7. (a) The power consumed by a certain processor when executing program A at a r

Question

7. (a) The power consumed by a certain processor when executing program A at a rate of 900 million instructions per second is 60 Watt. To reduce the power, the frequency of the processor's clock has been reduced by 10%, what will be the new execution rate of program A? WI (b) The program executes 2.8 billion instructions. Calculate the totol execution time of the program for the base and the reduced frequencies (c) What is the CPI of the processor when executing this program at the base frequency given that this frequency is 2 GHz? What will be the CPI when the frequency is lowered to 1.8 GHz?

Explanation / Answer

Solution:

a)

frequency and MIPS are directly proportional to each other so MIPS will increase by 10%.

so MIPS will become 900 * 0.9= 810 MIPS

b)

for 900 MIPS

Total execution time= 2.8 billion/900 MIPS= 3.11 seconds

for reduced frequency

Total execution time= 2.8 billion/810 MIPS= 3.457 seconds

c)

for 2 Ghz processor

cycle time= 1/(2 *10^9)= 0.5 ns

CPI= Total execution time/ (Instruction count * cycle time)

= 3.11/ (2.8 * 10^9 * 0.5 * 10^-9)

= 2.22

for 1.8 Ghz processor

cycle time= 1/(1.8 * 10^9)= 0.555556 ns

CPI= Total execution time/ (Instruction count * cycle time)

= 3.457 / (2.8 * 10^9 * 0.55556 * 10^-9)

= 2.22

CPI will be same

d)

CPI= 0.48 * 1 + 0.22 * 1.88 + 0.1 * 1.3 + 0.2 * 1.65

= 1.3536

then execution time will be 4 seconds

4= 3 billion/x

x= 4/ (3 *10^9)

MIPS= 1.3333 * 10^-9

Frequency= 1/1.3333 * 10^-9= 750018750.469 hz

= 0.7500 Ghz

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