Question

) Consider an 8-way associative 64 Kilo Byte cache with 32 byte cache lines. Assume memory addresses are 32 bits long.

a). Show how a 32-bit address is used to access the cache (show how many bits for Tag, Index and Byte offset).

b). Calculate the total number of bits needed for this cache including tag bits, valid bits and data

c). Translate the following addresses (in hex) to cache set number, byte number and tag (i) B2FE3053hex (ii) FFFFA04Ehex (iii) FFEEA0FFhex (iv) 0B0EF3056hex

Answer 1

= Block size & lines =2 lined hence spits are 32 B size of line 28 Bytes = hence & bit are required for Block off on no. of sets = 20 sets hence & bit are required for set No. Set No Index Block offset 16 bit . 8bitt 8 bits

② total no. of bith need? ith n total lines = 2 linen? index bit 6 total bite = 2 Trolid bit & tay bits + Calatabit) offret bit = 2 total bits *(1 +16+ (32 )* 2 7 = 2101504 C data bits 32 ③ we for have - 16 this 8 cache 8 Setro Block hence for (i) B 2 FE3053 each dixit in 4 bit repretentation hence plock offret set no tag = 53 30 F BLFE

Similarly FFFFAO AE tay S set No offaet F FEE AO FF E tt tag Set No Block . t (10) ROBO EF 3056 tay Set Block no ofset