Відео

vive la suisse !

Many thanks for taking the time to share your knowledge so articulately.

I would like to see a real example. In my case the line search begins from the current point in the opposite direction of the gradient. I must search along the line. I can just repeatably iterate along the line and evaluate the cost function until it no longer gets smaller. You are suggesting searching between two points but how far should the end point or 'low" point be from the current "high" point. You are assuming the boundary of the end point is known and will bracket the minimum point along the line search. What if it isn't?

Dear sir, Thank you for your series. I was doing some research in the field of trust region updates and found you video series. I have a question on this video, though. Let's say we are working in reals. By definition, a positive definite matrix is a SYMMETRIC matrix M \in R^N \times R^N s.t. x*Mx >0 \forall x \in R^N and ||x||>0. At 04:04 it is said, that the matrix D_k = inv(A + \tau I) and \tau \in R is such, that D_k is positive definite. Am I correct that such method will not work if A is not symmetric? Is there a remedy for that case? I know about using a diagonal or symmetrization of the matrix (M = 1/2(A+A*)). But are there any other good suggestions, apart from trust region type methods, especially if the Newton's method is applied to find a solution to the problem F(x) = 0, not the minimization problem? Thank you for your time in advance.

kroasan

Thank you Sir

I appreciate you deeply due to your amazing teaching.

great

Wooooooooooooooooooooow. Dear Professor, your explanations are brilliant and helpful. I also downloaded your book. Thank you so much for your services.

you saved my life!

Extremely concise and lucid explanation. Thank you Prof. Michel.

Dear Prof, is there any reason we set the upper bound for the nest parameter as 10?

love the french accent. cheers !

Don't understand customer behaviour. Customers dont go to a bar when it is closed. And i guess just leave when service takes too long.

Cristal clear

Vous êtes une référence à mes yeux ! Je connais votre chaîne depuis la fac et désormais je travaille dans un projet autour de la MDA / MDO et vos vidéos sont d’une grande aide, merci !

Dear Michel, Thank you so much for this insightful video. Can you make a video demonstrating choice modeling with latent variables in SPSS, Stata or other softwares? That would help us figure out how to integrate SEM with discrete choice modeling in the form of Structural choice modeling. Thank you so much again

Why are the gradients not perpendicular to the level sets in 2:23?

The video Survival of the fittest is not available on UA-cam anymore.

Well...math is not the strongest area of the French:). Wines and good food maybe.

Dr. Bierlaire you are the best. I have been involved with SP studies for the past 34 years from Park and Ride , LRT , BRT , early or late start time ( peak spreading) , risk averse propensity at signalised junctions. having conducted over 30K SP surveys myself on the past 34 years I always have had questions and never found transparent answers concerning theory and estimation but you are a super start who explains leaving nothing un-answered. A Big Thanks

Merci beaucoup !

thank you for the tutorial!

Hi Sir, thank you for your amazing videos. At 15:53 it might be a mistake in the slide as you show that the strata are define by x and i simultaneously, but it is written ESS which is a pure choice-based strategy (so only based in i). Is it not XESS there ? Thank you again for all your great work!

very good and intuitive explanation. Thank you so much sir! you make my learning a really wonderful experience

Thankyou.

Where is the part 3?

Thank you so much. So well explained, exactly what I needed!

Why I could not install biogeme through anaconda?

Monsieur a 2:38 nous avions les couts reduits de c6 = -1.25 et celui de c5 = -0.75, c'etait pas plus interessant de choisir c6 comme variable entrante vu qu'elle diminuera la fonction objectif plus que c5 ? Merci.

Hey please at 3:27 i didn't understand why N * dn is the same thing as the SUM of (Aj * dj) with j from m + 1 to n

using directional derivative to explain reduced cost is really intuitive!

not work on my pc how to correct it

Hi Sir, thank you for all your fantastic videos. How do you solve the system of equation at 6:39 to get the vector of probabilities to be in (perfect condition, partially damaged, seriously damaged, completely useless) = (5/8 , 1/4 , 3/32 , 1/32) ? Thank you very much

pourquoi on a choisi x2 = 2,5 et non pas x1 = 1,5

where can I get a dataset like you use in this system ?

Thank you for sharing, Professor.

Dear professor Michel,why am I unable to get the output for my model correct?Any help is greatly appreciated.Please see my code snippet. The model syntax # Import necessary libraries import pandas as pd import biogeme.biogeme as bio from biogeme.expressions import Beta import numpy as np import biogeme.models as models import biogeme.database as db import biogeme.version as ver # Load and reshape the dataset df = pd.read_excel('My_file’.xlsx') # Create a Biogeme database database = db.Database('My_file', df) globals().update(database.variables) ## Parameters to be estimated ASC_1 = Beta('ASC_1',0,None,None,0) ASC_2 = Beta('ASC_2',0,None,None,0) ASC_3 = Beta('ASC_3',0,None,None,1) B_Wsup = Beta('B_Wsup',0,-10000,10000,0) B_Wqua = Beta('B_Wqua',0,-10000,10000,0) B_Price = Beta('B_Price',0,-10000,10000,0) Specification of utility functions(the choice design is generic/unlabelled and hence ASC_1 and ASC_2 are expected to be similar or same, also ASC_3 corresponds to the third alternative which is the status quo in the choice alternatives).Also tried without lower and upper bound setting or setting equal to 0(zero) but the result didn't change. V1 = ASC_1 +\ B_Wsup*Wsup+\ B_Wqua*Wqua+\ B_Price*Price V2 = ASC_2 +\ B_Wsup*Wsup+\ B_Wqua*Wqua+\ B_Price*Price V3 = ASC_3 +\ B_Wsup*Wsup+\ B_Wqua*Wqua+\ B_Price*Price Association of alternatives with the utility functions V = {1:V1, 2:V2, 3:V3} Associate the availability conditions with the alternatives (in the dataframe Concept column is for availability ie Concept =1 for alternative 1,Concept = 2,for alternative 2 and Concept = 3 for the alternative 3,which is the status quo) av = {1: database.variables['Concept'] == 1, 2: database.variables['Concept'] == 2, 3: database.variables['Concept'] == 3} define the contribution to the loglikelihood of each observation (Chosen =1 if alternative 1 is chosen,=2 if alternative 2 is chosen and =3 if alternative 3 is chosen) logprob = models.loglogit(V,av,Chosen) Biogeme biogeme = bio.BIOGEME(database,logprob) biogeme.modelName = '01logit' Running the estimation results = biogeme.estimate() Get the result pandasResults = results.getEstimatedParameters() pandasResults Value Rob. Std err Rob. t-test Rob. p-value ASC_1 1.77649 0.0 1.797693e+308 0.0 ASC_2 0.00000 0.0 1.797693e+308 0.0 B_Price 0.00000 0.0 1.797693e+308 0.0 B_Wqua 0.00000 0.0 1.797693e+308 0.0 B_Wsup 0.00000 0.0 1.797693e+308 0.0

Thank you so much.

Really informative. I just don"t understand why the noise at @20:35 follows a normal distribution?

Does the line search algorithm described starting at 1:11 satisfy only the weak Wolfe condition on the curvature but not the strong condition (the one with absolute values)? Is there an easy modification of the algorithm if I would like to satisfy the strong condition on the curvature?

Oh my god! Thank you so much for making the video.

thank you for explaining the concept in simple terms. Really good video

Merci

Thank you from Vietnam !

great presentation, as usual... thank you Michel!

Bonjour monsieur, excellente vidéo mais je voulais simplement savoir pourquoi comme B est inversible alos B=B^(-1). Merci à vous.

Amazing explanation sir. Thank you.

great videos professor

good