What is the address breakdown for cache?
A computer uses 32-bit byte addressing. The computer uses a 2-way associative cache with a capacity of 32KB. Each cache block contains 16 bytes. Calculate the number of bits in the TAG, SET, and OFFSET fields of a main memory address.
What are the three fields in a direct mapped cache address?
In set associative cache mapping, a memory reference is divided into three fields: tag, set, and word, as shown below. As with direct-mapped cache, the word field chooses the word within the cache block, and the tag field uniquely identifies the memory address.
How do I calculate total cache size?
In a nutshell the block offset bits determine your block size (how many bytes are in a cache row, how many columns if you will). The index bits determine how many rows are in each set. The capacity of the cache is therefor 2^(blockoffsetbits + indexbits) * #sets. In this case that is 2^(4+4) * 4 = 256*4 = 1 kilobyte.
What is tag in cache memory?
tag – A unique identifier for a group of data. Because different regions of memory may be mapped into a block, the tag is used to differentiate between them. valid bit – A bit of information that indicates whether the data in a block is valid (1) or not (0). 2 Locating data in the cache.
How to calculate the set field of cache?
So Block offset field can be represented by 6 bits. Number of lines in the cache = C a c h e S i z e B l o c k S i z e =Number of blocks = 32 K blocks = 2 5 + 10 = 2 15 blocks. Number of sets in the cache = N u m b e r o f L i n e s S e t A s s o c i a t i v i t y = 2 15 blocks / 2 blocks per set = 2 14 sets.
How to calculate cache address in Excel calculator?
Once initially calculated, input a memory access address (in decimal) and press “Add Address” to show the set index for that address You can use the set index to see if addresses overlap for determining hits and misses Made by Connick Shields.
How to calculate the number of cache blocks?
Let’s assume the system is byte addressable. Then each cache block contains 8 words* (4 bytes/word)=32=2 5 bytes, so the offset is 5 bits. The index for a direct mapped cache is the number of blocks in the cache (12 bits in this case, because 2 12 =4096.) Then the tag is all the bits that are left, as you have indicated.
How to calculate the number of lines in the cache?
Number of lines in the cache = C a c h e S i z e B l o c k S i z e =Number of blocks = 32 K blocks = 2 5 + 10 = 2 15 blocks. Number of sets in the cache = N u m b e r o f L i n e s S e t A s s o c i a t i v i t y = 2 15 blocks / 2 blocks per set = 2 14 sets.