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xeze [42]
3 years ago
14

Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, a

nd C) in the instruction set. M1 has a clock rate of 80 MHz and M2 has a clock rate of 100 MHz. The average number of cycles for each instruction class and their frequencies (for a typical program) are as follows:______
a. Calculate the average CPI for each machine, M1, and M2.
b. Which implementation (M1 or M2) is faster?
c. Find the clock cycles required for both processors.
Computers and Technology
1 answer:
ANTONII [103]3 years ago
7 0

Answer:

a) the average CPI for machine M1 = 1.6

the average CPI for machine M2 = 2.5

b) M1 implementation is faster.

c) the clock cycles required for both processors.52.6*10^6.

Explanation:

(a)

The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4

= 1.6

The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4

= 2.5

(b)

The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6

Given 80MHz = 80 * 10^6

The average MIPS ratings for M1 = 80 x 10^6  / 1.6 x 10^6

= 50

Given 100MHz = 100 * 10^6

The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6

= 40

c)

Machine M2 has a smaller MIPS rating

Changing instruction set A from 2 to 1

The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)

and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.

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The correct answer would be details.

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What is output?
Alex777 [14]

Answer:

The output of this question is 21. As show in the image

The explanation is given in below

Explanation:

Let first write the question

C=1

sum = 0

while(C<10):

C=C+3

sum=sum + C

print(sum)

Now Focus on

while(C<10):

C=C+3

sum=sum + C

The value of C is initially 1

C=1+3

Sum= 0+4

In second loop the value of C will become 4

c=4+3

sum=4+7

In third loop the value of C will be 7

c=7+3

sum=11+10

so the answer is 11+10=21

7 0
3 years ago
Write only in C, not C++.
yawa3891 [41]

Answer:

#include <iostream>

using namespace std;

int main() {

  int k;

double d;

string s;

cin >> k >> d >> s;

cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }

                                                                   

Explanation:

k is int type variable that stores integer values.

d is double type variable that stores real number.

s is string type variable that stores word.

cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.

cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.

First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.

Then next line \n is used to produce a new line.

So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.

The program along with the output is attached.

7 0
3 years ago
Professor Gig A. Byte needs to store text made up of the characters A with frequency 6, B with frequency 2, C with frequency 3,
Luden [163]

Answer:

This is not true

Explanation:

The optimal Huffman code is used to encrypt and compress text files. It uses fixed-length code or variable-length code for encryption and compression of data.

The professor's character code is similar to Huffman's variable-length coding which uses variable length od binary digits to represent the word strings. The file size of the text file above is;

= 6 x 1 + 2 x 2 + 3 x 2 + 2 x 2 + 8 x 1 = 28 bits

This would be the same for both cases.

The encrypt would be the problem as the encoded and decoding of the characters B and E may cause an error.

8 0
3 years ago
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