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Linear Programming

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Linear programming A technique that allows decision makers to solve maximization and minimization problems where there are certain constraints that limit what can be done, given that all objectives and constraints are linear.

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DefinitionsDefinitions The objective function is the function to be maximized (or minimized) The constraints are given in inequalities. Both the objective and constraints are linear in choice variables

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ExampleExample Process 1: requires 3 machine-hours and 0.4 labor-hours per batch of cloth. Process 2: requires 2.5 machine-hours and 0.5 labor-hours per batch of cloth. Process 3: requires 5.25 machine-hours and 0.35 labor-hours per batch of cloth. The profit is $1, $0.9, and $1.1 with process 1, 2, and 3, respectively. Max L and K are 600 and 6,000, respectively.

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The objective is to maximize profit subject to resource constraints Maximize = 1.0Q 1 + 0.9Q 2 + 1.1Q 3 subject to the constraints: 3Q 1 + 2.5Q 2 + 5.25Q 3 ≦ 6000 (Machine hours) 0.4Q 1 + 0.5Q 2 + 0.35Q 3 ≦ 600 (Labor Hours) Q 1 ≧ 0; Q 2 ≧ 0; Q 3 ≧ 0

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Process rays Q 3 (5.25 / 0.35) Q 1 (3 / 0.4) Q 2 (2.5 / 0.5)

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Isoquants (line ABC, NOT line AC) A B C Q 3 (5.25 / 0.35) Q 1 (3 / 0.4) Q 2 (2.5 / 0.5)

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Feasible Set (6000, 600) Q 2 (2.5 / 0.5) Q 1 (3 / 0.4) Q 3 (5.25 / 0.35)

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Graphical Solution: A* A* Q 3 (5.25 / 0.35) Q 1 (3 / 0.4) Q 2 (2.5 / 0.5)

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Suppose that the company is no longer constrained by limits on the amount of Labor and capital. It can hire all of labor it wants at $12 per hour and all of the capital at $1.6 per machine-hour. Its problem is to choose that combination of processes which will produce, say, 400 units at minimum cost. Min C = 9.6 Q 1 + 10 Q 2 + 12.6 Q 3 Subject to: Q 1 + Q 2 + Q 3 ≧ 400 Subject to: Q 1 + Q 2 + Q 3 ≧ 400 Q 1 ≧ 0, Q 2 ≧ 0, Q 3 ≧ 0 Q 1 ≧ 0, Q 2 ≧ 0, Q 3 ≧ 0

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Optimal Solution: Cost Minimization Problem B* Iso-cost line Q 3 (5.25 / 0.35) Q 1 (3 / 0.4) Q 2 (2.5 / 0.5) Solution: All of the output are produced with process 1

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Simplex method A systematic process for comparing extreme point or corner solutions to linear programming problems

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Dual problems Every optimization problem has a corresponding problem called its dual If the primal problem is a maximization, the dual is a minimization (and vice versa) Solutions to the dual are shadow prices

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Shadow Prices Shadow prices tell you what would happen to the objective function if you relaxed the constraint by one unit. They show which types of capacity are bottlenecks, or effective constraints on output, since capacity that is underutilized receives a zero shadow price. More important, shadow prices indicate how much it would be worth to management to expand each type of capacity.

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Example Output 1 (Q 1 ): requires 3 units of capital and 2 units of labor to produce one unit of output 1. Output 2 (Q 2 ): requires 5 units of capital and 4 units of labor to produce one unit of output 2. The firm has 4,000 units of labor and 5,400 units of capital. The price of Q 1 and Q 2 are $50 and $80, respectively.

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Primal problem The primal problem is to find the value of Q 1 and Q 2 that maximize total revenue subject to the resource constraint. Max TR = 50 Q 1 + 80 Q 2 Subject to: 2Q 1 + 4 Q 2 ≦ 4,000 Subject to: 2Q 1 + 4 Q 2 ≦ 4,000 3Q 1 + 5 Q 2 ≦ 5,400 3Q 1 + 5 Q 2 ≦ 5,400 Q 1 ≧ 0, Q 2 ≧ 0 Q 1 ≧ 0, Q 2 ≧ 0

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Dual problem The dual problem is to find the value of P L and P K, which are the price of L and K, respectively, to minimize total value of resource available subject to the constraint. Min C = 4,000 P L + 5,400 P K Subject to: 2 P L + 3 P K ≧ 50 (*) 4 P L + 5 P K ≧ 80 (**) P L ≧ 0, P K ≧ 0

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Constraint (*) and (**) say that the total value of resources used in the production of a unit of output must be greater or equal to the price. P L and P K are the prices that a manager should be willing to pay for these resources. They are the opportunity cost of using theses resources. If a resource is not fully utilized, its shadow prices will be zero, since an extra unit of the resource would not increase total revenue (objective function).

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Max 50 Q 1 + 80 Q 2 S. t. 2 Q 1 + 4 Q 2 ≦ 4000 P L 3 Q 1 + 5 Q 2 ≦ 5400 P K ≦≦

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Slack Variables Slack variables (U L and U K ) represent the amounts of various inputs that are unused. 2Q 1 + 4 Q 2 + U L = 4,000 3Q 1 + 5 Q 2 + U K = 5,400 If the slack variable for an input turns out to be zero, this means that this input is fully utilized (U h = 0 implies P h > 0, h = L, K ) If it turns out to be positive, this means that some of this input is redundant (U h > 0 implies P h = 0, h = L, K ).

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How H. J. Heinz Minimize Its Shipping To determine how much ketchup each factory should send to each warehouse, Heinz has used linear programming techniques. The capacities of each factory, the requirements of each warehouse, and freight rates are given in the first table. The optimal daily shipment from each factory to each warehouse is shown in the second table. For example, all warehouse A’s ketchup should come from factory I.

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