 Hello friends, I am Mr. Sanjeev Vinay working as an assistant professor in Mechanical Engineering Department, Vulturen Institute of Technology, Singapore. In this video, I am explaining the graphical method which is used for solving the linear programming problem. At the end of the session, the learners will be able to formulate the linear programming problem as well will be able to determine the optimal solution to linear programming problem by graphical method. Graphical method is one of the simplest method which can be used to solve the linear programming problem as compared to other methods which are available like simplex method, dual simplex method, two phase method which are cumbersome whereas graphical method is very very simple. However, it has got the biggest drawback that it can be only used for the problems which are having only two variables that means the decision variables must be only two and that is the biggest problem. However, if the two variables are there involved in the linear program problem it can be very well solved by graphical method. Now, to explain the method let me consider an example of a manufacturing form which produces two products A and B and one unit of product A it contributes profit of rupees 30 whereas the product B contributes the profit of rupees 40. And each product is processed on two machines that is the requirement is the two machines lathe and milling machine and product A requires one hour of processing time on lathe and two hours of processing time on milling machine and similarly B requires one hour on lathe and one hour on milling machine. So, this is the requirement of each product per unit of the resources whereas these resources like lathe machine and milling machine they have got limited availability. So, per month lathe machine capacity available is 450 hours whereas the milling machine capacity available is 600 hours per month. So, by utilizing this available limited resource it is interested to know the number of units of product A and B to be produced per month so that the profit is maximum. So, this becomes an objective of solving the problem to derive the maximum profit by utilization of limited resources and that is why we have to convert into linear programming model. So, friends pause the video for a while and just recall how to formulate the LPP. In my previous video I explained an example how the LPP can be formulated. So, just you recall and try to formulate this problem as a LP model. Now, as far as the formulation is concerned we know that the decision variables are the number of units of product A to be produced so let it be expand variable and number of units of product B to be produced let it X2 variable so that the profit is maximum. Subjective function is maximization of profit and it is contributed by particular product A and B and that is why they become decision variables. So, we have identified these two as a decision variables and that is why with the help of these decision variables we can construct a linear equation for objective function which is to maximize the profit and that is why the maximization of profit Z as a function we get equation 30 X1 plus 40 X2 because X1 is the quantity of product A required to be produced as a variable and 30 rupees per unit contribution so total profit contributed by product A is 30 X1 whereas product B contributes 40 X2 40 rupees per unit contribution and that is why X2 unit if you decide then it becomes 40 X1 so total profit equation as a linear equation becomes 30 X1 plus 40 X2 so this is what a formulation of objective function as a linear equation now this is been subjected to the constraints so there are two constraints as a resource requirement one is a lathe machine constraint as you see over here so availability of this lathe is restricted and that is why it is a constraint so what is availability per month 450 hours and whereas the requirement for product A to produce X1 unit is X1 hours because it is 1 hour per unit so 1 into X1 1 X1 similarly product B also requires 1 hour of lathe machine time to manufacture 1 unit and that is why total time is X2 so linear equation for this constraint of lathe having 450 hours maximum which can be used as equal or one can use less than that so always this constraint is imposed as the sign less or equal so we can use maximum 450 hours or less than that for what for manufacturing product A quantity X1 and product B X2 so total is X1 plus X2 and that is why this becomes the first constraint equation as a linear equation for resource lathe similarly the other requirement is milling machine in which the maximum availability is 600 hours and that is why utilization of this resource is maximum up to 600 hours or less than that so once again the constraint is either less or equal whereas the contribution of product B for this capacity or machining hours is 2 X1 because 2 hours it requires for milling per unit so total requirement is 2 X1 on milling machine by product A and similarly for product B it is X2 because it requires 1 hour on milling machine so it becomes 1 X1 that is why 2 X1 plus X2 must be restricted to a value less or equal to 600 so this way we impose two constraint equations for this objective function and then we know that the product produced X1 and X2 is either 0 or more than 0 so that is why we add this restriction as X1 and X2 is greater or equal to 0 so this is the first part to convert given problem given to appropriate LP model. Now this problem can be solved by graphical method in which we draw a graph and the graph has got two axis which is X axis we use for X1 variable whereas X2 is shown by Y axis so X1 is taken along X and X2 is taken along Y axis and then we have to plot these constraints as a line so we consider only equality sign in this case and we can draw the line X1 plus X2 is equal to 450 as equation for first constraint. It is very simple that we put X1 and that is why X2 is 450 so the first point we can get is X1 0 X2 is 450 so this is one point on the line similarly putting X2 0 we can get X1 450 so this is another point on the line and we join the line through this point and that gives a line number 1. So any point on the line satisfy equality sign or any point below the line satisfy another constraint that is less so both sides can be satisfied by all the points on the line and below the line and that is what shown by the arrow so feasible solution lies on the line or below the line to satisfy this constant. Similarly we can draw another constraint as a line number 2 where we consider another equation 2 X1 plus X2 is equal to 600 so putting once again X1 X2 is 600 this point 0 600 and putting X2 0 X1 is 300 so 300 0 so this will be the line which can be drawn for second constraint so once again the same thing any point on the line or below the line satisfy both the signs less or equal so this is the way we have to draw two lines on the graph representing two constraints and we have to understand the feasible solution that points all the points we satisfy the particular line. Now we are interested to know the values of X1 and X2 or that particular point having some X1 and X2 we satisfy both the lines so if you want to satisfy both the lines we have to consider a point say A and E if I consider so this is the line number 1 on which any point on the line or below the line satisfy the constant 1 at the same time this A is below D and what is D? D is line number 2 and that is why any point on the line or below the line satisfy the constant and that is why any point on A and below the line satisfy both the constants similarly we play this logic that E and C is a second line any point on this line or below the line is the solution for that constant at the same time it is also below the line number 1 and that is why this is the area where all the points will satisfy both the constants and that is why any value of X1 and X2 in this area becomes a feasible solution and that is what is called the feasible solution region. However, we are not interested to any solution we are interested in the maximum solution because our objective function is maximum and naturally the maximum solution lies along the boundary any point within the boundary will have less value of X1 X2 corresponding to the point on the boundary and that is why I would interest it to find out the objective function as maximum value and that is why we consider the region feasible region and its vertices which is the boundary and at every point of boundary for example, A, E and C and O we calculate the profit Z with corresponding values of point as X1 and X2 and that gives us that if you look to this one A is the point at which the profit is maximum. So, profit is obtained is maximum at the A point and that is why its value is 18,000 rupees where number of units of A to be produced equal to 0 and number of units of product B is 450. So, this is the way we can find out the optimal solution for given problem in which we are deciding the quantity of product A and product B to be produced. So, that total profit is maximized 18,000 rupees these are my references.