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This example is going to talk about the multi objective decision making, and we are about to find out

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the Parata optimal front so far, the examples that we have analyzed with each other and are all referring

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to a single objective.

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So we had one objective function that we wanted to minimize it or maximize it and find out the optimal

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decisions, variables.

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And as you can see on this plot and suppose we want to maximize to different objective functions simultaneously.

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And, um, there are some solutions found, um, as you can see on this graph.

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So if I show you another solution on this space, the blue one.

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The dark blue one.

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So if I look at these these solutions, there are five solutions you can see here.

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Now, um, if I want to compare these solutions together, I can say that these this solution is better

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than the dark one because it's it is better than the dark one in terms of if or if one or two.

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And it has a solution in every aspect.

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OK, and also the white one, the white one is has has bigger or if one or two but the green one has

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a bigger of two but smaller one.

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So I can say that the solution can be dominated by this blue one and the white one.

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So if there is any solution that can dominate this guy, then we have to eliminate that and other solution.

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Consider this one.

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Now, if I want to compare the red one and the orange one, which one is dominating one?

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The red one.

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The red one is dominating the orange one.

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OK, so now I hope you understand the meaning of, uh, domination.

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So if I ignore these two new solutions that I showed you and just look at these for these solutions

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are not better than each other in every aspect.

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So if I improve one objective function, I'm deteriorating the other one.

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So the question is, how can I if I have a problem with two different objective functions, how can

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I find out these optimal solution?

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So but before I proceed to the coding and everything else, I want to ask you, why do we need multi

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objective optimizations?

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So can we combine the objective functions and then solve the augmented one in one shots, for example?

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Let me give you another example.

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So you want to build, um.

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Production plant, and there are some costs associated with that construction.

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There are some operating costs.

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There are some investment costs.

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So you can easily do the summation and then ask your decision making tool to find out the best decision

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for you.

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OK, um, but sometimes the objective functions cannot be added together.

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They are from different dimensions, for example, one of them is environmental, for example, pollutions,

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and the other one is cost.

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Some people say that the environmental aspects can be translated into cars.

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Sometimes it's true, sometimes it's not.

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Or there are some other aspects, for example, the beauty of something and the cost of something,

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you know, so these two cannot be added together because the dimensions are different or sometimes these

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are costs.

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But both are most of the objective functions are costs or benefits or whatever financial.

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But and the size of each is completely different with the other one.

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So the order of magnitude is not at the same level.

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So optimizing one of them is not putting any effect on the other one, you know, and sometimes the

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objective functions are the same.

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For example, they are both cost, but two different people are going to pay them, so.

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Adding them together and minimizing the total cost is not always beneficial for both.

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OK, so we have to find out some set of solutions, some set of non dominated solutions and give them

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to the decision maker and ask them to choose.

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And there is another thing.

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If we go with the weightism approach and add the different objective functions using different weights.

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Then we will find out.

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Uh, just one, um, objective function and, uh, one set of decision variables.

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OK, and what happens if that weighting factors change?

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So we have to run the optimization again, OK, but now we want to find out the set of solutions, as

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you can see here, those that are connected with dashed lines and in a way that if you're waiting for

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your change, you can change.

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Your solution is not needed to run the program again.

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And also, you can see that if you move toward any of those objective functions, what will happen to

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the other one?

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So if I am at the Orange Solution, if I want to go to the blue one, for example, improve it in terms

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of two, I want to increase the have to this is how much I will lose in terms of one.

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OK, so for simplicity, consider this very easy example.

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There are two constraints, or if one is something or if two is something else.

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And the range of variation for X1 and X2 are also a specified.

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OK, so let's have a look at the Python code, I import the required packages.

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Sometimes some of these packages are not really necessary, but there is no harm if you keep them right

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there, because maybe you need them in there during your coding.

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So am I.

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Specify the bounds of X1 or next to you can see here they are both non-negative real's.

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And also there are two objective functions.

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Or if one or two, they are defined as variables.

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And also equation one is something.

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Equation two, I will explain it later on.

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A question three is the other equation.

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So what is equation two.

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Equation two is and telling us that or if two is less than something, OK.

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So in order to solve that problem, just let me show you something else on the slides and then come

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back to the points.

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So if if you look at this graph and if I want to, for example, maximize the F one, what will happen?

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What is the obvious solution for that?

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The obvious solution is here.

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OK, but and every time you just maximize the fun, which is the capability of the promo package, you

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can only maximize or minimize the one objective function, OK, at the time.

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And if you want to maximize or F1, this is a solution that you will get.

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OK, but in order to be able to not into not trapping here, we add another constraint for o f two I

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can say or two should be and bigger than something.

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Hmm.

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Or smaller than something and then solve the problem.

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So I change that limit for off to and then try to optimize the other one.

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I could do it for the other objective function, the o f one and I could say OK, the minimum value

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or F one or a maximum value of OIF one is something.

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Now optimize the other um objective function.

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But how can I write it down in a pile.

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So this is the concern I was talking about.

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So I will limit the off to less than something and I will change that limit.

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OK iteratively and ask the panel to solve it.

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So as you can see here, this is strange.

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I do have more than one objective function that's fine in the definition phase, but when I want to

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solve that, I have to deactivate one of them.

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OK, so I put a limit for or of two and I deactivate objective two.

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OK, pay full attention here.

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And then I asked the Palomo to solve it.

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So what objective is being optimized.

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This is the active objective, the objective one which is o f one.

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So it is solved in this way.

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The maximum value of of one is found, OK?

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And how much is the Absolon at the beginning?

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It's a very big number.

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See, so actually there is no limit happening here, but and next time I will deactivate objective one

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and keep the objective to activate it.

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Hmm.

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And then resolve the problem again.

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So look at the objective functions.

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The first time, the objective function one is its maximum value, second time objective function two

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is has its maximum value.

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OK, so the range of that Parata optimal is found.

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Hmm.

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And the X1 annexure are found.

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But how should I continue?

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I have to do something about the debt limit.

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I already specified that the Epsilon.

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OK, so if I am talking about the oath to I should look at the valuation ratio of two.

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It is changing between zero point two to nine point fifty eight.

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So I can say and it can start from this quantity and go down up to here and then try to optimize the

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other objective function.

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OK, so look at here.

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At each point, I write down a for loop, and at each iteration I am updating the value of the Absolon,

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so at the first iteration it is of a minimum of one, OK?

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And at the final one it becomes a maximum of F.

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OK, so in this way, when I solve the problem and optimize it, I can easily find out the value for

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the other objective function.

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So if you can see here I am starting from zero point two and little by little I am increasing this Epsilon

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C.

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And try to optimize the other objective function.

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OK.

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And if you want to graphically show it, you can see it here.

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OK, so, um, the number of points depends on the, um, how many steps you have to specify for your

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programming.

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OK.

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And that's it.

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Thank you very much.