quantitative decision making 7 th ed. by lapin and whisler chapter 8: linear programming
TRANSCRIPT
Quantitative Decision Making 7th ed.By
Lapin and Whisler
Chapter 8: Linear Programming
The story behind Linear Programming
George B. Dantzig John von Neumann Leonid Kantorovich
DefinitionLinear programming is a mathematical
method that is used to establish a plan that efficiently allocates limited resources to achievement of a desired objective.
Linear programming is the process of maximizing or minimizing a linear function subject to a set of constraints.
Possible Applications for LPDeveloping a production schedule and inventory policy.
Establishing a portfolio that maximizes return.
Maximizing advertising effectiveness.
Minimizing total transportation costs.
Main Characteristics of LP ProblemsConcerned with maximizing or minimizing
some quantity.Restrictions or constraints that limit the
degree to which the objective can be pursued.
Only linear relationships are involved.
http://en.wikipedia.org/wiki/Linear_programming
Three Components to LP
1. Variables2. Constraints3. Objective Function Non-negativity conditions.
Example (p.263)The Redwood Furniture Company makes tables and chairs as part of its
line of patio furniture.
Resource
Unit Requirements Amount Available
Table Chair
Wood(board ft)
30 20 300
Labor(hours)
5 10 110
Unit Profit $6 $8How many tables and chairs should be made to maximize the total profit?
Redwood FurnitureProblem FormulationLet XT and XC denote the number of tables and
chairs to be made. (Define variables)
Maximize P = 6XT + 8XC (Objective function)
Subject to: (Constraints) 30XT + 20XC < 300 (wood)
5XT + 10XC < 110 (labor)
where XT and XC > 0 (non-negativity conditions)
Letting XT represent the horizontal axis and XC the vertical, the constraints and non-negativity conditions define the feasible solution region.
Feasible Solution Region for Redwood Furniture Problem
Graphing to Find Feasible Solution Region For an inequality constraint (with < or >),
first plot as a line: 30XT + 20XC = 300. Get two points. Intercepts are easiest:
Set XC = 0, solve for XT for horizontal intercept: 30XT + 20(0) = 300 => XT = 300/30 = 10
Set XT = 0, solve for XC for vertical intercept: 30(0) + 20XC = 300 => XC = 300/20 = 15
Above gets wood line. Do same for labor. Mark valid sides and shade feasible solution
region. Any point there satisfies all constraints and non-negativity conditions.
Graphing to Find Feasible Solution Region To establish valid side, pick a test point
(usually the origin). If that point satisfies the constraint, all points on same side are valid. Otherwise, all points on other side are instead valid.
Equality constraints have no valid side. The solution must be on the line itself.
Some constraint lines are horizontal or vertical. These involve only one variable and one intercept.
Finding Most Attractive Corner The optimal solution will always correspond to a
corner point of the feasible solution region. Because there can be many corners, the most
attractive corner is easiest to find visually. That is done by plotting two P lines for arbitrary
profit levels. Since the P lines will be parallel, just hold your
pencil at the same angle and role it in from the smaller P’s line toward the bigger one’s That is the direction of improvement.
Continue rolling until only one point lies beneath the pencil. That is the most attractive corner. (Problems can have two most attractive corners.)
Most Attractive Corner for Redwood Furniture Problem
Finding the Optimal Solution The coordinates of the most attractive
corner provide the optimal levels. Because reading from graph may be
inaccurate, it is best to solve algebraically. Simultaneously solving the wood and labor
equations, the optimal solution is:XT = 4 tables XC = 9 chairsP = 6(4) + 8(9) = 96 dollars
Note: supply the computed level of the objective in reporting the optimal solution.
Finding most attractive corner algebraically
Identify corner points: (0,0), (0,11), (10,0) (4,9)
Substitute into objective function and compare values:
P = 6XT + 8XC
(XT=0, XC=0) P=6(0)+8(0)=0
(XT=0, XC=11) P=6(0)+8(11)=88(XT=10, XC=0) P=6(10)+8(0)=60
(XT=4, XC=9) P=6(4)+8(9)=96
Feasible Solution Region for Redwood Furniture Problem