DM545 Linear and Integer Programming. Lecture 7 Revised Simplex Method. Marco Chiarandini

Transcription

1 DM545 Linear and Integer Programming Lecture 7 Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

2 Outline

3 Motivation Complexity of single pivot operation in standard simplex: entering variable O(n) leaving variable O(m) updating the tableau O(mn) Problems with this: Time: we are doing operations that are not actually needed Space: we need to store the whole tableau: O(mn) floating point numbers Most problems have sparse matrices (many zeros) sparse matrices are typically handled efficiently the standard simplex has the Fill in effect: sparse matrices are lost accumulation of Floating Point Errors over the iterations 3

4 Outline

5 Several ways to improve wrt pitfalls in the previous slide, requires matrix description of the simplex. max n j=1 c j x j n a ij x j b i i = 1..m j=1 x j 0 j = 1..n max c T x max{c T x Ax = b, x 0} Ax = b x 0 A R m (n+m) c R (n+m), b R m, x R n+m At each iteration the simplex moves from a basic feasible solution to another. For each basic feasible solution: = {1... m} basis N = {m m + n} A = [a 1... a m ] basis matrix A N = [a m+1... a m+n ] x N = 0 x 0 5

6 A N A 0 b c T N c T 1 0 Ax = A N x N + A x = b Theorem A x = b A N x N asic feasible solution A is non-singular x = A 1 b A 1A Nx N 6

7 for the objective function: z = c T x = c T x + c T N x N Substituting for x from above: z = c T (A 1 b A 1 A Nx N ) + c T N x N = = c T A 1 b + (ct N c T A 1 A N)x N Collecting together: x = A 1 b A 1 A Nx N z = c T A 1 b + (ct N ct A 1 A N)x N }{{} Ā In tableau form, for a basic feasible solution corresponding to we have: A 1 A N I 0 A 1 b c T N ct A 1 A N 0 1 c T A 1b We do not need to compute all elements of Ā

8 c T = [ ] c T N = [ 0 0 ] 8 Example max x 1 + x 2 x 1 + x 2 1 x 1 3 x 2 2 x 1, x 2 0 Initial tableau x1 x2 x3 x4 x5 z b max x 1 + x 2 x 1 + x 2 + x 3 = 1 x 1 + x 4 = 3 x 2 + x 5 = 2 x 1, x 2, x 3, x 4, x 5 0 After two iterations x1 x2 x3 x4 x5 z b asic variables x 1, x 2, x 4. Non basic: x 3, x 5. From the initial tableau: x 1 [ ] A = A N = 0 0 x = x 2 x3 x N = x x 4

9 Entering variable: in std. we look at tableau, in revised we need to compute: c T N ct A 1 A N 1. find y T = c T A 1 (by solving yt A = c T, the latter can be done more efficiently) 2. calculate c T N yt A N 9

10 Step 1: [ y1 y 2 ] y = [ ] [ ] = [ ] Step 2: [ ] [ ] = [ 1 2 ] 0 1 y T A = c T c T A 1 = yt c T N y T A N (Note that they can be computed individually: c j y T a j > 0) Let s take the first we encounter x 3 10

11 Leaving variable we increase variable by largest feasible amount θ R1: x 1 x 3 + x 5 = 1 x 1 = 1 + x 3 0 R2: x 2 + 0x 3 + x 5 = 2 x 2 = 2 0 R3: x 3 + x 4 x 5 = 2 x 4 = 2 x 3 0 x = x A 1 A Nx N x = x dθ d is the column of A 1 A N that corresponds to the entering variable, ie, d = A 1 a where a is the entering column 3. Find θ such that x stays positive: Find d = A 1 a (by solving A d = a) Step 3: = = d = 0 = x = 2 0 θ d 1 d 2 d 3 2 θ 0 = θ 2 x 4 leaves

12 So far we have done computations, but now we save the pivoting update. The update of A is done by replacing the leaving column by the entering column x 1 d 1 θ x = x 2 d 2 θ = 2 A = θ Many implementations depending on how y T A = c T and A d = a are solved. They are in fact solved from scratch. many operations saved especially if many variables! special ways to call the matrix A from memory better control over numerical issues since A 1 can be recomputed. 12

13 Outline

14 Solving the two Systems of Equations A x = b solved without computing A 1 (costly and likely to introduce numerical inaccuracy) Recall how the inverse is computed: For a 2 2 matrix [ ] a b A = c d For a 3 3 matrix a 11 a 12 a 13 A = a 21 a 22 a 23 a 31 a 32 a 33 the matrix inverse is A 1 = 1 [ ] T [ ] d c 1 d b = A b a ad bc c a the matrix inverse is + a22 a23 a 32 a 33 a 21 a 23 a 31 a 33 + a 21 a 22 a 31 a 32 A 1 = 1 A a12 a13 a 32 a 33 + a 11 a 13 a 31 a 33 a 11 a 12 a 31 a 32 + a12 a13 a 22 a 23 a 11 a 13 a 21 a 23 + a 11 a 12 a 21 a 22 T 14

15 Eta Factorization of the asis Let := A, kth iteration k be the matrix with col p differing from k 1 Column p is the a column appearing in k 1 d = a solved at 3) Hence: k = k 1 E k E k is the eta matrix differing from id. matrix in only one column = No matter how we solve y T k 1 = c T and k 1d = a, their update always relays on k = k 1 E k with E k available. Plus when initial basis by slack variable 0 = I and 1 = E 1, 2 = E 1 E 2 : k = E 1 E 2... E k eta factorization ((((y T E 1)E 2)E 3) )E k = c T, u T E 4 = c T, v T E 3 = u T, w T E 2 = v T, y T E 1 = w T (E 1(E 2 E k d)) = a, E 1u = a, E 2v = u, E 3w = v, E 4d = w 15

16 Solving y T k = c T also called backward transformation (TRAN) Solving k d = a also called forward transformation (FTRAN) E i matrices can be stored by only storing the column and the position If sparse columns then can be stored in compact mode, ie only nonzero values and their indices 19

17 More on LP Tableau method is unstable: computational errors may accumulate. Revised method has a natural control mechanism: we can recompute A 1 at any time Commercial and freeware solvers differ from the way the systems y T A = c T and A d = a are resolved 21

18 Efficient Implementations Dual simplex with steepest descent (largest increase) Linear Algebra: Dynamic LU-factorization using Markowitz threshold pivoting (Suhl and Suhl, 1990) sparse linear systems: Typically these systems take as input a vector with a very small number of nonzero entries and output a vector with only a few additional nonzeros. Presolve, ie problem reductions: removal of redundant constraints, fixed variables, and other extraneous model elements. dealing with degeneracy, stalling (long sequences of degenerate pivots), and cycling: bound-shifting (Paula Harris, 1974) Hybrid Pricing (variable selection): start with partial pricing, then switch to devex (approximate steepest-edge, Harris, 1974) A model that might have taken a year to solve 10 years ago can now solve in less than 30 seconds (ixby, 2002). 22