The document discusses cache mapping and different cache mapping techniques. It explains: - The physical address is divided into tag, index, and offset bits for mapping blocks to cache. - Fully associative mapping allows a block to map to any cache location, while set associative mapping groups blocks into sets within the cache. - Direct mapping dedicates a specific cache block to each main memory block based on the block offset. - Examples are given to illustrate cache hits and misses under direct mapping as memory blocks are accessed in sequence.

1. Cache mapping

3. • Number of blocks in main memory=
• Main memory size/ block size
• = 2 20 Bytes/ 2 4 Bytes

4. • Main memory = 220 Bytes
• number of bits in physical address= 20 bits.
• Block size= 16 Bytes
• = 2 4 Bytes
• number of bits in block offset = 4 bits.

5. • Cache memory contains 32 blocks = 2 5 blocks
• Number of bits in cache index = 5 bits
• Number of bits in tag : the remaining
• = 20 – (5+4)
• =11 bits

6. • ODB63 = 0000 1101 1011 0110 0011
• The first least significant bits are for the offset
• The next 5 bits are for cache index.
• 10110 is 22 in decimal
• ODBC will map (22 Mod 32) = 22 cache block
that is the block number 23 (blocks start from
0)

8. • In the figure below, the physical address in
only divided into offset and tag.
• A block in the main memory can map any
address in the cache memory.
• It is the fully associative mapping.
• When the cache is searched, all entries (block
in cache) are searched in parallel to retrieve
the data.

10. • Main memory size= 2 14 words
• number of bits in physical address = 14 bits
• Block size= 8 words
• = 23 words
•  number of bits in offset = 3 bits

11. • Number of lines in cache memory = 16 blocks
• 2-way set associative mapping means each set
contains 2 blocks
•  16/2 = 8 sets = 2 3 sets
• That is 3 bits for the set index
• The tag bits are the remaining tag=
• 14 – (3+3) = 8 bits

13. • Direct mapping case:
• Block size = 32 Bytes = 2 5 Bytes
• 5 bits offset
• Cache Memory size =64 Kbytes = 26* 210 Bytes
• Number of lines = CM size/ Block size
•  216 Bytes / 32 Bytes = 211 Bytes
•  11 bits for index
Tag 8 bits Index 11 bits Offset 5 bits

14. • Tag bits = the remaining
• = 24- (11+5)
• = 8 bits

15. • Fully associative case:
• Block size = 32 Bytes = 2 5 Bytes
• 5 bits offset
• In full associative, all the rest of bits is for the
tag since a block in Main Memory can be
mapped to any cache block
Tag 19 bits Offset 5 bits

16. • 4-way set associative :
• Block size = 32 Bytes = 2 5 Bytes
• 5 bits offset
• Cache Memory size =64 Kbytes = 26* 210 Bytes
• Number of lines = CM size/ Block size
•  216 Bytes / 32 Bytes = 211 lines
• Each set contains 4 blocks = 22 blocks

17. • So the number of sets is :
• 211/22=29
•  9 bits for set
• Tag bits = the remaining
• = 24- (9+5)
• = 10 bits
Tag 10 bits Index 9 bits Offset 5 bits

19. • Step 1 : read 30
• 30 Mod 5 = 0
• So we go to address zero and map the
memory block  miss
• Step 2 : read 31
• 31 Mod 5 = 1
• So we go to address one and map the memory
block again miss

20. • Step 3 : read 30
• It already exists at this position  hit
• Step 4 : read 26
• 26 Mod 5 = 1 (address location busy, we will
replace and we no longer have 31 in cache)
• So we go to address one and replace the
memory block  miss

21. • Step 5 : read 25
• 25 Mod 5 = 0 (address location busy, we will
replace and we no longer have 30 in cache)
• So we go to address zero and replace the
memory block  miss
• Step 6 : read 28
• 28 Mod 5 = 3
• So we go to address three and map the memory
block  miss

22. • Step 7 : read 23
• 23 Mod 5 = 3 (address location busy, we will
replace and we no longer have 28 in cache)
• So we go to address three and replace the
memory block  miss
• Step 7 : read 25
• Already in cache memory  hit

23. • Step 8: read 28
• 28 Mod 5 = 3 (address location busy, we will
replace and we no longer have 23 in cache)
• So we go to address three and replace the
block 23 by 28  miss