Discrete Optimization with Ordering Elena Fernndez Tecnical - PowerPoint PPT Presentation

Discrete Optimization with Ordering Elena Fernández Tecnical University of Catalonia Barcel-berg Visiting Institut für Informatik. Heidelberg Justo Puerto University of Seville Antonio Rodríguez-Chía University of Cádiz Aussois 2009

Discrete Optimization with ordering Discrete optimization problems where feasible solutions are sequences of elements which are ordered with respect to a priority (hierarchy) function. The cost of an element depends on its position on the sequence. • Multiperiod problems • Scheduling and sequencing problems • … � The simple ordering problem (SOP) � Ordered sequences � Some properties � The polyhedron of the SOP � The simple ordering problem with cardinality constraint � The simple ordering problem on an independence system � The ordered median spanning tree problem

Ordered sequences � Ground set: E= { e 1 , e 2 , …, e n }. N = {1, 2, …, n }. � c : E → ℝ order function s.t. c 1 := c ( e 1 ) ≥ c ( e 2 ) ≥ … ≥ c ( e n ):= c n � Feasible Solutions: sequences with at most p ≤ n elements which are ordered wrt function c . K = {1, 2, … p }. [ e j 1 , e j 2 , …, e jr ], r ≤ p , such that j i < j i +1 e j1 e j1 … … e j2 e j2 … … e jr e jr � E K : multiset with p copies of each element e ∈ E. { } F ⊆ E K , � with k 1 ≤ k 2 ≤ … ≤ k r = k k k 1 2 K r F e , e , , e j j j 1 2 r F ⊆ E K , ordered sequence ⇔ k i < k i +1 and j i < j i +1 , i =1 ,…, p -1 � � Additive objective function: The value of each element depends on its position in the sequence. d : E K ⎯→ ℝ k ⎯→ d j k e j ∑ F ⎯→ k d j ∈ F k e j

Example E = { e 1 , e 2 , e 3 , e 4 } ( c 1 ≥ c 2 ≥ c 3 ≥ c 4 ), p = 3. ⎛ ⎞ 2 0 5 ⎜ ⎟ ⎜ ⎟ 2 2 3 = d ⎜ ⎟ 3 0 1 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 1 1 2 F = { e 2 1 , e 2 2 , e 3 2 , e 1 3 } is not an ordered sequence Ordered sequences: 1 , e 3 2 } , F 1 = { e 2 F 2 = { e 2 2 } , F 3 = { e 1 3 } d ( F 1 ) = 2 , d ( F 2 ) = 2 , d ( F 3 ) = 5

Some properties of ordered sequences � I = (E K , F ) is an Independence System, where F = { F ⊆ E K : F is an ordered sequence}. F ⊆ E K ordered sequence ⇒ S ordered sequence, for all S ⊆ F . � ℓ ( F ) = max{ | S | : S ⊆ F , S is an ordered sequence}. F is an ordered sequence ⇔ ℓ ( F ) = | F |. � I = (E K , F ) is not a matroid For a given F , it is possible to find maximal ordered sequences S , T ⊆ F such that ℓ ( S ) ≠ ℓ ( T ),

The Simple Ordering Problem (SOP): Given E, c , p , d , to find an ordered sequence of maximum total weight with respect to d . d ( F ∗ ) := max d ( F ) s.t. | F | ≤ ℓ ( F ), for all F ⊆ E K An optimal solution may have any number of elements in the range [0 , p ]. − − ⎛ ⎞ ⎛ ⎞ 2 0 15 2 0 5 ⎛ ⎞ 2 0 5 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − − − ⎜ ⎟ ⎜ ⎟ 2 2 3 2 2 3 ⎜ ⎟ 2 2 3 = = d = d ⎜ ⎟ ⎜ ⎟ d ⎜ ⎟ − − 3 0 1 3 0 1 3 0 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − − − ⎝ ⎠ 1 1 2 ⎝ ⎠ 1 1 2 ⎝ ⎠ 1 1 2

The polyhedron of the SOP ⎧ k 1 if element e is selected = ⎨ j x jk ⎩ 0 otherwise P SOP = conv { x ∈ {0, 1} n×p : x ( F ) ≤ ℓ ( F ), for all F ⊆ E K } ∑ ≤ ∈ x 1 , k K � jk ∈ j N ∑ ≤ ∈ x 1 , j N � jk ∈ k K + ≤ ∀ ∈ ∀ ∈ > ∑ x x 1 , j N , k , k ' K , k ' k � ' ' jk j k ≤ ' j j + ≤ ∀ ∈ ∀ ∈ > ∑ ∑ x x 1 , j N , k , k ' K , k ' k j ' k j ' k ' ≥ ≤ j ' j j ' j + + ≤ ∀ ∈ ∀ ∈ > ∑ ∑ ∑ x x x 1 , j N , k , k ' K , k ' k j ' k jh j ' k ' ≥ < < ≤ j ' j k h k ' j ' j

∑ ≤ x 1 ∑ ≤ x 1 jk jk ∈ N j ∈ k K k ⋯ ⋯ j ⋯ ⋯ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ∑ + ∑ ≤ x x 1 ∑ + ∑ + ∑ ≤ x x x 1 j ' k j ' k ' j ' k jh j ' k ' ≥ < j ' j j ' j ≥ < < < j ' j k h k ' j ' j k k’ ⋯ k k’ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ⋯ j ⋯ ⋯ j ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

+ + ≤ ∀ ∈ ∀ ∈ > ∑ ∑ ∑ x x x 1 , j N , k , k ' K , k ' k j ' k jh j ' k ' ≥ < < ≤ j ' j k h k ' j ' j Not enough + + ≤ ∀ ∈ < ∀ ∈ ∑ ∑ ∑ x x x 1 j , j ' N , j ' j , k K ' ' ' jk hk j k ≤ < < > ' ' ' k k j h j k k E = { e 1 , e 2 , e 3 , e 4 } p = 4, x 11 = x 41 = x 32 = x 23 = x 14 = 1/3

k ’ ∈ H and j ′ ≤ j , then k ≤ k ′ . k , e j ’ Staircase: H ⊆ E K s.t. e j Maximal staircase: staircase not contained in any other ⋯ ⋯ ⋯ ⋯ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ∑ ≤ Valid inequality for the SOP that x 1 Staircase inequality jk generalizes all the previous ones ∈ H k e j ⇒ H maximal staircase undominated inequality Theorem: P SOP = { x ∈ [0 , 1] n×p : x ( H ) ≤ 1 , for all maximal staircase H ⊆ E K }

Separation of staircase inequalities Proposition: For a given y ∈ ℝ n × p such that 0 ≤ y jk ≤ 1, for all j ∈ N , k ∈ K , the separation problem for staircase inequalities can be solved in polynomial time by finding an s − t − path of maximum cost in N ( y ). 1/3 1/3 1/3 1/3 1/3 1/3 E = { e 1 , e 2 , e 3 , e 4 } p = 4, 0 0 0 1/3 1/3 s t x 11 = x 41 = x 32 = x 23 = x 14 = 1/3 1/3 0 0 1/3 1/3 N ( y ) = ( V y ∪ { s , t }, A ( y )) k ∈ E K : y jk > 0}: support of y ; � V y = { v j � s and t : fictitious source and sink. A ( y ) contains the following arcs: k ∈ V y . � One arc ( s , v j k ) of cost y jk , for each node v j k ʼ ) of cost y j ′ k ′ for each pair v j k ʼ ∈ V y , with j ≥ j ′ and k ≤ k ′ . One arc ( v j k , v j ʼ k , v j ʼ � k ∈ V y . � One arc ( v j k , t ) of cost zero, for each node v j

Theorem: The SOP can be solved in polynomial time.

The simple ordering problem with cardinality constraint (SOPC) Given E, c , p , d , to find an ordered sequence that contains exactly one element of each k ∈ K , of maximal total weight with respect to d . d ( F ∗ ) := max d ( F ) F ⊆ E K . s.t. | F | ≤ ℓ ( F ), | F |= p ∑ ∑ ≤ ∀ ≤ ∀ x 1 H maximal stair x 1 H maximal stair jk jk k ∈ e H k ∈ j e H j ∑ ∑ ∑ = ∈ jk = x 1 , k K x p jk ∈ ∈ ∈ j N k K j N Theorem: ∑∑ P SOPC = { x ∈ [0, 1] n × p : x ( H ) ≤ 1, for all maximal staircase H ⊆ E K , jk = }. x p ∈ ∈ k K j N Theorem: The SOPC can be solved in polynomial time.

The simple ordering problem on an independence system (SOPI) Given E, c , p , d + Independence System J =(E, H ) I = (E K , F ) Independence System induced by order function c J K = (E K , H K ) Independence System ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ k k k H ⎪ ⎪ ∈ H ∈ e e e 1 2 r ⎨ ⎬ ⎨ e e e ⎬ K K , , , , , , K ⎪ j j jr ⎪ ⎪ ⎪ j j jr ⎩ ⎭ ⎩ ⎭ 1 2 1 2 to find an ordered sequence of F which is an independent set of H K of maximum total weight with respect to d . ℒ ( F ) = max{| S | : S ⊆ F , S ∈ F ∩ H K }. d ( F ∗ ) := max d ( F ) | F | ≤ ℒ ( F ), F ⊆ E K . s.t. | F |= p P SOPI = conv { x ∈ { 0, 1 } n × p : x ( F ) ≤ ℒ ( F ), F ⊆ E K } P SOPIC = conv { x ∈ { 0, 1 } n × p : x ( F ) ≤ ℒ ( F ), F ⊆ E K , | F |= p }

Mathematical Programming Formulation of SOPI(C) ∑ ∑ k Max d x j jk ∈ ∈ k K j N ( ) ≤ ⊂ for all maximal staricase s . t . x H 1 H E ( 1 ) K ≤ ⊆ for all x ( F ) r ( F ) F E ( 2 ) K ( ) = x F p ( 3 ) { } ∈ ∈ ∈ for all x 0 , 1 j N , k K jk r (·): rank function on J K { x ∈ {0, 1} n × p : x satisfies (1) and (2)} has fraccional vertices E = { e 1 , e 2 , e 3 } edges of K 3 , p =3 and H given by forests in K 3 ⎛ ⎞ 1 0 1 . 5 ⎜ ⎟ x 11 = x 22 = x 13 = x 33 =1/2 z *=2.25 = ⎜ ⎟ d 0 1 0 ⎜ ⎟ ⎝ ⎠ 0 0 1

Some properties of P SOPI = conv { x ∈ { 0, 1 } n × p : x satisfies (1) and (2) } � Inequalities x ( F ) ≤ r ( F ) need not be facets of P SOPI E = { e 1 , e 2 , e 3 } edges of K 3 , p =3 and H given by forests in K 3 F = { e 2 1 , e 2 2 , e 3 3 } , r ( F ) = 2 x 11 + x 22 + x 33 ≤ 2 is dominated by x 11 + x 22 + x 32 + x 33 ≤ 2

More properties of P SOPI = conv { x ∈ {0, 1} n × p : x satisfies (1) and (2)} Proposition: Let H ⊂ E K be a maximal staircase such that k ∈ E K \ H ∃ e j ’ k ’ } ∈ F ∩ H K . for all e j k ’ ∈ H s.t. { e j k , e j ’ Then, x ( H ) ≤ 1 is a facet of P SOPI . ⋯ ⋯ ⋯ ⋯ ∶ ∶ ∶ ∶ ∶ ∶ ∶ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯