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: