Question

A direct mapped cache has 32 cache lines.Each cache line consists of 4 words, and each word is four bytes.The address bus consists of 16 bits.

How many bits are required for the tag in this direct-mapped cache?

Answer 1

Solution:

Given,

=>Direct mapped cache

=>Number of lines in the cache = 32

=>Number of words in cache = 4 words

=>Word size = 4 bytes

=>Address = 16 bits

The answer will be "7"

Explanation:

Direct mapped cache:

Tag

Cache index

Block offset

    7 bits                          5 bits                                    4 bits

Calculating size of line:

=>Size of line = 4 words

=>Size of line = 4*4 bytes as 1 word = 4 bytes

=>Size of line = 16 bytes

Calculating number of bits for block offset:

=>Number of bits for block offset = log₂(size of line in bytes)

=>Number of bits for block offset = log₂(16)

=>Number of bits for block offset = 4 bits

Caculating number of bits for cache index:

=>Number of bits for cache index = log₂(number of lines)

=>Number of bits for cache index = log₂(32)

=>Number of bits for cache index = 5 bits

Calculating number of bits for tag:

=>Number of bits for tag = total address bits - offset bits - cache index bits

=>Number of bits for tag = 16 bits - 4 bits - 5 bits

=>Number of bits for tag = 7 bits

I have explained each and every part with the help of statements attached to the answer above.