Consider a direct mapped cache of size $32 K B$ with block size $32$ bytes. The $C P U$ generates $32$ bit addresses. The number of bits needed for cache indexing and the number of tag bits are respectively

10, 17

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10, 22

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15, 17

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5, 17

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Solution

The correct option is A $10, 17$
$S i z e o f c a c h e = 32 K B$
$= 32 \times 2^{10} b y t e = 2^{5} \times 2^{10} b y t e$
$= 2^{15} b y t e = 15 b i t s$

$S i z e o f t a g = 32 - 15 = 17 b i t s$
$C a c h e i n d e x = L O - W O = 15 - 5 = 10 b i t s$

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