GATE CS 2021 Set 1 — Question 32
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Computer Organization & Architecture → Memory Hierarchy & Cache → Cache Numericals
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Question
Consider a computer system with a byte-addressable primary memory of size bytes. Assume the computer system has a direct-mapped cache of size 32 KB ( bytes), and each cache block is of size 64 bytes.
The size of the tag field is __________ bits.
The size of the tag field is __________ bits.
Correct answer
17 to 17
Solution
To find the size of the tag field in a direct-mapped cache, we first determine the total number of bits in the memory address and how they are partitioned.
In a direct-mapped cache, the number of index bits is bits.
1.Total Address Bits: The primary memory is bytes and is byte-addressable. Therefore, the physical address size is bits.
2.Block Offset Bits: The cache block size is 64 bytes, which is bytes. The number of bits required for the block offset is bits.
3.Index Bits: The total cache size is 32 KB, which is bytes.
The number of blocks (lines) in the cache is:In a direct-mapped cache, the number of index bits is bits.
4.Tag Bits: The physical address is divided into [Tag | Index | Offset].
Therefore, the size of the tag field is 17 bits.Continue learning with Success Tracker
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