L-5.11: Question Explanation on Paging | Memory Management | Operating System

This video contains Question Explanation on Paging.

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[priyankachatterjee]

If i can then I will give you likes. You are great sir.

[abhishekmohite]

- page table
- pages are stored in secondary memory
- frames are stored in ram, during processing
- offset no. is place from left, needed bytedata is stored in a page row
- logical address space (las) [ just a table of pointers ]
- local process level addressing for data storage, acc. to data stored in secondary memory
- contains in an entry :
page no. (in msb)
offset/bit no. (in lsb)
- physical address space (pas) [ just a table of pointers ]
- memory level addressing for data storage
- contains in an entry :
frame no. (in msb)
offset/bit no. (in lsb)
- page table maps page no. of LAS to frame no. of PAS for use by cpu

[kaushiksai]

The amount of effort you are putting to make the viewers understand the difference between the number of bits to represent the size and to calculate the size is outstanding brother...!! Awesome work..!!

[sakshidwivedi]

I like the way of your teaching...after years I have found a place where I have acquired good knowledge otherwise I was far from online teaching which different coaching institutes provide...They just keep looking at live chats and are not serious...but here you give a best example of online teaching...Keep it up...Thank you

[willturner]

Sir, Your channel is like a gift 🎁 in these lockdown days. Thank you sir🤴

[ShivAdiShakti]

Excellent Sir. We need teachers like you. Thank you so much. Please keep posting videos for NTA UGC NET COMPUTER SCIENCE. Your Videos are really helpful.

[aayansh]

12:20 I think only 3 bits are required to represent the value of pages in each page ie 263 = 8 bytes is the size of each page and not 8 bits to represent each page value.

[aluminakhan]

14 minute ki video hai, men ny sirf 3 minute ki hi suni hai aur like kr dia, bht acha smjhaty hn sir v.v .v.good excellent sir.

[lakshitasaini]

Awesome Explanation..I am obsessed with your teaching Style..Thank You So Much Sir..You are Great❤️

[poojachauhan]

Thank you sir for sharing your knowledge with us... You make these questions so simple...

[thedramaape-tizer]

Sir, in the 2nd question, at 12:27, isn't it 3 bits? Coz each frame is represented using 3 bits right? So the size of page table must be 16 * 3 right? Please reply sir.

[asktostranger]

Size of page table
Me confuse ho gayaa ismee
Size of page table = no of entries in page table * ( bits required to represent frame no)
Is it correct tell me??

[pratikbhatt]

Very Good and neat explanation. Keep it up like this (y)

[nabeelaashraf]

I have a confusion plz solve...
Size of a page table=16*3 isn't it? Because 3 bits are required to represent a frame as you said in your last video.

[poonamswarnkar]

Sir Infinite likes for you.I took Galvin and strted self study as i am an EC student thn i found ur lectures..Thankyou so much ❤❤❤❤❤❤❤❤❤❤Lots of love and respect ...Amazing Teacher .Besides teaching i found u as Wow person.

[sanawarali]

great sir g easiest method of teaching i ever seen, , love you sir from pakistan

[garimasrivastava]

Sir Please confirm the size of page table either it's (entries in page table * bits required to represent the page offset) or (entries in page table * no. of bytes can put into the page offset).

[harshgupta]

At 12:40 duration, it should be 16 X 3, because no of bits used to represent frame number = 3 and no of pages = 2^4 = 16.

[akshvirbrar]

Oh man the way you explain things and the environment you create is so good. I wish you had all the computer science courses on this channel and once again thank you for these uploads!!

[priyanshiarora]

Those who are confusing with byte addressable .E.g. If memory is byte addressable and Page size is 8 byte this means there are 8 blocks of 1byte .To represent 8 blocks we need 3 bits Lets take e.g. Logical Address is of 7 bit and memory is byte addressable, Page size is 8 byte to represent 8byte =3 bits rest 4bits represent 16 locations i.e. total pages .
.Lets suppose LA=1011011 FIRST 4 BITS ARE 1011=11TH PAGE and 011=3rd Byte which means from 8 BYTE OF A PAGE TAKE THE 4TH BYTE(if 0-7 then 3rd byte)