Answer:
a. 100%
b. 50%
c. 66.6%
Explanation:
a.
The memory references D6, 58, D8, 9A, DE and 5C all results in misses and the miss ratio is 100 (that is , 6 misses/ 6 references*100 = 100%)
b.
If fully associative cache is used, there will be a 3 cache misses with miss ratio of (3/6*100) = 50%.
c.
With a 2-way set-associative cache, 4 memory reference misses will occur, with a miss ratio of (4/6*100)= 66.6% misses.
C.cannot be viewed on a screen
Answer:
Here is c program:
#include<stdio.h>
#include<conio.h>
void main()
{
//Variable declaration
//array to hold 6 values you can change this as needed
float nums[6];
//integer i is for loop variable
int i;
//to hold maximum value
float max=0.0;
//clear screen
clrscr();
//iterate 6 times you can change as per your need
for(i=0;i<6;i++)
{
printf("Enter a number:");
scanf("%f",&nums[i]);
//check if entered value is greater than previous value
//if it is greater then assign it
if(nums[i]>max)
max = nums[i];
}
//print the value
printf("The largest number entered was %f",max);
getch();
}
Explanation:
