Question

a 4-way set - associative cache memory unit with a capacity of 16kb is built using a block size of 8 words. the word length is 32 bits. the size of the physical address space is 4gb.the number of bits for the tag field is

i am geting the answer 21 tag field bit length

but answer is 22

can any explain this question if answer is 22...

Answer 1

Assuming memory is word addressable

No of bits for offset=3

No of blocks=2^14/4*8=2^9

No  of sets =no of blocks/4

                               =2^9/4

                               =2^7

Bits for set=7

Bits for tag=32-7-3=22

If memory was byte addressable then tag bits=32-7-5=20

Comments

2^14/4*8

why did you use 4 here...plzz explain

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32 bit word =4 byte word

 Size of one block=4*8=32 bytes

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@pooja 2^14/4*8.. cache size in bits. Isnt it will be converted in bytes?

Answer 2

Memory is word adddressable

No. of words in physical memory = 4GB/4B = 1G

So no of bits required for Physical Address = 30 Bits

No. of bits required for block offset = 3 bits

No. of words in cache = 16KB/4B = 4KB

No. of blocks in cache = 4KB/8 = 512

No. of sets in cache = 512/4 = 128

So, no.of bits required to represent sets = 7 bits

Remaining bits are to represent TAG, so remaining bits are 30-(7+3) = 20 bits.

Comments

but according to key in gate 22 is correct answer...