NIELIT 2017 DEC Scientist B - Section B: 13

Question

Consider a non-pipelined machine with $6$ stages; the lengths of each stage are $\text{20ns, 10ns, 30ns,25ns, 40 ns}$ and $\text{15ns}$ respectively. Suppose for implementing the pipelining the machine adds $\text{5 ns}$ of overhead to each stage for clock skew and set up. What is the speed up factor of the pipelining system (ignoring any hazard impact)?

• A. $7$
• B. $14$
• C. $3.11$
• D. $6.22$

Comments

without pipeline = 140ns
with pipeline = 40ns+ 5ns= 45
speedup = 140 / 45 = 3.111

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option C is correct..speed up =without pipeline/ pipeline=140/45=3.11

Answer 1

Without pipeline For 1 instruction- 20ns+10ns+30ns+25ns+40ns+15ns =140ns

With ideal pipeline it will take 1 cycle time. Her 1cycle time is max(20+5=25, 15, 35, 30, 45, 20) = 45ns

Speedup= Time without pipelining/Time with pipeling = 140/45= 3.11

Answer: C

Answer 2

In implementation without pipeline, the total time for execution would be the  sum of the delays of all the six stages

Let it be denoted by  $T_{wp}$

$\therefore$  $T_{wp}$  $=$  $20+10+30+25+40+15$ $ns$   $=$  $140$ $ns$

In pipeline implementation , $5$ $ns$ overhead has to be added to individual stages in the pipeline and the execution time of pipeline would be the stage delay having the maximum value i.e.  $45$ $ns$ ($T_{p}$)

So, speedup would be:  $\frac{ T_{wp}}{T_{p}}$  $=$  $\frac{140}{45}=3.11$ $ns$

Option C is correct

Comments

The correct option is c i.e 3.11

First see what is given...so

 Non pipeline system:

#stages=6

Delays={20,10,30..15}

therefore with non pipeline simply add all the deals associated with each stage 20+10+30+25+40+15=140ns

For implementing pipeline system it is given to add 5ns to each stage , therefore we get are delays as:{25,15,35,30,45,20}

Take the largest delay that is 45ns,

Speedup=twp/to= 140/45=3.11

So here we have to remember when do we had to add the overhead.

Answer 3

Given data,
Non pipelined machine with 6 stages,
Length of each stage=20,10,30,25,40,15 ns.
Implementing the pipelining the machine adds each stage=5ns overhead.
Speed up factor of the pipelining system=?

Step-1: Non pipeline for 1 instruction to all stages=20+10+30+25+40+15 ns
=140 ns
Step-2: Per cycle adds 5ns overhead to each stage =25,15,35,30,45,20 ns
= 45 ns
Step-3: Speedup factor= Time non pipelining / Time with pipeline
= 140/45
= 3.11 ns

Answer 4

tp=max(all stages timing )+delay

tp=40+5=45

tnp=140

s=140/45=3.11