Dynamic Combinatorial Optimization David Adjiashvili, Sandro Bosio , - PowerPoint PPT Presentation

Dynamic Combinatorial Optimization David Adjiashvili, Sandro Bosio , Robert Weismantel Institute for Operations Research (IFOR), ETH Zurich 16th Combinatorial Optimization Workshop January 8-13, 2012, Aussois

Table of Contents Introduction: the Integer Knapsack 1 Dynamic Combinatorial Optimization 2 Complexity and Approximability 3 Conclusions and Future Work 4 2 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N a groundset A of object types a load coefficient τ a ∈ N for each object type a ∈ A a gain coefficient p a ∈ R for each object type a ∈ A Introduction: the Integer Knapsack 3 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N Knapsack ( T = 9) a groundset A of object types p 1 = 2 a load coefficient τ a ∈ N for each object type a ∈ A p 2 = 3 p 3 = 6 a gain coefficient p a ∈ R for each object type a ∈ A Groundset A Introduction: the Integer Knapsack 3 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N Knapsack ( T = 9) a groundset A of object types p 1 = 2 a load coefficient τ a ∈ N for each object type a ∈ A p 2 = 3 p 3 = 6 a gain coefficient p a ∈ R for each object type a ∈ A Groundset A The goal is to find an integer vector x such that the load capacity is satisfied: � a ∈ A τ a x a � T the gain � a ∈ A p a x a is maximized Introduction: the Integer Knapsack 3 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N Knapsack ( T = 9) a groundset A of object types p 1 = 2 a load coefficient τ a ∈ N for each object type a ∈ A p 2 = 3 p 3 = 6 a gain coefficient p a ∈ R for each object type a ∈ A Groundset A The goal is to find an integer vector x such that 8 the load capacity is satisfied: � a ∈ A τ a x a � T 9 the gain � 6 a ∈ A p a x a is maximized 10 Introduction: the Integer Knapsack 3 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N Knapsack ( T = 9) a groundset A of object types p 1 = 2 a load coefficient τ a ∈ N for each object type a ∈ A p 2 = 3 p 3 = 6 a gain coefficient p a ∈ R for each object type a ∈ A Groundset A The goal is to find an integer vector x such that 8 the load capacity is satisfied: � a ∈ A τ a x a � T 9 the gain � 6 a ∈ A p a x a is maximized 10 Classical results for the IK: is weakly NP-Hard [Ibarra and Kim, 1975] admits a pseudopolynomial-time algorithm admits an FPTAS [Lawler, 1977] Introduction: the Integer Knapsack 3 / 17

Integer Knapsack In the Integer Knapsack (IK), we are given: a knaspack with capacity T ∈ N Knapsack ( T = 9) a groundset A of object types p 1 = 2 a load coefficient τ a ∈ N for each object type a ∈ A p 2 = 3 p 3 = 6 a gain coefficient p a ∈ R for each object type a ∈ A Groundset A The goal is to find an integer vector x such that 8 the load capacity is satisfied: � a ∈ A τ a x a � T 9 the gain � 6 a ∈ A p a x a is maximized 10 Classical results for the IK: is weakly NP-Hard [Ibarra and Kim, 1975] admits a pseudopolynomial-time algorithm admits an FPTAS [Lawler, 1977] We are interested in the greedy approximation algorithms for IK Introduction: the Integer Knapsack 3 / 17

Greedy algorithms for IK Density-based heuristic: 1 choose the element type a ∈ A with maximal density p a τ a � � T 2 use a as many times as possible: x a = τ a Guarantee 2. Iterating doesn’t improve the guarantee. Introduction: the Integer Knapsack 4 / 17

Greedy algorithms for IK Density-based heuristic: 1 choose the element type a ∈ A with maximal density p a τ a � � T 2 use a as many times as possible: x a = τ a Guarantee 2. Iterating doesn’t improve the guarantee. Total-value heuristic: � � T 1 choose the element type a ∈ A with β -maximal total value p a τ a � � T 2 use a as many times as possible: x a = τ a Guarantee 1 . 691 [Kohli, Krishnamurti, 1992]. Iterating doesn’t improve the guarantee. Introduction: the Integer Knapsack 4 / 17

Greedy algorithms for IK Density-based heuristic: 1 choose the element type a ∈ A with maximal density p a τ a � � T 2 use a as many times as possible: x a = τ a Guarantee 2. Iterating doesn’t improve the guarantee. Total-value heuristic: � � T 1 choose the element type a ∈ A with β -maximal total value p a τ a � � T 2 use a as many times as possible: x a = τ a Guarantee 1 . 691 [Kohli, Krishnamurti, 1992]. Iterating doesn’t improve the guarantee. Combined heuristic: 1 run both (iterated) heuristics 2 choose the best solution Guarantee 1 . 5 [Kohli, Krishnamurti, 1995]. Introduction: the Integer Knapsack 4 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 � � x ∗ � a � ALG T a ∈ A τ a Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 � � � x ∗ max � x a � � a � � ALG T T a ∈ A a ∈ A τ a τ a � τ a x a � T a ∈ A x a ∈ N + a ∈ A Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 � � � x ∗ max � x a � � a � � ALG T T a ∈ A a ∈ A τ a τ a � τ a x a � T a ∈ A x a ∈ N + a ∈ A τ a � 0 a ∈ A Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 � � � x ∗ max � x a � � a � � ALG T 1 a ∈ A a ∈ A τ a τ a � τ a x a � 1 a ∈ A x a ∈ N + a ∈ A τ a � 0 a ∈ A Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 � � � x ∗ max � x a � � a � � ALG T 1 a ∈ A a ∈ A τ a τ a � 1 τ a x a � 1 � � 1 1 a ∈ A τ a x a ∈ N + a ∈ A τ a � 0 a ∈ A 1 / 2 1 / 3 1 / 4 1 / 5 1 / 6 1 / 7 1 / 8 1 / 9 1 1 1 1 1 1 1 1 1 τ a 9 8 7 6 5 4 3 2 Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 1 � � � � x ∗ max � x a = max k y k = α � � a � � ALG T 1 a ∈ A a ∈ A k � 1 τ a τ a 1 � � k + 1 y k < 1 q � 1 1 τ a x a � 1 � � 1 1 k � q a ∈ A τ a y k ∈ N + x a ∈ N + a ∈ A k � 1 τ a � 0 a ∈ A 1 / 2 1 / 3 1 / 4 1 / 5 1 / 6 1 / 7 1 / 8 1 / 9 1 1 1 1 1 1 1 1 1 τ a 9 8 7 6 5 4 3 2 Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 1 � � � � x ∗ max � x a = max k y k = α � � a � � ALG T 1 a ∈ A a ∈ A k � 1 τ a τ a 1 � � k + 1 y k < 1 q � 1 1 τ a x a � 1 � � 1 1 k � q a ∈ A τ a y k ∈ N + x a ∈ N + a ∈ A k � 1 τ a � 0 a ∈ A 1 / 2 α = 1 / 3 1 / 4 1 / 5 1 / 6 1 / 7 1 / 8 1 / 9 1 1 1 1 1 1 1 1 1 τ a 9 8 7 6 5 4 3 2 Introduction: the Integer Knapsack 5 / 17

Analysis of the total-value heuristic (sketch) Let g be the element with highest total value. For every a ∈ A : � T � T � � p a � ALG p a � p g = ALG and hence � � τ a τ g T τ a Let OPT = � a p a x ∗ a . Guarantee: OPT 1 1 1 � � � � x ∗ max � x a = max k y k = α � � a � � ALG T 1 a ∈ A a ∈ A k � 1 τ a τ a 1 � � k + 1 y k < 1 q � 1 1 τ a x a � 1 � � 1 1 k � q a ∈ A τ a y k ∈ N + x a ∈ N + a ∈ A k � 1 τ a � 0 a ∈ A 1 / 2 α = 1 1 / 3 1 / 4 1 / 5 1 / 6 1 1 / 7 1 / 8 1 / 9 2 1 1 1 1 1 1 1 1 1 τ a 9 8 7 6 5 4 3 2 Introduction: the Integer Knapsack 5 / 17