memory management

Question

Consider a direct mapped cache of size 16 KB and block size is 4 words. The word length is 16 bits. Find the number of bits needed for cache indexing if CPU generates 32 bit address

Answer 1

block size=4*2=8 bytes

no of blocks =16kB/8=2048

so for cache indexing 11 bits are required

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3 bit for offset (assuming byte addressing; 2 for word addressing)

and 32-14=18 for tag (19 if word addressing is used)

Comments

it should be word offset which needed only 2 bits...please confirm?

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2 in case of word addressing. But, in both cases index bits remain 11- we dont need to find offset/tag bits to calculate this.

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here which type of addressing we need to consider.

i know if not given anything then we have to use byte addressing.

but here word size is given...so byte addressing or word addressing ?

i learn from some source that index is (block offset + word offset)... means i'm confused.