COA | Cache | Problems on set associative mapping | Ravindrababu Ravula | Free GATE CS Classes

[@tarunsolanki9610]

In the 5th question, the cache size can be evaluated as follows: (Cache Memory Size/k-way)=2^(Number of bits for Set Offset + Block Offset) So we've 26 bits = 10(tag bits) + 16(Set Offset + Block Offset) Cache Memory Size = 2^16 * 2*2 = 256KB And similarly in 6th question Main Memory Size = 8MB

[@saideepakbejawada6773]

We can find the cache size in problem 5,also we can find main memory size in problem 6. No problem...you have presented a great stuff!

[@168Siri]

how??

[@Badrkhan2016]

I think every weekend in my class lecture I waste 90 minutes very technically to listen my teacher. I don't know how I can easily understand you lecture in 10 to 20 minutes except 90 minutes with extra tutorials. Thank you! again you spend your precious time for all of us.

[@maldinila]

feeling lucky i found this lecture.. thank you.

[@rahulranjan3273]

+Ravindrababu Sir, Is the following relation valid for set associative mapping- Tag Bits =(log ((MM Size/Cache Size)*Set-Associativity)) (Because blocks are mapped to sets instead lines as in direct mapping, log taken in base-2) From here we can find the cache size for problem 5 and MM Size for problem 6

[@manojputchala5798]

memory for 6th q can also be found like: Let- x=set Offset, y= line/block Offset -no of sets with x bits will be 2^x -for 2^x sets we need 2^(x+3) lines (i.e 1 set = 8 lines -2^x sets= 2^(x+3) lines) -memory for cache =no of lines*2^y -2^19=2^(x+3) *2^y -2^(x+3+y)=2^19 -applying log on b.s, x+y+3=19 -x+y=16 since we know x(set Offset)+y(lin/blockOffset)+7(Tag)=m(main memory bits ) therefore, m=x+y+7 m=23. 2^23=8mb.

[@vishalgupta5144]

Sir , i think we can find solution to question 5 and 6.Main memory size in 6th is 8MB.thanku

[@SajalJainprogrammer]

In ques 5 we can find out the cache memory size Main Memory size Physical address : 26 bits 4 way associative Tag : 10 bits Set Number + Block offset : 16 so cache size is 2^(16+2) = 2^(18) = 256 Kb I am doing it right ?

[@sandeep007734]

This seems like correct

[@mohdanas9857]

Thanks a lot sir. you made my concepts clear. You are a great teacher.

[@chessfolks3827]

You taught all things Very easiest way... The way you teaching awesome... Plz upload some videos of Gate problems of COA...

[@laharibasu9731]

Thank you Sir. The lecture was of great help.

[@sonalganvir8334]

This video is underrated! It deserves more views.

[@aryangawand7454]

Great video all Question in one video

[@darshanparmar1998]

in 5th qu. cache size is 256KB and In 6th qu. memory size is 8MB

[@WalkingtotheTruth]

youre right

[@abhishekkumar-os5zk]

how

[@muhammadsafiullah8428]

Yes you are right!

[@soldadopreciso]

In 6th question , the BlockSize would be between 2^0

[@100rabhkr_India]

In the second question, all is fine, we have tag bits=22, set bit =2 and page offset=10. When we map this in the Cache address. The division in the cache address is (| Tag bits | log2(N) | page offset |). In my understanding the total summation should match the cache memory size i.e 15 but tag bit itself is 22. log2(4)=2 and page offset is 8. Hence a total of 32. Can you please explain or is it something I am doing wrong!!!

[@RandomReview12]

Dude. You're a life saver

[@bhupendragothwal759]

Sir please upload the video on fork() system call, it is very difficult to understand , but your teaching way is very understable , so please upload the video sir

[@mahanteshmise6930]

In direct mapping main memory divided by cache memory is equal to tag size(There was a direct relation) we could find the cache memory given tag bit size and main memory. Why is that not true here? Correct me if I am wrong.

[@devakid2332]

Thank you Ravi ji really very much useful for us

[@VishwasuDeshpande]

In problem 1, you use Byte representation ( MM = 2^17) , similarly it should be 2^32 for second in bytes and not 2^35 which is in bits. Thus going by that, are the tag bits in first problem 4 bytes and not bits? I am little confused here.

[@mohdanas9857]

Since MM size, cache size and block size are taken in same dimensions, they will yield a dimensionless no. and to have the number of possible binary combinations u will require log2 ( that number ) and it will always come in bits not bytes.

[@shaiksonu8955]

Sir what about index sir? How to find index?

[@NakulMahale]

ok 1kb= 2 to the power of 10 i.e. 10 bits , 1mb=2^20,1gb=2^30 and same for other but how come 128kb = 17 bits . I request you to elaborate it .

[@abhishekpokala313]

128 = 2^7. kilo = 2^10. 128 kb = 2^7 * 2^ 10 = 2^ 17

[@ramyagarimella]

128KB means 17 bytes or bits???

[@sonnix31]

+Ramya Garimella 128 KB means 2^(17) bytes, which is represented by 17 bits

[@ramyagarimella]

i got that .thanks

[@paris737]

+Sonnix Jackson but if we go by the basics, 1byte = 8 bits, therefore, 128KB or 2^7 bytes = 2^10 bits. I don't understand how youve have written 2^17 bytes is represented by 17 bits?

[@sonnix31]

utkarsh srivastava In binary sistem the number y = 2^n is represented by n bits. Thats the basic concept. Also remember, the bits are used to to represent the number of bytes, so 2^n bytes are represented by n bits.

[@abdulazeem1607]

thanx sir ji...

[@wellbeing1539]

the last answer is wrong

[@reidiko341]

YOU ROCK!

[@wellbeing1539]

you missed the concept in the last two examples ...

[@shivashaw2220]

excellent ...

[@kaushilkundalia2197]

Thanks a lot sir !

[@blackarrow3138]

thank you sir

[@NaveenKumar-hd7vk]

lecture was awesome, but in the example 5 MM=64MB,TAG=10bits,4-way,cs=? you said it is not possible but it is possible. tag has 10 bits so if we remove 4 way that is 2 bits we get 8bits,so 26-8=18bits CS=18bits

[@vin_jha]

I think you are mistaken over here. See tag bits has nothing to do with number of lines in each set Also set number is log (total no. of sets) and not log(lines in each set)

[@nikhilnambiar7160]

If block size is variable u cnt verify the cache size

[@cl3375]

YASSS :) :)

[@BrajeshKumar-ez8zs]

loved it

[@bindumohan8216]

Thank you

[@AnkitaYadav-qp4wn]

#CFBR