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The Geometry of Scheduling

The Geometry of Scheduling. Nikhil Bansal (TU Eindhoven) Kirk Pruhs (U. Pittsburgh). The problem. Input: Jobs 1,…n. Job i has release time r i , size p i , Non-decreasing weight function w i (t): cost if i finishes at t.

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The Geometry of Scheduling

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  1. The Geometry of Scheduling Nikhil Bansal (TU Eindhoven) Kirk Pruhs (U. Pittsburgh)

  2. The problem Input: Jobs 1,…n. Job i has release time ri, size pi, Non-decreasing weight function wi(t): cost if i finishes at t. Goal: Find a schedule that minimizes the total cost. Allow preemption: A job can be interrupted. 2 1 3 time w1(t1) w2(t2) w3(t3)

  3. Examples 1. Weighted Flow time wi(t) = w(i) (t- ri) for t ¸ ri. 2. Weighted Tardiness wi(t) = 0 if t · di = w(i) (t – di) if t > di 3. Weighted flow time squared. …

  4. Previous Results 1. Weighted Flow Time: Extensive work O(log2 nP)[Chekuri-Khanna-Zhu 02] O(log W), O(log nP)[Bansal-Dhamdhere 03] (P: max/min job size, W: max/min job wt) 2. Weighted Tardiness: n-1 apx. if all ri =0[Cheng et al 05] (nothing for general release times) Various special cases studied. 3. Flow Time Squared: Only guarantees with resource augmentation[Bansal-Pruhs 03]

  5. Main Result

  6. Min r wr xr r : p 2 r cr xr¸ dp8 p xr2 {0,1} Capacitated Version Main Idea Reduce to a 2-d geometric covering problem (2RC) (lose only factor 4 in reduction) Scheduling is most naturally viewed as a packing problem. Give O(log log nP) approx for 2RC. 2-approx if unit demands and capacities (regular set cover)

  7. Outline 1. Reduction to 2RC 2. Reduce 2RC to other uncapacitated geometric problems (Knapsack Cover inequalities) 3. Use Geometric structure to beat O(log n)

  8. Reduction to 2RC

  9. Deadline Feasibility: Basics s t Interval I Job units Time slots Feasible Deadlines -> Schedule (use Earliest Deadline First)

  10. s t Interval I Our Problem: Alternate View

  11. rj dj s I t Alternate View Every interval I = (s,t) “demands” that Jobs with rj2 [s,t], dj > t have total size >= Size(J(I)) – |I|. Map I=[s,t] to point (s,t) with demand d(I) = Size(J(I)) – |I|. For job j with deadline dj Associate rectangle [0,rj] x [rj,dj] Weight = wj(dj) and capacity = pj. dj (s,t) Schedule is feasible iff each point’s demand satisfied. rj rj 0

  12. Geometric Formulation Input: 1) For each job j, several weighted rectangles [0,rj] x [rj,dj] , one for each possible deadline. 2) Points with demands. Problem: Find min wt collection of rectangles 1) Exactlyone rectangle per job 2) demand for all points is satisfied. Cheating: If two rectangles of same job used to satisfy demand of a point Can be removed by another trick. Painful constraint

  13. cost = 4 Make rectangles disjoint cost = 2 cost = 1 Rj(i) = [0,rj] x (dj,i-1,dj,i ] 2RC problem For job j, let dj,1,dj,2,… be times when cost doubles. Suffices to have those rectangles. A pure covering problem now! cost = 4 cost = 2 cost = 1 Rj(i) = [0,rj] x [rj,dj,i ]

  14. Capacitated Covering Problems

  15. Knapsack Cover Problem Min i wi xi i ci xi¸ B 0 <= xi <= 1 Has arbitrarily large gap (2 items of size B-1, one with wt. 0, other wt. large) Integrally: Need both items. LP: Picks item 1 completely, item 2 to extend 1/B-1 only. Just cover one point

  16. Knapsack Cover Problem Min i wi xi i ci xi¸ B 0 <= xi <= 1 Solution: Knapsack cover inequalities.  i 2 Sc min (ci, B-C(S)) xi¸ B-C(S) 8 S Thm[Carr, Fleischer, Leung, Philips 00]: Gives 2 approximation. Our problem: For each point, we add the knapsack cover inequalities. Just cover one point

  17. Priority Framework A general approach (based on KC inequalities) Given any capacitated covering problem suffices to solve • Priority version • Multicover version (no more capacities). Thm (Chakrabarty, Grant, Konemann 10): O( + ) approx. where  = approx for priority prob.,  for multi-cover version.

  18. Priority Framework Capacitated Covering(Set system S, capacities, demands) Priority Cover: Sets, items get a priority S covers i iff i 2 S and Prio(S) ¸ Prio(i) Note: capacity =1, demand = 1 Multi-Cover: Set system = S capacity =1, demand = integer as high priority

  19. What this means for us?

  20. y x z Need good approximations for: 1. Multi-cover version of the above. 2. Priority Set Cover: R covers p if Prio(R) >= Prio(p) Attach a new dimension to encode priority Point in 2d -> Point in 3d Rectangle -> Cuboid (s,t,prio)[0,r] x [r,d] x [0,prio(R)] Cover points in 3d by cuboids Touch x-y plane and x-z plane. O(log P) distinct heights

  21. Geometric Set Cover Extensive work. Most of it for unweighted case. Bronniman-Goodrich: Small  nets Imply good approximation. 1/ f(1/) implies f(OPT) approximation. Geometric objects have nets of size O(1/ log (1/)) Implies log(OPT) for geometric problems. Union Complexity: Clarkson-Varadarajan 07 (unweighted)

  22. Weighted Geometric Set Cover Union Complexity: Take k objects, look at their boundary (vertices,edges, holes). Scales as k h(k) Thm [Varadarajan’10]: exp(log*(n)) log (h(n)) approx. for weighted geometric set cover. Thm[Chan-Grant-Konemann-Sharpe’12]: O(log (h(n))) approx. O(k log log k) [Matousek et al 91] O(k log* k) [Aronov, de Berg, Ezra, Sharir 11] O(k) (k2)

  23. y x z Priority Problem Weighted Set Cover with cuboids So, [Chan et al’12] gives O(log log P) approx for priority problem. The union complexity of any k of our cuboids is O(k log P)

  24. The multi-cover version Implies O(1) approximation for multi-cover version

  25. Idea in a nutshell

  26. Concluding Remarks Main question: O(1) approx for the scheduling problem? Even for special cases eg. wt. flow time.

  27. Thank You!

  28. Weighted Geometric Set Cover Union Complexity: Take k objects, look at their boundary (vertices,edges, holes). Scales as k h(k) Thm [Varadarajan’10]: exp(log*(n)) log (h(n)) approx. for weighted geometric set cover. O(k log log k) [Matousek et al 91] O(k log* k exp((k)) [Ezra, Aronov, Sharir 11] O(k) (k2) Union complexity of k cuboids (for our type) is O(k log P)

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