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 size of the cache tag directory is
1. 160 Kbits
2. 136 Kbits
3. 40 Kbits
4. 32 Kbits

Answer 1

Correct Answer - Option 1 : 160 Kbits

The correct answer is option 1

Data:

Physical address space = 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

Virtual address = tag + set + Byte offset (in bits)

32 = tag + 11+ 5

∴ tag = 16 bits

16 bit address 2 valid bits, 1 modified bit and 1 replacement bit

So, total bits =20

So, Each cache tag directory entry contains = 20 × Number of blocks = 20 × 2¹³ bits

∴ 213 = 8 K bits

So, Each cache tag directory entry contains = 20 × 8 K bits=160 Kbits

So, the size of the cache tag directory is 160 Kbits