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alexandr1967 [171]
3 years ago
7

Which of the following calculations would evaluate to 12?

Computers and Technology
1 answer:
lidiya [134]3 years ago
8 0

Answer:

the answer for this is 3*(6+2)/2)

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A computer system has a 32KB, 8-way set associative cache, and the block size is 8 bytes. The machine is byte addressable, and p
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Answer:

Offset bits: 3-bits

Set number of cache: 12-bits

Tag bits: 7-bits

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Explanation:

Since the system is 32K so,

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

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2¹² are cache blocks

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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.

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