Question 8134 – Computer-Organization

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.

Correct Answer: A

Question 30 Explanation:

Given that it is a set associative cache, so the physical address is divided as following:
(Tag bits + bits for set no. + Bits for block offset)

In question block size has not been given, so we can assume block size 2^x Bytes.

The cache is of size 512KB, so number of blocks in the cache = 2¹⁹/2^x = 2^19-x

It is 8-way set associative cache so there will be 8 blocks in each set.
So number of sets = (2¹⁹ − x)/8 = 2¹⁶ − x

So number of bits for sets = 16−x

Let number of bits for Tag = T

Since we assumed block size is 2^x Bytes, number of bits for block offset is x.

So, T + (16−x) + x = 40

T + 16 = 40

T = 24

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