with(Optimization): with(plots):

Warning, the name changecoords has been redefined

Brett StrassnerMA515 Combinatorial Optimization HWProblem 4.5Running the GGMC LP for the original number of resource stock quantities yields the familiar solution given the original c vector and its associated optimal value. To set this up in Maple,make sure to add the option "maximize" since the LPSolve command defaults to a minimization problem. c := Vector([5, 4], datatype=float): A := Matrix([[1,2],[1,1],[2,1]], datatype=float): b := Vector([120,70,100], datatype=float): LPSolve(c, [A, b], assume=nonnegative, maximize);

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constraintsA:={x+2*y<=120, x+y<=70, 2*x+y<=100, x>=0, y>=0}: domainA:=inequal(constraintsA, x=0..100, y=0..100, optionsfeasible=(color=grey), optionsclosed=(color=black, thickness=2), optionsexcluded=(color=white) ):%;

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It is best to consider each sale of a single resource independently, as an LP altered only with respect to the one constraint. So, we will construct three new LPs.We represent the sale of one unit of labor by decreasing the labor component by one unit. Note that there is no change of either the solution or the optimal value.clabor := Vector([5, 4], datatype=float): A := Matrix([[1,2],[1,1],[2,1]], datatype=float): b := Vector([119,70,100], datatype=float): LPSolve(clabor, [A, b], assume=nonnegative, maximize);

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constraintsA:={x+2*y<=119, x+y<=70, 2*x+y<=100, x>=0, y>=0}: domainA:=inequal(constraintsA, x=0..100, y=0..100, optionsfeasible=(color=grey), optionsclosed=(color=black, thickness=2), optionsexcluded=(color=white) ):%;

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For the sale of a unit of wood, the wood constraint is lowered by one unit. This is represented graphically by transforming the b vector inward, toward the feasible region. This time, the optimal value has decreased by $3.00. This is the amount the buyer would have to pay GGMC for every unit of wood they would like to purchase in order for GGMC not to lose profit.cwood := Vector([5, 4], datatype=float): A := Matrix([[1,2],[1,1],[2,1]], datatype=float): b := Vector([120,69,100], datatype=float): LPSolve(cwood, [A, b], assume=nonnegative, maximize);

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constraintsA:={x+2*y<=120, x+y<=69, 2*x+y<=100, x>=0, y>=0}: domainA:=inequal(constraintsA, x=0..100, y=0..100, optionsfeasible=(color=grey), optionsclosed=(color=black, thickness=2), optionsexcluded=(color=white) ):%;

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The sale of a unit of metal also decreases the optimal value, but only by $1.00 this time. cmetal := Vector([5, 4], datatype=float): A := Matrix([[1,2],[1,1],[2,1]], datatype=float): b := Vector([120,70,99], datatype=float): LPSolve(cmetal, [A, b], assume=nonnegative, maximize);

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constraintsA:={x+2*y<=120, x+y<=70, 2*x+y<=99, x>=0, y>=0}: domainA:=inequal(constraintsA, x=0..100, y=0..100, optionsfeasible=(color=grey), optionsclosed=(color=black, thickness=2), optionsexcluded=(color=white) ):%;

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