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GRAPH BALANCING

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Scheduling on Unrelated Machines J1 J2 J3 J4 J5 M1 M2 M3

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M1 M2 M3 J3 J4 J1 J2 J5 J1 J2 J3 J4 J5 Scheduling on Unrelated Machines

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M1 M2 M3J1 J2J3 J4 J5 Makespan

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Restricted Assignment M1 M2M3 J1 J2 J3 J4 J5

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Graph Balancing Special case of Restricted Assignment Each job can be scheduled on at most 2 machines Machines vertices Jobs edges Assign dedicated loads to vertices Problem is to orient the edges

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Graph Balancing 22 4 3 5 2 1 6 5 6 5 5 2 1 2 2 1

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Graph Balancing Summary Given: weighted multigraph (V, E, p, q) V : Vertices Machines E : Edges Jobs e E, p e = processing time of job e v V, q v = dedicated load on vertex v Output: Orientation of edges :E V such that (e) e Load v = q v + e:v = (e) p e Objective: Minimize maximum load

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Optimization Decision Problem -relaxed decision procedure Is there any orientation with maximum load at most d ? Answer: “NO” Orientation with maximum load at most d. Binary search for d and scale everything appropriately

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2-approximation LP Find values x ev 0, for each e and v e, such that For each e E, u,v e: x eu + x ev = 1 For each v V: q v + e:v e x ev p e 1

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2-Relaxed Decision Procedure Solve LP in polynomial time. If not feasible, return “NO” If feasible, round solution using rotation and tree assignment After rounding, the maximum load is at most 2 For rounding, decompose the graph as Cycles Trees

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Rotation 1/2 1/4 1/3 1/2.3.5.4.5.4.7.5.6.4.5.125.1.2.3.25.15

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Rotation 1/2 1/4 1/3 1/2.1 0.3.7.9 1.3.4.6.7

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Rotation Increases the number of integral solutions. Breaks the fractional cycles. Unchanged after rotation (x eu + x ev ) for all edges e ( e:v e x ev p e ) for all vertices v Maximum load after rotation is 1. After all fractional cycles are broken, only fractional trees remain

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1 Tree Assignment

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2 1 2 1 Tree Assignment

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1 1 v`v` Integrality Gap

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>1/2 Big trees constraint

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1.75-approximation Find values x ev 0, for each e and v e, such that For each e E, u,v e: x eu + x ev = 1 For each v V: q v + e:v e x ev p e 1 For each T G B (graph induced on big edges)

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v v >1/2.1.1.1 0.3.7.9.9.9 1.3.4.6.7 Algorithm

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Invariants

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Algorithm Case 1: e is a big edge (p e >= 0.5) Increase in load = p e – x ev p e = x eu p e <= 0.75 Case 2: e is not a big edge (p e < 0.5) Increase in load = p e – x ev p e < p e < 0.5 Since x eu p e > 0.75, e is definitely a big edge For any vertex u’ in the tree T, the path joining u’ to v is also a subtree of G B By the tree constraint, x ev p e + x e’u’ p e’ >= p e + p e’ -1 p e’ – x e’u’ p e’ <= 1 – (p e - x ev p e ) Increase in load <= 0.25

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Integrality gap.25 1 v`v` 11 1 v`v` 1 1 1 v`v` 11 1 1 1

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