1 1 Slide Graphical solution A Graphical Solution Procedure (LPs with 2 decision variables can be solved/viewed this way.) 1. Plot each constraint as an equation and then decide which side of the line is feasible (if it’s an inequality). 2. Find the feasible region. 3. find the coordinates of the corner (extreme) points of the feasible region. 4. Substitute the corner point coordinates in the objective function 5. Choose the optimal solution

2 2 Slide Example 1: A Minimization Problem n LP Formulation Min z = 5 x x 2 Min z = 5 x x 2 s.t. 2 x x 2 > 10 s.t. 2 x x 2 > 10 4 x 1 - x 2 > 12 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 x 1, x 2 > 0 x 1, x 2 > 0

3 3 Slide Example 1: Graphical Solution n Graph the Constraints Constraint 1: When x 1 = 0, then x 2 = 2; when x 2 = 0, then x 1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 1: When x 1 = 0, then x 2 = 2; when x 2 = 0, then x 1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 2: When x 2 = 0, then x 1 = 3. But setting x 1 to 0 will yield x 2 = -12, which is not on the graph. Thus, to get a second point on this line, set x 1 to any number larger than 3 and solve for x 2 : when x 1 = 5, then x 2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. Constraint 2: When x 2 = 0, then x 1 = 3. But setting x 1 to 0 will yield x 2 = -12, which is not on the graph. Thus, to get a second point on this line, set x 1 to any number larger than 3 and solve for x 2 : when x 1 = 5, then x 2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. Constraint 3: When x 1 = 0, then x 2 = 4; when x 2 = 0, then x 1 = 4. Connect (4,0) and (0,4). The ">" side is above this line. Constraint 3: When x 1 = 0, then x 2 = 4; when x 2 = 0, then x 1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.

4 4 Slide Example 1: Graphical Solution n Constraints Graphed x2x2x2x2 x2x2x2x2 4x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 4x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 2x 1 + 5x 2 > 10 x1x1x1x1 x1x1x1x1 Feasible Region 4 (5,0) (16/5,4/5) (10/3, 2/3)

5 5 Slide Example 1: Graphical Solution n Solve for the Extreme Point at the Intersection of the second and third Constraints 4 x 1 - x 2 = 12 4 x 1 - x 2 = 12 x 1 + x 2 = 4 x 1 + x 2 = 4 Adding these two equations gives: Adding these two equations gives: 5 x 1 = 16 or x 1 = 16/5. 5 x 1 = 16 or x 1 = 16/5. Substituting this into x 1 + x 2 = 4 gives: x 2 = 4/5 Substituting this into x 1 + x 2 = 4 gives: x 2 = 4/5 n Solve for the extreme point at the intersection of the first and third constraints 2 x x 2 =10 2 x x 2 =10 x 1 + x 2 = 4 x 1 + x 2 = 4 Multiply the second equation by -2 and add to the first equation, gives 3x 2 = 2 or x 2 = 2/3 3x 2 = 2 or x 2 = 2/3 Substituting this in the second equation gives x1 = 10/3 pointZ (16/5, 4/5) 88/5 (10/3, 2/3) 18 (5, 0) 25

6 6 Slide Example 2: A Maximization Problem Max z = 5 x x 2 Max z = 5 x x 2 s.t. x 1 < 6 s.t. x 1 < 6 2 x x 2 < 19 2 x x 2 < 19 x 1 + x 2 < 8 x 1 + x 2 < 8 x 1, x 2 > 0 x 1, x 2 > 0

7 7 Slide Example 2: A Maximization Problem n Constraint #1 Graphed x 2 x 2 x1x1x1x1 x 1 < 6 (6, 0)

8 8 Slide Example 2: A Maximization Problem n Constraint #2 Graphed x 1 + 3x 2 < 19 x 2 x 2 x1x1x1x1 (0, 6 1/3 ) (9 1/2, 0)

9 9 Slide Example 2: A Maximization Problem n Constraint #3 Graphed x 2 x 2 x1x1x1x1 x 1 + x 2 < 8 (0, 8) (8, 0)

10 Slide Example 2: A Maximization Problem n Combined-Constraint Graph x 1 + 3x 2 < 19 x 2 x 2 x1x1x1x1 x 1 + x 2 < 8 x 1 < 6

11 Slide Example 2: A Maximization Problem n Feasible Solution Region x1x1x1x1 FeasibleRegion x 2 x 2

12 Slide Example 2: A Maximization Problem n The Five Extreme Points x1x1x1x1 FeasibleRegion (0, 19/3) (5, 3) (6, 2) (6, 0) (0, 0)

13 Slide Example 2: A Maximization Problem n Having identified the feasible region for the problem, we now search for the optimal solution, which will be the point in the feasible region with the largest (in case of maximization or the smallest (in case of minimization) of the objective function. n To find this optimal solution, we need to evaluate the objective function at each one of the corner points of the feasible region.

14 Slide Example 2: A Maximization Problem n Optimal Solution x1x1x1x1 x 2 x 2 Optimal Solution Point Z (0,0) 0 (6,0) 30 (6,2) 44 (5,3) 46 (0,19/3) 44.33

15 Slide Extreme Points and the Optimal Solution n The corners or vertices of the feasible region are referred to as the extreme points. n An optimal solution to an LP problem can be found at an extreme point of the feasible region. n When looking for the optimal solution, you do not have to evaluate all feasible solution points. n You have to consider only the extreme points of the feasible region.

16 Slide Feasible Region n The feasible region for a two-variable linear programming problem can be nonexistent, a single point, a line, a polygon, or an unbounded area. n Any linear program falls in one of three categories: is infeasibleis infeasible has a unique optimal solution or alternate optimal solutionshas a unique optimal solution or alternate optimal solutions has an objective function that can be increased without boundhas an objective function that can be increased without bound n A feasible region may be unbounded and yet there may be optimal solutions. This is common in minimization problems and is possible in maximization problems.

17 Slide Special Cases n Alternative Optimal Solutions In the graphical method, if the objective function line is parallel to a boundary constraint in the direction of optimization, there are alternate optimal solutions, with all points on this line segment being optimal. n Infeasibility A linear program which is overconstrained so that no point satisfies all the constraints is said to be infeasible. A linear program which is overconstrained so that no point satisfies all the constraints is said to be infeasible. n Unbounded For a max (min) problem, an unbounded LP occurs if it is possible to find points in the feasible region with arbitrarily large (small) Z

18 Slide Example with Multiple Optimal Solutions

19 Slide Example: Infeasible Problem n Solve graphically for the optimal solution: Max z = 2 x x 2 Max z = 2 x x 2 s.t. 4 x x 2 < 12 s.t. 4 x x 2 < 12 2 x 1 + x 2 > 8 2 x 1 + x 2 > 8 x 1, x 2 > 0 x 1, x 2 > 0

20 Slide Example: Infeasible Problem n There are no points that satisfy both constraints, hence this problem has no feasible region, and no optimal solution. x2x2x2x2 x1x1x1x1 4x 1 + 3x 2 < 12 2x 1 + x 2 >

21 Slide Example: Unbounded Problem n Solve graphically for the optimal solution: Max z = 3 x x 2 Max z = 3 x x 2 s.t. x 1 + x 2 > 5 s.t. x 1 + x 2 > 5 3 x 1 + x 2 > 8 3 x 1 + x 2 > 8 x 1, x 2 > 0 x 1, x 2 > 0

22 Slide Example: Unbounded Problem n The feasible region is unbounded and the objective function line can be moved parallel to itself without bound so that z can be increased infinitely. x2x2x2x2 x1x1x1x1 3x 1 + x 2 > 8 x 1 + x 2 >

23 Slide Solve the following LP graphically max 45 x x2x2 Objective Function s.t. 10 x 1 +  x x 2  x x 2  x x 2 0  2400 nonnegativity Are the LP assumptions valid for this problem? Structural constraints x 1 ≥ 0, x 2 ≥ 0 x 1  40 x 2  x 2 Where x 1 is the quantity produced from product Q, and x 2 is the quantity of product P A B C D E F

24 Slide The graphical solution (1)(1) Assignment: Complete the problem to find the optimal solution E F

25 Slide Possible Outcomes of an LP 1. Infeasible – feasible region is empty; e.g., if the constraints include x 1 + x 2  6 and x 1 + x 2  7 2. Unbounded -Max 15 x x 2 (no finite optimal solution) s.t. 3. Multiple optimal solutions - max 3 x x 2 s.t. x 1 + x 2  1 x 1, x 2  0 4. Unique Optimal Solution Note: multiple optimal solutions occur in many practical (real-world) LPs. x 1 + x 2  1 x 1, x 2  0