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Linear Programming. Linear programming A technique that allows decision makers to solve maximization and minimization problems where there are certain.

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Presentation on theme: "Linear Programming. Linear programming A technique that allows decision makers to solve maximization and minimization problems where there are certain."— Presentation transcript:

1 Linear Programming

2 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.

3 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

4 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.

5 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 

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

7 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)

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

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

10  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

11 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

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

13 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

14 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.

15 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.

16 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

17 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

18  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).

19 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 ≦≦

20 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 ).

21 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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