Segmented Paging

Question

Consider a system using segmented paging architecture, the segment is divided into 32K pages and each page is having 8K entries. The segment table is divided into 16K pages and each page is having 4K entries. The memory is byte addressable and page table entry size is 16-bits. The Physical Address space is 1GB. Evaluate :

1. The Page table size of segment.
2. The Page table size of segment table.

Comments

@DSMaurya what is the ans for this question?

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 “can somebody explain the meaning of line the segment is divided into 32K pages and each page is having 8K entries”

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$The\ segment\ is\ divided\ into\ 32 K pages\ and\ each\ page\ is\ having\ 8K\ entries$

@sachin486 This line gives us the number of pages each segment is having i.e here a segment is divided into 32K pages.

Further each page is having 8K entries  tells us the size of page and that is 8K i.e 2$^{13}$ B

 

For more understanding watch https://youtu.be/kGcr7jI1f7Y

Answer 1

PageTable Size of Segment:

Segment is divided into 32K pages. No of entries in page table = 2¹⁵ 

PTE = 2 B Thus PTS = 2¹⁵ * 2 B = 64 KB

Page table size of Segment table :

Segment table is divided into 16K pages. 

No of entries in the page table of Segment table = 2¹⁴ 

Size = 2¹⁴  * 2 B = 32 KB

Comments

@rahul

for first case: we have a segment having 32k entries so it is pointing to 32 k Page table each having 8k entries

to Total PTS = 32k * 8k *PTE =512MB

same for 2nd case we have 16k segments each with 4k entries so

total PTS = 16k * 4k * PTE=128MB

don't know correct or not?

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@ cse we want to store 32K entries and each page is storing 8k right. So no. Of tables is 4. So we get 4×8K×2B im this way we get 64 kb. This should be the proper method I think.

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And if we are having 4 pages storing page tables then we must use two level paging, isn't it???