Question

A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits

Answer 1

Sala Direct marned cache holes GOKB of useful leta block size = 32 byte & the add re) is 32-bit cache size = 6unB Block size = 32B i. main memory address format for direct mapped cache 32 bit Tag Block no.ar Bloch ofset Byte no ar Byte offset cache size ... No of Bloch in cache memory Blah size memory 64KB 32B - 2 1 - 2 20 Bloch offset - depends on Number of Bloch in came lo-bit Byte numbee (or) Block cize Byte offset ( depends G-bit lugz cao Inden Block numbee & By le number 17-bit is-bit Il + 6 = tag = 32-19 Š

bit i Numbee of bits for offset: :bit ho 17 2) Numbee of bits for Inden bits. re 15 3) Numbee of bib for Tag
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