Simple Linear Programming Model Katie Pease In partial fulfillment of the requirements for the Master of Arts in Teaching with a Specialization in the Teaching of Middle Level Mathematics in the Department of Mathematics. David Fowler, Advisor July 8, 2008
Linear Programming 2 Introduction I was given the task of studying a simple linear programming model and creating two of my own. Before I could create problems of my own I had to understand what linear programming was and how to model it. Linear programming is the process of finding a maximum or minimum of a linear objective function subject to a system of linear constraints. In a linear programming problem, we are trying to find the maximum or minimum of a linear objective function in the form ax+by+cz+. where a, b, c, etc. are the coefficients of the variables x, y, z, etc. For this type of linear programming a method called the simplex method can be used to solve the problem. For the case of this study I have chosen to focus on two variables so that I am able to solve the problems using the graphical method. In the graphical method I can graph the constraints, find their vertices, and then check the vertices with the objective expression to find the maximum or minimum. With this method I am able to then graph the objective function on top of the constraints graph to form a sort of flat roof showing why the fundamental theorem of linear programming is true. The Fundamental Theorem of Linear Programming is used to find the optimum values. The Fundamental Theorem, which also known as the Corner Point Theorem, says that a maximum or a minimum of a set of constraints based on an objective function will be found at a vertex. The proof of this theorem can be found at www.it.uu.se/edu/course/homepage/opt1/ht06/lectures/fundamentat_thm-2up.pdf - Example 1 One of the most valuable things about linear programming is that it is easily applicable to real life. Some of the applications of linear programming are in product mix planning, distribution networks, truck routing, staff scheduling, financial portfolios, and corporate restructuring. Here is the first linear programming problem I analyzed. It was drawn from The New World of Math, Fortune Magazine 1958. An oil refinery produces two products: jet fuel and gasoline. The profit for the refinery is $0.10 per barrel for jet fuel and $0.20 per barrel for gasoline. The following conditions must be met. 1. Only 10,000 barrels of crude oil are available for processing. 2. The refinery has a government contract to produce at least 1,000 barrels of jet fuel.
Linear Programming 3 3. The refinery has a private contract to produce at least 2,000 barrels of gasoline. 4. Both products are shipped in trucks, the delivery capacity of the truck fleet is 180,000 barrel-miles. 5. The jet fuel is delivered to an airfield 10 miles from the refinery. 6. The gasoline is transported 30 miles to the distributor. How much of each product should be produced for maximum profit? Let x represent the number of barrels of jet fuel and y represent the number of barrels of gasoline. Then the profit function is p(x,y) = 0.10x + 0.20y. The constraints are as follows x y x 1000 y 2000 10,000 10x 30y 180,000 The graphical method says we should first graph the constraints to find all the vertices. Figure 1A We can see that the vertices are (1000, 2000), (1000, 17000/3), (6000,4000), and (8000, 2000). The fundamental theorem of linear programming says that the maximum will be at
Linear Programming 4 one of these points, so I check the points with the profit function to find the maximum profit. Vertices (x,y) Profit p(x,y) = 0.10x + 0.20y (1000, 2000) $500 (1000, 17000/3) $1233.33 (6000, 4000) $1400 (8000, 2000) $1200 We can see that the vertex with the most profit is (6000, 4000). This means that to make the most profit, the company should produce 6000 barrels of jet fuel and 4000 barrels of gasoline. This amount of each would yield a profit of $1400. This profit can be shown as a flat roof over the graph of the constraints. Then the maximum profit occurs at the highest point of the flat roof. It is a good visual aid for showing that this vertex is indeed the absolute maximum over that particular region. Figure 1B Example 2 I have written two application problems that use linear programming to solve. Here is the first
Linear Programming 5 Mary works at a very popular fresh-squeezed lemonade stand on Myrtle Beach that sells regular lemonade and strawberry lemonade. To make a 20 oz. regular lemonade it takes 4 whole lemons. To make a 20 oz. strawberry lemonade it takes 3 whole lemons and 2 average sized strawberries. The owners figured they have a budget so that each morning a truck delivers 900 lemons to the stand and 408 strawberries to the stand. On each 20 oz. glass of regular lemonade a profit of $.95 is made. The stand is able to charge more for strawberry lemonades and a profit of $1.10 is made on each 20 oz. glass of that. The owners want Mary to sell at least twice the amount of strawberry lemonade each day as regular lemonade because more profit is made on the strawberry lemonade. They have also asked that at least 50 strawberry lemonades get sold each day because the strawberries spoil more quicklyr than the lemons. How many regular lemonades and how many strawberry lemonades should Mary sell each day to maximize profit? To solve this problem I first need to define the unknown. In this problem, x will represent the number of regular lemonades sold and y will represent the number of strawberry lemonades sold. The objective function in this situation is P =.95x + 1.10y. The constraints in this problem are the following. 4x 3y 900, which represents the limit on number of lemons available 2y 408, which represents the limit on number of strawberries available y 2x, which represents the fact that the owners want twice as many strawberry lemonades as regular lemonades y 50, which represents the fact that more than 50 strawberry lemonades must be sold x 0, which represents the fact that it is impossible to sell a negative number of regular lemonades These constraints can be seen in Figure 2A. A TI-84 calculator was used to create Figure 2A.
Linear Programming 6 Figure 2A I need to look at the vertices of Figure 2A to find the maximum profit possible. Vertex Point (x,y) Profit P=.95x+1.10y (0,50) $55 (25,50) $78.75 (90,180) $283.50 (72,204) $292.80 (0,204) $224.40 It is apparent that the number of each type of lemonade that needs to be sold to make the most profit is 72 regular lemonades and 204 strawberry lemonades. To model this graphically we can make the profit function be a flat roof on the graph of the constraints. This can be see in figure 2B, which was created using Google Sketchup.
Linear Programming 7 Figure 2B Example 3 The second problem that I wrote is similar to the preceding problem. A jewelry designer makes necklaces and earrings each week to sell at a local boutique. The boutique will buy between 10 and 40 necklaces each week. Based on last month s sales, each week the boutique bought at least twice the number of earrings as necklaces from the designer. The designer cannot make more than 50 pieces of jewelry per week. If the necklaces make the designer a profit of $25 each piece and the earrings make a profit of $10 each piece, how many necklaces and how many earrings should the designer sell to the boutique to maximize her profit? I will let a represent the number of necklaces sold and let b represent the number of pairs of earrings sold. Observe that the objective function for this problem is P = 25a+10b. The constraints are as follows 10 a 40, which represents the fact that the boutique will only buy between 10 and 40 necklaces
Linear Programming 8 b 0, which represents the fact that the boutique can t buy a negative number of pairs of earrings a b 50, which represents the fact that the designer can only make 50 pieces in a week 2 b a, which represents the fact that in the past the month, the boutique has bought at least twice the number of earrings as necklaces. I can now look at the graph of the constraints to find the vertices. They are graphed in Figure 3A, which was created using a TI-84 calculator. Figure 3A Based on the Fundamental Theorem of Linear Programming, I can look at the three vertices in Figure 3A to find the number of necklaces and earrings the designer needs to sell to the boutique to maximize profit. Vertex Point Profit Made (a,b) P(a,b) = 25a+10b (10,40) $650 (10,5) $300 (33.3, 16.6) $998.50 The designer should sell 33 necklaces and 16 earrings to the boutique to maximize her profit. I can model the profit function like a flat roof on top of the constraint function.
Linear Programming 9 This can be seen in Figure 3B. The program Google Sketchup was used to create Figure 3B. Figure 3B The roof of this figure shows that the maximum is at the point where the designer sells 33.3 necklaces and 16.6 earrings to the boutique. From this combination of pieces, they will make a profit of $998.50. Example 4 Every problem we have seen so far has used linear programming to find a maximum, but I wanted to also explore finding a minimum. The next problem will deal with finding a minimum, and it was drawn from the website http://www.purplemath.com/modules/linprog4.htm. In order to ensure optimal health (and thus accurate test results), a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than five ounces of food a day. Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce, and costs $0.20 per ounce. Food Y
Linear Programming 10 contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per ounce, at a cost of $0.30 per ounce. What is the optimal blend? Let x represent the number of ounces of Food X and y represent the number of ounces of food Y. The objective function is that c(x,y) = $0.20x + $0.30y. The constraints are as follows: I know I cannot buy a negative amount of food, so x 0 y 0 8x 12y 24, which represents the amount of fat grams needed 12x 12y 36, which represents the amount of carbohydrate grams needed 2x 1y 4, which represents the amount of protein grams needed x y 5, which represents the fact that the maximum number of ounces a rabbit should be fed is five I will now graph the constraints to find the vertices. These can be seen in Figure 4A, which was created using a TI-84 calculator. Figure 4A
Linear Programming 11 According to the Fundamental Theorem of Linear Programming, the minimum cost will occur at a vertex. I will now use the objective function and the vertices to find the minimum cost. Label Vertex (x,y) Cost c(x,y) = $0.20x + $0.30y A (0,4) $1.20 B (0,5) $1.50 C (1,2) $0.80 D (3,0) $0.60 E (5,0) $1.00 The optimal blend is actually not even a blend at all. The cheapest rabbit food that meets all the constraints is when the lab technician uses 3 ounces of Food X and none of the Food Y. This can be shown with a flat roof as the cost function on top of the constraints graph in Figure 4B. This figure was also created using the computer program Google Sketchup. Figure 4B This was an interesting situation because sometimes the best solution is not a mix of both types in a situation. The linear programming model will show the optimum amount of each, whether we are trying to find the maximum or the minimum.
Linear Programming 12 Conclusion The linear programming model has taught me another method of optimization. I have learned the most about linear programming by having to create my own real life problems that require linear programming to solve. It is possible to use linear programming with more than two variables, but I chose to limit this research to just two variables because it enabled me to use the flat roof technique. The flat roof graph gives a great visual of why the Fundamental Theorem of Linear Programming is true. The possible solutions for linear programming may involve finding a maximum, which we did in examples 1, 2, and 3. We can also use linear programming to find a minimum like in example 4. It is also possible for there to be no maximum or similarly no minimum. One type of problem where that is the case, is if the constraints produce an unbounded region. One argument that may be used for why linear programming is not a good means of optimization is that it only works for constraint functions that are lines. However, many functions can be approximated by lines, so the linear programming model can still be used in a variety of cases.
Linear Programming 13 References Larson, R., Boswell, L., Kanold, T., & Stiff, L. (2001). Algebra 2. Evanston, IL: McDougal Littell Inc. http://people.hofstra.edu/stefan_waner/realworld/summary4.html http://www.purplemath.com/modules/linprog4.htm http://www.sonoma.edu/users/w/wilsonst/courses/math_131/lp/default.html http://home.alltel.net/okrebs/page34.html http://mathforum.org/library/drmath/view/52852.html http://www.ltcconline.net/greenl/courses/103b/matrices/linprog.htm www.it.uu.se/edu/course/homepage/opt1/ht06/lectures/fundamentat_thm-2up.pdf -