The point P, is the furthermost point, since no area is seen further. Move the line (parallel to line L1) away from the origin to locate the furthermost point. The line L1 does not give the maximum profit because the furthermost point of the feasible area lies above the line L1. The objective function line contains all possible combinations of values of xl and x 2. Therefore, the co-ordinates for the objective function line are (0, 15), (10, 0) as indicated by dotted line L1 in Figure. Substituting x 1 = 0, we get x 2 = 15 and To locate the point, we need to plot the objective function (profit) line.Įquate the objective function for any specific profit value Z, The point that lies at the furthermost point of the feasible area will give the maximum profit. Now, the objective is to maximize the profit. If the points are substituted in all the equations, it should satisfy the conditions.Ģx 1 + 3x 2≤ 120 = 30 + 60 ≤ 120 = 90 ≤ 120 For example, let the solution point be (15,20) which lies in the feasible region. Now any point in the shaded portion will satisfy the constraint equations. This indicates that while considering both the constraints, the feasible region gets reduced further. When the second constraint is drawn, you may notice that a portion of feasible area is cut. Now the second constraints line is drawn. The no shaded portion can be seen is the feasible area shown in Figure (Note: If the constraint type is ≥ then the solution zone area lies away from the origin in the opposite direction). The inequality constraint of the first line is (less than or equal to) ≤ type which means the feasible solution zone lies towards the origin. The lines are drawn on a graph with horizontal and vertical axis representing boxes x 1 and x 2 respectively. The line 2x 1 + x 2 = 60 passes through co-ordinates (0, 60) (30, 0). In equation (1), put x 1 = 0 to get x 2 and vice versa (2)įind the co-ordinates of the lines by substituting x 1 = 0 and x 2 = 0 in each equation. (ii)Īs a first step, the inequality constraints are removed by replacing ‘equal to’ sign to give the following equations:Ģx 1 + 3x 2 = 120. The available machine-hours for each machine and the time consumed by each product are given.Ģx 1 + 3x 2≤ 120. Given profits on corrugated box and carton box are Rs. The objective is to maximize the profits. X 2 be the number of carton boxes to be manufactured Let x 1 be the number of corrugated boxes to be manufactured. To determine how many (number of) corrugated and carton boxes are to be manufactured. Determine the optimum quantities of the two boxes to maximize the profits. The available operating time is 120 minutes and 60 minutes for cutting and pinning machines. Each corrugated box requires 2 minutes for cutting and 3 minutes for pinning operation, whereas each carton box requires 2 minutes for cutting and 1 minute for pinning. The boxes undergo two major processes: cutting and pinning operations. The solution method of solving the problem through graphical method is discussed with an example given below.Ī company manufactures two types of boxes, corrugated and ordinary cartons. Though in real-life, the two variable problems are practiced very little, the interpretation of this method will help to understand the simplex method. Linear programming problems with two variables can be represented and solved graphically with ease. It also helps to understand the different terminologies associated with the solution of LPP. Graphical method to solve Linear Programming problem (LPP) helps to visualize the procedure explicitly.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |