# GATE | GATE-CS-2014-(Set-2) | Question 65

• Difficulty Level : Medium
• Last Updated : 29 Sep, 2021

A 4-way set-associative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is _____
(A) 5
(B) 15
(C) 20
(D) 25

Explanation:

In a k-way set associate mapping, cache memory is divided into sets, each of size k blocks.
Size of Cache memory = 16 KB
As it is 4-way set associative,K = 4
Block size B = 8 words
The word length is 32 bits.
size of Physical address space = 4 GB.
—————————————————

No of blocks in Cache Memory(N) = (size of cache memory / size of a block)

= (16*1024 bytes / 8*4 bytes) = 512 (as 1 word = 4 bytes)

No of sets(S) = (No of blocks in cache memory/ no of blocks in a set)

= N/K = 512/4 = 128

Now,size of physical address = 4GB = 4*(2^30) Bytes = 2^32 Bytes

These physical adresses are divided equally among the sets.

Hence, each set can access ((2^32)/128) bytes = 2^25 bytes = 2^23 words = 2^20 blocks

So, each set can access total of 2^20 blocks. So to identify these 2^20 blocks, each
set needs TAG bits of length 20 bits.

Hence option C.

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