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, fnd the number of cycles and the number of instructions. c. We are trying to reduce the execution time by 30of 20% in the CPI. What clock rate should we have to get this time reduction?

Respuesta :

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

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