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A 4-way set-associative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is _____

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Data:

1 W = 4 B (word length is 32 bits)

Physical address space (PS) = 4 G B = 230 W

Cache size (CS) = =16 KB = 214 B = 212 W

Block size (BS) = 8 W = 23 W

set associativity = 4

Formula:

number of bits = log2 n 

Number of lines in cache \(= \frac{{{\rm{CS}}}}{{{\rm{BS}}}}\)

Number of sets in cache \(= \frac{{number\;of\;lines}}{{set\;associativity}}\)

PAS = tag + set + block offset .....(in bits)

Calculation:

Number of lines in cache \(= \frac{{{2^{12}}}}{{{2^3}}} = {2^{9}}\)

Number of sets in cache \(= \frac{{{2^{9}}}}{4} = {2^7}\)

Tag

Set

Block Offset

x bits

7 bits

3 bits

30 = x + 7 + 3

x = 20

tag = 20 bits

The number of bits for the TAG field is 20.

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