CHAPTER 04.06: GAUSSIAN ELIMINATION: Naive Gauss Elimination: Round-off Error Issues: Example: Part 2 of 3
In this segment, we're going to continue talking about the example we are taking to illustrate the problems with getting large round off errors in Naive Gaussian method. Now we are not saying that that's going to happen every time, but it is more prone to it than the other methods.
So the example which we are doing is that we ended up with this at the end of the forward elimination steps, this is what we got, 20, 15, 10, 0, 0.001, 8.5, 0, 0, 23375, so we had three equations, three unknowns. We had three equations, three unknowns, and we are trying to solve for them, and this is what we obtained at the end of the forward elimination steps as the three equations, three unknowns, and what we want to do is we want to do back substitution now. We want to do back substitution to be able to show that what kind of values do we get for x1, x2, x3, and as we said that we are using five, in all of our operations, we are using five significant digits with chopping. So that's something which you have to keep in mind when you are doing forward elimination, that's what we did in the previous segment, and now here in the back substitution part.
So the first equation which I need to solve is the last equation, because that's why it's called back substitution, so I get 23375 x3 is equal to 23374, so in order to calculate x3 I will just divide the right-hand side by the coefficient of x3, and this is what I get, I get 0.99995, that's what I get. So again, keep in mind that this is the value of x3 which I get, 0.99995, and again, I've not used any more, I'm not showing you any more digits more than that, the reason why that is so is because we are using five significant digits with chopping, so we get x3 equal to 0.99995. Let's go ahead and see what x2 will turn out to be. So in order to calculate x2 I need to write down the second equation, which is 0.001 x2, plus 8.5 x3 is equal to 8.501, this is my second equation when I expand it. I already know what the value of x3 is, so I can calculate what the value of x2 is. So let me write down what x2 is, it will be 8.501, minus 8.5 x3, divided by 0.001. So this one here I get 8.501, minus 8.5 times 0.99995, and divided by 0.001, so that's the value which I have for x3. Now again, keep in mind that I cannot do this whole calculation on the calculator in one step, or . . . what I mean by one step is that you put in these numbers and whatever answer you get is the answer. No, you'll have to conduct this step first, use your five significant digits with chopping once you have multiplied the two, then do the subtraction, which is within five significant digits with chopping, and then do the division, with five significant digits with chopping, so each step has to be done separately, you get 8.501 minus this times this turns out to be 8.4995. So when you multiply this by this, you're going to get a different . . . you'll have some digits after this, but again, you are only using five significant digits with chopping, that's why I'm using one, two, three, four, five, divided by 0.001. Again, the subtraction between these two numbers has to be done with five significant digits with chopping, and that turns out to be 0.0015, now you're going to divide it by 0.001, and you're going to get 1.5 from there. So that's what turns out to be the value of x2 when you are conducting these steps by doing all the arithmetic operations, intermediate as well as final, we're using five significant digits with chopping, so 1.5 is the answer for x2. Let's go ahead and see what the value of x1 is. So if you look at the first equation, we have 20 x1, plus 15 x2, plus 10 x3 is equal to 8.501, that's what we have . . . not 8.501, it's 45, so that's the first equation which we have, and we're going to calculate x1 from here, x1 is 45, minus 15 times x2, minus 10 times x3, divided by 20. So we're going to substitute the value of x2, which is 1.5, the value of x3, which we got as 0.99995, divided by 20. Again, each of these operations have to be done by using five significant digits with chopping, so we have to be careful about what those numbers are, so you get 45, minus 22.5, and minus 9.9995, then divided by 20. Now I'm going to subtract this from 45, and whatever I get from there, within five significant digits, I'm going to subtract it from there, so that turns out to be 12.5, so each one has to be done separately, so I get 12.5, and then I divide it by 20, and I get . . . 12.500, and I get 0.625 as the answer for x1. So you can very well see now that by using these five significant digits with chopping, the value which I obtain for x1, x2, and x3 . . . x1, x2, and x3 which I obtain here, x1 is 0.625, x2 is 1.5, and this is 0.9995. The exact value which we have for this is actually 1, 1, and 1. So this is a good example to show that how Naive Gaussian method can . . . is prone to large round off errors. We have taken a specific example, but it doesn't happen all the time, but it's a good example to show that it can, even with . . . by taking five significant digits with chopping, which is a large number of significant digits which you are taking. And that's the end of this segment.