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Vera_Pavlovna [14]
2 years ago
15

Ld' is the instruction with the longest latency on the CPU from Section 4.4 (in RISC-V text). If we modified ld and sd so that t

here was no offset (i.e., the address to be loaded from/stored to must be calculated and placed in rs1 before calling ld/sd), then no instruction would use both the ALU and Data memory. This would allow us to reduce the clock cycle time. However, it would also increase the number of instructions, because many ld and sd instructions would need to be replaced with ld/add or sd/add combinations.
(a) What is the primary factor that influences whether a program will run faster or slower on the new CPU?
(b) Do you consider the original CPU a better overall design; or do you consider the new CPU a better overall design? Why?

Computers and Technology
1 answer:
antiseptic1488 [7]2 years ago
8 0

Answer:

See explaination

Explanation:

a)

The primary factors that influence the program's speed on a new CPU are given as:-

CPU Clock speed where the speed of the process instructions is being measured.

Multi-core which is when the transistors work faster than respective CPU.

Cache which helps to note that the transition of data is smooth and fast.

b)

So, there are two CPUs suppose the old one as 'A' and the modified one as 'B'.

'A' has following features---

It takes more time to execute program as it has more clock cycle time or we can say it has low clocking rate or low speed in terms of MHz or GHz. Clock rate is the inverse of Clock Cycle Time. When you will increase the clocking rate of CPU, it will get faster and then Clock Cycle Time will get reduced.

It has less instructions provided.

'B' has following features----

It takes less time to execute program as it has less clock cycle time or we can say it has high clocking rate or high speed in terms of MHz or GHz.

It has more instructions provided.

The performance of CPU depends upon 2 factors:

The number and types of instructions that are executed by the CPU

How fast the CPU can execute those instructions?

So, overall CPU B is better as it has less execution time than CPU A but the performance will always depend upon the number and type of instructions executed by the CPU so it may vary.

Please refer to attachment for instructions and formulas.

You might be interested in
Jim is writing a program to calculate the wages of workers in a teddy bear factory.
34kurt

Answer:

The algorithm is as follows;

1. Start

2. Input TeddyBears

3. Input Hours

4. WagebyTeddy = 2 * TeddyBears

5. WagebyHour = 5 * Hours

6. If WagebyHour > WagebyTeddy then

6.1 Print WagebyHour

7. Else

7.1. Print WagebyTeddy

8. Stop

Explanation:

The following variables are used;

TeddyBears -> Number of teddy bears made

Hours -> Number of Hours worked

WagebyTeddy -> Wages for the number of teddy bears made

WagebyHour -> Wages for the number of hours worked

The algorithm starts by accepting input for the number of teddy bears and hours worked from the user on line 2 and line 3

The wages for the number of teddy bears made  is calculated on line 4

The wages for the number of hours worked  is calculated on line 5

Line 6 checks if wages for the number of hours is greated than wages for the number of bears made;

If yes, the calculated wages by hour is displayed

Otherwise

the calculated wages by teddy bears made is displayed

3 0
2 years ago
You are given an array x of string elements along with an int variable n that contains the number of elements in the array. You
Slav-nsk [51]

Answer:

The code to this question can be given as:

Code:

int i,j,count_previous=0,count_next=0; //define variable

for (j=0; j<n; j++) //loop for array

{

if (x[0]==x[j]) //check condition

{

count_previous++; //increment value by 1.

}

}

for (i=0; i<n; i++) //loop

{

for (j=0; j<n; j++)

{

if (x[i]==x[j]) //check condition

{

count_next++; //increment value by 1.

}

}

if (count_previous>count_next) //check condition

{

mode=x[i-1];

}

else

{

mode=x[i];

count_previous=count_next; //change value.

count_next= 0 ; //assign value.

}

}

Explanation:

In the question it is define that x is array of the string elements that is already define in the question so the code for perform operation in the array is given above. In the code firstly we define the variable that is i, j, count_previous, count_next. The variable i,j is used in the loop and variable count_previous and count_next we assign value 0 that is used for check the values of array. Then we define the for loop in this loop we use conditional statement in the if block we check that array of zero element is equal to array of j value then the count_previous is increase by 1. Then we use nested loop It is also known as loop in a loop. In this first loop is used for array and the second loop is used for check array element.In this we use the condition that if array x of i value is equal to array x of j then count_next will increment by 1.Then we use another condition that is if count_previous is greater then count_next then mode of x is decrement by 1. else mode equal to array and count_previous holds the value of count_next and count_next is equal to 0.

8 0
2 years ago
Design a BCD-to-Gray code decoder. Your decoder will have 4 inputs: A, B, C and D, representing a 4-bit BCD code (A being the MS
ra1l [238]

Answer:

Binary to Gray Code Converter

The logical circuit which converts the binary code to equivalent gray code is known as binary to gray code converter. An n-bit gray code can be obtained by reflecting an n-1 bit code about an axis after 2n-1 rows and putting the MSB (Most Significant Bit) of 0 above the axis and the MSB of 1 below the axis.

The 4 bit binary to gray code conversion table is given in attached file.

3 0
2 years ago
Read 2 more answers
Assume your friend just sent you 32 bits of pixel data (just the 0s and 1s for black and white pixels) that were encoded after s
lara31 [8.8K]

Answer:

your friend just sent you 32 bits of pixel data (just the 0s and 1s for black and white pixels) that were encoded after sampling an image. Choose the two statements that are true.

------------------

The 32 bits of pixel data is enough to produce the image using the widget. Nothing else is needed.

------------------------

The digital image would be an exact copy of the analog image.

-----------------------------The correct width and height must be input into the pixelation widget to produce the image.

---------------------------The fact that only 32 bits were used to represent the image indicates relatively large sample squares were used. The digital image may vary from the analog image significantly

Explanation:

3 0
2 years ago
Write a program FindDuplicate.java that reads n integer arguments from the command line into an integer array of length n, where
kozerog [31]

Answer:

  1. public class FindDuplicate{
  2.    public static void main(String[] args) {
  3.        Scanner input = new Scanner(System.in);
  4.        int n = 5;
  5.        int arr[] = new int[n];
  6.        for(int i=0; i < arr.length; i++){
  7.            int inputNum = input.nextInt();
  8.            if(inputNum >=1 && inputNum <=n) {
  9.                arr[i] = inputNum;
  10.            }
  11.        }
  12.        for(int j =0; j < arr.length; j++){
  13.            for(int k = 0; k < arr.length; k++){
  14.                if(j == k){
  15.                    continue;
  16.                }else{
  17.                    if(arr[j] == arr[k]){
  18.                        System.out.println("True");
  19.                        return;
  20.                    }
  21.                }
  22.            }
  23.        }
  24.        System.out.println("False");
  25.    }
  26. }

Explanation:

Firstly, create a Scanner object to get user input (Line 4).

Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).

Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.

If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".

7 0
2 years ago
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