GATE CS 2025 Set 1 — Question 36
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Computer Organization & Architecture → Memory Hierarchy & Cache → Direct-Mapped Cache
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Question
Consider a memory system with bytes of main memory and bytes of cache memory. Assume that the processor generates -bit memory address, and the cache block size is bytes. If the cache uses direct mapping, how many bits will be required to store all the values? [Assume memory is byte addressable, , .]
Correct answer
(A) 6 × 2¹⁰
Solution
1.Identify Address Components: In a direct-mapped cache, the memory address is divided into three parts: Tag, Index, and Offset.
2.Calculate Offset Bits: The block size is bytes. Since the memory is byte-addressable, the number of offset bits is bits.
3.Calculate Index Bits:
- Cache size = bytes.
- Number of blocks in cache = blocks.
- Number of index bits = bits.
- Total address bits = bits.
- Tag bits = Total bits - (Index bits + Offset bits) = bits.
- Each block in the cache has one tag entry.
- Total bits for all tag values = Number of blocks Tag bits per block = .
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