Main Page

Welcome to MWiki

Theorem of the month

The centre method solves every solvable LP in [math]\displaystyle{ \mathcal{O}(\omega{\vartheta}^{2}) }[/math].

Proof and algorithm

Let [math]\displaystyle{ z := \grave{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, \check{r}], Ax - b \le \underline{r}_m, c - {A}^{T}y \le \underline{r}_n\} }[/math] have the radius [math]\displaystyle{ \check{r} := 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{ \underline{0}_{z} \in \partial P_{\check{r}} }[/math]. By the strong duality theorem, the LP min [math]\displaystyle{ \{ r \in [0, \check{r}] : (x, y)^T \in P_r\} }[/math] solves the 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}\,p_k + \text{max}\,p_k)/2 }[/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, \check{r}] : t \in {}^{\omega}\mathbb{R}_{> 0}, t(x_0, y_0)^T \in P_r\} }[/math] approximates [math]\displaystyle{ g }[/math] better and achieves [math]\displaystyle{ r \le \check{r}/\sqrt{\grave{z}} }[/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}({}_z\check{r} {}_e\check{r}dmn) }[/math] if it exists. Numbers of length [math]\displaystyle{ \mathcal{O}({\omega}) }[/math] can only be processed in [math]\displaystyle{ \mathcal{O}(\vartheta) }[/math] as is generally known.

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_z }[/math] for [math]\displaystyle{ p^* := p + wq }[/math] and [math]\displaystyle{ w \in {}^{\omega}\mathbb{R}_{\ge 0} }[/math] would be also to solve. If [math]\displaystyle{ \text{min}_k \Delta r_k r = 0 }[/math] follows, stop, otherwise repeat until min [math]\displaystyle{ r = 0 }[/math] or min [math]\displaystyle{ r > 0 }[/math] is sure. If necessary, constraints are temporarily relaxed by the same small modulus.[math]\displaystyle{ \square }[/math]

Recommended readings

Nonstandard Mathematics