GATE | GATE CS 2019 | Question 11

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?

(A) 24 bits and 0 bits
(B) 28 bits and 4 bits
(C) 24 bits and 4 bits
(D) 28 bits and 0 bits


Answer: (D)

Explanation: Given cache block size is 16 bytes, so block or word offset is 4 bits. Fully associative cache of size 16 kB, so line offset should be,

= cache size / block size
= 16 kB / 16 B
= 1 k 
= 1024
= 10 bits Line or Index Offset 

Tag bit size would be,

= processor address size - (line offset + word offset)
= 32 - 10 - 4
= 18 bits tag size 

Since, there no option matches, but if we assume that Line Offset is a part of Tag bits, therefore,

Tag bits = 18+10 = 28 bits
Line or Index offset = 0 bits (since fully associative cache memory),
Word or block offset = 4 bits 



Alternative way:
We know in fully associative mapping,

Line size = block size = frame size 

Number of bits in tag can be founded using given below formula

Number of Tag bits 
= Total number of bits in Physical Address - no of bits in Block offset 

Here number of bits in block offset is not given. It can be found using

ceil(log2 Cache block size) = ceil(log2 16) = 4 

So,

Number of Tag bits = 32-4 = 28 

No index bits is there in fully associative mapping,hence Index bits = 0

So, option (D) is correct.

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