please solve the problem fully! thank you! 1.5 4] Consider three different proce

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please solve the problem fully! thank you!

1.5 4] Consider three different processors P1, P2, and P3 executing the same instruction set. P1 has a 3 GHz clock rate and a CPI of 1.5. P2 has a 2.5 GHz clock rate and a CPI of 1.0. P3 has a 4.0 GHz clock rate and has a CPI of 2.2 a. Which processor has the highest performance expressed in instructions per second? b. If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions. c. We are trying to reduce the execution time by 30% but this leads to an increase of 20% in the CPI, what clock rate should we have to get this time reduction?

Explanation / Answer

(a) for p1:

Clock speed = 3 GHz (which means 3*109 cycles of instruction can be performed in a second)

3*109 cycles/second

cpi = 1.5

no. of instructions completed per second = (3*109)/1.5 = 2*109

for p2: Clock speed = 2.5 GHz (which means 2.5*109 cycles of instruction can be performed in a second)

2.5*109 cycles/second

cpi = 1.0

no. of instructions/ second = (2.5*109)/1.0 = 2.5*109

for p3: Clock speed = 4 GHz (which means 4*109 cycles of instruction can be performed in a second)

4*109 cycles/second

cpi = 2.2

no. of instructions/ second = (4*109)/2.2 = 1.82*109

So, p2 has the highest performance in instructions executed per second.

(b) For p1: in 10 seconds 3*1010 cycles can be executed.

Number of instructions in the program = (3*1010)/1.5 = 2*1010

For p2: in 10 seconds 2.5*1010 cycles can be executed.

Number of instructions in the program = (2.5*1010)/1.0 = 2.5*1010

For p3: in 10 seconds 4*1010 cycles can be executed.

Number of instructions in the program = (4*1010)/2.2 = 1.82*1010

(c) If we want to reduce the execution time by 30%, then the effective cycle rate should be increased by ((10-7)/3)*100% = 42.8%

As we know, CPI = Clock cycles / Number of instructions

Now, we got

1.2*CPI = (Clock cycles*1.42) / (Number of instructions)

So, for p1: 1.2*1.5 = (3*109*1.42)/Number of instructions

Number of instructions = (4.26*109)/1.8 = 2.36*109

Number of cycles = 4.26*109/second

So, the clock speed for p1 is 4.26 Ghz.

As the CPI has been increased by 20%, to keep up the same execution time the clock speed should be increased by 20%

So, the clock speed for p1 is (4.26*1.2) Ghz = 5.11 Ghz.

So, for p2: 1.2*1.0 = (2.5*109*1.42)/Number of instructions

Number of instructions = (3.55*109)/1.2 = 2.96*109

Number of cycles = 3.55*109/second

So, the clock speed for p2 is 3.55 Ghz.

As the CPI has been increased by 20%, to keep up the same execution time the clock speed should be increased by 20%

So, the clock speed for p2 is (3.55*1.2) Ghz = 4.26 Ghz.

So, for p3: 1.2*2.2 = (4*109*1.42)/Number of instructions

Number of instructions = (5.68*109)/2.64 = 2.15*109

Number of cycles = 5.68*109/second

So, the clock speed for p3 is 5.68 Ghz.

As the CPI has been increased by 20%, to keep up the same execution time the clock speed should be increased by 20%

So, the clock speed for p3 is (5.68*1.2) Ghz = 6.81 Ghz.

Hope this helps

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