DM545 Linear and Integer Programming Lecture 7 Revised Simplex Method Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark
Revised Simplex Method Outline Efficiency Issues 1. Revised Simplex Method 2. Efficiency Issues 2
Revised Simplex Method Motivation Efficiency Issues 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
Revised Simplex Method Outline Efficiency Issues 1. Revised Simplex Method 2. Efficiency Issues 4
Revised Simplex Method Revised Simplex Method Efficiency Issues Several ways to improve wrt pitfalls in the previous slide, requires matrix description of the simplex. n max c T x max { c T x | A x = b , x ≥ 0 } � max c j x j A x = b j = 1 n x ≥ 0 � a ij x j ≤ b i i = 1 .. m A ∈ R m × ( n + m ) j = 1 c ∈ R ( n + m ) , b ∈ R m , x ∈ R n + m x j ≥ 0 j = 1 .. n At each iteration the simplex moves from a basic feasible solution to another. For each basic feasible solution: • x N = 0 • B = { 1 . . . m } basis • x B ≥ 0 • N = { m + 1 . . . m + n } • A B = [ a 1 . . . a m ] basis matrix • A N = [ a m + 1 . . . a m + n ] 5
Revised Simplex Method Efficiency Issues A N A B 0 b c T c T 1 0 N B A x = A N x N + A B x B = b A B x B = b − A N x N Theorem Basic feasible solution ⇐ ⇒ A B is non-singular x B = A − 1 B b − A − 1 B A N x N 6
for the objective function: z = c T x = c T B x B + c T N x N Substituting for x B from above: z = c T B ( A − 1 B b − A − 1 B A N x N ) + c T N x N = B A − 1 B A − 1 = c T B b + ( c T N − c T B A N ) x N Collecting together: x B = A − 1 B b − A − 1 B A N x N B A − 1 B A − 1 z = c T B b + ( c T N − c T B A N ) x N � �� � ¯ A In tableau form, for a basic feasible solution corresponding to B we have: We do not need to A − 1 A − 1 B A N I 0 B b compute all elements of ¯ A B A − 1 B A − 1 c T N − c T − c T B A N 0 1 B b
Revised Simplex Method Example Efficiency Issues max x 1 + x 2 max x 1 + x 2 − x 1 + x 2 ≤ 1 − x 1 + x 2 + x 3 = 1 x 1 ≤ 3 x 1 + x 4 = 3 x 2 ≤ 2 x 2 + x 5 = 2 x 1 , x 2 ≥ 0 x 1 , x 2 , x 3 , x 4 , x 5 ≥ 0 Initial tableau After two iterations x 1 x 2 x 3 x 4 x 5 − z b x 1 x 2 x 3 x 4 x 5 − z b − 1 1 1 0 0 0 1 1 0 − 1 0 1 0 1 1 0 0 1 0 0 3 0 1 0 0 1 0 2 0 1 0 0 1 0 2 0 0 1 1 − 1 0 2 1 1 0 0 0 1 0 0 0 1 0 − 2 1 3 Basic variables x 1 , x 2 , x 4 . Non basic: x 3 , x 5 . From the initial tableau: − 1 1 0 1 0 x 1 � x 3 � A B = 1 0 1 A N = 0 0 x B = x 2 x N = x 5 0 1 0 0 1 x 4 � � � � c T c T B = 1 1 0 N = 0 0 8
Revised Simplex Method Efficiency Issues • Entering variable : in std. we look at tableau, in revised we need to compute: B A − 1 c T N − c T B A N 1. find y T = c T B A − 1 (by solving y T A B = c T B , the latter can be done B more efficiently) 2. calculate c T N − y T A N 9
Revised Simplex Method Efficiency Issues Step 1: − 1 1 0 � y 1 y 2 y 3 � � 1 1 0 � = y T A B = c T 1 0 1 B 0 1 0 − 1 0 1 − 1 � 1 1 0 � = B A − 1 c T = y T 0 0 1 0 B 1 1 − 1 2 Step 2: 1 0 � 0 0 � � − 1 0 2 � � 1 − 2 � = c T N − y T A N − 0 0 0 1 (Note that they can be computed individually: c j − y T a j > 0) Let’s take the first we encounter x 3 10
• 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 + 0 x 3 + x 5 = 2 x 2 = 2 ≥ 0 R3: − x 3 + x 4 − x 5 = 2 x 4 = 2 − x 3 ≥ 0 B − A − 1 x B = x ∗ B A N x N d is the column of A − 1 B A N that x B = x ∗ B − d θ corresponds to the entering variable, ie, d = A − 1 B a where a is the entering column 3. Find θ such that x B stays positive: Find d = A − 1 B a (by solving A B d = a ) Step 3: d 1 − 1 0 1 1 − 1 1 − 1 = = = θ ≥ 0 d 2 0 0 1 0 ⇒ d = 0 ⇒ x B = 2 − 0 d 3 1 1 − 1 0 1 2 1 2 − θ ≥ 0 = ⇒ θ ≤ 2 � x 4 leaves
Revised Simplex Method Efficiency Issues • So far we have done computations, but now we save the pivoting update. The update of A B is done by replacing the leaving column by the entering column x 1 − d 1 θ 3 − 1 1 1 = x ∗ B = x 2 − d 2 θ 2 A B = 1 0 0 2 0 1 0 θ • Many implementations depending on how y T A B = c T B and A B 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. B 12
Revised Simplex Method Outline Efficiency Issues 1. Revised Simplex Method 2. Efficiency Issues 13
Revised Simplex Method Solving the two Systems of Equations Efficiency Issues A B x = b solved without computing A − 1 B (costly and likely to introduce numerical inaccuracy) Recall how the inverse is computed: For a 2 × 2 matrix the matrix inverse is � d � d � T � a b � A − 1 = 1 1 � − c − b A = = c d | A | − b a ad − bc − c a For a 3 × 3 matrix the matrix inverse is T a 11 a 12 a 13 � � � � � � a 22 a 23 a 21 a 23 a 21 a 22 � � � � � � + � − � + A = a 21 a 22 a 23 � � � � � � a 32 a 33 a 31 a 33 a 31 a 32 � � � � a 31 a 32 a 33 � � � � � � 1 a 12 a 13 a 11 a 13 a 11 a 12 A − 1 = � � � � � � − � + � − � � � � � � | A | a 32 a 33 a 31 a 33 a 31 a 32 � � � � � � � � � � a 12 a 13 a 11 a 13 a 11 a 12 � � � � � � + � − � + � � � � � � a 22 a 23 a 21 a 23 a 21 a 22 � � � � 14
Revised Simplex Method Eta Factorization of the Basis Efficiency Issues Let A B = B , k th iteration B k be the matrix with col p differing from B k − 1 Column p is the a column appearing in B k − 1 d = a solved at 3) Hence: B k = B k − 1 E k E k is the eta matrix differing from id. matrix in only one column − 1 1 1 − 1 1 0 1 − 1 = 1 0 0 1 0 1 1 0 0 1 0 0 1 0 1 No matter how we solve y T B k − 1 = c T B and B k − 1 d = a , their update always relays on B k = B k − 1 E k with E k available. Plus when initial basis by slack variable B 0 = I and B 1 = E 1 , B 2 = E 1 E 2 · · · : B 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 B , v T E 3 = u T , w T E 2 = v T , y T E 1 = w T B , ( E 1 ( E 2 · · · E k d )) = a , E 1 u = a , E 2 v = u , E 3 w = v , E 4 d = w 15
Revised Simplex Method LU factorization Efficiency Issues Worth to consider also the case of B 0 � = I : B k = B 0 E 1 E 2 . . . E k eta factorization (((( y T B 0 ) E 1 ) E 2 ) · · · ) E k = c T B ( B 0 ( E 1 · · · E k d )) = a We need an LU factorization of B 0 16
Revised Simplex Method LU Factorization Efficiency Issues To solve the system A x = b by Gaussian Elimination we put the A matrix in row echelon form by means of elementary row operations. Each row operation corresponds to multiply left and right side by a lower triangular matrix L and a permuation matrix P . Hence, the method: A x = b L 1 P 1 A x = L 1 P 1 b L 2 P 2 L 1 P 1 A x = L 2 P 2 L 1 P 1 b . . . L m P m . . . L 2 P 2 L 1 P 1 A x = L m P m . . . L 2 P 2 L 1 P 1 b thus U = L m P m . . . L 2 P 2 L 1 P 1 A triangular factorization of A where U is an upper triangular matrix whose entries in the diagonal are ones. (if A is nonsingular such triangularization is unique) [see numerical example in Va sc 8.1] 17
We can compute the triangular factorization of B 0 before the initial iterations of the simplex: L m P m . . . L 2 P 2 L 1 P 1 B 0 = U We can then rewrite U as U = U m U m − 1 . . . , U 1 Hence, for B k = B 0 E 1 E 2 . . . E k : L m P m . . . L 2 P 2 L 1 P 1 B k = U m U m − 1 . . . U 1 E 1 E 2 · · · E k Then y T B k = c T B can be solved by − 1 U m · · · E k first solving: B k = ( L m P m · · · L 1 P 1 ) � �� � � �� � U L (((( y T U m ) U m − 1 ) · · · ) E k = c T B y L − 1 U = c and then replacing y T by (( y T L m P m ) · · · ) L 1 P 1 w U = c w = y L − 1 = ⇒ y = L w
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