Sir x,y>= 0 is not part of the discussion .it simply states that you have to have a value greater that or equal to 0. Or we have to consider (0,0) as a point
In the problem just previous the last problem when D4 is calculated for the point (14/5, 0, 14/10; -14/10), the element a33 of Bordered Hessian matrix is not 2(as you show in the screen), it should be -4/5.Then sign of D4 will be positive. It matches neither the condition for maximum or minimum. Then what will be the answer?
Thank you, sir.It is very well explained
Very helpful ..
Thank you @Dr.Harish Garg
Thanks for watching
very helpful video
THANK U SIR
Thanks
Very well explained!
For the second to last example can we use the Hessian matrix to determine the max and min of the set of points?
did you get to know the answer?
Very nice video, sir last algorithm Roa 3 , is remaining, please make the video.thank you very much
Sir when to use bordered hessian or normal hessian??
Sir x,y>= 0 is not part of the discussion .it simply states that you have to have a value greater that or equal to 0. Or we have to consider (0,0) as a point
In the 2 nd problem, H matrix should be 4*4 as m+n= 3 + 1(no of variable + no of constraints)
That is only done to find the border hessian, in method 2, you just find the normal hessian of the objective function alone.
In the last problem, the lagrangian function is f(x) - :\h(x).
There must be a -ve sign instead of +ve sign.
In the problem just previous the last problem when D4 is calculated for the point (14/5, 0, 14/10; -14/10), the element a33 of Bordered Hessian matrix is not 2(as you show in the screen), it should be -4/5.Then sign of D4 will be positive. It matches neither the condition for maximum or minimum. Then what will be the answer?
Will it be considered a saddle point?
Hello sir..
We need lecturer notes for Jacobian Method
Nice Sir,
Keep watching
@@DrHarishGarg Sure Sir,
what should we do if answer of one is -ve and another is +ve ......... this will be max or min
pls give the answer
All the other case which not satisfy the given condition then it is saddle point.
Sir what if principal minors of NLP problems are all positive (in my question ∆3 and ∆4 are +ve)????
All other cases which does not met the conditions (as described in the lecture for maxima or minima) is said to be saddle point.
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