Question

The width of the physical address on a machine is 40 bits. The width of the tag field in a 512 KB 8-way set associative cache is______ bits.

Answer 1

Data:

Physical address (PA) = 40 bits

Cache Memory Size (CS) = 512 KB = 2¹⁹ B

Block Size (BO) = 2^x B

Set associativity = 8

Formula:

Formula to find number of bits in tag in set associative mapping

PA = tag + set + BO

\({\rm{number\;of\;set}} = \frac{{{\rm{CS}}}}{{{\rm{S}} \times {\rm{BO}}}}\)

Calculation:

\({\rm{number\;of\;set}} = \frac{{{2^{19}}}}{{{2^3} \times {2^{\rm{x}}}}} = {2^{19 - 3 - x}}\)

PA = tag + set + BO

40 = tag + 19 – 3 – x 

∴ tag = 24 bits

Important Points:

Assume: Memory is byte addressable