Question:
The question is not complete. The question to answer was not added. See below the possible question and the answer.
a. How many blocks are in the cache with this new arrangement?
b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.
C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?
Answer:
(a) Number of blocks = 512 blocks
(b) Tag is 18
(c) Total size of the cache = 8388608 bytes
Explanation:
a .
block size = 32 bytes
cache size = 16384 bytes
No.of blocks = 16384 / 32
No.pf blocks = 512 blocks
b.
Total address size = 32 bits
Address bits = Tag + Line index +block offset
Block Size = 32 bytes.
So block size = 25 bytes.
Hence Offset is 5
No . of Cache blocks = 512 blocks = 29 blocks
Hence line offset is 9
We know that Address bits = Tag + Line index +block offset
So , 32 =tag+9+5
tag = 32-(9+5)
So Tag is 18
c.
Data bits = 32 bits
Tag=18 bits
Valid bit is 1 bit
so Total cache size = 25+218+20
= 223
=8388608 bytes
