Comprehension A computer has a 256 KByte, 4-way set associative, write back data cache with block size of 32 Bytes. The processor sends 32 bit address

Question

The number of bits in the tag field of an address is
1. 11
2. 14
3. 16
4. 27

Answer 1

Correct Answer - Option 3 : 16

The correct answer is option 3

Data:

Physical address (PA) = 2³² bytes

Block size = 32 B = 25 B

Number of bits for block size = 5

cache size =256 KByte = 2¹⁸ B

set associative = 4 -way 

Calculation:

number of bits = ⌈log2 n⌉ 

Number of blocks = \(\frac{cache\;size}{block\; size}\) =\({2^{18} B\over 2^5B}=2^{13}\)

Number of sets in cache = \({\frac{number\; of\; blocks}{set\; associativity}} = {2^{13}\over 2^2}=2^{11}\)

Number of bits in set =11

PA = tag + set + offset

32 = tag + 11+ 5

∴ tag = 16 bits

So, the number of bits in the tag field of an address is 16 bits