Accessing Cache

Accessing Cache

Total number of bits needed for a cache
-size of tag field is
                -32-(n+m+2)
-the total number of bits in direct-mapped cache
                -2ⁿx(block size+ tag size+ valid field size)
Block size = 2^m words (2^m+5 bits), 1 bit for valid field size
Total number of bits in cache is
-2ⁿ x (2^m x 32 + (32-n-m-2) +1)
=2ⁿ x (2^m x32 +31-n-m)

Example
How many total bit are required for a direct mapped cache with 16kb of data and 4-word blocks, assuming a 32-bit address?
-16kb is 4K (2¹²) words
-with block size of 4 words (2²), 1024 (2¹⁰) blocks remaining for cache size
-each block has 4 x 32= 128 bits data plus a tag, which is 32 – 10 – 2 – 2 bits
-plus a valid bit
-total cache size = 2¹⁰ x (4 x 32+ (32 – 10 – 2 – 2) + 1)
                                = 2¹⁰ x 147
                                =147 Kbits