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
