You have explained in such a way that we don't need to memorize anything. The logical progression of this lecture is phenomenal. Dil Se Shukriya Bhaiya
Most of the time , i don't skip aid after 4 second when youtube allow me to skip , i just let it run so that i can pay fee indirectly to this great content creator..love this series ❤️
Sir you have worked really hard to make best and easily understandable note. I have never seen notes which I can understand just by reading by the way I finished this whole playlist in just 2 days.
@@lakshaykumar7518 i'm a new student sir, i give my heartiest thanks to you. OS was always a favourite subject, but my college professor is rude enough for me to not understand a single thing. but from your lectures, i'm close to completing OS (placement-wise) within one month.
Lakho bacho present future me OS ka problem solve ho gaya.. Best content of OS in any platform any professor .... 🔥🔥🔥🔥 Bhaiya DBMS ka bhi kuchh karo.. Bahut problem hai usme kuchh smjh nhi aa rha.. 🙏🙏
If the size of the logical address space is 2^m, and a page size is 2^n bytes, then the high-order m − n bits of a logical address designate the page number, and the n low-order bits designate the page offset.
What if the process size is of 24KB, then first page(for 0-16KB) will completely occupy the frame, but the second page(17-24KB) will occuply half frame, and when the internal fragmentation of some number of such pages is combined, will it not cause external fragmentation? As the space will be empty and the total empty space will also be enough to accomodate a process of 16 KB.
I'm a little confused between the thread and the page. what is the difference b/w pages and threads? if threads and pages are different so can we also divide threads into pages. By the way Thank you so much for your efforts. it really matters a lot.
the usage of bytes is very much wrong here, this created a lot of confusion as we learnt that 1 byte=8 bits and then all of sudden you write 64 bytes will require 6 bits, it should be , the 64byte addressable memory will require 6 bits to represent the logical address space
yaa. actually he mugged up theree. what he meant to say was , to represent 64 addresses in the logical address we need 6 bits, because 2^6 is 64 , so we can represent at max 64 byte addressable space in 6 bits
Heyy guys !! I have made complete syllabus Operating System handwritten notes, I am selling it at a very low price. Does anyone want to buy it ? I got A+ in my exams by studying from these notes so you can rely on them. Let me know
You have explained in such a way that we don't need to memorize anything. The logical progression of this lecture is phenomenal. Dil Se Shukriya Bhaiya
Thankyou bhaiya Roz aap hmare liye itni mehnat krte ho ❤️
You have explained in such a way that we don't need to memorize anything. The logical progression of this lecture is phenomenal. Dil Se Shukriya Bhaiya
But we will surely have to revise to remember all the information clearly.
Most of the time , i don't skip aid after 4 second when youtube allow me to skip , i just let it run so that i can pay fee indirectly to this great content creator..love this series ❤️
bro use adblocker to reduce ads and watch lecture distraction free
I use youtube vanced so can't do that 😢 . And also sponsored is auto skip
Attendance marked
Lecture watching with notes 😀
Sir you have worked really hard to make best and easily understandable note. I have never seen notes which I can understand just by reading by the way I finished this whole playlist in just 2 days.
this lecture was 🔥... mazza aa gya bhaiya
thank you so much for all the effort, It is HIGHLY appreciated
Thanks buddy!
Keep learning 😊
@@lakshaykumar7518 i'm a new student sir, i give my heartiest thanks to you. OS was always a favourite subject, but my college professor is rude enough for me to not understand a single thing. but from your lectures, i'm close to completing OS (placement-wise) within one month.
division of process into pages
page : process ka part
page size == Frame size {main memory ka segment} [alloted based on archietecture ]
Thank you so much bhaiya you are seriously weapon for us ❤ thanks you for helping .....
Your support are enough for us
Thank you so much bhaiya😄 excited for live session 😇😊
this playlist is very very helpful..... enjoying too much without spend any money.
aag laga diye bhaiyaji
Present Bhaiya. OS ki course ki consistency bahut sahi hai
Lakho bacho present future me OS ka problem solve ho gaya..
Best content of OS in any platform any professor ....
🔥🔥🔥🔥
Bhaiya DBMS ka bhi kuchh karo..
Bahut problem hai usme kuchh smjh nhi aa rha..
🙏🙏
Thanks buddy!
Keep learning 😊
Sir your videos are very good and clear crystall and i got like 23 out 24 through your videos only👍👍👍
Thank you soo much bhaiya..pehli baar paging puraa samjh aaya hai🙏🏻
such a great in depth for this topic . wow. intresting
Thankyou bhaiya Roz aap hmare liye itni mehnat krte ho
kaafi In-Depth lecture tha.... Hope these questions are important
theyre not
@@pranav288 so what are the important ones..please emphasize a little bit
paging is important ....it came in my interview from foxconn
Love babbar bhaiya and os vale bhaiya.
Dil se thank you ❤
naam to bol do laskhay h
QUALITY OF CONTENT 🔥🔥
If the size of the logical address space is 2^m, and a page size is 2^n bytes, then the high-order m − n bits of a logical address designate the page
number, and the n low-order bits designate the page offset.
Boht acha smghaye sir
Amazing explanation🔥
Thank you!!!
we can easily understand your lecture thank you so much 💯💯💯💯💯💯💯💯💯
you are better then apna college wali
Bahi Dil se thank you.
thank you for the best explanation
Thanks a lot. Amazing lectures.
What if the process size is of 24KB, then first page(for 0-16KB) will completely occupy the frame, but the second page(17-24KB) will occuply half frame, and when the internal fragmentation of some number of such pages is combined, will it not cause external fragmentation? As the space will be empty and the total empty space will also be enough to accomodate a process of 16 KB.
Super video Sir
Thanks bhaiya for your superb lectures
Day 26 Done ✓
Great Work!
Maine Video Pura dekha
Thankyou so much bhaiya
Awesome video 🔥
TLB is simply works like DP.
mazza aagaya bhaiya
consistency++
Thanks for this informative lesson
Thank you for your hard work❤
Always superb ☺️
Very hi nice explanation.😊
mza aa gya bhai shahb😉😉😉😉😉
To represent 64 bytes, we require 6*3=18 bits.
Bhaiya jee, Kab Linux kernel ko explain kere GE, I am very excited,
I'm a little confused between the thread and the page. what is the difference b/w pages and threads? if threads and pages are different so can we also divide threads into pages.
By the way Thank you so much for your efforts. it really matters a lot.
Bhaiya Computer Networks ka playlist be lao, please.
Thank you guru ji ❤️
logical addresh ma frame kitni bit ka hona chya ya explain kra do?
Memoization is used in case of TLB
Fabulous
Thanks
You are awesome dude !!
Does in memory 1 byte is represented through 1 bit?
the usage of bytes is very much wrong here, this created a lot of confusion as we learnt that
1 byte=8 bits and then all of sudden you write 64 bytes will require 6 bits, it should be , the 64byte addressable memory will require 6 bits to represent the logical address space
tu hi pdhale na fir lodu
great
Bhaiya moj kr dii
Thank you sir😊
Dil se thanks !
06/07/24 paging tlb cleared++ Lakshay bhaiyaa
Thanku bhaiya
WoW . . .
Bhaiya DSA ki videos constant dallo plzz
How are we fixing the 2 bits for page number, what if a process has more than 4 pages?
Sir, I don't understand how paging is a non-contiguous memory allocation if the size of the page is fixed.
Bro because we are adjusting data in RAM non- contiguously..
the bit representation of 25 is 11001 but you had written 011001
Simply bol do ki TLB is just like DP array 😂
How many videos are left in DSA course??
correction
addresses should be 0-127
and 0-63
❤❤❤❤❤
how did we convert 01 to 111 in frame 7 at 21:19
Love you
Approximately how many videos are more on the way until the completion of this course ?
4-5 videos!
Bhiya plz continue DSA placement course
bhaiya ne dp ka real life example dikhlaya hai #tlb
nice observation
attendance marked
Reach++
20 din baad koi nya person aane wala tha, koi nyi series start honey wali thi but it's been more than 20 days but still no update regarding that.
Day 26 done... 07/06/24
Bhaiya DSA waali playlist kaafi slow ho gyi hai
buiteefuul..
Day19
Sir 25 kaise Aya please
Plzzzzz make subtitles available plzzzzzzz🙏🙏🙏🙏🙏🙏🙏
I am excited giveaway participate and supported coad help by Babar
64 byte is 512 bits then how do we get 6 bit?? @16:42 Confusing
yaa. actually he mugged up theree. what he meant to say was , to represent 64 addresses in the logical address we need 6 bits, because 2^6 is 64 , so we can represent at max 64 byte addressable space in 6 bits
To watch add free content first download it amd watch it undisturbed
Bhaiya c++ ka video dalo nah pls bhaiya
C++ playlist plzz
op
bump
Heyy guys !! I have made complete syllabus Operating System handwritten notes, I am selling it at a very low price. Does anyone want to buy it ?
I got A+ in my exams by studying from these notes so you can rely on them. Let me know
What does over head means?
naukri dilwado sir
apse hi padha hai sab
IS PURE SERIES KA SBSE KHRAB VIDEO HAI
You have explained in such a way that we don't need to memorize anything. The logical progression of this lecture is phenomenal. Dil Se Shukriya Bhaiya
Present Bhaiya. OS ki course ki consistency bahut sahi hai
Thank you so much bhaiya you are seriously weapon for us ❤ thanks you for helping .....
Your support are enough for us ❤
Thanks buddy!
Keep learning 😊