There are four classes of instructions, A, B, C, and D. The clock rate and CPI o

Question

There are four classes of instructions, A, B, C, and D. The clock rate and CPI of each implementation are given in the following table.

Clock Rate

CPI Class A

CPI Class B

CPI Class C

CPI Class D

P1

2.5 GHz

1

2

3

3

P2

3 GHz

2

2

2

2

Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, what CPU time is? What is the global CPI for each implementation? Find the clock cycles required in both cases.

Clock Rate

CPI Class A

CPI Class B

CPI Class C

CPI Class D

P1

2.5 GHz

1

2

3

3

P2

3 GHz

2

2

2

2

Explanation / Answer

Answer:

There are four classes of instructions, A, B, C, and D. The clock rate and CPI of each implementation are given in the following table.

  

Clock Rate

CPI Class A

CPI Class B

CPI Class C

CPI Class D

P 1

2.5 GHz

1

2   

3

3

P2

3 GHz

2

2

2

2

Given that a program has 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D

For processor P1:

The clock rate(R) = 2.5 GHz

Instruction count(I)=106

The clock per Instruction (CPI)=1*10/100+2*20/100+3*50/100+3*20/100=1*0.1 + 2*0.2 + 3*0.5 + 3*0.2=

0.1+0.4+1.5+0.6=2.6

So, execution time = (1/R)*I*CPI= =(1/2.5) * 10-9 * 106 * 2.6 sec = 0.4* 10-9 *106 * 2.6 sec.=110.24 * 10-9 sec

CPU time is the execution time of the program. For processor P1 CPU time is 110.24*10-9 sec.

For processor P2:

The clock rate(R) = 3 GHz

Instruction count(I)=106

The clock per Instruction (CPI)=2*10/100+2*20/100+2*50/100+2*20/100=2*0.1 + 2*0.2 + 2*0.5 + 2*0.2=

0.2+0.4+1+0.4=2.0

So, execution time = (1/R)*I*CPI= =(1/3) * 10-9 * 106 * 2.0 sec = 0.3333* 10-9 *106 * 2.0 sec.=70.66 * 10-9 sec

CPU time for processor P2 is 70.66*10-9 sec.

Global CPI for each implementation:

For processor P1:

Instruction count (I)=106

The total number of clock cycles required = (1*10/100+2*20/100+3*50/100+3*20/100)*106 = (0.1+0.4+1.5+0.6)*106 = 2.6*106

Global CPI = (2.6*106)/106 = 2.6

So, the Global CPI for first implementation (P1) is 2.6

For processor P2:

Instruction count (I)=106

The total number of clock cycles required = (2*10/100+2*20/100+2*50/100+2*20/100)*106 = (0.2+0.4+1.0+0.4)*106 = 2.0*106

Global CPI = (2.0*106)/106 = 2.0=2

So, the Global CPI for second implementation (P2) is 2

Total clock cycles for each implementations:

For processor P1:

Instruction count (I)=106

The total number of clock cycles required = (1*10/100+2*20/100+3*50/100+3*20/100)*106 = (0.1+0.4+1.5+0.6)*106 = 2.6*106=275.6

Total clock cycles required for first implementation (P1) = 275.6

For processor P2:

Instruction count (I)=106

The total number of clock cycles required = (2*10/100+2*20/100+2*50/100+2*20/100)*106 = (0.2+0.4+1.0+0.4)*106 = 2.0*106 =212

Total clock cycles required for second implementation (P2) = 212

  

Clock Rate

CPI Class A

CPI Class B

CPI Class C

CPI Class D

P 1

2.5 GHz

1

2   

3

3

P2

3 GHz

2

2

2

2

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