Question

A direct-mapped cache has 4 blocks and each block holds four bytes of data. The memory system is byte-addressable. Determine if each of the memory references below is a hit (H) or miss (M). You assume the cache is initially empty and memory references are given in decimal. 12 21 13 32 27 23 34 19 34 23 36 39 REFE RENC E H/M2 I

Answer 1

No of cache Blocks = 4 blocks = 2² // means 2 bits needed for index in cache block

Block size = 4 byte = 2² // means 2 bits needed for byte offset

6 bits...... Tagl 2 bits) Cache Block index (2 bits) Offset (2 bits) Cache Direct Mapping

Cache Block No Tag Match Miss/Hit Direct Mapp address (6 bits) Tag Index (2 bits) (2 bit) (cache block no) 00 11 Offset (2 bits) 1 00 + Memory Memory Reference Reference (In Binary) 12 001100 21 010101 13 001101 32 100000 27 011011 2 01 01 01 00 3 11 01 4 10 00 00 Initial access Initial access Yes Initial access Initial access yes yes No 5 01 10 11 6 23 010111 01 01 Block 3 Block 1 Block 3 Block 0 Block 2 Block 1 Block 0 Block 0 Block o Block 1 Block 1 Block 1 11 miss miss hit miss miss hit hit miss miss hit miss hit 7 34 100010 10 00 10 8 19 010011 01 00 11 Here we bring 19 associated block into cache 9 34 10 00 10 No 100010 010111 10 23 01 01 11 00 yes No 11 36 100100 10 01 Here we bring 36 associated block into cache 12 39 100111 10 01 11 yes MISS Ratio = 7/12) =0.583 HIT Ratio = (5/12) =0.416 Total Hits = 05 Total Miss 07

Note: In direct map cache First we go to particular cache block then we compare TAG bits if tag bits are match with already stored data in cache then it is called HIT, otherwise it is called MISS

No of hits = 05   

No of Miss =07

Total memory reference = 12

HIT Ratio = (5/12) =0.416

Miss Ratio = (7 /12) =0.583