Paging

Question

Suppose you have a computer system with a 48-bit logical address, page size of 16KB and 4 bytes per page table entry. If we have a 48MB program such that the entire program and all necessary page tables are in memory. Assume that each page table at diff level fits in a single page.How much memory is used by program, including its page tables?

why do we go for more levels of paging when one level of paging is enough for this 48Mb process the solution splits the logical  address space of 48 bits into parts ??

how do we decide that split ???

Answer 1

Number of pages for 48 MB program = $\frac{48 MB}{16 KB}$= 3K

Now if we have single level paging then split will be of 34|14

so we have $2^{32}$ entries in Page table, size of page is 16 KB here page table  will not fit in one single page so we have to break it

Now for two level paging

Number of entries in single page = $\frac{16KB}{4B}$ = 4 K so in single page we can have maximum 4 K entries

and Number of entries require for 48 MB is 3 K entries so all the entries of process will be single page table .

So outer level we will have single page table having 4 K entries and corresponding to it we wll have single inner level page table having 1 entry.

Thus for 2 level paging we will split it as 22|12|14

But again this is not possible as inner level page table does not into 1 single page.

So we divide it into 3 level paging

outer most level we will have single page table having 4 K entries

middle level will also have single page table having 4 K entry

inner level will have single level page table 1K entry

Breaking will be 10|12|12|14

Here all the page table fits in one single page hence further breaking is not required

Comments

@Tesla  bro here the size of the process is 48Mb and hence it is having 48*2^20 bytes and 48*2^20/2^14 pages  which is 3k pages and hence the next level page table should contain 3k pointers to this each of size 4 bytes which is 3k*4=12 bytes and this itself fits in one page and hence one level of paging is enough why do we go for multiple levels of paging here ???? here the logical address space is 48 bits which means we need to have a process of 256 Tb but our process is just 48 Mb is this question correct ??

is logical address space for the computer or for the process ??

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3K*4=12 KB not 12 Bytes which can be fitted into one single page

so let us say you are correct and we have single level paging, now a process of an process of 96 MB comes it will have 6K pages so page table entry size = 6K*4B=24 KB so now we cannot fit single page table in one page is 24 KB>16 KB, so it means that entire process is not supported.

which means despite of having enough logical space we are not able to utilize it, hence single level paging is not possible, when  we decide level of paging and bits required for it we take into consideration various factors not just only process size

what you are saying is with respect to single process only

in single level paging maximum file size supported is 64 MB

where as in 3 level paging we can support process of size upto 16 GB

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@ Tesla bro can u suggest me any material where we can read some content about this generalised logical address space for a machine  other than that uc texas assignment where did u read about this concept of generalised address space i feel a bit confused as this is the only problem puzzling me from many days ..  what i thoought till now is logical address space for a system is not fixed and it is adjusted according to the process as it is the space as far as the processor is concerned it depends on the size of the process alone it acceses word by word ..please suggest some material where this concept is there bro

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well there is no single source to learn paging, But this can help

Paging from galvin

Paing from stalling

and paging from NPTEL P.K biswas

follow these in same order and then solve uc texas question