Question

16 KB, 4-way set-associative cache, 32-bit address, byte addressable memory, 32-byte cache blocks/lines

1. How many tag bits?
2. Where would you find the word at address 0x200356A4?

Answer 1

Cache size = 16kB =  14 bit

Block size = 32B = 5 bit

No of sets = 4 = 2 bit

No of cache lines = 14 -(5+2) = 7 bit

No of tag bit = 32 - (#set bit) - (#block bit)

        .              = 32 - 7 - 5 = 20 bit

 

 

0x200356A4 = (<0010 0000 0000 0011 0101> <0110 101> <0 0100>)

=(<tag no><set no><block no>)

Set no = 0110101 = 53th set

Block no = 00100 = 4th block

Comments

yes it wil be  53rd set :)

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Corrected !!!

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Not Block no. That is Word/Byte no.