Consider a direct mapping cache of size 32KB with block size of 32 bytes. The CPU generates 32 bit addresses. find out the no of bits required for physical memory addressing, cache addressing and tag bits.

Below is a series of byte addresses in a system with 32 bit words. Assuming a direct-mapped cache with 4-word blocks and a total size of 32 words that is initially empty, (a) label each reference in the list as a hit or a miss and (b) show the entire history of the cache, including tag and data. Byte Address Byte Address (Hexadecimal) (Binary) Hit/Miss 76 0000 0000 0111 0110 1E3 0000 0001 1110 0011 815 0000 1000 0001 0101 141 0000 0001 0100 0001 170 0000 0001 0111 0000 5E1 0000 0101 1110 0001 320 0000 0011 0010 1100 B10 0000 1011 0001 0000 175 0000 0001 0111 0101 583 0000 0101 1000 0011 1FF 0000 0001 1111 1111 7A 0000 0000 0111 1010 2B2 0000 0010 1011 0010 5E4 0000 0101 1110 0100 816 0000 1000 0001 0110 438 0000 0100 0011 1000

3. The table below represents five lines from a cache that uses fully associative mapping with a block size of 8. Identify the address of the shaded data, 0xE6, first in binary and then in hexadecimal. The tag numbers and word id bits are in binary, but the content of the cache (the data) is in hexadecimal.                   Word id bits Tag 000 001 010 011 100 101 110 111 ------------------------------------------ 1011010 10 65 BA 0F C4 19 6E C3 1100101 21 76 CB 80 D5 2A 7F B5 0011011 32 87 DC 91 E6 3B F0 A6 1100000 43 98 ED A2 F7 4C E1 97 1111100 54 9A FE B3 08 5D D2 88

Suppose a computer using fully associative cache has 224 words of main memory and a cache of 512 blocks, where each cache block contains 16 words. How many blocks of main memory are there? What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? To which cache block will the memory reference 17042416 map?

Consider a 32-bit physical memory space and a 32 KiB 2-way associative cache with LRU replacement. You are told the cache uses 5 bits for the offset field. Write in the number of bits in the tag and index fields. Tag length in bits = Index length in bits =