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

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

Which processor has the highest peak clock frequency?

  1. $P1$
  2. $P2$
  3. $P3$
  4. $P4$
asked in CO & Architecture by Veteran (106k points)
edited by | 1.6k views

3 Answers

+26 votes
Best answer
frequency $=\dfrac{1}{ \text{max(time in stages)}}$
for $P_3$, it is $\dfrac{1}{1}$ GHz

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

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

for $P_4$, it is $\dfrac{1}{1.1} \text{GHz}$
answered by Active (3.6k points)
edited by
+7 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) 

answered by Loyal (7.6k points)
+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.
answered by (249 points)
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