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If the page size is 512 bytes and the paging table has 64 elements, each 11 bit. Howmany bits is the physical address then? - What does '11 bit' mean in this case? Is it the number of the frame (2^11) at the physical address or something else? It's for my exam tomorrow.

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A page of 512bytes requires 9 bits for intra page addressing (assuming addressing is in byte units). An element in the paging table is presumable a page address, which makes 11bits mean there are 2048 such pages possible. Assuming all these assumptions are correct for your case, and the pages are non-overlapping - there should be a total of 20 bits for the address (11+9). Which makes the 64 elements an irrelevant parameter for your question.

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Thank u so so much. :) – user66900 Feb 11 '11 at 14:06

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