Applications of Max-Algebra to Diagonal Scaling of …web.mat.bham.ac.uk/P.Butkovic/My papers/Beamer...

Applications of Max-Algebra to Diagonal Scalingof Matrices

P.Butkoviµc (University of Birmingham)H.Schneider (University of Wisconsin)

Contents

1 MAX-ALGEBRA

De�nition and basic propertiesAn application in mixed-integer programming

2 MATRIX SCALING

Diagonal Scaling and variantsDiagonal Scaling: Finding ALL solutions using max-algebraScaling to Diagonal DominanceFull Term Rank Scaling

Contents

1 MAX-ALGEBRA

De�nition and basic properties

An application in mixed-integer programming

2 MATRIX SCALING


Contents

1 MAX-ALGEBRA


2 MATRIX SCALING


Contents

1 MAX-ALGEBRA


2 MATRIX SCALING

Diagonal Scaling and variants

Diagonal Scaling: Finding ALL solutions using max-algebraScaling to Diagonal DominanceFull Term Rank Scaling

Contents

1 MAX-ALGEBRA


2 MATRIX SCALING

Diagonal Scaling and variantsDiagonal Scaling: Finding ALL solutions using max-algebra

Scaling to Diagonal DominanceFull Term Rank Scaling

Contents

1 MAX-ALGEBRA


2 MATRIX SCALING

Diagonal Scaling and variantsDiagonal Scaling: Finding ALL solutions using max-algebraScaling to Diagonal Dominance

Full Term Rank Scaling

Contents

1 MAX-ALGEBRA


2 MATRIX SCALING


1.MAX-ALGEBRADe�nitions and basic properties

Max-Plus

a� b = max(a, b)a b = a+ b

a, b 2 R := R[ f�∞g(ε = �∞, e = 0)

Max-Times

a� b = max(a, b)a b = aba, b 2 R+

(ε = 0, e = 1)�R,max,+

�� (R+,max, .)

a� ε = a = ε� aa ε = ε = ε aa e = a = e a


Extension to matrices and vectors:

A� B = (aij � bij )A B =

�∑�k aik bkj

�α A = (α aij )

diag(d1, ..., dn) =

0BBBBBBB@

d1. . . ε

. . .

ε. . .

dn

1CCCCCCCAI = diag(e, ..., e)A I = A = I A


For A 2 Rn�n :

A A ... A| {z }k -times

= A(k )

Γ(A) = A� A2 � ...� An ... metric matrix∆(A) = I � Γ(A)

DA = (N, f(i , j); aij > εg, (aij ))... associated digraphA is called irreducible if DA is strongly connected.


For A 2 Rn�n :


= A(k )

Γ(A) = A� A2 � ...� An ... metric matrix

∆(A) = I � Γ(A)



For A 2 Rn�n :


= A(k )




For A 2 Rn�n :


= A(k )


DA = (N, f(i , j); aij > εg, (aij ))... associated digraph

A is called irreducible if DA is strongly connected.


For A 2 Rn�n :


= A(k )




MAX-TIMES ALGEBRA

The permanent:

maper(A) = ∑π

�∏i

ai ,π(i ) = maxπ

∏iai ,π(i )

ap(A) =

(π 2 Pn ;maper(A) = ∏

iai ,π(i )

)

Maximal cycle mean:

max(i1,...,ik )

ai1 i2ai2 i3 ...aik i1k| {z }

λ(A)

= max fλ; (9x 6= ε)A x = λ xg


The eigenspace of A 2 Rn�n :

V (A) = fx 6= ε; (9λ)A x = λ xg

Theorem

Let A 2 Rn�n

be irreducible. Then λ(A) is the unique eigenvalueof A. If moreover, aii = λ(A) = e for all i 2 N = f1, ..., ng then

V (A) = fΓ(A) u; u 2 Rn+g .

1.MAX-ALGEBRAAn application in mixed-integer programming

System of Dual Inequalities (SDI):

aij � xi � xj (i , j 2 N)

Mixed-Integer Solution to SDI (MISDI):

lj � xj � uj (j 2 N)

xj integer (j 2 J)A = (aij ) 2 Rn�n, u, l 2 Rn and J � N = f1, ..., ng are givenFor solving MISDI we need to know ALL solutions to SDI






xj integer (j 2 J)

A = (aij ) 2 Rn�n, u, l 2 Rn and J � N = f1, ..., ng are givenFor solving MISDI we need to know ALL solutions to SDI






xj integer (j 2 J)A = (aij ) 2 Rn�n, u, l 2 Rn and J � N = f1, ..., ng are given

For solving MISDI we need to know ALL solutions to SDI






xj integer (j 2 J)A = (aij ) 2 Rn�n, u, l 2 Rn and J � N = f1, ..., ng are givenFor solving MISDI we need to know ALL solutions to SDI


The systemaij � xi � xj (i , j 2 N)

is equivalent to

maxj2N

(aij + xj ) � xi (i 2 N)

In max-algebra:

∑j2N

�aij xj � xi (i 2 N)

A x � x (1)

How to �nd ALL solutions to (1)?

2. MATRIX SCALINGDiagonal Scaling

A x � x

In max-times:x�1i aijxj � 1 (i , j 2 N)

In conventional notation:

X�1AX � E

where X = diag(x1, ..., xn), x1, ..., xn > 0 and

E =

0B@ 1 � � � 1.... . .

...1 � � � 1

1CA


A x � xIn max-times:

x�1i aijxj � 1 (i , j 2 N)

In conventional notation:

X�1AX � E


E =

0B@ 1 � � � 1.... . .

...1 � � � 1

1CA


A x � xIn max-times:

x�1i aijxj � 1 (i , j 2 N)In conventional notation:

X�1AX � E


E =

0B@ 1 � � � 1.... . .

...1 � � � 1

1CA

2. MATRIX SCALINGDiagonal Scaling and variants

Diagonal Scaling (DS): Given A 2 Rn�n+ , �nd a diagonal

matrix X with positive diagonal such that

X�1AX � E

Generalizations/variants:

C � X�1AX � B

X�1AX = B

C (k ) � X�1A(k )X � B(k ) k = 1, ..., s

Solvability and solution methods have been known (Fiedler,Pták, Engel, Schneider, Saunders, Hershkowitz)Max-algebra: E¢ ciently describes ALL solutions




X�1AX � E


C � X�1AX � B

X�1AX = B

C (k ) � X�1A(k )X � B(k ) k = 1, ..., s

Solvability and solution methods have been known (Fiedler,Pták, Engel, Schneider, Saunders, Hershkowitz)

Max-algebra: E¢ ciently describes ALL solutions




X�1AX � E


C � X�1AX � B

X�1AX = B

C (k ) � X�1A(k )X � B(k ) k = 1, ..., s

Solvability and solution methods have been known (Fiedler,Pták, Engel, Schneider, Saunders, Hershkowitz)Max-algebra: E¢ ciently describes ALL solutions

2. MATRIX SCALINGDiagonal Scaling: Finding all solutions using max-algebra

A x � x () x � A x = x () (I � A) x = x

Γ(I � A) = ∆(A)For A 2 Rn�n

+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem

The following are equivalent for any A 2 Rn�n+ :

λ(A) � 1There exists a positive vector x satisfying ∆(A) x = xThere exists a positive vector x satisfying A x � x


A x � x () x � A x = x () (I � A) x = xΓ(I � A) = ∆(A)

For A 2 Rn�n+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem




A x � x () x � A x = x () (I � A) x = xΓ(I � A) = ∆(A)For A 2 Rn�n

+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem





+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem


λ(A) � 1

There exists a positive vector x satisfying ∆(A) x = xThere exists a positive vector x satisfying A x � x



+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem


λ(A) � 1There exists a positive vector x satisfying ∆(A) x = x

There exists a positive vector x satisfying A x � x



+ irreducible

A x � x () x = ∆(A) u, u � 0

Theorem




Theorem

For any A 2 Rn�n+ and x 2 Rn

+ the following are equivalent:

A x � x , x > 0

x = ∆(A) u for some u > 0x = ∆(A) u for some u � 0


Theorem



A x � x , x > 0x = ∆(A) u for some u > 0

x = ∆(A) u for some u � 0


Theorem



A x � x , x > 0x = ∆(A) u for some u > 0x = ∆(A) u for some u � 0


Theorem

LetQ = ∑�

k A(k )/B (k ) �∑�

k C(k )/A(k )

and X = diag(x1, ..., xn), x1, ..., xn > 0. Then

C (k ) � X�1A(k )X � B (k ) k = 1, ..., s

if and only if x = ∆(Q) u for some u > 0.

2. MATRIX SCALINGScaling to Diagonal Dominance

A = (aij ) 2 Rn�n+ is called diagonally dominant if for all i 2 N

aii = maxjaij = max

jaji

Scaling to Diagonal Dominance (SDD): Given A 2 Rn�n+ ,

with positive diagonal, �nd a diagonal matrix X with positivediagonal such that X�1AX is diagonally dominant.

Theorem

Let A = (aij ) 2 Rn�n+ be a matrix with positive diagonal,

L = diag(a11, ..., ann) and Q = L�1 A� A L�1.

SDD for A exists if and only if λ(Q) � 1.If λ(Q) � 1 then X�1AX is diagonally dominant,X = diag(x), if and only if x = ∆(Q) u for some u > 0.



aii = maxjaij = max

jaji



Theorem


L = diag(a11, ..., ann) and Q = L�1 A� A L�1.SDD for A exists if and only if λ(Q) � 1.

If λ(Q) � 1 then X�1AX is diagonally dominant,X = diag(x), if and only if x = ∆(Q) u for some u > 0.



aii = maxjaij = max

jaji



Theorem


L = diag(a11, ..., ann) and Q = L�1 A� A L�1.SDD for A exists if and only if λ(Q) � 1.If λ(Q) � 1 then X�1AX is diagonally dominant,X = diag(x), if and only if x = ∆(Q) u for some u > 0.

2. MATRIX SCALINGFull Term Rank Scaling

Full Term Rank Scaling (FTRS): GivenA 2 Rn�n

+ , α1, ..., αn, β1, ..., βn > 0, �nd diagonal matricesX ,Y with positive diagonal such that for some permutationπ 2 Pn the matrix B = XAY = (bij ) satis�es

1 B has row maxima α1, ..., αn2 B has column maxima β1, ..., βn3 bi ,π(i ) is both a row and column maximum for every i

If FTRS exists then it exists for π 2 Pn if and only ifπ 2 ap(A)If FTRS exists for π 2 Pn then αi = βπ(i ) for all i 2 N




1 B has row maxima α1, ..., αn

2 B has column maxima β1, ..., βn3 bi ,π(i ) is both a row and column maximum for every i





1 B has row maxima α1, ..., αn2 B has column maxima β1, ..., βn

3 bi ,π(i ) is both a row and column maximum for every i






If FTRS exists then it exists for π 2 Pn if and only ifπ 2 ap(A)

If FTRS exists for π 2 Pn then αi = βπ(i ) for all i 2 N


A method for �nding FTRS:

1 Find a π 2 ap(A) and check αi = βπ(i ) for all i 2 N

2 Permute the rows (or cols) of A to A0 so that id 2 ap(A0)3 Scale the rows (or cols) of A0 to C so that cii = αi for alli 2 N

4 Set L = diag(c11, ..., cnn) and Q = L�1 C � C L�15 If λ(Q) > 1 then FTRS does not exist for A, stop6 Set x = ∆(Q) u for any u > 0 and X = diag(x)7 Set D = X�1CX and apply the inverse permutation to therows (or cols) of D to obtain B



1 Find a π 2 ap(A) and check αi = βπ(i ) for all i 2 N2 Permute the rows (or cols) of A to A0 so that id 2 ap(A0)

3 Scale the rows (or cols) of A0 to C so that cii = αi for alli 2 N




1 Find a π 2 ap(A) and check αi = βπ(i ) for all i 2 N2 Permute the rows (or cols) of A to A0 so that id 2 ap(A0)3 Scale the rows (or cols) of A0 to C so that cii = αi for alli 2 N





4 Set L = diag(c11, ..., cnn) and Q = L�1 C � C L�1

5 If λ(Q) > 1 then FTRS does not exist for A, stop6 Set x = ∆(Q) u for any u > 0 and X = diag(x)7 Set D = X�1CX and apply the inverse permutation to therows (or cols) of D to obtain B




4 Set L = diag(c11, ..., cnn) and Q = L�1 C � C L�15 If λ(Q) > 1 then FTRS does not exist for A, stop

6 Set x = ∆(Q) u for any u > 0 and X = diag(x)7 Set D = X�1CX and apply the inverse permutation to therows (or cols) of D to obtain B




4 Set L = diag(c11, ..., cnn) and Q = L�1 C � C L�15 If λ(Q) > 1 then FTRS does not exist for A, stop6 Set x = ∆(Q) u for any u > 0 and X = diag(x)

7 Set D = X�1CX and apply the inverse permutation to therows (or cols) of D to obtain B

Applications of Max-Algebra to Diagonal Scaling of …web.mat.bham.ac.uk/P.Butkovic/My papers/Beamer...

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