Q.1. Consider a machine with a byte addressable main memory of 216 bytes and block size of 16 bytes. Assume that a direct mapped cache consisting of 16 lines used with this machine. (a) Show the format of the main memory address.
(b) Suppose the byte with address 0001 1010 0110 0001 is stored in the cache.
(c) What are the addresses of other bytes stored along with it? (d) How many total bytes of memory can be stored in the cache? (e) Into what line would bytes with each of the following
addresses be stored? i) 1101 0000 0001 1101 ii) 0001 0010 0011 1101 iii) 1010 1010 1010 1010 iv) 1001 0001 1111 0101 Solution :
2^16 of main memory means 16 bit address of main memory 0000 0000 0000 0000 to 1111 1111 1111 1111
Block size of 16 Byte : which can be represented in 4 bits 0000 to 1111 Cache has 16 lines means 16 blocks of main memory can at most reside in cache.
So, cache can address total 8 bits.
Show the format of the main memory address.
Tag – 8bits Line – 4 bits WORD – 4 bits
Suppose the byte with address 0001 1010 0110 0001 is stored in the cache.
What are the addresses of other bytes stored along with it? Make the high 12 bit constant and lower 4 bit from 0000 to 1111 0001 1010 0110 0000 0001 1010 0110 0001 0001 1010 0110 0010 … … . 0001 1010 0110 1110 0001 1010 0110 1111
How many total bytes of memory can be stored in the cache? 2^8 = 256 Bytes
Into what line would bytes with each of the following addresses be stored?
i) 1101 0000 0001 1101 - 1st Li ne (0001) ii) 0001 0010 0011 1101 - 3rd Line(0011) iii) 1010 1010 1010 1010 – 9th Line(1010) iv) 1001 0001 1111 0101 - 15th Line (1111) Considering line starts from 0 to 15.
Q.1 Consider a machine with a byte addressable main memory of 216 bytes and block size of 16 bytes. Assume that a direct
mapped cache consisting of 64 lines is used with this machine. (a) How is a 16-bit memory address divided into tag, line
number and byte number?
(b) How is a 16-bit memory address divided if the mapping used is fully set
associative?
(c) How is a 16-bit memory address divided if the mapping used is 4-way set associative?
(d) Why is a tag also stored in the cache? Solution :
2^16 bytes of main memory means 16 bit address
Block size of 16 byte means – 4 bits are used to represent the memory location witin the block.
64 lines of cache means, 64 blocks of main memory can atmost reside in cache.
(a) How is a 16-bit memory address divided into tag, line number and byte
number?
(b) How is a 16-bit memory address divided if the mapping used is fully set
associative?
TAG = 12 bits Word = 4 bit
(c) How is a 16-bit memory address divided if the mapping used is 4-way set associative?
Total lines in cache is 64.
So in vase of 4 way set associative mapping we have 64/4 ( 2^6/2^2) == 16 ( 2^4 )
4 bits are used as SET bits
TAG (16-4-4) = 8 SET BIT = 4 WORD= 4
(d) Why is a tag also stored in the cache?
There are various address of main memory that maps to cache, The tag bits are used to uniquely identify a particular address.
Q.1 Consider a machine with a byte addressable main memory of 216 bytes and block size of 8 bytes. Assume that a direct mapped cache consisting of 32 lines is used with this machine.
(a) How is a 16-bit memory address divided into tag, line number and byte
number?
(b) Into what line would a byte with following address be stored?
0001 0001 0001 1011
(c) Suppose the byte with address 0001 1010 0001 1010 is stored in the cache.
What are the addresses of other bytes stored along with it? (d) How many total bytes of memory can be stored in the cache? (e) Why is a tag also stored in the cache?
Solution
Bits to represent address of main memory = 16 bits
Block size = 8 byte ( 2^3) , so 3 bits are used to identify memory location within the block.
Cache of 32 lines ( 2^5) , 5 bits are used to represent the line. (a) How is a 16-bit memory address divided into tag, line number and byte
number?
TAG ( 16-5-3) = 8 LINE = 5 bits WORD - 3 bits
(b) Into what line would a byte with following address be stored?
0001 0001 0001 1011 - Line 3 (00011)
(c) Suppose the byte with address 0001 1010 0001 1010 is stored in the cache.
What are the addresses of other bytes stored along with it? 0001 1010 0001 1000 0001 1010 0001 1001 0001 1010 0001 1010 0001 1010 0001 1011 0001 1010 0001 1100 0001 1010 0001 1101 0001 1010 0001 1110 0001 1010 0001 1111
How many total bytes of memory can be stored in the cache? 2^8 = 256 bytes
(e) Why is a tag also stored in the cache?
There are various address of main memory that maps to cache, The tag bits are used to uniquely identify a particular address.
Consider a memory system that uses a 32- bit address to address at the byte level, plus a cache memory that uses a 64-byte line size. a) Assume an associative cache and determine the following: (i) Address format
(ii) Number of blocks in main memory (iii) Number of addressable units
b) Assume a 4-way Associative cache with a tag field in the address of 9 bits and
determine the following: (i) Address format
(ii) Number of lines in a set
(iii) Number of lines in cache memory
c) If cache design is changed to 8-way Associative then what will be the address
format?
d) Draw the block diagram to show how processor’s requests are interpreted in 4-way
Associative cache design i.e. in (b).
Solution
Number of Bits to represent main memory address : 32 Block/Line Size : 64 byte ( 2^6)
Assume an associative cache and determine the following: (i) Address format
TAG = 26 bits WORD = 6 bits
(ii) Number of blocks in main memory (2^32)/2^6 = ( 2^26 blocks)
(iii) Number of addressable units 2^32 addressable units.
b) Assume a 4-way Associative cache with a tag field in the address of 9 bits and
determine the following: (i) Address format
