GATE CSE 2014 | Set 3 | Question: 9

Question

Consider the following processors (ns stands for nanoseconds). Assume that the pipeline registers have zero latency. 

• $\text{P1:}$ Four-stage pipeline with stage latencies $\text{1 ns, 2 ns, 2 ns, 1 ns}$. 
• $\text{P2:}$ Four-stage pipeline with stage latencies $\text{1 ns, 1.5 ns, 1.5 ns, 1.5 ns}$. 
• $\text{P3:}$ Five-stage pipeline with stage latencies $\text{0.5 ns, 1 ns, 1 ns, 0.6 ns, 1 ns}$. 
• $\text{P4:}$ Five-stage pipeline with stage latencies $\text{0.5 ns, 0.5 ns, 1 ns, 1 ns, 1.1 ns}$. 

Which processor has the highest peak clock frequency?

• A. $\text{P1}$
• B. $\text{P2}$
• C. $\text{P3}$
• D. $\text{P4}$

Comments

What does the ‘peak’ in the ‘highest peak clock frequency’ signify? As per the below answers and comments, it seems that we are considering all the stages to have uniform delay (which I believe is generally the case). Then, what’s the need for the word ‘peak’ as pointed above?

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Max Frequency=>Min cycle time

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Find max Tp from each processor , then whichever is having least Tp it will have highest freq

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$ \ Clock \ frequency \ \cong \ Clock \ rate \ \cong \frac{1 }{ Cycle \ time}  $

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Answer 1

frequency $=\dfrac{1}{ \text{max(time in stages)}}$
for $P_3$, it is $\dfrac{1}{1} = 1 \; \textsf{GHz}$

for $P_1$, it is $\dfrac{1}{2} = 0.5\; \textsf{GHz}$

for $P_2$, it is $\dfrac{1}{1.5} = 0.67\;\textsf{GHz}$

for $P_4$, it is $\dfrac{1}{1.1} = 0.90 \;\textsf{GHz}$

Correct Answer: $C$

Comments

How this can be best answer without explaination.. I didn't understand the solution.

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what doubt do you have @ahthneeuhl

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In the Question, we are asked to determine the processor with the highest peak clock frequency, given the stage latencies of different pipelined processors. However, I'm unclear about how we are directly comparing frequency with stage latencies.

Since we know that the frequency is inversely proportional to the cycle time, and the stage latencies represent the time each stage of the pipeline takes to complete, isn't it incorrect to directly comment on the frequency based only on the stage latencies?

Answer 2

Lesser the time-period of the clock, higher the frequency of the clock (i.e) number of clock-cycles per second.

To get the processor with the highest clock-frequency find the processor with the lowest time-period.

P1 : Time period = max ( 1 ns, 2 ns, 2 ns, 1 ns ) = 2 ns 

P2 : Time period = max ( 1 ns, 1.5 ns, 1.5 ns, 1.5 ns ) = 1.5 ns

P3 : Time period = max (0.5 ns, 1 ns, 1 ns, 0.6 ns, 1 ns ) = 1 ns

P4 : Time period = max ( 0.5 ns, 0.5 ns, 1 ns, 1 ns, 1.1 ns ) = 1.1 ns

Time-period of P3 < P4 < P2 < P1

So clock frequency of P3 > P4 > P2 > P1.

So P3 has the highest clock-frequency.  Option C) 

Comments

First-line itself made me realize my silly mistake. The time period is inversely proportional to frequency.

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Better than above (so called best) solution

Answer 3

Here , time for pipeline P1= 2 ns, and Frequency = 1/tp = 1/ 2 ghz= 0.5 ns.

          Time for p2=1.5 ns, similarly fp = 0.66 ns

          same for p3= 1 ns ,so fp = 1 ns and  p4= 1.1 ns , so fp =0.909 ns .

Hence , My ans will be P3.

Answer 4

Just find that which pipeline had the lowest clock genration time and it will give the highest clock frequency

 

Bcoz clock genration time is inversly propotional to the clock frequency  

Comments

🐐

Answer 5

F = 1/T

Clock time of the following processors
P1 = 2 ns
P2 = 1.5 ns
P3 = 1 ns
P4 = 1.1 ns

So, P3 is taking least time hence P3 has highest clock frequency

Option C

Answer 6

Generally, one stage of pipeline is executed in one clock cycle.
So, here we have different processor with different stage times. We will consider maximum time of stage for that processor as time for all stages

Max stage time => single cycle time => frequency (1/Time period of one cycle)