Okay, so may we do better? The answer is definitely yes. Or if the answer is no, then we are not going to have this lecture, right? The answer is definitely yes. Let's see how to do this. We know the decision variable set changes from X1, X2 to X1, X2 and X3. All right. Now we have one additional variable. So there is one solution, which is X1 star, X2 star and zero. This basically means 2, 2, 0 in our example. Okay, this is certainly feasible. Why is that? Because all we are doing is to set X3 to be zero, if X3 is zero pretty much that means there is no such thing. And we only consider X1 and X2. If all the constraints, all the constraint coefficients are still there, then this is of course feasible. Okay, so our original optimal solution is definitely a new candidate for your new optimal solution. So this may serve as an initial searching point. Okay, initial basic feasible solution, in some sense. Okay, so what we are trying to do is that now, if we have this option which still ignores the new option, then all we need to do is to consider whether we want to add this new option. Okay, it's always possible that your new options so bad, it consumes so much resources, and it requires so many consumptions and it can be sold at a low price. If that's the case, we should not consider product 3. If we should not consider, then the current solution would remain optimal. And then we're done. Otherwise, all we need to do is to increase this currently non basic variables. And once we do that, we keep doing that until one basic variable becomes zero. Okay? And now you should be able to know how to do this. All you need is to calculate the reduced cost for X3. Okay. X3 is currently 0, currently non basic. You are considering whether to increase X3. That means you need to calculate its reduced cost. If it's reduced, causes negative in this example. Then we enter it because this is a maximization problem. So let's go back to our original problem, original problem is here, new problems here, we have a new column for X3 and now let's do this by considering the reduced cost. We want to calculate the reduced cost for X3, somehow that require us to recall our memory about our famous formula CB transpose, AB inverse, AN minus CN. Okay. This is a formula that gives us all the shadow, all the reduced cost once we determine a basis and a set of non basic variables. Okay. So somehow, if we focus on one, just one non basic variables, then this may be further reduced to AJ and a CJ. Okay. I hope you still have some memory about this. If you are not, maybe you want to do some review by looking at the previous lectures. Okay, so now let's look at our basis for the original problem. The basis the optimal basis is X1, X2. Because they are put into the basis, they are positive. Now we have a new decision variable, X3. We know it's zero, and that means it is non basic currently. And don't forget that you have 2 constraints right. You have 2 resources in the 2 constraints. So for any basic feasible solution, you have only 2 basic variables. So your new one must be non basic. Naturally. So to solve the new problem, we start from the basis, X1, X2 and the set of non basic variables now is expanded to include X3. Okay, so we are having a basic feasible solution. All we need to do is to use this B and N to calculate our CB transpose, AB inverse, AJ minus CJ. Where J means three. Because in this particular example is X3 and then pretty much we know how to do this. Okay. We don't need to do those things from scratch. So because tableau is an easier representation or easier visualization, let's used tableau to go through the process. Initially, we have this. Okay, this is our optimal tableau. All right, so you can see the identity metrics here. So now we want to ask what's going to happen if we have the new column? So the new column is here in the optimal tableau. Okay. We don't know what are the values, but we are able to calculate them. So we want to calculate the values in that column. The vector of constraints is AB inverse AN. Okay, according to our previous lectures. So all you need to do is to calculate AB inverse, AJ, where AJ here is the coefficient column for X3. So once we do the calculation, we know this part should be negative one. Okay, we'll have the connection here. Also, we want to calculate CB transpose. AB inverse, AN minus CN. All right. In this particular case, we focus on J. J is three. And CJ means eight. So all the calculations here tells you that, tells us that this is negative seven. Okay, so once we have all of this, we are able to go from this tableau and then do some more iterations, and then we're done. We are going to get our new optimal solution, pretty much if you consider this process and the previous process which goes from the scratch. We are omitting the first few iterations. Previously, we did a few iterations to get to here. No, we don't need to do that. It doesn't matter whether previously you have two iterations or you have two hundred iterations or two million iterations. We only need to go from here and to go from here, all you need to do is to calculate this new column, right? And you just invoke our formulas two times, one for the bottom numbers, one for the reduced cost. And don't forget that you don't need to work with AN and the CN. You just need to work with AJ and CJ. So everything now makes so clear and it's so easy. All you need to do is to simple some simple evaluation, and then you obtain this new column and then you know how to do. Okay. The last thing I want to remind you is that in fact, we don't need to really use the tableau because we already know that the metrics representation is a more natural and the easier way to do that. So you use the metrics to calculate this number to determine whether you need to go further. If you need calculate these numbers as the denominators, do the ratio test and then you know which variable to enter, which variable to leave, and then you follow the process will introduce to you. You are able to solve this problem. So a very quick conclusion here is that when you have a new activity, actually, you don't need to do the old things from scratch. You just go from your initial original optimal solution. Do a few iterations and then you will be done. You will get to your new optimal solution in a much quicker time. So then you will be able to equip yourself to do sensitivity analysis because there are all kinds of what if questions. What if our new problem is like new products like this? What if our new products like that, for example, previously we say $8? No consumption for resource one. No consumption for resource two, now we are able to ask, well, what's going to happen if it can be sold only at $5? Do it again. Well, what if this actually requires 0.5? Do it again. So there are many what if questions to answer. If you all need to solve all of them from scratch, that's too much time. So having this technique we just introduced to you would be beneficial if you really need to do those new activities.