Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below:The page size is 4 KB(1 KB = 210 bytes ) and page table entry size at every level is 8 bytes. A process P is currently using 2 GB(1 GB = 230 bytes) virtual memory which is mapped to 2 GB of physical memory. The minimum amount of memory required for the page table of P across all levels is _____________ KB.a)4108b)4110c)3108d)4008Correct answer is option 'A'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below:

The page size is 4 KB(1 KB = 2¹⁰ bytes ) and page table entry size at every level is 8 bytes. A process P is currently using 2 GB(1 GB = 2³⁰ bytes) virtual memory which is mapped to 2 GB of physical memory. The minimum amount of memory required for the page table of P across all levels is _____________ KB.

a)
4108
b)
4110
c)
3108
d)
4008

Correct answer is option 'A'. Can you explain this answer?

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Most Upvoted Answer

Consider a three-level page table to translate a 39-bit virtual addres...

Given,

Virtual address (VA) = 39 bits
Page size = 4 KB
Physical address (PA) = 2 GB
Page table entry size (PTE) = 8 B
Three level pages tables with address division (9, 9, 9, 12)

Three level pages tables with address division (9, 9, 9, 12) means:

9 most significant bits for indexing into the level−1 (outer level).
9 bits for the level−2 index.
9 bits for the level−3 index.
12 bits for the offset within a page.

The entries of the level−1 page table are pointers to a level−2 page table, the entries of the level−2 page table are pointers to a level−3 page table, and the entries of the level−3 page table is PTEs that contain actual frame number where our desired word resides.

9 bits for a level means 2⁹ entries in the one-page table of that level.

For our process P:

P is using 2 GB of its VM. The rest of its VM is unused.

2 GB VM will have 2 GB/4 KB = 2¹⁹ Pages.

But level −3 page table has only 2⁹ entries. So, one-page table of level −3 can point to 2⁹ pages of VM only. So, we need 2¹⁰ level−3 page tables of process P.

So, at level−3, we have 2¹⁰page tables. So, we need 2¹⁰ entries in level−2. But level −2 page table has only 2⁹entries, so, one-page table of level −2 can only point to 2⁹ page tables of level−3. So, we need 2 level−2 page tables.

So, we need 1 level−1 page table to point to level−2 page tables.

So, for process P, we need only 1 level−1 page table, 2 level−2 page tables, and 2¹⁰ level−3 page tables.

Note that, all the page tables, at every level, have the same size which is 2⁹ × 8 B = 2¹² B = 4 KB (because every page table at every level has 2⁹ entries and page table entry size at every level is 8 B ).

So, in total, we need 1 + 2 + 2¹⁰ page tables (1 level−1,2 level−2,2¹⁰ level−3, and each page table size is 4 KB.

So, total page tables size,

= 1027 × 4 KB

= 4108 KB

Hence, the correct option is (A).

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Community Answer

Consider a three-level page table to translate a 39-bit virtual addres...

Given,

Virtual address (VA) = 39 bits
Page size = 4 KB
Physical address (PA) = 2 GB
Page table entry size (PTE) = 8 B
Three level pages tables with address division (9, 9, 9, 12)

Three level pages tables with address division (9, 9, 9, 12) means:

9 most significant bits for indexing into the level−1 (outer level).
9 bits for the level−2 index.
9 bits for the level−3 index.
12 bits for the offset within a page.

9 bits for a level means 2⁹ entries in the one-page table of that level.

For our process P:

P is using 2 GB of its VM. The rest of its VM is unused.

2 GB VM will have 2 GB/4 KB = 2¹⁹ Pages.

But level −3 page table has only 2⁹ entries. So, one-page table of level −3 can point to 2⁹ pages of VM only. So, we need 2¹⁰ level−3 page tables of process P.

So, we need 1 level−1 page table to point to level−2 page tables.

So, for process P, we need only 1 level−1 page table, 2 level−2 page tables, and 2¹⁰ level−3 page tables.

So, in total, we need 1 + 2 + 2¹⁰ page tables (1 level−1,2 level−2,2¹⁰ level−3, and each page table size is 4 KB.

So, total page tables size,

= 1027 × 4 KB

= 4108 KB

Hence, the correct option is (A).

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Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below:The page size is 4 KB(1 KB = 210 bytes ) and page table entry size at every level is 8 bytes. A process P is currently using 2 GB(1 GB = 230 bytes) virtual memory which is mapped to 2 GB of physical memory. The minimum amount of memory required for the page table of P across all levels is _____________ KB.a)4108b)4110c)3108d)4008Correct answer is option 'A'. Can you explain this answer?

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