Question

A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size. Index/Set → 29 = 512 blocks + s = 9 bits Next, determine the offset through the use of the block size. Offset → 2b = 256 bytes + b = 8 bits Finally, you know that you have 128 address bits, meaning that an address must be 128 bits long. Nine of those digits will be used by the set and 8 digits will be used by the offset. The remaining will be used for the tag. Total Address Bits = Tag + Index + Offset 128 Bits = Tag + 9 bits + 8 bits Tag = 111 bits

Answer 1

The whole Solution is correct but as we have to calculate Number of blocks in Cache so we should be

Number of blocks in cache = $\frac{cachesize}{block size}$

so, we have Cache size =131,072 bytes and block size = 256 bytes

So,

#blocks in cache=$\frac{131,072}{256}$ = 512 blocks.

Here answer matches means 128,000 in the solution may be misprinted by mistake.It should be 131,072 not 128,000