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Question

how its 4?

Comments

It is 8.?

but u take tag bits but given is tag size.

Answer 1

 <-------------------------16 bits--------------------------------------->

9

set

3

no. of lines = cache size / block size = ${}\frac{2^{10}}{2^3} = 2^7$

sets = lines/associativity = ${}\frac{2^{7}}{2^x} = 2^{7-x}$

so, 16 = 9 +(7-x) +3

$\therefore$ x= 3

associativity = $2^x = 2^3 =8$