GATE CSE 1997 | Question: 3.10, ISRO2008-57, ISRO2015-64

Question

Dirty bit for a page in a page table

• A. helps avoid unnecessary writes on a paging device
• B. helps maintain LRU information
• C. allows only read on a page
• D. None of the above

Comments

Source: Galvin 9th edition, chapter 9(9.4.1)

---

please if someone, can we relate this with the ditry page that is generated when a write happens in the write back cache policy.

---

yes, but in write back we are writing in the main memory when using write back cache, but here the write is for frames that are being written in the hard disk as part of demand paging, so the destination is diff but the concept same, please let me know if I'm wrong anywhere

Answer 1

The dirty bit allows for a performance optimization. A page on disk that is paged in to physical memory, then read from, and subsequently paged out again does not need to be written back to disk, since the page hasn't changed. However, if the page was written to after it's paged in, its dirty bit will be set, indicating that the page must be written back to the backing store answer: (A)

Comments

Yes, in a virtual memory system, paging is the act of transferring pages between physical memory and backing store (usually disk).

---

I am not quite comfortable with the phrase ‘paged out’ in the above answer. Moreover, the question doesn’t seem to say anything about the presence/absence of virtual memory in the concerned system, so we don’t know whether swap space is present or not in the same. Nevertheless, let’s assume that swap space is indeed present in the system, and proceed with our arguments.

If the dirty bit is not set, then that means that the corresponding page in memory is an exact copy of the page on disk, and in which case we can simply overwrite that page with some other page currently required. The next time the former page is needed, we load it from its original disk location.

If the dirty bit is set, then I believe the system should behave as follows, though I found no supporting resources/evidences/counterevidences (if you find any, kindly post it down below) – then the page is swapped out onto the swap space, and reloaded from the same as per need.

That is how I believe we can argue as to why option \((A)\) is correct.

---

A paging device is the storage device used to hold pages when they are not in main memory—typically the disk/SSD (swap space or backing store).

Dirty bit indicates whether a page has been modified after being loaded into memory. If the dirty bit is 0, the page need not be written back to the paging device during replacement. Hence, the dirty bit helps avoid unnecessary writes on the paging device.

Answer 2

ans is A...to perform any write operation on page firstly we check the dirty bit status...dirty bit will be set when page will be modified(suppose s=1 set condition) and it will remain s=1(set) until the page modified will write back into the memory..if page was not modified then it will not change the dirty bit value i.e.s=0...now  suppose we perform write operation on page then its dirty bit become s=1..and suppose after some time we were performing write operation on that page then firstly we check the dirty bit value here its value is 1 it means page was modified..so firstly we write that page to memory and then perform any modification by this way we were avoiding unnecessary write operation on paging device

Comments

You mentioned

dirty bit value here its value is 1 it means page was modified..so firstly we write that page to memory and then perform any modification by this way we were avoiding unnecessary write operation on paging device

If we need to perform 100 writes on a page then every time we will write it to memory is wastage. Although your answer is correct but i think we can perform write on the page as many times as we want.Its just when we are about to replace page we will check if dirty bit=1,then write to memory back else dont.If every time before write i check dirty bit=1,then i write page to memory and then do any modification then how is it avoding unnecessary write to memory?

---

true, when we do replacement only, so that the final modifications reflected back in memory.

Answer 3

Suppose if we don't have "Dirty but concept" then whenever page fault occurs, the page has to be written back first in secondary memory in order to assure consistency ( even though the page may or may not be modified ).

Now if we have "Dirty bit concept" then by examining it's status it can be concluded whether the page should be written back in secondary memory or not whenever page fault occurs. In this manner "Dirty bit helps avoid unnecessary writes on paging device"

So (A) is correct 

Answer 4

Option A: Its right. Dirty bit helps us to identify whether the victim page which will be replaced has been modified or not after loading it to the Physical Memory From Hard Disk. If it has been modified, then we need to write the page to Hard Disk to ensure in future due to some mishaps, data doesn't gets lost. But if the page hasn't been modified since its transfer to the Physical Memory, no need to write it to the Hard Disk. Thus it helps to "avoid unnecessary writes"

Option B: LRU doesn't need Dirty Bit. Dirty Bit doesn't give us any info as to when was the last time a page was used, which is a crucial info for LRU. That's why this is wrong.

Option C: Dirty Bit hasn't got any control as far as reading from the page or writing to the page is concerned. So, this is wrong.