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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 GaussJordan Method Matrix Manipulations Postmultiplying a matrix by unit column vector
Be = B Premultiplying 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 ,..., e1 , y k , e+1 , ..., em ) F 1 = E = (e1 , e2 , ..., e1 , 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 (wAc)dk>0. Thus k is eligible to enter the basis Subproblem: bounded subproblem is bounded: optimal solution xk (wAc)dj 0 for all extreme directions DantzigWolfe 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 (wAc)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 righthandside. 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.
 Fall '09
 XinChen

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