Transcript Chapter 02

Chapter 2 - Linear Programming: Basic Concepts
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Linear Programming – An Overview
A Maximization Model Example
Model Formulation
Graphical Solutions of Linear Programming Models
A Minimization Model Example
Irregular Types of Linear Programming Models
Characteristics of Linear Programming Problems
1 - Chap 02
Linear Programming - An Overview
• Objectives of business firms frequently include maximizing profit or
minimizing costs.
• Linear programming is an analysis technique in which linear algebraic
relationships represent a firm’s decisions given a business objective
and resource constraints.
• Steps in application:
1- Identify problem as solvable by linear programming.
2- Formulate a mathematical model of the unstructured problem.
3- Solve the model.
4- Sensitivity Analysis.
2 - Chap 02
A Maximization Model Example
Problem Definition
• Product mix problem - Beaver Creek Pottery Company
• The firm is planning to produce two products, Bowl and Mug.
• Question: How many bowls and mugs should be produced in order to
maximize profits given labor and materials constraints?
• Product resource requirements and unit profit:
Product
Bowl
Mug
Availability
Resource Requirements
Labor
Clay
(hr/unit)
(lb/unit)
1
4
2
40 hr per day
3
Unit Profit ($)
40
50
120 pounds per day
3 - Chap 02
A Maximization Model Example
Decision Variables:
x1=number of bowls to produce/day
x2= number of mugs to produce/day
Objective function
maximize Z = $40x1 + 50x2
where Z= profit per day
Resource Constraints:
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
Non-negativity Constraints:
x10; x2  0
4 - Chap 02
Complete Linear Programming Model:
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Daily profit
Labor constraint
Clay constraint
Non-negativity
5 - Chap 02
Model Components and Formulation
• Decision variables: mathematical symbols representing levels of
activity of a firm .
• Objective function: a linear mathematical relationship describing an
objective of the firm, in terms of decision variables, that is maximized
or minimized.
• Functional Constraints: restrictions placed on the firm by the
operating environment stated in linear relationships of the decision
variables.
– Resource Limitation
– Minimum Requirement
– Fixed Requirement
( <= )
( >= )
(=)
• Parameters: numerical coefficients and constants used in the objective
function and constraint.
6 - Chap 02
Feasible/Infeasible Solutions
• A feasible solution does not violate any of the constraints:
Example x1= 5 bowls
x2= 10 mugs
Daily profit Z = $40 x1 + 50x2= $700
Labor constraint check:
1(5) + 2(10) = 25 < 40 hours, within constraint
Clay constraint check:
4(5) + 3(10) = 50 < 120 pounds, within constraint
• An infeasible solution violates at least one of the constraints:
Example x1 = 10 bowls
x2 = 20 mugs
Daily profit Z = $1400
Labor constraint check:
1(10) + 2(20) = 50 > 40 hours, violates constraint
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General Form of Linear Programming Problem
The usual and most intuitive form of describing a linear programming
problem. It consists of the following three parts:
A linear function to be maximized
Maximize Z = c1x1 + c2x2 + c3x3 + c4x4
Problem constraints of the following form
a11x1 + a12x2 + a13x3 + a14x4 ≤ b1 Constraint type 1
a21x1 + a22x2 + a23x3 + a24x4 ≥ b2 Constraint type 2
a31x1 + a32x2 + a33x3 + a34x4 = b3 Constraint type 3
Non-negative variables
For xj ≥ 0
8 - Chap 02
Graphical Solution of Linear Programming Models
• Graphical solution is limited to linear programming models containing
only two decision variables. (Can be used with three variables but only
with great difficulty.)
• Graphical methods provide visualization of how a solution for a linear
programming problem is obtained.
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Graphical Solution of a Maximization Model
Both Constraints
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
Graph of both model Constraints
10 - Chap 02
Graphical Solution of a Maximization Model
Feasible Solution Area
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
The feasible solution area constraints
11 - Chap 02
Graphical Solution of a Maximization Model
Objective Function = $800
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
Objective function line for Z = $800
12 - Chap 02
Graphical Solution of a Maximization Model
Alternative Objective Functions
Z=$800, $1200, $1600 = $40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
Alternative objective function lines for profits, Z, of $800, $1,200, and $1,600
13 - Chap 02
Graphical Solution of a Maximization Model
Optimal Solution
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
Identification of optimal solution point
14 - Chap 02
Graphical Solution of a Maximization Model
Corner Point Solutions
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
Solutions at all corner points
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Graphical Solution of a Maximization Model
Optimal Solution for New Objective Function
maximize Z=$70x1 + 20x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
The optimal solution with Z = 70x1 1 20x2
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Slack Variables
• Standard form requires that all constraints be in the form of equations.
• A slack variable is added to a  constraint to convert it to an equation
(=).
• A slack variable represents unused resources.
• A slack variable contributes nothing to the objective function value.
• If a constraint has zero slack, the constraint is binding.
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Complete Linear Programming Model in Standard Form
maximize Z=$40x1 + 50x2 + 0s1 + 0s2
subject to
1x1 + 2x2 + s1 = 40
4x1 + 3x2 + s2 = 120
x1,x2,s1,s2  0
where x1 = number of bowls
x2 = number of mugs
s1, s2 are slack variables
Solutions at points A, B, and C with slack
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A Minimization Model Example
Problem Definition
• Two brands of fertilizer available - Super-gro, Crop-quick.
• Field requires at least 16 pounds of nitrogen and 24 pounds of
phosphate.
• Super-gro costs $6 per bag, Crop-quick $3 per bag.
• Problem : How much of each brand to purchase to minimize total cost
of fertilizer given following data ?
Brand
Super-gro
Crop-quick
Chemical Contribution
Nitrogen (lb/bag) Phosphate (lb/bag)
2
4
4
3
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A Minimization Model Example
Model Construction
Decision variables
x1 = bags of Super-gro
x2 = bags of Crop-quick
The objective function:
minimize Z = $6x1 + 3x2
where $6x1 = cost of bags of Super-gro
3x2 = cost of bags of Crop-quick
Model constraints:
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
x1, x2  0 (non-negativity constraint)
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A Minimization Model Example
Complete Model Formulation and Constraint Graph
Complete model formulation:
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x1, x2  0
Constraint lines for fertilizer model
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A Minimization Model Example
Feasible Solution Area
Complete model formulation:
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x1, x2  0
Feasible solution area
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A Minimization Model Example
Optimal Solution Point
Complete model formulation:
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x1, x2  0
The optimal solution point
23 - Chap 02
A Minimization Model Example
Surplus Variables
• A surplus variable is subtracted from a  constraint to convert it
to an equation (=).
• A surplus variable represents an excess above a constraint
requirement level.
• Surplus variables contribute nothing to the calculated value of the
objective function.
• If a constraint has zero surplus, the constraint is binding.
• Subtracting surplus variables in the farmer problem constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)
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A Minimization Model Example
Graphical Solutions
minimize Z = $6x1 + 3x2 + 0s1 + 0s2
subject to
2x1 + 4x2 - s1 = 16
4x1 + 3x2 - s2 = 24
x1, x2, s1, s2  0
Graph of the fertilizer example
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Irregular Types of Linear Programming Problems
• The general rules do not apply for some linear programming models.
• Special types of problems include those with:
1. Multiple optimal solutions
2. Infeasible solutions
3. Unbounded solutions
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Multiple Optimal Solutions
Objective function is parallel to a
constraint line:
maximize Z=$40x1 + 30x2
subject to
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
x1, x2  0
where x1 = number of bowls
x2 = number of mugs
Graph of a problem with multiple optimal solutions
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An Infeasible Problem
Every possible solution violates
at least one constraint:
maximize Z = 5x1 + 3x2
subject to
4x1 + 2x2  8
x1
4
x2  6
x1 ,
x2  0
Graph of an infeasible problem
28 - Chap 02
An Unbounded Problem
Value of objective function
increases indefinitely:
maximize Z = 4x1 + 2x2
subject to
x1  4
x2  2
x1, x2  0
Graph of an unbounded problem
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Characteristics of Linear Programming Models
• Proportionality - The rate of change (slope) of the objective function
and constraint equations is constant.
• Additivity - Terms in the objective function and constraint equations
must be additive.
• Divisibility -Decision variables can take on any fractional value and are
therefore continuous as opposed to integer in nature.
• Certainty - Values of all the model parameters are assumed to be
known with certainty (non-probabilistic).
• Non-negativity – All variables should be non-negative.
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