Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 1024-byte 4-way set associative cache using 32-bit memory addresses and 8-byte cache blocks.

The answer is The Cache Sets (S) = 32, Tag bits (t)=24, Set index bits(s) = 5 and Block offset bits (b) = 3

Explanation:

Solution

Given Data:

Physical address = 32 bit (memory address)

Cache size = 1024 bytes

Block size = 8 bytes

Now

It is a 4 way set associative mapping, so the set size becomes 4 blocks.

Thus

Number of blocks = cache size/block size

=1024/8

=128

The number of blocks = 128

=2^7

The number of sets = No of blocks/set size

=128/4

= 32

Hence the number of sets = 32

←Block ←number→

Tag → Set number→Block offset

←32 bit→

Now, =

The block offset = Log₂ (block size)

=Log₂⁸ = Log₂^2^3 =3

Then

Set number pc nothing but set index number

Set number = Log₂ (sets) = log₂³² =5

The remaining bits are tag bits.

Thus

Tag bits = Memory -Address Bits- (Block offset bits + set number bits)

= 32 - (3+5)

=32-8

=24

So,

Tag bits = 24

Therefore

The Cache Sets = 32

Tag bits =24

Set index bits = 5

Block offset bits = 3

Note: ←32 bits→

Tag 24 → Set index  5→Block offset 3

answer: a, c

divide any of the y values by x values and you will always get 3 is the most simple way i can put it for you.