MCDAandMOO,HansurLesse,September17–212007 Lecture4:MultiobjectiveCombinatorialOptimizationMatthiasEhrgott InternationalDoctoralSchoolAlgorithmicDecisionTheory:MCDAandMOO

International Doctoral School Algorithmic Decision Theory: MCDA and MOO

Lecture 4: Multiobjective Combinatorial Optimization

Matthias Ehrgott

Department of Engineering Science, The University of Auckland, New Zealand Laboratoire d’Informatique de Nantes Atlantique, CNRS, Universit´ e de Nantes,

France

MCDA and MOO, Han sur Lesse, September 17 – 21 2007

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Mathematical Formulation

min z (x) = Cx subject to Ax = b

x ∈ {0, 1} ⁿ

x ∈ {0, 1} ⁿ −→ n variables, i = 1, . . . , n

C ∈ Z ^p×n −→ p objective functions, k = 1, . . . , p A ∈ Z ^m×n −→ m constraints, j = 1, . . . , m

Combinatorial structure: paths, trees, flows, tours, etc.

min z (x) = Cx subject to Ax = b

x ∈ {0, 1} ⁿ

x ∈ {0, 1} ⁿ −→ n variables, i = 1, . . . , n

C ∈ Z ^p×n −→ p objective functions, k = 1, . . . , p A ∈ Z ^m×n −→ m constraints, j = 1, . . . , m

Combinatorial structure: paths, trees, flows, tours, etc.

Mathematical Formulation

min z (x) = Cx subject to Ax = b

x ∈ {0, 1} ⁿ

x ∈ {0, 1} ⁿ −→ n variables, i = 1, . . . , n

C ∈ Z ^p×n −→ p objective functions, k = 1, . . . , p A ∈ Z ^m×n −→ m constraints, j = 1, . . . , m

Combinatorial structure: paths, trees, flows, tours, etc.

min z (x) = Cx subject to Ax = b

x ∈ {0, 1} ⁿ

x ∈ {0, 1} ⁿ −→ n variables, i = 1, . . . , n

C ∈ Z ^p×n −→ p objective functions, k = 1, . . . , p A ∈ Z ^m×n −→ m constraints, j = 1, . . . , m

Combinatorial structure: paths, trees, flows, tours, etc.

Feasible Sets

X = {x ∈ {0, 1} ⁿ : Ax = b}

feasible set in decision space Y = z(X ) = {Cx : x ∈ X } feasible set in objective space conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

X = {x ∈ {0, 1} ⁿ : Ax = b}

feasible set in decision space Y = z(X ) = {Cx : x ∈ X } feasible set in objective space conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Y = C ( X )

Feasible Sets

X = {x ∈ {0, 1} ⁿ : Ax = b}

feasible set in decision space Y = z(X ) = {Cx : x ∈ X } feasible set in objective space conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv(C(X)) + R

^p=

Individual minima

z k (ˆ x) 5 z k (x) for all x ∈ X Lexicographic optimality (1) z (ˆ x) 5 lex z(x) for all x ∈ X Lexicographic optimality (2) z ^π (ˆ x) 5 lex z ^π (x) for all x ∈ X and some permutation z ^π of (z ₁ , . . . , z _p )

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Lexicographic Optimality

Individual minima

z k (ˆ x) 5 z k (x) for all x ∈ X Lexicographic optimality (1) z (ˆ x) 5 lex z(x) for all x ∈ X Lexicographic optimality (2) z ^π (ˆ x) 5 lex z ^π (x) for all x ∈ X and some permutation z ^π of (z ₁ , . . . , z _p )

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Individual minima

z k (ˆ x) 5 z k (x) for all x ∈ X Lexicographic optimality (1) z (ˆ x) 5 lex z(x) for all x ∈ X Lexicographic optimality (2) z ^π (ˆ x) 5 lex z ^π (x) for all x ∈ X and some permutation z ^π of (z ₁ , . . . , z _p )

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Efficient Solutions

Weakly efficient solutions X _wE : there is no x with z (x) < z (ˆ x) z (ˆ x) is weakly nondominated Y _wN := z(X _wN )

Efficient solutions X E :

there is no x with z (x) ≤ z (ˆ x) z (ˆ x) is nondominated

Y N := z (X E )

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Weakly efficient solutions X _wE : there is no x with z (x) < z (ˆ x) z (ˆ x) is weakly nondominated Y _wN := z(X _wN )

Efficient solutions X E :

there is no x with z (x) ≤ z (ˆ x) z (ˆ x) is nondominated

Y N := z (X E )

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Efficient Solutions

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X _NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv(C(X)) + R

^p=

λ =

⁵₆

,

¹₆

Efficient Solutions

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X _NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv(C(X)) + R

^p=

λ =

¹₂

,

¹₂

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X _NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv(C(X)) + R

^p=

λ =

¹₃

,

²₃

Efficient Solutions

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X _NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv(C(X)) + R

^p=

λ =

²₃

,

¹₃

Supported efficient solutions X _SE : There is λ > 0 with λ ^T C x ˆ 5 λ ^T Cx for all x ∈ X

C x ˆ is extreme point of conv(Y ) + R

^p₌

→ X

SE1

C x ˆ is in relative interior of face of conv(Y ) + R

^p₌

→ X

SE2

Nonsupported efficient solutions X _NE : C x ˆ is in interior of conv(Y ) + R ^p ₌

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

conv( C ( X )) + R

ⁿ=

Nonsupported Effcient Solutions

Shortest path:

¹

2

4 3 (2,0)

(0,2) (3,3)

(2,0)

(0,2)

Y _N = 4

0

, 3

3

, 0

4

Spanning tree:

1 2

3 (3,0)

(1,1)

(0,3)

Y _N =

4 1

,

3 3

,

1 4

Assignment:

1 2 3

1 2 3

(3,0)

(2,2) (0,3) (2,2)

(2,0) (0,2)

(2,2)

(0,2) (2,0)

Y _N = 7

0

, 4

4

, 0

7

Hansen 1979:

x ¹ , x ² ∈ X E are equivalent if Cx ¹ = Cx ²

Complete set: ˆ X ⊂ X E such that for all y ∈ Y _N there is x ∈ X ˆ with z(x) = y

Minimal complete set contains no equivalent solutions

Maximal complete set contains all equivalent solutions

X

_E

X

_SE

X

_{N E}

X

_SE1

X

_E

m

X

_SE2

X

_{N E}

m

X

_SE1m

X

_SE2m

Classification of Efficient Sets

Hansen 1979:

x ¹ , x ² ∈ X E are equivalent if Cx ¹ = Cx ²

Complete set: ˆ X ⊂ X E such that for all y ∈ Y _N there is x ∈ X ˆ with z(x) = y

Minimal complete set contains no equivalent solutions

Maximal complete set contains all equivalent solutions

X

_E

X

_SE

X

_{N E}

X

_SE1

X

_E

m

X

_SE2

X

_{N E}

m

X

_SE1m

X

_SE2m

Hansen 1979:

x ¹ , x ² ∈ X E are equivalent if Cx ¹ = Cx ²

Complete set: ˆ X ⊂ X E such that for all y ∈ Y _N there is x ∈ X ˆ with z(x) = y

Minimal complete set contains no equivalent solutions

Maximal complete set contains all equivalent solutions

X

_E

X

_SE

X

_{N E}

X

_SE1

X

_E

m

X

_SE2

X

_{N E}

m

X

_SE1m

X

_SE2m

Classification of Efficient Sets

Hansen 1979:

x ¹ , x ² ∈ X E are equivalent if Cx ¹ = Cx ²

Complete set: ˆ X ⊂ X E such that for all y ∈ Y _N there is x ∈ X ˆ with z(x) = y

Minimal complete set contains no equivalent solutions

Maximal complete set contains all equivalent solutions

X

_E

X

_SE

X

_{N E}

X

_SE1

X

_E

m

X

_SE2

X

_{N E}

m

X

_SE1m

X

_SE2m

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

MOCO Problems Are Hard

Decision problem: Given b ∈ Z ^p : Does there exist x ∈ X such that Cx 5 b?

Counting problem: Given b ∈ Z ^p : How many x ∈ X satisfy Cx 5 b?

How many efficient solutions (nondominated points) do exist?

Knapsack: Given a ¹ , a ² ∈ Z ⁿ and b ₁ , b ₂ ∈ Z , does there

exist x ∈ {0, 1} ⁿ such that (a ¹ ) ^T x 5 b 1 and (a ² ) ^T x = b 2 ?

Knapsack is NP-complete and #P-complete

Decision problem: Given b ∈ Z ^p : Does there exist x ∈ X such that Cx 5 b?

Counting problem: Given b ∈ Z ^p : How many x ∈ X satisfy Cx 5 b?

How many efficient solutions (nondominated points) do exist?

Knapsack: Given a ¹ , a ² ∈ Z ⁿ and b ₁ , b ₂ ∈ Z , does there

exist x ∈ {0, 1} ⁿ such that (a ¹ ) ^T x 5 b 1 and (a ² ) ^T x = b 2 ?

Knapsack is NP-complete and #P-complete

MOCO Problems Are Hard

Decision problem: Given b ∈ Z ^p : Does there exist x ∈ X such that Cx 5 b?

Counting problem: Given b ∈ Z ^p : How many x ∈ X satisfy Cx 5 b?

How many efficient solutions (nondominated points) do exist?

Knapsack: Given a ¹ , a ² ∈ Z ⁿ and b ₁ , b ₂ ∈ Z , does there

exist x ∈ {0, 1} ⁿ such that (a ¹ ) ^T x 5 b 1 and (a ² ) ^T x = b 2 ?

Knapsack is NP-complete and #P-complete

Decision problem: Given b ∈ Z ^p : Does there exist x ∈ X such that Cx 5 b?

Counting problem: Given b ∈ Z ^p : How many x ∈ X satisfy Cx 5 b?

How many efficient solutions (nondominated points) do exist?

Knapsack: Given a ¹ , a ² ∈ Z ⁿ and b ₁ , b ₂ ∈ Z , does there

exist x ∈ {0, 1} ⁿ such that (a ¹ ) ^T x 5 b 1 and (a ² ) ^T x = b 2 ?

Knapsack is NP-complete and #P-complete

MOCO Problems Are Hard

Decision problem: Given b ∈ Z ^p : Does there exist x ∈ X such that Cx 5 b?

Counting problem: Given b ∈ Z ^p : How many x ∈ X satisfy Cx 5 b?

How many efficient solutions (nondominated points) do exist?

Knapsack: Given a ¹ , a ² ∈ Z ⁿ and b ₁ , b ₂ ∈ Z , does there

exist x ∈ {0, 1} ⁿ such that (a ¹ ) ^T x 5 b 1 and (a ² ) ^T x = b 2 ?

Knapsack is NP-complete and #P-complete

Observation

Multiobjective combinatorial optimization problems are NP-hard,

#P-complete, and intractable.

min

n

X

i=1

c _i ^k x _i k = 1, . . . , p

subject to x i ∈ {0, 1} i = 1, . . . , n

The Unconstrained MOCO Problem

Observation

Multiobjective combinatorial optimization problems are NP-hard,

#P-complete, and intractable.

min

n

X

i=1

c _i ^k x _i k = 1, . . . , p

subject to x i ∈ {0, 1} i = 1, . . . , n

Does there exist x ∈ {0, 1} ⁿ such that (c ¹ ) ^T x 5 d 1 and (c ² ) ^T x 5 d 2 ?

With an instance of Knapsack c ¹ := a ¹ , d 1 = b 1 ,

c ² := −a ² , d 2 := −b ₂ is a parsimonious transformation

With c _i ^k := (−1) ^k 2 ⁱ⁻¹ it holds Y = Y _N

The Unconstrained MOCO Problem

Does there exist x ∈ {0, 1} ⁿ such that (c ¹ ) ^T x 5 d 1 and (c ² ) ^T x 5 d 2 ?

With an instance of Knapsack c ¹ := a ¹ , d 1 = b 1 ,

c ² := −a ² , d 2 := −b ₂ is a parsimonious transformation

With c _i ^k := (−1) ^k 2 ⁱ⁻¹ it holds Y = Y _N

Does there exist x ∈ {0, 1} ⁿ such that (c ¹ ) ^T x 5 d 1 and (c ² ) ^T x 5 d 2 ?

With an instance of Knapsack c ¹ := a ¹ , d 1 = b 1 ,

c ² := −a ² , d 2 := −b ₂ is a parsimonious transformation

With c _i ^k := (−1) ^k 2 ⁱ⁻¹ it holds Y = Y _N

Multiobjective Shortest Path Problem

NP-hard: Hansen 1979

• • • •

(a

¹1

, 0) (a

¹2

, 0)

(0, a

²1

) (0, a

²2

)

v

0

v

1

v

2

v

n

Exponentially many efficient paths

• • • • • • • •

(1,0) (2,0) (2ⁿ⁻¹2 ,0)

(0,1) (0,0) (0,2) (0,0) (0,2ⁿ⁻¹2 ) (0,0)

. . .

v1 v2 v3 v4 v5 vn−2 vn−1 vn

NP-hard: Hansen 1979

• • • •

(a

¹1

, 0) (a

¹2

, 0)

(0, a

²1

) (0, a

²2

)

v

0

v

1

v

2

v

n

Exponentially many efficient paths

• • • • • • • •

(1,0) (2,0) (2ⁿ⁻¹2 ,0)

(0,1) (0,0) (0,2) (0,0) (0,2ⁿ⁻¹2 ) (0,0)

. . .

v1 v2 v3 v4 v5 vn−2 vn−1 vn

Other Examples

Assignment (NP-hard: Serafini 1986, #P-hard: Neumayer 1994)

Spanning tree (Intractable: Hamacher and Ruhe 1994)

Network flow (Intractable: Ruhe 1988)

Assignment (NP-hard: Serafini 1986, #P-hard: Neumayer 1994)

Spanning tree (Intractable: Hamacher and Ruhe 1994)

Network flow (Intractable: Ruhe 1988)

Other Examples

Assignment (NP-hard: Serafini 1986, #P-hard: Neumayer 1994)

Spanning tree (Intractable: Hamacher and Ruhe 1994)

Network flow (Intractable: Ruhe 1988)

Intractable: X E , even Y SN , can be exponential in the size of the instance

Number of nondominated points for biobjective shortest path and assignment problems (Raith 2007, Przybylski 2006)

Shortest path Assignment Nodes Edges |Y _N | n |Y _N |

4,902 19,596 6 10 13

4,902 19,596 1,594 20 82

3,000 33,224 15 40 243

14,000 153,742 17 60 470

330,386 1,202,458 21 80 671

330,386 1,202,458 24 100 947

Number of Efficient Solutions

Intractable: X E , even Y SN , can be exponential in the size of the instance

Number of nondominated points for biobjective shortest path and assignment problems (Raith 2007, Przybylski 2006)

Shortest path Assignment Nodes Edges |Y _N | n |Y _N |

4,902 19,596 6 10 13

4,902 19,596 1,594 20 82

3,000 33,224 15 40 243

14,000 153,742 17 60 470

330,386 1,202,458 21 80 671

330,386 1,202,458 24 100 947

Empirically often

|X _NE | grows exponentially with instance size

|X _SE | grows polynomially with instance size

but this depends on numerical values of C

Number of Efficient Solutions

Empirically often

|X _NE | grows exponentially with instance size

|X _SE | grows polynomially with instance size

but this depends on numerical values of C

Empirically often

|X _NE | grows exponentially with instance size

|X _SE | grows polynomially with instance size

but this depends on numerical values of C

Number of Efficient Solutions

0 0.05 0.1 0.15 0.2 0.25 0.3

50 100 150 200 250 300 350 400 450 500

nb SE par rap. NE

instance

Bi-KP Problems : proportion de supportees

A B C D

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

05 10 15 20 25 30 35 40 45 50

prop SE par rap. NE

instance Bi-AP Problems : prop SE par rap. NE

A B C D

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Digraph G = (V, A) with arc costs c _ij ^k , k = 1, . . . , p, (i, j ) ∈ A Given origin s ∈ V , destination t ∈ V find efficient paths from s to t:

P min ∈P

X

(i,j )∈P

c ij

where P is set of all s -t paths Assume that all c _ij ^k = 0 Proposition

Let P _st be an efficient path from s to t. Then any subpath P _uv

from u to v , where u and v are vertices on P _st is an efficient path

from u to v .

The Multiobjective Shortest Path Problem

Digraph G = (V, A) with arc costs c _ij ^k , k = 1, . . . , p, (i, j ) ∈ A Given origin s ∈ V , destination t ∈ V find efficient paths from s to t:

P min ∈P

X

(i,j )∈P

c ij

where P is set of all s -t paths Assume that all c _ij ^k = 0 Proposition

Let P _st be an efficient path from s to t. Then any subpath P _uv

from u to v , where u and v are vertices on P _st is an efficient path

from u to v .

Digraph G = (V, A) with arc costs c _ij ^k , k = 1, . . . , p, (i, j ) ∈ A Given origin s ∈ V , destination t ∈ V find efficient paths from s to t:

P min ∈P

X

(i,j )∈P

c ij

where P is set of all s -t paths Assume that all c _ij ^k = 0 Proposition

Let P _st be an efficient path from s to t. Then any subpath P _uv

from u to v , where u and v are vertices on P _st is an efficient path

from u to v .

The Multiobjective Shortest Path Problem

Digraph G = (V, A) with arc costs c _ij ^k , k = 1, . . . , p, (i, j ) ∈ A Given origin s ∈ V , destination t ∈ V find efficient paths from s to t:

P min ∈P

X

(i,j )∈P

c ij

where P is set of all s -t paths Assume that all c _ij ^k = 0 Proposition

Let P _st be an efficient path from s to t. Then any subpath P _uv

from u to v , where u and v are vertices on P _st is an efficient path

from u to v .

Concatenations of efficient paths need not be efficient!

1

2

3

4 (0, 10) (0, 10)

(1, 9) (1, 9)

(1, 15)

1-3 is efficient, 3-4 is efficient, 1-3-4 is not

The Multiobjective Shortest Path Problem

Concatenations of efficient paths need not be efficient!

1

2

3

4 (0, 10) (0, 10)

(1, 9) (1, 9)

(1, 15)

1-3 is efficient, 3-4 is efficient, 1-3-4 is not

For a labelling algorithm we need

Sets of nondominated labels at each node

Store all costs, the predecessor, the label number at the predecessor, the number of the current label

A list of permanent and temporary labels

Make sure that a permanent label defines an efficient path:

Choose the lexicographically smallest label from temporary list Lemma

If P ₁ and P ₂ are two paths between nodes s and t and

c(P ₁ ) ≤ c (P ₂ ) then c(P ₁ ) < _lex c (P ₂ ).

The Multiobjective Shortest Path Problem

For a labelling algorithm we need

Sets of nondominated labels at each node

Store all costs, the predecessor, the label number at the predecessor, the number of the current label

A list of permanent and temporary labels

Make sure that a permanent label defines an efficient path:

Choose the lexicographically smallest label from temporary list Lemma

If P ₁ and P ₂ are two paths between nodes s and t and

c(P ₁ ) ≤ c (P ₂ ) then c(P ₁ ) < _lex c (P ₂ ).

For a labelling algorithm we need

Sets of nondominated labels at each node

Store all costs, the predecessor, the label number at the predecessor, the number of the current label

A list of permanent and temporary labels

Make sure that a permanent label defines an efficient path:

Choose the lexicographically smallest label from temporary list Lemma

If P ₁ and P ₂ are two paths between nodes s and t and

c(P ₁ ) ≤ c (P ₂ ) then c(P ₁ ) < _lex c (P ₂ ).

The Multiobjective Shortest Path Problem

For a labelling algorithm we need

Sets of nondominated labels at each node

Store all costs, the predecessor, the label number at the predecessor, the number of the current label

A list of permanent and temporary labels

Make sure that a permanent label defines an efficient path:

Choose the lexicographically smallest label from temporary list Lemma

If P ₁ and P ₂ are two paths between nodes s and t and

c(P ₁ ) ≤ c (P ₂ ) then c(P ₁ ) < _lex c (P ₂ ).

For a labelling algorithm we need

Sets of nondominated labels at each node

Store all costs, the predecessor, the label number at the predecessor, the number of the current label

A list of permanent and temporary labels

Make sure that a permanent label defines an efficient path:

Choose the lexicographically smallest label from temporary list Lemma

If P ₁ and P ₂ are two paths between nodes s and t and

c(P ₁ ) ≤ c (P ₂ ) then c(P ₁ ) < _lex c (P ₂ ).

Algorithm (Multiobjective label setting algorithm)

Input: A digraph G = (V, A) with p arc costs.

Initialization: Create label L = (0, . . . , 0, 0, 0, 1) at node s and let T L := {L}.

While T L 6= ∅ do

Let label L = (c

¹

, . . . , c

^p

, v

h

, l , k) of node v

i

be the lexicographically smallest label in T L.

Remove L from T L and add it to PL.

For all v

j

∈ V such that (v

i

, v

j

) ∈ A do

Create label L

⁰

= (c

¹

+ c

¹

(v

i

, v

j

), . . . , c

^p

+ c

^p

(v

i

, v

j

), v

i

, k, t) as the next label at node v

j

and add it to T L.

Delete all temporary labels of node v

j

dominated by L

⁰

, delete L

⁰

if it is dominated by another label of node v

j

.

End for.

End while.

Use the predecessor labels in the permanent labels to recover all efficient paths from s to other nodes of G.

Output: All efficient paths from node s to all other nodes of G.

1

2

3

4

5

6 (10

; 4

; 2

; 10)

(6

;

1

;

18

;

10)

(6;2;10;10)

(0;10;12;1)

(4

; 0

; 0

; 3)

(1;4;8;1)

(5;1;3;7)

(10

;

1

;

1

;

1)

(6

; 0

; 0

; 6)

Multiobjective Label Setting Algorithm

1 2

3

4

5 6 1

1

2 9

6 4

5 7

9 2

3 10

75

3 10















 0 0 0 0 0 0 1

















1

0















 10

4 2 10

1 1 1

















4

1















 12

3 28 20 3 1 2

















6

2















 20 16 24 24 3 2 3

















9















 6 1 18 10 1 1 1

















2

1















 14 14 14 14 4 2 2

















9

5















 16

6 29 21 4 1 3

















7















 12

6 29 18 5 1 1

















7

3















 10 14 14 11 2 1 2

















5

4















 12 13 40 21 2 2 3

















6















 20 19 25 22 5 2 4

















10















 7 5 26 11 3 1 1

















3

2















 15 18 22 15 3 2 2

















10

9















 13

5 26 17 5 1 1

















8

3















 20 15 15 12 4 2 2

















11

5















 22

7 30 19 4 1 3

















7















 21 18 22 21 5 2 4

















11

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Multiobjective Label Correcting Algorithm

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Multiobjective Label Correcting Algorithm

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Multiobjective Label Correcting Algorithm

Label setting fails if negative arc lengths are permitted Negative cycles C

Case 1: If P

a∈C

c

_a^k

< 0 and P

a∈C

c

_a^j

> 0 for j 6= k there are infinitely many efficient paths

Case 2: If P

a∈C

c

a

≤ 0 there is no efficient path A label correcting algorithm is required

Let L(i , k ) be set of labels at node i in iteration k

Input: A digraph G = (V, A) with p arc costs.

Initialization: Set d

ii

:= (0, . . . , 0) for i = 1, . . . , n.

Set d

ij

:= (∞, . . . , ∞) if v

i

6= v

j

and (v

i

, v

j

) ∈ A. / Set d

ij

:= (c

¹

(v

i

, v

j

), . . . , c

^p

(v

i

, v

j

)) otherwise.

Set L(i , 1) := {d

1i

}, i = 1, . . . , n.

For k := 1 to n − 1 do For i := 1 to n do

L(i , k + 1) := min

n

S

j=1

˘ d

ji

+ l

j^k

: l

j^k

∈ L(j, k) ¯

End for.

If L(i, k + 1) = L(i, k) for all i = 1, . . . , n then If k = n − 1 then STOP, a negative cycle exists.

STOP End for.

Output: All efficient paths from node v

1

to all other nodes.

Multiobjective Label Correcting Algorithm

1

2

3

4 (1,−1,2) (−1,0,3)

(0,3,−2) (5,3,3)

(−1,2,4) ^dij=

0 B B B B B B B B B B B B B B B B B

@ 0

@ 0 0 0

1 A

0

@ 1

−1 2

1 A

0

@ 0 3

−2 1 A

0

@

∞

∞

∞ 1 A 0

@

∞

∞

∞ 1 A

0

@ 0 0 0

1 A

0

@

−1 2 4

1 A

0

@

−1 0 3

1 A 0

@

∞

∞

∞ 1 A

0

@

∞

∞

∞ 1 A

0

@ 0 0 0

1 A

0

@ 5 3 3

1 A 0

@

∞

∞

∞ 1 A

0

@

∞

∞

∞ 1 A

0

@

∞

∞

∞ 1 A

0

@ 0 0 0

1 A

1 C C C C C C C C C C C C C C C C C A

1

2

3

4 0

0

0

!

1

1

2

!

0

3

2

! 1

1

1

!

Multiobjective Label Correcting Algorithm

1

2

3

4 0

0

0

!

1

1

2

!

min (

0

3

2

!

; 1

1

2

!

+ 1

2

4

!)

= (

0

3

2

!

; 0

1

6

!) min

(

1

1

2

!

+ 1

0

3

!

; 0

3

2

!

+ 5

3

3

!)

= (

0

1

5

!

; 5

6

1

!)

1

2

3

4 0

0

0

!

1

1

2

!

(

0

3

2

!

; 0

1

6

!) min

(

0

1

5

!

; 5

6

1

!

; 0

1

6

!

+ 5

3

3

!)

= (

0

1

5

!

; 5

6

1

!)

The Multiobjective Spanning Tree Problem

Graph G = (V, A) with edge costs c _ij ^k , k = 1, . . . , p; (i, j ) ∈ E Find efficient spanning trees of G:

T min ∈T

X

[i,j ]∈T

c _ij

where T is set of all spanning trees of G

Graph G = (V, A) with edge costs c _ij ^k , k = 1, . . . , p; (i, j ) ∈ E Find efficient spanning trees of G:

T min ∈T

X

[i,j ]∈T

c _ij

where T is set of all spanning trees of G

The Multiobjective Spanning Tree Problem

Theorem (Hamacher and Ruhe 1994)

T efficient spanning tree of G

1

Let e ∈ E(T ) be an edge of T . Let (V(T ₁ ), E(T ₁ )) and (V (T 2 ), E(T ₂ )) be the two connected components of G \ {e }. Let

C (e) := {f = (v i , v j ) ∈ E : v i ∈ V (T 1 ), v j ∈ V(T ₂ )} be the cut defined by deleting e. Then c (e) ∈ min{c (f ) : f ∈ C (e )}.

e

T₁ T₂

edges inC(e)

Theorem (Hamacher and Ruhe 1994)

T efficient spanning tree of G

1

Let f ∈ E \ E(T ) and let P (f ) be the unique path in T connecting the end nodes of f . Then c(f ) ≤ c (e ) does not hold for any e ∈ P (f ).

f

P(f)

The Multiobjective Spanning Tree Problem

Algorithm (Prim’s spanning tree algorithm) Input: A graph G = (V, E) with p edge costs.

T ₀ := {({1}, ∅)}

For k := 1 to n − 1 do

T _k := {E (T ) ∪ {e _j } : T ∈ T _k−1 , e j ∈ argmin{c (e ) = c ([v _i , v _j ]) : v _i ∈ V (T ), v _j ∈ V \ V(T )}}

End for

T n−1 := argmin{c(T ) : T ∈ T n−1 }

Output: T n−1 , all efficient spanning trees of G.

1 2

3 4 1

4

0

2

1

3

3

1 2

0

1

The Multiobjective Spanning Tree Problem

1 2

3 4 1

4

0

2

1

3

3

1 2

0

1

3

1 2

3 4 1

4

0

2

1

3

3

1 2

0

1 2

3

1 2

3

The Multiobjective Spanning Tree Problem

1 2

3 4 1

4

0

2

1

3

3

1 2

0

1 2

3

4 1

2

3 4

1 2

3

4 1

2

3 4

1 2

3 4 1

4

0

2

1

3

3

1 2

0

1 2

3

4 1

2

3 4

1 2

3

4 1

2

3 4

Overview

1 Formulation and Definitions of Optimality

2 Complexity

3 The Multiobjective Shortest Path and Spanning Tree Problems

4 The Two Phase Method

5 Scalarization

6 Branch and Bound

7 Conclusion

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

The Two Phase Method with 2 Objectives

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

The Two Phase Method with 2 Objectives

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

The Two Phase Method with 2 Objectives

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

The Two Phase Method with 2 Objectives

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Phase 1: Compute X _SE(1)

1

Find lexicographic solutions

2

Recursively:

Calculate λ Solve min

x∈X

λ

^T

Cx Phase 2: Compute X _NE

1

Solve by triangle

2

Use neighborhood (wrong)

3

Use constraints (bad)

4

Use variable fixing (possible)

5

Use ranking (good)

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10

Enumeration Problems

Finding maximal complete set:

Enumeration to find all optimal solutions of min

_x∈X

λ

^T

Cx Enumeration to find all x ∈ X

NE

with Cx = y ∈ Y

ND

Finding minimal complete set:

Enumeration to find X

SE2

205 210 215 220 225 230 235

195 200 205 210 215 220 225

z2

z1 ( 206,223 ) ( 207,222 ) ( 208,221 ) ( 209,220 ) ( 210,219 ) ( 211,218 ) ( 199,230 )

( 220,209 ) ( 218,211 ) ( 202,227 )

( 203,226 ) ( 204,225 )

( 212,217 ) ( 213,216 ) ( 214,215 ) ( 215,214 ) ( 216,213 )

z(x)∈ZSN1 z(x)∈ZSN2

λ= 0.5666

=0.5 λ

=0.4918 λ