisfied for all \(x\in X{\setminus }\{x_0\}\). Assume that \(x_0\) is not a solution of (IVOP). Then there exists an \(\bar{x}\in X{\setminus }\{x_0\}\) such that \(F(x_0)\le _{mw}F(\bar{x})\). Then, \(f_U(x_0)+f_L(x_0)\le f_U(\bar{x})+f_L(\bar{x})\) and \(f_U(\bar{x})-f_L(\bar{x})\le f_U(x_0)-f_L(x_0)\) for all \(\bar{x}\in X{\setminus }\{x_0\}\). Because F is symmetric valued on X, we obtain \(f_U(\bar{x})+f_L(x_0)\le 0\). This contradicts with (4). Therefore, \(x_0\) is a solution of (IVOP). \(\square \)

Corollary 2

Let the interval-valued function F be symmetric valued on X and \(x_0\in X\). Then,

  1. (i)

    \(x_0\) is a weak solution of (IVOP) if and only if \(f_U(x)+f_L(x_0)\ge 0\) for all \(x\in X\),

  2. (ii)

    \(x_0\) is a strongly solution of (IVOP) if and only if \(f_U(x_0)+f_L(x)\le 0\) for all \(x\in X\).

Proof

Similar to the proof of Theorem 2. \(\square \)

4.1 Scalarization

Interval programming problems, which are symmetric valued on constraint set, are expressed as a scalar programming problem in this subsection.

We can write \(f_L(x_0)>f_L(x)\) or \(f_U(x)>f_U(x_0)\) instead of \(f_U(x)+f_L(x_0)>0\) for all \(x\in X{\setminus }\{x_0\}\) in Theorem 2 because F is symmetric valued on X. Similarly, \(f_U(x)\ge f_U(x_0)\) or \(f_L(x_0)\ge f_L(x)\) can be replaced instead of \(f_U(x)+f_L(x_0)\ge 0\) for all \(x\in X\) in Corollary 2 (i). Furthermore, we can also represent \(f_U(x_0)+f_L(x)\le 0\) with \(f_U(x_0)\le f_U(x)\) or \(f_L(x)\le f_L(x_0)\) for all \(x\in X\) in Corollary 2 (ii).

Scalar programming problems \((\hbox {SOP}_L)_{\max }\), \((\hbox {SOP}_L)_{\min }\), \((\hbox {SOP}_U)_{\max }\) and \((\hbox {SOP}_U)_{\min }\) are defined as:

$$\begin{aligned} \left\{ \begin{array}{l} \max f_L(x) \\ \ {\hbox {s.t.}} \ x\in X, \end{array}\right. \left\{ \begin{array}{l} \min f_L(x) \\ \ {\hbox {s.t.}} \ x\in X, \end{array} \right. \left\{ \begin{array}{l} \max f_U(x) \\ \ {\hbox {s.t.}} \ x\in X, \end{array} \right. \left\{ \begin{array}{l} \min f_U(x) \\ \ {\hbox {s.t.}} \ x\in X, \end{array} \right. \end{aligned}$$

respectively. Relationships between scalar programming problems and interval programming problems are introduced in the following results.

Theorem 3

Let the interval-valued function F be symmetric valued on X and \(x_0\in X\). Then, \(x_0\) is a solution of (IVOP) if and only if \(x_0\) is the unique solution of \((\hbox {SOP}_L)_{\max }\).

Proof

Assume that \(x_0\) is a solution of (IVOP). Then we have \(f_U(x)+f_L(x_0)>0\) for all \(x\in X{\setminus }\{x_0\}\) from Theorem 2. Because F is symmetric valued on X, we have \(f_U(x)=-f_L(x)\) for all \(x\in X\). Then, we get \(f_L(x_0)>f_L(x)\) for all \(x\in X{\setminus }\{x_0\}\). This means \(x_0\) is the unique solution of \((\hbox {SOP}_L)_{\max }\).

Conversely, let \(x_0\) be the unique solution of \((\hbox {SOP}_L)_{\max }\). Then, we have

$$\begin{aligned} f_L(x)<f_L(x_0) \end{aligned}$$
(5)

for all \(x\in X{\setminus }\{x_0\}\). Assume that \(x_0\) is not a solution of (IVOP). Then, there exists an \(\bar{x}\in X\) such that \(F(x_0)\le _{mw}F(\bar{x})\) and \(F(\bar{x})\ne F(x_0)\). So, we get \(f_L(\bar{x})\ge f_L(x_0)\) for all \(\bar{x}\in X{\setminus }\{x_0\}\). This contradicts with (5). Then, \(x_0\) is also solution of (IVOP). \(\square \)

Corollary 3

Let the interval-valued function F be symmetric valued on X and \(x_0\in X\). Then,

  1. (i)

    \(x_0\) is a solution of (IVOP) iff \(x_0\) is the unique solution of \((\hbox {SOP}_U)_{\min }\),

  2. (ii)

    \(x_0\) is a weak solution of (IVOP) iff \(x_0\) is a solution of \((\hbox {SOP}_L)_{\max }\),

  3. (iii)

    \(x_0\) is a weak solution of (IVOP) iff \(x_0\) is a solution of \((\hbox {SOP}_U)_{\min }\),

  4. (iv)

    \(x_0\) is a strongly solution of (IVOP) iff \(x_0\) is a solution of \((\hbox {SOP}_U)_{\max }\),

  5. (v)

    \(x_0\) is a strongly solution of (IVOP) iff \(x_0\) is a solution of \((\hbox {SOP}_L)_{\min }\).

Proof

It can be proved similar to Theorem 3. \(\square \)

4.2 Vectorization

In this subsection, we will define a vector-valued function from \(\mathcal {I}\) to \(\mathbb R^2\) in order to replace (IVOP) with a vector programming problem.

Definition 3

Let \(A=[a_1,a_2]\in \mathcal {I}\). Then vector-valued function \(v:\mathcal {I}\rightarrow \mathbb R^2\) is defined by setting

$$\begin{aligned} v(A)=(a_1+a_2,a_1-a_2)^\mathrm{T}. \end{aligned}$$

In the following proposition, relationships between order relation on intervals and vector-valued function are given.

Proposition 1

Let \(A=[a_1,a_2],B=[b_1,b_2]\in \mathcal {I}\). Then, the following properties hold:

  1. (i)

    \(A\le _{mw}B \Longleftrightarrow v(A)\preceq v(B)\),

  2. (ii)

    \(A<_{mw}B\Longleftrightarrow v(A)\prec v(B)\),

  3. (iii)

    \(A\le _{mw}0\Longleftrightarrow v(A)\preceq (0,0)^\mathrm{T}\),

  4. (iv)

    \(A<_{mw}0\Longleftrightarrow v(A)\prec (0,0)^\mathrm{T}\).

Proof

  1. (i)

    \(\begin{array}{ll} A\le _{mw}B &{} \Leftrightarrow \left\{ \begin{array}{c} a_1+a_2\le b_1+b_2 \\ \text {and} \\ b_2-b_1\le a_2-a_1 \end{array} \right\} \Leftrightarrow \left\{ \begin{array}{c} b_2+b_1-a_1-a_2\ge 0 \\ \text {and} \\ a_2-a_1-b_2+b_1\ge 0 \end{array} \right\} \\ &{} \Leftrightarrow \begin{array}{l} (b_2+b_1-a_1-a_2, a_2-a_1-b_2+b_1)^\mathrm{T}\in \mathbb R^2_+ \end{array} \\ &{} \Leftrightarrow \begin{array}{l} v(B)-v(A)\in \mathbb R^2_+ \end{array} \Leftrightarrow v(A)\preceq v(B). \end{array}\)

  2. (ii-iv)

    It can be proved similar to (i).

In the rest of the study, \((\hbox {VOP}_v)\) vector programming problem is defined as:

$$\begin{aligned} (\hbox {VOP}_v)\left\{ \begin{array}{ll} \min v(F(x)) &{} \\ \ x\in X. &{} \end{array} \right. \end{aligned}$$

Because v(F(x)) is a vector-valued function, to determine the solutions of \((\hbox {VOP}_v)\) it will be used Definition 2.

Relationships between the solution of (IVOP) and solution of \((\hbox {VOP}_v)\) are obtained in the following results.

Theorem 4

\(x_0\in X\) is a solution of (IVOP) if and only if \(x_0\) is a solution of \((\hbox {VOP}_v)\).

Proof

Assume that \(x_0\) is a solution of (IVOP). We have

$$\begin{aligned} F(x_0)\nleqslant _{mw}F(x)\Longleftrightarrow [f_L(x_0),f_U(x_0)] \nleqslant _{mw}[f_L(x),f_U(x)], \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}\). Then,

$$\begin{aligned} f_L(x_0)+f_U(x_0)\nleqslant f_L(x)+f_U(x) \text { or } f_U(x)-f_L(x)\nleqslant f_U(x_0)-f_L(x_0), \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) So,

$$\begin{aligned} f_L(x)+f_U(x)-f_L(x_0)-f_U(x_0)\ngeqslant 0 \text { or }f_U(x_0)-f_L(x_0)-f_U(x)+f_L(x)\ngeqslant 0, \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}.\) That is,

$$\begin{aligned} v(F(x))-v(F(x_0))= & {} (f_L(x)+f_U(x)-f_L(x_0)-f_U(x_0),f_U(x_0)-f_L(x_0)\nonumber \\&-f_U(x)+f_L(x))\not \in \mathbb R^2_+ \end{aligned}$$

for all \(x\in X{\setminus }\{x_0\}\). Therefore, \(v(F(x_0))\npreceq v(F(x))\) for all \(x\in X{\setminus }\{x_0\}\). This gives that \(x_0\) is a solution of \((\hbox {VOP}_v)\). \(\square \)

Corollary 4

Let \(x_0\in X\). Then,

  1. (i)

    \(x_0\) is a weak solution of (IVOP) if and only if it is a weak solution of \((\hbox {VOP}_v)\),

  2. (ii)

    \(x_0\) is a strongly solution of (IVOP) if and only if it is a strongly solution of \((\hbox {VOP}_v)\).

Proof

This can be proven similarly to Theorem 4.

Example 1

Let interval-valued function \(F:\mathbb R\rightarrow \mathcal {I}\) be defined as

$$\begin{aligned} F(x)=\left[ -|x|,|x|\right] \end{aligned}$$

for all \(x\in \mathbb R\). Consider the basic global interval programming problem

$$\begin{aligned} ({\hbox {IVOP}}) \left\{ \begin{array}{ll} \max F(x) &{} \\ {\hbox {s.t.}} \ x\in \mathbb R. &{} \end{array} \right. \end{aligned}$$

It is clear that the solution of the problem is 0, and interval-valued function is symmetric valued on \(\mathbb R\). Then, the necessary conditions of Theorem 1 and Theorem 2 are satisfied.

Let us consider the following scalar programming problem

$$\begin{aligned} (\hbox {SOP}_L)_{\max }\left\{ \begin{array}{l} \max f_L(x) \\ \ {\hbox {s.t.}} \ x\in \mathbb R, \end{array}\right. \end{aligned}$$

where \(f_L(x)=-|x|\) because \(F(x)=\left[ -|x|,|x|\right] \) for all \(x\in \mathbb R\). Because 0 is unique solution of \((\hbox {SOP}_L)_{\max }\), we can say again that 0 is a solution of (IVOP) from Theorem 3.

Let us consider the following vector programming problem

$$\begin{aligned} (\hbox {VOP}_v)\left\{ \begin{array}{l} \min v(F(x))\\ \ x\in \mathbb R, \end{array} \right. \end{aligned}$$

where the vector-valued function \(v:\mathcal {I}\rightarrow \mathbb R^2\) is defined as

$$\begin{aligned} v(F(x))=\left( f_L(x)+f_U(x),f_L(x)-f_U(x)\right) ^\mathrm{T}=\left( 0,-2|x|\right) ^\mathrm{T}, \end{aligned}$$

where \(f_L(x)=-|x|\) and \(f_U(x)=|x|\) because \(F(x)=\left[ -|x|,|x|\right] \) for all \(x\in \mathbb R\). It is clear that 0 is a solution of \((\hbox {VOP}_v)\), from Theorem 4 we can say again that 0 is a solution of (IVOP).

5 Conclusion

In this paper, we study symmetric interval optimization problems, and a real life example is given. Also, necessary and sufficient optimality conditions are obtained for interval programming and symmetric valued interval programming problems. We use the maximal order relation defined by Ishibuchi and Tanaka [7] in this work. Necessary, sufficient and Karush–Kuhn–Tucker type optimality conditions can be obtained for the interval programming problems, which is have the symmetric valued function on the constraint set, given by the other relations in [4, 7, 18].