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
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Best way is to put your exact question! And do "advanced search"
Hope I helped and if you need more info or another question, then feel free to ask me!
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