Examples for optimization subject to inequality constraints, Kuhn-Tucker
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- เผยแพร่เมื่อ 18 พ.ย. 2024
- Two examples for optimization subject to inequality constraints, Kuhn-Tucker necessary conditions, sufficient conditions, constraint qualification
Errata: At 17:32, (6/5, 8/5) is the only candidate point; (6/5, -8/5) does not satisfy the equation 4-2\lambda_1 y=0. At 46:55, (0,0) is not a candidate point, since it contradicts y larger than 2. At 50:56, it should be +26, not -26.
Watching mathematics for economists' video as a control engineering student for tomorrow's optimization exam. Life is so strange :D
Replying to a control engineering student's comment on a economist's video for mathematics to ignore tomorrow's computer science exam on optimization. Life's strange.
this is so natural because this kuhn-tucker method and some other optimizing stuffs are used samely in both fields. economics and finance are came from natural science in some degrees.
Given that all optimization and equation solving happens using computers with these algorithms built in anyway, it's redundant to teach the math anymore to anyone but those who want to specialize in optimization techniques.
An intuition and a set of guidelines on when to use what is more than enough
watching this video as a computer science student. strange?
Watching mathematics for economists' video as a chemical engineering student for my thesis subject which I don't understand anything, and watching english videos because there's not videos in spanish. :'D
Currently took Optimization Theory course, and your tutorial is the only thing that helped me to understand the material, thank you!
You have no idea how much this helped. You are a SAVIOUR. Thank you.
This is so helpful, I couldn't get a better explanation for my kuhn tucker conditions
Speechless 🙊! You saved me, I didn't understand this concept. I watched so many TH-cam videos. Then I stumbled upon this one. Thanks a gazillion.
I can't thank you enough :) simple yet concise!
at 47:02, you got y =0, but the constraint was for y> 2 therefore (0, 0 ) shouldn't be a candidate point
Thanks for catching. Check the errata in the video description, there were a few more that slipped through my net.
Thank you for your service. I subscribed. Greetings from Colombia.
Thank you . You 'save a soul'.
Extremely helpful video. Thanks a lot
Excellent video, thank you so much for posting it!
Thank you for posting
the lagrange conditions are wrong? dont you think
Minute 47: How can (0,0) be a candidate point if Y is supposed to be strictly greater than two?
Great video, very grateful!
Amazing, thank you very much.
The best vid by far
Great lecture......thank you so much.
How do you know the function is concave in 23rd minute? Shouldn't we use bordered hessian matrix?
should not Lambda >=0 at the beginning?
is that because it is either slack or binding.
for slack, it doesn't satisfy the equality condition in constraint, hence the lambda will be 0 anyway
whereas for binding, it satisfies the equality condition and the gradient of the maximized function at solution will be a linear combination of all constraints gradient and each scalar will be in effect, hence lambda be >0.
so in short, it is either lambda = 0 or > 0
Amazing vid! few questions, is it right that you made 2 small mistakes in the end or maybe im wrong. Just for my personal clarification. In the last case of your second example isn't y=0 in contradiction with the condition that y>2? And when your simplifying the langrange in order to check the sufficiently, isn't it +26 instead of -26? Well, it were just some doubts, really great and useful vid!
Thanks for catching these. Indeed, at 46:55, (0,0) is not a candidate point, since it clearly contradicts y>2. At 50:56, it should be +26, not -26.
At the first problem,isn't y=-8/5 wrong?If you substite negative y and lambda1=5/4 in the second lagrange equation(dL/dy),it won't give us 0.The second lagrange equation results in 0 only for y=8/5.Doesn't this mean that (6/5,-8/5) is not a critical point?Maybe I'm wrong,but I'm curious.The video is very helpful by the way thanks a lot.
Thanks for catching. Indeed, only (6/5,8/5) is a critical point.
paynehunter .
what happened in 40:40? i used the general formula and got something waaaay different
Very good video
Thank you 😊
What happens if they are linearly dependent vectors for all values of x,y or how are the constraint qualifications formed if we only have one constraint since we get points and not vectors??
Sir, Thanks Great job
You channel is really instructive. I also went ahead to download your lecture notes but found only your lecture note on linear algebra. Can you kindly include that of optimization? Thank you.
why it's minus with constraint and not plus (Lagrangean)? i mean 3x+4y-lambda1*(const1)-lambda2*(const2)? why not 3x+4y+lambda1*(const1)+lambda2*(const2). but I know the result will be the same with 4 cases conditions but they just opposite
See my comment to Happy Turtle some time ago.
Hello, if we have 3 constants , but only two variables (x1,x2), do we add 3rd lamda
In 9:33 he writes 3- 4/sqt3 + λ2 instead of 3- (4/sqt3)*x + λ2
Because x=1, look at the conditions
Thanks
Thanks for the video. By any chance,can you upload your working?
Glad to hear it helps. I'm not sure I understand what you mean by my working.
What if you got a case where there is no multiplier is positive? what is the approach?
See my answer to Happy Turtle some 3 years ago further down in the comments.
can you explain why in some books we find that lambdas must be negative in the max case and positive in the min case ?
There are many possible ways to write optimization problems subject to inequality constraints, and it can get confusing at times. I like to write things this way:
max f(x) s.t. g(x) = b,
with Lagrangian L = f + lambda (g-b), g-b >= 0, lambda >= 0.
Why did I change minus to plus in L?
min f s.t. g >= b
is equivalent to
max -f s.t. -g = 0 in the max-case, and in the min-case, this derivative is -lambda = 0), but increasing b by one unit leads to a lambda-increase in the max-objective function, and a lambda-decrease in the min-objective function. Another reason to write things this way is the standard formulation of the Lagrange multiplier theorem: grad f = lambda grad g, that is, grad f - lambda grad g = 0.
Of course you can do it differently, and in both cases write
L = f + lambda (g-b),
then lambda = 0 in the min-case. I guess this is what you found in a book.
thanks for full reply
@9:50 how did the x attached to -2*lambda1 disappear? was this a mistake?
putting the value of x=1.
ty so much , pdf pls :)
the lagrangian should be + lambda
Hi, thanks for the tutorial. It's really great! I just want to comment at 38:00 you write 3x - λ1 = 0, However the second equation should be 3y - λ1 = 0 or λ1 = 3y. This makes the calculation on the next substitution to be cubic formula because dL/dx = 4x+3y-λ1x = 0 to be 4x+3y-3xy = 0. Thanks and have a great day!
at 17:15 , there shouldn't be two candidate points because (6/5, -8/5) will imply a negative lambda from equation 2 , right? so there should be only one point (6/5,8/5) ... please respond to my query.
Thanks for catching. Indeed only (6/5, 8/5) is a critical point.
T H A N K Y O U!!
at 13:21 you forgot the 2 in the denominator.
This probably makes the rest of the problem wrong.......
That one cancels with the coefficient 2 in 2\lambda_1 y.
I don't speak english xD :"c
💚💚💚💚💚💚💚💚💚 thanks a lot sir
2puissanse 2 + 2puissance 2 > 4 !!!!
You jumped in the vid. The solution (2,2) is from the second problem, and you put it into the constraint of the first problem.
are mad or............?
Your solution is wrong, because the point (2,2) is not in the feasible region, at (2,2) the inequality constraints do not satisfied. The correct answer is the point (1, sqrt(3)) for max, and (1,-sqrt(3)) for min.
You probably jumped in the vid. Note there are two different problems being solved back to back.
0q
tamal
u r wrong
So sad to see what economics has become.
Economics is about human behaviour.
Keynesianism and trying to make economics a scientific field is why economics today is pointless.
Very sad to see.
Thank you for this comment. You have put forward quite an interesting idea. In some capacity, I agree with you. However, the Kuhn Tucker technique on optimization is not especially useful in the cases you mention (i.e. explaining human behavior and economic trends). It is, however, very useful in quantifiable cases of profit optimization, efficiency optimization. It can be very useful to shipping companies, energy providers and charities. Please let me know what you think.
@@TheInavsayo if it is very useful to the REAL WORLD then why not teach that.
I need examples.
Start applying theory with examples.
Majority of theory is bs.
@@qeoo6578 Perhaps a career in applied economics is what you're looking for.
@@qeoo6578 that is general for linear optimization problems and provably so. it is way more general than whatever economics you are into. so calling it bs is extremely dumb
Sad indeed. Economics has been bastardized into a subfield of math. And then people complain economics has become stale and uncreative
Dear sir, I realize your mother tongue is not English but please say ''REGARDLESS", or ''IRRESPECTIVE", NOT "IRREGARDLESS". Thank you.
It is synonymous. although there's a controversy about that, you can still understand the context so what is the matter. Also you said sir, so South asia? I don't see any brits here complaining about the word usage, so why should we, speaker of english as second language, complain about such trivial matter?
How do you know the function is concave in 23rd minute? Shouldn't we use bordered hessian matrix?