Consider the operation of a machine with the data path given below. Suppose that loading the ALU input registers takes 5 nsec, running the ALU takes 10 nsec, and storing the result back in the register scratchpad takes 5 nsec. What’s the maximum number of MIPS this machine is capable of with pipelining with the three execution stages?

Answers

Answer 1
Answer:

Answer:

The load instructions in the ALU input registers take the 5 nsec, and tuns the ALU and this takes the 10 nsec and thus stores the result back into the scratchpad registers and this takes 5 nsec. The data path cycle in it is 20 nsec.

The total time is 20 nsec for one cycle.

To calcualte the MIPS, divide one second with 20 nsec.

Millions of instructions per second (MIPS) = (1*10^9 nsec)/20 nsec = 50,000,000 nsec

Therefore, the maximum number of MIPS this machine is capable of in the absence of pipelining is 50 MIPS.

Explanation:


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Answers

Answer:

import java.util.*;

public class Country

{

  public static void main(String args[])

  {

      char ch,temp;

      int flag = 0;

      String country;

      ArrayList<String> countries = new ArrayList<String>();

      Scanner sc = new Scanner(System.in);

      do

      {

          System.out.println("enter the country you have visited:\t");

          country = sc.next();

          for(int i=0;i<countries.size();i++)

          {

              if(countries.get(i).equals(country))

              {

                  System.out.println("you have already entered the country");

                  flag = 1;

                  break;      

              }

          }

          if(flag == 0)

          {

              countries.add(country);

              flag = 0;

          }

          System.out.println("want to add another country(y/n):\t");

          ch = sc.next().charAt(0);

      }while(ch!='n');

      Collections.sort(countries);

      System.out.println("Countries you have visited:\t"+countries);

      System.out.println("Total number of contries visited:"+countries.size());

     

  }

}

Explanation:

What are the differences betweenCONS, LIST, and APPEND?

Answers

Answer:

These all are constructors.

CONS

(CONS A B)

makes a pair like this:   (A . B)  

In other words, A is the CAR of the pair; B is the CDR of the pair.

For  example:

(CONS 4 5) ==> (4 . 5)

(CONS 4 '(5 6)) ==> (4 5 6)

[The pair (4 . (5 6)) is the same thing as (4 5 6)].

APPEND  

(APPEND A B)

makes a new list by replacing the final nil in A with the list B. A and  

B must be proper lists.

For example:

(APPEND '(4) '(5 6)) ==> (4 5 6)

This takes any amount of number and put in this order

LIST  

In this ,it will return a list whose elements are value of arguments in the order as it appeared in the LIST.It can take any amount of parameters

For example,

(LIST 4) ==> (4)

(LIST 4 5) ==> (4 5)

(LIST 4 5 '(6 7)) ==> (4 5 (6 7))

(LIST (+ 5 6)(* 5 6)) ==> (8 9)

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Answers

Answer:

To edit an existing pattern, double-click the pattern in the pattern swatch, or select an object containing the pattern and choose Object > Pattern > Edit Pattern

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Answer:

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Answer:

Hi Samantha, i have a work with you.

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Answers

Answer:

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Explanation:

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Answers

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