Download presentation

Presentation is loading. Please wait.

Published byGraciela Chelton Modified over 2 years ago

1
GRAPH BALANCING

2
Scheduling on Unrelated Machines J1 J2 J3 J4 J5 M1 M2 M3

3
M1 M2 M3 J3 J4 J1 J2 J5 J1 J2 J3 J4 J5 Scheduling on Unrelated Machines

4
M1 M2 M3J1 J2J3 J4 J5 Makespan

5
Restricted Assignment M1 M2M3 J1 J2 J3 J4 J5

6
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

7
Graph Balancing 22 4 3 5 2 1 6 5 6 5 5 2 1 2 2 1

8
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

9
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

10
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

11
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

12
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

13
Rotation 1/2 1/4 1/3 1/2.1 0.3.7.9 1.3.4.6.7

14
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

15
1 Tree Assignment

16
2 1 2 1 Tree Assignment

17
1 1 v`v` Integrality Gap

19
>1/2 Big trees constraint

20
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)

21
v v >1/2.1.1.1 0.3.7.9.9.9 1.3.4.6.7 Algorithm

22
Invariants

23
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

24
Integrality gap.25 1 v`v` 11 1 v`v` 1 1 1 v`v` 11 1 1 1

Similar presentations

Presentation is loading. Please wait....

OK

Integer Programming (정수계획법)

Integer Programming (정수계획법)

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Converter pub to ppt online viewer Ppt on 4 stroke petrol engine Ppt on first conditional sentence Ppt on grease lubrication pump Ppt on 3d printing pen Ppt on effect of global warming on weather we like it or not Ppt on meeting etiquettes meaning Authority of the bible ppt on how to treat Ppt on introduction to object-oriented programming tutorial Ppt on marie curie nobel