International Doctoral School Algorithmic Decision Theory: MCDA and MOO Lecture 2: Multiobjective Linear Programming Department of Engineering Science, The University of Auckland, New Zealand Laboratoire d Informatique de Nantes Atlantique, CNRS, Université de Nantes, France MCDA and MOO, Han sur Lesse, September 17 21 2007
Overview 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
Overview Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
Formulation and Example Solving MOLPs by Weighted Sums Variables x R n Objective function Cx where C R p n Constraints Ax = b where A R m n and b R m min {Cx : Ax = b, x 0} (1) X = {x R n : Ax = b, x 0} is the feasible set in decision space Y = {Cx : x X } is the feasible set in objective space
Formulation and Example Solving MOLPs by Weighted Sums Variables x R n Objective function Cx where C R p n Constraints Ax = b where A R m n and b R m min {Cx : Ax = b, x 0} (1) X = {x R n : Ax = b, x 0} is the feasible set in decision space Y = {Cx : x X } is the feasible set in objective space
Formulation and Example Solving MOLPs by Weighted Sums Variables x R n Objective function Cx where C R p n Constraints Ax = b where A R m n and b R m min {Cx : Ax = b, x 0} (1) X = {x R n : Ax = b, x 0} is the feasible set in decision space Y = {Cx : x X } is the feasible set in objective space
Formulation and Example Solving MOLPs by Weighted Sums Variables x R n Objective function Cx where C R p n Constraints Ax = b where A R m n and b R m min {Cx : Ax = b, x 0} (1) X = {x R n : Ax = b, x 0} is the feasible set in decision space Y = {Cx : x X } is the feasible set in objective space
Formulation and Example Solving MOLPs by Weighted Sums Example ( ) 3x1 + x min 2 x 1 2x 2 subject to x 2 3 3x 1 x 2 6 x 0 ( 3 1 C = 1 2 ) A = ( 0 1 1 0 3 1 0 1 ) b = ( 3 6 )
Formulation and Example Solving MOLPs by Weighted Sums Example ( ) 3x1 + x min 2 x 1 2x 2 subject to x 2 3 3x 1 x 2 6 x 0 ( 3 1 C = 1 2 ) A = ( 0 1 1 0 3 1 0 1 ) b = ( 3 6 )
Formulation and Example Solving MOLPs by Weighted Sums 4 3 2 1 0 x 2 x 3 X x 1 x 4 0 1 2 3 4 Feasible set in decision space
Formulation and Example Solving MOLPs by Weighted Sums 0-1 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Cx 1 Cx 4-3 -4 Y -5-6 -7 Cx 2-8 -9 Cx 3-10 Feasible set in objective space
Formulation and Example Solving MOLPs by Weighted Sums Definition Let ˆx X be a feasible solution of the MOLP (1) and let ŷ = C ˆx ˆx is called weakly efficient if there is no x X such that Cx < C ˆx; ŷ = C ˆx is called weakly nondominated ˆx is called efficient if there is no x X such that Cx C ˆx; ŷ = C ˆx is called nondominated ˆx is called properly efficient if it is efficient and if there exists a real number M > 0 such that for all i and x with c T i x < c T i ˆx there is an index j and M > 0 such that c T j x > c T j ˆx and c T i ˆx c T i x c T j x c T j ˆx M
Formulation and Example Solving MOLPs by Weighted Sums Definition Let ˆx X be a feasible solution of the MOLP (1) and let ŷ = C ˆx ˆx is called weakly efficient if there is no x X such that Cx < C ˆx; ŷ = C ˆx is called weakly nondominated ˆx is called efficient if there is no x X such that Cx C ˆx; ŷ = C ˆx is called nondominated ˆx is called properly efficient if it is efficient and if there exists a real number M > 0 such that for all i and x with c T i x < c T i ˆx there is an index j and M > 0 such that c T j x > c T j ˆx and c T i ˆx c T i x c T j x c T j ˆx M
Formulation and Example Solving MOLPs by Weighted Sums Definition Let ˆx X be a feasible solution of the MOLP (1) and let ŷ = C ˆx ˆx is called weakly efficient if there is no x X such that Cx < C ˆx; ŷ = C ˆx is called weakly nondominated ˆx is called efficient if there is no x X such that Cx C ˆx; ŷ = C ˆx is called nondominated ˆx is called properly efficient if it is efficient and if there exists a real number M > 0 such that for all i and x with c T i x < c T i ˆx there is an index j and M > 0 such that c T j x > c T j ˆx and c T i ˆx c T i x c T j x c T j ˆx M
Formulation and Example Solving MOLPs by Weighted Sums Definition Let ˆx X be a feasible solution of the MOLP (1) and let ŷ = C ˆx ˆx is called weakly efficient if there is no x X such that Cx < C ˆx; ŷ = C ˆx is called weakly nondominated ˆx is called efficient if there is no x X such that Cx C ˆx; ŷ = C ˆx is called nondominated ˆx is called properly efficient if it is efficient and if there exists a real number M > 0 such that for all i and x with c T i x < c T i ˆx there is an index j and M > 0 such that c T j x > c T j ˆx and c T i ˆx c T i x c T j x c T j ˆx M
Formulation and Example Solving MOLPs by Weighted Sums 0-1 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Cx 1 Cx 4-3 -4 Y -5-6 -7 Cx 2-8 -9 Cx 3-10 Nondominated set
Formulation and Example Solving MOLPs by Weighted Sums 4 3 2 1 0 x 2 x 3 X x 1 x 4 0 1 2 3 4 Efficient set
Overview Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
Formulation and Example Solving MOLPs by Weighted Sums Let λ 1,, λ p 0 and consider LP(λ) p min λ k ck T x = min λt Cx k=1 subject to Ax = b x 0 with some vector λ 0 (Why not λ = 0 or λ 0?) LP(λ) is a linear programme that can be solved by the Simplex method If λ > 0 then optimal solution of LP(λ) is properly efficient If λ 0 then optimal solution of LP(λ) is weakly efficient Converse also true, because Y convex
Formulation and Example Solving MOLPs by Weighted Sums Let λ 1,, λ p 0 and consider LP(λ) p min λ k ck T x = min λt Cx k=1 subject to Ax = b x 0 with some vector λ 0 (Why not λ = 0 or λ 0?) LP(λ) is a linear programme that can be solved by the Simplex method If λ > 0 then optimal solution of LP(λ) is properly efficient If λ 0 then optimal solution of LP(λ) is weakly efficient Converse also true, because Y convex
Formulation and Example Solving MOLPs by Weighted Sums Let λ 1,, λ p 0 and consider LP(λ) p min λ k ck T x = min λt Cx k=1 subject to Ax = b x 0 with some vector λ 0 (Why not λ = 0 or λ 0?) LP(λ) is a linear programme that can be solved by the Simplex method If λ > 0 then optimal solution of LP(λ) is properly efficient If λ 0 then optimal solution of LP(λ) is weakly efficient Converse also true, because Y convex
Formulation and Example Solving MOLPs by Weighted Sums Let λ 1,, λ p 0 and consider LP(λ) p min λ k ck T x = min λt Cx k=1 subject to Ax = b x 0 with some vector λ 0 (Why not λ = 0 or λ 0?) LP(λ) is a linear programme that can be solved by the Simplex method If λ > 0 then optimal solution of LP(λ) is properly efficient If λ 0 then optimal solution of LP(λ) is weakly efficient Converse also true, because Y convex
Formulation and Example Solving MOLPs by Weighted Sums Let λ 1,, λ p 0 and consider LP(λ) p min λ k ck T x = min λt Cx k=1 subject to Ax = b x 0 with some vector λ 0 (Why not λ = 0 or λ 0?) LP(λ) is a linear programme that can be solved by the Simplex method If λ > 0 then optimal solution of LP(λ) is properly efficient If λ 0 then optimal solution of LP(λ) is weakly efficient Converse also true, because Y convex
Formulation and Example Solving MOLPs by Weighted Sums Illustration in objective space 2 4 6 8 10 12 14 10 8 6 4 2 0 2 Y = CX Cx 1 Cx 2 Cx 3 Cx 4 y 1 y 2 {(λ 1 ) T Cx = α 1 } {(λ 2 ) T Cx = α 2 } {(λ 3 ) T Cx = α 3 } λ 1 = (2, 1), λ 2 = (1, 3), λ 3 = (1, 1)
Formulation and Example Solving MOLPs by Weighted Sums Illustration in objective space 2 4 6 8 10 12 14 10 8 6 4 2 0 2 Y = CX Cx 1 Cx 2 Cx 3 Cx 4 y 1 y 2 {(λ 1 ) T Cx = α 1 } {(λ 2 ) T Cx = α 2 } {(λ 3 ) T Cx = α 3 } λ 1 = (2, 1), λ 2 = (1, 3), λ 3 = (1, 1)
Formulation and Example Solving MOLPs by Weighted Sums y R p satisfying λ T y = α define a straight line (hyperplane) Since y = Cx and λ T Cx is minimised, we push the line towards the origin (left and down) When the line only touches Y nondominated points are found Nondominated points Y N are on the boundary of Y Y is convex polyhedron and has finite number of facets Y N consists of finitely many facets of Y The normal of the facet can serve as weight vector λ
Formulation and Example Solving MOLPs by Weighted Sums y R p satisfying λ T y = α define a straight line (hyperplane) Since y = Cx and λ T Cx is minimised, we push the line towards the origin (left and down) When the line only touches Y nondominated points are found Nondominated points Y N are on the boundary of Y Y is convex polyhedron and has finite number of facets Y N consists of finitely many facets of Y The normal of the facet can serve as weight vector λ
Formulation and Example Solving MOLPs by Weighted Sums y R p satisfying λ T y = α define a straight line (hyperplane) Since y = Cx and λ T Cx is minimised, we push the line towards the origin (left and down) When the line only touches Y nondominated points are found Nondominated points Y N are on the boundary of Y Y is convex polyhedron and has finite number of facets Y N consists of finitely many facets of Y The normal of the facet can serve as weight vector λ
Formulation and Example Solving MOLPs by Weighted Sums y R p satisfying λ T y = α define a straight line (hyperplane) Since y = Cx and λ T Cx is minimised, we push the line towards the origin (left and down) When the line only touches Y nondominated points are found Nondominated points Y N are on the boundary of Y Y is convex polyhedron and has finite number of facets Y N consists of finitely many facets of Y The normal of the facet can serve as weight vector λ
Formulation and Example Solving MOLPs by Weighted Sums y R p satisfying λ T y = α define a straight line (hyperplane) Since y = Cx and λ T Cx is minimised, we push the line towards the origin (left and down) When the line only touches Y nondominated points are found Nondominated points Y N are on the boundary of Y Y is convex polyhedron and has finite number of facets Y N consists of finitely many facets of Y The normal of the facet can serve as weight vector λ
Formulation and Example Solving MOLPs by Weighted Sums Question: Can all efficient solutions be found using weighted sums? If ˆx X is efficient, does there exist λ > 0 such that ˆx is optimal solution to min{λ T Cx : Ax = b, x 0}? Lemma A feasible solution x 0 X is efficient if and only if the linear programme max e T z subject to Ax = b Cx + Iz = Cx 0 x, z 0, (2) where e T = (1,, 1) R p and I is the p p identity matrix, has an optimal solution (ˆx, ẑ) with ẑ = 0
Formulation and Example Solving MOLPs by Weighted Sums Question: Can all efficient solutions be found using weighted sums? If ˆx X is efficient, does there exist λ > 0 such that ˆx is optimal solution to min{λ T Cx : Ax = b, x 0}? Lemma A feasible solution x 0 X is efficient if and only if the linear programme max e T z subject to Ax = b Cx + Iz = Cx 0 x, z 0, (2) where e T = (1,, 1) R p and I is the p p identity matrix, has an optimal solution (ˆx, ẑ) with ẑ = 0
Formulation and Example Solving MOLPs by Weighted Sums Question: Can all efficient solutions be found using weighted sums? If ˆx X is efficient, does there exist λ > 0 such that ˆx is optimal solution to min{λ T Cx : Ax = b, x 0}? Lemma A feasible solution x 0 X is efficient if and only if the linear programme max e T z subject to Ax = b Cx + Iz = Cx 0 x, z 0, (2) where e T = (1,, 1) R p and I is the p p identity matrix, has an optimal solution (ˆx, ẑ) with ẑ = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Proof LP is always feasible with x = x 0, z = 0 (and value 0) Let (ˆx, ẑ) be optimal solution If ẑ = 0 then ẑ = Cx 0 C ˆx = 0 Cx 0 = C ˆx There is no x X such that Cx Cx 0 because (x, Cx 0 Cx) would be better solution x 0 efficient If ˆx 0 efficient there is no x X with Cx Cx 0 there is no z with z = Cx 0 Cx 0 max e T z 0 max e T z = 0
Formulation and Example Solving MOLPs by Weighted Sums Lemma A feasible solution x 0 X is efficient if and only if the linear programme min u T b + w T Cx 0 subject to u T A + w T C 0 w e u R m (3) has an optimal solution (û, ŵ) with û T b + ŵ T Cx 0 = 0 Proof The LP (3) is the dual of the LP (2)
Formulation and Example Solving MOLPs by Weighted Sums Lemma A feasible solution x 0 X is efficient if and only if the linear programme min u T b + w T Cx 0 subject to u T A + w T C 0 w e u R m (3) has an optimal solution (û, ŵ) with û T b + ŵ T Cx 0 = 0 Proof The LP (3) is the dual of the LP (2)
Formulation and Example Solving MOLPs by Weighted Sums Theorem A feasible solution x 0 X is an efficient solution of the MOLP (1) if and only if there exists a λ R p > such that for all x X λ T Cx 0 λ T Cx (4) Note: We already know that optimal solutions of weighted sum problems are efficient
Formulation and Example Solving MOLPs by Weighted Sums Theorem A feasible solution x 0 X is an efficient solution of the MOLP (1) if and only if there exists a λ R p > such that for all x X λ T Cx 0 λ T Cx (4) Note: We already know that optimal solutions of weighted sum problems are efficient
Formulation and Example Solving MOLPs by Weighted Sums Proof Let x 0 X E By Lemma 4 LP (3) has an optimal solution (û, ŵ) such that û T b = ŵ T Cx 0 (5) û is also an optimal solution of the LP { } min u T b : u T A ŵ T C, (6) which is (3) with w = ŵ fixed There is an optimal solution of the dual of (6) { } max ŵ T Cx : Ax = b, x 0 (7)
Formulation and Example Solving MOLPs by Weighted Sums Proof Let x 0 X E By Lemma 4 LP (3) has an optimal solution (û, ŵ) such that û T b = ŵ T Cx 0 (5) û is also an optimal solution of the LP { } min u T b : u T A ŵ T C, (6) which is (3) with w = ŵ fixed There is an optimal solution of the dual of (6) { } max ŵ T Cx : Ax = b, x 0 (7)
Formulation and Example Solving MOLPs by Weighted Sums Proof Let x 0 X E By Lemma 4 LP (3) has an optimal solution (û, ŵ) such that û T b = ŵ T Cx 0 (5) û is also an optimal solution of the LP { } min u T b : u T A ŵ T C, (6) which is (3) with w = ŵ fixed There is an optimal solution of the dual of (6) { } max ŵ T Cx : Ax = b, x 0 (7)
Formulation and Example Solving MOLPs by Weighted Sums Proof Let x 0 X E By Lemma 4 LP (3) has an optimal solution (û, ŵ) such that û T b = ŵ T Cx 0 (5) û is also an optimal solution of the LP { } min u T b : u T A ŵ T C, (6) which is (3) with w = ŵ fixed There is an optimal solution of the dual of (6) { } max ŵ T Cx : Ax = b, x 0 (7)
Formulation and Example Solving MOLPs by Weighted Sums Proof By weak duality u T b ŵ T Cx for all feasible solutions u of (6) and for all feasible solutions x of (7) We already know that û T b = ŵ T Cx 0 from (5) x 0 is an optimal solution of (7) Note that (7) is equivalent to { } min ŵ T Cx : Ax = b, x 0 with ŵ e > 0 from the constraints in (3)
Formulation and Example Solving MOLPs by Weighted Sums Proof By weak duality u T b ŵ T Cx for all feasible solutions u of (6) and for all feasible solutions x of (7) We already know that û T b = ŵ T Cx 0 from (5) x 0 is an optimal solution of (7) Note that (7) is equivalent to { } min ŵ T Cx : Ax = b, x 0 with ŵ e > 0 from the constraints in (3)
Formulation and Example Solving MOLPs by Weighted Sums Proof By weak duality u T b ŵ T Cx for all feasible solutions u of (6) and for all feasible solutions x of (7) We already know that û T b = ŵ T Cx 0 from (5) x 0 is an optimal solution of (7) Note that (7) is equivalent to { } min ŵ T Cx : Ax = b, x 0 with ŵ e > 0 from the constraints in (3)
Formulation and Example Solving MOLPs by Weighted Sums Proof By weak duality u T b ŵ T Cx for all feasible solutions u of (6) and for all feasible solutions x of (7) We already know that û T b = ŵ T Cx 0 from (5) x 0 is an optimal solution of (7) Note that (7) is equivalent to { } min ŵ T Cx : Ax = b, x 0 with ŵ e > 0 from the constraints in (3)
Overview Multiobjective Linear Programming The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Modification of the Simplex algorithm for LPs with two objectives min ( (c 1 ) T x, (c 2 ) T x ) subject to Ax = b x 0 (8) We can find all efficient solutions by solving the parametric LP { } min λ 1 (c 1 ) T x + λ 2 (c 2 ) T x : Ax = b, x 0 for all λ = (λ 1, λ 2 ) > 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Modification of the Simplex algorithm for LPs with two objectives min ( (c 1 ) T x, (c 2 ) T x ) subject to Ax = b x 0 (8) We can find all efficient solutions by solving the parametric LP { } min λ 1 (c 1 ) T x + λ 2 (c 2 ) T x : Ax = b, x 0 for all λ = (λ 1, λ 2 ) > 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example We can divide the objective by λ 1 + λ 2 without changing the optima, ie λ 1 = λ 1/(λ 1 + λ 2 ), λ 2 = λ 2/(λ 1 + λ 2 and λ 1 + λ 2 = 1 or λ 2 = 1 λ 1 LPs with one parameter 0 λ 1 and parametric objective c(λ) := λc 1 + (1 λ)c 2 { } min c(λ) T x : Ax = b, x 0 (9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example We can divide the objective by λ 1 + λ 2 without changing the optima, ie λ 1 = λ 1/(λ 1 + λ 2 ), λ 2 = λ 2/(λ 1 + λ 2 and λ 1 + λ 2 = 1 or λ 2 = 1 λ 1 LPs with one parameter 0 λ 1 and parametric objective c(λ) := λc 1 + (1 λ)c 2 { } min c(λ) T x : Ax = b, x 0 (9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Let B be a feasible basis Recall reduced cost c N = c N cb T B 1 N Reduced cost for the parametric LP c(λ) = λ c 1 + (1 λ) c 2 (10) Suppose ˆB is an optimal basis of (9) for some ˆλ c(ˆλ) 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Let B be a feasible basis Recall reduced cost c N = c N cb T B 1 N Reduced cost for the parametric LP c(λ) = λ c 1 + (1 λ) c 2 (10) Suppose ˆB is an optimal basis of (9) for some ˆλ c(ˆλ) 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Let B be a feasible basis Recall reduced cost c N = c N cb T B 1 N Reduced cost for the parametric LP c(λ) = λ c 1 + (1 λ) c 2 (10) Suppose ˆB is an optimal basis of (9) for some ˆλ c(ˆλ) 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Let B be a feasible basis Recall reduced cost c N = c N cb T B 1 N Reduced cost for the parametric LP c(λ) = λ c 1 + (1 λ) c 2 (10) Suppose ˆB is an optimal basis of (9) for some ˆλ c(ˆλ) 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Let B be a feasible basis Recall reduced cost c N = c N cb T B 1 N Reduced cost for the parametric LP c(λ) = λ c 1 + (1 λ) c 2 (10) Suppose ˆB is an optimal basis of (9) for some ˆλ c(ˆλ) 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Case 1: c 2 0 From (10) c(λ) 0 for all λ < ˆλ ˆB is optimal basis for all 0 λ ˆλ Case 2: There is at least one i N with c 2 i < 0 there is λ < ˆλ such that c(λ) i = 0 λ c 1 i + (1 λ) c 2 i = 0 λ( c 1 i c 2 i ) + c2 i = 0 c2 i λ = c i 1 c2 i Below this value ˆB is not optimal
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example I = {i N : c 2 i < 0, c 1 i 0} λ := max i I c 1 i c i 2 c i 2 (11) ˆB is optimal for all λ [λ, ˆλ] As soon as λ < λ new bases become optimal Entering variable x s has to be chosen where the maximum in (11) is attained for i = s
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example I = {i N : c 2 i < 0, c 1 i 0} λ := max i I c 1 i c i 2 c i 2 (11) ˆB is optimal for all λ [λ, ˆλ] As soon as λ < λ new bases become optimal Entering variable x s has to be chosen where the maximum in (11) is attained for i = s
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example I = {i N : c 2 i < 0, c 1 i 0} λ := max i I c 1 i c i 2 c i 2 (11) ˆB is optimal for all λ [λ, ˆλ] As soon as λ < λ new bases become optimal Entering variable x s has to be chosen where the maximum in (11) is attained for i = s
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example I = {i N : c 2 i < 0, c 1 i 0} λ := max i I c 1 i c i 2 c i 2 (11) ˆB is optimal for all λ [λ, ˆλ] As soon as λ < λ new bases become optimal Entering variable x s has to be chosen where the maximum in (11) is attained for i = s
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example I = {i N : c 2 i < 0, c 1 i 0} λ := max i I c 1 i c i 2 c i 2 (11) ˆB is optimal for all λ [λ, ˆλ] As soon as λ < λ new bases become optimal Entering variable x s has to be chosen where the maximum in (11) is attained for i = s
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Algorithm (Parametric Simplex for biobjective LPs) 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 let B be an optimal basis Phase II: Solve the LP (9) for λ = 1 starting from basis B found in Phase I yielding an optimal basis ˆB Compute à and b Phase III: While I = {i N : c 2 i < 0, c 1 i 0} = c 2 i λ := max i I c i 1 c2 i n o c i s argmax i I : 2 c n i 1 c2 i o b r argmin j B : j, à à js > 0 js Let B := (B \ {r}) {s} and update à and b End while Output: Sequence of λ-values and sequence of optimal BFSs
Overview Multiobjective Linear Programming The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Example LP(λ) min ( 3x1 + x 2 ) x 1 2x 2 subject to x 2 3 3x 1 x 2 6 x 0 min (4λ 1)x 1 + (3λ 2)x 2 subject to x 2 + x 3 = 3 3x 1 x 2 + x 4 = 6 x 0
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Use Simplex tableaus showing reduced cost vectors c 1 and c 2 Optimal basis for λ = 1 is B = {3, 4}, optimal basic feasible solution x = (0, 0, 3, 6) Start with Phase 3
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Use Simplex tableaus showing reduced cost vectors c 1 and c 2 Optimal basis for λ = 1 is B = {3, 4}, optimal basic feasible solution x = (0, 0, 3, 6) Start with Phase 3
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Use Simplex tableaus showing reduced cost vectors c 1 and c 2 Optimal basis for λ = 1 is B = {3, 4}, optimal basic feasible solution x = (0, 0, 3, 6) Start with Phase 3
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Iteration 1: c 1 3 1 0 0 0 c 2-1 -2 0 0 0 x 3 0 1 1 0 3 x 4 3-1 0 1 6 λ = 1, c(λ) = (3, 1, 0, { 0), B 1 = {3, } 4}, x 1 = (0, 0, 3, 6) I = {1, 2}, λ = max 1 3+1, 2 1+2 = 2 3 s = 2, r = 3
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Iteration 2 c 1 3 0-1 0-3 c 2-1 0 2 0 6 x 2 0 1 1 0 3 x 4 3 0 1 1 9 λ = 2/3, c(λ) = (5/3, { 0, 0, } 0), B 2 = {2, 4}, x 2 = (0, 3, 0, 9) I = {1}, λ = max 1 3+1 = 1 4 s = 1, r = 4
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Iteration 3 c 1 0 0-2 -1-12 c 2 0 0 7/3 1/3 9 x 2 0 1 1 0 3 x 1 1 0 1/3 1/3 3 λ = 1/4, c(λ) = (0, 0, 5/4, 0), B 3 = {1, 2}, x 3 = (3, 3, 0, 0) I =
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight values λ 1 = 1, λ 2 = 2/3, λ 3 = 1/4, λ 4 = 0 Basic feasible solutions x 1, x 2, x 3 In each iteration c(λ) can be calculated with the previous and current c 1 and c 2 Basis B 1 = (3, 4) and BFS x 1 = (0, 0, 3, 6) are optimal for λ [2/3, 1] Basis B 2 = (2, 4) and BFS x 2 = (0, 3, 0, 9) are optimal for λ [1/4, 2/3], and Basis B 3 = (1, 2) and BFS x 3 = (3, 3, 0, 0) are optimal for λ [0, 1/4] Objective vectors for basic feasible solutions: Cx 1 = (0, 0), Cx 2 = (3, 6), and Cx 3 = (12, 9)
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Values λ = 2/3 and λ = 1/4 correspond to weight vectors (2/3, 1/3) and (1/4, 3/4) Contour lines for weighted sum objectives in decision are parallel to efficient edges 2 3 (3x 1 + x 2 ) + 1 3 ( x 1 2x 2 ) = 5 3 x 1 1 4 (3x 1 + x 2 ) + 3 4 ( x 1 2x 2 ) = 5 4 x 2 4 3 2 x 2 x 3 X Feasible set in decision space and efficient set 1 x 1 x 4 0 0 1 2 3 4
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Values λ = 2/3 and λ = 1/4 correspond to weight vectors (2/3, 1/3) and (1/4, 3/4) Contour lines for weighted sum objectives in decision are parallel to efficient edges 2 3 (3x 1 + x 2 ) + 1 3 ( x 1 2x 2 ) = 5 3 x 1 1 4 (3x 1 + x 2 ) + 3 4 ( x 1 2x 2 ) = 5 4 x 2 4 3 2 x 2 x 3 X Feasible set in decision space and efficient set 1 x 1 x 4 0 0 1 2 3 4
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Values λ = 2/3 and λ = 1/4 correspond to weight vectors (2/3, 1/3) and (1/4, 3/4) Contour lines for weighted sum objectives in decision are parallel to efficient edges 2 3 (3x 1 + x 2 ) + 1 3 ( x 1 2x 2 ) = 5 3 x 1 1 4 (3x 1 + x 2 ) + 3 4 ( x 1 2x 2 ) = 5 4 x 2 4 3 2 x 2 x 3 X Feasible set in decision space and efficient set 1 x 1 x 4 0 0 1 2 3 4
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Weight vectors (2/3, 1/3) and (1/4, 3/4) are normal to nondominated edges 0-1 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Cx 1 Cx 4-3 -4 Y -5-6 -7 Cx 2-8 -9 Cx 3-10 Objective space and nondominated set
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Example Algorithm finds all nondominated extreme points in objective space and one efficient bfs for each of those Algorithm does not find all efficient solutions just as Simplex algorithm does not find all optimal solutions of an LP min (x 1, x 2 ) T subject to 0 x i 1 i = 1, 2, 3 Efficient set: {x R 3 : x 1 = x 2 = 0, 0 x 3 1}
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Example Algorithm finds all nondominated extreme points in objective space and one efficient bfs for each of those Algorithm does not find all efficient solutions just as Simplex algorithm does not find all optimal solutions of an LP min (x 1, x 2 ) T subject to 0 x i 1 i = 1, 2, 3 Efficient set: {x R 3 : x 1 = x 2 = 0, 0 x 3 1}
The Parametric Simplex Algorithm Biobjective Linear Programmes: Example Example Algorithm finds all nondominated extreme points in objective space and one efficient bfs for each of those Algorithm does not find all efficient solutions just as Simplex algorithm does not find all optimal solutions of an LP min (x 1, x 2 ) T subject to 0 x i 1 i = 1, 2, 3 Efficient set: {x R 3 : x 1 = x 2 = 0, 0 x 3 1}
Overview Multiobjective Linear Programming A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples 1 Multiobjective Linear Programming Formulation and Example Solving MOLPs by Weighted Sums 2 The Parametric Simplex Algorithm Biobjective Linear Programmes: Example 3 A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples min{cx : Ax = b, x 0} Let B be a basis and C = C C B A 1 B A and R = C N How to calculate critical λ if p > 2? ( ) 3 1 At B 1 : C N =, λ 1 2 = 2/3, λ = (2/3, 1/3) T and λ T C N = (5/3, 0) T ( At B 2 : C 3 1 N = 1 2 λ T CN = (0, 5/4) T ), λ = 1/4, λ = (1/4, 3/4) T and Find λ R p, λ > 0 such that λ T R 0 (optimality) and λ T r j = 0 (alternative optimum) for some column r j of R
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma If X E then X has an efficient basic feasible solution Proof There is some λ > 0 such that min x X λ T Cx has an optimal solution Thus LP(λ) has an optimal basic feasible solution solution, which is an efficient solution of the MOLP
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma If X E then X has an efficient basic feasible solution Proof There is some λ > 0 such that min x X λ T Cx has an optimal solution Thus LP(λ) has an optimal basic feasible solution solution, which is an efficient solution of the MOLP
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma If X E then X has an efficient basic feasible solution Proof There is some λ > 0 such that min x X λ T Cx has an optimal solution Thus LP(λ) has an optimal basic feasible solution solution, which is an efficient solution of the MOLP
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Definition 1 A feasible basis B is called efficient basis if B is an optimal basis of LP(λ) for some λ R p > 2 Two bases B and ˆB are called adjacent if one can be obtained from the other by a single pivot step 3 Let B be an efficient basis Variable x j, j N is called efficient nonbasic variable at B if there exists a λ R p > such that λ T R 0 and λ T r j = 0, where r j is the column of R corresponding to variable x j 4 Let B be an efficient basis and let x j be an efficient nonbasic variable Then a feasible pivot from B with x j entering the basis (even with negative pivot element) is called an efficient pivot with respect to B and x j
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Definition 1 A feasible basis B is called efficient basis if B is an optimal basis of LP(λ) for some λ R p > 2 Two bases B and ˆB are called adjacent if one can be obtained from the other by a single pivot step 3 Let B be an efficient basis Variable x j, j N is called efficient nonbasic variable at B if there exists a λ R p > such that λ T R 0 and λ T r j = 0, where r j is the column of R corresponding to variable x j 4 Let B be an efficient basis and let x j be an efficient nonbasic variable Then a feasible pivot from B with x j entering the basis (even with negative pivot element) is called an efficient pivot with respect to B and x j
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Definition 1 A feasible basis B is called efficient basis if B is an optimal basis of LP(λ) for some λ R p > 2 Two bases B and ˆB are called adjacent if one can be obtained from the other by a single pivot step 3 Let B be an efficient basis Variable x j, j N is called efficient nonbasic variable at B if there exists a λ R p > such that λ T R 0 and λ T r j = 0, where r j is the column of R corresponding to variable x j 4 Let B be an efficient basis and let x j be an efficient nonbasic variable Then a feasible pivot from B with x j entering the basis (even with negative pivot element) is called an efficient pivot with respect to B and x j
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Definition 1 A feasible basis B is called efficient basis if B is an optimal basis of LP(λ) for some λ R p > 2 Two bases B and ˆB are called adjacent if one can be obtained from the other by a single pivot step 3 Let B be an efficient basis Variable x j, j N is called efficient nonbasic variable at B if there exists a λ R p > such that λ T R 0 and λ T r j = 0, where r j is the column of R corresponding to variable x j 4 Let B be an efficient basis and let x j be an efficient nonbasic variable Then a feasible pivot from B with x j entering the basis (even with negative pivot element) is called an efficient pivot with respect to B and x j
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples No efficient basis is optimal for all p objectives at the same time Therefore R always contains positive and negative entries Proposition Let B be an efficient basis There exists an efficient nonbasic variable at B
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples No efficient basis is optimal for all p objectives at the same time Therefore R always contains positive and negative entries Proposition Let B be an efficient basis There exists an efficient nonbasic variable at B
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples No efficient basis is optimal for all p objectives at the same time Therefore R always contains positive and negative entries Proposition Let B be an efficient basis There exists an efficient nonbasic variable at B
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples It is not possible to define efficient nonbasic variables by the existence of a column in R with positive and negative entries Example R = ( 3 2 2 1 ) λ T r 2 = 0 requires λ 2 = 2λ 1 λ T r 1 0 requires λ 1 0, an impossibility for λ > 0
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Lemma Let B be an efficient basis and x j be an efficient nonbasic variable Then any efficient pivot from B leads to an adjacent efficient basis ˆB Proof x j efficient entering variable at basis B there is λ R p > with λ T R 0 and λ T r j = 0 x j is nonbasic variable with reduced cost 0 in LP(λ) Reduced costs of LP(λ) do not change after a pivot with x j entering Let ˆB be the resulting basis with feasible pivot and x j entering Because λ T R 0 and λ T r j = 0 at ˆB, ˆB is an optimal basis for LP(λ) and therefore an adjacent efficient basis
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples How to identify efficient nonbasic variables? Theorem Let B be an efficient basis and let x j be a nonbasic variable Variable x j is an efficient nonbasic variable if and only if the LP has an optimal value of 0 max e t v subject to Rz r j δ + Iv = 0 z, δ, v 0 (12) (12) is always feasible with (z, δ, v) = 0
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples How to identify efficient nonbasic variables? Theorem Let B be an efficient basis and let x j be a nonbasic variable Variable x j is an efficient nonbasic variable if and only if the LP has an optimal value of 0 max e t v subject to Rz r j δ + Iv = 0 z, δ, v 0 (12) (12) is always feasible with (z, δ, v) = 0
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Proof By definition x j is an efficient nonbasic variable if the LP min 0 T λ = 0 subject to R T λ 0 (r j ) T λ = 0 I λ e (13) has an optimal objective value of 0, ie if it is feasible (13) is equivalent to min 0 T λ = 0 subject to R T λ 0 (r j ) T λ 0 I λ e (14)
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Proof By definition x j is an efficient nonbasic variable if the LP min 0 T λ = 0 subject to R T λ 0 (r j ) T λ = 0 I λ e (13) has an optimal objective value of 0, ie if it is feasible (13) is equivalent to min 0 T λ = 0 subject to R T λ 0 (r j ) T λ 0 I λ e (14)
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Proof The dual of (14) is max e T v subject to Rz r j δ + Iv = 0 z, δ, v 0 (15)
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Need to show: ALL efficient bases can be reached by efficient pivots Definition Two efficient bases B and ˆB are called connected if one can be obtained from the other by performing only efficient pivots Theorem All efficient bases are connected
A Multiobjective Simplex Algorithm Multiobjective Simplex: Examples Need to show: ALL efficient bases can be reached by efficient pivots Definition Two efficient bases B and ˆB are called connected if one can be obtained from the other by performing only efficient pivots Theorem All efficient bases are connected