Question

Assume the following about a computer with a cache: .. The memory is byte addressable. • Memory accesses are to 1-byte words (not to 4-byte words). .. Addresses are 8 bits wide. .. The cache is 2-way associative cache (E=2), with a 2-byte block size (B=2) and 4 sets (5=4). • The cache contents are as shown below (V="Valid"): Set #Way #0 Way #1 V=1;Tag=0x12; Data = v=1;Tag=0x10; Data = Ox39 0x00 0x26 Ox63 V=1;Tag=0x09; Data = v=1;Tag=0x11; Data = Oxe8 Oxf6 0x79 Oxb5 v=1;Tag=0x01: Data = v=1;Tag=0xOc; Data = 0x5f Ox12 Oxda Ox7f V=1;Tag=0x16; Data = v=1;Tag=0x1c Data = 0x6a Oxf7 |0x3c Oxab Assume that memory address 0x4a has been referenced by a load instruction. Indicate the cache entry accessed and the cache byte value returned in hex. Indicate whether a cache miss occurs. If there is a cache miss, enter"-" for the "Cache Byte Returned". For values that need a hexadecimal value, do not enter leading zeros even if leading zeros are shown in the value above. Cache block Offset (CO) OX Cache set index (CI) Ox Cache tag (CT) OX Cache hit (Y/N)? Cache byte returned Ox

Answer 1

Memory address size = 8 bits

Cache associativity = 2

#sets in cache = 4

Cache block size = 2 bytes

So, cache block offset = log2(block size) = log2(2) = 1 bit

Cache Set Index = log2(number of sets)= log2(4) = 2 bits

Cache tag = Address size - block offset - set index = 8 - 1 - 2 = 5 bits

Memory Address is represented as below

Tag	Set Index	Block Offset
5	2	1

memory address = 0x4a = 0100 1010

• Cache Tag = 01001 = 0x09
• Cache Set index = 01 = 0x1
• Cache block offset = 0 = 0x0
• Cache hit = Yes
  • All entries in Set index 1 is looked for any matche with tag value 0x9. If found e check for valid bit V. if V=1 data is returned based on block offset.
• Cache byte returned = 0xe8
  • ​​​​​​​Each cache entry contains 2 byte. as block offset is 0, 0th byte is returned.