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andrew11 [14]
3 years ago
13

Suppose a program is running on a distributed-memory multiprocessor. There are 1,000 instructions in the program and 80% of inst

ruction references hit in the local memory and 20% of instruction references involve the remote communication in the remote memory. 10 ns time is required for the remote communication for each instruction reference. If the CPU clock rate is 4 GHZ and CPI is 0.5, what is the running time of this program?
Computers and Technology
1 answer:
Licemer1 [7]3 years ago
5 0

Answer:

2125 ns.

Explanation:

First of all Execution Time = Number of Instructions * CPI * Clock Cycle

Number of Instructions = 1000

CPI = 0.5

Clock Cycle = 1/clock rate = 1/4GHz = 0.25 * 10-9 s

So Execution Time = 1000 * 0.5 * 0.25 * 10-9

Execution Time = 125 * 10-9 s

Execution Time = 125 ns

Now 20% of the instructions take 10 ns extra time for remote communication.

1 instruction takes 10ns

so 20% of 1000 = 200 instructions will take 200*10 = 2000ns

So Total Execution Time = 125+2000 = 2125 ns.

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Answer:

Soft Reset

Explanation:

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Hence, a SOFT RESET occurs when you reset a mobile device but retain your installed applications and personal settings

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Answer:

B. Animal

Explanation:

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3 years ago
Given two Strings String s1 = "11223351638791377189728193"; String s2 = "983763978669976862724569282405742578"; and String varia
inn [45]

Answer:

import java.math.BigInteger;  

public class Main {  

public static void main(String[] args)  {  

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

BigInteger num1 = new BigInteger(s1);  

BigInteger num2 = new BigInteger(s2);  

BigInteger product = num1.multiply(num2);  

String s3 = product.toString(10);

System.out.println(s3);  

}  

}  

Explanation:

Because s1 and s2 contain large integers, the program need the biginteger module in other to compute the products of s1 and s2 when converted to integers.

This line initializes the two string variables s1 and s2

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

The next two lines convert the two strings to BigInteger. This allows the program to multiply large integers

<em> BigInteger a = new BigInteger(s1);   </em>

<em>BigInteger b = new BigInteger(s2);  </em>

<em />

This computes the product of the two big integers

BigInteger product = num1.multiply(num2);  

This converts the product to String s3

String s3 = product.toString(10);

This prints s3

System.out.println(s3);  

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3 years ago
The following checksum formula is widely used by banks and credit card companies to validate legal account numbers: d0 + f(d1) +
Aleksandr [31]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to import Scanner class

public class ISBN

{   public static void main(String[] args)  { // start of main() function body

   Scanner s = new Scanner(System.in); // creates Scanner object

//prompts the user to enter 10 digit integer

   System.out.println("Enter the digits of an ISBN as integer: ");    

   String number = s.next(); // reads the number from the user

   int sum = 0; // stores the sum of the digits

   for (int i = 2; i <= number.length(); i++) {

//loop starts and continues till the end of the number is reached by i

          sum += (i * number.charAt(i - 1) ); }

/*this statement multiplies each digit of the number with i and adds the value of sum to the product result and stores in the sum variable*/

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 {System.out.println("The ISBN number is " + number + remainder); }  }  }  

Explanation:

This program takes a 10-digit integer as a command line argument and uses Scanner class to accept input from the user.

The for loop has a variable i that starts from 2 and the loop terminates when the value of i exceeds 10 and this loop multiplies each digit of the input number with the i and this product is added and stored in variable sum. charAt() function is used to return a char value at i-1.

This is done in the following way: suppose d represents each digit:

sum=d1 * 1 + d2 * 2 + d3 * 3 + d4 * 4 + d5 * 5 + d6 * 6 + d7 * 7 + d8 * 8 + d9 * 9

Next the mod operator is used to get the remainder by dividing the value of sum with 11 in order to find the checksum and stores the result in remainder variable.

If the value of remainder is equal to 10 then use X for 10 and the output will be the 10 digits and the 11th digit checksum (last digit) is X.

If the value of remainder is not equal to 10, then it prints a valid 11-digit number with the given integer as its first 10 digits and the checksum computed by sum % 11 as the last digit.  

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Which view allows you to make changes to the content of your presentation?
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