 In the last lecture we were solving goal programming problems using the graphical method for those problems that had only two decision variables and the rest of the variables were the deviation variables. Now today we will take an example that has more than two decision variables and more deviation variables and try to solve this using the simplex algorithm. So the objectives are minimize eta1 plus eta2 plus rho3 and then rho4 and eta5 subject to these set of constraints. Now because eta1, eta2 and rho3 are part of the objective function, the objective function has three parts. The first part contains these three which implies that these three constraints are the rigid constraints. So we start the simplex algorithm by considering this objective function and only the rigid constraints to begin with. So we can set up the simplex table like this which will have x1, x2, y1, y2, eta1, eta2, eta3, rho1, rho2, rho3 and a right hand side. So we can start the simplex table this way with when we write this x1 plus y1 plus eta1 minus rho1 equal to 20, x2 plus y2 plus eta2 minus rho2 equal to 20, 4x1 plus 3x2 plus eta3 minus rho3 equal to 90. Now we could start with this and we want to minimize eta1 plus eta2 plus rho3. So this is what we want to minimize, so this is how your first simplex table will look like. Because every constraint is an equation and every constraint has an eta or a rho, simply starting with eta1, eta2, eta3, the etas will always qualify to be initial basic variables and etas will have an identity matrix associated among themselves. For example, if you can have eta1, 1 0 0, 0 1 0, 0 0 1, you can do that and then your first simplex table will look like this with a 1 for eta1 with a 1 for eta2 and a 0 for eta3 with value equal to 40 here on the right hand side and then all these are 0s. So you calculate your Cj minus Zj or Zj minus Cj, you could calculate any one of them but you should remember that it is a minimization problem and also remember that we have not converted it to a maximization problem, we are solving the minimization problem as it is. So if you evaluate Cj minus Zj the usual way, you will get 1 into 1 1, so this will become minus 1, this will also become minus 1, this will become 0, this will become minus 1, this will also become minus 1, this will become 0, this will also become 0, you will get a plus 1 here, you will get a plus 1 here and you will get a 1 here and 40. So it is a minimization problem, you have calculated Cj minus Zj, so for a minimization problem what will happen? A negative Cj minus Zj will enter, so you could enter either this or this or this or this and so on and maybe you can start with this and then compute your Theta's and proceed in your simplex iteration till you reach the optimum solution. For want of time and because of our own familiarity, we will not go through the simplex iterations, we will only give you the optimum table and then see something interesting from the optimum table. So the optimum table alone will look like this, this will actually happen after three iterations and you will finally get something like this, you will have X1, Y1, X2 with 0, 0 and 0, you will have 1, 0, Y1, you will have 0, 1, 0, X2, you have 0, 0, 1, Y2, minus 3 by 4, plus 3 by 4, 1, eta 1, 0, 1, 0, eta 2, minus 3 by 4, 3 by 4, 1, eta 3, 1 by 4, minus 1 by 4, 0, rho 1, 0, minus 1, 0, rho 2, 3 by 4, minus 3 by 4, minus 1 and rho 3, minus 1 by 4, 1 by 4, 0, 15 by 2, 25 by 2 and 20 with Z equal to 0, the Cj minus Zj values will be 0.