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In the following section, we solve linear programmes (LPs). The well-known simplex method is described, together with the polynomial normal method. Smale's 9th problem is demonstrated to be unsolvable. We may generalise the normal method to (non-) convex programmes (with vector-valued functions). The diameter theorem for polytopes is proven. It is shown that (mixed) integer LPs are polynomial.

Proof and algorithm: Let M := {x ∈ ^{κ}ℝ^{n} : Ax ≤ b, b ∈ ^{κ}ℝ^{m}, A ∈ ^{κ}ℝ^{m×n}, m, n ∈ ^{κ}ℕ*} be the feasible domain of the LP max {c^{T}x : c ∈ ^{κ}ℝ^{n}, x ∈ M}. By taking the dual or setting x := x^{+} - x^{-} with x^{+}, x^{-} ≥ 0, we obtain x ≥ 0. We first solve max {-z : Ax - b ≤ (z, …, z)^{T} ∈ ^{κ}ℝ^{m}, z ≥ 0} to obtain a feasible x when b ≥ 0 does not hold. Initial and target value are z := |min {b_{1}, ..., b_{m}}| resp. z = 0 and we begin with x := 0 as in the first case. Pivoting if necessary, we may assume that b ≥ 0.

Let i, j, k ∈ ^{κ}ℕ* and let a_{i}^{T} the i-th row vector of A. If c_{j} ≤ 0 for all j, the LP is solved. If for some c_{j} > 0, a_{ij} ≤ 0 for all i, the LP is positively unbounded. Otherwise, we divide all a_{i}^{T}x ≤ b_{i} by ||a_{i}|| and all c_{j} and a_{ij} by the minimum of |a_{ij}| such that a_{ij} ≠ 0 for each j. This will be reversed later. If necessary, renormalise by ||a_{i}||. This yields good runtime performance even on strongly deformed polytopes.

In each step, we can remove multiple constraints and such with a_{i} ≤ 0, since they are redundant (avoidable by adding an extra slack variable in each case). The second case is analogous. If in both cases b_{i} = 0 and a_{i} ≥ 0 for some i, then the LP has maximum 0 and solution x = 0 if b ≥ 0, otherwise it has no solutions. In each step, for each c_{j} > 0 and non-base variable x_{j}, we select the minimum ratio b_{k}/a_{kj} for a_{ij} > 0.

The variables with * are considered in the next step. The next potential vertex is given by x_{j}* = x_{j} + b_{k}/a_{kj} for feasible x*. To select the steepest edge, select the pivot a_{kj} corresponding to x_{j} that maximises c^{T}(x* - x)/||x* - x|| i.e. c_{j}|c_{j}|/(1 + Σ a_{ij}^{2}) in the k-th constraint. If there are multiple maxima, select max c_{j}b_{k}/a_{kj} or alternatively the smallest angle min Σ c_{j}*/||c*||, according to the rule of best pivot value.

If there are more than n values b_{i} equal 0 and we cannot directly maximise the objective function, we relax (perturb) the constraints with b_{i} = 0 by the same, minimal modulus. These do not need to be written into the tableau: We simply set b_{i} = ||a_{i}||. If another multiple vertex is encountered, despite this being unlikely, simply increase the earlier b_{i} by ||a_{i}||.

The cost of eliminating a multiple vertex, after which we revert the relaxation, corresponds to the cost of solving an LP with b = 0. The same task is potentially required at the end of the process when checking whether the LP admits any other solutions if at least two of the c_{j} are 0. Along the chosen path, the objective function increases (effectively) strictly monotonically. We can then simply calculate c_{j}*, a_{ij}* and b_{i}* using the rectangle rule.

In the worst-case scenario, the simplex method is not strongly polynomial despite the diameter formula for polytopes (see below) under any given set of pivoting rules, since an exponential "drift" can be constructed with Klee-Minty or Jeroslow polytopes, or others, creating a large deviation from the shortest path by forcing the selection of the least favourable edge. This is consistent with existing proofs. The result follows.⃞

Theorem: The normal method solves the LP in O(mnh_{0}/Δh_{⌀}) with Δh_{⌀} as the average loss of the *height* h assuming h_{0} := max(max(Σ x_{j})/√n, max(Σ y_{i})/√m), if possible.

Proof and algorithm: We compute the LP min {h ∈ ^{κ}ℝ_{≥0} : ±(c^{T}x - b^{T}y) ≤ h, 0 ≤ x ∈ ^{κ}ℝ^{n}, 0 ≤ y ∈ ^{κ}ℝ^{m}, Ax - b ≤ (h, …, h)^{T} ∈ ^{κ}ℝ^{m}, c - A^{T}y ≤ (h, …, h)^{T} ∈ ^{κ}ℝ^{n}} via the (dual) programme min {b^{T}y : 0 ≤ y ∈ ^{κ}ℝ^{m}, A^{T}y ≥ c} for the (primal) programme max {c^{T}x : c ∈ ^{κ}ℝ^{n}, x ∈ M_{≥0}}. Initial and target value of the *height* h are |min {b_{1}, ..., b_{m}, -c_{1}, …, -c_{n}}| resp. 0. To obtain h* as h_{0} from h, we normalise ±(c^{T}x - b^{T}y) ≤ h, Ax ≤ b and A^{T}y ≥ c. Let v := (x^{T}, y^{T})^{T} ∈ ^{κ}ℝ^{m+n}.

We begin with v := 0 and successively define a few times all v_{k}* := (max v_{k} + min v_{k})/2. Then, we solve the two-dimensional LPs min h_{k} in h_{k}, v_{k} ∈ ^{κ}ℝ_{≥0}. We compute a favourable direction d ∈ ^{κ}ℝ^{m+n} with d_{k} := Δv_{k} Δh_{k}/min Δh_{k}. Thereafter, we solve the two-dimensional LP min h in h, w ∈ ^{κ}ℝ_{≥0} with v* := v + wd. We always continue with the v* that belongs to the smallest h resp. h_{k} reached so far. We solve the two-dimensional LPs via bisection method.

We basically may extrapolate h und v in O(m + n). When we cannot leave v, we relax all constraints except v ≥ 0 a bit and undo that at the end of one step of an iteration in O(mn). We compute b_{k} = a_{k}^{T}x/√n = -max(Σ x_{j})/√n reaching its minimum for a_{k} = -(1, ..., 1)^{T}. An analogous computation for y yields h_{0} as stated above. Beginning with v ≠ 0 is mutatis mutandis possible. By the strong duality theorem, the claim follows.⃞

Corollary: Every linear system (LS) of equations Ax = b with x ∈ ^{κ}ℝ^{n} may be solved in O(mnh_{0}/Δh_{⌀}), provided that a solution exists, where a_{i} ≠ 0 must hold for b_{i} ≠ 0 and x = 0 solves the case b = 0.

Proof: We write Ax = b as Ax ≤ b and -Ax ≤ -b where x = x^{+} - x^{-} with x^{+}, x^{-} ≥ 0 and solve min h with h ∈ ^{κ}ℝ_{≥0} and Ax - b ≤ (h, …, h)^{T} ∈ ^{κ}ℝ^{m}, -Ax + b ≤ (h, …, h)^{T} plus h_{0} := max |b_{i}|/||a_{i}||.⃞

Remark: This is only really interesting for a small h_{0}/Δh_{⌀}.

Theorem: Smale's 9th problem is unsolvable. Thus, there is no strongly polynomial algorithm to solve an LP or a system of linear inequalities (SLI).

Proof: We normalise and relax the SLI by (h, …, h)^{T} ∈ ^{κ}ℝ^{n} with h ≥ 0 so that 0 ∈ ^{κ}ℝ^{n} solves it and min h is the corresponding LP. Let C be n-dimensional axes of coordinates (of equal length) inscribed in a polytope of the LP. We can linear often juxtapose this C along a "diagonal", changing its direction again and again, so that extrapolation is not possible, until the minimum for h = 0 is reached, if the LP can be solved.

The minimum h_{k} of every iteration step is assumed always to be located above the centre of C and is computed for all axes of C in two-dimensional cuts. The length of the axes of C may change here. The linear factor k ∈ ^{κ}ℕ* is proportional to the extent of the polytope along the "diagonal" passed and hence dependent on the size of the input of the initial problem, even if we would align C always differently.⃞

Theorem: Every convex programme min {f_{1}(x) : 0 ≤ x ∈ ^{κ}ℝ^{n}, (f_{2}(x), …, f_{m}(x))^{T} ≤ 0} where the f_{i} ∈ ^{κ}ℝ are convex posynomials is strongly polynomial and may be solved by the normal method and Newton's method, which handles the f_{i}, in O(ph_{0}/Δh_{⌀}), assuming that it is solvable, where p ∈ ^{κ}ℕ* denotes the number of operands x_{j} of the f_{i} and the objective function f_{1} is linearised.

Proof: The claim follows from the existence of the normal method.⃞

Remarks: The normal method is numerically very stable, since the initial data are barely altered, and currently the fastest known worst-case LP-solving algorithm. It may also be applied to (mixed) integer problems and branch-and-bound, in particular for nonconvex optimisation. It can be better paralleled than the simplex method. It can easily be extended to convex programmes with vector-valued or other convex f_{i}.

Diameter theorem for polytopes: The diameter of a n-dimensional polytope defined by m constraints with m, n ∈ ^{κ}ℕ* is at most max(2(m - n), 0).

Proof: Each vertex of a (potentially deformed) hypercube is formed by at most n hyperplanes. If we complete the polytope by adding or removing hyperplanes, the claim follows for the chosen path, since each step adds at most two additional edges. This theorem can be extended to polyhedra analogously by dropping the requirement of finiteness.⃞

Further thoughts: Gomory or Lenstra cuts can find an integer solution of the original problem in polynomial time if we additionally assume that a, b, and c are integers wlog and that m and n are fixed. By dualising, a full-dimensional LP may be obtained as described before within the normal method. This shows that the problem of (mixed) integer linear programming is not NP-complete:

Theorem: (Mixed) integer LPs may be solved in polynomial time.⃞

code of simplex method and data file example

© 2008-2017 by Boris Haase

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