Using Fractional Primal-Dual to Schedule Split Intervals with Demands Reuven Bar-Yehuda (CS, Technion) Dror Rawitz (CRI, University of Haifa) – p.1
Scheduling t -intervals with Demands The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 – p.2
Scheduling t -intervals with Demands The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j : t -interval demand , d j ∈ [0 , 1] weight , w j – p.2
Scheduling t -intervals with Demands The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j : t -interval demand , d j ∈ [0 , 1] weight , w j A schedule S : � ∀ p ∈ R , d j ≤ 1 j : p ∈ j ∈ S – p.2
Scheduling t -intervals with Demands The problem: Set of jobs that require the utilization of a resource Amount of resource is normalized to 1 Job j : t -interval demand , d j ∈ [0 , 1] weight , w j A schedule S : � ∀ p ∈ R , d j ≤ 1 j : p ∈ j ∈ S � Goal: max w j S j ∈ S – p.2
2 -intervals Example j 1 d 1 = 0 . 7 j 2 d 2 = 0 . 2 j 3 d 3 = 0 . 4 j 4 d 4 = 0 . 8 j 5 d 5 = 0 . 5 S = { j 1 , j 4 } is a feasible solution S = { j 1 , j 2 , j 4 } is a maximal solution (with respect to set inclusion) – p.3
b b b b b b b b b Scheduling t -intervals with Demands LP relaxation: � max w j x j j � s.t. d j x j ≤ 1 ∀ p ∈ R j : p ∈ j x j ∈ [0 , 1] ∀ j – p.4
Scheduling t -intervals with Demands LP relaxation: � max w j x j j � s.t. d j x j ≤ 1 ∀ p ∈ � R Right-end-points j : p ∈ j x j ∈ [0 , 1] ∀ j Observation: It is enough to consider right end-points of segments b b b b b b b b b – p.4
Scheduling t -intervals with Unit Demands Observation: Scheduling t -intervals with unit demands reduces to IS on t -intervals graphs Poly-time solvable for t = 1 (IS on interval graphs) – p.5
Scheduling t -intervals with Unit Demands Observation: Scheduling t -intervals with unit demands reduces to IS on t -intervals graphs Poly-time solvable for t = 1 (IS on interval graphs) Positive result [BHNSS02]: 2 t -approximation algorithm using a novel extension of the local ratio technique , called fractional local ratio – p.5
Scheduling t -intervals with Unit Demands Observation: Scheduling t -intervals with unit demands reduces to IS on t -intervals graphs Poly-time solvable for t = 1 (IS on interval graphs) Positive result [BHNSS02]: 2 t -approximation algorithm using a novel extension of the local ratio technique , called fractional local ratio Negative results [BHNSS02]: APX-hard (even on t -union graphs) Cannot be approximated within O ( t/ log t ) unless P = NP – p.5
Scheduling t -intervals with Demands Observation: NP-hard even when t = 1 , since knapsack is the special case where all intervals intersect – p.6
Scheduling t -intervals with Demands Observation: NP-hard even when t = 1 , since knapsack is the special case where all intervals intersect Our results: 6 t -approximation algorithm that uses a new extension of primal-dual schema, called fractional primal-dual Fractional Primal-Dual ≡ Fractional Local Ratio Contiguous allocation of a non-fungible resource: Bi-criteria approx alg. that computes 4 t -approx solutions that may need up to 4 times the given amount of resource – p.6
Scheduling t -intervals with Demands Observation: NP-hard even when t = 1 , since knapsack is the special case where all intervals intersect Our results: 6 t -approximation algorithm that uses a new extension of primal-dual schema, called fractional primal-dual Fractional Primal-Dual ≡ Fractional Local Ratio Contiguous allocation of a non-fungible resource: Bi-criteria approx alg. that computes 4 t -approx solutions that may need up to 4 times the given amount of resource In this talk: we focus on fractional primal-dual – p.6
Primal-Dual Algorithms — Maximization � n � m ( P ) max j =1 w j x j ( D ) min i =1 b i y i � n � m s.t. j =1 a ij x j ≤ b i ∀ i s.t. i =1 a ij y i ≥ w j ∀ j x j ≥ 0 ∀ j y i ≥ 0 ∀ i w T x ∗ b T y ∗ w T x ∗ I – p.7
Primal-Dual Algorithms — Maximization � n � m ( P ) max j =1 w j x j ( D ) min i =1 b i y i � n � m s.t. j =1 a ij x j ≤ b i ∀ i s.t. i =1 a ij y i ≥ w j ∀ j x j ≥ 0 ∀ j y i ≥ 0 ∀ i b T y Idea: w T x ∗ b T y ∗ Find an integral primal solution x w T x ∗ and a dual solution y s.t. w T x ≥ b T y/r I w T x ⇒ w T x ≥ b T y/r ≥ OPT ( P ) /r ≥ OPT /r = b T y/r – p.7
Primal-Dual Algorithms — Maximization � n � m ( P ) max j =1 w j x j ( D ) min i =1 b i y i � n � m s.t. j =1 a ij x j ≤ b i ∀ i s.t. i =1 a ij y i ≥ w j ∀ j x j ≥ 0 ∀ j y i ≥ 0 ∀ i b T y Idea: w T x ∗ b T y ∗ Find an integral primal solution x w T x ∗ and a dual solution y s.t. w T x ≥ b T y/r I w T x ⇒ w T x ≥ b T y/r ≥ OPT ( P ) /r ≥ OPT /r = b T y/r Question: How do we find such solutions? – p.7
Primal-Dual Schema Idea: Find an integral primal x and dual y that satisfy: � m Primal: ∀ j, x j > 0 ⇒ i =1 a ij y i = w j b i /r ≤ � n Relaxed Dual: ∀ i, y i > 0 ⇒ j =1 a ij x j ≤ b i – p.8
Primal-Dual Schema Idea: Find an integral primal x and dual y that satisfy: � m Primal: ∀ j, x j > 0 ⇒ i =1 a ij y i = w j b i /r ≤ � n Relaxed Dual: ∀ i, y i > 0 ⇒ j =1 a ij x j ≤ b i � m � n n m n m RD ≥ 1 P � � � � � � y i w j x j = a ij y i x j = a ij x j b i y i r j =1 j =1 i =1 i =1 j =1 i =1 – p.8
Primal-Dual Schema Idea: Find an integral primal x and dual y that satisfy: � m Primal: ∀ j, x j > 0 ⇒ i =1 a ij y i = w j b i /r ≤ � n Relaxed Dual: ∀ i, y i > 0 ⇒ j =1 a ij x j ≤ b i � m � n n m n m RD ≥ 1 P � � � � � � y i w j x j = a ij y i x j = a ij x j b i y i r j =1 j =1 i =1 i =1 j =1 i =1 Question: How do we find such solutions? – p.8
Primal-Dual Schema Idea: Find an integral primal x and dual y that satisfy: � m Primal: ∀ j, x j > 0 ⇒ i =1 a ij y i = w j b i /r ≤ � n Relaxed Dual: ∀ i, y i > 0 ⇒ j =1 a ij x j ≤ b i � m � n n m n m RD ≥ 1 P � � � � � � y i w j x j = a ij y i x j = a ij x j b i y i r j =1 j =1 i =1 i =1 j =1 i =1 Question: How do we find such solutions? In each step we use only good constraints: b i /r ≤ � n j =1 a ij x j ≤ b i ( ∀ maximal solution) y i > 0 = ⇒ x contains elements that are not over paid : � m x j = 1 = ⇒ i =1 a ij y i = w j – p.8
Primal-Dual vs. Fractional Primal-Dual Standard Primal-Dual: In each iteration we consider a single valid constraint [BT95,BR01] Solution is compared to an optimal solution with respect to this constraint The superposition of these “local optima” may be significantly worse than the “global optimum” – p.9
Primal-Dual vs. Fractional Primal-Dual Standard Primal-Dual: In each iteration we consider a single valid constraint [BT95,BR01] Solution is compared to an optimal solution with respect to this constraint The superposition of these “local optima” may be significantly worse than the “global optimum” Fractional Primal-Dual: An optimal fractional solution x ∗ is used as a yardstick In each iteration we consider a single constraint that is satisfied by x ∗ but may be invalid – p.9
Fractional Primal-Dual Algorithms A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x ∗ Let P ′ be the LP that consists of the objective function of P , and the constraints used ⇒ x ∗ is a feasible solution of P ′ – p.10
Fractional Primal-Dual Algorithms A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x ∗ Let P ′ be the LP that consists of the objective function of P , and the constraints used ⇒ x ∗ is a feasible solution of P ′ ⇒ OPT ( P ) ≤ OPT ( P ′ ) – p.10
Fractional Primal-Dual Algorithms A new LP: Let P be the original LP relaxation Algorithm constructs constraints that are satisfied by x ∗ Let P ′ be the LP that consists of the objective function of P , and the constraints used ⇒ x ∗ is a feasible solution of P ′ ⇒ OPT ( P ) ≤ OPT ( P ′ ) The dual: Algorithm constructs a solution y ′ of D ′ (dual of P ′ ) It finds an integral solution x s.t. w T x ≥ b T y ′ /r ⇒ w T x ≥ b T y ′ /r ≥ OPT ( P ′ ) /r ≥ OPT ( P ) /r ≥ OPT /r – p.10
Fractional Complementary Slackness Conds. Find x and y ′ such that � m Primal: i =1 a ij y ′ ∀ j, x j > 0 ⇒ i = w j c i /r ≤ � n ∀ i, y ′ Relaxed Dual: i > 0 ⇒ j =1 a ij x j and � n j =1 a ij x ∗ j ≤ c i – p.11
Fractional Complementary Slackness Conds. Find x and y ′ such that � m Primal: i =1 a ij y ′ ∀ j, x j > 0 ⇒ i = w j c i /r ≤ � n ∀ i, y ′ Relaxed Dual: i > 0 ⇒ j =1 a ij x j and � n j =1 a ij x ∗ j ≤ c i In this case: � m � n � � n n m m i ≥ 1 � � � � � � a ij y ′ y ′ c i y ′ w j x j = x j = a ij x j i i r j =1 j =1 i =1 i =1 j =1 i =1 – p.11
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