MadeEasy Test Series: Operating System - Memory Management

Question

Consider a system using segmented paging architecture.The segment is divided into 1K pages.Each page having 128 entries.Segment table is divided into 512 pages and each page having 1024 entries.The size of page table entries are 2 words.If 15 bits are required to represent the frames of physical memory,the size of page table of segment table is ___________(in words).[assume memory is word addressable]

The answer given is 1024 while I am getting 2048.

Can anyone provide a detailed explanation if I am wrong?

Answer 1

Given Word Addressable (means every word can be identified uniquely)

given :segment table is divided into 512 pages and each page contain 1024 entries

but each page table length is (2word) therefore 2^9 indexing must be there to identify each word uniquely

therefore no of bits for indexing page table is 9 means 2^9=512

and 512 *2words =1024 words is the size of page table

 

 

 

may be ,i too wrong ,but if u find the reason please share

Comments

According to me since there are 1K pages in a segment,so the no. of entries in the page table will be 1K.The page table entry is 2 words length so the size of the page table (because the question asks for the size of the page table of the segment table)  will be 1024*2=2048.

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According to me as well answer should be 2048