Constant-factor approximation algorithms for the minmax regret - PowerPoint PPT Presentation

Constant-factor approximation algorithms for the minmax regret problem Juan Pablo Fern´ andez G. ın Cra 87 N o 30 - 65, Colombia Universidad de Medell´ e-mail : jpfernandez@udem.edu.co Adviser : Eduardo Conde Universidad de Sevilla, Espa˜ na. Doc-course Mayo 21, 2010

Introduction and existing results 1 The minmax regret approach for parameter optimization problems. Results from general complexity Some approximated algorithms of constant factor General Result of 2-approximation 2 Applications 3 The sequencing problem n / 1 // F . Finding compromise solution in a linear multiple objective problem. Facility Location under uncertain demand. Bibliography

Introduction and existing results 1 The minmax regret approach for parameter optimization problems. Results from general complexity Some approximated algorithms of constant factor General Result of 2-approximation 2 Applications 3 The sequencing problem n / 1 // F . Finding compromise solution in a linear multiple objective problem. Facility Location under uncertain demand. Bibliography

Introduction and existing results 1 The minmax regret approach for parameter optimization problems. Results from general complexity Some approximated algorithms of constant factor General Result of 2-approximation 2 Applications 3 The sequencing problem n / 1 // F . Finding compromise solution in a linear multiple objective problem. Facility Location under uncertain demand. Bibliography

Definition Given an optimization problem with parameter cost function Opt ( w ) = min x ∈ X F ( x , w ) where the parameter w ∈ W an hyperrectangle in R n and X ⊆ R n is a compact feasible set. How can i choose x under unknown scenario w ?

Definition Definition (minmax regret criterion) The minimization of the maximum absolute regret problem can be expressed as min x ∈ X Z ( x ) where Z ( x ) = max w ∈ W R ( x , w ) the worst-case regret and R ( x , w ) = F ( x , w ) − min y ∈ X F ( y , w ) the regret assigned to the feasible solution x under scenario w .

Robust? 1 Let x ⋆ be a minmax regret solution. 2 if w H was the scenario that take place after the decision x ⋆ has been implemented. � w H � 3 Let y H be the solution of Opt then F ( x ⋆ , w H ) − F ( y H , w H ) ≤ ǫ where ǫ = Z ( x ⋆ ).

Minmax regret complexity I In [4], it is described one of the classical combinatorial problem as Definition Elements of relative robust shortest path problem (RRSPP): G = ( V , A ), directed arc weighted graph. V , node set, | V | = n . A , arc set, | A | = m . � � l ij , l ij , ( i , j ) ∈ A . Lengths (weights) of the arcs are intervals which express ranges of possible realizations of lengths. No probability distribution is assumed for arc lengths.

Minmax regret complexity I Definition Elements of RRSPP: � � Length l w ij ∈ l ij , l ij is assigned for each ( i , j ) ∈ A , is called a scenario w , where l w ij denotes the length of arc ( i , j ) in scenario w . P , denotes the set of all the paths in G from o to d . � l w l w p = ij denotes the length of a path p ∈ P in scenario w . ( i , j ) ∈ p W , denote the set of possible scenarios.

Minmax regret complexity I applying the minmax regret concept: Definition Element of RRSPP: d w p = l w p − l w p ⋆ ( w ) the regret for the path p in scenario w , where p ⋆ ( w ) ∈ P is the shortest path in scenario w . w ∈ W d w Z p = max p is the maximum regret. w ∈ W l w p − l w min p ∈ P Z p = min p ∈ P max (1) p ⋆ ( w ) can equivalently define the problem RRSPP.

Minmax regret complexity I For the problem (1), it is proved: Theorem 1 (1) is NP-hard. 2 Decision- (1) is NP-complete, even if G is restricted to a planar acyclic graph with node degree three. 3 (1) is NP-hard, even if G is restricted to a planar acyclic graph with node degree three.

Minmax regret complexity II In [6], it is described another combinatorial problem as Definition Elements of minimizing the total flow time in a scheduling problem with interval data (MTFT) via minmax regret criterion: J , | J | = n , n ≥ 2, set of jobs that have to be processed on a single machine. The machine cannot process more than one job at any time. � � p k = � p k , p k , J k ∈ J . Then the processing times are intervals which express ranges of possible processing times for the jobs. p w k ∈ � p k processing time of jobs J k ∈ J is called a scenario .

Minmax regret complexity II Definition Elements of MTFT via minmax regret criterion: W being the Cartesian product of all � p k . The set of all scenarios. π = ( π (1) , . . . , π ( n )), a schedule of job. Π, the set of all feasible schedules. The total flow time in π under w is n � ( n − k + 1) p w F ( π, w ) = π ( k ) . (2) k =1

Minmax regret complexity II applying the minmax regret concept: Definition Element of MTFT: R ( π, w ) = F ( π, w ) − F ⋆ ( w ) the regret assigned to the schedule π in scenario w , where F ⋆ ( w ) = min y ∈ Π F ( y , w ) is the flow for the shortest processing time schedule under the scenario w . Z ( π ) = max w ∈ W R ( π, w ) is the maximum regret. min π ∈ Π Z ( π ) (3) The minmax regret version of Problem MTFT.

Minmax regret complexity II Definition (Problem ROB1) Problem ROB1 is the special case of problem (3) where all intervals of p k + p k uncertainty have the same center, that is, is the same for all J k ∈ J . 2 Definition Let J l , J k ∈ J be jobs. Job J l is wider than job J k if � p k ⊂ � p l .

Minmax regret complexity II Definition For any job J k ∈ J and schedule π ∈ Π, let q ( π, J k ) = min { n − π ( k ) , π ( k ) − 1 } A permutation π ∈ Π is called uniform if for any J l , J k ∈ J , if J l is wider than J k , then q ( π, J l ) ≥ q ( π, J k ).

Minmax regret complexity II Theorem 1 if the number of jobs n is even, then any uniform permutation is an optimal solution to problem ROB1 (and therefore problem ROB1 with even number of jobs is solvable in O ( n log n ) time). 2 Problem ROB1 with odd number of jobs is NP-hard. 3 Problem (3) is NP-hard; it remains NP-hard even if the number of jobs is even.

Some approximated algorithms of constant factor Definition (Elements of the problem.) E = { e 1 , e 2 , . . . , e n } a finite set. Φ ⊆ 2 E a feasible solutions set. � c e = [ c e , c e ], e ∈ E a range of possible values of the cost. w = ( c w e ) e ∈ E a particular vector assignment of costs c w e to elements e ∈ E is called scenario . W being the Cartesian product of all � c k . The set of all scenarios.

Special combinatorial optimization Definition (Problem formulation.) F ( χ, w ) = � c w e . Its cost function for a given solution χ ∈ Φ, e ∈ χ under a fixed scenario w ∈ W . R ( χ, w ) = F ( χ, w ) − F ⋆ ( w ) the regret assigned to feasible solution χ in scenario w , where F ⋆ ( w ) = min y ∈ Φ F ( y , w ) is the value of the cost of the optimal solution under scenario w . Z ( χ ) = max w ∈ W R ( χ, w ) is the maximum regret. min χ ∈ Φ Z ( χ ) (4)

Special combinatorial optimization Using the worst case characterization, we obtain bound for Z ( χ ) and then Theorem � � c e + c e Let M be the solution of min x ∈ Φ F ( x , w ) where w = e ∈ E . Then 2 for every χ ∈ Φ it holds Z ( M ) ≤ 2 Z ( χ ) . In particular, if χ ⋆ is the solution of (4), then Z ( M ) ≤ 2 Z ( χ ⋆ ) . M is known as the mid-point solution, and this w is the mid-point scenario.

Classical formulation of sequencing We return to the problems of n jobs to be processed on a single machine, but now, we consider it with precedence constrains. Definition It is used the following notation. n jobs for being processing in only one machine. The subscripts i refers to job J i . The subscripts k refers to position which is processed a particular job. The following data pertain to job J i . 1 p i the processing time of the job J i . � 1 if J i is proccesed in the position th-k 2 x ik = 0 otherwise

Classical formulation of sequencing Definition 3 C i is the time to finish the processing of the job J i . 4 C i ( k ) time of completion of the job J i in the th- k process. calculating � n The completion time of the job J i is C i ( k − 1) + p i x ik . And i =1

Classical formulation of sequencing: The above 2-approximation results can not be applied. Integer programming � n � k � n min p i x ij k =1 j =1 i =1 subject to n � x ik = 1 for i = 1 , . . . , n . k =1 k − 1 � x qk − x pj ≤ 0 for p , q such that job J p precedes job J q . j =1 x ik ∈ { 0 , 1 } for i , k = 1 , . . . , n .

Sequencing optimization problem For the sake of simplicity, we will denote by i π the position occupying by job J i in the schedule π . So, the total flow time function becomes � n ( n − i π + 1) p w F ( π, w ) = i i =1 and

Sequencing optimization problem Property For any two feasible schedules π, σ and scenario w ∈ W , 1 � n ( i σ − i π ) p w F ( π, w ) − F ( σ, w ) = i . i =1 2 � � Z ( π ) ≥ ( i σ − i π ) p i + ( i σ − i π ) p i { i : i σ > i π } { i : i σ < i π } 3 � � Z ( σ ) ≤ Z ( π ) + ( i π − i σ ) p i + ( i π − i σ ) p i { i : i π > i σ } { i : i π < i σ }