<span>Units ..................................................</span>
Answer:A) SQS
Explanation: Simple Queue Service (SQS) is Amazon based service that deals in queue that decouples the micro-services infrastructure or applications.It has the benefit of maintaining of the security, eliminating complexity, transmitting the data reliably etc.
It also provides the service of the storing the message, storing it and other functioning without the integrity in its security or any other resource need.Thus, option(a) is the correct option.
Answer:
Check the explanation
Explanation:
// include the necessary packages
import java.io.*;
import java.util.*;
// Declare a class
public class DataReader
{
// Start the main method.
public static void main(String[] args)
{
// create the object of scanner class.
Scanner scan = new Scanner(System.in);
// Declare variables.
boolean done = false;
boolean done1 = false;
float sum = 0;
double v;
int count = 0;
// start the while loop
while (!done1)
{
// start the do while loop
do
{
// prompt the user to enter the value.
System.out.println("Value:");
// start the try block
try
{
// input number
v = scan.nextDouble();
// calculate the sum
sum = (float) (sum + v);
}
// start the catch block
catch (Exception nfe)
{
// input a character variable(\n)
String ch = scan.nextLine();
// display the statement.
System.out.println(
"Input Error. Try again.");
// count the value.
count++;
break;
}
}
// end do while loop
while (!done);
// Check whether the value of count
// greater than 2 or not.
if (count >= 2)
{
// display the statement on console.
System.out.println("Sum: " + sum);
done1 = true;
}
}
}
}
Sample Output:
Value:
12
Value:
12
Value:
ten
Input Error. Try again.
Value:
5
Value:
nine
Input Error. Try again.
Sum: 29.0
Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
