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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

1.

2^(34−n−11) x 2^11

 

2.

2^24 /2^(34−n−11)

 

3.

2^24 / 2^(34−n)

 

4.

2^(n−11) x 2^11
in Operating System by (333 points) | 190 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

=2^24/2^(34-n).
by Active (1.6k points)

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