LP formulation and solution

Editor's Notes

• #3 Then the inequalities are plotted on a X-Y plane. Once we have plotted all the inequalities on a graph the intersecting region gives us a feasible region. The feasible region explains what all values our model can take. And it also gives us the optimal solution.
• #5 To solve this problem, first we gonna formulate our linear program. Formulation of Linear Problem
• #6 To solve this problem, first we gonna formulate our linear program. Formulation of Linear Problem
• #7 Step 3: Writing the constraints  1. It is given that the farmer has a total budget of US$10,000. The cost of producing Wheat and Barley per hectare is also given to us. We have an upper cap on the total cost spent by the farmer. So our equation becomes: 100X + 200Y ≤ 10,000  2. The next constraint is, the upper cap on the availability on the total number of man-days for planning horizon. The total number of man-days available are 1200. As per the table, we are given the man-days per hectare for Wheat and Barley. 10X + 30Y ≤ 1200 3. The third constraint is the total area present for plantation. The total available area is 110 hectares. So the equation becomes, X + Y ≤ 110
• #8 Step 3: Writing the constraints  1. It is given that the farmer has a total budget of US$10,000. The cost of producing Wheat and Barley per hectare is also given to us. We have an upper cap on the total cost spent by the farmer. So our equation becomes: 100X + 200Y ≤ 10,000  2. The next constraint is, the upper cap on the availability on the total number of man-days for planning horizon. The total number of man-days available are 1200. As per the table, we are given the man-days per hectare for Wheat and Barley. 10X + 30Y ≤ 1200 3. The third constraint is the total area present for plantation. The total available area is 110 hectares. So the equation becomes, X + Y ≤ 110
• #9 Step 3: Writing the constraints  100X + 200Y ≤ 10,000  10X + 30Y ≤ 1200 X + Y ≤ 110 Since we know that X, Y ≥ 0. We will consider only the first quadrant. To plot for the graph for the above equations, first I will simplify all the equations. 100X + 200Y ≤ 10,000 can be simplified to X + 2Y ≤ 100 by dividing by 100. 10X + 30Y ≤ 1200 can be simplified to X + 3Y ≤ 120 by dividing by 10. The third equation is in its simplified form, X + Y ≤ 110. Plot the first 2 lines on a graph in first quadrant (like shown below) The optimal feasible solution is achieved at the point of intersection where the budget & man-days constraints are active. This means the point at which the equations X + 2Y ≤ 100 and X + 3Y ≤ 120 intersect gives us the optimal solution. The values for X and Y which gives the optimal solution is at (60,20). To maximize profit the farmer should produce Wheat and Barley in 60 hectares and 20 hectares of land respectively. The maximum profit the company will gain is, Max Z = 50 * (60) + 120 * (20) =  US$5400
• #10 Plot the first 2 lines on a graph in first quadrant (like shown below) The optimal feasible solution is achieved at the point of intersection where the budget & man-days constraints are active. This means the point at which the equations X + 2Y ≤ 100 and X + 3Y ≤ 120 intersect gives us the optimal solution. The values for X and Y which gives the optimal solution is at (60,20). To maximize profit the farmer should produce Wheat and Barley in 60 hectares and 20 hectares of land respectively. The maximum profit the company will gain is, Max Z = 50 * (60) + 120 * (20) =  US$5400
• #11 Step 9: Once the model is saved click on Data tab then click solve. The optimal solution and values are displayed in the corresponding cells. The optimal minimum cost is US$0.90. Sara should consume 3 units of Food Item 2 and 1 unit of Food Item 3 for the required nutrient content at the minimum cost. This is solves our linear program.