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Viceroy: Scalable Emulation of Butterfly Networks For Distributed Hash Tables

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## Viceroy: Scalable Emulation of Butterfly Networks For Distributed Hash Tables

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### Viceroy: Scalable Emulation of Butterfly Networks For Distributed Hash Tables

By: Dahlia Malkhi, Moni Naor & David Ratajzcak

Nov. 11, 2003

Presented by Zhenlei Jia

Nov. 11, 2004

Acknowledgments

Some of the following slides are adapted from the slides created by the authors of the paper

Outline

- Outline
- DHT Properties
- Viceroy
- Structure
- Routing Algorithm
- Join/Leave
- Bounding In-degree: Bucket Solution
- Fault Tolerance
- Summary

DHT

- What’s DHT
- Store (key, value) pairs
- Lookup
- Join/Leave
- Examples
- CAN, Pastry, Tapestry, Chord etc.

DHT Properties

- Dilation
- Efficient lookup, usually O(log(n))
- Maintenance cost
- Support dynamic environment
- Control messages, affected servers
- Degree
- Number of opened connections
- Servers impacted by node join/leave
- Heartbeat, graceful leave

DHT Properties (cont.)

- Congestion:
- Peers should share the routing load evenly
- Load (of a node): the probability that it is on a route with random source and destination.
- If path length = O(log(n)) then on average, each node is on n2 x O(log(n))/n = O(nlog(n)) routes. Average load = O(nlogn)/n2 = O(log(n))/n

Intuition

- Route is a combination of links of appropriate size
- Chord: Each node has ALL log(n) links
- Viceroy
- Each node has ONE of the long-range links
- A link of length 1/2k points to a node has link of length 1/2k+1

Chrod

110

000

001

010

100

101

111

Level 1

Level 2

Level 3

Level 4

A Butterfly Network- Each node has ONE of the long-range links
- A link of length 1/2k points to a node has link of length 1/2k+1
- Nodes “share” each other’s long link
- Routing
- Route to root
- Route to right group
- Route to right level
- Path: O(log(n))
- Degree: O(1)

1001

1011

1101

1111

0001

0010

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1

Level 1

Level 2

Level 3

A Viceroy network- Ideally, there should be log(n) levels
- There is not a global counter
- Later, we will see how a node can estimate log(n) locally

Structure: Nodes

- Node
- Id: 128 bits binary string, u
- Level: positive integer, u.level
- Order of ids
- b1b2…bk ∑i=1…k bi/2i
- Each node has a SUCCESSOR and a PREDECESSOR

SUCC(u), PRED(u)

- Node u stores the keys k such that u≤k<SUCC(u)

0

1

PRED(x)

x

SUCC(x)

Structure: Nodes- Lemma 2.1

Let n0 = 1/d(x, SUCC(x)), then w.h.p. (i.e. p>1-1/n1+e) that

log(n)-log(log(n))-O(1) <log(n0) ≤3log(n)

- Node x selects level from 1…log(n0) uniformly randomly

Structure: Links

- A node u in level k has six out links
- 2 x Short: SUCCESSOR ,PREDECESSOR
- 2 x Medium: (left) closest level-(k+1) node whose id matches u.id[k] and is smaller than u.id.
- 1 x Long: the closest level-(k+1) node with prefix u1…uk-1(1-uk)(?)

u1…uk-1(1-uk)uk+1…uw*

where w=log(n0)-log(log(n0))

- 1 x Parent: closest level-(k-1) node
- Also keeps track of in-bound links

1001

1011

1101

1111

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Short link

Long link, cross over about 1/2k

Matches u[w] except kth bit. (11*)

Short link

Level 1

Medium link

Matches x[k]0*

Matches 1*

Wrong!

Level 2

Parent link, to level k-1

Level 3

Structure: LinksRouting: Algorithm

LOOKUP(x, y):

Initialization: set cur to x

Proceed to root: while cur.level > 1:

cur = cur.parent

Greedy search:

if cur.id ≤ y < SUCC(cur).id, return cur.

Otherwise, choose m from links of cur that minimize d(m, y), move to m and repeat.

Demo: http://www.cs.huji.ac.il/labs/danss/anatt/viceroy.html

Routing: Analysis (2)

- Expected path length = O(log(n))
- log(n ) to `level-1’ node
- log(n ) for traveling among clusters
- log(n ) for final local search

Routing: Theorems

- Theorem 4.4

The path length from x to y is O(log(n)) w.h.p.

- Proof is based on several lemmas
- Lemma 4.1

For every node u with a level u.level < log(n)-log(log(n)), the number of nodes between u and u.Medium-left (Medium-right), if it exists, is at most 6log2(n) w.h.p.

Routing: Theorems (2)

- Lemma 4.2

In the greedy search phase of a lookup of value Y from node x, let the j’th greedy step vj, for 1 ≤ j ≤ m, be such that vj is more than O(log2(n)) nodes away from y. Then w.h.p. node vj is reached over a Medium or Long link, and hence satisfies vj.level = j and vj[j] = Y[j].

- m = log(n)-2loglog(n)-log(3+e)
- W.h.p. within m steps, we are n/2m = 6log2(n) nodes away from the destination

Routing: Theorems (3)

- Lemma 4.3

Let v be a node that is O(log2(n)) nodes away from the target y. Then w.h.p., within O(log(n)) greedy steps that target y is reached from v.

- Theorem 4.4

The total length of a route from x to y is O(log(n)) w.h.p.

- Theorem 4.6

Expected load on every node is O(log(n)/n).

The load on every node is log2(n)/n w.h.p.

- Theorem 4.7

Every node u has in-degree O(log(n)) w.h.p.

Join: Algorithm

- Choose identifier: select a random 128 bits x1x2…x128
- Setup short links: invoke LOOKUP(x), let x’ be the result node. Insert x between x’ and x’.SUUCESSOR.
- Choose level: let k be the maximal number of matching prefix bits between x and either SUCC(x) or PRED(x), choose level from 1…k.
- Set parent link: If SUCC(x) has level x.level-1, set x.parent to it. Otherwise, move to SUCC(x) and repeat.
- Set long link: p = x1…xk-1(1-xk)xk+1…xw

Invoke LOOKUP(p), stop after a node at level x.level+1 and matches p

is reached.

Join: Algorithm (cont.)

6. Set medium links: Denote p = x1x2…xx.level. If SUCC(x) has prefix p and level x.level+1, set x.Medium-right link to it. Otherwise, move the SUCC(x) and repeat.

7.Set inbound links: Denote p = x1x2…xx.level.

Set inbound Medium links: Following SUCC links, so long as successor y has a prefix p and a level different from x.level, if y.level = x.level-1, set y.Medium-left to x.

Set inbound long links: Following SUCC links, find y that has a prefix matches p and has level x.level. Take any inbound links that is closer to x than y.

Set inbound parent links: Following PRED link, find y such that y.level = x.level+1. Repeat until meet a node in same level as x.

Set Medium link: O(lg2n) w.h.p

p = x1x2…xk (01)

If y[k] != p: stop

If y[k]=p and y.level=k+1:

set Medium link

Otherwise, move to succ(y)

0101

0111

1001

1011

1101

1111

0001

0010

0011

0100

0110

1000

1110

0

1

Lookup(x)

Level 1

X

Level 2

Set inbound long links:

Following short links, find y such that y[k]=x[k] and y.level = x.level, check y’s inbound links.

STOP

Level 3

Set long link

P = x1…xk-1(1-xk)…xw stop at level k+1?

In this case, find 00*

Join: ExampleSet Parent link:

Following SUCC link, find a node has level k-1.

0111

Join: Analysis

- LOOKUP takes O(log(n)) messages w.h.p.
- Travels on short links during link setting phase is O(lg2n) w.h.p.
- A Medium link is within 6log2(n) nodes from x w.h.p.
- Similar for others
- Theorem 5.1:

A JOIN operation by a new node x incurs expected O(log(n)) number of messages, and O(log2(n)) messages w.h.p.

The expected number of nodes that change their state as a result of x’s join is constant, and w.h.p is O(log(n)).

Because node x has O(log(n)) in-degrees w.h.p.

Similar results holds for LEAVE.

Bounding In-degrees

- Theorem 4.7

Every node has expected constant in-degree, and has O(log(n)) in-degree w.h.p.

- In-degree=# of servers affected by join/leave
- How to guarantee constant in-degree?
- Bucket solution
- A background process to balance the assignment of levels

Level k

Bucket Solution: Intuition~log(n)

- Node x has log(n) in-degree, assuming Medium Right

x

- Too many nodes at level k-1;

Too few nodes at level k

- Improve the level selection procedure

1001

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0011

0110

1101

1110

1111

0001

0100

1000

0010

0

1

Bucket Solution- The name space is divided into non-overlapped buckets.
- A bucket contains m nodes, where log(n) ≤m ≤ clog(n), for c>2.
- In a buckets, levels are NOT assigned randomly
- For each 1≤j≤log(n), there are 1…c nodes at level j in each bucket
- In(x) < 7c (?? 2c)

Maintaining Bucket Size

- n can be accurately estimated
- When bucket size exceeds clog(n), the bucket is split into two equal size buckets.
- When bucket size drops below log(n), it is merged with a neighbor bucket.

Further more, if the merged bucket is greater than log(n)x(2c+2)/3, the new bucket is split into two buckets.

(c+1)/3 > 1 since c>2

- Buckets are organized into a ring, which can be merged or split with O(1) message.

Maintain Level Property

- Node join/leave without merging or splitting O(1)
- Join: size < clog(n), choose a level that has less that c nodes
- Leave: If it is the only node in its level, find another level that has two nodes, reassign level j to one of them.
- Bucket merge or split may result in a reassignment of the levels to all nodes in the bucket(s) O(log(n))
- Merging/splitting are expensive, but they do not happen very often
- After a merging or splitting of buckets, at least log(n) (c-2)/3 JOIN/LEAVE must happen in this bucket until another merging or splitting of this bucket is performed

Amortized Overhead = c/((c-2)/3) = O(1) for c>2

New bucket size

clog(n)

d1

d2

Log(n)

min(c/2lgn, (c+1)/3lgn)

max(c/2lgn, (2c+2)/3lgn)

d1, d2 > (c-2)/3

Amortized analysisFault Tolerance

- Viceroy has no built in support for fault tolerance
- Viceroy requires graceful leave
- Leaves are NOT the same as failures
- Performance is sensitive to failure
- External techniques:
- Thickening Edges
- State Machine Replication

Related Works

- De Bruijn Graph Based Network
- Distance halving
- D2B
- Koorde
- Others
- Symphony (Small world model)
- Ulysses (ButterFly, log(n), log(n)/loglogn)

Summary

- Constant out-degree
- Expected constant in-degree
- O(log(n )) w.h.p.
- O(1) with bucket solution
- O(log(n )) path length w.h.p
- Expected log(n )/n load:
- O(log2(n)/n) w.h.p.
- Weakness/improvements:
- Not Locality Aware
- No Fault Tolerance Support
- Due to the lack of flexibility of ButterFly network

Question

Photo by Peter J. Bryant

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