import java.util.Scanner;
public class JavaApplication83 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter Strings: ");
String word1 = scan.nextLine();
String word2 = scan.nextLine();
String newWord = "";
if (word1.length() == word2.length()){
for (int i = 0; i < word1.length(); i++)
{
newWord += word1.charAt(i) +""+word2.charAt(i);
}
}
else{
newWord = "error";
}
System.out.println(newWord);
}
}
I hope this helps!
Answer:
a) the average CPI for machine M1 = 1.6
the average CPI for machine M2 = 2.5
b) M1 implementation is faster.
c) the clock cycles required for both processors.52.6*10^6.
Explanation:
(a)
The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4
= 1.6
The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4
= 2.5
(b)
The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6
Given 80MHz = 80 * 10^6
The average MIPS ratings for M1 = 80 x 10^6 / 1.6 x 10^6
= 50
Given 100MHz = 100 * 10^6
The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6
= 40
c)
Machine M2 has a smaller MIPS rating
Changing instruction set A from 2 to 1
The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)
and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.
C. Insert
Insert will let you inseet different things into your presentation.
Answer:
Step 1: Identify the Problem
Step 2: Review the Literature. ...
Step 3: Clarify the Problem Step 4: Clearly Define Terms and Concepts.
