Problem Solving:

Mapping an Address to a Multiword Cache Block

1) Consider a cache with 64 blocks with 64 blocks and a block size 16 bytes. What block number does byte address 1200 map to?

Ans: The block is given by (block address) modulo (Number of cache blocks)

Where the address of the block is Byte address/Bytes per clock

Notice that his block addresss is the block containing all addresses between (Byte address/Bytes per clock)*Bytes per clock and ( Byteaddresss/ Bytes per block) *Bytes per block+ (Bytes per block - 1)

Thus, with 16 bytes per clock, byte address 1200 is block address (1200/16)=75 which maps to cache block number (75 modulo 64)=11

Bits in a Cache

2) How many total bits are required for a direct-mapped cache with 64kb of data and one-word blocks, assuming a 32-bit adress?

Ans: We know that 64 KB is 64K words, which is 2 to the power 14 words, and, with a block size of one word, 2 to the power 14 blocks. Each block hs 32 bits of data plus a tag, which is 32-14-2 bits, plus a valid bit. Thus, the total cache size is (2 to the power 14)*49= 784 bits or 98KB for a 64-KB cache. For this cache, the total number of bits in the cache is over 1.5 times as many as needed just for the storage of the data.