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‫Welcome back.

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‫Let's now solve the exercise from the previous video and see if there are solutions that minimize the

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‫cost function while respecting all the constraints, that was our first question.

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‫And then in the second question, we have to justify the answer for the first question.

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‫So here you have your double bar matrix small have transposed your matrix and then the H vector transposed.

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‫If we say that our delta you global equals Delta Delta zero end Delta A0, then if you look at this

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‫h double bar matrix, then you already know that your horizon period equals one.

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‫That's because you have two inputs and only one timestamp, so you only have your zero and zero.

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‫You don't have here Delta, Delta One, Delta A1, etc. So this vector is two by one two rows in one

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‫column and then your h double bar matrix.

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‫It's two by two.

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‫And so from that, you know that your horizon equals one.

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‫And so this is how your cost function looks like J equals one half times delta.

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‫You global transposed.

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‫Then your h double bar matrix, then your delta.

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‫Your global plus small f transposed times, your delta, you global.

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‫We can, of course, expand it.

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‫And so you will have one half times.

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‫This is your delta, your global transposed.

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‫This is your h double bar matrix.

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‫Your Delta.

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‫Your global plus your small f transposed.

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‫And your delta.

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‫You global.

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‫Now, just to mention something, if instead of this, I had written Delta You Global Equals and then

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‫Delta Delta zero and Delta Delta One, then that means that I would have had only one input, but two

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‫time stamps, which means that my horizon period would have been two.

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‫Then instead of Delta A0, I would have had Delta Delta One.

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‫However, from the point of view of mathematics, it's exactly the same thing for you.

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‫There is a different meaning, but mathematically your cost function has two independent variables.

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‫Either it's Delta Delta.

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‫Zero and Delta zero or Delta Delta zero in Delta Delta One.

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‫For the solver, it's all the same.

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‫And so if I now write the cost function out completely, then this is what you will get.

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‫You will have one half times 100 Delta Delta zero squared plus one half Delta A0 squared.

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‫That's the first term.

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‫And then for the second term, I will write plus zero times Delta and Delta zero plus zero times Delta

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‫A0.

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‫In other words, you will have 50 times Delta Delta zero squared plus one half Delta A0 squared, and

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‫that's your cost function.

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‫Now, if you look at it and we don't consider any constraints whatsoever, then can you tell me the

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‫values for Delta Delta?

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‫Zero and Delta A0 that will give you the smallest j possible?

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‫Well, in this case, it's quite easy.

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‫Your jammin without the constraints it will be located at Delta.

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‫You Global equals zero and zero.

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‫So your Delta Delta zero equals zero gradients and Delta A0, it equals zero meters per second squared.

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‫That will give you the smallest J.

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‫Possible J equals 50 times zero squared plus one half times zero squared.

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‫It will give me zero.

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‫Any other combination of Delta Delta zero and Delta A0 will give you a positive j.

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‫Even if Delta Delta zero and Delta A0 are negative values, since you square your independent variables

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‫in the cost function.

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‫Even if, let's say, your Delta Delta zero equals minus one end Delta a zero equals minus two.

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‫Then you would have J equals.

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‫50 times.

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‫Minus one squared plus one half times minus two squared.

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‫It will be 50 plus two equals 52 and well, 52 is greater than zero.

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‫However, what if small f transposed was this one instead?

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‫In that case, your cost function would be all that.

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‫Now you see that this is your small f transpose here.

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‫Instead of a zero and zero, you have three and minus four.

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‫And so all that will equal 50 times Delta Delta zero squared plus one half times Delta A0 squared plus

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‫three times Delta Delta zero minus four times Delta A0.

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‫In this case, it's a different thing.

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‫The solution without the constraints would be the following I will take the gradient of my cost function.

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‫It will be the partial j.

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‫With respect to Delta on Delta zero and the partial of J.

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‫With respect to Delta A0.

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‫So this will be the first element and this will be the second element.

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‫That's because in the first element, I only take the derivative with respect to Delta and to zero.

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‫And in the second element, I only take the derivative with the respect to Delta A0.

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‫And so you know that you have to make the gradient equal to zero.

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‫So all this, it will be zero.

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‫That means that this element here, I will write it down like this.

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‫And then this element here it will be written down like this.

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‫And so here you will find that 100 times the R 2.0 equals minus three.

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‫And here you will find that Delta A0 equals four.

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‫And so from here, you can find that Delta Delta zero equals minus three over 100.

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‫And so this is your solution minus three over one hundred and four.

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‫So your delta you global that minimizes your cost function.

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‫It's this vector here.

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‫And now we can calculate our G men.

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‫It equals 50 times minus three over 100 squared plus one half times four squared plus three times minus

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‫three over 100 minus four times four.

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‫And that gives you minus eight point zero for five.

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‫That's your game.

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‫In any other combination of Delta Delta zero or Delta A0 would give you a bigger G.

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‫Essentially, you could have achieved the same result with this formula from the previous section.

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‫So that was the case for when your small f transposed was three and minus four.

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‫But in our case, it was zero and zero.

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‫Therefore, this was your cost function.

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‫The gradient of J now equals again the partial of J.

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‫With respect to Delta Delta zero and then the partial J.

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‫With respect to Delta A0, that will give you 100 times Delta and Delta zero.

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‫And then the second term here, your Power two will go in front.

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‫And so you will simply have Delta A0, then you make them equal to zero.

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‫So from here, you can see that Delta Delta zero equals zero over 100, so that's just zero.

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‫And so that's why this is your solution that gives you mean that equals zero.