The following figure shows the address that is going from a certain processor to a direct-mapped cache. The address is divided into fields. The index of the first bit and the last bit of each field is written below it. Calculate the size of the data that is stored in the cache, in Kibytes, and the total number of bits within the cache, in Kibits.
TAG = 31 - 16
INDEX = 15 - 6
BLOCK OFFSET = 5 - 2
BYTE OFFSET = 1 - 0
Homework Answers
ANSWER:--
GIVEN THAT:--
-
Block offset bits = 0 to 5 =>6 bits.
Therefore block size = 2^6 = 64 B
Index bits = 6 to 15 => 10 bits
Therefore total # of block inside cache = 2^10= 1K blocks.
Therefore size of data stored in cache = Total # of block inside cache * size of each block = 1k*64 = 64K Byte = 62.5Kibytes
Tag bits = 16 to 31 = 16 bits.
Therefore total # of bits for one block entry of cache = 16 tag bits + 1 valid bit + 512 data bits = 529 bits
Therefore total # of bits within cache = 529 * 1K = 529Kbits = 529*1024 = 541696 bits = 529Kibits.
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