Linear Programming Part 2

1. Linear ProgrammingPart 2 By Anita Lee-Post

2. LP solution methods • Graphical solution method • Corner-point solution method • Excel Solver solution method

3. Graphical solution method Limited to 2 decision variables problems • Assign X1 and X2 to the vertical and horizontal axes • Plot constraint equations • Determine the area of feasibility • Plot the objective function • Find the optimum point

4. LP example #1 Minimize Cost = 5X1 + 3X2 Subject to: 100X1 + 100X2 >= 4000 200X1 + 400X2 >= 10000 200X1 + 100X2 >= 5000 X1, X2 >= 0

5. Graphical solution method Step 1. Assign decision variables to axes X1 X2

6. Graphical solution method continued Step 2. Plot constraint 100X1+100X2 >= 4000 • Plot the line 100X1 + 100X2 = 4000  (0, 40), (40,0) • Determine the constraint region  (0, 0) < 4000, thus, the constraint region is to the right of the line X1 40 X2 (0,0) 40

7. Graphical solution method continued Step 2. Plot constraint 200X1+400X2 >= 10000 • Plot the line 200X1 + 400X2 = 10000  (0, 25), (50,0) • Determine the constraint region  (0, 0) < 10000, thus, the constraint region is to the right of the line X1 50 40 X2 (0,0) 25 40

8. Graphical solution method continued Step 2. Plot constraint 200X1+100X2 >= 5000 • Plot the line 200X1 + 100X2 = 5000  (0, 50), (25,0) • Determine the constraint region  (0, 0) < 5000, thus, the constraint region is to the right of the line X1 50 40 25 X2 (0,0) 25 40 50

9. Graphical solution method continued Step 3. Determine the area of feasibility The area of feasibility is found by intersecting the three constraint regions X1 50 Area of feasibility 40 25 X2 (0,0) 25 40 50

10. Graphical solution method continued Step 4. Plot the objective function:cost = 5X1 + 3X2 • Assume an arbitrary cost, e.g., 150 • Plot the line 5X1 + 3X2 = 150  (0, 50), (30,0) X1 Area of feasibility 30 X2 (0,0) 50 Cost = 150

11. Graphical solution method continued Step 5. Find the optimal point • The region to the left of the iso-cost line (Cost=150) is represented by Cost <= 150  Move the iso-cost line leftwards until it reaches the most extreme point of the area of feasibility. • The extreme point is the optimal point  (10, 30) X1 Area of feasibility 30 (10,30) X2 Cost = 150 50

12. LP example #2 Maximize Profit = 40X1 + 30X2 Subject to: X1 <= 400 X2 <= 700 X1 + X2 <= 800 X1 + 2X2 <= 1000 X1, X2 >= 0

13. Graphical solution method X1 1000 Following the 5-step process, the optimal point is found to be X1=400, X2=300 800 X1+2X2=1000 X2=700 X1+X2=800 (400,300) X1=400 400 300 Profit = 12000 Area of feasibility X2 400 500 700 800

14. Corner-point solution method Use in conjunction with the graphical solution method to pinpoint the optimal point algebraically • Find the co-ordinates of each corner point of the feasible region by simultaneously solving the equations of a pair of intersecting lines • Substitute the values of the co-ordinates in the objective function

15. LP example #1 X1 Optimal solution A B C D X2

16. LP example #2 X1 Optimal solution A B X1=400 X1 + 2X2 = 1000 D C X2