Correct Answer - Option 4 : 28 bits and 0 bits
Given:
Cache size = 16 KB = 214 B
\({\log _2}{2^{14}}\) = 14 bit
Cache block size = 16 B = 24 B
\({\log _2}{2^{4}}\) = 4 bit (offset)
Physical address = 232 B = 4 GB //32-bit address
To find:
Number of bits in tag field
Number of bits in Index field
Explanation:
In Fully Associative Mapping:
Physical address:
Physical Address = Tag + block offset
32 = Tag + 4 // use number of bit
Tag = 28 bit
In Fully Associative Mapping, index fields is included in tag field there is no separate field for index in fully Associative Mapping. Hence number of bit in index field is zero
Note:
KB is Kilo Byte (2
10 B)