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Question: Consider a system using  a segmented paging architecture. The segment is divided into 8k pages each of size 2k words . The segment table is divided into 256 k pages each of size 512 words. The page table entry size requires 64 bits . The frame number requires 22 bits then calculate logical address (LA). Note: 1 Word = 1 Byte

Options Are:

 

         A. 51 bits
         B. 61 bits
         C. 33 bits
         D. 64 bits  

 

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Correct Answer : (A) 51 bits

This question is using both segmented paging and paged segmentation. 

Segmented Paging – Perform paging on each segment as the segment size may be too large, i.e. offset within the segment is divided into two parts ==> page identity and page offset.

Paged Segmentation – Perform paging on segment table as the segment table size may be too large, i.e. segment identity is divided into two parts ==> page identity and page offset.

Now, lets look at how the logical address will look in each case differently:

Now, in the question we are given following:

  • “The segment is divided into 8k pages each of size 2k words” – This refers to Segmented Paging.
  • “The segment table is divided into 256 k pages each of size 512 words” – This refers to Paged Segmentation.

Now, we are given that each segment is divided into 8K pages, so page identity(P2) = 13 and size of each page is 2K words, so offset within page(d2) = 11

Also, we are given that segment table is divided into 256K pages, so page identity(P1) = 18 and size of each page is 512 words, so offset within page(d1) = 9

Now, the structure of Logical address will look as follows:

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