Answer:
See explanation
Explanation:
Given:
Processors:
P1
P2
P3
Clock Rate of Processors:
Clock rate of P1 = 3 GHz
Clock rate of P2 = 2.5 GHz
Clock rate of P3 = 4.0 GHz
CPI of Processors:
Cycles per instruction = CPI of P1 = 1.5
Cycles per instruction = CPI of P2 = 1.0
Cycles per instruction = CPI of P3 = 2.2
To find:
a) Which processor has the highest performance expressed in instructions per second
Solution:
Performance = clock rate / CPI
The performance of processor depends on instructions count and CPU time.
As we know that
CPU time = Instructions * Cycles Per Instruction / clock rate
CPU time = Instructions * CPI / clock rate
Instructions per second = Instruction count / CPU time
As:
CPU time = Instructions * CPI/ clock rate
Instructions/CPU time = clock rate/CPI
Instructions per second = clock rate / CPI
IPS = clock rate/ CPI
Hence
Performance = Clock rate / CPI = clock rate/ CPI
Compute Performance of P1:
Performance for P1 = IPS of P1 = clock rate of P1 / CPI of P1 = 3 GHz / 1.5 = 2
As we know that 1 GHz = 10⁹ Hz. So:
Performance of P1 expressed in instructions per second is 2 x 10⁹
Compute Performance of P2:
Performance for P2 = IPS of P2= clock rate of P2 / CPI of P2 = 2.5 GHz / 1.0 = 2 .5
As we know that 1 GHz = 10⁹ Hz. So:
Performance of P2 expressed in instructions per second is 2.5 x 10⁹
Compute Performance of P3:
Performance for P3 = IPS of P3= clock rate of P3 / CPI of P3 = 4.0 GHz / 2.2 = 1.82
As we know that 1 GHz = 10⁹ Hz. So:
Performance of P3 expressed in instructions per second is 1.82 x 10⁹
From the above computed performances of each processor it can be seen that Processor 2 (P2) has the highest performance expressed in instructions per second i.e. 2.5 x 10⁹
b) find the number of cycles and the number of instructions.
Given:
processors each execute a program in 10 seconds, So,
CPU time = 10 sec
Solution:
Compute number of cycles:
As we know that:
CPU time = cycles count / clock rate = clock cycles/clock rate
So
clock cycles = CPU time x clock rate
Compute number of cycles of P1:
clock cycles = 10 x 3 GHz
= 30
As we know that 1 GHz = 10⁹ Hz. So:
clock cycles of P1 = 3 x 10¹⁰
Compute number of cycles of P2:
clock cycles = 10 x 2.5 GHz
= 25
As we know that 1 GHz = 10⁹ Hz. So:
clock cycles of P2 = 2.5 x 10¹⁰
Compute number of cycles of P3:
clock cycles = 10 x 4.0 GHz
= 40
As we know that 1 GHz = 10⁹ Hz. So:
clock cycles of P3 = 4 x 10¹⁰
Now as we know that:
Instructions per second = Instruction count / CPU time
IPS = IC + CPU time
So to find number of instructions:
instruction count = Instructions per second x CPU time
Compute number of instructions of P1:
instructions of P1 = Instructions per second of P1 x CPU time
= 2 x 10⁹ x 10
= 2 x 10¹⁰
Compute number of instructions of P2:
instructions of P2 = Instructions per second of P2 x CPU time
= 2.5 x 10⁹ x 10
= 2.5 x 10¹⁰
Compute number of instructions of P3:
instructions of P3 = Instructions per second of P3 x CPU time
= 1.82 x 10⁹ x 10
= 1.82 x 10¹⁰
c) What clock rate should we have to reduce the execution time by 30%
As we know
CPU time = Execution time = instructions x CPI / clock rate
We have to find new clock rate to reduce execution time by 30%
This means we have to find:
New Execution Time = 70% of Old Execution Time
According to formula of Execution time:
instructions(new) x CPI(new) / clock rate(new) = 0.7 [instructions(old) x CPI(old) / clock rate(old)]
As the instructions(new) = instructions(old)
So,
CPI(new) / clock rate(new) = 0.7 [CPI(old) / clock rate(old)]
When trying to reduce the execution time by 30%, this leads to an increase of 20% in the CPI.
CPI(new) = 1.2 CPI(old)
New CPI of P1:
CPI(new P1) = 1.2 CPI(old P1)
= 1.2 x 1.5
CPI(new P1) = 1.8
New CPI of P2:
CPI(new P2) = 1.2 CPI(old P2)
= 1.2 x 1.0
CPI(new P2) = 1.2
New CPI of P3:
CPI(new P3) = 1.2 CPI(old P3)
= 1.2 x 2.2
CPI(new P3) = 2.6
1.2 / clock rate (new) = 0.7 / clock rate(old)
So new clock rate is computed as:
clock rate (new) = (1.2 / 0.7 ) x clock rate(old)
clock rate (new) = 1.71 x clock rate(old)
clock rate (new) = 1.71 x clock rate(old)
Hence the clock rate should be increased by 71% approx.
Now new clock rate for each processor is:
clock rate (new) for P1 = 3 GHz x 1.71 = 5.13 GHz
clock rate (new) for P2 = 2.5 GHz x 1.71 = 4.27 GHz
clock rate (new) for P3 = 4.0 GHz x 1.71 = 6.84 GHz