These slides explains set-associative memory mapping and fully associative memory mapping with problem.

1. Associative Memory and set-
associative memory mapping
Ms. Snehalata Agasti
CSE department

2. Fully-Associative Mapping
 In direct mapping, though cache memory block is vacant still conflict misses
occurs.
 To overcome this problem Tag bit and cache-offset field is combined.
 So main memory block can be stored anywhere.
 Physical address is divided into two parts.
 Word-offset
 Tag
Tag Word-offset

3. Problem using filly-associative mapping
 Cache memory size= 64KB
Block size= 32B
Number of bits given for main memory addressing= 32
Find number of bits required for tag and word-offset?
 Solution: -
Block size=32B
Block offset= log225=5
Tag= 32-5=27
Tag = 27 Word-offset = 5
32

4. Set-associative Mapping
Draw back in direct mapping:-
 Compulsory miss occurs.
 conflict miss occurs.
 Cache memory could not be used effectively.
Draw back in fully-associative mapping:-
 Compulsory miss occurs.
 Capacity miss occurs.
 To over come the loopholes present in both mapping Set-associative mapping
technique is used.

5. Contd…
 Cache lines are grouped into sets.
 Particular block of main memory is mapped to particular set of cache lines.
 Within the set, block can be mapped to the free cache lines.
 To find the set number :-
set number= main memory block number % Number of sets in cache.
 Physical address is divided into three parts.
 Block-offset
 Set-offset
 Tag
Tag Set-offset Block-offset

6. Problem using set-associative mapping
 Cache memory size=64KB
Main memory size=4GB
Block size=32B , 4-way associative
Find tag, set-offset and word-offset?
 Solution:- Number of blocks in cache = size of cache memory / Block size
=64KB / 32B = 2KB =21 x 210 =211
Number of sets in cache = no of blocks in cache / associativity
= 2KB/4 = 211 / 24 = 29
Set offset = log229 =9

7. Contd…
Block size =32B
Number of bits required for word offset= log225=5
Size of main memory =4GB= 22 x 230 = 232
Number of bits required for memory addressing
= log2232 =32
Tag = 32-(set-offset + word-offset)
= 32 – (9+5) =18
Tag=18 Set-offset=9 Block-offset=5

8. Tag-directory size computation
 Tag-directory size = Tag x number of blocks
=18 x 211
= 36 x 210
= 36B
 Tag-directory size= (tag + number of extra bits given) x (number of blocks)
 [if in question it is given that number of dirty_bits=1 and number of modified
bit =2]
 Tag-directory size = (18 + 1+ 2) x 211
= 21 x211
= 42 x 2 10
= 42KB