1 Geometrical solution of Linear Programming Model

2 Example 1: Production of glasses A machine can produce 100 cases of juice glasses in six hours, or 100 cases of cocktail glasses in five hours. It operates only 60 hours per week. The week’s production is kept in own stockroom where he has an effective capacity of 15,000 cubic feet. A case of juice glasses requires 10 cubic feet of storage space, while a case of cocktail glasses requires 20 cubic feet. The contribution of the juice glasses is $5.00 per case; the only customer available will not accept more than 800 cases per week. The contribution of the cocktail glasses is $4.50 per case and there is no limit on the amount that can be sold. How many cases of each type of glass should be produced each week in order to maximize the total contribution? Create the mathematical model. Z max = 500x x2 6x1 + 5x2 ≤ 60 10x1 + 20x2 ≤ 150 x1 ≤ 8 x1 ≥ 0, x2≥ 0

3 Graphical Representation of the Decision Space Each line in the figure is represented by a constraint expressed as an equality. The arrows associated with each line show the direction indicated by the inequality sign in each constraint. The set of values of the decision variables x1 and x2 that simultaneously satisfy all the constraints indicated by the shaded area are the feasible production possibilities or feasible solutions to the problem. Among these feasible production alternatives, we want to find the values of the decision variables x1 and x2 that maximize the resulting contribution = optimal solution

4 Graphical solution of the LP Model 1.Graphical Representation of the feasible solution: construction the lines and the planes corresponding with the constrains Finding the intersection 2.Finding an Optimal Solution: Choose the level of z. construction the straight line corresponding with the goal function with the chosen righ hand side. Construction of parallel line that intercepts the farthest point from the origin within the feasible region.

5 Graphical representation of the feasible region

6 Finding an Optimal Solution z = 500x x2 If z is held fixed at a given constant value, this expression represents a straight line. Let ´s put: z=4000 the line goes through the points [0;8,8] and [8;0] Z=5000 the line goes through the points [0;11,1] and [10;0] Note that the slope of this straight line is constant, independent of the value of z. As the value of z increases, the resulting straight lines move parallel to themselves in a northeasterly direction away from the origin At the point labeled P1, the line intercepts the farthest point from the origin within the feasible region, and the contribution z cannot be increased any more.

7 Finding the optimal solution

8 Shadow Prices on the Constraints The shadow price on a particular constraint represents the change in the value of the objective function per unit increase in the right hand-side value of that constraint. For example, suppose that the number of hours of molding-machine capacity was increased from 60 hours to 61: 6x1 + 5x2 = 61 10x1 + 20x2 = 150

9 The new values of the decision variables are x1 = 6 5/7 and x2 = 4 1/7 and the new value of the objective function is: z = 500x x2 = / /7= = 5221,429 The old value of z= 5142,857 The difference: 5221, ,86 = 78,57 The shadow price associated with production capacity is $78,57 per additional hour of production time. It implies that it would be profitable to invest up to $78,57 each week to increase production time by one hour.

10 Changes in the Coefficients of the Objective Function Suppose we consider the contribution per one hundred cases of six-ounce juice glasses, and determine the range for that coefficient such that the optimal solution remains unchanged. Production slope ≤Objective slope ≤Storage slope. Suppose that the objective function were z = 540x x2. It would be parallel to the line determined by the production-capacity constraint; and all levels of the decision variables lying on the line segment joining the points labeled P1 and P2

11 Objective function coincides with a constrain

12 An unbounded solution Attempting to solve our linear program with ‘‘greater than or equal to’’ constraints instead of ‘‘less than or equal to’’ constraints: Maximize z = 500x x2 subject to: 6x1 + 5x2 ≥ 60, 10x1 + 20x2 ≥ 150, x1 ≥ 8, x1 ≥ 0, x2 ≥ 0.

13 An unbounded solution.

14 An infeasible group of constraints. An infeasible linear program might result from a poorly formulated problem, or from a situation where requirements exceed the capacity of the existing available resources. An error was made that resulted in the following constraint being included in the formulation: 6x1 + 5x2≥ 80

15 An infeasible group of constraints.

16 Example 2: Paintings Suppose we sell reproductions of two diferent paintings. We can sell any number of reproductions of painting 1 for $3 each, and any number of reproductions of painting 2 for $2 each. We have only a limited amount of paint: we have 16 units of blue paint, 8 units of green paint, and 5 units of red paint. A reproduction of painting 1 requires 4 units of blue, 1 unit of green, and 1 unit of red. A reproduction of painting 2 requires 2 units of blue, 2 units of green, and 1 unit of red. How many reproductions of each painting should we create to maximize our revenue? Let x1 denote the number of reproductions of painting 1 that we create, and x2 the number of reproductions of painting 2.

17 Mathematical model maximize z=3x1 + 2x2 subject to 4x1 + 2x2 ≤ 16 x1 + 2x2 ≤ 8 x1 + x2 ≤ 5 x1 ≥ 0; x2 ≥ 0

18 Graphical solution Optimal solution: x1=3, x2=2;z = 13

19 Example 2: Modification Now suppose we modify the problem slightly, by reducing the available blue paint from 16 to 15 units. This results in the following linear program: maximize z=3x1 + 2x2 subject to 4x1 + 2x2 ≤ 15 x1 + 2x2 ≤ 8 x1 + x2 ≤ 5 x1 ≥ 0; x2 ≥ 0

20 Graphical solution

21 Integer programming Now, in the context of our painting example, this is not a very reasonable solution: how are we going to sell half a painting? Realistically, we need additional integrality constraints: x1 and x2 should both be integers. maximize z=3x1 + 2x2 subject to 4x1 + 2x2 ≤ 15 x1 + 2x2 ≤ 8 x1 + x2 ≤ 5 x1 ≥ 0; integer x2 ≥ 0; integer

22 Graphical representation with integrality constraints.

23 Mixed Integer programming maximize z=3x1 + 2x2 subject to 4x1 + 2x2 ≤ 15 x1 + 2x2 ≤ 8 x1 + x2 ≤ 5 x1 ≥ 0; x2 ≥ 0; integer

24 Graphical representation with one integrality constraint.

25 Overwiev Any assignment of values to the variables is called a solution. A solution is feasible if it respects all the constraints optimal if it maximizes the objective among feasible solutions. A program may have no feasible solutions. If it is possible to make the objective as large as we like, then the program is unbounded.