in CO and Architecture edited by
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3 votes
3 votes

in CO and Architecture edited by
669 views

4 Comments

i got  2.56.
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800/(1*3*5 + 99*3) = 800/313 = 2.56... :o

I didn't use efficiency ! what to do with it?
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i think here no use of efficiency
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In the case of non_pipelined processor  ...100cycle  ..  cycle time=8ns

for pipeline cycle time=3ns

first instrcution will take 5 cycle and remaining 99 will take 1cyle total time :- (5+99)*104*3=312

speedup=800/312=2.56

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2 Answers

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For Pipelined system

There are five stages : S1  S2  S3  S4  S5 with time delays of 2ns, 2ns, 2ns, 3ns, 3ns. Also, the maximum buffer delay is 1ns.

Therefore, T for pipeline = (3+1)ns = 4ns.

Also, Efficiency = 0.8 so, $\frac{1}{S} = 0.8$ or S = $\frac{1}{0.8}$

So, total time taken for 100 instructions = $\left ( 100 + 5 - 1 \right )\frac{1}{0.8}4 = 520$ ns

 

For non-pipelined system

Total time taken = $\left ( 100 \times \left ( 2 + 2+ 2+ 3+ 3 \right ) \right ) = 1200$ ns

 

Speedup = $\frac{1200ns}{520ns} = 2.31$

3 Comments

how you are deciding pipeline cycle time as 4  ,buffer delay in case of stage of delay 3 is zero

It should be 3
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i have the same doubt! why did you took 4 instead of 3?
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Cycle time will be 3 for pipeline.. Think logically rather than using formulas..
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For pipelined processor

the maximum stage delay is 3ns hence in pipelined processor 1 instruction will take 3ns.

if efficiency of pipeline is 80% means that 80 out of 100 instruction will take 3 ns to complete and 20 instruction will take 5 cycle hence 15 ns to complete therefore

total time taken for 100 instructions in pipeline = 80 x 3 + 20 x 15 => 540ns

For non-pipelined processor

Total time taken = 100×(2+2+2+3+3)=>1200 ns

 

Speedup = 1200ns / 540ns = 2.22

 

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