# 9th Max-Planck Advanced Course on the Foundations of Computer Science (ADFOCS) Primal-Dual Algorithms for Online Optimization: Lecture 1

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1 9th Max-Planck Advanced Course on the Foundations of Computer Science (ADFOCS) Primal-Dual Algorithms for Online Optimization: Lecture 1 Seffi Naor Computer Science Dept. Technion Haifa, Israel

2 Introduction Online algorithms and competitive analysis Deterministic Randomized Review of important techniques Randomized rounding Dual fitting

3 What is an Online Algorithm? Input is given in pieces over time, each piece is called a request Request sequence: { ¾ = ¾ 1 ; ¾ 2 ; : : : ; ¾ n ; : : : } Upon arrival of request ¾ i : Online algorithm has to serve the request Previous decisions for requests ¾ 1,, ¾ i-1 cannot be changed

4 Performance Evaluation: Competitive Factor How to evaluate performance of online algorithm A? For every request sequence ¾ = ¾ 1,, ¾ n : compare cost of A to the cost of an optimal offline algorithm that knows the request sequence in advance Competitive factor of online algorithm A is if For every request sequence ¾ = ¾ 1,, ¾ n :

5 Example 1:The Ski Rental Problem Buying costs \$B Renting costs \$1 per day Problem: Number of ski days is not known in advance each ski day is a request served by buying or renting Goal: Minimize the total cost.

6 Ski Rental: Analysis Online Algorithm: rent for m days and then buy What is the optimal choice of m? m=b If # of ski days B, cost(online) = OPT If # of ski days > B, cost(online) 2OPT ) Competitive ratio = 2

7 Example 2:The Paging Problem Universe of of n pages Cache of size k n Request sequence of pages: 1, 6, 4, 1, 4, 7, 6, 1, 3, If requested page is in cache: no penalty. Else, cache miss! load requested page into cache, evicting some other page. Goal: minimize number of cache misses. Question: which page to evict in case of a cache miss?

8 Paging: Analysis Online Algorithm LRU (Least Recently Used): Upon cache miss, evict the page whose last access was earliest (least recently used page). Theorem 1: The competitive factor of LRU is k. Theorem 2: The competitive factor of any (deterministic) paging algorithm k.

9 Proof of Theorem 1 Phase 1 Phase 2 time In each phase: LRU has precisely k misses (OPT and LRU start from the same initial configuration) Claim: Each phase has requests to k different pages Proof: If the first miss in a phase is due to page x, then all k-1 remaining pages in the cache will be evicted before x, since x has a higher priority.

10 Proof of Theorem 1 (contd.) x page that caused first miss in phase S pages that caused rest of cache misses in phase + page that caused first miss in next phase By claim, S {x} k Since page x 2 OPT s cache, OPT has a cache miss p number of full phases cost of OPT p cost of LRU k(p+1)

11 Randomization Online algorithm A uses random bits r=r 1, r 2, Expected cost of A on ¾: Exp r [A(¾,r)] A is -competitive if, ¾: Oblivious adversary: knows online algorithm A, request sequence ¾, but not the outcome of the random bits r

12 Example: Randomized Paging Algorithm Marking Algorithm: Each requested page is marked Page miss: evict one of the unmarked pages, chosen uniformly at random When all pages are marked, unmark them Theorem 1: The competitive factor of the Marking Algorithm is 2H k Theorem 2: The competitive factor of any randomized paging algorithm is H k =(logk)

13 Set Cover Elements: U ={1,2,,n} Sets: S 1,,S m (each S i µ {1,2,,n}) Each set S i has cost c i Goal: find a min cost collection of sets that cover U set cover can be formulated as integer/linear program: x i - indicator variable for choosing set S i mx minimize c i x i i=1 for every element j: x i 2 f0; 1g X ijj2s i x i 1 Relaxation: 0 x i 1 LP can be solved in poly time LP provides a lower bound on optimal integral solution!!!

14 Rounding a Fractional Solution (1) For each S i : 0 x i 1 For each element j: i x i 1 (summed over x i, j 2 S i ) Randomized Rounding: For each set S i : pick it to the cover with probability x i Analysis: Exp[cost of cover] = i c i x i = LP cost Pr[element j is not covered] = Conclusion: probability of covering element j is at least a constant!

15 Rounding a Fractional Solution (2) Amplify probability of success: Repeat experiment clogn times so that Pr[element j is not covered] Analysis: Pr[some element is not covered] Exp[cost of cover] = O(logn) i c i x i = O(logn)(LP cost) Conclusion: approximation factor is O(logn)

16 Set Cover: Greedy Algorithm Initially: C is empty While there is an uncovered element: Add to C the set S i minimizing c i /(# new elements covered) Analysis: via dual fitting Primal: covering Dual: packing

17 Dual Fitting (1) Primal solution is feasible Dual solution: if element j is covered by set S i then y j = c i /(# new elements covered by S i ) In the iteration in which S i is picked: primal = dual = c i since cost of S i is shared between the new elements Thus, cost of primal solution = cost of dual solution Is the dual solution feasible? Almost, but not quite

18 Dual Fitting (2) For set S: suppose the order in which elements in S are covered is e 1,, e k When element e i is covered, Thus, Dividing dual variables by H(n) ¼ logn yields a feasible solution Greedy algorithm is an O(logn)-approximation: primal dual H(n)

19 The Online Primal-Dual Framework Introduction: covering The ski problem The online set cover problem

20 Buying costs \$B Renting costs \$1 per day Back to Ski Rental Problem: Number of ski days is not known in advance Goal: Minimize the total cost.

21 Ski Rental Integer Program x 1 - Buy 0 - Don't Buy z i 1 - Rent on day i 0 - Don't rent on day i Subject to: min Bx k i1 z i For each day i: xz i 1 xz, {0,1} i

22 Ski Rental Relaxation P: Primal Covering min Bx For each day i: Online setting: k i1 z i xz i 1 xz, i 0 D: Dual Packing max k i1 y For each day i: k i1 Primal: New constraints arrive one by one. y i B i yi Requirement: Upon arrival, constraints should be satisfied. Monotonicity: Variables can only be increased. 1

23 Ski Rental Algorithm P: Primal Covering min Bx For each day i: k i1 z i xz i 1 xz, i 0 Initially x 0 Each new day (new constraint): if x<1: z i 1-x x x(1+ 1/B) + 1/(c*B) y i 1 D: Dual Packing max k i1 y For each day i: i k i1 - c later. y 1 i y i B

24 Analysis of Online Algorithm Proof of competitive factor: 1. Primal solution is feasible. 2. In each iteration, ΔP (1+ 1/c)ΔD. 3. Dual is feasible. Conclusion: Algorithm is (1+ 1/c)-competitive Initially x 0 Each new day (new constraint): if x<1: z i 1-x x x(1+ 1/B) + 1/(c*B) y i 1 - c later.

25 Analysis of Online Algorithm 1. Primal solution is feasible. If x 1 the solution is feasible. Otherwise set: z i 1-x. 2. In each iteration, ΔP (1+ 1/c)ΔD: If x 1, ΔP =ΔD=0 Otherwise: Change in dual: 1 Change in primal: BΔx + z i = x+ 1/c+ 1-x = 1+1/c Algorithm: When new constraint arrives, if x<1: z i 1-x x x(1+ 1/B) + 1/c*B y i 1

26 Analysis of Online Algorithm 3. Dual is feasible: Need to prove: k i1 We prove that after B days x 1 x is a sum of geometric sequence a 1 = 1/(cB), q = 1+1/B B y i B B B x cb 1 c 1 1 B B Algorithm: When new constraint arrives, if x<1: z i 1-x x x(1+ 1/B) + 1/c*B y i 1 1 c 1 1 e1 B 1 1 e c e 1 B

27 Randomized Algorithm X: 0 1 X 1 X 2 X 3 X 4 Choose d uniformly in [0,1] Buy on the day corresponding to the bin d falls in Rent up to that day Analysis: Probability of buying on the i-th day is x i Probability of renting on the i-th day is at most z i

28 Going Beyond the Ski Problem Ski problem: coefficients in the constraint matrix belong to {0,1} What can be said about general constraint matrices with coefficients from {0,1}?

29 The Online Set-Cover Problem elements: e 1, e 2,, e n set system: s 1, s 2, s m costs: c(s 1 ), c(s 2 ), c(s m ) Online Setting: Elements arrive one by one. Upon arrival elements need to be covered. Sets that are chosen cannot be unchosen. Goal: Minimize the cost of the chosen sets.

30 Online Set-Cover: Lower Bound elements: 1,,n µ n pn sets: all subsets of cardinality n cost: unit cost Adversary s strategy: While possible: pick an element that is not covered (# of elements offered n) Competitive ratio: n (cost of online: = n, cost of OPT = 1) µ p npn But, n ¼ log. So polylog(m,n) is not ruled out.

31 Set Cover Linear Program P: Primal Covering ss min c( s) x( s) e E x( s) 1 s es D: Dual Packing ee max ye ( ) s S y( e) c( s) es Online setting: Primal: constraints arrive one by one. Requirement: each constraint is satisfied. Monotonicity: variables can only be increased.

32 Primal-Dual Algorithms We will see two algorithms: Discrete algorithm generalizing ideas from the ski problem Continuous algorithm

33 Set Cover Discrete Algorithm P: Primal Covering ss min c( s) x( s) e E x( s) 1 s es D: Dual Packing ee max ye ( ) s S y( e) c( s) es Initially x(s) 0 When new element arrives, while y(e) y(e)+1. s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

34 Analysis of Online Algorithm Proof of competitive factor: 1. Primal solution is feasible. 2. In each iteration, ΔP 2ΔD. 3. Dual is (almost) feasible. Conclusion: We will see later. Initially x(s) 0 When new element e arrives, while y(e) y(e)+1. s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

35 Analysis of Online Algorithm 1. Primal solution is feasible. We increase the primal variables until the constraint is feasible. Initially x(s) 0 When new element e arrives, while y(e) y(e)+1. s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

36 Analysis of Online Algorithm 2. In each iteration, ΔP 2ΔD. In each iteration: ΔD = 1 xs ( ) 1 P = c(s) xs ( ) c(s) s es s es c( s) mc( s) Initially x(s) 0 When new element e arrives, while y(e) y(e)+1. s es x(s) 1/m 2 s es s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

37 Analysis of Online Algorithm 3. Dual is (almost) feasible: We prove that: s S, y(e) c(s)o(log m) If y(e) increases, then x(s) increases (for e in S). es x(s) is a sum of a geometric series: a 1 = 1/[mc(s)], q = (1+ 1/c(s)) Initially x(s) 0 When new element e arrives, while y(e) y(e)+1. s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

38 Analysis of Online Algorithm After c(s)o(log m) rounds: c( s) O(log m) 1 11/ cs ( ) 1 xs ( ) mc( s) 11/ c( s) 1 c( s) O(log m) 11/ cs ( ) 1 m 1 We never increase a variable x(s)>1! Initially x(s) 0 When new element e arrives, while y(e) y(e)+1. s es xs ( ) 1: s es x( s) x( s) 11/ c( s) 1/ mc( s)

39 Conclusions The dual is feasible with cost 1/O(log m) of the primal. The algorithm produces a fractional set cover that is O(log m)-competitive. Remark: no online algorithm can perform better (in the worst case).

40 Set Cover Continuous Algorithm Initially x(s) 0, y(e) 0 When new element e arrives: While s es xs ( ) 1: Increase variable y(e) continuously For each s e 2 s,.

41 Analysis of Online Algorithm Proof of competitive factor: 1. Primal solution is feasible 2. In the iteration corresponding to 2 ln(1 3. Dual solution feasible

42 Analysis of Online Algorithm 1. Primal solution is feasible. We increase the primal variables until the constraint is feasible.

43 Analysis of Online Algorithm 2. In 2 ln(1 = 1 (variable y(e) is increased continuously)

44 Primal Change

45 Analysis of Online Algorithm 3. Dual is feasible: We prove that x(s) 1 (otherwise s satisfies the violated constraint for e) Hence,

46 The primal is feasible Conclusions The dual is feasible The ratio between primal change and dual change is 1/O(log m) The algorithm produces a fractional set cover which is O(log m)-competitive.

47 Discrete vs. Continuous Both algorithms are essentially the same: as long as c(s) is not too small. (c(s) 1) Description of discrete algorithm is simpler Analysis of continuous algorithm is simpler

48 Summary: Key Idea for Primal-Dual Update Primal: Min i c i x i Dual: Max t b t y t Step t, new constraint: New variable y t a 1 x 1 + a 2 x a j x j b t + b t y t in dual objective x i (1+ a i /c i ) x i (mult. update) y t y t + 1 (additive update) primal cost = = Dual Cost

49 Online Randomized Rounding What about an integral solution? Round fractional solution: For set s, choose it with probability Δx(s) when incrementing variable x(s) Repeat O(logn) times to amplify success probability Competitive ratio is O(logm logn) Can be done deterministically online [AAABN03].

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