Let's continue. Now let's consider our next issue, which is a case that we will use to talk about your simplex. Previously, what we were trying to do is that when we have linear program, we are considering one additional column. Or if you think about production planning, we're talking about a new activity. But now what we are doing is that sometimes we have new constraints. For example, here we are doing production. Now maybe we have a new constraint, something like this. Some people say we can do at most one unit of Product 1, for example. There can be all reasons for you to have a new constraint. Maybe there is another critical resource that is needed for producing some part of your products. Maybe there is a new regulation saying that your total supply can not be too much, for example. Anyway, when this happens, obviously we need to do another sensitivity analysis. We need to understand what's the impact of this new constraint on our optimal solution. Do we need to do any adjustments for it? The most intuitive way to consider the impact is to plug in the original optimal solution into that new constraint. Suppose the original optimal solution satisfies these constraints, then obliviously ED would be optimum. The only thing that we need to worry about is when it is not feasible. In our example here, because we try to produce two units of Product 1, but we are required to produce at most one unit. This is infeasible, then we need to do something else. For this particular example, what we need to do is that we will consider a new row. Somehow, our previous example when we have a new activity gives us some hint saying that maybe what we should do is to try to look at the tableau again. Why not? Because we are able to construct that new tableau. Let's try to do this. Our original ultimo tableau is here. Something you are familiar with, with a new constraint which is x_1 less than or equal to 1, may be we may construct our new tableau. Obviously, our new tableau would have one additional new row. Somehow we need to rely on metrics representation to find it because we know here we have basic columns, non-basic columns, and so on and so on. We somehow need to apply laws AB inverse AM blah blah blah, or to construct all these numbers here. If we are able to do that, maybe we get some more hints. Here we basically have some parts, correlated or basic columns. Some parts correlate to non-basic columns and we have a right-hand side. We're going to apply formulas for them. But maybe your first issue is about whether your new slack variable should be a basic variable or not, so why do we have a new slack variable? Well, when we add a new constraint, obviously the new constraint according to our formulation, is less than or equal to xy is less than or equal to 1. We need to make it equation if we are talking about standard form, so we will introduce S_3 as our new variable. Here because we have now three constraints, so we need three basic variables. Previously when we have that capital B, capital N, we have two basic variables and the two non-basic variables. If now we need three basic variables, naturally, this new one should be basic. It also makes sense because when you cannot satisfy this constraint, somehow there must be a gap between your solution and that constraint. In that case you must have a slack, you must have a gap there. That's S_3 is our basic variable. Then our capital B would become x_1, x_2, and x_3. Capital N remains to be S_1 S_2. When we have that, then we are almost done. We will write down CB CN AB AN and the right hand side B, we can see that this slake S_3 contributes this 0 to CB and also contributes this column 001 to AB. Then we will be able to calculate AB inverse, AB inverse AN, AB inverse B, and so on and so on. Obviously, all this can be done by just applying our matrix representation. Of course, you also know that there are some parts that we don't really need to calculate at this moment so this is just a double check. This is also just a double-check. They don't really change at all. What may be changed is AB inverse AN and AB inverse B, because now we need one additional row. You may see that here. Those negative two and one, and the negative one would enter our tableau. For the remaining numbers, they are just formed according to that identity matrix that we need to have for basic columns. This completes our calculation for constructing the new tableau. This is where we are when we are having these new additional constraint at our objective, our original optimal solution. This looks good. But unfortunately, we are unable to go from here to do a few iterations. Previously when we have one additional column, what we look at is the reduced cost. If the reduced cost is negative, we know what to do. We do some iterations and then we're done. But here, what is negative is the right-hand side. Previously we have no experience talking about the negative right-hand side during simplex iterations, because this does not happen. This is actually when we have a negative term. This is actually an invalid the simplex tableau. Previously, when we are running simplest method, if all your calculation is correct, a right-hand side value cannot be negative. When we have a negative number here, well, we know we need to fix it somehow. But the thing is that for this infeasible basic feasible solution, for this infeasible basis, we need to somehow find a way to solve it, but we don't know how to do that. We need some more other ideas to do that. That's something we will tell you later.
