Understanding Parallel Computing: Amdahl's Law

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More cores mean better performance, right? That's not what Amdahl says. Learn one of the foundations of parallel computing in "Amdahl's Law." Prepare yourself for math. And lawn mowing.
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Amazing introduction explained with wisdom and enthusiasm. Thank you!

BrandNewByxor
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I like that part. boo-hoo it's doesn't!

rhlmshr
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Video not particularly focused but it was helpful and you have an excellent accent. ☻
Thank you for the upload.

inadaizz
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Man, I know this was in 2011, but jesus you'd think the camera quality would be better coming from a magazine company.
Nonetheless, it's nice to see Amdahl's Law verbally explained like this.

neux
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Gustafson's law (weak scaling) might be interesting to discuss, as some people see it as more relevant than Amdahl's Law.

lennyogarc
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Watch it in 720p or in 144p.. u get the same quality.. *_DAMN_*

hirakmondal
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@xpeedX because you increase the road 2 cars can go over the road so you double the speed. but if you increase the single treat performance you could increase the clock = speed or you could increase the instructions/clock look at it as using a car that has 6 seats instead of 4. and HT is just getting all the cars full instead of leaving one guy whit one car. turbo is like increasing the of the roads when there are a low amount of cars at the road.

themassau
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Hi Douglas, I am too bored to read all that text about this law and verbally it sound simple and nice explanation, thanks for that...and kind of personal comment...you kind of look like "al pacino" and sound like him too...you know that ?

achuou
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@xpeedX i think you forgot that it is (somthing)+p/n so the n can only make p/n zero and the equation that is left whit like 1000core would be 1/(1-p). so you can only speed up than by making more of the core parallel or increasing the single thread performance.
look at it as driving form A to B but you got a speed limit of 120km/h (and you can't go faster) the distance between it is 240km so it takes 2 hours. no matter what you do. but you can get more cars trough the road by laying more roads

themassau
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I've found a Bitwise method for multiplying 2 numbers that does not require addition steps. My method can be done sequentially 2 bits at a time but it also can be done in parallel with each compute cell working on 2bits of each number due to the fact that in each step of the process you don't require any sequential information passing thus each of the 2 bit processes can be done independently of ever other 2 bit process. Although It seems as if I can achieve multiplication in a bit level parallelisable way I have still yet to achieve the same ability for addition. I suppose in my case the sequential part would be the 3 steps required to perform the multiplication. It would be wired after completion of my work I was to find that I had to use multiplication steps in order to achieve bitwise parallelization of addition functionality.

analodimripe
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Brilliant video, really helpful, thank you!

katef
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This happened to me rendering videos! :'( At least I know why I only got a three minute gain after adding another processor. Also explains why my cores only run at 30% rendering DVD's. Glad I found this video. Lucky the power isn't wasted tho, Unreal Engine still gives the cores a serious thrashing.

ReubenAStern
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Sry I have to write this.
As a practitioner uploading a 360p video as hd is just fail :D

phonix
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You really remind me of Saul Goodman from Breaking Bad

unholyjebus
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thank you sir.. I got the concept very clearly now... :) It was really helpful...

shahronak
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That black shirt makes you look chiseled!

ThuyNguyen-buge
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hi, my name is frigider and i AM mathematically challenged.

otdosa
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great video!! but i have a question regarding the division of labor.so i have a code of which the 60% can be handled with parallel computing.that 60% is actualy 20% with K proccesors and 40% with N. K + N adds to a total of 5.The division of labor should be K = 3 and N = 2 right?

hellzor
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his voice is like rick from rick and morty and he looks similar

ΚλέαρχοςΤσίγκρος-ηκ
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Thank you very much, it really helped :-)

rendlikk