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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?

  1. $\text{P1}$
  2. $\text{P2}$
  3. $\text{P3}$
  4. $\text{P4}$
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Best answer
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47 votes
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: $C$
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11 votes

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) 

4 votes
4 votes
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.
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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  

Answer:

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