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Consider a computer system with 48-bit logical address space and 2Kbytes page size. The system has 1 Gbytes of physical memory. How many entries are there in an inverted page table?

Adam Konieska
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sh H
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  • Possible duplicate of [Calculating Page Table Size](http://stackoverflow.com/questions/16323890/calculating-page-table-size) – Alex Jun 13 '16 at 11:40

2 Answers2

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Number of entries in inverted page table = physical address space / page size

Physical address space= 1GB =2^30 Page size = 2KB= 2^11

Number of entries =2 ^ (30-11) =2^19 =512K entries.

Rupsingh
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As it can be inferred from its name, if size of physical memory is 2^m frame then size of inverted page table must be 2^m. In your specific case, size (number of entries) of inverted page table is = 2^30. The inverted page table is used by all processes,which is consist of(process-id , page number, control bits ) Benefits: 1)Reduce overhead of creating page table for each process 2)nverted Page tables are sorted in order of frame number but the memory look-up takes place with respect to the virtual address, so, it usually takes a longer time to find the appropriate entry but often these page tables are implemented using hash data structures for a faster lookup. for more >>> https://www.geeksforgeeks.org/inverted-page-table-in-operating-system/

MSD
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