 Satish Thallange, Assistant Professor, Department of Civil Engineering, Vyalchand Institute of Technology, Solapur. In today's session, we are going to see minimization of the objective function of the linear problem using graphical method. In today's session, the learner is going to determine the solution of linear problem problem for minimizing of the objective function using the graphical method. Now, let us see what is the linear problem. The linear problem problem is a mathematical modeling technique in which the following terms are involved. Here, the objective function is a first term which is defining the maximize and minimize case. Second is the n decision variables presents is the non-negative variables which are subjected to the set of constraints expressed by linear equations. It may be equality or inequality equations. The object of the linear problem problem is to maximize the profit benefits production of goods, etc. Second, to minimize the time loss and the wastage of resource, etc. Finally, it is planning and making of decision about resource allocation. Now, the particular linear problem problem can be solved by following method that is graphical method. It is also known as two variable method. Second, simplex method, third, begin method and last is two phase method. Now, let us see what is the graphical method. The graphical method is suitable for solving the linear problem problem having only two variables. It is going to find out the feasible region satisfying most of the constraints present in the problem and it helps to find out the highest and lowest point on the graph which is going to give the optimum solution according to the case. These are the steps involved in the solving of the linear problem problem by the graphical method. Now, let us start to solve the particular problem of minimize case. This is a problem in which the multiple having the water system for the distribution. Here, they are distributing the water by the gravitational method but they are facing the problem in the distribution. So, they are finally interested to place the mechanical pumps for the distribution and they are interested to find out the pressure required, minimum pressure required by the both the methods. You can say gravitational method as well as mechanically pump that water distribution should be properly. They are saying that they have three stations housing is state A, B and C and these are the particularly pressures with the gravitational method and the mechanical method for each station. Now, let us start to convert the particular problem in the LPP form. As we have seen in the first step our objective is to minimize the pressure required in the municipality for the water distribution. X1 and X2 are the two non-negative variables of gravitational method and the mechanical booster pumps. As it is a case of minimizing the pressure or to find out the minimum pressure, minimize case is equal to that is equal to X1 plus 8 X2. In the first step we are observing that there are the set of constraints for the each stations A, B and C. These are the particularly constraints for the A, B and C. Finally, we have to give the restriction to the non-negative variables. This is a plot of the particular problem in the graphical method. Here there are the three lines which are highlighting the three constraints that is of housing state A, housing state B and housing state C. As we are seeing that each state having the restriction of the pressure. In the first step they are saying that the pressure should not be of 24 kilo Newton per meter square and for housing state B the pressure should not be more than 132 kilo Newton per meter square and for the state C it should not be less than 12 kilo Newton per meter square. With the help of the particular equation that are inequality equation which are going to be converted to equality to get the values of X1 and X2 by obtaining these values we are going to plot these lines. Now we have to finalize the areas. This is a slide which is showing the areas or the arrows of two lines are towards the origin and one of the arrow is away from the origin. The first and second constraints having the limitations of the pressure. So they have maximum so they are towards the origin and the third they are saying that minimum 12 kilo Newton per meter square pressure is required that is why the arrows are away from the origin. When we are observing this particular graph we are getting three areas area X area Y and area Z. Out of these three areas area X is only satisfying the first constraint that is of housing state A and second area Y it is satisfying the all three constraints that is housing state A housing state B and housing state C and the particular area Z is only satisfying the third constraint we can say the area Z is only satisfying the one constraint. Here when we see this particular area only area Y is satisfying all the constraints that is why it is a feasible region. Now let us see the corner points of the feasible region to get the optimal solution. Once we see points on the boundary of the feasible region B, C, E, F and the point P are the points which are lying on the boundaries of feasible region. Now let us get the values or you can say the coordinates values of the particular points here B, C, E, F and P are the points of the feasible region having the coordinate values as shown on the slide. Now let us place these values in the objective function. This is a table which is showing you the various point B, C, P, E, F and its coordinates values. These values are going to be placed in the objective function and we will get the value of Z. When we observe this particular column of Z value we are observing that for the point B we are getting Z is equal to 96, for C we are getting Z is equal to 37.7, for P it is getting 66, for E it is getting 24 and for F it is getting 48. Here as we are interested to minimize the pressure or we are interested to find out the minimum pressure for the multiple water supply line we have to select such a point which is satisfying all the constraint of housing state A, B and C. When we are seeing this column we are saying that we are getting that minimum water pressure we are observing here it is 24, but this is a point E which is only satisfying the third you can say which is only satisfying the third constraint. So, we have to select such a point which is satisfying all the constraint. Here when we observe this graph P is a point see as I am shown here point of intersection P which is satisfying all the constraint of housing state A, B and C and the Z value here it is 66 kilo Newton per meter square. Now, finally we are getting the optimum solution that is a P is a point which is giving the optimum solution of minimum pressure required in the multiple T to supply the water. Here it is a point which is having the coordinate value 15 and 4.5. When we place this in the particular objective function we are getting 66 kilo Newton which is the minimum pressure required for the multiple T. Now, let us see the particular comparison. In maximize case we are going to maximize the particular benefits and here we are going to minimize the images like wastage of resource loss etcetera. Here in the maximize case resource limitations are always there, but in the minimize case resource limitations may or may not be there. The feasible region is most of the time it is bonded here the feasible region may or may not be bonded. Here we are going to find out the optimum point by selecting within the bonded region or you can say feasible region or on the boundary of the feasible region. Here similarly we are going to find out the optimum point which is lying within the bonded or unbonded feasible region. Here optimum point are the points which are far away from the origin and in the minimize case the optimum point are closer to the origin. Finally, there is one more check the iso profit line is used in the maximize case and the iso loss line is used in the minimize case to get the optimum solution. Select the correct answers for the question. Hope so you have selected the correct answer. These are the references for the today's session. Thank you.