Linear Pipelining

7/27/2019 Linear Pipelining

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Principles of Linear Pipelining


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In pipelining, we divide a task into set of

subtasks.

The precedence relation of a set of subtasks

{T1, T2,, Tk} for a given task T implies that the

same task Tj cannot start until some earlier

task Ti finishes.

The interdependencies of all subtasks form

the precedence graph.


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With a linear precedence relation, task Tj

cannot start until earlier subtasks { Ti} for all

(i < j) finish.

A linear pipeline can process subtasks with a

linear precedence graph.


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A pipeline can process successive

subtasks if

Subtasks have linear precedence order Each subtasks take nearly same time to

complete


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Basic Linear Pipeline

L: latches, interface between different stages

of pipeline

S1, S2, etc. : pipeline stages


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It consists of cascade of processing stages. Stages:Pure combinational circuits

performing arithmetic or logic operations over

the data flowing through the pipe. Stages are separated by high speed interface

latches.

Latches : Fast Registers holding intermediateresults between stages

Information Flow are under the control of

common clock applied to all latches


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The flow of data in a linear pipeline having four stagesfor the evaluation of a function on five inputs is asshown below:


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The vertical axis represents four stages

The horizontal axis represents time in units of

clock period of the pipeline.


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Clock Period () for the pipeline

Let i be the time delay of the circuitry Si and

t1 be time delay of latch.

Then the clock period of a linear pipeline is

defined by

The reciprocal of clock period is called clockfrequency (f = 1/) of a pipeline processor.

111

max ttt mi

k

i


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Performance of a linear pipeline

Consider a linear pipeline with k stages. Let T be the clock period and the pipeline is initially

empty.

Starting at any time, let us feed n inputs and wait tillthe results come out of the pipeline.

First input takes k periods and the remaining (n-1)

inputs come one after the another in successive

clock periods.

Thus the computation time for the pipeline Tp is

Tp = kT+(n-1)T = [k+(n-1)]T


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Performance of a linear pipeline

For example if the linear pipeline have fourstages with five inputs.

Tp = [k+(n-1)]T = [4+4]T = 8T


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Performance Parameters

The various performance parameters of

pipeline are :

1. Speed-up

2. Throughput

3. Efficiency


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Speedup

Speedup is defined asSpeedup = Time taken for a given computation by a non-pipelined functional unit

Time taken for the same computation by a pipelined version

Assume a function of k stages of equal

complexity which takes the same amount oftime T.

Non-pipelined function will take kT time for one

input. Then Speedup = nkT/(k+n-1)T = nk/(k+n-1)


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Speed-up

For e.g., if a pipeline has 4 stages and 5 inputs,

its speedup factor is

Speedup = ?

The maximum value of speedup is

Lt [Speedup] = ?

n


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Speed-up

The maximum value of speedup is

Lt [Speedup] = k

n


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Efficiency

It is an indicator of how efficiently theresources of the pipeline are used.

If a stage is available during a clock period,

then its availability becomes the unit ofresource.

Efficiency can be defined as

ncomputatiothatduringavailableunitstimestageofnumberTotal

ncomputatioduringusedactuallyunitstimestageofNumber=Efficiency


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Efficiency

No. of stage time units = nk

there are n inputs and each input uses k stages.

Total no. of stage-time units available

= k[ k + (n-1)]

It is the product of no. of stages in the pipeline (k)

and no. of clock periods taken for

computation(k+(n-1)).


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Efficiency

Thus efficiency is expressed as follows:

The maximum value of efficiency is

11-nknk

Efficiency

nk

n

k

?1

nk

nLtEfficiencyLtnn


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Efficiency

Efficiency is minimum when n = 1.

Efficiency = ?

For k = 4 and n = 5, Efficiency = ?


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Throughput

It is the average number of results computed

per unit time.

For n inputs, a k-staged pipeline takes

[k+(n-1)]T time units

Then,

Throughput = n / [k+n-1] T = nf / [k+n-1]

where f is the clock frequency


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Throughput

The maximum value of throughput is

Lt [Throughput] = ?

n


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Throughput

The maximum value of throughput is

Lt [Throughput] = f

n

Throughput = Efficiency x Frequency


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Problem

Consider the execution of a program of 15000instructions by a linear pipeline processor with a

clock rate of 25MHz. Assume that the instruction

pipeline has 5 stages and that one instruction is

issued per clock cycle. The penalties due to branchinstructions and out-of-sequence executions are

ignored

a) Calculate the speedup factor as compared with

non-pipelined processor

b) What are the efficiency and throughput of this

pipelined processor


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Linear Pipelining

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