Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
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hope this helps
Wait. What? What does that even mean?
Answer:
void printC()
{
int i, j;
for (i = 0; i < 4; i++) //i indicate row number. Here we have 5 rows
{
printf("C"); //print C for every row
for (j = 0; j < 6; j++) //j indicate column number. Here we have 7 Rows
{
if (i == 0 || i == 4) //For first and last row
printf("C"); //print 'CCCCCCC'
else if (i = 1|| i= 3) //for Second forth row
printf("C + +"); //print 'C + +'
else if (i = 2) For second row
printf("C +++++"); //print 'C +++++'
else
continue; //to jump to next iteration
}
printf("\n"); // print in next line
}
}
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