Q1) Given a memory of 16k and a cache memory of 1024 bytes with block size 128 bytes. The system uses Direct mapping. A- How many blocks available in both physical memory and cache memory  B- How the address will be split to indicate tag, line (block no) and offset C- Calculate the cache line number and offset in cache that will contain the content of memory address 230.

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A:

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A: Solution ;   Main memory = 256MB Main memory in bits = log2(256M) = log2(228) = 28 bits

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Q: 5. Find the Cache line number in Direct Cache Mapping, if full memory address: 16FFFC and Cache is…

A: Given:

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Q: 2. 2. Suppose a computer using direct-mapped cache has 2³2 bytes of byte-addressable main memory and…

A: Given Data Main Memory Size = 232 Bytes Number of Cache Blocks = 1024 = 210 Block Size = 32 Bytes =…

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