Contents
- 1 How many bits are in the tag set and offset fields?
- 2 How many bits does a tag field block field and field?
- 3 How are tag fields calculated?
- 4 How many bits are needed to represent the tag set and word fields?
- 5 How to calculate the number of tag, index and offset bits of different?
- 6 How many bits are used for the tag and block fields?
How many bits are in the tag set and offset fields?
Since there are 16 bytes in a cache block, the OFFSET field must contain 4 bits (24 = 16). To determine the number of bits in the SET field, we need to determine the number of sets. Each set contains 2 cache blocks (2-way associative) so a set contains 32 bytes.
How many bits does a tag field block field and field?
Therefore 8 bits are needed to identify the set number. For the 64-Mbyte main memory, a 26-bit address is needed. Main memory consists of 64-Mbyte/16 bytes = 222 blocks. Therefore, the set plus tag lengths must be 22 bits, so the tag length is 14 bits and the word field length is 4 bits.
How many bits are used for the tag?
Of those 7 bits, 2 bits are needed for the block offset, 2 bits for the set index, and 3 bits for the tag.
How many bits are used for the offset?
14 bits
The offset field requires 14 bits to address 16 KB.
How are tag fields calculated?
You need 6 bits for the offset within a block. You need 10 bits to identify one of the 1,024 possible blocks in the cache. That’s 16 bits in total. Therefore the tag needs to be 32 bits – 16 bits = 16 bits.
How many bits are needed to represent the tag set and word fields?
The main memory consists of 16,384 blocks and each block contains 256 eight bit words. How many bits are required for addressing the main memory? How many bits are needed to represent the TAG, SET and WORD fields?…GO Book for GATECSE 2022.
| tags | tag:apple |
|---|---|
| force match | +apple |
| views | views:100 |
| score | score:10 |
| answers | answers:2 |
How many bits are required for page number and how many for offset?
There’s a virtual address which is generated by the CPU and consists of Virtual page number (20 bits) and page offset (12 bits).
How is offset page calculated?
First get page offset by calculating log2(page size in bytes). In your example, page size is 16 KBytes, so log2(16*2^10) is 14; that is, page offset is 14 bits.
How to calculate the number of tag, index and offset bits of different?
Then you work from the bottom up. 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.)
How many bits are used for the tag and block fields?
(Whereas under set association, the tag is used to search through a set of blocks for a hit: the set elements are associated with addresses via the tag field. The cache has four blocks, because it holds eight words, but pairs of words are considered blocks. So the set/block part of the address requires two bits.
How are tag, block, and offset fields used?
So no set -> block associative lookup is required, and the set field of the address can be called a block field. This field directly determines the block to which the address maps, hence the “direct mappped” designation. The tag is used to determine whether a given block in the cache is a “hit” for the address, or holds data for some other address.
How to determine the number of bits in the set field?
To determine the number of bits in the SET field, we need to determine the number of sets. Each set contains 2 cache blocks (2-way associative) so a set contains 32 bytes. There are 32KB bytes in the entire cache, so there are 32KB/32B = 1K sets. Thus the set field contains 10 bits (2 10 = 1K).