Please asnwer this quesion ASAP How many total bits are requ

Please asnwer this quesion ASAP

How many total bits are required for a direct-mapped cache with 2048 entries each holding an 8 word (of 32 bits or 4 bytes) block, assuming a 32-bit address? Discuss with the help of a diagram the logic to check a hit or miss (that is, presence or absence) of a required memory block in the cache described in 1(a), and access a specific word of the block in case of a hit. Label all the fields in a cache word.

Solution

Q1) a )

We know that 64KB is 64K/4 = 16K words, which is 2¹⁰ x 2 ⁴ = 2 ¹⁴ words, and with a block size of one word,

2 ¹⁴ blocks.

Each block has 32 bits of data plus a tag, which is 32 – 14 – 2 bits, plus a valid bit.

Thus the total cache size is 2 ¹⁴ x (32 + (32 – 14 – 2) + 1) = 2 ¹⁴ x 49 = 784 x 2 ¹⁰ = 784 Kbits

or 784K/8 = 98KByte for a 64-KByte cache.

For this cache, the total number of bits in the cache is over 1.5 times as many needed just for the storage of

the data.