In the following, we pay more attention to generate a range reduction method for reducing the investigated outcome space region of the denominators in which there does not contain the global optimal solution for the problem (SAR), and to use this method as an accelerating tool for accelerating the computational speed of the proposed outcome space algorithm for the problem (SAR).

By the assumption that the denominator *g*_{i}(*x*) ≠ 0 for ∀ *x* ∊ Λ, and the continuity of fractional function $\frac{f{i}_{}\left(x\right)}{g{i}_{}\left(x\right)}$, we can get that *g*_{i}(*x*) > 0 or *g*_{i}(*x*) < 0. Therefore, without loss of generality, we can always assume that *g*_{i}(*x*) > 0, *i* = 1, 2, . . ., *T*; *g*_{i}(*x*) < 0, *i* = *T* + 1, *T* + 2, ..., *p*. Besides, if *f*_{i}(*x*) < 0 for some *i* ∊ {1, 2, . . ., *p*}, by using the technique proposed in [20], *f*_{i}(*x*) ≥ 0 always can be satisfied. Thus, without loss of generality, we can always suppose that *f*_{i}(*x*) ≥ 0, and set

$${L}_{i}^{0}=\underset{x\text{\hspace{0.17em}}\in \text{\hspace{0.17em}}\wedge}{\mathrm{min}}{g}_{i}\left(x\right),{U}_{i}^{0}=\underset{x\text{\hspace{0.17em}}\in \text{\hspace{0.17em}}\wedge}{\mathrm{max}}{g}_{i}\left(x\right),i=1\text{,}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,p,$$

and denote the initial rectangle

$${Y}^{0}=\left\{y\in {R}^{p}|{L}_{i}^{0}\le {y}_{i}\le {U}_{i}^{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}p\right\},$$

then the problem (SAR) can be transformed into the equivalent problem (Q) as follows.

$$Q\left(Y{0}^{}\right):\{\begin{array}{ll}\mathrm{max}\hfill & {H}_{0}\left(x,y\right)={\displaystyle \sum _{i=1}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}\hfill \\ s.t.\hfill & \begin{array}{l}{H}_{i}\left(x,y\right)={y}_{i}-{g}_{i}\left(x\right)\ge 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}T,\\ {H}_{i}\left(x,y\right)={y}_{i}-{g}_{i}\left(x\right)\le 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=T+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}T+2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}p,\\ x\in \wedge ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\in {Y}^{0}.\end{array}\hfill \end{array}$$

The key equivalence results for the problem (SAR) and the *Q*(*Y*^{0}) are discussed in the following theorem.

**Theorem 2.1:** *If* (*x** , *y**) *is a global optimum point of the problem Q*(*Y*^{0}), *then* *x** *is also a global optimum point of the problem (SAR). Conversely, if x** *is a global optimum point of the problem (SAR), then* (*x**, *y**) *is a global optimum point of the problem **Q*(*Y*^{0}), where y* = *g*_{i}(*x**), *i* = 1, 2, ..., *p*.

**Proof:** *The conclusion is obvious, therefore, it is omitted.□*

By Theorem 2.1, in order to globally solve the problem (SAR), we may globally solve the problem *Q*(*Y*^{0}) instead.

For each rectangle ${Y}^{k}=\left\{y\in {R}^{p}|{L}_{i}^{k}\le {y}_{i}\le {U}_{i}^{k},i=1,2,\dots ,p\right\}\subseteq {Y}^{0}$, the notations and functions of this paper are introduced as follows:

$$\begin{array}{l}{f}_{i}\left(x\right)={\displaystyle \sum _{j=1}^{n}{c}_{ij}{x}_{j}+{e}_{i};}\hfill \\ {H}_{0}^{U}\left(x\right)={\displaystyle \sum _{i=1}^{p}(}{\displaystyle \sum _{j=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{ij}>0}^{n}\frac{{c}_{ij}}{{L}_{i}^{k}}{x}_{j}}+{\displaystyle \sum _{j=1,{c}_{ij}<0}^{n}\frac{{c}_{ij}}{{U}_{i}^{k}}{x}_{j})+{\displaystyle \sum _{i=1,{e}_{i}>0}^{p}\frac{{e}_{i}}{{L}_{i}^{k}}+{\displaystyle \sum _{i=1,{e}_{i}<0}^{p}\frac{{e}_{i}}{{U}_{i}^{k}};}}}\hfill \\ {H}_{0}^{L}\left(x\right)={\displaystyle \sum _{i=1}^{p}(}{\displaystyle \sum _{j=1,{c}_{ij}>0}^{n}\frac{{c}_{ij}}{{U}_{i}^{k}}{x}_{j}+{\displaystyle \sum _{j=1,{c}_{ij}<0}^{n}\frac{{c}_{ij}}{{L}_{i}^{k}}{x}_{j})+{\displaystyle \sum _{i=1,{e}_{i}>0}^{p}\frac{{e}_{i}}{{U}_{i}^{k}}+{\displaystyle \sum _{i=1,{e}_{i}<0}^{p}\frac{{i}_{}}{{L}_{i}^{k}};}}}}\hfill \\ \begin{array}{l}{H}_{i}^{U}\left(x\right)={U}_{i}^{k}-{g}_{i}\left(x\right),{H}_{i}^{L}\left(x\right)={L}_{i}^{k}-{g}_{i}\left(x\right),i=1,2,\dots ,T;\\ {H}_{i}^{L}\left(x\right)={L}_{i}^{k}-{g}_{i}\left(x\right),{H}_{i}^{U}\left(x\right)={U}_{i}^{k}-{g}_{i}\left(x\right),i=T+1,T+2,\dots ,p.\end{array}\hfill \end{array}$$

**Theorem 2.2:** *Consider the functions* ${H}_{i}^{U}\left(x\right),{H}_{i}^{L}\left(x\right),{H}_{i}\left(x,y\right)$ *for any* *x* ∊ Λ, *y* ∊ *Y*^{k} ⊆ *Y*^{0}, *where* *i* = 0, 1, 2, ..., *p*. *Then the following two statements are valid*.(i) ${H}_{i}^{U}\left(x\right)\ge {H}_{i}\left(x,y\right)\ge {H}_{i}^{L}\left(x\right),i=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}p.$.(ii) ${\mathrm{lim}}_{\left|\right|}{}_{{U}^{k}-{L}^{k}\left|\right|\to 0}\left[{H}_{i}^{U}\left(x\right)-{H}_{i}\left(x,y\right)\right]={\mathrm{lim}}_{\left|\right|{U}^{k}-{L}^{k}\left|\right|\to 0}\left[{H}_{i}\left(x,y\right)-{H}_{i}^{L}\left(x\right)\right]=0.$.

**Proof:** *The proof can be easily given, here it is omitted.□*

Based on the Theorem 2.2, for any *Y*^{k} ⊆ *Y*^{0}, we can establish the corresponding linear relaxation program problem LRP(*Y*^{k}) of the *Q*(*Y*^{k}) as follows, which can provide a reliable upper bound for the global optimum value of the problem *Q*(*Y*^{k}).

$$LRP\left({Y}^{k}\right):\{\begin{array}{ll}\mathrm{max}\hfill & {H}_{0}^{U}\left(x\right)={\displaystyle \sum _{i=1}^{p}(}{\displaystyle \sum _{j=1,{c}_{ij}>0}^{n}\frac{{c}_{ij}}{{L}_{i}^{k}}{x}_{j}}+{\displaystyle \sum _{j=1,{c}_{ij}<0}^{n}\frac{{c}_{ij}}{{U}_{i}^{k}}{x}_{j}})+{\displaystyle \sum _{i=1,{e}_{i}>0}^{p}\frac{{e}_{i}}{{L}_{i}^{k}}+{\displaystyle \sum _{i=1,{e}_{i}<0}^{p}\frac{{e}_{i}}{{U}_{i}^{k}}}}\hfill \\ \begin{array}{c}\text{s}\text{.t}\text{.}\\ \\ \end{array}\hfill & \begin{array}{l}{H}_{i}^{U}\left(x\right)={U}_{i}^{k}-{g}_{i}\left(x\right)\ge 0,i=1,2,\dots ,T,\\ {H}_{i}^{L}\left(x\right)={L}_{i}^{k}-{g}_{i}\left(x\right)\le 0,i=T+1,T+2,\dots ,p,\\ x\in \wedge .\end{array}\hfill \end{array}$$

For any rectangle *Y*^{k} ⊆ *Y*^{0}, in the process of algorithm iteration and still of interest, we want to recognize whether or not *Y*^{k} contains global optimal solution of the *Q*(*Y*^{0}). The proposed new outcome space range reduction method aims at replacing the rectangle *Y*^{k} = [*L*^{k}, *U*^{k}] with a smaller rectangle $\ddot{Y}=\left[\ddot{L},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ddot{U}\right]$ without deleting any global optimal solution of the *Q*(*Y*^{0}).

In the following, we suppose without loss of generality that *LB* is a currently known lower bound of the global optimal value for the *Q*(*Y*^{0}) at the iteration k, and that *v*(*Y*^{k}) is the maximum value of the *H*_{0}(*x*, *y*) in *Y*^{k} and Λ, and set

$$\begin{array}{l}{l}_{i}^{0}=\underset{x\in \wedge}{\mathrm{min}}{f}_{i}\left(x\right),{u}_{i}^{0}=\underset{x\in \wedge}{\mathrm{max}}{f}_{i}\left(x\right),i=1,2,\dots ,p,\\ {\overline{UB}}^{k}={\displaystyle \sum _{i=1}^{T}\frac{{u}_{i}^{0}}{{L}_{i}^{k}}+{\displaystyle \sum _{i=T+1}^{p}\frac{{l}_{i}^{0}}{{L}_{i}^{k}}},}\\ {\beta}_{i}^{k}=\{\begin{array}{l}\frac{{u}_{i}^{0}}{\underset{\_}{LB}-{\overline{UB}}^{k}+\frac{{u}_{i}^{0}}{{L}_{i}^{k}}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\dots ,T,\hfill \\ \frac{{l}_{i}^{0}}{\underset{\_}{LB}-{\overline{UB}}^{k}+\frac{{l}_{i}^{0}}{{L}_{i}^{k}}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=T+1,T+2,\dots p.\hfill \end{array}\end{array}$$

**Theorem 2.3:** *For any sub-rectangle* ${Y}^{k}={\left({Y}_{i}^{k}\right)}_{p\times 1}={\left[{L}_{i}^{k},{U}_{i}^{k}\right]}_{p\times 1}\subseteq {Y}^{0}$*, we have the following conclusions*:

(i) *If* ${\overline{UB}}^{k}<\underset{\_}{LB}$*, then there exists no global optimal solution of the Q*(*Y*^{0}) *in Y*^{k}.

(ii) *If* ${\overline{UB}}^{k}\ge \underset{\_}{LB}$*, then, for each s ∊* {1, 2, ..., *p*}*, there exists no global optimal solution of the Q*(*Y*^{0}) in ${\underset{\_}{Y}}^{k}$*, where*

$${\underset{\_}{Y}}^{k}={\left({\underset{\_}{Y}}_{i}^{k}\right)}_{p\times 1}\subseteq {Y}^{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}with\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\underset{\_}{Y}}_{i}^{k}=\{\begin{array}{l}{Y}_{i}^{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i\ne s,i=1,2,\dots ,p,\hfill \\ ({\beta}_{i}^{k},{U}_{i}^{k}]\cap {Y}_{i}^{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=s\in \left\{1,2,\dots ,p\right\},\hfill \end{array}$$

**Proof:** *(i) **If* ${\overline{UB}}^{k}<\underset{\_}{LB}$, then we have

$$\upsilon \left({Y}^{k}\right)=\underset{y\in {Y}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=1}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}\le {\displaystyle \sum _{i=1}^{T}\frac{{u}_{i}^{0}}{{L}_{i}^{k}}+{\displaystyle \sum _{i=T+1}^{p}\frac{{l}_{i}^{0}}{{L}_{i}^{k}}}={\overline{UB}}^{k}<\underset{\_}{LB}},$$

therefore, there exists no global optimal solution of the problem *Q*(*Y*^{0}) in *Y*^{k}.

(ii) If ${\overline{UB}}^{k}\ge \underset{\_}{LB}$, then we can get the following results.

For any *s* ∊ {1, 2, ..., *T*}, for ∀ *x* ∊ Λ and $y\in {\underset{\_}{Y}}^{k}$, since $0\le {l}_{i}^{0}\le {f}_{i}\left(x\right)\le {u}_{i}^{0},i=1,2,\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}p;\text{\hspace{0.17em}}0\le {L}_{i}^{k}\le \text{\hspace{0.17em}}{y}_{i}\le {U}_{i}^{k},i=1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}p,i\ne s;0\le {\beta}_{s}^{k}<{y}_{s}\le {U}_{s}^{k};$; we can follow that

$$\begin{array}{ll}\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=1}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}\hfill & \le \underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=1,i\ne s}^{T}\frac{{f}_{i}\left(x\right)}{{y}_{i}}+\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}\text{\hspace{0.17em}}\frac{{f}_{s}\left(x\right)}{{y}_{s}}+\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=T+1}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}}\hfill \\ \hfill & <{\displaystyle \sum _{i=1,i\ne s}^{T}\frac{{u}_{i}^{0}}{{L}_{i}^{k}}+\frac{{u}_{s}^{0}}{{\beta}_{s}^{k}}+{\displaystyle \sum _{i=T+1}^{p}\frac{{l}_{i}^{0}}{{L}_{i}^{k}}}}\hfill \\ \hfill & ={\overline{UB}}^{k}-\frac{{u}_{s}^{0}}{{L}_{s}^{k}}+\frac{{u}_{s}^{0}}{{\beta}_{s}^{k}}\hfill \\ \hfill & =\underset{\_}{LB}.\hfill \end{array}$$

Therefore, there exists no any global optimization solution of the *Q*(*Y*^{0}) in ${\underset{\_}{Y}}^{k}$.

Using the same proving method, for any *s* ∊ {*T* + 1, *T* + 2, ..., *p*}, and for ∀ *x* ∊ Λ and $y\in {\underset{\_}{Y}}^{k}$, since $0\le {l}_{i}^{0}\le {f}_{i}\left(x\right)\le {u}_{i}^{0},i=1,2,\dots ,p;\text{\hspace{0.17em}}0\le {L}_{i}^{k}\le {y}_{i}\le {U}_{i}^{k},\text{\hspace{0.17em}}i=1,2,\dots ,p,i\ne s;0\le {\beta}_{s}^{k}<{y}_{s}\le {U}_{s}^{k};$; we can follow that

$$\begin{array}{ll}\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=1}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}\hfill & \le \underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=1}^{T}\frac{{f}_{i}\left(x\right)}{{y}_{i}}+\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}{\displaystyle \sum _{i=T+1,i\ne s}^{p}\frac{{f}_{i}\left(x\right)}{{y}_{i}}}+\underset{y\in {\underset{\_}{Y}}^{k},x\in \wedge}{\mathrm{max}}\frac{{f}_{s}\left(x\right)}{{y}_{s}}}\hfill \\ \hfill & <{\displaystyle \sum _{i=1,i\ne s}^{T}\frac{{u}_{i}^{0}}{{L}_{i}^{k}}+{\displaystyle \sum _{i=T+1,i\ne s}^{p}\frac{{l}_{i}^{0}}{{L}_{i}^{k}}}+\frac{{l}_{s}^{0}}{{\beta}_{s}^{k}}}\hfill \\ \hfill & ={\overline{UB}}^{k}-\frac{{l}_{s}^{0}}{{L}_{s}^{k}}+\frac{{l}_{s}^{0}}{{\beta}_{s}^{k}}\hfill \\ \hfill & =\underset{\_}{LB}.\hfill \end{array}$$

Therefore, there exists no any global optimization solution of the *Q*(*Y*^{0}) over ${\underset{\_}{Y}}^{k}$.□

By the Theorem 2.3, we can construct the new range reduction method to cut away a part of outcome space region of the denominators in which the global optimal solution of the *Q*(*Y*^{0}) does not exist. Assume that a sub-rectangle

$${Y}^{k}={(}_{{Y}_{i}^{k}}\subseteq {Y}^{0}with{Y}_{i}^{k}=\left[{L}_{i}^{k},{U}_{i}^{k}\right]$$

will be reduced or deleted, then according to the Theorem 2.3, the checked rectangle *Y*^{k} should be replaced by the new sub-rectangle

$$\stackrel{\mathrm{..}}{Y}={(}_{{\stackrel{\mathrm{..}}{Y}}_{i}}with{\stackrel{\mathrm{..}}{Y}}_{i}=\left[{L}_{i}^{k},{\beta}_{i}^{k}\right]{\displaystyle \cap {Y}_{i}^{k},i\in \{1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots ,p\}.}$$

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