# A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes. Assume that the main memory is byte addressable an

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A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes. Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the Tag and the Index fields respectively in the addresses generated by the processor?

1. 24 bits and 0 bits
2. 28 bits and 4 bits
3. 24 bits and 4 bits
4. 28 bits and 0 bits

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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)

To find:

Number of bits in tag field

Number of bits in Index field

Explanation:

In Fully Associative Mapping:

 Tag Block offset

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 (210 B)