Next, we will talk about one thing called the tableau representation. Pretty much is a way to simplify the calculation for the simplex method and to let us concentrate on the concepts. That's also some chances for us to see a little bit more examples. Pretty much what we were trying to do is to play with the linear systems. Originally we have several constraints. They are equalities and then we add one constraint which represents the objective role. Once we do other things, we get a system and then we do it repeatedly again, again and again with some row operations. Let me use this section to give you a summary with simple notations. When we update those systems, typically we don't need to write down variables. Just like the Gaussian elimination. The Gaussian elimination is to play with systems of equations. Pretty much what we do is to eliminate those variables when we do all those calculations. All we need to do is to writing down those coefficients because there is no need to repeatedly writing down all those variables. The tableau approach of running a simplex method is again to ignore all those variables. Also if you go back to the previous slides, you will see that the column with z actually never changes so we would also exclude that from a tableau. Our initial system, if it looked this way, what we will do is that we will focus on these parts, we exclude the column for z, and we record only numbers. We would copy all the numbers to our tableau and then may be the only variable that I want to leave it in the tableau is a representation or a record for basic variables. I'm going to specifically remind myself, what are the basic variables, in which row do we have each basic variable, and what's their corresponding values. When we do this, when we abbreviate all the systems into tableaus, each tableau represents a set of linear equations. The basic columns here would contain zero in the zero's row, is here, and identity matrix in the remaining rows. The identity metrics associated each row with a basic variable. That's how you see for this particular row, you'll know that basic variable is x_3, because x_3 is identified in the identity matrix as one. For the second one you'll know is x_4 because that one indicates the variable is x_4. Then a negative number in the zeros row of a non-basic column means that variable can enter, for this particular maximization problem. Basically, if we are running all the simplex iterations with tableaus, pretty much we are eliminating the column for z and eliminating all those variables and then record just numbers into our tableau. Also, we have the basic variables with us, so that we don't get confused about which basic variables are we talking about. Also, we may still remember that when we have a system like this, first we look at the zeros row coefficients, and then once we identify one that is negative, we know this should be an entering variable, we may pick that as an entering variable. Then we do some ratio test. Here, the ratio test would be carried according to 6, 8 and 1, 2. Once we do the ratio test and find the smallest ratio according to this one, then we know we get one column to enter, that's for x_1, one row to leave, that's for x_4, so x_1 should enter, x_4 should leave, and their intersection is [inaudible] here. That number is considered as the pivot. The pivot here has nothing to do with the pivot you just did in your linear algebra. These particular number simply tells us that we should make it one, all other numbers above it or below it, zero, to make it something like this. We are going to have a new column for the identity matrix for the basic columns by doing row operations. Once we do row operations, we get to the second tableau. This is a negative number, we do the ratio test, we get the minimum ratio. This should be our pivot, and then we do it again to make it one all others zero. Then here we get to our third tableau. Everything is non-negative at the zeros role so we're done. This is our optimal tableau, which corresponds to the optimal basis or optimal solution. Optimal solution is x_1 should be ten over three, x_2 should be four over three, and our z value is thirty-two over three.
