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© 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
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