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Consider a byte addressable virtual memory system with 34-bit addresses where the first 23 bits are used as a page number, and the last 11 bits is the offset. Suppose the system using two-level paging, with first n bits of the address used as an index into the first-level page table. Assume that a typical program uses 16MB memory. An expression in terms of n that gives the number of second level page-tables used by the typical program is given by


2^(34−n−11) x 2^11



2^24 /2^(34−n−11)



2^24 / 2^(34−n)



2^(n−11) x 2^11
asked in Operating System by (311 points) | 151 views

1 Answer

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Consider the division |n|23-n|11|(Keep this in mind.)

The number of bits that have been used in second level page table for indexing pages of the program = 23-n

The size of the program covered by one frame of second level page table = 2^(23-n) * 2^11Bytes

The number of frames required to cover 16 MB of the program = 16MB / 2^(23-n)*2^11

answered by Active (1.6k points)

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