This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Economics 146 – Fall, 2007 Linear Programming – Part I Last revision: 10/8/07. A Small Example – A Blending Model. In order to feed his collection of tortoises Darwin can purchase two kinds of reptile pellet feeds. Each tortoise requires at least 60, 84, and 72 milligrams of nutritional elements A, B, and C, respectively, per day. The nutritional contents (milligrams/gram) and cost (cents/gram) of a gram of each of the two feeds are given in the following table. A B C cost Feed 1 3 7 3 10 Feed 2 2 2 6 4 Darwin (being a true academic) would like to choose the least expensive way of providing an adequate diet for tortoises by using a combination of the two pellet feeds. (It is clear that he could provide adequate diets by just using enough of either feed.) So, he needs to consider all possible diets that satisfy the nutritional requirements and choose from that set a diet of minimal cost. Let y 1 be the number of grams of Feed 1 and y 2 be the number of grams of Feed 2 to be used in the daily diet. A priori, y 1 and y 2 must both be nonnegative. The amount of nutrient A in this diet would be 3 y 1 + 2 y 2 . The requirement for nutrient A is that 3 y 1 + 2 y 2 ≥ 60 Similarly, the requirements for nutrient B and C are 7 y 1 + 2 y 2 ≥ 84 and 3 y 1 + 6 y 2 ≥ 72 respectively. The cost function to be minimized is f ( y 1 ,y 2 ) = 10 y 1 + 4 y 2 Thus, Darwin’s problem is to find numbers y 1 and y 2 so as to minimize 10 y 1 + 4 y 2 (1) subject to 3 y 1 + 2 y 2 ≥ 60 (2) 7 y 1 + 2 y 2 ≥ 84 (3) 3 y 1 + 6 y 2 ≥ 72 (4) and y 1 ≥ ,y 2 ≥ (5) One way to solve this problem is to do so geometrically. The steps will be carried out in class. We graph the constraint set as the intersection of the three sets specified by the three linear inequalities in the first quadrant. Then we draw some of the level curves of the objective function. The level curves of the function form the family of lines determined by 1 10 y 1 + 4 y 2 = c where c is a constant. Note that all the lines have the same slope 10 4 and the lines move to the left as c decreases. Each of the level curves consist of points that give the same value for the cost function 10 y 1 + 4 y 2 . Consequently, we want the line furthest to the left that intersects the constraint region that we have drawn. The line through the point (6 , 21) is the line we seek. So, 6 grams of Feed 1 and 21 grams of Feed 2 is the optimal solution (i.e. cheapest adequate diet) and gives the value of the minimal cost diet to be (10)(6) + (4)(21) = 144 cents. Another Small Example – A Production Model. A manufacturer makes two kinds of boats – family row boats and sporty canoes. The manufacturing process for each involves (i) molding by a pressing machine and (ii) a followup by hand finishing. These are quality boats and the factory will sell as many as are produced. It takes 50 pounds of aluminum, 6 minutes of machine time, and 3 hours of finishing labor to produce a row boat. It takes 30 pounds of aluminum, 5 minutes of machine time, and 5 hours of finishing labor to produce...
View
Full
Document
 Fall '07
 farmer
 Economics, Optimization, Standard form, Mj

Click to edit the document details