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How many search points are needed in each iteration of simplex search for finding minimum of an unconstrained N- variable optimization problem...

Question

How many search points are needed in each iteration of simplex search for finding minimum of an unconstrained N- variable optimization problem

How many search points are needed in each iteration of simplex search for finding minimum of an unconstrained N- variable optimization problem



Answers

Describe an algorithm that interchanges the values of the variables $x$ and $y,$ using only assignments. What is the minimum number of assignment statements. What is the this?

Okay. We know that if we're looking at this graft, we have a ship that looks like this, and this is on the X axis is You can see so four coordinate points we find the vergis is zero car 02 times zero plus zero equals 01 comma 02 times one plus zero equals two to calm itude two times two plus two equals sex. Zero coma for two times zero plus four equals four. Now we know the minima ax. Come. Why is when X equals zero y equals zero and has a value of zero? No, The minimum for X equals two y equals two has a value of sex. So the answer is the men is zero, and the max is sex for this question.

Where are we looking at? The objective function Z equals X plus five Y. With the following five constraints. We're looking for a maximum or minimum value of Z within the constraints given. So I just typed in the constraints into the Desmond's grafting, calculates a great user friendly free graphing calculator. When you're typing them in. If you need a less than or equal to or greater than or equal to you, just type in the less than or greater than inequality from your keyboard and then hit the equal sign on your keyboard. Okay, so our first constraint is the X. Is greater than or equal to zero. So it's on the right side. Our next constraint, is that why is also greater than equal to zero? So it's in our first quadrant. Our next constraint gives us this little triangle down here that we can work with snips off a little piece of that and and it cuts off another piece of it. So the next thing that we're looking for is the vertex is of of our quadrilateral where all of them are intersecting. These are what we're gonna plug into our function to find the maximum minimum for into our objective function. So I'm going to copy these courtesies so we can work with them. Science 0.50 and 00 is our last for a C. The last word text. Then we're going to just use the scientific calculator. Any calculate role work? We're gonna plug them into our objective function that I just copied over here. So we Z. Is X plus five Y. Again so far. Second vertex C. Is X plus five. Y. For a third vertex C. Is X plus five Y. And again see it's X plus five Y. Let's just attach those values for our objective function back to the verses. They came from five and zero. Okay, so now we're going to look for a maximum minimum. So there last step we're looking at the Z values 25 is our largest see value. So that's gonna be a maximum maximum occurs at 05 And our minimum is zero. This is gonna be our lowest suit value.

We're using linear programming to solve optimization problems. We're trying to find where Z. In our objective function is the lowest and the smallest given a bunch of constraints. So the objective function working with for this problem is four X plus five Y. And the constraints are these five constraints listed. We want to visualize the constraints. You wanna graph them so we can find the region that they're all intersecting. I like to use Desmond's too good. Free user friendly graph. But any graphing calculator or software will work when you're typing into Dismas. If you need an or equals to inequality types the less than or greater than sign on your keyboard and type the equal sign on your keyboard. Okay, let's look at those constraints. So X. Is positive greater than equal to zero. Why is also go to the equal to zero? So, we're working in this first quadrant, Then let's add in our other three constraints. We've got a little triangle now that they're all intersecting in. We cut a little piece off and cut more off. So now we're down to this quadrilateral kind of highlight the corner so we can see it a little better. We're down to this quadrilateral. This is the area contained by our constraints that we're going to be evaluating with our objective function. The maximum and minimum for Z will occur. One of these vortices. Our job is to figure out which one. So let's list. Harvard is is just on a scratch paper so we can keep them handy. 05. It's over text. 7.3 repeating 3.167 is a vertex. 00, I was a vertex and 10.50's over tax. The third step is to evaluate our objective function for the different vertex is. So I'm just using a scientific calculator and you will work And I'm gonna start plugging in. So this .05, that's our first vertex. I'm gonna plug that in for X. and Y. So four x. was five y. Let's do our next point for X plus five. Why then four X plus five? Y. And for X plus five Y. Again. Okay, so let's look at this. Let's just kind of copy R. Z values. When I plugged in 05, I got out 25 When I plugged in my seven and a 3rd and 3.167, I got out 45.167. Or we can make that into a fraction if you prefer. It's not much different. Okay, And then we got out when we plugged in zero, we got out zero. And when we plugged in 10.50, we got up 42. So now we're looking for a maximum and minimum and we look at these points and the Z values. This is our highest value. So this is our maximum maximum occurs at seven another And 3.167. And the maximum is a little more than 45. And our minimum, yeah, Is at the .00100.

We are going to be using linear programming to solve optimization problems. We have the objective functions equals four x plus y. We are looking for the maximum minimum values for Z. Given the list of five constraints. So our first step is to visualize these constraints. We want to graph them and see where they intercept. I'm using the dez most graphing calculator. It's free user friendly. One of my favorites, by the way. If you're using this calculator and you type in an or equals to inequality, you type in the less than or greater than symbol on your keyboard and then hit the equal sign on your keyboard. Let's take a look at these constraints. So X is greater than or equal to zero. So we've got a positive X. Y is also greater than or equal to zero. So, working in our first quadrant here, we're looking for where they overlap. We add in our other constraints. At this point, you've got a little triangle that's mapped out. Cut off a little piece of art triangle and cut off a little bit more. So we've got this quadrilateral down here that we're working with. The next step is to find the verdict is of the graph, the minimum or maximum foresee in this area will be at one of the vertex is So we want to find those vortices were these lines are overlapping where they're intersecting and I'm just gonna copy them onto our scratch paper here. So one vertex 05 7.3, repeating For 167, 10.50 and 00 mm. Now that we have our four vortices, one of them will be a maximum. One of them will be a minimum. Because this is a bounded area. I need to plug them into our objective function. So I'm just going to use Desmon scientific calculator here And use this objective function that we're gonna be plugging into. Its objective function is four X. Plus. Why? So for this first point I'm plugging in four times zero plus five And I got out five. Let's forget our other points. So for X plus five. No sorry for X plus Y. Okay, perfect. And then four X. Plus Why? And four x. plus y. Okay. So now let's copy over our Z values Attached to the vertex they go with. So when I plugged in 05, I got out five for objective function. When I plugged in seven and 3rd 3.67, I got out 32.5 with rounding and I plugged in 10.50. I got out 42. And when I plugged in 00 I got out zero. Now we're going to compare our Z values were looking for the maximum and the minimum Our maximum is the largest city value that we have. Our maximum occurs at the 10.50 where Z is 42, and our minimum occurs when We're at the zero and Z is zero.


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