Calories per oz. Price per oz Corn Wheat

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1 Donald Wittman Lecture 1 Diet Problem Consider the following problem: Corn costs.6 cents an ounce and wheat costs 1 cent an ounce. Each ounce of corn has 10 units of vitamin A, 5 calories and 2 units of protein. Each ounce of wheat has 4 units of vitamin A, 5 calories and 6 units of protein. Minimum daily requirements for a chicken are 20 units of vitamin A, 20 calories and 12 units of protein. Find the least cost mix of feed. It is helpful to put this information in chart form. Units of Vit. A per oz Calories per oz Units of protein per oz Price per oz Corn Wheat In order to solve the problem, we need to rewrite the problem in terms of an objective function and constraints. c is corn and w is wheat. Minimize.6c + w This is the objective function Vitamin A 10c + 4w 20 This and the following three lines are the constraints Calories 5c + 5w 20 Protein 2c + 6w 12 c 0 w 0 These are the non-negativity conditions Intuition is developed if we use a graphical analysis (See Figure 1.1). We first consider the Vitamin A constraint. In drawing the constraint lines it is usually, but not always, best to start where one or the other variable value is 0. If w = 0, then c = 2 will satisfy the Vitamin A constraint exactly; If c = 0, then w = 4 will satisfy the Vitamin A constraint exactly. 1

2 Connecting these two points with a strait line, we get the Vitamin A constraint line. Any point on this line or above (to the right) will provide an adequate amount of Vitamin A. By a similar process, the other constraints (solid lines) can be also be drawn. We also have the non-negativity constraints. One cannot add negative amounts of corn to the food mix. Therefore the inputs are in the positive quadrant, on or above the horizontal axis and on or to the right of the vertical axis. w 5 FIGURE 1.1 THE DIET PROBLEM 4 3 feasible set 2 Objective Function 1 Vita Cal Protein c The dotted line is an isocost curve. Isocost curves are a set of parallel lines based on the cost equation.6 + w. The lowest isocost curve in the feasible set is chosen. The isocost curve (the dotted line) is chosen in the following way: Set w = 0 and choose c within the range of the graph (0-6). I chose 5. If w = 0 and c = 5, then total cost =.6(5) + 1(0) = 3. I need to find another point on the same isocost curve. 2

3 Letting c = 0, then.6 (0) + w = 3 is another point on the same isocost curve. So if c = 0 and w = 3, total cost is again 3. This isocost is not the minimal isocost curve that is feasible. One could shift it down in a parallel fashion so that it goes through the intersection of the calorie and protein constraints. At that point the calorie and protein constraints hold with strict equality and the vitamin constraint is non-binding. That is, we have more than enough vitamins. We use the graph to find the optimal basic feasible solution. The calorie and protein constraints are binding. The vitamin constraint is not part of the optimal feasible solution since vitamins are not binding. 5c + 5w = 20 Calorie constraint 2c + 6w =12 Protein constraint Solve using Cramer's Rule: c = = = = 3 w = = = = 1 Cost =.6c + w =.6 (3) + 1 = = 2.8 What happens if we "loosen" the vitamin constraint by one unit (10c + 4w 19)? Are costs reduced? No. We already had more than enough vitamins, so reducing the vitamin requirement does not reduce costs. 3

4 Ralston-Purina and other animal feed manufactures face similar cost minimization problems every day. The only difference is that there are many more choices than corn and wheat and many more constraints. Saving just a penny per pound can mean tens of millions of dollars more in profits per year. 4

5 Donald Wittman Lecture 2 Production Problem Pool table production: Golden Eagle table Excell Table Amount Available Resource Hard wood (board feet) Slate (sheets) Plastic (squares) Felt (square feet) Labor (hours) NET PROFIT $1000 $600 Converting this information into a constrained maximization problem: MAX PROFIT = 1000 GE EX SUBJECT TO 20 GE + 15 EX, 600 Hardwood 2 GE + 1 EX 70 slate 20 GE + 30 EX 1000 plastic 80 GE + 40 EX 2000 felt 40 GE + 30 EX 2000 labor GE 0 Non-negativity constraints EX 0 Drawing the diagram: Looking at first constraint -- hardwood constraint: If EX = 0 then GE = 30 will satisfy the constraint exactly. So (30, 0) is one point on constraint line. 5

6 If GE = 0, then EX = 40 will satisfy the constraint exactly, So (0, 40) is another point on the hardwood constraint line. Connecting these two points, we get the Hardwood constraint. We follow a similar procedure for the other constraints. 6

7 With regard to the objective function. We first choose a number on the graph, say (20,0). Thus the value of the objective function (profit) is 20*1000 = $20,000. In order to find the iso-profit line, we find another point where the profit is $20,000. In this case it is (0, EX) or 600 EX = Equivalently, EX =33 1/3. This is the dotted line. We then shift the line up in a parallel fashion as far as we can as long as all the constraints are satisfied. 70 excell slate FIGURE 2.1 PRODUCTION PROBLEM felt labor 20 plstc Objective Function 10 feasible set hw Golden Eagle It is easy to see that the labor and slate constraints are non-binding. The objective function is steeper than the hardwood constraint. Thus parallel shifts upward of the objective function (isoprofit line) to hire profit will intersect the hardwood and felt constraints. The plastic constraint will be non-binding. 7

8 Thus we are left with two equations in two unknowns: 20 GE + 15 EX = 600 Hardwood 80 GE + 40 EX = 2000 felt Solving using Cramers rule: GE= = = = XL= = = = 20 PROFIT = 15 (1000) + 20 (600) = $15,000 + $12,000 = $27,000 What happens to profit if more slate were available? Profit would not increase as slate is not a binding constraint. So the manufacturer would pay nothing for adding slate to its inventory. 8