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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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kondor19780726 [428]

Answer:

Transition  section helps us to move from one shot to the next.

Explanation:

Synopsis: This tells actually what is the story is all about. We can call that as a “short description about the story”.

Sketch: It is the drawing window, where we pictorially represent the story.

Transition: This actually tells us about the next move.

Shot description: We can consider a “shot” as one of the scene in the story. So, it shot contain image and its description.

Shot Sequence: This is for “Pre-visualizing” video.

Among all the choice, Transition option takes the write definition.

5 0
3 years ago
Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards
konstantin123 [22]

Answer:

for(i = 0 ; i < NUM_VALS; ++i)

{

   cout << courseGrades[i] << " ";

}

cout << endl;

for(i = NUM_VALS-1 ; i >=0 ; --i)

{

   cout << courseGrades[i] << " ";

}

cout << endl;

Explanation:

The first loop initializes i with 0, because we have to print the elements in order in which the appear in the array. We print each element, adding a space (" ") character at its end. After the loop ends, we add a new line using endl.

The second loop will print the values in a reverse order, so we initialize it from NUM_VALS-1, (since NUM_VALS = 4, and array indices are 0,1,2,3). We execute the loop till i >= 0, and we print the space character and new line in a similar way we executed in loop1.

4 0
3 years ago
What is the output of this code? import java.util.HashSet; class A { public static void main(String[ ] args) { HashSet set = new
kompoz [17]

Answer:

The code will give an error that is "At least one public class is required in main file".

Explanation:

In the given code if we do not use the public access modifier to the class. It will give an error so, the correct code to this question as follows:

Program:

import java.util.HashSet; //import package

public class A  //define class as public.

{

   public static void main(String[ ] args) //define main method.

   {

       HashSet set = new HashSet(); //creating hashset object.

       set.add("A"); //add alphabet in hashset

       set.add("B"); //add alphabet in hashset

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       System.out.print("Size of HashSet is :"set.size()); //print the size of hashset.

   }

}

Output:

Size of HashSet is : 3

Explanation of the program:  

  • In the above program, we define a public class that is "A" and inside the class, we define the main method.  
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There are three types of value for money. Which of the following is not a method of value?
AlladinOne [14]
The answer to this question is c
4 0
2 years ago
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masha68 [24]

Answer: Please find below the answer along with explanation.

Explanation:

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The throughput, instead, refers to the actual data rate achieved in a given communications channel, taking into account the different channel impairments.

For instance, in a LAN segment that uses the original Ethernet 802.3 standard (CSMA/CD), a frequent occurrence of collisions can take down the actual data rate from the theoretical 100 Mbps to a very lower figure, i.e., 5 Mbps.

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