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## How many bits are in a tag?

20 bits
Address format: Tag = 20 bits; Line = 6 bits; Word = 6 bits. Number of addressable units = 232 bytes; number of blocks in main memory = 226; Number of lines in cache = 26 = 64; size of tag = 20 bits.

## How many bits are in a block?

Each block hs 32 bits of data plus a tag, which is 32-14-2 bits, plus a valid bit.

## How many bits wide is the tag field?

How many bits wide is the tag field? Bits in the tag field = (32 address bits) – (4 bits to select line) – (4 bits to select word/byte) = 24 bits Page 3 D. Briefly explain the purpose of the one-bit V field associated with each cache line.

## How do you calculate the number of tag bits?

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 many bytes is a word?

2 bytes
A byte is eight bits, a word is 2 bytes (16 bits), a doubleword is 4 bytes (32 bits), and a quadword is 8 bytes (64 bits).

## What are tag bits?

The tag bits are compared with the tags of all cache lines present in selected set. If the tag matches any of the cache lines, it is a cache hit and the appropriate line is returned. If the tag doesn’t match any of the lines, then it is a cache miss and the data is requested from next level in the memory hierarchy.

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## 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.

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## How do I find my tag field?

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.

## Is a word 4 bytes?

Fundamental Data Types A byte is eight bits, a word is 2 bytes (16 bits), a doubleword is 4 bytes (32 bits), and a quadword is 8 bytes (64 bits).

## How to calculate the number of bits in the tag field?

Total number of address bits = 16 Therefore, Number of bits in the Tag field = 16 – 6 – 5 = 5. For a given 16-bit address, the 5 most significant bits, represent the Tag , the next 5 bits represent the Block , and the 6 least significant bits represent the Word .

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## How to calculate the number of tag sets?

The answer is – Each block is 32 bytes (8 words), so we need 5 offset bits to determine which byte in each block – Direct-mapped => number of sets = number of blocks = 4096 => we need 12 index bits to determine which set.

## How many bits does a block take to address the cache?

Since the cache is direct-mapped, the size of the set is 256 bits (?) = 32 byte and therefore it takes 5 bits to address the memory, so the number of bits that are not the tag is 5 (?) = 27 bits are used for the tag (?) Since a line is 2 words = 64 bits (?) = 8 byte then the offset takes 3 bytes and therefore the block takes 2 bytes (?).

## What is the correct number of bits in physical address?

= Number of bits in physical address – (Number of bits in set number + Number of bits in block offset) = 2 13 x (16 bits + 2 valid bits + 1 modified bit + 1 replacement bit) For part-01, Option (C) is correct. For part-02, Option (A) is correct.