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Problem 7-26 Presentation Piraeus, Greece - Olive Yield: Pruning vs. non-Pruning Methods. Pruning Yield More trees per acre Increased output, smaller olives One barrel = 5 hours labor on 1 acre One barrel sells for $20. Non-pruning Yield One barrel = 2 hours labor on 2 acres
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Problem 7-26 PresentationPiraeus, Greece - Olive Yield: Pruning vs. non-Pruning Methods Pruning Yield More trees per acre Increased output, smaller olives One barrel = 5 hours labor on 1 acre One barrel sells for $20 Non-pruning Yield One barrel = 2 hours labor on 2 acres One barrel sells for $30 Grower will produce no more than 40 barrels of “pruned” olives. Grower has 250 hours of labor available. Grower has 150 acres available for growing. Jan White - BUS340
Use graphical Linear Programming to find: • The maximum possible profit • The best combination of barrels of pruned and regular olives • The number of acres that the olive grower should devote to each growing process Jan White - BUS340
Equations: Labor Hours and Acres Available for Use Labor Hours: Pruned Olives Hours = 5P Unpruned Olives Hours = 2U 5P + 2U <= 250 Acres Available: Pruned Olives Acres = 1P Unpruned Olives Acres = 2U 1P + 2U <= 150 Other Constraints: P<= 40 barrels Jan White - BUS340
Equations and Solutions Labor: When U = 0 5P + 2(0) = 250, so 5P = 250, then P = 50 When P = 0 5(0) + 2U = 250, so 2U = 250, then U = 125 Acres: When U = 0 1P + 2(0) = 150, then P = 150 When P = 0 1(0) + 2U = 150, so 2U = 150, then U = 75 Use the points derived from the above equations to graph the lines for labor hours and acres and to display the “Feasible Region” Jan White - BUS340
Graph of the Equations and “Feasible Region” Pruned Olives “P” axis To find the coordinates of the point of maximum profit, the constraint equations must be solved simultaneously. The equations are: 5P + 2U <= 250, and 1P + 2U <= 150 (150, 0) 100 “Corner Point”, the possible point of maximum profit within the “Feasible Region” as outlined by the constraint equations (50,0) Feasible Region (0,125) 10 Unpruned Olives (Not to Scale) “U” axis 100 (0, 75) 10 Jan White - BUS340
Solving to find the solutions for maximum profit, barrels of olives to be produced and amount of acreage to be allocated for growing methods To isolate one variable (P) for solution, multiply the second equation by -1, so -1(1P + 2U = 150), which yields -P - 2U = -150. This equation is then added to the first. -P - 2U = -150 5P + 2U = 250 which solves as 4P = 100, so P = 25. This solution for P is now substituted back into the first equation to solve for U. 5(25) + 2U = 250, so 125 + 2U = 250, 2U = 125. U = 62.5 These solutions give you the coordinates for maximum profit, (25, 62.5) which is the number of barrels of each type of olive to be grown (25 pruned olives and 62.5 unpruned). Maximum profit (the Objective) can be solved (z = 20P + 30U) by substituting in the number of barrels of each, so 20(25) + 30(62.5) = z. So, z = $500 + $1875 = $2375 The amount of acreage to be used for each growth method can be solved as 1(25) + 2(62.5) <= 150, so 25 + 125 <= 150. This is true, so 25 acres will be used for pruned olives, and 125 acres for unpruned olives. Jan White - BUS340