Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the

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Question

Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the processor generates 20-bit memory address, and the cache block size is 16 bytes. If the cache uses direct mapping, how many bits will be required to store all the tag values?

Accepted Answer

Correct Answer: (A) 6 × 210Concept — Direct-Mapped Cache Address Partition: In direct-mapped cache, each memory address is partitioned into three fields:| Tag | Index (Cache line number) | Block Offset |Step 1 — Block offset bits:Block size = 16 bytes = 2⁴ bytes → Block offset = 4 bits.Step 2 — Number of cache lines (index bits):Cache size = 16 KB = 214 bytes. Number of cache lines = Cache size / Block size = 214 / 24 = 210 = 1024 lines → Index = 10 bits.Step 3 — Tag bits:Total address bits = 20 (given). Tag bits = Total − Index − Offset = 20 − 10 − 4 = 6 bits.Step 4 — Total bits to store all tag values:Each cache line stores one tag. Number of cache lines = 210. Each tag is 6 bits.Total tag storage = 210 lines × 6 bits/line = 6 × 210 bitsWhy other options are wrong:Option B (8 × 210): Would imply 8 tag bits — incorrect; tag = 6 bits.Option C (212 = 4096): Does not match 6 × 1024 = 6144.Option D (214): Equals the number of bytes in cache, not the tag storage.

Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the processor generates 20-bit memory address, and the cache block size is 16 bytes. If the cache uses direct mapping, how many bits will be required to store all the tag values?

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Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the processor generates 20-bit memory address, and the cache block size is 16 bytes. If the cache uses direct mapping, how many bits will be required to store all the tag values?

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