Question

A computer system has a 32KB, 8-way set associative cache, and the block size is 8 bytes. The machine is byte addressable, and physical addresses generated by the CPU are 22 bits. Calculate the number of bits in the TAG, SET, and OFFSET fields.

Answer 1

Answer:

Offset bits: 3-bits

Set number of cache: 12-bits

Tag bits: 7-bits

22-bit physical address

Explanation:

Since the system is 32K so,

=2⁵.2¹⁰

=2¹⁵

As we know that it is 8-way set associative so,

=2¹⁵/2³

=2¹⁵⁻³

=2¹²

2¹² are cache blocks

22-bit physical address

Off-set bits are 3 as they are calulated from 8-way set associative information.

Set number of cache : 12-bits

For tag-bits:

Add off-set bits and cache bits and subtract from the total bits of physical address.

=22 - (12+3)

=22 - 15

=7