The size of virtual address is 32,now page size is 4KB given or,number of bits for offset is 12(2^12=4KB).
Page(20bits) |
OFFSET(12bits) |
Now for TLB header is as:
TAG(y bits) |
SET(x bits) |
OFFSET(12bits) |
Now the set is given as 4 way set associative,and it can hold 128 entries,
Now to divide 128 entries into 4 sets,number of bits required to do so:
set bits(x)=$128(or,2^7)/(4(or,2^2)))$
hence sets bits would be: 5
therefore:
$y+5+12=32$
or,$y=15 bits$
Hence option C is correct