In this section, we will present a new parametric linearization approach for constructing the parametric linear programming relaxation problem of QP. The detailed deriving process of the parametric linearization approach is given as follows. Without loss of generality, we assume that *Y* = {(*y*_{1}, *y*_{2}, *. . .*, *y*_{n})^{T} ϵ *R*^{n} : *l*_{j} ≤ *y*_{j} ≤ *u*_{j}, *j* = 1, *. . .*, *n*}⊆ *Y*^{0}, = (γ_{jk})_{n}_{×n} ϵ *R*^{n}^{×n} is a symmetric matrix, and _{γjk} ϵ {0, 1}.

For convenience in expression, for any *y* ϵ *Y*, for any *j* ϵ {1, 2, *. . .*, *n*}, *k* ϵ {1, 2, *. . .*, *n*}, *j* ≠ *k*, we define

$$\begin{array}{l}{y}_{k}({\gamma}_{kk})={l}_{k}+{\gamma}_{kk}({u}_{k}-{l}_{k}),\\ {y}_{k}(1-{\gamma}_{kk})={l}_{k}+(1-{\gamma}_{kk})({u}_{k}-{l}_{k}),\\ {f}_{kk}(y)={y}_{k}^{2},\\ {\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})=[{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}({\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})],\\ {\overline{f}}_{kk}(y,Y,{\gamma}_{kk})=[{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}(1-{\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})],\\ {y}_{j}({\gamma}_{jk})={l}_{j}+{\gamma}_{jk}({u}_{j}-{l}_{j}),\\ {y}_{k}({\gamma}_{jk})={l}_{k}+{\gamma}_{jk}({u}_{k}-{l}_{k}),\\ {y}_{j}(1-{\gamma}_{jk})={l}_{j}+(1-{\gamma}_{jk})({u}_{j}-{l}_{j}),\\ {y}_{k}(1-{\gamma}_{jk})={l}_{k}+(1-{\gamma}_{jk})({u}_{k}-{l}_{k}),\\ ({y}_{j}+{y}_{k})({\gamma}_{jk})=({l}_{j}+{l}_{k})+{\gamma}_{jk}({u}_{j}+{u}_{k}-{l}_{j}-{l}_{k}),\\ ({y}_{j}+{y}_{k})(1-{\gamma}_{jk})=({l}_{j}+{l}_{k})+(1-{\gamma}_{jk})({u}_{j}+{u}_{k}-{l}_{j}-{l}_{k}),\\ {f}_{jk}(y)={y}_{j}{y}_{k}=\frac{({y}_{j}+{y}_{k}{)}^{2}-{y}_{j}^{2}-{y}_{k}^{2}}{2},\\ {\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})=\frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\},\\ {\overline{f}}_{jk}(y,Y,{\gamma}_{jk})=\frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\}.\end{array}$$It is obvious that

$${y}_{k}(0)={l}_{k},{y}_{k}(1)={u}_{k},({y}_{j}+{y}_{k})(0)={l}_{j}+{l}_{k},({y}_{j}+{y}_{k})(1)={u}_{j}+{u}_{k}.$$**Theorem 2.1**. *For any k* ϵ {1, 2, *. . .*, *n*}, *for any y* ϵ *Y, then we have*:

$$\begin{array}{rl}& {\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})\le {f}_{kk}(y)\le {\overline{f}}_{kk}(y,Y,{\gamma}_{kk}),\\ & [{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\le {y}_{j}^{2}\le [{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})],\\ & {y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\le {y}_{k}^{2}\le [{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})],\\ & ({y}_{j}+{y}_{k}{)}^{2}\le [({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})],\\ & ({y}_{j}+{y}_{k}{)}^{2}\ge [({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})],\end{array}$$(1)*and*

$${\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})\le {f}_{jk}(y)\le {\overline{f}}_{jk}(y,Y,{\gamma}_{jk}).$$(2)*(ii) The following limitations hold*:

$$\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{kk}(y)-{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})]=0,$$(3)$$\begin{array}{c}\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{kk}(y,Y,{\gamma}_{kk})-{f}_{kk}(y)]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[{y}_{j}^{2}-\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]-{y}_{j}^{2}]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[{y}_{k}^{2}-\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]-{y}_{k}^{2}]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]-({y}_{j}+{y}_{k}{)}^{2}]=0,\\ \underset{\parallel u-l\parallel \to 0}{lim}[({y}_{j}+{y}_{k}{)}^{2}-\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[({y}_{j}+{y}_{k})-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}]=0,\end{array}$$(4)$$\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{jk}(y)-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})]=(0)$$(5)*and*

$$\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{jk}(y,Y,{\gamma}_{jk})-{f}_{jk}(y)]=0.$$(6)*Proof*. (i) By the mean value theorem, for any *y* ϵ *Y*, there exists a point *ξ*_{k} = *αy*_{k} + (1 − *α*)*y*_{k}(γ_{kk}), where *α* ϵ [0, 1], such that

$${y}_{k}^{2}=[{y}_{k}({\gamma}_{kk}){]}^{2}+2{\xi}_{k}[{y}_{k}-{y}_{k}({\gamma}_{kk})].$$If γ_{kk} = 0, then we have

$${\xi}_{k}\ge {l}_{k}={y}_{k}({\gamma}_{kk})\text{and}{y}_{k}-{y}_{k}({\gamma}_{kk})={y}_{k}-{l}_{k}\ge 0.$$If γ_{kk} = 1, then it follows that

$${\xi}_{k}\le {u}_{k}={y}_{k}({\gamma}_{kk})\text{and}{y}_{k}-{y}_{k}({\gamma}_{kk})={y}_{k}-{u}_{k}\le 0.$$Thus, we can get that

$$\begin{array}{lll}{f}_{kk}(y)& =& {y}_{k}^{2}\\ & =& [{y}_{k}({\gamma}_{kk}){]}^{2}+2{\xi}_{k}[{y}_{k}-{y}_{k}({\gamma}_{kk})]\\ & \ge & [{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}({\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})]\\ & =& {\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk}).\end{array}$$Similarly, if γ_{kk} = 0, then we have

$${\xi}_{k}\le {u}_{k}={y}_{k}(1-{\gamma}_{kk})\text{and}{y}_{k}-{y}_{k}({\gamma}_{kk})={y}_{k}-{l}_{k}\ge 0.$$If γ_{kk} = 1, then it follows that

$${\xi}_{k}\ge {l}_{k}={y}_{k}(1-{\gamma}_{kk})\text{and}{y}_{k}-{y}_{k}({\gamma}_{kk})={y}_{k}-{u}_{k}\le 0.$$Thus, we can get that

$$\begin{array}{lll}{f}_{kk}(y)& =& {y}_{k}^{2}\\ & =& [{y}_{k}({\gamma}_{kk}){]}^{2}+2{\xi}_{k}[{y}_{k}-{y}_{k}({\gamma}_{kk})]\\ & \le & [{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}(1-{\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})]\\ & =& {\overline{f}}_{kk}(y,Y,{\gamma}_{kk}).\end{array}$$Therefore, for any *y* ϵ *Y*, we have that

$${\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})\le {f}_{kk}(y)\le {\overline{f}}_{kk}(y,Y,{\gamma}_{kk}).$$From the inequality (1), replacing γ_{kk} by γ_{jk}, and replacing *y*_{k} by *y*_{j}, we can get that

$$[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\le {y}_{j}^{2}\le [{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})].$$From the inequality (1), replacing γ_{kk} by γ_{jk}, we can get that

$$[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\le {y}_{k}^{2}\le [{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})].$$From (1), replacing γ_{kk} and *y*_{k} by γ_{jk} and (*y*_{j} + *y*_{k}), respectively, we can get that

$$\begin{array}{c}({y}_{j}+{y}_{k}{)}^{2}\le [({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[({y}_{j}+{y}_{k})-({y}_{j}+{y}_{k})({\gamma}_{jk})],\\ ({y}_{j}+{y}_{k}{)}^{2}\ge [({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[({y}_{j}+{y}_{k})-({y}_{j}+{y}_{k})({\gamma}_{jk})].\end{array}$$From the former several inequalities, it is easy to follow that

$$\begin{array}{lll}{f}_{jk}(y)& =& {y}_{j}{y}_{k}=\frac{({y}_{j}+{y}_{k}{)}^{2}-{y}_{j}^{2}-{y}_{k}^{2}}{2}\\ & \ge & \frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ & & -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ & & -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\},\\ & =& {\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})\end{array}$$and

$$\begin{array}{lll}{f}_{jk}(y)& =& {y}_{j}{y}_{k}=\frac{({y}_{j}+{y}_{k}{)}^{2}-{y}_{j}^{2}-{y}_{k}^{2}}{2}\\ & \le & \frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ & & -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ & & -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\},\\ & =& {\overline{f}}_{jk}(y,Y,{\gamma}_{jk}).\end{array}$$Therefore, we have

$${\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})\le {f}_{jk}(y)\le {\overline{f}}_{jk}(y,Y,{\gamma}_{jk}).$$(ii) Since

$$\begin{array}{lll}{f}_{kk}(y)-{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})& =& {y}_{k}^{2}-\{[{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}({\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})]\}\\ & =& ({y}_{k}-{y}_{k}({\gamma}_{kk}){)}^{2}\\ & \le & ({u}_{k}-{l}_{k}{)}^{2},\end{array}$$(7)we have

$$\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{kk}(y)-{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})]=0.$$Also since

$$\begin{array}{lll}{\overline{f}}_{kk}(y,Y,{\gamma}_{kk})-{f}_{kk}(y)& =& [{y}_{k}({\gamma}_{kk}){]}^{2}+2{y}_{k}(1-{\gamma}_{kk})[{y}_{k}-{y}_{k}({\gamma}_{kk})]-{y}_{k}^{2}\\ & =& ({y}_{k}({\gamma}_{kk})+{y}_{k})({y}_{k}({\gamma}_{kk})-{y}_{k})\\ & & +2{y}_{k}(1-{\gamma}_{kk})({y}_{k}-{y}_{k}({\gamma}_{kk}))\\ & =& [{y}_{k}-{y}_{k}({\gamma}_{kk})][2{y}_{k}(1-{\gamma}_{kk})-{y}_{k}({\gamma}_{kk})-{y}_{k}]\\ & =& [{y}_{k}-{y}_{k}({\gamma}_{kk})][{y}_{k}(1-{\gamma}_{kk})-{y}_{k}({\gamma}_{kk})]\\ & & +[{y}_{k}-{y}_{k}({\gamma}_{kk})][{y}_{k}(1-{\gamma}_{kk})-{y}_{k}]\\ & \le & 2({u}_{k}-{l}_{k}{)}^{2}.\end{array}$$(8)Therefore, it follows that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{kk}(y,Y,{\gamma}_{kk})-{f}_{kk}(y)]=0.$$From the limitations (3) and (4), replacing γ_{kk} and *y*_{k} by γ_{jk} and *y*_{j}, respectively, we have

$$\underset{\parallel u-l\parallel \to 0}{lim}[{y}_{j}^{2}-\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}]=0$$and

$$\underset{\parallel u-l\parallel \to 0}{lim}[[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]-{y}_{j}^{2}]=0.$$From the limitations (3) and (4), replacing γ_{kk} by γ_{jk}, it follows that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{y}_{k}^{2}-\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}]=0$$and

$$\underset{\parallel u-l\parallel \to 0}{lim}[[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]-{y}_{k}^{2}]=0.$$By the limitations (3) and (4), replacing γ_{kk} and *y*_{k} by γ_{jk} and (*y*_{j} + *y*_{k}), respectively, we can get that

$$\underset{\parallel u-l\parallel \to 0}{lim}[[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]-({y}_{j}+{y}_{k}{)}^{2}]=0$$and

$$\underset{\parallel u-l\parallel \to 0}{lim}[({y}_{j}+{y}_{k}{)}^{2}-\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}+2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}]=0.$$From the inequalities (7) and (8), we have

$$\begin{array}{lll}{f}_{jk}(y)-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})& =& {y}_{j}{y}_{k}-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})\\ & =& \frac{({y}_{j}+{y}_{k}{)}^{2}-{y}_{j}^{2}-{y}_{k}^{2}}{2}-\frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}\\ & & +2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ & & -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ & & -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\},\\ & =& \frac{1}{2}[\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}(1-{\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}-{y}_{j}^{2}]\\ & & +\frac{1}{2}[\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}(1-{\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}-{y}_{k}^{2}]\\ & & +\frac{1}{2}\{({y}_{j}+{y}_{k}{)}^{2}-\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}\\ & & +2({y}_{j}+{y}_{k})({\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ & \le & ({u}_{j}-{l}_{j}{)}^{2}+({u}_{k}-{l}_{k}{)}^{2}+\frac{1}{2}({u}_{k}+{u}_{j}-{l}_{j}-{l}_{k}{)}^{2}.\end{array}$$Thus, we can get that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{jk}(y)-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})]=0.$$Also from the inequalities (7) and (8), we get that

$$\begin{array}{lll}{\overline{f}}_{jk}(y,Y,{\gamma}_{jk})-{f}_{jk}(y)& =& {\overline{f}}_{jk}(y,Y,{\gamma}_{jk})-{y}_{j}{y}_{k}\\ & =& \frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}\\ & & +2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]\}\\ & & -\{[{y}_{j}({\gamma}_{jk}){]}^{2}+2{y}_{j}({\gamma}_{jk})[{y}_{j}-{y}_{j}({\gamma}_{jk})]\}\\ & & -\{[{y}_{k}({\gamma}_{jk}){]}^{2}+2{y}_{k}({\gamma}_{jk})[{y}_{k}-{y}_{k}({\gamma}_{jk})]\}\}\\ & & -\frac{({y}_{j}+{y}_{k}{)}^{2}-{y}_{j}^{2}-{y}_{k}^{2}}{2}\\ & =& \frac{1}{2}\{\{[({y}_{j}+{y}_{k})({\gamma}_{jk}){]}^{2}\\ & & +2({y}_{j}+{y}_{k})(1-{\gamma}_{jk})[{y}_{j}+{y}_{k}-({y}_{j}+{y}_{k})({\gamma}_{jk})]-({y}_{j}+{y}_{k}{)}^{2}\}\end{array}$$Thus, it follows that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{jk}(y,Y,{\gamma}_{jk})-{f}_{jk}(y)]=0.$$Without loss of generality, for any *Y* = [*l*, *u*] b *Y*^{0}, for any parameter matrix = (γ_{jk})_{n}_{×n}, for any *y* ϵ *Y* and *i* ϵ {0, 1, *. . .*, *m*}, we let

$${\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk})=\left\{\begin{array}{ll}{p}_{kk}^{i}{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk}),& \text{if}{p}_{kk}^{i}>0,\\ {p}_{kk}^{i}{\overline{f}}_{kk}(y,Y,{\gamma}_{kk}),& \text{if}{p}_{kk}^{i}<0,\end{array}\right.$$$${\overline{f}}_{kk}^{i}(y,Y,{\gamma}_{kk})=\left\{\begin{array}{ll}{p}_{kk}^{i}{\overline{f}}_{kk}(y,Y,{\gamma}_{kk}),& \text{if}{p}_{kk}^{i}>0,\\ {p}_{kk}^{i}{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk}),& \text{if}{p}_{kk}^{i}<0,\end{array}\right.$$$${\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})=\left\{\begin{array}{ll}{p}_{jk}^{i}{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk}),& \text{if}{p}_{jk}^{i}>0,j\ne k,\\ {p}_{jk}^{i}{\overline{f}}_{jk}(y,Y,{\gamma}_{jk}),& \text{if}{p}_{jk}^{i}<0,j\ne k,\end{array}\right.$$$${\overline{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})=\left\{\begin{array}{ll}{p}_{jk}^{i}{\overline{f}}_{jk}(y,Y,{\gamma}_{jk}),& \text{if}{p}_{jk}^{i}>0,j\ne k,\\ {p}_{jk}^{i}{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk}),& \text{if}{p}_{jk}^{i}<0,j\ne k.\end{array}\right.$$$${F}_{i}^{L}(y,Y,\gamma )=\sum _{k=1}^{n}({d}_{k}^{i}{y}_{k}+{\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}))+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk}).$$$${F}_{i}^{U}(y,Y,\gamma )=\sum _{k=1}^{n}({d}_{k}^{i}{y}_{k}+{\overline{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}))+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\overline{f}}_{jk}^{i}(y,Y,{\gamma}_{jk}).$$**Theorem 2.2**. *For any y* ϵ *Y* = [*l*, *u*] ⊆ *Y*^{0}*, for any parameter matrix* = (γ_{jk})_{n}_{×n}*, for any i* = 0, 1, *. . .*, *m*, *we can get the following conclusions*:

$${F}_{i}^{L}(y,Y,\gamma )\le {F}_{i}(y)\le {F}_{i}^{U}(y,Y,\gamma ),$$$$\underset{\parallel u-l\parallel \to 0}{lim}[{F}_{i}(y)-{F}_{i}^{L}(y,Y,\gamma )]=0$$*and*

$$\underset{\parallel u-l\parallel \to 0}{lim}[{F}_{i}^{U}(y,Y,\gamma )-{F}_{i}(y)]=0.$$*Proof*. (i) From (1) and (2), for any *j*, *k* ϵ {1, *. . .*, *n*}, we can get that

$${\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk})\le {p}_{kk}^{i}{y}_{k}^{2}\le {\overline{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}),$$(9)and

$${\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})\le {p}_{jk}^{i}{y}_{j}{y}_{k}\le {\overline{f}}_{jk}^{i}(y,Y,{\gamma}_{jk}).$$(10)By (9) and (10), for any *y* ϵ *Y* ⊆ *Y*^{0}, we have that

$$\begin{array}{lll}{F}_{i}^{L}(y,Y,\gamma )& =& \sum _{k=1}^{n}({d}_{k}^{i}{y}_{k}+{\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}))+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})\\ & \le & \sum _{k=1}^{n}{d}_{k}^{i}{y}_{k}+\sum _{k=1}^{n}{p}_{kk}^{i}{y}_{k}^{2}+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{p}_{jk}^{i}{y}_{j}{y}_{k}={F}_{i}(y)\\ & \le & \sum _{k=1}^{n}({d}_{k}^{i}{y}_{k}+{\overline{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}))+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\overline{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})\\ & =& {F}_{i}^{U}(y,Y,\gamma ).\end{array}$$Therefore, we obtain that

$${F}_{i}^{L}(y,Y,\gamma )\le {F}_{i}(y)\le {F}_{i}^{U}(y,Y,\gamma ).$$(ii)

$$\begin{array}{lll}{F}_{i}(y)-{F}_{i}^{L}(y,Y,\gamma )& =& \sum _{k=1}^{n}{d}_{k}^{i}{y}_{k}+\sum _{k=1}^{n}{p}_{kk}^{i}{y}_{k}^{2}+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{p}_{jk}^{i}{y}_{j}{y}_{k}-[\sum _{k=1}^{n}{d}_{k}^{i}{y}_{k}\\ & & +\sum _{k=1}^{n}{\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk})+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})]\\ & =& \sum _{k=1}^{n}[{p}_{kk}^{i}{y}_{k}^{2}-{\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk})]\\ & & +\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}[{p}_{jk}^{i}{y}_{j}{y}_{k}-{\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk})]\\ & =& \sum _{k=1,{p}_{kk}^{i}>0}^{n}{p}_{kk}^{i}[{f}_{kk}(y)-{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})]\\ & & +\sum _{k=1,{p}_{kk}^{i}<0}^{n}{p}_{kk}^{i}[{f}_{kk}(y)-{\overline{f}}_{kk}(y,Y,{\gamma}_{kk})]\\ & & +\sum _{j=1}^{n}\sum _{k=1,k\ne j,{p}_{jk}^{i}>0}^{n}{p}_{jk}^{i}[{f}_{jk}(y)-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})]\\ & & +\sum _{j=1}^{n}\sum _{k=1,k\ne j,{p}_{jk}^{i}<0}^{n}{p}_{jk}^{i}[{f}_{jk}(y)-{\overline{f}}_{jk}(y,Y,{\gamma}_{jk})].\end{array}$$From (3)-(6), we can obtain that $\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{kk}(y)-{\underset{\_}{f}}_{kk}(y,Y,{\gamma}_{kk})]=0,\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{kk}(y,Y,{\gamma}_{kk})-{f}_{kk}(y)]=0,$ $\underset{\parallel u-l\parallel \to 0}{lim}[{f}_{jk}(y)-{\underset{\_}{f}}_{jk}(y,Y,{\gamma}_{jk})]=0\phantom{\rule{thickmathspace}{0ex}}\text{and}\phantom{\rule{thickmathspace}{0ex}}\underset{\parallel u-l\parallel \to 0}{lim}[{\overline{f}}_{jk}(y,Y,{\gamma}_{jk})-{f}_{jk}(y)]=0.$

Therefore, we obtain that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{F}_{i}(y)-{F}_{i}^{L}(y,Y,\gamma )]=0.$$Similarly to the proof above, we can get that

$$\underset{\parallel u-l\parallel \to 0}{lim}[{F}_{i}^{U}(y,Y,\gamma )-{F}_{i}(y)]=0.$$The proof is completed.

By Theorem 2.2, we can establish the following parametric linear programming relaxation problem (PLPRP) of QP over *Y*:

$$\mathbf{(}\mathbf{P}\mathbf{L}\mathbf{P}\mathbf{R}\mathbf{P}\mathbf{)}:\left\{\begin{array}{ll}\mathrm{m}\mathrm{i}\mathrm{n}& {F}_{0}^{L}(y,Y,\gamma ),\\ \mathrm{s}.\mathrm{t}.& {F}_{i}^{L}(y,Y,\gamma )\le {\beta}_{i},i=1,\dots ,m,\\ & y\in Y=\{y:l\le y\le u\}.\end{array}\right.$$where

$${F}_{i}^{L}(y,Y,\gamma )=\sum _{k=1}^{n}({d}_{k}^{i}{y}_{k}+{\underset{\_}{f}}_{kk}^{i}(y,Y,{\gamma}_{kk}))+\sum _{j=1}^{n}\sum _{k=1,k\ne j}^{n}{\underset{\_}{f}}_{jk}^{i}(y,Y,{\gamma}_{jk}).$$Based on the above parametric linearization process, we know that the PLPRP can provide a reliable lower bound for the minimum value of QP in the region *Y*. In addition, Theorem 2.2 ensures that the PLPRP will sufficiently approximate the QP as ∥*u* − *l*∥ → 0, and this ensures the global convergence of the proposed branch-and-bound algorithm.

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