Question

2. A computer uses a memory with addresses of 8 bits. (What's the size of the MM?) This computer has a 16-byte cache with 4 bytes per block. (How many blocks in the cache?) The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. a. What's the format of a memory address as seen by the cache ? Tag ? bits Block ? bits Offset ? bits b. The current cache is as below. Note: Cache contents use word address(HEX) to represent Tag Contents 1110 Cache Contents represented by address EO Contents Cache Contents represented by address 14 Block 0 Block 1 0001 S3 Block 2 1011 Block 3 0110 a. Describe what will happen when the word at 4E is requested. (Hint: Point out when there is Miss and where block is loaded into cache b. Describe what will happen when the word 4F is requested

Answer 1

Memory address = 8 bits. Main memory size = 28256 B, Coche size = 16B Block size= AB 4B Index offset - 16B . 4 log 77= 2 bits. Block offset : 4B = [109247 2bits. Tag bit = 8-(2+2) = 4 bits. @ what will happen when 4E is requested. 4€ - 0100 Dio Block number . Tag bits of block 3 presently is 0110. when request to 4E is made it will result 2 in miss The word 4E will be bought in 4th block of cache and tag bit will be changed to 0100.

b) What will happen when word 4F is request. 4F - 10100 Oui Tag bit of 4th block is 0100 @ present after word 4 € is bought in cache! Thus after 4F is request, it will a hit since it is already present in cache.
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