I've got a cache for a MIPS 64 bit architecture, 2-way of 4KB per block and lines of 16 words (for every block). I have to determine:
[ TAG | Index | Offset ]
Bit sizes for every field
How many bit is the entire memory made of?
I've got no clue if what I've done is correct since I don't have the results, since we are working with 64bit MIPS and not 32 I thought that a word would have been made out of 8 bytes instead of 4.
I've calculated the numbers of blocks by doing 4KB / (16 words * 8 byte) = 32
Then I've done 32 / 2-way= 16 --> 4 bit of INDEX
Then I got word and byte offset as follows:
16 words --> 4 bit word offset
8 byte --> 3 bit byte offset
Finally I calculated the TAG as 64 - 4(index) - 4(word offset) - 3 (byte offset) = 53 bit
And the total number of bit would have been 53 tag + 64 + 1 validate = 118 which multiplied by the number of blocks(32) would have given 3776 bit.
I'm not completely sure about the result, can someone help me with this?
Thanks.
Part of the question is not very clear, I'll try to make a best guess.
Assumption : Cache size is 4KB, word is 64 bits, memory is byte addressing.
Information : Cache line is 16 words wide, and its a 2-way set assoc cache.
Cacheline size = 16 word * 64 bits (8 byte) = 128 byte
Number of cachelines = 4 KB/128 B = 32
Number of blocks = 32 / 2 (2-way) = 16
As the memory is byte addressable, least 3 bits of the address can be ignored.
Next 4 bytes will be used to offset in a word in the 16 word cacheline.
Next bit of the address is ignored because its 2-way assoc
As there are 16 blocks, next 4 bits will be used to index a block.
That leaves 64 - (4 + 1 + 4 + 3) = 52 bits for the TAG.
TAG - 52 bits [bits 12 to 63]
Index - 4 bits [bits 8 to 11]
Offset - 4 bits [bits 3 to 6]
Its bit hard to answer to the second question as a cache would typically have extra bits per cacheline, like valid bits etc. For the sake of simplicity lets assume this cache only has a TAG per cacheline. This gives us 52 bits * 16 blocks * 2 lines in a block = 1664 bits = 208 bytes.
Therefore the whole cache size would be
= 4 KB + 208 B
= 4096 B + 208 B
= 4304 Bytes