Question

Please calculate the total number of bits needed for the cache with listed below for a direct-mapped cache.

Size of Cache Data: 32KB

Size of Cache Block: 2 Words

Answer 1

For direct mapping , CPU generated address is divided into : Tag field , Line field and Word offset

Given data memory size of cache = 32 KB

Assuming the memory to be byte addressable (ie. 1 word size = 1B)

No. of cache lines = Data memory size of cache/ Data size of one cache line

                           = 32 KB /2B = 2¹⁴ lines

Then number of bits needed to represent cache line = log ₂ (2¹⁴) = 14 bits

Number of bits needed to represent a word in a line = log₂ 2 = 1 bit

No. of bits needed for tag = 32-14-1 =17 bits

For a given line, we will have the same tag. So one tag is associated with one line in case of direct mapping

So size of the tag memory = No. of tag bits * no.of lines = 17 * 2¹⁴ bits = 279 K bits

data memory size of cache = 32 KB = 256 K bits

So total memory needed for cache = 256 K bits + 279 K bits = 535 K bits