hanlecture2 me

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Multiobjective Linear ProgrammingBiobjective LPs and Parametric Simplex

Multiobjective Simplex Method

International Doctoral School Algorithmic Decision

Theory: MCDA and MOO

Lecture 2: Multiobjective Linear Programming

Matthias Ehrgott

Department of Engineering Science, The University of Auckland, New Zealand

Laboratoire dInformatique de Nantes Atlantique, CNRS, Universite de Nantes,

France

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

Matthias Ehrgott MOLP I
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Overview

1 Multiobjective Linear ProgrammingFormulation and ExampleSolving MOLPs by Weighted Sums

2 Biobjective LPs and Parametric SimplexThe Parametric Simplex AlgorithmBiobjective Linear Programmes: Example

3 Multiobjective Simplex MethodA Multiobjective Simplex AlgorithmMultiobjective Simplex: Examples

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Formulation and ExampleSolving MOLPs by Weighted Sums

Overview




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Variables x Rn

Objective function Cx where C Rpn

Constraints Ax=bwhere A Rmn and b Rm

min {Cx :Ax=b, x 0} (1)

X ={x Rn :Ax=b, x 0}

is thefeasible set in decision space

Y ={Cx :xX}

is thefeasible set in objective space


M l i bj i Li P i
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Variables x Rn



min {Cx :Ax=b, x 0} (1)



Y ={Cx :xX}



M lti bj ti Li P i
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Variables x Rn



min {Cx :Ax=b, x 0} (1)



Y ={Cx :xX}



Multiobjective Linear Programming
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Variables x Rn



min {Cx :Ax=b, x 0} (1)



Y ={Cx :xX}



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Example

min

3x1+x2x1 2x2

subject to x2 3

3x1 x2 6

x 0

C =

3 11 2

A=

0 1 1 03 1 0 1

b=

36




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Example

min

3x1+x2x1 2x2

subject to x2 3

3x1 x2 6

x 0

C =

3 11 2

A=

0 1 1 03 1 0 1

b=

36


Multiobjective Linear ProgrammingF l i d E l
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0 1 2 3 4

0

1

2

3

4

X

x1

x2 x3

x4

Feasible set in decision space


Multiobjective Linear ProgrammingF l ti d E l
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j g gBiobjective LPs and Parametric Simplex



0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

0

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

Y

Cx1

Cx2

Cx3

Cx4

Feasible set in objective space


Multiobjective Linear ProgrammingFormulation and Example
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j g gBiobjective LPs and Parametric Simplex



Definition

Let xXbe a feasible solution of the MOLP (1) and let y=Cx.x is calledweakly efficientif there is no xX such thatCx0 such that for all i and x with cTi x0 such that cTj x>cTj x and

cTi x cTi x

cTj x cTj x

M.


Multiobjective Linear ProgrammingFormulation and Example
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Biobjective LPs and Parametric SimplexMultiobjective Simplex Method


Definition


cTi x cTi x

cTj x cTj x

M.


Multiobjective Linear ProgrammingBi bj i LP d P i Si l

Formulation and Example
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Definition


cTi x cTi x

cTj x cTj x

M.


Multiobjective Linear ProgrammingBi bj ti LP d P t i Si l



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Definition


cTi x cTi x

cTj x cTj x

M.


Multiobjective Linear ProgrammingBiobjecti e LPs a d Pa a et ic Si le



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pSolving MOLPs by Weighted Sums

0 1 2 3 4 5 6 7 8 9 10 11 12 1 3 14

0

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

Y

Cx1

Cx2

Cx3

Cx4

Nondominated set



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pSolving MOLPs by Weighted Sums

0 1 2 3 4

0

1

2

3

4

X

x1

x2 x3

x4

Efficient set





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Solving MOLPs by Weighted Sums

Overview






Formulation and ExampleS l i MOLP b W i h d S
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Let 1, . . . , p 0 and consider

LP() min

pk=1

kcTk x= min

TCx

subject to Ax = b

x 0

with some vector 0 (Why not = 0 or 0?)

LP() is a linear programme that can be solved by theSimplex method

If >0 then optimal solution ofLP() isproperly efficient

If 0 then optimal solution ofLP() isweakly efficient

Converse also true, because Y convex



Formulation and ExampleS l i MOLP b W i ht d S
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object e s a d a a et c S p eMultiobjective Simplex Method



LP() min

pk=1

kcTk x= min

TCx

subject to Ax = b

x 0








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j pMultiobjective Simplex Method



LP() min

pk=1

kcTk x= min

TCx

subject to Ax = b

x 0








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Multiobjective Simplex MethodSolving MOLPs by Weighted Sums


LP() min

pk=1

kcTk x= min

TCx

subject to Ax = b

x 0










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Multiobjective Simplex MethodSolving MOLPs by Weighted Sums


LP() min

pk=1

kcTk x= min

TCx

subject to Ax = b

x 0






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Multiobjective Simplex Methodg y g

y Rp satisfyingTy= define a straight line (hyperplane)

Since y=Cx and TCxis minimised, we push the linetowards the origin (left and down)

When the line only touches Ynondominated points are foundNondominated points YNare on the boundary ofY

Yis convex polyhedron and has finite number of facets. YNconsists of finitely many facets ofY. The normal of the facetcan serve as weight vector




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j p








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j p

Question: Can all efficient solutions be found using weighted sums?If xX is efficient, does there exist >0 such that x is optimal

solution to min{TCx :Ax=b, x 0}?

Lemma

A feasible solution x0 X is efficient if and only if the linearprogramme

max eTzsubject to Ax = b

Cx+Iz = Cx0

x, z 0,

(2)

where eT = (1, . . . , 1) Rp and I is the p p identity matrix, hasan optimal solution(x, z) with z= 0.




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Lemma



Cx+Iz = Cx0

x, z 0,

(2)





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Lemma



Cx+Iz = Cx0

x, z 0,

(2)





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Proof.LP is always feasible with x=x0, z= 0 (and value 0)

Let (x, z) be optimal solution

If z= 0 then z=Cx0 Cx= 0 Cx0 =Cx

There is no xX such that CxCx0

because (x, Cx0

Cx)would be better solution x0 efficient

If x0 efficient there is no xX with CxCx0

there is no z with z=Cx0 Cx0

max eT

z 0 max eT

z= 0


Multiobjective Linear ProgrammingBiobjective LPs and Parametric SimplexMultiobjective Simplex Method



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Proof.

LP is always feasible with x=x0, z= 0 (and value 0)




because (x, Cx0




max eT

z 0 max eT

z= 0





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Proof.





because (x, Cx0




max eT

z 0 max eT

z= 0



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Proof.





because (x, Cx0




max eT

z 0 max eT

z= 0



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Proof.





because (x, Cx0




max eT

z 0 max eT

z= 0



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Proof.





because (x, Cx0




max eT

z 0 max eT

z= 0



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Proof.





because (x, Cx0




max eT

z 0 max eT

z= 0



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Lemma


min uTb+wTCx0

subject to uTA+wTC 0w e

u Rm(3)

has an optimal solution(u, w) with uTb+ wTCx0 = 0.

Proof.

The LP (3) is the dual of the LP (2)



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Lemma


min uTb+wTCx0

subject to uTA+wTC 0w e

u Rm(3)

has an optimal solution(u, w) with uTb+ wTCx0 = 0.

Proof.

The LP (3) is the dual of the LP (2)



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Theorem

A feasible solution x0 X is an efficient solution of the MOLP (1)if and only if there exists a Rp> such that

TCx0 TCx (4)

for all xX .

Note: We already know that optimal solutions of weighted sum

problems are efficient



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Theorem

A feasible solution x0 X is an efficient solution of the MOLP (1)if and only if there exists a Rp> such that

TCx0 TCx (4)

for all xX .

Note: We already know that optimal solutions of weighted sum

problems are efficient



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Proof.

Let x0 XE

By Lemma4LP (3) has an optimal solution (u,w) such that

uTb=wTCx0 (5)

u is also an optimal solution of the LP

min

uTb:uTA wTC

, (6)

which is (3) with w= w fixed

There is an optimal solution of the dual of (6)

max

wTCx :Ax=b, x 0

(7)



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Proof.

Let x0 XE


uTb=wTCx0 (5)


min

uTb:uTA wTC

, (6)



max

wTCx :Ax=b, x 0

(7)




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Proof.

Let x0 XE


uTb=wTCx0 (5)


min

uTb:uTA wTC

, (6)



max

wTCx :Ax=b, x 0

(7)




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Proof.

Let x0 XE


uTb=wTCx0 (5)


min

uTb:uTA wTC

, (6)



max

wTCx :Ax=b, x 0

(7)



Multiobjective Simplex MethodFormulation and ExampleSolving MOLPs by Weighted Sums
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Proof.

By weak duality uTb wTCx for all feasible solutions uof(6) and for all feasible solutions x of (7)

We already know that uTb=wTCx0 from (5)

x0 is an optimal solution of (7)

Note that (7) is equivalent to

min

wTCx :Ax=b, x 0

with w e>0 from the constraints in (3)



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Proof.





min

wTCx :Ax=b, x 0




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Proof.





min

wTCx :Ax=b, x 0




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Proof.





min

wTCx :Ax=b, x 0




Multiobjective Simplex MethodThe Parametric Simplex AlgorithmBiobjective Linear Programmes: Example

Overview
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Overview



3 Multiobjective Simplex Method

A Multiobjective Simplex AlgorithmMultiobjective Simplex: Examples



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Modification of theSimplex algorithmfor LPs with twoobjectives

min

(c1)Tx, (c2)Tx

subject to Ax = b

x

0

(8)

We can find all efficient solutions by solving the parametric LP

min

1(c1)Tx+2(c

2)Tx :Ax=b, x 0

for all = (1, 2)> 0



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Modification of theSimplex algorithmfor LPs with twoobjectives

min

(c1)Tx, (c2)Tx

subject to Ax = b

x

0

(8)

We can find all efficient solutions by solving the parametric LP

min

1(c1)Tx+2(c

2)Tx :Ax=b, x 0

for all = (1, 2)> 0




The Parametric Simplex AlgorithmBiobjective Linear Programmes: Example


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We can divide the objective by 1+2 without changing theoptima, i.e. 1=1/(1+2),

2=2/(1+2 and

1+2= 1 or

2= 1 1

LPs with one parameter 0 1 and parametric objective

c() :=c1 + (1 )c2

min

c()Tx :Ax=b, x 0

(9)






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We can divide the objective by 1+2 without changing theoptima, i.e. 1=1/(1+2),

2=2/(1+2 and

1+2= 1 or

2= 1 1

LPs with one parameter 0 1 and parametric objective

c() :=c1 + (1 )c2

min

c()Tx :Ax=b, x 0

(9)




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LetBbe a feasible basis

Recall reduced cost cN =cN cTBB

1N

Reduced cost for the parametric LP

c() =c1 + (1 )c2 (10)

Suppose Bis an optimal basis of (9) for some

c() 0






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1N


c() =c1 + (1 )c2 (10)


c() 0




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1N


c() =c1 + (1 )c2 (10)


c() 0




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1N


c() =c1 + (1 )c2 (10)


c() 0




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1N


c() =c1 + (1 )c2 (10)


c() 0




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Case 1: c2 0

From (10) c() 0 for all


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Case 1: c2 0



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Case 1: c2 0



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Case 1: c2 0



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Case 1: c2 0



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Case 1: c2 0



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Case 1: c2 0



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Case 1: c2 0



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I={i N : c2i


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I={i N : c2i


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I={i N : c2i


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I={i N : c2i


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Input: Data A, b, C for a biobjective LP.

Phase I: Solve the auxiliary LP for Phase I using the Simplex algorithm. If the optimal value is positive, STOP, X =. Otherwise letB be an optimalbasis.

Phase II: Solve the LP (9) for= 1 starting from basisB found in Phase Iyielding an optimal basisB. ComputeA andb.

Phase III: WhileI={i N : c2

i 0

o.

LetB:= (B \ {r}) {s} and updateA andb.

End while.

Output: Sequence of -values and sequence of optimal BFSs.




The Parametric Simplex Algorithm

Biobjective Linear Programmes: Example

Overview


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2

Biobjective LPs and Parametric SimplexThe Parametric Simplex AlgorithmBiobjective Linear Programmes: Example








Example
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Example

min

3x1+x2x1 2x2

subject to x2 3

3x1 x2 6x 0

LP()

min (4 1)x1 + (3 2)x2subject to x2 + x3 = 3

3x1 x2 + x4 = 6x 0.





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Use Simplex tableaus showing reduced cost vectors c1 and c2

Optimal basis for = 1 is B={3, 4}, optimal basic feasible

solution x= (0, 0, 3, 6)Start with Phase 3





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Iteration 1:

c1 3 1 0 0 0

c2 -1 -2 0 0 0

x3 0 1 1 0 3

x4 3 -1 0 1 6

= 1, c() = (3, 1, 0, 0),B1 ={3, 4}, x1 = (0, 0, 3, 6)

I={1, 2}, = max

13+1

, 21+2

= 2

3

s= 2, r= 3

Matthias Ehrgott MOLP I Multiobjective Linear Programming



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Iteration 2

c1 3 0 -1 0 -3

c2 -1 0 2 0 6

x2 0 1 1 0 3

x4 3 0 1 1 9

= 2/3, c() = (5/3, 0, 0, 0),B2 ={2, 4}, x2 = (0, 3, 0, 9)

I={1}, = max

13+1

= 1

4

s= 1, r= 4




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Iteration 3

c1 0 0 -2 -1 -12

c2 0 0 7/3 1/3 9

x2 0 1 1 0 3

x1 1 0 1/3 1/3 3

= 1/4, c() = (0, 0, 5/4, 0),B3 ={1, 2}, x3 = (3, 3, 0, 0)I=





Weight values 1 = 1 2 = 2/3 3 = 1/4 4 = 0


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Weight values = 1, = 2/3, = 1/4, = 0

Basic feasible solutions x1

, x2

, x3

In each iteration c() can be calculated with the previous andcurrent c1 and c2.

BasisB1 = (3, 4) and BFS x1 = (0, 0, 3, 6) are optimal for [2/3, 1].

BasisB2 = (2, 4) and BFS x2 = (0, 3, 0, 9) are optimal for [1/4, 2/3], and

BasisB3 = (1, 2) and BFS x3 = (3, 3, 0, 0) are optimal for [0, 1/4].

Objective vectors for basic feasible solutions: Cx1 = (0, 0),Cx2 = (3, 6), and Cx3 = (12, 9)





Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0
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Weight values 1, 2/3, 1/4, 0


, x2

, x3










Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0


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Weight values 1, 2/3, 1/4, 0


, x2

, x3










Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0
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g , / , / ,


, x2

, x3










Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0
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g , / , / ,

Basic feasible solutions x

1

, x

2

, x

3










Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0
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g , / , / ,


1

, x

2

, x

3










Weight values 1 = 1, 2 = 2/3, 3 = 1/4, 4 = 0
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/ /


1

, x

2

, x

3










Values = 2/3 and = 1/4 correspond to weight vectors(2/3, 1/3) and (1/4, 3/4)

C
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Contour lines for weighted sum objectives in decision are

parallel to efficient edges

2

3(3x1+x2) +

1

3(x1 2x2) =

5

3x1

1

4 (3x

1+x

2) +

3

4 (x

1 2x

2) =

5

4x

2

0 1 2 3 4

0

1

2

3

4

X

x1

x2

x3

x4

Feasible set in decision space andefficient set






C li f i h d bj i i d i i
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2

3(3x1+x2) +

1

3(x1 2x2) =

5

3x1

1

4 (3x

1+x

2) +

3

4 (x

1 2x

2) =

5

4x

2

0 1 2 3 4

0

1

2

3

4

X

x1

x2

x3

x4







C li f i h d bj i i d i i
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2

3(3x1+x2) +

1

3(x1 2x2) =

5

3x1

1

4 (3x

1+x

2) +

3

4 (x

1 2x

2) =

5

4x

2

0 1 2 3 4

0

1

2

3

4

X

x1

x2

x3

x4



Biobjective LPs and Parametric Simplex




Weight vectors (2/3, 1/3) and (1/4, 3/4) are normal tonondominated edges


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0 1 2 3 4 5 6 7 8 9 10 1 1 1 2 1 3 1 4

0

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

Y

Cx1

Cx2

Cx3

Cx4

Objective space and nondominated set





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Algorithm findsall nondominated extreme pointsin objectivespace andone efficient bfsfor each of those

Algorithmdoes not find all efficient solutionsjust as Simplexalgorithm does not find all optimal solutions of an LP

Example

min (x1, x2)T

subject to 0 xi 1 i= 1, 2, 3

Efficient set: {x R3 :x1=x2= 0, 0 x3 1}







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Example

min (x1, x2)T







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Example

min (x1, x2)T






A Multiobjective Simplex Algorithm

Multiobjective Simplex: Examples

Overview
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min{Cx :Ax=b, x 0}


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{ }

LetBbe a basis and C =C CBA1B Aand R= CN

How to calculate critical ifp>2?

AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

T

CN = (5/3, 0)T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T

Find Rp, >0 such that TR 0 (optimality)andTrj = 0 (alternative optimum)for some column rj ofR






min{Cx :Ax=b, x 0}
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AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

T

CN = (5/3, 0)T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T







min{Cx :Ax=b, x 0}
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AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

T

CN = (5/3, 0)T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T







min{Cx :Ax=b, x 0}
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AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

T

CN = (5/3, 0)T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T







min{Cx :Ax=b, x 0}
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AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

TCN = (5/3, 0)

T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T







min{Cx :Ax=b, x 0}
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AtB1: CN =

3 11 2

, = 2/3, = (2/3, 1/3)T and

TCN = (5/3, 0)

T

AtB2: CN =

3 11 2

, = 1/4, = (1/4, 3/4)T and

TCN = (0, 5/4)T







Lemma


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Lemma

IfXE= then X has an efficient basic feasible solution.

Proof.

There is some >0 such that minxXTCxhas an optimal

solution

Thus LP() has an optimal basic feasible solution solution,which is an efficient solution of the MOLP





Lemma
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Lemma


Proof.


solution






Lemma


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Lemma


Proof.


solution






Definition

1 A feasible basis B is calledefficient basisifBis an optimal
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basis of LP() for some Rp>.

2 Two basesBand Bare calledadjacentif one can be obtainedfrom the other by a single pivot step.

3 LetBbe an efficient basis. Variable xj,j N is called

efficient nonbasic variableatB if there exists a Rp

> suchthat TR 0 and Trj = 0, where rj is the column ofRcorresponding to variable xj.

4 LetBbe an efficient basis and let xjbe an efficient nonbasicvariable. Then a feasible pivot from Bwith xjentering the

basis (even with negative pivot element) is called an efficientpivotwith respect to Band xj





Definition

1 A feasible basis B is calledefficient basisifBis an optimal


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Definition

1 A feasible basis B is calledefficient basisifBis an optimalp


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Definition

1 A feasible basis B is calledefficient basisifBis an optimalp


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No efficient basis is optimal for all pobjectives at the sametime

ThereforeRalways contains positive and negative entries

Proposition

LetBbe an efficient basis. There exists an efficient nonbasicvariable atB.




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Proposition





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Proposition






It is not possible to define efficient nonbasic variables by the


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p yexistence of a column in Rwith positive and negative entries

Example

R= 3 2

2 1

Tr2 = 0 requires 2= 21

Tr1 0 requires1 0, an impossibility for >0





Lemma

LetBbe an efficient basis and xjbe an efficient nonbasic variable.Then any efficient pivot fromBleads to an adjacent efficient basis
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B.

Proof.

xjefficient entering variable at basis B

there is Rp

> with

T

R 0 and

T

r

j

= 0 xjis nonbasic variable with reduced cost 0 in LP()

Reduced costs of LP() do not change after a pivot with xjentering

Let Bbe the resulting basis with feasible pivot and xj entering

Because TR 0 and Trj = 0 at B, Bis an optimal basisfor LP() and therefore an adjacent efficient basis





Lemma



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B.

Proof.


there is Rp

> with

T

R 0 and

T

r

j



Let Bbe the resulting basis with feasible pivot and xj entering






Lemma

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B.

Proof.


there is Rp

> with

T

R 0 and

T

r

j



Let Bbe the resulting basis with feasible pivot and

xj entering






Lemma

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B.

Proof.


there is Rp

> with

T

R

0 and

T

r

j



Let

Bbe the resulting basis with feasible pivot andxj entering






Lemma

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B.

Proof.


there is Rp

> with

T

R

0 and

T

r

j



Let Bbe the resulting basis with feasible pivot and xj

entering






Lemma

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B.

Proof.


there is Rp

> with

TR0 and

Trj




entering






Lemma

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B.

Proof.


there is Rp

> with

TR0 and

Trj




entering






How to identify efficient nonbasic variables?

Th


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TheoremLetBbe an efficient basis and let xj be a nonbasic variable.Variable xj is an efficient nonbasic variable if and only if the LP

max etvsubject to Rz rj+Iv = 0

z, , v 0(12)

has an optimal value of 0.

(12) is always feasible with (z, , v) = 0





How to identify efficient nonbasic variables?

Th
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TheoremLetBbe an efficient basis and let xj be a nonbasic variable.Variable xj is an efficient nonbasic variable if and only if the LP

max etv

subject to Rz rj+Iv = 0z, , v 0

(12)

has an optimal value of 0.

(12) is always feasible with (z, , v) = 0





Proof.

By definition xjis an efficient nonbasic variable if the LP

min 0T = 0T
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subject to RT 0(rj)T = 0

I e

(13)

has an optimal objective value of 0, i.e. if it is feasible(13) is equivalent to

min 0T = 0subject to RT 0

(rj

)T

0I e

(14)





Proof.

By definition xjis an efficient nonbasic variable if the LP

min 0T = 0T
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subject to RT 0(rj)T = 0

I e

(13)

has an optimal objective value of 0, i.e. if it is feasible(13) is equivalent to

min 0T = 0subject to RT 0

(rj

)T

0I e

(14)





Proof
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Proof.The dual of (14) is

max eTvsubject to Rz rj+Iv = 0

z, , v 0.

(15)





Need to show: ALL efficient bases can be reached by efficient

pivots


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pivots

Definition

Two efficient bases Band Bare calledconnectedif one can beobtained from the other by performing only efficient pivots.

Theorem

All efficient bases are connected.






pivots
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pivots

Definition


Theorem







pivots
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pivots

Definition


Theorem






Proof.

Band Btwo efficient bases

, Rp> such that Band Bare optimal bases for LP() and

LP()
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LP()Parametric LP ( [0, 1]) with objective function

c() = TC+ (1 )TC (16)

Assume Bis first basis (for = 1)

After several pivots get an optimal basis Bfor LP()

Since = + (1 ) Rp> for all [0, 1] all bases areoptimal for LP() for some Rp>, i.e. efficient

IfB=B, doneOtherwise obtain B from Bby efficient pivots: they arealternative optima for LP()





Proof.



LP()
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c() = TC+ (1 )TC (16)









Proof.



LP()
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c() = TC+ (1 )TC (16)









Proof.



LP()
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c() = TC+ (1 )TC (16)









Proof.



LP()


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c() = TC+ (1 )TC (16)





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Proof.



LP()


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( )Parametric LP ( [0, 1]) with objective function

c() = TC+ (1 )TC (16)

Assume Bis first basis (for = 1)After several pivots get an optimal basis Bfor LP()







Proof.



LP()


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( )Parametric LP ( [0, 1]) with objective function

c() = TC+ (1 )TC (16)

Assume Bis first basis (for = 1)After several pivots get an optimal basis Bfor LP()







3 casesX =, infeasibilityX = butXE=, no efficient solutionsX =, XE=

result in three phase multiobjective Simplex algorithmS { T A I b }
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p j p gPhase I: Solve min{eTz :Ax+Iz=b, x 0, z 0}If optimal value is nonzero, X =Otherwise find bfs ofAx=b, x 0 from optimal solution

Phase II: Find efficient bfs by solving appropriate LP()Note: LP() can be unbounded even ifXE=Solve min{uTb+wTCx0 :uTA+wTC 0, w e}If unbounded then XE =Otherwise find optimal wand solve

min{w Cx :Ax=b, x 0}Optimal bfs x1 exists and is efficient bfs for MOLPPhase III: Starting from x1 find all efficient bfs by efficientpivots, even with negative pivot elements






result in three phase multiobjective Simplex algorithmPh I S l i { T A I b 0 0}
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p j p gPhase I: Solve min{eTz :Ax+Iz=b, x 0, z 0}If optimal value is nonzero, X =Otherwise find bfs ofAx=b, x 0 from optimal solution

Phase II: Find efficient bfs by solving appropriate LP()

Note: LP() can be unbounded even ifXE=Solve min{uTb+wTCx0 :uTA+wTC 0, w e}If unbounded then XE =Otherwise find optimal wand solve







result in three phase multiobjective Simplex algorithmPh I S l i { T A + I b 0 0}
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Phase I: Solve min{eTz :Ax+Iz=b, x 0, z 0}If optimal value is nonzero, X =Otherwise find bfs ofAx=b, x 0 from optimal solution









result in three phase multiobjective Simplex algorithmPh I S l i { T A + I b 0 0}
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Phase I: Solve min{eTz :Ax+Iz=b, x 0, z 0}If optimal value is nonzero, X =Otherwise find bfs ofAx=b, x 0 from optimal solution








Algorithm (Multicriteria Simplex Algorithm.)

Input: Data A, b, C of an MOLP.

Initialization: SetL1:=, L2:=.
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Phase I: Solve the LPmin{eTz :Ax+Iz=b, x, z 0}.If the optimal value of this LP is nonzero, STOP,X =.Otherwise let x0 be a basic feasible solution of the MOLP.

Phase II: Solve the LPmin{uTb+wTCx0 :uTA+wTC 0, w e}.If the problem is infeasible, STOP,XE =.Otherwise let(u, w) be an optimal solution.Find an optimal basisBof the LPmin{wTCx :Ax=b, x 0}.

L1:={B},L2:=.





Algorithm

Phase III:

WhileL1 =

ChooseB inL1, setL1 :=L1\ {B}, L2 :=L2 {B}.ComputeA, b, and R according toB.

N N
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EN :=N.For all j N.

Solve the LPmax{eTv :Ry rj+Iv= 0; y, , v 0}.If this LP is unboundedEN :=EN \ {j}.

End forFor all j EN.

For all i B.

IfB = (B \ {i}) {j} is feasible andB L1 L2then L1 :=L1 B

.

End for.End for.

End while.

Output: L2.





Example

There can be exponentially many efficient bfs


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min xi i= 1, . . . , nmin xi i= 1, . . . , n

subject to xi 1 i= 1, . . . , nxi 1 i= 1, . . . , n.

n variables, m= 2n constraints, p= 2n objective functions

all 2n extreme points of the feasible set are efficient





Example

There can be exponentially many efficient bfs
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min xi i= 1, . . . , nmin xi i= 1, . . . , n

subject to xi 1 i= 1, . . . , nxi 1 i= 1, . . . , n.

n variables, m= 2n constraints, p= 2n objective functions

all 2n extreme points of the feasible set are efficient




A Multiobjective Simplex AlgorithmMultiobjective Simplex: Example

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