Set Associative Mapping - Solved Examples (Part 1)

COA: Set Associative Mapping - Solved Examples (Part 1)
Topics discussed:
1. Finding the P.A. Split for Set Associative Mapping.
2. Finding the number & type of Comparators needed for Set Associative Mapping.
3. Derivation of Cache Size formula.
4. Calculating the size of Main Memory using the Cache Size formula.
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Axol x Alex Skrindo - You [NCS Release]
#COAByNeso #ComputerOrganizationAndArchitecture #SetAssociativeMapping

[@manojharshavardhan2385]

I'm following all the videos from this CA tutorial. It's fantastic 👍👍

[@harshrai2845]

Your videos are awesome.
They are very helpful to me....
Thanks alot😃

[@peargaming9579]

Very awesome examples Sir ji...

[@gatecomputerscience1484]

Priceless lectures ❤️❤️❤️❤️

[@mymoviemania1]

you are the one who is priceless

[@Jeba-Princy]

Thankssss a lotttt

[@sakshammaggo8926]

why did he multiply 2^13 no of lines with tag bits, I mean tag bits won't be associated with every line of the cache no?

[@sossisan664]

Awesooome

[@gatecomputerscience1484]

Please upload PPT as well

[@uttamkarmakarece3534]

Please upload more videos

[@gatecomputerscience1484]

Please upload ppt brother please

[@samarthtandale9121]

Also pls someone confirm that value of x=5 and y=10 ...

[@gatecomputerscience1484]

Upload ppt brother please

[@mohammadj8830]

please solve this question for me:
Find the number of tags for cache of size 8KByte and 4 lines per set . The main
memory contains 4K blocks of 128 words each

[@luuu_na35]

This is my solution, please correct me if I'm wrong:
* Given:
No. of Memory Blocks = 4K blocks
Size of each Memory Block = 128 words
Cache size = 8KB
4 lines per set
----------------------------
* Solution:
P.A.S = No. Blocks * Block Size = 4K * 128 words = 2^12 * 2^7 = 2^19 Bytes
=> P.A.S = 2^19 -> P.A bits = 19
=> Block Size = 2^7 -> Offset = 7 bits
Set size = K line(s) * Block Size = 4 * 2^7 = 2^9 Bytes
No. of Sets = Cache Size / Set Size = 2^13 / 2^9 = 2^4 Sets -> Set bits = 4
* P.A bits = Tag bits + Set bits + Offset
=> Tag bits = P.A bits - (Set bits + Offset) = 19 - (4 + 7) = 8

[@cricketindiaOnly]

​@@luuu_na35it is correct.
I got the same

[@shivanshkumar3772]

Sir what if set is odd!!

[@good-tn9sr]

!!!

[@Zeehan_Haider]

It is impossible because the computer is the world of 0 and 1.

[@skybuck2000]

My laptop has L3 cache size: 3 MB and dual core, now what ? :)

[@skybuck2000]

Also it's 12 way set associative, now what ? :)

[@skybuck2000]

Toshiba Laptop L670.

[@skybuck2000]

I guess it has to be calculated in full, the 2 to the power of trick no longer applies.

[@alifirfan7267]

good informations ! thanks . keep it up 🫠

[@alifirfan7267]

done subscribe

[@sandipdas007]

No offense,but we know simple addition.👍