Direct Memory Mapping - Solved Examples
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- เผยแพร่เมื่อ 2 ต.ค. 2024
- COA: Direct Memory Mapping - Solved Examples
Topics discussed:
For Direct-mapped caches
1. How to calculate P.A. Split?
2. How to find out Tag Directory size?
3. Significance of Tag bits & unique method of calculating the number of Tag bits.
4. How to find out Cache size?
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Music:
Axol x Alex Skrindo - You [NCS Release]
#COAByNeso #ComputerOrganizationAndArchitecture #DirectMemoryMapping
My confusion is never ending on why these youtubers can explain these difficult stuff in very clear and concise way, while our university lectures suck at it.
exactly lol we get better quality for free than we get by paying a huge amount of money lol
man this seems to happen in every country, mine isn't exception too. The lecturer is atrocious!
Having worked as a teacher professionally, I can tell you that most teachers do not (a) find the time/energy to make their educational content in this way (I can imagine each short video on this channel took 2 hours to make i.e. a university lecture = ~8 short videos = 16 hours of effort vs 1.5 hours of plain lecturing) and more importantly (b) fail to place themselves in the shoes of the learner i.e. as a beginner (this is called expert blindness).
In practice you need someone crazy enough/talented enough to make content like this --- imagine taking 2-4 hours out of your day to make this powerpoint a 10 minutes video? and/or an economy of scale.
@@willemhekman1788 well said
Wow, this is an amazing lecture... far better than the three 1h 30m lectures that my professor gives. Once I start my career, I can't wait to donate to this channel. It's been SUCH a lifesaver for me!
Same thought!
Actually same ......
Hum sb fees dete h college m pdhte youtube ss haii.... uske baad bhi clg balo ko placement nhi diye jte... etne paise ke baad bhi...
For the last one, while I at first failed monumentally, after rewatching the video (and hopefully understanding better) I realized (to me at least) there's a simpler way to find the cache size.
You are told the # of tag bits is 10. So you know you need 2^10 different tags, and by the same token that the cache fits 2^10 times into the MM. So you get this equation 2^34/2^x = 2^10. Since you know algebra, you know you can convert the system to (just use exponent laws) 34-x = 10. Solving for x you get 24.
third one can be done by 2^#tag bits= MM size/ Cache size , we know MM size and Cache size
my head hurts
Screw it. Im'm going to fail
+2
Us b
i dont know why but the concepts are not clear might be too much vagueness is there..i have been watching neso for a long time now but only this time i m left with the doubts.
Consider a main memory of the size 64 kB with each word being of 8 bits (one byte) only and a direct mapping Cache memory of size 4 kB also having data word size of 8 bits. Find the following :
(i) What is the size of tag and index fields of cache ?
(ii) In what location of Cache, hexadecimal address to main memory (AABB) (if exists in cache) will be located ?
what's answer of this question, can anyone tell..?
Great explanation in just 10mins.
keep it up.
I recommended this lecture series to my children's XD
Thanks for covering every kind of questions 🙂🙂🙂🙂❤️❤️❤️❤️🙂🙂🙂🙂
Totally confused very confusing
fabulous!! very clearly understood in less time!! thank you neso!!
Amazing🙏
Awesome video and teacher. Thank you so much for these easy-to-understand videos.
There's one thing I haven't understood, isn't 1Kb the same as 1000 bytes, and not 1024 like in your examples? I thought 1Kib was 1024 bytes
example 3: 10 tag bits..... So 2*34/2^10 = 2^24 =16MB. no need other info i think
Very informative and amazing lecture. Can you please suggest a good book also for computer architecture and to solve more numerical based problems like this
8:07 Consider a main memory of the size 64 kB
with each word being of 8 bits (one byte) only and a direct mapping Cache memory of size 4 kB also having data word size of
8 bits. Find the following :
(i) What is the size of tag and index fields of cache ?
(ii) In what location of Cache, hexadecimal address to main memory (AABB) (if exists in cache) will be located ?
can anyone solve this question..?
@Nesco Academy can you solve it..?
Without knowing block size, I am gonna assume arbitrarily that it is 1kB. So 10 lines for block offset, and 4 lines in cache. So AABB will be in line#1 of cache since AABB ends with 01.
If you take a different block size, AABB will still end up in line#1 (01) but there may be more lines in the cache.
Right on time for my exams
where can i find your course ppt or pdf ? can you share it? thanks
So basically, the number of caches that can be fit in main memory = number of blocks that can fit in a cache
Please complete lecture video on Compiler design plsss sir really need
I didn't get the concept of tag directory. Can anyone help me out?
the same
I love TH-cam porofessors. They always save my poor life ... Thank you so much for your effort sir
so tag bits represent the log(times the cache would fit in the ram). if ram is 4 times the cache size, then the tag bits are 2. in above case, 2^10 times x cache size = ram size. therefore. cache size is ram size/2^10= 16 x 2^10 x 2^10 x 2^10 bytes. = 16 MB.
you should have taken one example for word size != 0
that would be much helpfull
Thank you so much!!!
MM size = 4GB =2^2*2^30Bytes=2^32bytes=2^32*2^3bits=2^35 bits right?
have the same question man
Bow down 🙇🏻♂️ to him ❤
Thank you, You all are really blessed. Thank you, one thousand times Thank You.
You saved me, LOVE YOU ❤️
First viewer.. 👍🏾
you made this so easy to understand, thank you sir!!
👍
Unbelievable explanation. The best one so far
Got it
❤❤❤
WORD is 1 or 2 Bytes?
Thanks for the video sir ....
❤️
👍
Thank you sir❤
disliked because of cringe voice. taking lecture at 3 in night for exam at 8 and this boy sound so cheerful
hates off
hates off
Its 'hats off' brothers 😭
8:44
Hey! I noticed one thing that size of cache is equal to the no. of blocks in MM. Is this true for every computer?
This playlist is progressing too slow.If possible please fast a little bit.
You're too quick, slow down