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Performances of List Scheduling for set partition problems

Performances of List Scheduling for set partition problems. 吳邦一歐秀慧 樹德科技大學 資訊工程系 報告人：歐秀慧. Set partition problems. Given: a set of positive numbers (not necessary distinct) Divide them into m parts Objective functions: Minimize the maximum part Minimize the sum of squares (L 2 metric)

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Performances of List Scheduling for set partition problems

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  1. Performances of List Scheduling for set partition problems 吳邦一歐秀慧 樹德科技大學 資訊工程系 報告人：歐秀慧

  2. Set partition problems • Given: a set of positive numbers (not necessary distinct) • Divide them into m parts • Objective functions: • Minimize the maximum part • Minimize the sum of squares (L2 metric) • others

  3. S= 3 3 2 1 1 5 5 The incoming number is put into the currently smallest subset. LS Method: M1= M2= M3=

  4. S= 1 3 5 2 1 3 5 It first sorts the numbers in decreasing order. LPT Method: M1= M2= M3= 1 5 5 1 3 2 3 The incoming number is put into the currently smallest subset.

  5. Previous results • Min-max • LS: at most 2-1/m time of the optimal. • LPT: at most 4/3-1/(3m) time of the optimal. • L2 metric: • LPT: a25/24 worse case upper bound • Tight bound unknown.

  6. Our resultsThe worse-case performance ratios of LS for the following problems: • Maximizing the minimum part (max-min problem):the tight bound ism • Maximizing the sum of the smallest K parts for any 1  k  m (max-k-min problem) :the tight bound is m/k • Minimizing the sum of the largest k parts for any 1  k  m (min-k-max problem) :the tight bound is 2-m/k • Minimizing the ratio of the largest to the smallest part (min-ratio problem) :the tight bound is m+1

  7. LS Property: Xi : the last element in Qi • For anyQ iand Qj in the LS partition, w (Q i)-x iw (Q j) , Qj Qj Q i Q i x i x i

  8. The max-min problem : the tight bound is m The worst cases: • C*min  m CLsmin

  9. m xi x1 S Qm Q1 Qi

  10. If xi > • In the optimal partition, xi must be in a part of singleton • The min part will not be reduced if we replacing xi with • We can assume that • No element is larger than

  11. The min part is always less than or equal to the mean Theorem: The tight bound of LS for the max-min partition is m

  12. The Max-k-Min problem Lemma : 在任意的partition中, 最大的k塊必不小於最大的k個elements之和 Proof: by induction. (否則最大的k個數要放到哪裡?) Corollary : 對任意的m-partition, 若 s1 是 s中任意 m – k個elements, 則最小的k塊之和 w*k w (s) – w (s1)

  13. max-k-min problem : the tight bound is m/k Qj Q i X i

  14. The tight bound is m / k

  15. Future Research • The tight bound of LS/LPT for L2 metric • The Minimum sum of squares problem • Good approximation algorithm for the min-ratio problem

  16. Thank you

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