© 2003 Anita Lee-Post Linear Programming Part 2 By Anita Lee-Post

© 2003 Anita Lee-Post LP solution methods Graphical solution method Corner-point solution method Excel Solver solution method

© 2003 Anita Lee-Post Graphical solution method Limited to 2 decision variables problems 1.Assign X 1 and X 2 to the vertical and horizontal axes 2.Plot constraint equations 3.Determine the area of feasibility 4.Plot the objective function 5.Find the optimum point

© 2003 Anita Lee-Post LP example #1 Minimize Cost = 5X 1 + 3X 2 Subject to: 100X X 2 >= X X 2 >= X X 2 >= 5000 X 1, X 2 >= 0

© 2003 Anita Lee-Post Graphical solution method Step 1. Assign decision variables to axes X1 X2

© 2003 Anita Lee-Post Graphical solution method continued Step 2. Plot constraint 100X X 2 >= Plot the line 100X X 2 = 4000  (0, 40), (40,0) 2.Determine the constraint region  (0, 0) < 4000, thus, the constraint region is to the right of the line X1 X2 40 (0,0)(0,0)

© 2003 Anita Lee-Post Graphical solution method continued Step 2. Plot constraint 200X X 2 >= Plot the line 200X X 2 =  (0, 25), (50,0) 2.Determine the constraint region  (0, 0) < 10000, thus, the constraint region is to the right of the line X1 X (0,0)(0,0) 40 25

© 2003 Anita Lee-Post Graphical solution method continued Step 2. Plot constraint 200X X 2 >= Plot the line 200X X 2 = 5000  (0, 50), (25,0) 2.Determine the constraint region  (0, 0) < 5000, thus, the constraint region is to the right of the line X1 X (0,0)(0,0)

© 2003 Anita Lee-Post Graphical solution method continued Step 3. Determine the area of feasibility The area of feasibility is found by intersecting the three constraint regions X1 X (0,0)(0,0) Area of feasibility

© 2003 Anita Lee-Post Graphical solution method continued Step 4. Plot the objective function: cost = 5X 1 + 3X 2 1.Assume an arbitrary cost, e.g., Plot the line 5X 1 + 3X 2 = 150  (0, 50), (30,0) X1 X2 (0,0)(0,0) 50 Area of feasibility 30 Cost = 150

© 2003 Anita Lee-Post Graphical solution method continued Step 5. Find the optimal point 1.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. 2.The extreme point is the optimal point  (10, 30) X1 X2 (10,30) 50 Area of feasibility 30 Cost = 150

© 2003 Anita Lee-Post LP example #2 Maximize Profit = 40X X 2 Subject to: X 1 <= 400 X 2 <= 700 X 1 + X 2 <= 800 X 1 + 2X 2 <= 1000 X 1, X 2 >= 0

© 2003 Anita Lee-Post Graphical solution method Following the 5-step process, the optimal point is found to be X 1 =400, X 2 =300 X1 X2 (400,300) 700 Area of feasibility 400 Profit = X1=400 X2= X1+X2= X1+2X2=1000

© 2003 Anita Lee-Post Corner-point solution method Use in conjunction with the graphical solution method to pinpoint the optimal point algebraically 1.Find the co-ordinates of each corner point of the feasible region by simultaneously solving the equations of a pair of intersecting lines 2.Substitute the values of the co-ordinates in the objective function

© 2003 Anita Lee-Post LP example #1 X1 X2 A B C D Corner PointIntersecting linesCoordinatesCost=5X1 + 3X2 A X2=0 200X X2=10000 X1=50, X2=0Cost=250 B 100X X2 = X X2=10000 X1=30, X2=10Cost=180 C 100X X2 = X X2 = 5000 X1=10, X2=30Cost=140* D X1=0 200X X2 = 5000 X1=0, X2=50Cost=150 Optimal solution

© 2003 Anita Lee-Post LP example #2 X1 X2 X1=400 Corner PointIntersecting linesCoordinatesProfit=40X1 + 30X2 A X2=0 X1=400 X1=400, X2=0Profit=16000 B X1=400 X1 + 2X2 = 1000 X1=400, X2=300Profit=25000 C X1=0 X1 + 2X2 = 1000 X1=0, X2=500Profit=15000 D X1=0 X2=0 X1=0, X2=0Profit=0 AB CD X1 + 2X2 = 1000 Optimal solution