1. Linear Programming Graphical Solution

2. Graphical Solution to an LP Problem This is easiest way to solve a LP problem with two decision variables. If there are more than two decision variables, it is not possible to plot the solution on two dimensional graph. However, it provides us an useful insight how the other approaches work. So it is worthwhile to learn how it works. To find the optimal solution to an LP problem we must first identify a set of feasible solution (a region of feasible solution)

3. First step in doing so to plot of the problem’s constraints on a graph. X 1 is usually plotted as the horizontal axis X 2 is usually plotted as the vertical axis Because of the non-negativity constraint we are always working on the first quadrant x1x1 x2x2 This axis represents the constraint x 2 ≥ 0 This axis represents the constraint x 1 ≥ 0

4. Example 1 Giapetto’ Woodcarving, Inc., Subject to

5. The feasible region has been graphed. We may proceed to find the optimal solution to the problem. The optimal solution is the point lying in the feasible region that produce the highest profit. There are two different approaches to find it. -Isoprofit line method (isocost line) - Corner point method

7. Feasible Region D G F E Hx1x1 x2x2 Corner Point Solution Method The second approach to find the optimal solution to LP problem is corner point method. An optimal solution to LP problem lies at a corner point of (extreme point of) the feasible region. Hence it is only necessary to find the value of them. From the graph the problem is five sided polygon with five corner or extreme points. These points are labeled H, E, F, G, D. So we can find the coordinates of each corner and test the profit levels.

8. Feasible Region D G F E Hx1x1 x2x2 Carpentry Cons. Demand Cons. Finishing Cons. Point H (0, 0) Z = 3(0) + 2(0) = 0 Point E (40, 0) Z = 3(40) + 2(0) = 120 Point F (40, 20) Z = 3(40) + 2(20) = 160 Note that: at Point F Cons.2 and Cons.3 intersect So 2x 1 + x 2 = 100 x 1 = 40 x 1 = 40, x 2 = 20 Point G (20, 60) Z = 3(20) + 2(60) = 180 * optimal solution 2x 1 + x 2 = 100 x 1 + x 2 = 80 x 1 = 20, x 2 = 60 Point D (0, 80) Z = 3(0) + 2(80) = 160

9. The optimal solution to this problem (point G) x 1 = 20 x 2 = 60 Z = 180 [ Z = 3x 1 + 2x 2 ⇒ 3(20) + 2(60) = $120] If we substitute the optimal values of decision variables into the left hand side of the constraints. 2x 1 + x 2 ≤ 100 ⇒ 2(20) + 60 = 100 S 1 = 100 – 100 = 0 x 1 + x 2 ≤ 80 ⇒ 20 + 60 = 80 S 2 = 80 – 80 = 0 x 1 ≤ 40 ⇒ 20 S 3 = 40 – 20 = 20 S: Slack variable: represent the amount of resource unused S 1 = 0 means decision variables use up the resources completely S 2 = 0 S 3 = 20 20 units of resource is left over. Slack variable → [≤] less than or equal to

10. Once the optimal solution to the LP problem it is useful to classify the constraint. A constraint is binding constraint if left hand side and right hand side of it are equal when the optimal values of decision variables are substituted into the constraint. A constraint is nonbinding constraint if the left hand side and right hand side of constraint are not equal when the optimal values of decision variables are substituted in to the constraint. Left hand sideRight hand side Constraint 1: 2x 1 + x 2 ≤ 100 2(20) + 60 = 100100Binding Constraint 2: x 1 + x 2 ≤ 80 20 + 60 = 8080Binding Constraint 3: x 1 ≤ 40 20 40Not binding The other classification of resource ResourceSlack valueStatus Finishing hours 0Scarce Carpentry hours 0Scarce Demand 20Abundant

11. Example 2 Advertisement ( Winston 3.2, p 61) Subject to

13. Example 3 Giapetto’ Woodcarving, Inc., (changed) Subject to

15. Example 4 Giapetto’ Woodcarving, Inc., (changed) Subject to

16. Example 4 Giapetto’ Woodcarving, Inc., (changed) Subject to

17. Sensitivity Analysis

18. Shadow Prices Shadow price of resource i measures the marginal value of this resource.