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A Metaheuristic Optimization Algorithm for

Binary Quadratic Problems

Otto Nissfolk

CENTER OF EXCELLENCE IN

OPTIMIZATION AND SYSTEMS ENGINEERING

ÅBO AKADEMI UNIVERSITY

(2)

I

Problem Formulation

.

The Quadratic Assignment Problem

.

QAP with rank-1 flow matrix

.

Convex QAP with rank-1 flow matrix

.

Metaheuristic Algorithm

I

Testproblems

.

Taixxc

(3)

min

x∈X

n

¼

i=1

n

¼

j=1

n

¼

k=1

n

¼

l=1

f

ik

d

jl

x

ij

x

kl

+

n

¼

i=1

n

¼

j=1

c

ij

x

ij

X

= {x |

n

¼

j=1

x

ij

= 1

i ∈ N

n

¼

i=1

x

ij

= 1

j ∈ N

x

ij

∈ {0

, 1}

i

, j ∈ M }

(4)

Problem Formulation: QAP with rank-1 flow matrix

4 | 16

n

¼

i=1

n

¼

j=1

n

¼

k=1

n

¼

l=1

f

ik

d

jl

x

ij

x

kl

= trace(DXFX

T

)

= trace(DXqq

T

X

T

)

= trace(q

T

X

T

DXq)

= trace(Xq

T

DXq)

= trace(y

T

Dy) = y

T

Dy

(5)

Problem Formulation: QAP with rank-1 flow matrix

4 | 16

n

¼

i=1

n

¼

j=1

n

¼

k=1

n

¼

l=1

f

ik

d

jl

x

ij

x

kl

= trace(DXFX

T

)

F=qq

T

= trace(q

X

DXq)

= trace(Xq

T

DXq)

= trace(y

T

Dy) = y

T

Dy

(6)

Problem Formulation: QAP with rank-1 flow matrix

4 | 16

n

¼

i=1

n

¼

j=1

n

¼

k=1

n

¼

l=1

f

ik

d

jl

x

ij

x

kl

= trace(DXFX

T

)

F=qq

T

= trace(DXqq

T

X

T

)

= trace(q

T

X

T

DXq)

= trace(Xq

T

DXq)

(7)

n

¼

i=1

n

¼

j=1

n

¼

k=1

n

¼

l=1

f

ik

d

jl

x

ij

x

kl

= trace(DXFX

T

)

F=qq

T

= trace(DXqq

T

X

T

)

= trace(q

T

X

T

DXq)

= trace(Xq

T

DXq)

= trace(y

T

Dy) = y

T

Dy

(8)

min

x∈X

,y∈‘

n

y

T

Dy

subject to

y

i

=

n

¼

j=1

x

ij

q

j

∀ i

n

¼

i=1

y

i

=

n

¼

j=1

q

j

(9)

min

x∈X

, y,z∈‘

n

y

T

(D + Diag(u))y − u

T

z

subject to

y

i

=

n

¼

j=1

x

ij

q

j

∀ i

z

i

=

n

¼

j=1

x

ij

q

j

2

∀ i

n

¼

i=1

y

i

=

n

¼

j=1

q

j

(10)

min x

T

(D + diag(u))x − u

T

x

subject to

n

¼

i=1

x

i

= k

iteration constraint

x

iter

= 1

(11)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(12)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(13)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

2

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(14)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

2

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

3

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(15)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

2

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

3

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

r1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(16)

x

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

2

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

3

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

r1

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

x

r

2

=

1

0

1

1

0

0

0

1

0

0

0

1

1

0

(17)

T

rstu

=

max

v

,w∈{−1,0,1}

1

(r − t + nv)

2

+ (s − u + nw)

2

f

ij

=

(

1

if i ≤ m and j ≤ m

0

otherwise

d

ij

= d

n(r−1)+s n(t−1)+u

= T

rstu

(18)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

BK

(19)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

5

BK

(20)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

5

10

BK

(21)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

5

10

12

BK

(22)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

5

10

12

16

BK

(23)

10

0

10

1

10

2

10

3

10

4

10

5

10

6

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Time in seconds

Objective

value

Objective function value vs. time

1

5

10

12

16

18

BK

(24)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

BK

(25)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

5

BK

(26)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

5

10

BK

(27)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

5

10

12

BK

(28)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

5

10

12

16

BK

(29)

10

0

10

1

10

2

10

3

10

4

4.47

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

1

5

10

12

16

18

BK

(30)

10

0

10

1

10

2

4.48

4.49

4.50

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

20

BK

(31)

10

0

10

1

10

2

4.48

4.49

4.50

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

20

22

BK

(32)

10

0

10

1

10

2

4.48

4.49

4.50

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

20

22

24

BK

(33)

10

0

10

1

10

2

4.48

4.49

4.50

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

20

22

24

26

BK

(34)

10

0

10

1

10

2

4.48

4.49

4.50

·10

7

Number of iterations

Objective

value

Objective function value vs. iteration count

20

22

24

26

28

BK

(35)

10

5

10

6

4.48

4.49

4.50

·10

7

Time

Objective

value

Objective function value vs. time

20

BK

(36)

10

5

10

6

4.48

4.49

4.50

·10

7

Time

Objective

value

Objective function value vs. time

20

22

BK

(37)

10

5

10

6

4.48

4.49

4.50

·10

7

Time

Objective

value

Objective function value vs. time

20

22

24

BK

(38)

10

5

10

6

4.48

4.49

4.50

·10

7

Time

Objective

value

Objective function value vs. time

20

22

24

26

BK

(39)

10

5

10

6

4.48

4.49

4.50

·10

7

Time

Objective

value

Objective function value vs. time

20

22

24

26

28

BK

(40)

1

10

100

1000

10000

4.48

4.49

4.50

4.51

4.52

4.53

·10

7

Number of iterations

Objective

value

Solution spread

min

mean

max

(41)

Alain Billionnet, Sourour Elloumi, and Marie-Christine Plateau.

Improving the performance of standard solvers for quadratic 0-1 programs by a tight convex

reformulation: The qcr method.

Discrete Appl. Math., 157:1185–1197, March 2009.

R.E. Burkard, E. Cela, P.M. Pardalos, and L.S. Pitsoulis.

Handbook of Combinatorial Optimization, volume 3.

1998.

C. S. Edwards.

A branch and bound algorithm for the koopmans-beckmann quadratic assignment problem.

Combinatorial Optimization II, 13:35–52, 1980.

Tjalling C. Koopmans and Martin Beckmann.

Assignment problems and the location of economic activities.

Econometrica, 25(1):pp. 53–76, 1957.

Otto Nissfolk, Ray Pörn, Tapio Westerlund, and Fredrik Jansson.

A mixed integer quadratic reformulation of the quadratic assignment problem with rank-1

matrix.

In Iftekhar A. Karimi and Rajagopalan Srinivasan, editors, 11th International Symposium on

Process Systems Engineering, volume 31 of Computer Aided Chemical Engineering, pages

360 – 364. Elsevier, 2012.

(42)

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