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| − | It follows <math>0{\upharpoonright}_z \in \partial P_{\rho}</math>. By the strong duality theorem, LP min <math>\{ r \in [0, \rho] : (x, y)^T \in P_r\}</math> solves LPs max <math>\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}</math> and min <math>\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}</math>. Its solution is the geometric centre <math>g</math> of the polytope <math>P_0</math>. For <math>p_k^* := \text{min}\,\check{p}_k + \text{max}\,\check{p}_k</math> and <math>k = 1, ..., \grave{z}</math> approximate <math>g</math> by <math>p_0 := (x_0, y_0, r_0)^T</math> until <math>||\Delta p||_1</math> is sufficiently small. The solution <math>t^o(x^o, y^o, r^o)^T</math> of the two-dimensional | + | It follows <math>0{\upharpoonright}_z \in \partial P_{\rho}</math>. By the strong duality theorem, LP min <math>\{ r \in [0, \rho] : (x, y)^T \in P_r\}</math> solves LPs max <math>\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}</math> and min <math>\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}</math>. Its solution is the geometric centre <math>g</math> of the polytope <math>P_0</math>. For <math>p_k^* := \text{min}\,\check{p}_k + \text{max}\,\check{p}_k</math> and <math>k = 1, ..., \grave{z}</math> approximate <math>g</math> by <math>p_0 := (x_0, y_0, r_0)^T</math> until <math>||\Delta p||_1</math> is sufficiently small. The solution <math>t^o(x^o, y^o, r^o)^T</math> of the two-dimensional LP min <math>\{ r \in [0, \rho] : t \in {}^{\omega}\mathbb{R}_{> 0}, t(x_0, y_0)^T \in P_r\}</math> approximates <math>g</math> better and achieves <math>r \le \check{\rho}</math>. |
Revision as of 00:55, 1 December 2023
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Theorem of the month
Intex methods solve almost every solvable LP with inter-/extrapolations in [math]\displaystyle{ \mathcal{O}({\vartheta}^3) }[/math].
Proof and algorithm
Let [math]\displaystyle{ z := m + n }[/math] and [math]\displaystyle{ d \in [0, 1] }[/math] the density of [math]\displaystyle{ A }[/math]. First, normalise and scale [math]\displaystyle{ {b}^{T}y - {c}^{T}x \le 0, Ax \le b }[/math] as well as [math]\displaystyle{ {A}^{T}y \ge c }[/math]. Let [math]\displaystyle{ P_r := \{(x, y)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{z} : {b}^{T}y - {c}^{T}x \le r \in [0, \rho], Ax - b \le r{\upharpoonright}_m, }[/math] [math]\displaystyle{ c - {A}^{T}y \le r{\upharpoonright}_n\} }[/math] have the radius [math]\displaystyle{ \rho := s|\min \; \{b_1, ..., b_m, -c_1, ..., -c_n\}| }[/math] and the scaling factor [math]\displaystyle{ s \in [1, 2] }[/math].
It follows [math]\displaystyle{ 0{\upharpoonright}_z \in \partial P_{\rho} }[/math]. By the strong duality theorem, LP min [math]\displaystyle{ \{ r \in [0, \rho] : (x, y)^T \in P_r\} }[/math] solves LPs max [math]\displaystyle{ \{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\} }[/math] and min [math]\displaystyle{ \{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\} }[/math]. Its solution is the geometric centre [math]\displaystyle{ g }[/math] of the polytope [math]\displaystyle{ P_0 }[/math]. For [math]\displaystyle{ p_k^* := \text{min}\,\check{p}_k + \text{max}\,\check{p}_k }[/math] and [math]\displaystyle{ k = 1, ..., \grave{z} }[/math] approximate [math]\displaystyle{ g }[/math] by [math]\displaystyle{ p_0 := (x_0, y_0, r_0)^T }[/math] until [math]\displaystyle{ ||\Delta p||_1 }[/math] is sufficiently small. The solution [math]\displaystyle{ t^o(x^o, y^o, r^o)^T }[/math] of the two-dimensional LP min [math]\displaystyle{ \{ r \in [0, \rho] : t \in {}^{\omega}\mathbb{R}_{\gt 0}, t(x_0, y_0)^T \in P_r\} }[/math] approximates [math]\displaystyle{ g }[/math] better and achieves [math]\displaystyle{ r \le \check{\rho} }[/math].
Repeat this for [math]\displaystyle{ t^o(x^o, y^o)^T }[/math] until [math]\displaystyle{ g \in P_0 }[/math] is computed in [math]\displaystyle{ \mathcal{O}({}_2\rho^2dmn) }[/math] if it exists. Solving all two-dimensional LPs [math]\displaystyle{ \text{min}_k r_k }[/math] by bisection methods for [math]\displaystyle{ r_k \in {}^{\omega}\mathbb{R}_{\ge 0} }[/math] and [math]\displaystyle{ k = 1, ..., z }[/math] in [math]\displaystyle{ \mathcal{O}({\vartheta}^2) }[/math] each time determines [math]\displaystyle{ q \in {}^{\omega}\mathbb{R}^k }[/math] where [math]\displaystyle{ q_k := \Delta p_k \Delta r_k/r }[/math] and [math]\displaystyle{ r := \text{min}_k \Delta r_k }[/math]. Let simplified [math]\displaystyle{ |\Delta p_1| = ... = |\Delta p_{z}| }[/math]. Here min [math]\displaystyle{ r_{\grave{z}} }[/math] for [math]\displaystyle{ p^* := p + wq }[/math] and [math]\displaystyle{ w \in {}^{\omega}\mathbb{R}_{\ge 0} }[/math] may be solved, too. If [math]\displaystyle{ \text{min}_k \Delta r_k r = 0 }[/math] follows, stop computing, otherwise repeat until min [math]\displaystyle{ r = 0 }[/math] or min [math]\displaystyle{ r > 0 }[/math] is sure.[math]\displaystyle{ \square }[/math]