Complex Problems on Instructions (COA) | Marathon Session | GATE CSE/IT 2021 Exam

[@pragatipriya2555]

Sir I m not attending live lecture or classroom lecture but still it's toooo much interactive.. I m questioning in mind and suddenly you start discussing same 🙏 hats off to you

[@TrendWatchersTV12]

Did u completed the course?

[@yeshitamotwani2701]

One of the Best Playlists for GATE Preparation. Thank you so much, sir. You explain all the concepts and questions in a really very easy to understand manner.

[@aditya-bl5xh]

9:20

[@harshthakur9890]

Thanks 🙏

[@moonwalkingdancer8369]

thnks bro

[@SecretEscapist]

Thanks, Brother

[@PandeyJii9632]

Best teacher ever seen. What a concept sir. I never have praised anybody in my life but u.

[@saumyapatel9523]

Aap bahot achchhe se concept explain karte h sir @_@.Aap ke motivating words aur expressions char chand lga dete h _/\_........thankyou sir

[@asktostranger8296]

One of the best teacher I watched in my entire life

[@alpishjain1317]

i understood every single question...great teacher!!

[@sarthakbharadwaj5576]

That energy level 🔥

[@varun_immadisetti]

Best session sir, Thank you.

[@Pruthvikajaykumar]

These videos are really helpful. Thank You so much

[@janapalaswathi4262]

I became a big fan of you sir..Ur teaching is amazing..thank you so much sir..🤗

[@laharibasu9731]

Great session Sir. Your energy is always that keeps me going. Keep motivating 💕.

[@ashukashyap7142]

Option D is correct for GPK

[@krishnakumar-kh5vo]

it's incorrect , it will be option number B

[@SecretEscapist]

Lovely collection of questions, I have solved all of them 🙂 Start Time: 9:20 The answer to the GPK (Question 16) is: Option (B)

[@T.Rajalakshmi]

Can u pls explain tha answer for 16th question

[@259_parthpatidar9]

amazing sir questions solve hore he ab

[@ruchithamithinti9754]

pls provide ans for last question full doubt it will be helpful

[@mozart8782]

hw question ans is B part

[@LegitGamer2345]

your energy is on point !

[@Adityasharma-oe8zp]

5th question i done on my own😍😍😍

[@mohamedshaker7528]

in Question 13 the answer is (logx + 2m + 14) becase when converting the kilo 2^10 we aslo need to convert the byte to bits ?!

[@rizwan-io]

The system is byte-addressable. If you still feel confused take m as 0 and see you have 2^10 combinations for 10 address where each address is having 1 byte

[@infinitebeauties7253]

in Q8. answer is 28 why is it so?? 28 is the bits in which adresses can be put so for no. of addreses lines 2^28 please correct me ....

[@abhishektanwar8576]

Question 9 , In this question the number of bits for memory address must be 20 bits instead of 17 as sir forgot to convert Bytes to Bits which will add 3 into the 17.

[@ydayanandareddy7283]

I think sir done the right.Because it is a byte addressable.For every address it stores 1 byte not 1 bit.

[@shivamjain5248]

Yes sir has done correctly

[@TrendWatchersTV12]

Every address in memory will store 1B and therefore 128KB/1B = Number of addresses = 128K To represent 128^K we need 17bits (2^17=128K)

[@PandeyJii9632]

We don't convert byte to bits.

[@ayushjain2559]

question no 4 mei if the memory were byte addressable then why we take 8 bits in multiplcation with 2 raised to power 11.?? please tell if anyone knows

[@akshaykushwaha7003]

Beautifully done ♥️

[@csforcommonsense3360]

Hi

[@takshpatel8109]

Thank you sir

[@ashishietlucknow7230]

yes i m ready

[@darkexodus6404]

Q 16. Ans is B

[@leviackermen7672]

can you pls give the solution..it will be very helpful

[@juhikumari5131]

Sir,can you provide notes of marathon class.

[@moonwalkingdancer8369]

why in question 8 answer is 28?? can anyone explain me this

[@fortamils]

Hi,day-3

[@AmanSingh-dy4pp]

any one watching please help me In question 4 there are 3 addresses [as instruction is of 3 addresses instruction type.] size of each address is 11 as per the question. so total no of addresses will be 2 ^11. my doubt is as there are 3 addresses total no of addresses in memory should be 3*2^11. so please clear my doubt.....I Know there would be a simple reason but i cant figure out so help anyone

[@ASHISHSINGH-kr8oy]

You are lil bit confused bro....

[@abhinavgarg85]

3 addresses instructions can store 3 different addresses from the 2^11 combinations generated by 11 bits. Watch the instructions lectures from the playlist again.

[@sandipkumarswain277]

@@abhinavgarg85 sir... As the instruction size is 38 bits then it will take 2 word to store 1 instruction address, as the word size is less than tha instruction size that is 32 bit... So the answer will be 2^11*64...

[@abhinavgarg85]

@@sandipkumarswain277 Here we can see that the no. of bits for addresses are 11. Hence total no. of addresses are 2^11 and since the system is word addressable we can store one word at all these 2^11 addresses and hence 2^11 * 32 bits = 2^11 * 4 bytes = 8KB of memory. Yes you are correct that 2 words will be required to store an instruction but how does that effect the memory of the system ? It completely depends on the no. of addresses and the size that can be stored on each address.

[@piku3632]

@@abhinavgarg85 in question 4 ,i have another doubt if memory is word addressable with 32 bit then how instruction could be of 38bit even instrustion store in memory if we store instruction in memory then each instruction will have two address then programe counter store next adrress omly then how instruction will fetch plz help anyone know the ans

[@explorewithsingh98]

Your fan sir

[@Adityasharma-oe8zp]

We have converted 128Kbytes to bits without multiplying 2^3....how??

[@TrendWatchersTV12]

If we have 2^17 addresses and each address have 1bytes of data then memory size will 128KB Thats because of byte addressable.

[@IITKHARAGPUR_CSE]

after listening this I forgot my IIT professors

[@ydayanandareddy7283]

Can anyone explain question no.12

[@259_parthpatidar9]

q16> B ara he , shi he ya glt???

[@TrendWatchersTV12]

Sahi

[@future8060]

sir aapke language se lagta hai ki aap Uttar Pradesh se hai

[@040_priyanshukarmakar2]

Q. 15 solution is not right, we have to divide the 0 address into {(i-2a)+ 2a} bits.

[@jharkhandsongs1344]

That 40 mb speed isp is bhut bada jhhol they are using this strategy

[@ashishietlucknow7230]

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