How are tag bits calculated in cache?

How are tag bits calculated in cache?

The remaining bits are used for the tag. If ℓ is the length of the address (in bits), then the number of tag bits is t = ℓ − b − s. If the requested address is not found in the cache, then it will be brought in from memory and placed there.

How do you calculate memory tag size?

If memory is byte addressable and addresses are 16 bits then how many bytes are used for the tag? We have 16-3 = 13 bits left for the tag and word. Because the question says that memory is byte addressable, I think that this means that the word is 8 bits (= 1 byte) long, and thus the tag is 16-3-8 = 5 bits long.

How do I find my tag and index bits?

1 Answer

1. Calculate the size of each address in m bits. If main memory has 2048 bytes, then we have 2048=2^m unique addresses.
2. Calculate bit offset n from the number of bytes in a block. 64 bytes/8 blocks = 8 bytes per block.
3. Calculate the set index s.
4. Finally, we know the number of tag bits is T=m-s-n.

How do you calculate the number of index bits?

Number of index bits Since log2(1) is 0, there are no index bits. For Direct Mapped, CS is equal to CL, the number of cache lines, so the number of index bits is log2(CS) === log2(CL). For n-way Associative CS = CL ÷ n: log2(CL ÷ n) index bits.

What is the total cache size?

The “size” of the cache is the amount of main memory data it can hold. This size can be calculated as the number of bytes stored in each data block times the number of blocks stored in the cache.

How to calculate the size of a cache tag?

As it is 4 way set associative, each set contains 4 blocks, number of sets in a cache is : (2^5)/2^2 = 2^3 sets are there. so from these we got to know that 3 bits are required for adressing set offset. As it is byte addressable , to address a byte offset we require 7 bits as we have 128 (2^7) bytes in each block.

How to calculate the number of tag and index fields?

1) A direct-mapped cache with 4096 blocks/lines in which each block has 8 32-bit words. How many bits are needed for the tag and index fields, assuming a 32-bit address?

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 tag 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. As the cache gets more associative but stays the same size there are fewer index bits and more tag bits.