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andrew11 [14]
4 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]4 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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Write a program that takes a date as input and outputs the date's season. The input is a string to represent the month and an in
Ulleksa [173]

Answer:

The problem here is you are comparing month names with == , in case of strings there is a difference between comparing with == and .equals. So I have changed that code please check now

Program:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

String inputMonth;

int inputDay;

System.out.print("Enter month:");

inputMonth = scnr.next();

System.out.print("Enter day of month");

inputDay = scnr.nextInt();

if((inputMonth.equalsIgnoreCase("march")&&inputDay>19)||inputMonth.equalsIgnoreCase("april")||inputMonth=="may"||(inputMonth.equalsIgnoreCase("june")&&inputDay<21))

{

System.out.println("spring");

}

else if((inputMonth.equalsIgnoreCase("june")&&inputDay>20)||inputMonth.equalsIgnoreCase("july")||inputMonth=="august"||(inputMonth.equalsIgnoreCase("september")&&inputDay<22))

{

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}

else if((inputMonth.equalsIgnoreCase("september")&&inputDay>21)||inputMonth.equalsIgnoreCase("october")||inputMonth.equalsIgnoreCase("november")||(inputMonth.equalsIgnoreCase("December")&&inputDay<21))

{

System.out.println("autumn");

}

else if((inputMonth.equalsIgnoreCase("december")&&inputDay>20)||inputMonth.equalsIgnoreCase("january")||inputMonth.equalsIgnoreCase("february")||(inputMonth.equalsIgnoreCase("march")&&inputDay<20))

{

System.out.println("winter");

}

else

{

System.out.println("invalid");

} }

}

Output:

3 0
3 years ago
PLEASE HELP
Anika [276]

Answer: The answer would be 60 kilometers.

4 0
3 years ago
Read 2 more answers
Prompt
madam [21]

Have you ever tried to learn a new language or do you have friends who've had that experience?

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3 0
3 years ago
This program will calculate the rise in ocean levels over 5, 10, and 50 years, Part of the program has been written for you. The
GREYUIT [131]

Answer:

Here is the C++ program:

#include <iostream>  //to use input output functions

using namespace std;  //to identify objects cin cout

int main(){  //start of main method

     

   double risingLevel;  //declares a double type variable to hold rising level

   cin>>risingLevel;  //reads the value of risingLevel from user

   

   cout<<"Level: "<<risingLevel<<endl;  //displays the rising level

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl;  //displays the rise in ocean level over 5 years

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl;  //displays the rise in ocean level over 10 years

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl;  //displays the rise in ocean level over 50 years

}

Explanation:

The program works as follows:

Lets say the user enters rising level 2.1 then,

risingLevel = 2.1

Now the first print (cout) statement :

cout<<"Level: "<<risingLevel<<endl;  prints the value of risingLevel on output screen. So this statement displays:

Level: 2.1          

Next, the program control moves to the statement:

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl;

which computes the rise in ocean levels over 5 years by formula:

risingLevel * 5 = 2.1 * 5 = 10.5

It then displays the computed value on output screen. So this statement displays:

Years: 5, Rise: 10.5                                                                                                                          Next, the program control moves to the statement:

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl;

which computes the rise in ocean levels over 10 years by formula:

risingLevel * 10 = 2.1 * 10 = 21

It then displays the computed value on output screen. So this statement displays:

Years: 10, Rise: 21                                                                                      

Next, the program control moves to the statement:

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl;

which computes the rise in ocean levels over 50 years by formula:

risingLevel * 50 = 2.1 * 50 = 105

It then displays the computed value on output screen. So this statement displays:

Years: 50, Rise: 105                                                                                                                          Hence the output of the entire program is:

Level: 2.1                                                                                                                     Years: 5, Rise: 10.5                                                                                                           Years: 10, Rise: 21                                                                                                            Years: 50, Rise: 105

The screenshot of the program and its output is attached.

4 0
3 years ago
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Artemon [7]

Answer:

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Explanation:

4 0
3 years ago
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