Communication operations - PowerPoint PPT Presentation

About This Presentation
Title:

Communication operations

Description:

... is r words per second, then each word takes time tw=1/r to traverse the link. ... forward' routing, when a message is traversing a path with multiple links, each ... – PowerPoint PPT presentation

Number of Views:51
Avg rating:3.0/5.0
Slides: 26
Provided by: IgorPo
Category:

less

Transcript and Presenter's Notes

Title: Communication operations


1
Communication operations
  • Efficient Parallel Algorithms
  • COMP308

2
Communication time
  • Communication requires 3 costs
  • 1. Static start up time (ts)
  • It is the time required to handle a message at
    the sending processor
  • 2. Per-hop time (th) with l the Links that the
    message passes
  • It is take a finite amount of time to reach the
    next processor in its path after a message leaves
    a processor.
  • 3. Per-word transfer time (tw) with m the
    bytes
  • If the channel bandwidth is r words per second,
    then each word takes time tw1/r to traverse the
    link.

3
There are 2 main communication schemes
4
store and forward vs cut-through
  • In store and forward routing, when a message is
    traversing a path with multiple links, each
    intermediate node on the path forwards the
    message to the next node after it has received.
  • In cut-through routing an intermediate nodes
    does not wait for the entire message to arrive
    before forwarding it.
  • A tracer is first sent from the source to the
    designation node to establish a connection.
  • Once a connection has been established, the flits
    are sent one after the other. All flits follows
    the same path in a dovetailed fashion.
  • As soon as a flit is received at an intermediate
    node, the flit is passed on to the next node.

5
One to All Broadcast
  • Initially, only the source processor has the data
    of size m that need to be broadcast. At the end
    of the termination of the procedure, there are P
    copies of the initial data, one residing at each
    processor.

6
Broadcast on ring (Store and Forward)
If the sender sends the messages consecutively to
the p-1 other processors, it takes p-1 steps. By
optimisation, we can reduce this to p/2
steps. Eg. a 8-processor ring requires 4 steps
7
NS diagram for broadcast on ring
8
Ring network, Cut-Through routing
  • With cut-through routing, messages can be sent
    faster to nodes that are multiple hops away in
    the network. By using this, we send the message
    first to the outermost node.

In general, in a p-processor ring the source
processor first sends the data to the processor
at distance p/2, then both processors sends the
message to the processors at distance of p/4 in
the same direction, then to p/8, etc.
9
Broadcast on mesh (Store and Forward)
Most of the optimised communication algorithms on
a mesh are simple extensions of their ring
counterparts, by consecutively applying the ring
algorithm on each dimension of the mesh.
10
Hypercube
  • The regular binary structure of the hypercube
    plays an important role in optimising
    communication.
  • Here, a broadcast is performed by sending the
    message along each dimension at each step. This
    results in log p or d steps.
  • It can be proved easily that log p is the minimal
    number of steps for every network.

11
Hypercube
  • Important properties of the networks
  • Small degree,
  • Small diameter,
  • Regular recursive structure,
  • Easy way to embed trees, etc
  • Hypercube two nodes connected if they are
    differ precisely on one bit

12
Hypercube two nodes connected if they are
differ precisely on one bit
13
(No Transcript)
14
(No Transcript)
15
(No Transcript)
16
Broadcast on hypercube (SF)
17
Broadcast on ring (Cut-Through )
18
Broadcast on mesh (C-T)
19
Broadcast on binary tree (C-T)
20
Gossiping
All-to-All Communication
21
Gossiping on Ring (Store and Forward)
22
Gossiping on Mesh (Store and Forward)
23
Gossiping on Hypercube (SF)
24
Gossiping on Ring (and Mesh)Cut-Through Routing
  • Each process sends m(p-1) words of data because
    it has an m-word packet for every other processor
  • The average distance that an m word packet
    travels is
  • Since there are p processors, each performing the
    same type of communication, the total traffic on
    the network is
  • The total number of communication channels in the
    network to share this load is p.

Hence this procedure cannot be improved by using
CT routing
25
Gossiping on Hypercube (CT routing)
Write a Comment
User Comments (0)
About PowerShow.com