Econ 172A - Slides from Lecture 2

Transcription

1 Econ 205 Sobel Econ 172A - Slides from Lecture 2 Joel Sobel September 28, 2010

2 Announcements 1. Sections this evening (Peterson 110, 8-9 or 9-10). 2. Podcasts available when I remember to use microphone. 3. Textbook on reserve at SSH Library. 4. Look at Supplementary Formulation Problems

3 DIET PROBLEM 1. Given: A list of different foods. A list of different nutrients. The unit price of each food. The minimum daily requirement of each nutrient. The nutrient contribution of each food. 2. Find the cheapest way to minimize all nutritional requirements.

4 BASIC DATA 1. n different kinds of food. 2. p j price per unit of jth food. 3. m different nutrients. 4. nutritional requirement of Nutrient i is c i. 5. A is technology (a ij is the amount of the ith nutrient in one unit of the jth food).

5 INFORMALLY 1. Foods: lettuce, peanut butter, bread, apple juice. F j, the jth food, is one of these. 2. Nutrients: Vitamin B12, iron, calcium,.... N i, the ith nutrient, is one of these. 3. Everything has units: 3.1 prices dollars per unit of food 3.2 nutrient requirements: units of nutrient. 3.3 a ij : units of nutrient per unit of food

6 Step 1: Identify Variables. What are you looking for? You are looking for amounts of food. Variables are quantities of each of the n foods. These are unknowns and need names. Let x j be the number of units of food j purchased. You want to find x = (x 1,..., x n ).

7 Entire Problem. The problem is to find x to solve: min p x subject to Ax c and x 0. In practice, you will be given values for the parameters of the problem (A, p, and c) and then would go ahead and try to find a numerical solution.

8 WHAT DOES SOLUTION LOOK LIKE? Suppose each nutrient is contained in some food. Then: problem is feasible (you can satisfy constraints). Not unbounded (cost is non-negative). Hence: Expect a solution.

9 MORE 1. Will you satisfy all nutritional constraints exactly? 2. What if there are many more nutritional constraints than foods?

10 Another Problem All nutrients available in pill form. Someone offers to sell you c i units of nutrient i. This person can set unit prices of pills. You buy if it is cheaper to buy pills than to supply the nutritional requirements indirectly through food. Pill seller s goal: Set prices to maximize the amount he can make selling you nutrients subject to constraint that you would rather buy pills than food.

11 VARIABLES? y = (y 1,..., y i,..., y m ), where y i is the price charged for a pill that supplies one unit of nutrient i.

12 Objective The pill seller wants to maximize her profit. He sells c. If he can charge the prices y, then he earns c y.

13 Constraints. What does it mean for the pills to be cheaper than food? First food: Supplies a i1 units of nutrient i. Cost of nutrient i supplied by food 1 if purchased by pills instead: a i1 y i. Cost of the nutrients in food one: m i=1 a i1y i. Constraint: In order for the (nutrients in the) pills to be cheaper than (the nutrients in) food one, it must be that m a i1 y i p 1. i=1 And so: for each j = 1,..., n, m a ij y i p j. i=1

14 PUTTING IT TOGETHER Put the constraints together and we have the pill seller s problem: Find y = (y 1,..., y m ) to solve: max c y subject to ya p and y 0. Warning for experts: I write: ya p for A t y p.

15 Comparison Pill seller s problem is just a contrived way to practice problem formulation. But, the two problems are related. Same data: A, c, and p. Comparable values. Value of Diet Problem (min cost) Value of Pill Problem (max profit). Why? The constraints in the pill problem guarantee that pills are cheaper than food.

16 In Fact 1. When you solve pill and diet problems values will be equal. 2. Prices in pill problem describe true value of nutrients.

17 Graphing Linear Inequalities in the Plane 1. Two variable LPs can be solved graphically. 2. You need to know two things: Graph linear inequalities in the plane (you probably did this in high school) Figure out the relationship between these points and the objective function.

18 Graph Line For example: 2x 1 + x 2 = 2, (x 1, x 2 ) = (1, 0) and (x 1, x 2 ) = (0, 2) are on the line. Connect the dots.

19 Graph Halfplane The inequality 2x 1 + x 2 2, consists of all of the points above and to the right of the straight line. In general: inequalities are satisfied by points on one side of the line. To determine which set consists of the point that satisfies the inequality, test by checking an arbitrary point not on the line. For example, (x 1, x 2 ) = (0, 0) does not satisfy the inequality 2x 1 + x 2 2. Hence the set of points that satisfies the inequality consists of the points on the side of the line 2x 1 + x 2 = 2 that does not contain (0, 0).

20 Many Constraints For example, the set determined by the five inequalities 2x 1 + x 2 2 2x 1 + x 2 2 4x 1 + x 2 8 x 0. This is region bounded by the quadrilateral pictured. (The four corners are (0, 2), (1, 0), (2, 0), and (1, 4).)

21 Econ 205 Sobel Picture 1,4 x 0 = 9 0,2 x 0 = 2 1,0 2,0

22 Comments 5 inequalities? The first three lines describe one inequality each. The fourth line describes two: x 1 0 and x 2 0. If you have five inequalities, you would expect the feasible set of have five sides. This set has only four because the constraint that x 1 0 is redundant. If you satisfy the other four constraints, then you automatically satisfy x 1 0. Flawed intuition: you should have as many variables as equations to have a system that makes sense. Not true here. Reasons: 1. Inequalities not equations. 2. You want large feasible set.

23 Corners In the example, the feasible set has four corners. These corners are determined by the intersection of pairs of constraints, solved as equations. That is, (0, 2) is the solution to (1, 4) is the solution to 2x 1 + x 2 = 2 2x 1 + x 2 = 2, 2x 1 + x 2 = 2 4x 1 + x 2 = 8, (1, 0) is the solution to 2x 1 + x 2 = 2 and x 2 = 0, and (2, 0) is the solution to 4x 1 + x 2 = 8 and x 2 = 0.

24 More Generally The feasible region of a linear programming problem has corners determined by solving subsets of the constraints as equations. Once you have these corners, you get the feasible set by connecting the dots and identifying the region that satisfies all of the constraints. The feasible set may be empty. Replace the constraint that x 1 0 with one that said that x 1 1. The feasible set may be unbounded. That is, it may go out forever in one or more directions. (Having no constraints is perfectly ok.) The only way to have a problem that has an unbounded solution is to have an unbounded feasible set.

25 Graphical Solutions To solve LP graphically: 1. Graph feasible set. If empty, stop (problem is not feasible). If non empty, 2. Find solution if it exists. (Solution must exist if feasible set is bounded. It might exist otherwise.) 3. Graph a level set of the objective function. Level set of f : {x : f (x) = c}. Level sets of linear functions in the plane are lines. 4. Adjust level set so that it intersects feasible set. 5. Increase value of objective function until the greatest possible intersection.

26 SUMMARY 1. Graph feasible set. If feasible set is empty, then stop. The problem is infeasible. Otherwise continue. 2. Graph a level set of the objective function. 3. Shift the level set (parallel movement) until it intersects the feasible region. 4. Continue to shift the level set until it reaches the maximum value the intersects the feasible region.

27 Comments 1. Follow the same steps for a minimization problem, taking care to move the objective function in the opposite direction. 2. You know which direction increases the objective function value by drawing two level sets and comparing (the direction of increase never changes). 3. In the example, the level set x 1 + 2x 2 = 9 lies above and to the right of the level set x 1 + 2x 2 = 2; you always increase the objective function (in this example) by moving up and to the right.