Given that: for direct mapping technique;
- Main memory size= $2^{32}$ byte $\implies PAS=log_22^{32}=32$ bit
- Cache memory size(CM’s) = $32KB \implies 2^{15}$ B
- Block size/word offset = $64$ byte
we know that $N=\frac{CM’S}{B’S}$
$\therefore$ $N=\frac{2^{15}}{2^6}=2^9B$
so CLO= $\log_2N\implies \log_22^9=9$ bit
Block size= $\log_2B’S\implies \log_22^6=6$ bit
$\therefore$ tag bit= PAS-(CLO+B’S)
$\implies$Tag=32-(9+6) bit
$\implies$Tag= 17 bits.
so tag bit size in given problem is $17$ bits.
Note: same concept asked in GATE CSE 2017 Set 2 | Question: 53