2. 6. Linear Programming (Graphical Method) Objectives: 1.More than one solution 2.Unbounded feasible region 3.Examples Refs: B&Z 5.2.

3. For a general linear programming problem, the optimal solution (if it exists) will occur at at least one corner point. So to calculate the maximum value of P we can simply evaluate P = 50,00x + 20,000y at all corner points of the feasible region. The corner points are:(0,90),(0,30),(30,0). Evaluating the objective function at these points: P(0,90) = 50,000(0) + 20,000(90) = 1,800,000 P(0,30) = 50,000(0) + 20,000(30) = 600,000 P(30,0) = 50,000(30) + 20,000(0) = 1,500,000 So this confirms our earlier result.

4. Sometimes the maximum value will occur at two points. This means that the objective function is parallel to a boundary of the feasible region. max P = 12x + 4y subj to x + y ≤ 600 6x + 2y ≤ 1,800 x ≥ 0y ≥ 0 6x+2y=1800 300 900 300 600 x+y=600 feasible region 200 100 12x+4y=1200 12x+4y=2400 increasing P all points here are optimal 12x+4y=3600 The maximum value of P is 3,600. This occurs at all points on the boundary of the feasible region which intersect the line 12x + 4y = 3,600 Example 2. 12x+4y=3600

5. In the previous examples we were concerned with finding a maximum value for the objective function. We can just as easily determine a minimum value for the objective function. In Example 1, the minimum value is 600,000 and it occurs at the point (0,30). The techniques we use to find the minimum value of a function are exactly the same as those we use to find the maximum value of a function What happens when the feasible region is not bounded?

6. Unbounded feasible regions Example 3:Sketch the region defined by the inequalities 2x + 3y ≥ 12 3x + 2y ≥ 12 x ≥ 0 y ≥ 0. 2x+3y=12 3x+2y=12 4 4 6 6 In this example the feasible region is unbounded. The corner points are(0,6)(6,0)and we will calculate the remaining corner point using Gaussian elimination. (see ohp slide for details)

7. We are now going to minimize the function on this region. P = 2x + 2y P (0,6) = 12,P (6,0) = 12,P (12/5, 12/5) = 24/5 + 24/5 = 48/5. It appears that the minimum value of P is48/5. Since we are trying to minimize the function we follow our lines of constant slope in the direction of decreasing P. As we suspected, the optimal solution is (12/5, 12/5) and P* = 48/5. Let’s check the corner points first. 2x+3y=123x+2y=12 4 4 6 6 2x+2y=16 2x+2y=8 decreasing P increasing P

8. What happens now if we try to maximize P = 2x + 2y? If we return to our earlier calculation, we see that P = 12 at the points (0,6) and (6,0) (and hence at all points on the line segment joining them). But is this the maximum value of P? Following the lines of constant slope in the direction of increasing P, we see that we can increase the value of P as much as we like. In this case there is no maximum value. So this problem has no optimal solution. 2x+3y=123x+2y=12 4 4 6 6 2x+2y=16 2x+2y=8 decreasing P increasing P

9. Example 4: 1. Max P = 1/10 x + y Subj to4x + 3y ≥ 24 3x + 5y ≥ 30 x ≥ 0y ≥ 0 By drawing a diagram of the feasible region and sketching the objective function for various values of P, we can see that the feasible region is unbounded and we can increase P as much as we like. So there is no optimal solution. 10 8 6 6 4x+ 3y =24 3x+ 5y =30 1/10x+y=1 1/10x+y=2 inc P 1/10x+y=8

10. The feasible region is bounded below, so the problem will have a solution. To find it, we evaluate the objective function at the corner points. Example 4: 2. Min P = 1/10 x + y Subj to4x + 3y ≥ 24 3x + 5y ≥ 30 x ≥ 0y ≥ 0 10 8 6 6 4x+ 3y =24 3x+ 5y =30 1/10x+y=1 1/10x+y=2 dec P (see slide for details)

11. Example 4: 3. Max P = 1/10 x + y Subj to4x + 3y ≤ 24 3x + 5y ≤ 30 x ≥ 0y ≥ 0 10 8 6 6 4x+ 3y =24 3x+ 5y =30 The feasible region is now bounded so we will be able to find an optimal solution. We evaluate the function at the new corner points feasible region

12. The corner points are (0,6), (0,0), (6,0), (30/11,48/11). P(0,6) = 1/10 (0) + 6 = 6 P(30/11,48/11) = 1/10 (30/11) + 48/11 = 51/11 P(6,0) = 1/10 (6) + 0 = 6/10 P(0,0) = 1/10 (0) + 0 = 0 So the optimal solution is (0,6), P* = 6

13. We don’t need to do any further calculations. 10 8 6 6 4x+ 3y =24 3x+ 5y =30 feasible region Example 4: 4. Min P = 1/10 x + y Subj to4x + 3y ≤ 24 3x + 5y ≤ 30 x ≥ 0y ≥ 0 The optimal solution is (0,0) and P*=0.

14. You should now be able to complete Example Sheet 2 from the Orange Book.