Paging: Practice Questions | L 31 | Operating System | GATE 2022 CSE

[@UnacademyComputerScience]

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[@rajbopche7992]

1. Bits in page offset = log( P ) bits ........ [ log is base 2 ] 2. No of pages in process = L / P 3. Bits for page no = log( L/P ) bits .........[ log is base 2 ] 4. No of frames in physical memory = F/P 5. Bits for frames = log( F/P ) bits ..........[ log is base 2 ] 6. Page Table Size = ( L/P ) * E bytes Am I missing anything?

[@lalishansh]

Thanks

[@itsmytechnology9285]

👍👍

[@pooja8466]

Yah your gf's birthday date 😁

[@akshat_jainn]

@@pooja8466 you're his gf😂

[@Rajesh-iv3yw]

@@pooja8466 😂

[@akshaykushwaha7003]

Start 8:05

[@rajibdutta1234]

Confused because of bits and bytes. Sir, Sometimes you convert the bytes into bits and sometimes you use bytes as it is. Please someone help me to understand.

[@rishabhchakraborty2628]

When to talk about size we consider Bytes or Words......When we take Address we consider bits.

[@rajibdutta1234]

@@rishabhchakraborty2628 Thanks.😊

[@ykku-mb6ns]

​@@rishabhchakraborty2628 yha to question ne bits or bytes alag likh rkha h

[@akankshagupta1826]

Sir, in 6th question the first line does not specifically tells about the page size. I thought it to be page number.

[@ektarajput9546]

In ques 6 it's says implement 8kb pages it's should be consider no. Of pages insted of page size

[@akshat_jainn]

You got it wrong

[@kruthikhm6379]

8kb is 8 * 2^10 bytes, how can page number be in bytes

[@PopeyejawwwdP]

@@kruthikhm6379 in 4th qustion when it said 4k pages , it was considered as no.of pages and page size of 256 ... then why 8k is considered as page size instead of no .of pages?

[@AshishSinwer-o7e]

@@PopeyejawwwdP because here byte is written after kilo(K) , check -- 8kB ....

[@pankajgautam4964]

Sir aur subjects ki class kyu nhi ho rahi h

[@soumyadeepbarik5711]

In Q3 in PAGE TABLE , we have to represent FRAME as ENTRY && number of such ENTRY = the PAGEs present in a PROCESS , hence it would be 10 BYTES instead of 5 BYTES

[@soumyadeepbarik5711]

@Shivali Agarwal Please clarify it mam .

[@pallavibhardwaj5836]

Same doubt

[@mailareppapujar6875]

Sir what is the difference between extra bits and displacement bits.

[@ABugWithTheBigDreams]

same

[@mailareppapujar6875]

Okay thanks sir.

[@codelikeapro9724]

thoda der se smjh aya pr aa gya... lecture thankyou sir for awesome lecture!!

[@sandeepshivamreddy5684]

Sir, Questions are mistaken i respect of no.of pages, pagesize in examples

[@rupamdas7232]

sir how logical address space is used as logical address ? both are different

[@takshpatel8109]

Thanks for lecture

[@kashmirimonimahanta1799]

In the first question, 1entry size is 4bytes so the page table size must be (2^5)*(2^15)... Please correct me if I am wrong

[@ashishsrivastava2417]

You was right😅

[@thugism1545]

Both are correct. Your answer is in bits, sir's answer is in bytes.

[@sauravkumar7508]

In the question 5 LAS should be 2^15 bits And PAS should be 2^19 bits na? Anyone

[@_sf_editz1870]

S bro

[@techbay5850]

Q6 is wrong there will be physical address instead of physical address space

[@techbay5850]

In question 6 physical address space is 36 bit it means physical memory or main memory size is 36 bit??

[@samaymaan8014]

Kbhi sir aap alag btate ho pysical address or physical address space ko to aaapne Q.6 me 32 bit physical address space ko aap physical address bna rhe ho physical address space to 2^19*2^13Bytes hona chahiye jaisa aapne pdaya h

[@shrunkhalawankhede2611]

this concept was little bit tough

[@funnynature1794]

Sir, In Question 1: The page size is 2KB= 2¹¹Bytes=2¹⁴ bits. How can yoy divide bytes to bits. The question 1 is wrong. Can anybody tell me if I am right or not. @unacademy computer science, please tell if I am right or not.

[@utkarshajoshi1511]

samee doubt... apko answer mila??

[@ykku-mb6ns]

​@@utkarshajoshi1511same doubt chutiya bna rha h gnta ka experience h rtta maar ke aata h

[@thugism1545]

in every bit of address one byte is stored. that's why.

[@baizeofficial484]

correction : in question 4, d is 11 because in question 256 bytes are given instead of bits nd 256 byte require 11 bits to represent in memory

[@nobody6150]

sometimes use your calc bro

[@hemanth992]

assume question is byte addressable

[@swayamsharma7614]

homeWork --> 1) log(p) 2) L/P 3) log(L/P) 4) F/P 5) F-log(p) 6) (L/P)*E Correct me if i'm wrong :)

[@Chavan-zg8wb]

Bro 5th one If no of frames = F/P Then bits per each frame = log(F/P) base 2 = log(F)-log(P)

[@ayushsinghrathore360]

🥺

[@soumyadeepbarik5711]

It would be 36 in QNO 6 as byte no +page no = 13+23

[@UnacademyComputerScience]

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[@sixth_sense_working]

4k pages should no be equal to 4000 pages how it can be 2^12 bits

[@tatu_indranil]

Here whenever there is a K take it as 1024 and not 1000 then you'll get corrected answer

[@parakhindustries]

.

[@rudranshraghav319]

Rajput samrat Mihir Bhoj Pratihaar

[@rudranshraghav319]

Jai Rajputana