Simons Symposium on Approximation Algorithms, February 26, 2014

Transcription

1 Solving Optimization Problems with Diseconomies of Scale via Decoupling Konstantin Makarychev Microsoft Research Maxim Sviridenko Yahoo Labs Simons Symposium on Approximation Algorithms, February 26, 2014

2 What is a diseconomy of scale? 6 Cost of Resources

3 What is a diseconomy of scale? 6 Cost of Resources

4 What is a diseconomy of scale? Cost of Energy used for Computing Energy consumption grows as x q as a function of speed x

5 Example: Energy Efficient Routing Given: graph G set of demand pairs (s i, t i, d i ) Goal: route d i units of unsplittable flow from s i to t i So as to minimize energy: e E c e x e q

6 Example: Energy Efficient Routing Given: graph G set of demand pairs (s i, t i, d i ) Goal: route d i units of unsplittable flow from s i to t i So as to minimize energy: e E c e x e q

7 Example: Energy Efficient Routing Given: graph G set of demand pairs (s i, t i, d i ) Goal: route d i units of unsplittable flow from s i to t i So as to minimize energy: e E c e x e q = 5 1 q q

8 What s known? Arbitrary d i : Andrews, Fernández Anta, Zhang, Zhao O(#demands + log d max ) approximation All d i are the same: Andrews, Fernández Anta, Zhang, Zhao O(1) approximation Bampis, Kononov, Letsios, Lucarelli, Sviridenko P q q approximation We give P q q approximation for the general case

9 Graph of P q q

10 LP Relaxation P i is the set of s i t i paths variable λ p for all p P i constraints: λ p p Pi λ p = d i i λ p [0, d i ] Let Λ be the set of feasible λ s.

11 LP Relaxation P i is the set of s i t i paths variable λ p for all p P i constraints: p Pi λ p = d i i λ p [0, d i ] Minimize e c e p:e p λ p q Subject to λ Λ 2 λ p = 1 n Let Λ be the set of feasible λ s. cost = 2n 1 n 2 = 2/n

12 LP Relaxation Local Distributions of Paths Minimize e c e F q,e (λ) subject to λ Λ Let F q,e = min E Y D e e D e i Y q i, e where D e is a distribution over r.v. {Y e i }; e Y i is the amount of flow s i t i going via e E Y e i = p Pi :e p λ p Y e i {0, d i }

13 Rounding Algorithm LP: Algorithm A: Minimize e c e F q,e (λ) subject to λ Λ where F q,e = min D e E Y e D e p:e p Y p e q. For each i, pick a random path p: s i t i with probability λ p /d i. Pick paths p independently for all i.

14 Analysis Algorithm A: For each i, pick a random path p: s i t i with probability λ p /d i. Pick paths p independently for all i. Analysis: Compare LP and ALG edge by edge. For every edge e: Let X i = amount of flow from s i to t i routed via e by A. All X i are independent. alg. cost e = c e i X i e q.

15 Analysis We need to compare E[alg. cost e ] = E[c e i X i e q ] LP. cost e = E[c e i Y i e q ] Each X i e has the same distribution as Y i. All X i e are independent. But Y i e are not independent

16 Decoupling Inequality (de la Peña 90) If Y 1,, Y n jointly distributed nonnegative r.v. s; X 1,, X n independent nonnegative r.v.; Each X i has the same distribution as Y i ; Then for some absolute constant C q. X X n q C q Y Y n q

17 Decoupling Inequality (de la Peña 90) If Y 1,, Y n jointly distributed nonnegative r.v. s; X 1,, X n independent nonnegative r.v.; Each X i has the same distribution as Y i ; Then X X n q C q Y Y n q De la Peña, Ibragimov, Sharakhmetov 03: C q q 2 for q (1,2]; C q q P q q for q 2.

18 Decoupling Inequality (de la Peña 90) If Y 1,, Y n jointly distributed nonnegative r.v. s; X 1,, X n independent nonnegative r.v.; Each X i has the same distribution as Y i ; Then X X n q C q Y Y n q We show that C q q = P q q, where P is a Poisson r.v. with parameter 1.

19 What is the Poisson distribution? P λ is Poisson with parameter λ P P = k = λk e λ. P λ = number of chosen points in the Poisson process: N points; each chosen w.p. λ/n k!

20 What is the Poisson distribution? P λ is Poisson with parameter λ P P = k = λk e λ. P λ = number of chosen points in the Poisson process: N points; each chosen w.p. λ/n k!

21 Decoupling Inequality (de la Peña 90) If Y 1,, Y n jointly distributed nonnegative r.v. s; X 1,, X n independent nonnegative r.v.; Each X i has the same distribution as Y i ; Then X X n q C q Y Y n q We show that C q q = P q q, where P is a Poisson r.v. with parameter 1.

22 Better Language? Yes Convex Stochastic Order

23 Convex Order Def: X cx Y if for every convex φ: Eφ X Eφ(Y). Basic Properties: cx is a property of distribution: X = st Z, X cx Y Z cx Y If X cx Y, Y cx Z X cx Z If X cx Y, then αx + β cx αy + β If X cx Y, then EX = EY We can verify only for φ with φ 0 = 0

24 Example: Random Walk X = W τ1 Y = W τ2 X cx Y τ 1 τ

25 Example: B cx P B Bernoulli r.v. with parameter p; P Poisson r.v. with parameter p Then, B cx P Proof: φ convex with φ 0 = φ P l P φ B = l(b) φ t Eφ P El P = l EP = l EB = El B = Eφ(B) l t

26 Lemma I If X, Y, Z are independent r.v s and X cx Y, Then Z + X cx Z + Y. Proof: Eφ Z + X = E E φ Z + X Z] cx E E φ Z + Y Z] = Eφ Z + Y Since for any fixed z, Eφ z + X cx Eφ z + Y.

27 Lemma II X 1,, X n are independent r.v. s; Y 1,, Y n are independent r.v. s; X i cx Y i for all i Then X X n cx Y Y n Proof: X 1 + X 2 + X 3 + X 4 cx X 1 + X 2 + X 3 + Y 4 cx X 1 + X 2 + Y 3 + Y 4 cx X 1 + Y 2 + Y 3 + Y 4 cx Y 1 + Y 2 + Y 3 + Y 4

28 Decoupling Inequality If Y 1,, Y n jointly distributed nonnegative r.v. s; X 1,, X n independent nonnegative r.v.; Each X i has the same distribution as Y i ; Then X X n cx P (Y Y n ) where P is a Poisson r.v. with parameter 1 independent of Y i s.

29 Proof of de la Peña s Inequality X X n Y Y n q q = E X X q n q q q = E Y Y n E X X n q E P Y Y n q = E P q E Y Y n q

30 Why Poisson? X 1,, X n : independent Bernoulli r.v. s with parameter 1 n; Y 1,, Y n : Pick a random i [n], let Y i = 1; Y j = 0 for j i; Then So X X n P Y Y n = 1 X X n P(Y Y n )

31 Special Case of the Theorem X 1,, X n are independent Bernoulli random variables with E X i = 1/n; Y 1,, Y n are Bernoulli random variables s.t. i Y i = 1 (always) Then Y Y n cx X X n cx P Y Y n 1 cx X X n cx P

32 Special Case of the Theorem X 1,, X n are independent Bernoulli random variables; Y 1,, Y n are Bernoulli random variables s.t. i Y i = 1 (always) α 1,, α n are non-negative numbers Then α 1 Y α n Y n cx α 1 X α n X n cx P α 1 Y 1 + +α n Y n

33 General Case We compare i Y i = y Y χ Y = y i y i y Y B y i y i y Y i B i y y i y = (y 1,, y n ) Y i X i B y Bernoulli: P B y = 1 = P Y = y

34 General Case Let Y be the support of Y = (Y 1,, Y n ). Then Y i = y Y χ Y = y y i cx y Y B i y y i where B i y is a Bernoulli with parameter P Y = y.

35 General Case Let Y be the support of Y = (Y 1,, Y n ). Then X i = st Y i = y Y χ Y = y y i cx y Y B i y y i where B i y is a Bernoulli with parameter P Y = y.

36 General Case Let Y be the support of Y = (Y 1,, Y n ). Then X i = st Y i = y Y χ Y = y y i cx y Y B i y y i where B i y is a Bernoulli with parameter P Y = y. Thus, i X i cx i y Y B i y y i = y Y ( i B i y y i ) Now, y Y( i y i ) B y cx P y Y ( i y i ) χ Y = y = P i Y i

37 Lemma: i y i B i i y i B Rescale y s.t. y i = 1. Then, E φ i y i B i i y i E φ B i = i y i E φ B = E φ B

38 Special Case of the Theorem X 1,, X n are independent Bernoulli random variables; Y 1,, Y n are Bernoulli random variables s.t. i Y i = 1 (always) α 1,, α n are non-negative numbers Then α 1 Y α n Y n cx α 1 X α n X n cx P α 1 Y 1 + +α n Y n

39 Proof: α 1 Y α n Y n cx α 1 X α n X n Consider a convex function φ with φ 0 = 0 φ is superadditive: φ a + b φ a + φ(b) φ α 1 X α n X n φ α 1 X φ α n X n Eφ i α i X i i Eφ α i X i Eφ i α i Y i = i φ(α i )P(Y i = 1) = i Eφ α i Y i

40 Proof: α 1 X α n X n cx P(α 1 Y α n Y n ) Let P i be a Poisson r.v. with parameter P(X i = 1), then α 1 X α n X n cx α 1 P α n P n P = P 1 + P n is a Poisson r.v. with parameter 1.

41 Proof: α 1 P α n P n cx P(α 1 Y α n Y n ) Let P i be a Poisson r.v. with parameter P(X i = 1), then α 1 X α n X n cx α 1 P α n P n P = P 1 + P n is a Poisson r.v. with parameter 1.

42 Proof: α 1 P α n P n cx P(α 1 Y α n Y n ) Let P = P P n. Condition on P = k > 0: E φ i α i P i ) P = k = E φ( i P i k α ik) P = k E i P i k φ α ik P = k = i φ(α i k) E P i k P = k = i φ(α i k) E P i = i φ(α i k) E Y i = E φ P i α i Y i ) P = k

43 Conclusion Introduced a new framework for problems with diseconomies of scale : Energy minimization L p -minimization Found the exact constant in de la Peña s inequality. Generalized it to other convex and concave functions.

44 X X n cx P (Y Y n ) Thank you!