Consider a logical address space of 256 pages with a 4-kb page size, mapped onto a physical memory of 64 frames. (a) how many bits are required in the logical address? (b) how many bits are required in the physical address?
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Well, we have 4 kb
1 kb needs 10 bits to be adressed, 4kb just adds 2 more bits so it's 12 bits, and we need 8 bits for each page so we need around 20 bits for the logical address, this means it's around 1 mb
64 frames just requires 6 additional bits, so, 26 bits are required for the physical address, of 64mb
1 kb needs 10 bits to be adressed, 4kb just adds 2 more bits so it's 12 bits, and we need 8 bits for each page so we need around 20 bits for the logical address, this means it's around 1 mb
64 frames just requires 6 additional bits, so, 26 bits are required for the physical address, of 64mb
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Answer:
20 bits are required in the logical address
18 bits are required in the physical address
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