Christoph Lenzen, STOC 2011
What is Load Balancing?
optimizing
storage
utilization
work sharing
hashing
low-congestion
routing
Christoph Lenzen, STOC 2011
An Example of Parallel Load Balancing
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•
•
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fully connected network
small & equal bandwidths (one message/round)
n nodes need to send/receive up to n messages
minimize number of rounds
Christoph Lenzen, STOC 2011
An Example of Parallel Load Balancing
•
•
•
•
fully connected network
small & equal bandwidths (one message/round)
n nodes need to send/receive up to n messages
minimize number of rounds
Christoph Lenzen, STOC 2011
An Example of Parallel Load Balancing
?
?
but...
?
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•
•
•
fully connected network
small & equal bandwidths (one message/round)
n nodes need to send/receive up to n messages
minimize number of rounds
Christoph Lenzen, STOC 2011
Abstraction: Parallel Balls-into-Bins
• distribute n balls into n bins
=
=
• replace knowledge by randomization!
• n instances (one for each receiver)
Christoph Lenzen, STOC 2011
Naive Approach: Fire-and-Forget
• throw balls uniformly independently at random (u.i.r.)
• max. load (ln n / ln ln n) with high probability (w.h.p.)
Christoph Lenzen, STOC 2011
The Power of Two Choices (e.g. Azar et al., SIAM J. of Comp.‘99)
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inspect two bins and decide
take least loaded
max. load (ln ln n) w.h.p.
d choices: (ln ln n / ln d ) w.h.p.
(ln ln n / d ) possible Vöcking,
JACM‘92
...but not parallel!
Christoph Lenzen, STOC 2011
The Parallel Power of Two Choices
• strongest upper bound: Stemann, SPAA‘96
max. load 


ln n / ln ln n
• tight for constant r

1/ r
 in r rounds

Adler et al., STOC‘95
• ...and certain algorithms:
– non-adaptive
(fix bins to contact in advance)
– symmetric
(all choices uniform)
Christoph Lenzen, STOC 2011
An Adaptive Algorithm
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≈
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<
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contact one bin
every bin accepts one ball
n/e balls remain (w.h.p.)
contact 2<e bins
n/e2 balls remain (w.h.p.)
contact k<e2 bins, and so on
Christoph Lenzen, STOC 2011
The Power of the Tower
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termination in log*n  O(1) rounds
log*(2x )  log*( x)  1
cap # contacted bins at log n
log* x
x
O(n) total messages w.h.p.
0
1
O(1) in exp. (for each ball and bin)
1
2
can enforce max. load 2
2
4
tolerates message loss & faulty bins
3
16
4
65,536
5
≈1020,000
Christoph Lenzen, STOC 2011
Optimality?
• for symmetric algorithms:
–
–
–
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many balls in symmetric trees for (1  o(1)) log*(n) rounds
balls cannot contact many bins w/o incurring  (n) messages
if balls in such a tree terminate root gets expected load  (1)
many such trees => max. load  (1) w.h.p.
root
Christoph Lenzen, STOC 2011
No Faster Solution Possible, unless...
• bin loads of  (1) are accepted,
place (1  o(1))ns
balls at once
• bins have "identities" known to all balls
•  (n) messages are used
Christoph Lenzen, STOC 2011
Exploiting Asymmetry/Bin ID‘s
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subsets of  (1) bins
contact random “leader” bin
“leader” bins distribute balls in their subset
can place (1  o(1))n balls right away
continue with previous algorithm
max. load 3 in O(1) rounds
?
?
?
#1
#2
#3
Christoph Lenzen, STOC 2011
How to Use  (n) Messages
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balls “coordinate” constant fraction of bins
each ball contacts  (1) bins
(1  o(1))n balls find coordinated bins
coordinators assign O(1) balls to “their” bins
proceed with symmetric algorithm
X XX
X
 (1)
 (1)
Christoph Lenzen, STOC 2011
Summary
• optimal symmetric solution
– max. load 2, O(n) mess., log*n  O(1) rounds
• constant-time if:
– global enumeration of bins
–  (n) messages
–  (1) max. load
Christoph Lenzen, STOC 2011
...hey, What Happened to the Original Problem?!
• can be solved in O(1) rounds
• can be used to sort n 2 keys in O(1) rounds
Patt-Shamir and Teplitsky, PODC‘11
Christoph Lenzen, STOC 2011
Christoph Lenzen, STOC 2011