# Linear Programming

In the following section, Set Theory, Topology and Nonstandard Analysis are presupposed. The exponential simplex and the polynomial centre method solve linear programmes (LPs).

Diameter theorem for polytopes: The diameter of an $$n$$-dimensional polytope defined by $$m$$ constraints for $$m, n \in {}^{\omega}\mathbb{N}_{\ge 2}$$ is at most $$2(m + n – 3)$$.

Proof: At most $$\acute{m}$$ hyperplanes can be assembled into an incomplete cycle of dimension 2 and there are at most $$n – 2$$ alternatives sidewards in the remaining dimensions. Overcoming every minimal distance requires at most two edges and yields the factor 2.$$\square$$

Remark: Dropping the requirement of finiteness, the theorem can be extended to polyhedra analogously.

Definition: Let $$\vartheta := \omega\,{}_{e}\omega, \underline{u}_n := (u, …, u)^T \in{}^{\omega}\mathbb{R}^{n}$$ and $${|| \cdot ||}_{1}$$ the 1-norm. If $$x \in X \subseteq {}^{\omega}\mathbb{R}^{\omega}$$ is passed to the next step, it is followed by $${}^{*}$$ where $$\Delta x := {x}^{*} – x$$. The unit vector of $$x$$ is $${_1}{x} := x/||x||$$, where $$_1{0}$$ is undefined. Let $$x, y \in X$$ in $$(x, y, …)^T$$ be row vectors. If a method requires computation time in seconds and memory in bits of $$\mathcal{O}({\omega}^{\mathcal{O}(1)})$$, it is polynomial. If one of both quantities is $$\mathcal{O}({e}^{|\mathcal{O}(\omega)|})$$, the method is exponential. Let the eigenproduct (yet: determinant) of a square matrix be the product of its eigenvalues.$$\triangle$$

Theorem: The simplex method is exponential.

Proof and algorithm: Let $$P := \{x \in {}^{\omega}\mathbb{R}^{n} : Ax \le b, b \in {}^{\omega}\mathbb{R}^{m}, A \in {}^{\omega}\mathbb{R}^{m \times n}, m, n \in {}^{\omega}\mathbb{N}^{*}\}$$ be the feasible domain of the LP max $$\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in P\}$$. The dual of the latter gives $${x}^{*} \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}$$. Putting $$x := {x}^{+} – {x}^{-}$$ with $${x}^{+}, {x}^{-} \ge 0$$ yields $${x}^{*} \in {}^{\omega}\mathbb{R}_{\ge 0}^{2n}$$. Solving max $$\{-h \in {}^{\omega}\mathbb{R}_{\le 0} : Ax – b \le \underline{h}_n\}$$ yields $$x \in P_{\ge 0}$$ when $$b \ge 0$$ does not hold. Let $$|\text{min } \{{b}_{1}, …, {b}_{m}\}|$$ be the initial value for $$h$$ and 0 its target value. Start with $$x := 0$$. Only one pivot step causes that $${b}^{*} \ge 0$$. Let $$i, j, k \in {}^{\omega}\mathbb{N}^{*}$$ and let $${a}_{i}^{T}$$ the $$i$$-th row vector of $$A$$.

If $${c}_{j} \le 0$$ for all $$j$$, the LP is solved. If for some $${c}_{j} > 0$$ also $${a}_{ij} \le 0$$ for all $$i$$, the LP is positively unbounded. If for some $${c}_{j} = 0$$ also $${a}_{ij} \le 0$$ for all $$i$$, drop $${c}_{j}$$ and $${A}_{.j}$$ as well as $${b}_{i}$$ and $${a}_{i}$$, but only when $${a}_{ij} < 0$$ holds. The inequality $${a}_{ij}{x}_{j} \ge 0 > {b}_{i}$$ for all $$j$$ has no solution, too. If necessary, divide all $${a}_{i}^{T}x \le {b}_{i}$$ by $$||{a}_{i}||$$ and all $${c}_{j}$$ and $${a}_{ij}$$ by the minimum of $$|{a}_{ij}|$$ such that $${a}_{ij} \ne 0$$ for each $$j$$. This will be reversed later. If necessary, renormalise by $$||{a}_{i}||$$. Redundant constraints (with $${a}_{i} \le 0$$) may always be removed.

For each $${c}_{j} > 0$$ and non-base variable $${x}_{j}$$ select min $$\{{b}_{k}/{a}_{kj} : {a}_{ij} > 0\}$$ to obtain a next $${x}^{*} \in P_{\ge 0}$$ where $${x}_{j}^{*} = {x}_{j} + {b}_{k}/{a}_{kj}$$. To select the steepest edge, pick the pivot $${a}_{kj}$$ corresponding to $${x}_{j}$$ that maximises $${c}^{T}{_1}{\Delta x}$$ or $${c}_{j}^{2}/\grave{v}_{j}^{2}$$ with $$v_j := ||{A}_{.j}||$$ in the $$k$$-th constraint. Multiple maxima allow to use the rule of best pivot value max$${}_{k,j} {c}_{j}{b}_{k}/{a}_{kj}$$ or (slower) the smallest angle min $${{_{(1)}}(1, …, 1)}^{T}{_1}{c}^{*}$$. If always $$c^Tx^* = c^Tx$$ holds, perturb, which means relax, the constraints with $${b}_{i} = 0$$ by the same, minimal modulus.

For $${b}_{i} := ||{a}_{i}||$$, the modulus need not to be written into the tableau. If another multiple vertex is encountered, despite this being unlikely, simply increase the earlier $${b}_{i}$$ by $$||{a}_{i}||$$. Leaving a multiple vertex, after which the relaxation is reverted, may require to solve an LP with $$c > 0$$ and $$b = 0$$. Along the chosen path, the objective function increases otherwise strictly monotonically.

Eventually, $${c}_{j}^{*}, {a}_{ij}^{*}$$ and $${b}_{i}^{*}$$ can be simply computed using the rectangle rule1cf. Vanderbei, Robert J.: Linear Programming; 3rd Ed.; 2008; Springer; New York, p. 63. Despite the diameter theorem for polytopes, the simplex method is not polynomial under any given set of pivoting rules in the worst-case scenario: An exponential “drift” for e. g. Klee-Minty or Jeroslow polytopes may provide a large deviation from the shortest path by forcing the selection of an unfavourable edge for every step. The result follows in accordance with the state of research.$$\square$$

Theorem: The centre method solves every solvable LP in $$\mathcal{O}(\omega{\vartheta}^2)$$.

Proof and algorithm: First, normalise and scale $${b}^{T}y – {c}^{T}x \le 0, Ax \le b$$ as well as $${A}^{T}y \ge c$$. Let $$d \in [0, 1]$$ the density of $$A$$ and $$P_r := \{(x, y)^T : x \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}, y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m},{b}^{T}y – {c}^{T}x \le r \in [0, \check{r}], Ax – b \le \underline{r}_m, c – {A}^{T}y \le \underline{r}_n\}$$ for the radius $$\check{r} := s|\min \; \{b_1, …, b_m, -c_1, …, -c_n\}|$$ and the scaling factor $$s \in [1, 2]$$. Putting the number $$z := \grave{m} + n$$ implies $$\underline{0}_{\acute{z}} \in \partial P_{\check{r}}$$. By the strong duality theorem2loc. cit., p. 60 – 65 the LP min $$\{r \in [0, \check{r}] : (x, y)^T \in P_r\}$$ solves the LPs max $$\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}$$ and min $$\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}$$.

Its solution is the geometric centre $$g$$ of the polytope $$P_r$$. For $$p_k^* := (\text{min}\,p_k + \text{max}\,p_k)/2$$ and $$k = 1, …, z$$ approximate $$g$$ by $$p_0 := (x_0, y_0, r_0)^T$$ until $$||\Delta p||_1$$ is sufficiently small. The solution $$t^o(x_1, y_1, r_1)^T$$ of the two-dimensional LP min $$\{r \in [0, \check{r}] : t \in {}^{\omega}\mathbb{R}_{> 0}, t(x_0, y_0)^T \in P_r\}$$ approximates $$g$$ better and achieves $$r \le \check{r}/\sqrt{z}$$. Repeat this for $$t^o(x_1, y_1)^T$$ until $$g \in P_0$$ is computed in $$\mathcal{O}({}_z\check{r} {}_e\check{r}dmn)$$ if it exists. Finally, numbers of length $$\mathcal{O}({\omega})$$ can only be processed in $$\mathcal{O}(\vartheta).\square$$

Remarks: If the centre method is optimised for distributed computing in $${}^{\nu}\mathbb{R}^{\nu}$$, its runtime only amounts to $$\mathcal{O}(1)$$. It is also well-suited for (mixed) integer problems and (non-) convex (Pareto) optimisation (according to nature3cf. Vasuki, A: Nature-Inspired Optimization Algorithms; 1st Ed.; 2020; CRC Press; Boca Raton). Rounding errors can be kept small by using a modified Kahan-Babuška-Neumaier summation. Loops may be parallelised. The transfer to complex numbers is easy. All of this holds also for that what follows.

Conclusion: The LP max $$\{{||x – {x}^{o}||}_{1} : {c}^{T}x = {c}^{T}{x}^{o}, Ax \le b, x – {x}^{o} \in {[-1, 1]}^{n}, x \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}\}$$ can determine for the first solution $${x}^{o}$$ a second one in $$\mathcal{O}(\omega{\vartheta}^2)$$ if any, where $${y}^{o}$$ may be treated analogously.$$\square$$

Conclusion: The LP max $$\{\nu|\lambda_j| + ||x_j||_1: Ax_j = \lambda_j x_j, |\lambda_j| \in [0, r_j], x_j \in {[-1, 1]}^{n*}\}$$ can determine for every $$j = 1, …, n$$ the eigenvalue $$\lambda_j \in {}^{\omega}\mathbb{R}$$ and the eigenvector $$x_j \in {}^{\omega}\mathbb{R}^{n}$$ of the matrix $$A \in {}^{\omega}\mathbb{R}^{n \times n}$$ in $$\mathcal{O}(\omega{\vartheta}^2).\square$$

Conclusion: The LP min $$\{r \in [0, s \, \text{max } \{|{b}_{1}|, …, |{b}_{m}|\}] : \pm(Ax – b) \le \underline{r}_m\}$$ can determine an $$x \in {}^{\omega}\mathbb{R}^{n}$$ of every solvable linear system (LS) $$Ax = b$$ in $$\mathcal{O}(\omega{\vartheta}^2)$$. The LPs max $$\{{x}_{j} : Ax = 0\}$$ yield all solutions to the LS. The matrix $$A$$ is regular if and only if the LP max $$\{{||x||}_{1} : Ax = 0\} = 0.\square$$

Conclusion: Let $${\alpha }_{j} := {A}_{.j}^{-1}$$ for $$j = 1, …, n$$ concerning the matrix $${A}^{-1} \in {}^{\omega}\mathbb{R}^{n \times n}$$ and let $${\delta}_{ij}$$ the Kronecker delta. A regular $$A$$ has an eigenproduct $$\ne 0$$ and allows every LS $${A \alpha }_{j} = {({\delta}_{1j}, …, {\delta}_{nj})}^{T}$$ to be solved in $$\mathcal{O}(\omega{\vartheta}^2).\square$$

Corollary: Every solvable convex programme min $$\{{f}_{1}(x) : x \in {}^{\omega}\mathbb{R}^{n}, {({f}_{2}(x), …, {f}_{m}(x))}^{T} \le 0\}$$ where the $${f}_{i} \in {}^{\omega}\mathbb{R}$$ are convex functions for $$i = 1, …, m$$ may be solved by the centre method and two-dimensional bisection or Newton’s methods in polynomial runtime, if the number of operands $${x}_{j}$$ of the $${f}_{i}$$ is $$\le {\omega}^{\nu-3}$$ and if an $$x$$ exists4cf. Bertsekas, Dimitri P.: Nonlinear Programming; 3rd Ed.; 2016; Athena Scientific; Belmont, p. 589 ff. so that $${f}_{i}(x) < 0$$ for all $$i > 1.\square$$

Please, input linear programme (separators spaces, last row
objective function, first column right-hand sides as in the example): not the number of the correct decimal places matters, but the time complexity and the number of steps of the simplex method!