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trasher [3.6K]
1 year ago
5

Read the attached paper titled A Survey of Coarse-Grained Reconfigurable Architecture and comment on it. Make sure to specifical

ly address these points:_____.
a- What are the assumptions made about CGRA? What is it and why CGRA?
b- How does it fit in the ILP, TLP, DLP, RLP paradigm? Does it embrace all or some?
c- What are the major challenges in its design?
d- What type of advantages are claimed? For what workloads?
Computers and Technology
1 answer:
Pani-rosa [81]1 year ago
5 0

Answer:

lol

Explanation:

I NEED POINTS

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Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

               = 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 0
2 years ago
An intruder with malicious intent breaks into an office and steals a hard drive containing sensitive information about the compa
solong [7]

Answer:

firewalls and encryption

Explanation:

the company placed a firewall and an encryption code to the hard drive as a precaution against theft and malicious intent

3 0
2 years ago
Define a function compute_gas_volume that returns the volume of a gas given parameters pressure, temperature, and moles. Use
jeyben [28]

Question:

Define a function compute_gas_volume that returns the volume of a gas given parameters pressure, temperature, and moles. Use the gas equation PV = nRT, where P is pressure in Pascals, V is volume in cubic meters, n is number of moles, R is the gas constant 8.3144621 ( J / (mol*K)), and T is temperature in Kelvin.

Answer:

This solution is implemented in C++

double compute_gas_volume(double P, double T, double n){

   double V = 8.3144621 * n * T/P;

   return V;

}

Explanation:

This line defines the function, along with three parameters

double compute_gas_volume(double P, double T, double n){

This calculates the volume

   double V = 8.3144621 * n * T/P;

This returns the calculated volume

   return V;

}

To call the function  from the main, use:

<em>cout<<compute_gas_volume(P,T,n);</em>

<em />

<em>Where P, T and n are double variables and they must have been initialized</em>

5 0
2 years ago
In the middle of the iteration, how should a team handle requirement changes from the customer? (1 correct answer)
Rina8888 [55]

The answer is 1 because it is

5 0
2 years ago
Read 2 more answers
There are N bulbs numbered from 1 to N, arranged in a row. The first bulb is plugged into the power socket and each successive b
Ksivusya [100]

Answer:

The code is given below with appropriate comments

Explanation:

// TestSolution class implementation

import java.util.Arrays;

public class TestSolution

{

  // solution function implementation

  public static int solution(int[] arr)

  {

      // declare the local variables

      int i, j, count = 0;

      boolean shines;

     

      // use the nested loops to count the number of moments for which every turned on bulb shines

      for (i = 0; i < arr.length; i++)

      {

          shines = true;

          for (j = i + 1; j < arr.length && shines; j++)

          {

              if (arr[i] > arr[j])

                  shines = false;

          }

          if (shines)

              count++;

      }

      // return the number of moments for which every turned on bulb shines

      return count;

     

  } // end of solution function

 

  // start main function

  public static void main(String[] args)

  {

      // create three arrays named A, B, and C

      int[] A = {2, 1, 3, 5, 4};

      int[] B = {2, 3, 4, 1, 5};

      int[] C = {1, 3, 4, 2, 5};

     

      // generate a random number N within the range range[1..100000]

      int N = 1 + (int)(Math.random() * 100000);

     

      // create an array named D of size N

      int[] D = new int[N];

     

      // fill the array D with the distinct random numbers within the range [1..N]

      int i = 0;

      while(i < N)

      {

          int num = 1 + (int)(Math.random() * N);          

          boolean found = false;

          for(int j = 0; j < i && !found; j++)

          {

              if(D[j] == num)

                  found = true;

          }

         

          if(!found)

          {

              D[i] = num;

              i++;

          }

      }          

     

      // print the elements and number of moments of the arrays A, B, and C

      System.out.println("Array A: " + Arrays.toString(A) + " and Moments: " + solution(A));

      System.out.println("Array B: " + Arrays.toString(B) + " and Moments: " + solution(B));

      System.out.println("Array C: " + Arrays.toString(C) + " and Moments: " + solution(C));

     

      // print the size and number of moments of the array D

      System.out.println("Size(N) of Array D: " + N + " and Moments: " + solution(D));

     

  } // end of main function

} // end of TestSolution class

3 0
2 years ago
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