 Hello friends, I am Mr. Sanjubi Naik working as an assistant professor in Mechanical Engineering Department, Walton Institute of Technology, Singapore. In this video, I am explaining the graphical method to solve the linear programming problem in continuation of my earlier video. At the end of this session, learners will be able to determine the optimal solution to linear programming problem by graphical method. So, let me consider another example over here which is having mixed constraints. It means that the constraints if you look over here, one of them is having less or equal sign whereas remaining two are having greater or equal sign and that is why it is called the mixed constraint problem. So, when LPP is having mixed constraint, let us see how it is solved by graphical method. As it is having two variables, we can very well use the graphical method to solve this problem. And as we know that we have to construct the graph where we plot this constraint as an equation of a line and we plot that line considering equality sign. So every equation is considered or every constraint is considered just by equality sign. So first constraint if you take here x1 plus 2 x2 is equal to 40 and to plot this line on the graph, what we do is we assume x1 0 and x2 is 20 then we get one point on the line. Similarly, if you assume x2 is 0, we get another point x1 is 40 and that is another point on the line. So joining this a and b, we can plot the line 1 on the graph. Similarly we have to consider the constraint equation 2 as a line 2 and constraint equation 3 as a line 3. So let line 2 is the second constraint 3x1 plus x2 which is equal to 30. So we can establish the same way the points c and d for this line and the line 3 will be for third constraint 4x1 plus 3x2 is equal to 60. We can establish 2 points e and f. So by joining these points we get third line. So now we consider a graph having axis x for x1 variable and having axis y for x2 variable. So we take x2 all this axis and x1 along this axis and then we can plot these lines. As a line 1 we know it is passing through point a and b. So joining a and b we can get this as a line number 1 whereas line number 2 which is a constraint 2 is between c and d. So the points obtained are c and d for this and the third line we can draw for e and f. So e and a points merging at same because the values are same for 0 and 20 and that is why e and f will be third line. So this is the way first we have to draw all the 3 lines corresponding to 3 constraint equations where we are considering only equality sign. Now we have to look to other constraint because all the equations are having not only equals and some of them are less or equal. For example line number 1 which is a, b line it is having less or equal sign. So any point on the line or below the line is satisfying that constraint. Similarly second equation is a c, d line and that is why any point on the line but the constraint was greater or equal. That means any point on the line or above the line will satisfy that equation and c will be third equation is satisfied by e f line. Any point on the line once again it is having greater or equal sign if you look to this given problem. So any point on the line or above the line. So that means we have to consider that particular region which satisfy all the equations and there we have to search for optimal solution. So as you know we have performed this establishing the feasible region for the problem. Now I just appeal you that recall how to identify the feasible region. So in my previous video we have solved example where feasible region is identified. So try to identify feasible region for this problem also. Now look here the feasible region is identified just as I said for first constraint it was less or equal sign for that constraint. So any point on the line or below the line is the solution for this constraint. We satisfy this constraint. If I consider second line that is c, d line any point on the line or above the line because it was greater or equal when it is greater or equal equal means minimum and greater means more than that. So it is on the line or above the line will be the solution space and as well as for third equation also it was greater or equal. And that is why any point on this line or above the line is the solution. So now it gives me a common area this where any point in this area satisfy all these equations all three equations and that we call as a feasible solution space or feasible solution region. So we get a solution to the problem in this region as possible solution or feasible solution. But we know that we are not interested in any of the solution we are interested in the minimum solution. So always minimum solution or maximum solution lies on the boundary. And that is why we have to search for the solution at the boundary points that is for point which are been provided at g, h. So that is what region we define g, h, f and b. So these four points we have to check for the solution. And what is the solution we expect as optimization minimum value of z. So this is the problem where we can get minimum value of z by searching the solution at the vertices of this region. And that is been obtained by considering once again the value of z at different points on the boundary. So if you substitute g value as x1 and x2, 4 and 18 which is intersection of two lines which can be easily calculated by Sametian's equation we can calculate this intersection point. Similarly h is also intersection between two lines which we can find out. And substituting this x1 and x2 in this z function we get the value of z. What is needed over here is minimum solution. So we can easily identify from this what is the minimum solution. So from above of the graph and its feasible region we can calculate the minimum value of z which is 240 which is occurring at point h where the value of x1 is naturally 6 and 18. And that gives optimum solution to a problem under consideration by graphical method. So let me consider one more example where we are finding that the constraints are same type that is less or equal, less or equal and less or equal. And that is why in this case the objective function is maximized of z as 8x1 plus 5x2 subject to constraints less or equal 500 x1 is less or equal to 150 and x2 is less or equal to 250. And that is why we have to plot these three lines on the graph. So plotting these lines on the graph as usual we consider equality sign. So 2x1 plus x2 is 500 so it gives two points a as 0, 500 and b is 250, 0 then x2 second line is x1, 150 that is irrespective of x2 so it depends only on x1 and that is 150. And that is why it is a line parallel to x2. So whatever value of x2 x1 is fixed so it will be line parallel to x2. Similarly line 3 for third constraint it is x2 is fixed as 250 so this line will be parallel to x1 axis, line will be parallel to x1 axis. So this is the way we have to draw the lines for three constraints considering once again x1 along x and x2 along y axis. The first line is ab point which is joint any point on the line or below the line is the solution we know that because it is less or equal. The second line is a line parallel to x2 axis any point on the line or below because constraint is once again less or equal whereas for third line any point on the line or below the line is the solution because constraints are less or equal. And this is the way we can see that the solution is lying in this region which is a common to all three equations. So all three constraints are satisfied by this space as a solution and that is what we call as a feasible solution region ODEFC. So now work left out is to decide the objective function value which is maximized z. So we want maximum value over here and that is why we consider all these points on the vertices by considering it is x1 and x2 value putting in this equation of z. So we can derive these values and looking to this particular table we can identify that according to a feasible region space at point E we get the maximum solution of z. So at maximum value of z is obtained at point E where it is 125 and 250 x1 is 125 x2 is 250 is the solution for given problem. So this is the way we can find out optimal solution for this problem by using graphical method. These are my references. Thank you.