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Graphing LP problems

Graphing LP problems. OVERVIEW. Have you ever heard the saying about a picture saying 1000 words, or something like that? (Okay, I don’t have the words exactly correct, but I am in the ball park, right?)

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Graphing LP problems

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  1. Graphing LP problems

  2. OVERVIEW Have you ever heard the saying about a picture saying 1000 words, or something like that? (Okay, I don’t have the words exactly correct, but I am in the ball park, right?) Well, in linear programming problems it is often felt that using graphs can be an aid for you. An you have to make the graphs! PROBLEM – I do not want you to have to learn how to make graphs on the computer, or to have you scan in your hand made graphs and e-mail them to me. In this section I want to show you how to give me information when you are asked to do graphs.

  3. Example Here is our basic linear programming problem. We want to Maximize profit = 7x1 + 5x2, but we have constraints 4x1 + 3x2 <= 240 2x1 + 1x2 <= 100, Where x1 stands for tables and x2 stands for chairs and <= means less than or equal to. The point is we want to make as much profit as we can by making tables and chairs, but we face constraints that give us road blocks to overcome. Let’s see the constraints in the same graph on the next slide.

  4. Example Recall we said the feasible region would be made up of the trapizoinkal (is that a word?) area BCD and the two axes back to the origin, in this example. X2 A B C X1 D E

  5. Example – vertical intercept Our two constraints again are 4x1 + 3x2 <= 240 carpentry 2x1 + 1x2 <= 100 painting How do we know which line has vertical intercept A and which has B? To find out: 1) Work with the equality part of each line, 2) Set x1 = 0 in each line, and 3) Solve for x2 in each line. So, in carpentry 4(0) + 3x2 = 240, or x2 = 240/3 = 80, and in painting 2(0) + 1x2 = 100, or x2 = 100. Thus point A in the graph is (0, 100) and point B is (0, 80). So, in a graphical sense you only have to report on point B on the vertical axis. In fact you should report Vertical axis intercept (0, 80)

  6. Example – horizontal intercept How do we know which line has horizontal intercept D and which has E? To find out: 1) Work with the equality part of each line, 2) Set x2 = 0 in each line, and 3) Solve for x1 in each line. So, in carpentry 4x1 + 3(0) = 240, or x1 = 240/4 = 60, and in painting 2x1 + 1(0) = 100, or x1 = 50. Thus point D in the graph is (50, 0) and point E is (60, 0). So, in a graphical sense you only have to report on point D on the horizontal axis. In fact you should report horizontal axis intercept (60, 0)

  7. Example – where the constraints cross Next I want to show in Excel how to find point C, where the constraints cross. Now think about our two constraints in equality form and in matrix form 4x1 + 3x2 = 240 4 3 x1 = 240 2x1 + 1x2 = 100 2 1 x2 100 Now go to the Excel file to see how to solve this part. Then report point C in our graph as Where the constraints cross is (30, 40)

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