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if there are only 28 address lines and each carries one bit how there can be any more data? I don't read 28 address lines all told, but 28 address lines going into one specific memory device. where are the 4 address lines to carry [the 4 bits out of 32 that are used to activate the memory device]? There are two answers on the conceptual level: A> Those ...


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You got it a little mix up and wrong Page number = logical address / page size in this case is $\frac{21205}{1024} = 20$ Offset = logical address mod page size in this case is $21205 \mod 1024 = 725$ Sources: http://www.yorku.ca/pkashiya/cse1520/Paged%20memory%20technique.htm http://www2.cs.uregina.ca/~hamilton/courses/330/notes/memory/paging.html


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