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[M] Compare two methods for finding the steady-state vector $\mathbf{q}$ of a regular stochastic matrix $P :(1)$ computing $\mathbf{q}$ as in Example $5,$ or $(2)$ computing $P^{k}$ for some large value of $k$ and using one of the columns of $P^{k}$ as an approximation for q. IThe Study Guide describes a program nulbasis that almost automates method $(1) . ]$ Experiment with the largest random stochastic matrices your matrix program will allow, and use $k=100$ or some other large value. For each method, describe the time you need to enter the keystrokes and run your program. (Some versions of MATLAB have commands flops and tic $\ldots$ toc that record the number of floating point operations and the total elapsed time MATLAB uses.) Contrast the advantages of each method, and state which you prefer.

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Calculus 3

Chapter 4

Vector Spaces

Section 9

Applications to Markov Chains

Vectors

Campbell University

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

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In mathematics, a vector (…

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[M] Construct a random $4 …

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$A$ is an $m \times n$ mat…

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06:04

Hello, Room rate. We'll come back. Okay. I just finished doing number five on page 2 63 Chapter four, Section nine of Linear Algebra by Lay Lay and McDonald's tradition. And that's interesting because I'm doing now number 22 which is based on number five. Now what we found was in number five that if M is given as 0.1 point 6.94 and minus one is negative 10.9 positive 0.6 possible 9.6, which won't have an inverse, so we can't solve it the normal way. So what we did was we augmented with zero matrix to buy 10 matrix vector and try and think out student elimination, but were only able to do two of the four steps. But we got one and negative two thirds and zero in the first column, which means, see if I be X minus one x minus two thirds of sorry one times x one minus two thirds of X two equals zero and X being a um, what do they call it? A steady state vector is probabilistic. So the components that up to one So we solved in equation such that x one minus two thirds of x two equals zero and x one plus x two equals one and we've got the city state vector 0.4 point 62 fits in three fits right now. What's question 22? Well, as luck would have it, it's a it's saying we'll show that this answer that you just got I'm gonna separate me up with a little bit. Show that the same answer this answer you got is basically the limit off the mark of chains, which is, in other words, applying the matrix M two X and then the matrix em to the next X and the Matrix into my ex to in the text three into x for and see what you get. Okay, so let's do that. So here's a question. 22 on the same page, Page 29 to 63 in chapter four, Section nine. Okay. And, um, what we're gonna do is show that the answer to number five that we just got, uh, should be the limit of a Markov chain, which means just keep on playing m two x one and then take X to that answer X to play em two x two and then that answer x three Apply em two x three and so forth They keep on going. Okay, so let's see what happens if we do that. So let's make up some arbitrary initial X. Let's say X of zero. I'm just gonna make up something. Okay, let's make it a matrix. But what it really is is a to buy one, um, vector with. Just start with something arbitrary. How about 0.5 and 0.5? Whatever, that's what matter and then show. Let's show that make sure everything is is kosher. Here. Let's see. So x zero equals zero. Just make sure you set it up correctly and yeah, okay, you go. And what I want to do is see what happens as I apply em to that. And then again and again and again and again, right? Okay. So let's see what that iss. So let's show, um m times x zero. And what do we get? So that's m times X zero. All right, and we have them to find above EMS was 0.1 point 6.9 point four, which is just a statistic stochastic matrix, which we found the city state vector for Right. Okay, so let's run that. Hang on. Let's evaluate that and see what we get. Kevin, where you on we go? 8.35 point 65 Okay, so now we take that answer if I m to it. Are we doing a Aren't we doing em off M m times m times, like zero. So that's m squared. Oh, all right. So what I'm gonna do is I'm just gonna show the powers I'm gonna say instead of that, let's do, like, let's do a little loop. Okay, so how about four n in range? Let's start with one. These air experts will tell em to the Want to zoom to the three. Let's go to I don't know, 10. Now. Remember the Range command, if you just did Range 10 gives you 10 images from zero through nine. But do 1 to 10. It drops to zero, but it still stops at nine. So there's gonna do X one. It's 1 to 9 colon. And what I'm gonna do is I'm gonna show I'm gonna show em to the end. Let's see what happens. Alrighty. So 0.35 point 65.42 point 75.575 blah, blah, blah, blah, blah. It looks like it's getting darn close. 2.4 point six. Isn't it already? All right, let's get slightly more fancy here. How about if we actually print out the value of and they were talking about? So how about we put a quote here, Comma, come a quote. So it'll print up the end value that we're talking about and let's go, Let's do this. How about we start at 10 and go to 100 by 10 increments of 10? So will do 10. The experts will be 10 2030 40. Now, the question thing goes on and says, Okay, what if you do 100 and time it see how long it takes? But that's if you're using maple or Mathematica or, um, what else? Math lab math. I was really good from agencies, right on a PC with with, you know, limited number. Of course. Here we're running in the Google Compute platform, and it it was so fast. There's no point in timing this, um, this is online in the cloud, right? So it's sort of like you're running on a supercomputer. So this is pretty fast. And look 0.4 point six basically right. But now they wanted you to do 102 100 or let's try that. So let's do powers of hundreds. So how about let's start at 100? Let's go to 1000. That's increment by 100. No, that's what happens. Look how Francis is ridiculous. So there's no point in timing. This is just almost instantaneous. It's probably slower than one through nine, you know, But they go, and that's about as good as it's gonna get. I mean, there you go, it Z not diverging from 0.4 point six. There's some Randolph Air, probably at the at the end. There. That's probably not. And then a convergence at the end. There's like crazy Randolph error, playing with binary numbers and so forth. All right, so I think we covered this question. All right, so Question 22 it is done, and really all you had to do was those last few lines. If you want to take a look, number five have another video for number five more in detail for number five, so please take a look at that. All right, so there's the option Number five. It was helpful for over 22 and I'll just show the code for 22. All right, then. Right. Hope that was helpful. Uh, good luck with your maxilla.

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