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Consider a system with logical address space of 4GB and page size is 8KB, physical address space is 256MB. How many bits required to access an entry in single level page table and in inverted page table respectively?

 

A. 19, 15

B. 15, 15

C. 15, 19

D. 19, 19

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Given:

Logical address space =4GB =232        B

Physical address space =256MB =228 B

Page size is 8KB =213  B   (=    Frame size )

Solution:

#pages in LAS=232-13=219

#frames in PAS=228-13=215

Answer Option A :19,15

No.of entries in single level page table is no. of pages & No.of entries in inverted page table is no. of frames.

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