Question

1. Consider a simple paging system with the following parameters: 2³² bytes of physical memory; page size of 2¹⁰ bytes; 2¹⁶ pages of logical address space.

1. 1. How many entries in the page table?
2. 2. How many bits in each page table entry? Assume each page table entry contains a valid/invalid bit.

Answer 1

A) physical memory size =2^32 bytes

page size = 2^10 bytes

logical address space : 2^16 bytes

A)

Entries in page table = no of pages in virtual memory

                                = 2^16

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B)frames in physical memory = size of physical memory/size of the page frame

                                            =2^32/2^10

                                           =2^22

no of bits in physical memory=22

bits in each page table entry when each page table entry contains a valid/invalid bit. =22+1

=23