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
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
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