The thing is, actually because we know x_2 has some impact on x_1 and x_3, maybe we should fix the system or update the system before we try to do our second move. The idea is the following. Previously, when we have these two variables, non-basic, and then this x_3 and x_4 are basic, this seems not to be too difficult. Why is that? Because at that moment, if you increase x_1 or if you increase x_2, it has some impact on x_3 and x_4, but because all your basic variables only appear in one column, you don't need to look at too much constraints. You just need to take a look at one constraint to consider the impact on that basic variable, because that basic variable does not appear in any other constraints so the evaluation would be easy. Regarding the impact if the basic variables does not appear in the zeros row. You also don't need to consider how the non-basic variable would affect these basic ones. You just need to look at the coefficient and then you are done. In other words, in summary, what we plan to do is that before we evaluate the impact of increasing x_2 in the second move, we want to do something for the basic columns. We hope for basic variables we have an identity matrix at row 1 to row n, and also for the zeros row, we hope it would be a 0 vector. If that's the case, then all the things would be easily done. Going back to our original formulation, if we are having this system and we are talking about x_1, which is 0, 0, 6, 8. For the basic columns, indeed, we have the identity matrix and the 0 vector in the objective function. Now if x_1 enters and x_4 leaves, what we want to do is we now should have a new basis, which is x_1 and x_3. So we need to update the system so that the system becomes some equivalent thing such that for x_1, it only appear in the second row and for x_3, it only appear in the first row. Well, how do you know that it should be x_1, be the second row, and the x_3 in the first row? Obviously, it is because you are x_3 is already occupying your first row. So your new entering variable x_1 should only occupy the second row. There must be one row that is originally covered by original basic variable. Now this basic variable would leave a new basic variable should enter, so the newly added basic variable somehow must occupy the row that was originally occupied by the leaving variable. Everything makes sense. How may we somehow do something on this system to make it this way? Well, elementary row operations. Because we have some equations, we want to get another set of equations that is equivalent to the original one without changing any solution, but getting forms in this way. That's row operations. What we need to do is to removing all other values for x_1 in x_1's column so all we need to do is to multiply the second constraint by 1.5. That's going to make these two becoming one, and then we need to change all other numbers, of course and then we would try to eliminate this x_1 in your first row. We would multiply the second row by negative one, and then add that into the first row. Once we do that, your x_1 would disappear and there are some impact on the other variables. Finally, we may want to multiply the second constraint by one and add that into your zeros row. So here, when we are saying that you would like to multiply something by one or by negative one, you really need to consider whether you are talking about the values before you multiply this by 1.5 or after you do that by 1.5. If you multiply 2 by 1.5, then this number actually becomes one. This becomes 1.5, this becomes 1.5, this becomes 4. Once you do that, afterwards, when you want to eliminate that x_1 in your first row, you multiply your second row by negative one. And then once you do that, if you want to eliminate your x_1 in the zeros row, actually you should multiply that number by two instead of one. But that's, there's some arithmetic s and that basically follows the principle of row operations. After you do all these steps, your system becomes this one. Then updating the system would also gives us objective value where our new basic variables are 4 and 2 and our z_2 will still be eight. Now let's take a look at how this new system would help us. We are looking at this new solution 4, 0, 2, 0. So x_1 and x_3 layout basic 4 and the 2. z2 is eight, no problem. Now what we are doing is that one of x_2 and x_4 may enter. If x_2 enters, we can see that this is negative, so z would go up. That's great. If x_4 enters, however, that simply decrease your objective value z. That's bad. So very quickly we may see, okay, so for x_2, that's a good entering variable because the coefficient is negative. Then once x_2 enters, your row 1 says that it has some impact on your x_3, right? For row two, you know, x_2 has some impact on x_1. Then we all know how to do the calculations. All we need to do is to look at the entering column, 3 over 2 and 1 over 1 and the right-hand side, 2, 4. All you need to do is to do these ratios and then pick the smaller ratios or the smallest ratios in general. When x_2 goes up, x_3 would become 0 sooner than x_1. So it is x_3, that would leave the basis. So you are basic variables becomes x_1 and x_2 naturally, because x_2 is entering and x_3 is leaving, that's how it remains to be x_1 and x_2. So in particular, we are talking about the process. The idea is still the same. Having one to enter, having one to leave is just that. If we want, all the later evaluations can be easily obtained, then we need to fix the system to maintain an identity matrix for basic variables in the constraints, and a 0 vector, for basic variables in the objective row. Once we are done with that, all the remaining things may be easily carried out.
