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The Santa Claus Problem(Maximizing the minimum load on unrelated machines)

Nikhil Bansal (IBM)

Maxim Sviridenko (IBM)

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The Santa Claus Problem

m kids n toys [ n ¸ m]

Kid i has value pij (¸ 0) for toy j

Toys distributed among kids

Happiness of kid i = Sum of pij of toys it gets (additive utilities)

Goal: Maximize mini=1,…,m Happiness(i)(Given target T, ensure every kid has happiness ¸ T)

Natural notion of fair allocation,proposed in game theoretic setting by [Lipton et al 04]

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Problem definition

Scheduling Problem (unrelated machine scheduling)

Kid (i) = Machine Toy (j) = Job

pij = processing time of job j on machine i.

Goal: Maximize the minimum load (will use this terminology )

Restricted assignment case:

Job has same size

Can only be placed on subset of machines ( pij 2 {pj,0} )

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Approximation Algorithms

An efficient (polynomial time) algorithm that is comparable to optimum solution on every input I

approximation algorithm A

1) A(I) · Opt(I) for every instance I

2) A runs in time polynomial in size of I

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Previous work (Makespan minimization)Extensive work on minimizing makespan (max. load)(our problem is maximize min load)

2-approx for unrelated parallel machines (arbitrary pij)and restricted assignment [Lenstra Shmoys Tardos 90]

Lots of work on various other special cases…

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Previous Work (max min problem)If n = m : Matching techniques (poly time)

(n-m+1) approximation [Bezakova Dani 05]

Can achieve Opt – pmax [Bezakova Dani 05]

(pmax = maxij pij bad approx if Opt ¼ pmax)

(1-1/k) fraction of machines get ¸ Opt/k [Golovin 05]

Restricted Assignment Setting: pij 2 {pj, 0}

Big/small good case (pj 2 {1,X} ) : O(n1/2) [Golovin 05]

No < 2 approx, unless P=NP (follows from LST 90)

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LP Formulation: 1Assignment Formulation: xij =1 if job j on machine i

Max T

j pij xij ¸ T 8 i 2 Machines (machine load >= T)

i xij · 1 8 j 2 Jobs (a job used at most once)

Valid Integer programming formulation (xij 2 {0,1})

LP relaxation useless: Infinite Gap

m machines, 1 job of size x (on all machines)

LP: x/m IP = 0 Big jobs trick LP.

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Job Truncation Idea (Lenstra, Shmoys, Tardos)In schedule w/ value T, can truncate sizes to T

p’ij = min(pij,T)

j p’ij xij ¸ T 8 machines i

i xij · 1 8 jobs j

Good News: T – pmax [Bezakova Dani 05]

pmax = max job size

T

Find largest T for which LP is feasible

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Job Truncation for max min ?Bad news: (m) gap (even for restricted assignment case)

Instance: m small jobs, numbered 1,2,…,m

Job i : size 1 on machine i, and size 0 elsewhere

m-1 bigs jobs (size m on every machine)

m=31 2

1 2 3

M1

LP: m

IP: 1M2 M3

Small jobs

Big jobs

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Configuration LP

Target solution value T.

Set of jobs S is valid configuration for machine i, if

size of S on i is ¸ T Let C(i) = all valid sets for i

Variable for each valid configuration S for machine i

1 config. per machine

Job used · once

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Exponentially many variablesSeparation oracle for dual : knapsack problem.

1/2 1/4 1/4 0.4 0.6

Each machine assigned 1 unit of configurations

Each job has at most 1 unit of appearances

T

View of Configuration LP solution

Width = amount of configuration

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Our Results

1) O( log log m / log log log m) approximation for restricted assignment case (pij 2 {pj,0} )

(previous best n-m+1, LST LP has (m) gap)

(m1/2), (n1/2) integrality gap for configuration LP

when pij arbitrary

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This Talk

Restricted Assignment case:

1) Describe O(log n / log log n) approx (randomized rounding)

2) Improvement to

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Restricted Assignment: Basic ProblemOpt – pmax known: Hard case when pmax ¼ Opt

Assume jobs of size Opt (big) or 1 (small)

One big job suffices to load a machine.

Basic Problem: Match big jobs to machines s.t. remaining machines can be loaded with small jobs.

1 2 3 4

Pink on 1 Green on 4Good !

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Restricted Assignment: Basic ProblemOpt – pmax known: Hard case when pmax ¼ Opt

Assume jobs of size Opt (big) or 1 (small)

One big job suffices to load a machine.

Basic Problem: Match big jobs to machines s.t. remaining machines can be loaded with small jobs.

1 2 3 4

Pink on 2 Green on 4 Bad !

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Main Idea of Proof

Transform LP solution to more structured one (only O(1) loss). (Pseudo) Rounding Procedure: Match big jobs to machines,and place smalls on remaining machines 1) Load on each machine Opt2) Small jobs used up to times.

Easy but key Fact: Place smalls s.t. each machine has load Opt/)

How small can we make ??

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Structural LemmaMachines

Big Jobs

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Structural Lemma

1) |Ji | = |Mi| -1.

2) Ji can be placed on any |Mi|-1 machines in Mi.

M1 M3M2

J1 J2 J3

Machines

Big Jobs

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Proof Sketch: Structural Lemma

machines

Big jobs

0.3 0.5 Remove cycles,Tree with leaves as machines

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Reduced Problem

In each group choose 1 machine to place smalls (bigs handled automatically)

For each group: We have one unit of small configurations distributed among the machines.

M1

.4 .3 .3.2 .5 .3 .1 .5 .4J1

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Reduced Problem

Randomized rounding: View these as probabilities. Choose a machine (to place smalls) at random for each group.

Fractional congestion: O(1)Chernoff Bounds: Worst case congestion O(log n / log log n)

M1

.4 .3 .3.2 .5 .3 .1 .5 .4J1

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Congestion with Outliers

Do not care about worst case congestion.

Suppose throw away 50% small jobs from each machine, and get low congestion on remaining jobs.

Congestion with outliers problem

We show: Can throw away small fraction from each machine, and get congestion

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First attempts

Randomized Rounding with alterations: Job occurs O(1) times in expectation. Throw away high frequency jobs

Problem: Correlation!

Power of many choices: Choose O(1) sets at random instead of 1, and then pick best. Does not work either…

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Our Approach

Reduce Congestion with outliers to worst case congestion minimization on small sets.

Random restriction of small jobss.t. each set has ¼ polylog m jobs

Find low worst case congestion solution on this instance.

Key result: Whp, solution with low congestion on restricted instance is also low congestion with outliers on original instance.

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Small sets have low worst case congestionTheorem [Leighton Lu Rao Srinivasan 01]:

If sets of size k can get O(log k / log log k) congestion.

Non-trivial: Repeated application of Lovasz Local Lemma

In our setting k = polylog (m), gives

approximation

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Concluding Remarks

Huge gaps remain in our understanding.

Arbitrary pij : O(n) approx, vs. hardness factor of 2

Can we use the (n1/2) integrality gap instance to show better hardness?

For restricted assignment, is the right answer O(1) ?

Suffices to prove congestion with outliers property.

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Questions ?

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Attempt 1

For each Ci, choose sets at random,

Throw away elements with high congestion

(hopefully not too many)

Bad when sets correlated

Either Sil = Si’l’ or Sil Å Si’l’ = ;

If one element congested,Whole set congested.

Whp congestion log m / log log m )

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Correlated Sets

Independent sets : Random works wellCorrelated sets: Other techniques

Need to be more subtle.

Either Sil = Si’l’ or Sil Å Si’l’ = ;

Fractional Matching Between Sets & super-machines

Can find perfect machingCongestion = 1 !!

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Fact (Leighton et al 2001)

If each set has cardinality polylog m, then can find a choice of sets s.t. congestion = O( log log m / log log log m)

[Constructive version of Lovasz Local Lemma](intuitively, low dependence between sets)

In our case: Sets have big cardinality, but allow throwing elements out.

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Bad choices

Call a choice of sets bad, if no matter how you delete upto 50% elements, congestion ¸ c log log m / log log log m.

Main Result: Good choice always exists and can be found in poly time.

Given a good choice of sets, the elements to throw can be found via a max flow problem. (proof skipped)

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AlgorithmTake a random sample of elements.Consider restriction of sets of these elementsEach set has about poly log m jobs.

Choose sets with O(log log m/ log log log m) congestion (Leighton et al)

Theorem: The same choice of sets works for original set system.

Proof: For every bad choice of sets in original system => high congestion in sampled system.

Naïve counting does not work. Exponentially many choices of sets.Notion of core of a bad choice function. VC-dimension type arguments.

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Configuration LP is strong !

Example: m-1 bigs jobs (size m on every machine)

m small jobs, call then 1,2,…,m

Small Job i has size 1 on machine i and 0 elsewhere.

IP Opt =1 Configuration LP infeasible for T > 1.

Feasible configuration for any machine i

(has to contain a big)

Infeasible: m machines but only m-1 big jobs.

b b

s

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Robustness of Small JobsRecall: Assignment LP gives T – pmax

Suppose pmax < T/10 (integral solution: T – T/10)

Suppose xij only satisfy a relaxed inequality

j pij xij ¸ T 8 machines ii xij · 1 8 jobs j

j pij xij ¸ Ti xij · 3

Can still round it toT/3 - pmax

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Big jobs are not robust3 machines. 1 job of size T.

Opt = 0 Relaxed LP has value T.

(big jobs are delicate)

We will transform LP solution (get more structure)

s.t. Small jobs can occur upto O(1) times.

After Rounding: Match big jobs to machines,

and smalls on remaining machines

1) Load on each machine T

2) Small used up to times.

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An easy case

Suppose big jobs occupy < 3/4 of any machine. Great!

Hard case: If 1- o(1) machine used by bigs, very little by smalls.

.01.33 .33 .33

1/2 1/4 1/4 1/2 1/4 1/4

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Structural Lemma

100 machines form this group.

99 Big jobs (can be placed “anywhere” on these)

One unit of small configurations.

.01.33 .33 .33 .01.33 .33 .33 .01.33 .33 .33

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Can ignore bigs

.01.33 .33 .33 .01.33 .33 .33 .01.33 .33 .33Super-machine

Choose one “configuration” from each super-machine