# Decomposition - Revised Simplex Original Simplex worked on...

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Unformatted text preview: Revised Simplex Original Simplex worked on tableaus Great for hand computations but terrible for implementation Revised Simplex can obtain all Tableau entries as required less computation Connection to Gauss-Jordan Method Matrix Manipulations Post-multiplying a matrix by unit column vector Be = B Pre-multiplying matrix by unit row vector? e t B = B Post multiplying col. vector with unit row vector t yk e = 0 y 0 k Premultiplying col. Vector by unit row vector et yk = yk Basis Update Replacing a column B in the basis with Ak B = B + ( Ak - B ) e rank one matrix t 5 B = 2 1 5 B = 2 1 4 3 1 1 2 and A.k = 1 = 1 3 1 1 4 3 1 - 5 1 2 + 1 - 2 (1 0 0 ) 1 1 3 - 1 5 4 3 - 4 0 0 1 4 3 = 2 1 2 + - 1 0 0 = 1 1 2 1 1 1 2 0 0 3 1 1 New Basis from Old t B = B + ( Ak - B ) e mmm mm rank one matrix t = B + ( Ak - Be )e t t = B - Bee + A.k e ( ) = B( I - e e + y e ) m mmm m m m t t = B I - ee + B -1 A.k e t t k elementary matrix Differs from the identity matrix in exactly one column = B I - 0 0 0 1 0 0 0 + 0 y 0 k = BF Basis Inverse (B )-1 = F -1 B -1 b (new) = (B )-1b = F -1 B -1b = F -1(b ) F = (e1 , e2 ,..., e-1 , y k , e+1 , ..., em ) F -1 = E = (e1 , e2 , ..., e-1 , k , e+1 , ..., em ) -yik /yk where ik = 1/yk i i= To find E we store only k and position If initial basis is Identity after t iterations we get ( B t ) -1 = Et Et -1 E1 I Product Form of Inverse Product Form of Inverse ( B t ) -1 = Et Et -1 E1 I Used in older codes Eta file stored the sequence of matrices Et Et -1 E1 Only y k and were stored, eta was calculated as required Had advantages due to tape storage Necessity is the mother of invention Btran c ( B ) = c Et Et -1 E1 T Bt T Bt t -1 c Et = c + (c k - c e )e T Bt T Bt T Bt T Bt t c = c T Bt 1 Bt L c Bt L c m Bt c Et = c T Bt 1 Bt L cB t k L c m Bt Ftran ( B ) Ak = Et Et -1 E1 Ak t t E1 Ak = ( I + (k 1 - e )e ) Ak t = Ak + (k 1 - e )e Ak th -1 A ik + ik 1 A k = k1 A k i i = Ak = element of Ak i ik 1 A k A - A i = k k1 k Tableau Method IxB + B -1 NxN = B -1b T T cB ( IxB + B -1 NxN ) = cB B -1b T z - cB xB - c T x N = 0 N z 0z Pivot Row - 0 xB + IxB xBr 0 0 1 0 + (cB B -1 N - c N ) x N + B -1 Nx N x Bm 0 0 0 1 xN j L L L y1 j yrj ymj L L L = cB B -1b = xN k L L L L y1k yrk ymk B -1b Pivot Column z z 1 xB 0 xB 0 xB 0 1 r m xB1 0 1 0 0 RHS cB B _1b b1 Pivot Element br bm L L L L L L L L L z j - c j L z k - ck Example min s.t. - 12 x1 x1 x1 4 x1 x1 , - 9 x2 + x3 x2 + x2 + 2 x2 x2 , + x4 + x5 x3 , x4 , x5 , + x6 x6 = = = = 1000 1500 1750 4800 0 1 0 1 0 0 Initial Basis : B = 0 0 0 0 1000 -1 1500 0 0 b = B b= 1750 4800 0 0 0 0 0 B = 1 0 0 1 ( ) -1 ( ) T B w = c B = [ 0 0 0 0] B 0 -1 ( ) 0 -1 = [ 0 0 0 0] [ w0 A j - c j ] = [ -c j ] = [12,9] K = (1, 2) Select x1 to enter 1 0 1000 1750 4800 0 -1 = min y1= B A1 = ; , , = 1000 1 1 1 4 4 ( ) blocking variable is x3 ; 1 1 0 - 1 -yik /yk i 1 = i1 = = ; 1 = 1 - 1/yk i = 1 - 4 1 1 0 0 0 1 0 0 -1 -1 0 b 1 = B1 B1 = E1 B 0 = ; 1 0 1 0 - -4 0 0 1 ( ) ( ) ( ) 1000 -1 1500 b= 750 800 T w0 = cB B -1 = [ -12 0 0 0] B1 ( ) -1 = [ -12 0 0 0] [ w1 A j - c j ] = [0 - (-9), -12 - 0] = [9, -12] Review of Revised Simplex Method z z B B N 1 0 x 0 I x N N RHS BT -1 c -z -1 -c B b -1 x B N B b RHS BT -1 BASIS INVERSE T -w -1 -c B b -1 B B b An Example Example continued Example continued Example continued Example continued Example continued Example continued Example continued Example Continued Decomposition Principle Master Problem BASIS INVERSE T T RHS - (w , ) -1 B Optimality Conditions Subproblem To find a negative reduced cost, we only need to solve the following subproblem Subproblem: unbounded the subproblem is unbounded find an extreme direction dk such that (wA-c)dk>0. Thus k is eligible to enter the basis Subproblem: bounded subproblem is bounded: optimal solution xk (wA-c)dj 0 for all extreme directions Dantzig-Wolfe Decomposition Algorithm: I INITIALIZATION STEP BASIS INVERSE RHS -(w, ) -1 B MAIN STEP 1. Solve the following subproblem If the optimal objective value is zero, stop (optimal); otherwise, go to Step 2 if the subproblem is unbounded go to Step 3 if the subproblem is bounded Main Step 2 the subproblem is unbounded find an extreme direction dk such that (wA-c)dk>0. Thus k is eligible to enter the basis Main Step 3 subproblem is bounded: optimal solution xk Main Step 4 Pivot at yrk where the index r is determined as follows: This updates the dual variables, the basis inverse, and the right-hand-side. After pivoting, delete the column of k, and repeat Step 1. Remarks This procedure is sometimes referred to as a Column Generation Scheme If the master constraints are of the inequality type, then we must check the reduced cost for nonbasic slack variables in addition to solving the subproblem. For a master constraint i of the type having an associated slack variable The main advantage is storage requirement Computationally, one can usually achieve a convex combination of X that brings the resulting solution to within 1- 5% of optimality fairly quickly. However, after that, the progress are observed to be slow. A Lower Bound One Example Block Diagonal Structure Reformulation Optimality Conditions Subproblems Lower Bound Bender's Decomposition Technique Bender's Decomposition Technique Relaxed Master Program Lagrangian Relaxation Technique Lagrangian dual to problem P An Example Example continued Dual Program Lagrangian Dual Problem Step 1 Step 2 Step 3 ...
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## This note was uploaded on 09/08/2010 for the course IESE IE411 taught by Professor Xinchen during the Fall '09 term at University of Illinois, Urbana Champaign.

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