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In a $64$- bit machine, with $2$ GB RAM, and $8$ KB page size, how many entries will be there in the page table if its is inverted?

1. $2^{18}$
2. $2^{20}$
3. $2^{33}$
4. $2^{51}$
| 2.8k views

RAM 2 GB =231 B

Page size = 8 KB = 213 B

Number of Entries in inverted page table = 231 / 213 =218

by Veteran (119k points)
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@shivani

This is the reason why we are using inverted page table .

For 64 bit, we have 264 / 8kb = 251 single page table entries , so the size of page table will be for 4 bytes entry size is 253 . A very big page table to store per process .

Hence , we use inverted page table to store entry per physical frame which are 218 and take each entry to be 10 bytes suppose = 24

Then size of page table will be approx 2 MB.which is very less as compared to previous.

In case of inverted, entry size is upto 16 bytes and for simple paging it is upto 4 bytes.

+1
yes tnks @Kapilp
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@kapil bro "In case of inverted, entry size is upto 16 bytes and for simple paging it is upto 4 bytes" this is fixed ?? or u assume it
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@kapil - Can you please explain how you got entry size as 4B for simple paging and 16B for inverted paging?

Inverted page table is a global page table maintained by the operating system for all the processes. There is just one page table in the entire system, implying that additional information needs to be stored in the page table to identify page table entries corresponding to each process.

here no of pages are possible  = 231 / 213 =218

for ex by Active (4.9k points)