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Regret Minimization and Job Scheduling Yishay Mansour Tel Aviv University

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Talk Outline Regret Minimization – External Regret minimization Motivation algorithm – Internal Regret minimization Motivation – Regret Dynamics External Regret Job Scheduling and Regret Minimization – Model – Stochastic case

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Decision Making under uncertainty Online algorithms – Stochastic models – Competitive analysis – Absolute performance criteria A different approach: – Define “reasonable“ strategies – Compete with the best (in retrospect) – Relative performance criteria

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5 Routing Model: Each day 1. select a path from source to destination 2. observe the latencies. – Each day diff values Strategies: All source-dest. paths Loss: The average latency on the selected path Performance Goal: match the latency of best single path

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6 Financial Markets: options Model: stock or cash. Each day, set portfolio then observe outcome. Strategies: invest either: all in stock or, all in cash Gain: based on daily changes in stock Performance Goal: Implements an option ! CASH STOCK

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7 Machine learning – Expert Advice Model: each time 1. observe expert predictions 2. predict a label Strategies: experts (online learning algorithms) Loss: errors Performance Goal: match the error rate of best expert – In retrospect 1 2 3 4 1 0 1 1

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8 Parameter Tuning Model: Multiple parameters. Strategies: settings of parameters Optimization: any Performance Goal: match the best setting of parameters

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9 Parameter Tuning Development Cycle – develop product (software) – test performance – tune parameters – deliver “tuned” product Challenge: can we combine – testing – tuning – runtime

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10 Regret Minimization: Model Actions A={1, …,N} Time steps: t ∊ { 1, …, T} At time step t: – Agent selects a distribution p t (i) over A – Environment returns costs c t (i) ε [0,1] Adversarial setting – Online loss: l t (on) = Σ i c t (i) p t (i) Cumulative loss : L T (on) = Σ t l t (on)

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11 External Regret Relative Performance measure: – compares to the best strategy in A The basic class of strategies Online cumulative loss : L T (on) = Σ t l t (on) Action i cumulative loss : L T (i) = Σ t c t (i) Best action: L T (best) = MIN i {L T (i) }=MIN i {Σ t c t (i)} External Regret = L T (on) – L T (best)

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External Regret Algorithm Goal: Minimize Regret Algorithm: – Track the regrets – Weights proportional to the regret Formally: At time t – Compute the regret to each action Y t (i)= L t (on)- L t (i), and r t (i) = MAX{ Y t (i),0} p t+1 (i) = r t (i) / Σ i r t (i) – If all r t (i) = 0 select p t+1 arbitrarily. R t = and ΔR t = Y t - Y t-1

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External Regret Algorithm: Analysis R t = and ΔR t = Y t - Y t-1 LEMMA: ΔR t ∙ R t-1 = 0 Σ i (c t (i) – l t (on)) r t-1 (i) = Σ i c t (i)r t-1 (i)– Σ i l t (on)r t-1 (i) Σ i l t (on) r t-1 (i) = [Σ i c t (i) p t (i) ]Σ i r t-1 (i) = Σ i c t (i)r t-1 (i) LEMMA: R t-1 RtRt

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14 External regret: Bounds Average regret goes to zero – No regret – Hannan [1957] Explicit bounds – Littstone & Warmuth ‘94 – CFHHSW ‘97 – External regret = O(log N + √Tlog N)

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16 Dominated Actions Model: action y dominates x if y always better than x Goal: Not to play dominated actions Goal (unknown model): The fraction of times we play dominated actions is played is vanishing.3.8.9.3.6 Cost Action y Cost Action x.2.4.7.1.3

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Internal/Swap Regret Internal Regret – Regret(x,y) = ∑ t: a(t)=x c t (x) - c t (y) – Internal Regret = max x,y Regret(x,y) Swap Regret – Swap Regret = ∑ x max y Regret(x,y) Swap regret ≥ External Regret – ∑ x max y Regret(x,y) ≥ max y ∑ x Regret(x,y) Mixed actions – Regret(x,y) = ∑ t (c t (x) - c t (y))p t (x)

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Dominated Actions and Regret Assume action y dominates action x – For any t: c t (x) > c t (y)+δ Assume we used action x for n times – Regret(x,y) > δ n If SwapRegret < R then – Fraction of time dominated action used – At most R/δ

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19 Calibration Model: each step predict a probability and observe outcome Goal: prediction calibrated with outcome – During time steps where the prediction is p the average outcome is (approx) p.3.5.3.5 predictionsoutcome Calibration:.3.5 1/3 1/2 Predict prob. of rain

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Calibration to Regret Reduction to Swap/Internal regret: – Discrete Probabilities Say: {0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0} – Loss of action x at time t: (x – c t ) 2 – y*(x)= argmax y Regret(x,y) y*(x)=avg(c t |x) – Consider R(x,y*(x))

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21 Internal regret No internal regret – [Foster & Vohra], [Hart & Mas-Colell] Based on the approachability theorem [Blackwell ’56] – Explicit Bounds – [Cesa-Bianchi & Lugasi ’03] Internal regret = O(log N + √T log N) – [Blum & Mansour] Swap regret = O(log N + √T N)

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Regret: External vs Internal External regret – You should have bought S&P 500 – Match boy i to girl i Internal regret – Each time you bought IBM you should have bought SUN – Stable matching 22 Limitations: - No state - Additive over time

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23 [Even-Dar, Mansour, Nadav, 2009]

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Routing Games s1s1 t1t1 t2t2 s2s2 f 1, L f 1, R f 2, T f 2, B f2f2 f1f1 Atomic –Finite number of players –Player i transfer flow from s i to t i f 1,L f 2,T Latency on edge e = L e (f 1,L + f 2,T ) e Cost i = p ε (s i, t i ) Latency(p) * flow i (p) Splittable flows

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Cournot Oligopoly [Cournot 1838] Best response dynamics converges for 2 players [Cournot 1838] –Two player ’ s oligopoly is a super-modular game [Milgrom, Roberts 1990] Diverges for n 5 [Theocharis 1960] X Y Cost 1 (X)Cost 2 (Y) Market price Overall quantity Firms select production level Market price depends on the TOTAL supply Firms maximize their profit = revenue - cost Xy P

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Resource Allocation Games Each advertiser wins a proportional market share $5M $10M$17M $25M Advertisers set budgets: ‘s allocated rate = 5+10+17+25 25 f ( U = ) - $25M Utility: –Concave utility from allocated rate –Quasi-linear with money The best response dynamics generally diverges for linear resource allocation games

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Properties of Selfish Routing, Cournot Oligopoly and Resource Allocation Games 1.Closed convex strategy set 2.A (weighted) social welfare is concave 3.The utility of a player is convex in the vector of actions of other players R There exists 1,…, n > 0 Such that 1 u 1 (x) + 2 u 2 (x)+…+ n u n (x) Socially Concave Games

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The relation between socially concave games and concave games Zero Sum Games Socially Concave Games Concave Games Concave Games [ Rosen 65] The utility of a player is strictly concave in her own strategy A sufficient condition for equilibrium uniqueness Normal Form Games (with mixed strategies) Unique Nash Equilibrium Atomic, splittable routing Resource Allocation Cournot

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The average action and average utility converge to NE Theorem 1: The average action profile converges to NE Player 1: Player 2: Player n : Day 1Day 2Day 3Day T Average of days 1…T - Nash equilibrium Theorem 2: The average daily payoff of each player converges to her payoff in NE If each player uses a procedure without regret in socially concave games then their joint play converges to Nash equilibrium:

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Convergence of the “ average action ” and “ average payoff ” are two different things! Here the average action converges to ( ½, ½ ) for every player s t t s But the average cost is 2, while the average cost in NE is 1 s t On Even Days On Odd Days

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The Action Profile Itself Need Not Converge s t t s On Even Days On Odd Days

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32 Correlated Equilibrium CE: A joint distribution Q Each time t, a joint action drawn from Q – Each player action is BR Theorem [HM,FV]: Multiple players playing low internal (swap) regret converge to CE.3.8.9.3.6 Action x Action y.2.4.7.1.3

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33 [Even-Dar, Klienberg, Mannor, Mansour, 2009]

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34 Outline Job scheduling vs. online learning – similarities – differences Model & Results General algorithm – calibration based Simple makespan algorithm

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35 Job Scheduling: Motivating Example Load Balancer users servers GOAL: Minimize load on servers

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36 Online Algorithms Job scheduling – N unrelated machines machine = action – each time step: a job arrives – has different loads on different machines algorithm schedules the job on some machine – Given its loads – Goal: minimize the loads – makespan or L 2 Regret minimization – N actions machines – each time step First, algorithm selects an action (machine) Then, observes the losses – Job loads – Goal: minimize the sum of losses

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37 Modeling Differences: Information Information model: – what does the algorithm know when it selects action/machine Known cost: – First observe costs then select action – job scheduling Unknown cost: – First select action then observe costs – Regret Minimization

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38 Modeling Differences: Performance Theoretical Performance measure: – comparison class job scheduling: best (offline) assignment regret minimization: best static algorithm – Guarantees: job scheduling: multiplicative regret minimization: additive and vanishing. Objective function: – job scheduling: global (makespan) – regret minimization: additive.

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39 Formal Model N actions Each time step t algorithm ON – select a (fractional) action: p t (i) – observe losses c t (i) in [0,1] Average losses of ON – for action i at time T: ON T (i) = (1/T) Σ t

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40 Formal Model Static Optimum: – Consider any fixed distribution α Every time play α – Static optimum α * - minimizes cost C Formally: – Let α ◊ L = (α(1)L(1), …, α(N) L(N)) Hadamard (or Schur) product. – best fixed α * (L) = arg min α C(α ◊ L ) where L T (i) = (1/T) Σ t c t (i) – static optimality C * (L) = C(α * (L) ◊ L)

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41 Example Two machines, makespan: observed loads α*(L) L1L1 L2L2 final loads L1L1 L2L2 4 2 ( 1/3, 2/3)4/3

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42 Our Results: Adversarial General General Feasibility Result: – Assume C convex and C * concave includes makespan and L d norm for d>1. – There exists an online algorithm ON, which for any loss sequence L: C(ON) < C*(L) + o(1) – Rate of convergence about √N/T

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43 Our Results: Adversarial Makespan Makespan Algorithm – There exists an algorithm ON – for any loss sequence L C(ON) < C*(L) + O(log 2 N / √T) Benefits: – very simple and intuitive – improved regret bound Δ Two actions

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44 Our Results: Adversarial Lower Bound We show that for many non-convex C there is a non-vanishing regret – includes L d norm for d<1 Non-vanishing regret ratio >1 There is a sequence of losses L, such that, C(ON) > (1+γ) C * (L), where γ>0

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45 Preliminary: Local vs. Global time B1B1 B2B2 BkBk Low regret in each block Overall low regret

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46 Preliminary: Local vs. Global LEMMA: – Assume C convex and C* concave, – Assume a partition of time to B i – At each time block B i regret at most R i Then: C(ON)-C * (L) ≤ Σ i R i

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47 Preliminary: Local vs. Global Proof: C(ON) ≤ Σ C(ON(B i )) C is convex Σ C*(L(B i )) ≤ C*(L) C* is concave C(ON(B i )) – C*(L(B i )) ≤ R i low regret in each B i Σ C(ON(B i )) – C*(L(B i )) ≤ Σ R i C(ON) – C*(L) ≤ Σ R i QED Enough to bound the regret on subsets.

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48 Example t=1 t=2 M1M1 M2M2 arrival losses static opt α*=(1/2,1/2) cost = 3/2 M1M1 M2M2 local opt α*: (1/3,2/3) (2/3,1/3) cost = 4/3 M1M1 M2M2 M1M1 M2M2 global offline opt: (0,1) (1,0) cost = 1

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Stochastic case Assume that each action’s cost is drawn from a joint distribution, – i.i.d over time steps Theorem (makespan/L d ) – Known distribution Regret =O(log T /T) – Unknown distributions Regret = O( log 2 T /T)

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Stochastic case: Each time t the costs are drawn from a joint distribution, – i.i.d over time steps, not between actions INTUITION: Assume two actions (machines) Load Distribution: – With probability ½ : (1,0) – With probability ½ : (0,1) Which policy minimizes makespan regret?! Regret components: – MAX(L(1),L(2)) = sum/2 +|Δ|/2 – Sum=L(1)+L(2) & Δ=L(1)-L(2)

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Stochastic case: Static OPT Natural choice (model based) – Select always action ( ½, ½ ) Observations: – Assume T/2+Δ times (1,0) and T/2-Δ times (0,1) – Loads (T/4+ Δ/2, T/4-Δ/2) – Makespan = T/4+ Δ/2 > T/4 – Static OPT: T/4 – Δ 2 /T < T/4 W.h.p. OPT is T/4-O(1) sum=T/2 & E[|Δ|]= O(√T) – Regret = O(√T)

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Can we do better ?!

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Stochastic case: Least Loaded Least loaded machine: – Select the machine with the lower current load Observation: – Machines have same load (diff ≤ 1): |Δ| ≤ 1 – Sum of loads: E[sum] = T/2 – Expected makespan = T/4 Regret – Least Loaded Makespan LLM=T/4 ± √T – Regret =MAX{LLM-T/4,0} = O(√T) Regret considers only the “bad” regret

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Can we do better ?!

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Stochastic case: Optimized Finish Algorithm: – Select action ( ½, ½ ) For T-4√T steps – Play least loaded afterwards. Claim: Regret =O(T 1/4 ) – Until T-4 √T steps (w.h.p) Δ < 2√T – Exists time t in [T-4 √T,T]: Δ=0 & sum = T/2 + O( T 1/4 ) From 1 to t: regret = O(T 1/4 ) From t to T: Regret = O(√(T-t)) = O(T 1/4 )

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Can we do better ?!

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Stochastic case: Any time An algorithm which has low regret for any t – Not plan for final horizon T Variant of least loaded: – Least loaded weight: ½ + T -1/4 Claim: Regret = O(T 1/4 ) – Idea: Regret = max{(L 1 + L 2 )/2 – T/4,0} + Δ Every O(T 1/2 ) steps Δ=0 Very near (½, ½)

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Can we do better ?!

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Stochastic case: Logarithmic Regret Algorithm: – Use phases – Length of phases shrink exponentially T k = T k-1 /2 and T 1 = T/2 Log T phases – Every phase cancels deviations of previous phase Deviation from the expectation Works for any probabilities and actions ! – Assuming the probabilities are known.

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Can we do better ?!

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Stochastic case: Unknown distributions Basic idea: – Learn the expectations – Have super-phases increase over time B r = 2 B r-1 – In each super-phase run the logarithmic regret Using the observed expectations in the past – Total number of phases log 2 T The bound on the regret O(log 2 T /T)

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Stochastic case Assume that each action’s cost is drawn from a joint distribution, – i.i.d over time steps Theorem (makespan/L d ) – Known distribution Regret =O(log T /T) – Unknown distributions Regret = O( log 2 T /T)

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Summary Regret Minimization – External – Internal – Dynamics Job Scheduling and Regret Minimization – Different global function – Open problems: Exact characterization Lower bounds

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65 Makespan Algorithm Outline: – Simple algorithm for two machines Regret O(1/√T) simple and almost memory-less – Recursive construction: Given three algorithms: two for k/2 actions and one for 2 actions build an algorithm for k actions Main issue: what kind of feedback to “propagate”. Regret O(log 2 N /√T) – better than the general result.

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66 Makespan: Two Machines Intuition: – Keep online’s loads balanced Failed attempts: – use standard regret minimization In case of unbalanced input sequence L, algo. will put most of the load on single machine – use optimum to drive the probabilities Our approach: – Use the online loads not opt or static cumulative loads

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67 Makespan Algorithm: Two actions At time t maintain probabilities p t,1 and p t,2 = 1-p t,1 Initially p 1,1 = p 1,2 = ½ At time t: Remarks: – uses online loads – Almost memory-less Δ

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68 Makespan Algorithm: Analysis View the change in probabilities as a walk on the line. 0 ½ 1

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69 Makespan Algorithm: Analysis Consider a small interval of length ε Total change in loads: – identical on both started and ended with same Δ Consider only losses in the interval – local analysis Local opt is also in the interval Online used “similar” probability – loss of at most ε per step

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70 Makespan Algorithm: Analysis Simplifying assumptions: – The walk is “balanced” in every interval add “virtual” losses to return to initial state only O(√T) additional losses – relates the learning rate to the regret – Losses “cross” interval’s boundary line needs more sophisticated “bookkeeping”. – make sure an update affects at most two adjacent intervals. – Regret accounting loss in an interval additional “virtual” losses.

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71 Makespan: Recursive algorithm Recursive algorithm: A3 A1 A2

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72 Makespan: Recursive The algorithms: – Algorithms A1 and A2: Each has “half” of the actions – gets the actual losses and “balances” them each work in isolation. – simulating and not considering actual loads. – Algorithm A3 gets the average load in A1 and A2 – balances the “average” loads. A3 A1 A2 AVG(l t,i q t,i ) AVG(l t,i q’ t,i ) l t,1 …. l t,k/2 ….

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73 Makespan: Recursive algorithm Input to A3: average load A3 A1 A2 AVG(l t,1 q t,i ) AVG(l t,1 q’ t,i )

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74 Makespan: Recursive The combined output : A3 A1 A2 p2p2 p1p1 xx q 1, … q 1 p 1, …, q k/2, … q k/2 p 2, …, l 1, … l k/2, … AVG(l t,1 q t,i ) AVG(l t,k/2 q t,i )

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75 Makespan: Recursive Analysis (intuition): – Assume perfect ZERO regret just for intuition … – The output of A1 and A2 completely balanced The average equals the individual loads – maximum=average=minimum – The output of A3 is balanced the contribution of A1 machines equal to that of A2 Real Analysis: – need to bound the amplification in the regret.

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