Question

32 bytes of memory. 16 bytes of 2-way set-associative cache, where blocks can go anywhere within the set. Block is 2 bytes, set in cache is two blocks. Populate memory starting with upper-case letters, then 0-5. Hint- with full associativity in the set: each block has its own set of Tag bits in the cache. Memory is not organized by sets, though blocks get assigned to sets, and load in the cache per set.
1) Break down the addressing: Tag | Set |block offset ___ __ __
Explain and show what happens when the CPU generates these addresses to load from memory: 2) 10010 3) 00011 4) 10011 5) 10100 6) 10101 7) 11011

Answer 1

Answer: first no.of bits regcured to deste address = log, Csize of cache) = log 32=5679 no.of bits regadser to denote block offset = log2 (block site) clog 2-bit no. of uses - Cache size / block size = 32-R 16 likes no. of sets - no. of lines (P-coay=16/2=8 set no. of bits nescused to decote set: 109, no. of sets) = 1092(8)=3 bits now, no. of bits regcused to denote tag- 5- (no. of bies for set & no. of for block off get ): 5-4-1 now, first adddress 10010 the first bit (MSB)-1-tag. the 2nd to 4th bit from left-Set = 001 = Set, the last bit from left-block offset 0

second adexess 00011 - the first bit (MSB)=0-tag the 2nd to 4th bet from left-Set 200=set, then Cast bit from left- block offset=21 and address 100115 the first bit (MSB)=1= tag the 2nd to 4th bit from left: set=001 - set 1, since both & the blocks & set lare foll. so here LRD replace- ment policy is applied & thens first to come wree be replaced then last ble from left = block offsets 4th address 10100:- the frost bit (MSB)=1=tag. the 2nd to 4th bit fren left =get = 010-Sets the last bit from left-block offset to

5th acele ess 1010 0B the first blt (183)=1=tag the and to gen bit from left-set. 2010-set 2 then last bit from left: block offset=1 oth address 11013 the first bit (M8B) 4-tag the 2012 to 4th bit form left = Set=101=sets block off set-1 the last bit from left DO AA NODOD 0200000 ly set # - Aadress ofy