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Introduction to Management Science
8th Edition
by
Bernard W. Taylor III
Chapter 6
Linear Programming: Model
Formulation and Graphical Solution
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
1
Chapter Topics
Model Formulation
A Maximization Model Example
Graphical Solutions of Linear Programming Models
A Minimization Model Example
Irregular Types of Linear Programming Models
Characteristics of Linear Programming Problems
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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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:
Identify problem as solvable by linear programming.
Formulate a mathematical model of the unstructured
problem.
Solve the model.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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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
Constraints - restrictions placed on the firm by the
operating environment stated in linear relationships of the
decision variables.
Parameters - numerical coefficients and constants used in
the objective function and constraint equations.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Definition
A Maximization Model Example (1 of 2)
Product mix problem - Beaver Creek Pottery Company
How many bowls and mugs should be produced to
maximize profits given labor and materials constraints?
Product resource requirements and unit profit:
Product
Resource Requirements
Labor
Clay
Profit
(hr/unit)
(lb/unit)
($/unit)
Bowl
1
4
40
Mug
2
3
50
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Definition
A Maximization Model Example (2 of 3)
Resource
Availability:
40 hrs of labor per day
120 lbs of clay
Decision
Variables:
x1 = number of bowls to produce per day
x2 = number of mugs to produce per 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
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Definition
A Maximization Model Example (3 of 3)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to:
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Feasible Solutions
A feasible solution does not violate any of the constraints:
Example x1 = 5 bowls
x2 = 10 mugs
Z = $40x1 + $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
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Infeasible Solutions
An infeasible solution violates at least one of the
constraints:
Example x1 = 10 bowls
x2 = 20 mugs
Z = $1400
Labor constraint check:
1(10) + 2(20) = 50 > 40 hours, violates constraint
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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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.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Coordinate Axes
Graphical Solution of Maximization Model (1 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.1
Coordinates for Graphical Analysis
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Labor Constraint
Graphical Solution of Maximization Model (2 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Find the boundary first,
where 1x1 + 2x2 = 40
Figure 6.1
Graph of Labor Constraint
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Labor Constraint Area
Graphical Solution of Maximization Model (3 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Determine which side is
allowed, by checking if the
easiest choice, x1 = 0 = x2
satisfies the inequality.
Figure 6.3
Labor Constraint Area
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Clay Constraint Area
Graphical Solution of Maximization Model (4 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.4
Clay Constraint Area
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Both Constraints
Graphical Solution of Maximization Model (5 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.5
Graph of Both Model Constraints
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Feasible Solution Area
Graphical Solution of Maximization Model (6 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
T: violates both constraints;
S: violates the 1st constraint;
R: feasible.
Figure 6.6
Feasible Solution Area
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Objective Solution = $800
Graphical Solution of Maximization Model (7 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Plot the profit function, Z, for
A sample value, Z = $800.
Figure 6.7
Objection Function Line for Z = $800
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Alternative Objective Function Solution Lines
Graphical Solution of Maximization Model (8 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Plot the profit function for a
few more sample values.
Z increases
Figure 6.8
Alternative Objective Function Lines
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Optimal Solution
Graphical Solution of Maximization Model (9 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.9
Identification of Optimal Solution
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Optimal Solution Coordinates
Graphical Solution of Maximization Model (10 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
The point B is determined as
the common solution to
4x1 + 3x2 = 120
x1 + 2x2 = 40
Solve this system of equations…
Figure 6.10
Optimal Solution Coordinates
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Corner Point Solutions
Graphical Solution of Maximization Model (11 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.11
Profit at Each Corner Point
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Optimal Solution for New Objective Function
Graphical Solution of Maximization Model (12 of 12)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 6.12
Optimal Solution with Z’ = 70x1 + 20x2
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Slack Variables
Standard form requires that all constraints be in the form of
equations.
A slack variable is added to an inequality (≤ or ≥) constraint
to convert it to an equation (=).
A slack variable represents unused resources.
A slack variable contributes nothing to the objective function
(total profit, cost, etc.) value.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Linear Programming Model
Standard Form
Max Z = 40x1 + 50x2
subject to:1x1 + 2x2 + s1 = 40
4x1 + 3x2 + s2 = 120
x1, x2, s1, s2  0
Where:
s2
x1 = number of bowls
x2 = number of mugs
s1, s2 are slack variables
Figure 6.13
Solution Points A, B, and C with Slack
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Definition
A Minimization Model Example (1 of 7)
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 ?
Chemical Contribution
Nitrogen
(lb/bag)
Phosphate
(lb/bag)
Super-gro
2
4
Crop-quick
4
3
Brand
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Definition
A Minimization Model Example (2 of 7)
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)
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Model Formulation and Constraint Graph
A Minimization Model Example (3 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 6.14
Graph of Both Model Constraints
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Feasible Solution Area
A Minimization Model Example (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 6.15
Feasible Solution Area
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Optimal Solution Point
A Minimization Model Example (5 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 6.16
Optimum Solution Point
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Surplus Variables
A Minimization Model Example (6 of 7)
A surplus variable is subtracted from an inequality (≥ or ≤)
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.
Subtracting slack variables in the farmer problem
constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Graphical Solutions
A Minimization Model Example (7 of 7)
Minimize Z = $6x1 + $3x2 + 0s1 + 0s2
subject to:
2x1 + 4x2 – s1 = 16
4x1 + 3x2 – s2 = 24
x1, x2, s1, s2  0
Figure 6.17
Graph of Fertilizer Example
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Irregular Types of Linear Programming Problems
For some linear programming models, the general rules do
not apply.
Special types of problems include those with:
Multiple optimal solutions
Infeasible solutions
Unbounded solutions
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Multiple Optimal Solutions
Beaver Creek Pottery Example
Objective function is parallel
to a constraint line.
Maximize Z=$40x1 + 30x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Where:
x1 = number of bowls
x2 = number of mugs
Figure 6.18
Example with Multiple Optimal Solutions
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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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
Figure 6.19
Graph of an Infeasible Problem
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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An Unbounded Problem
Value of objective function
increases indefinitely:
Maximize Z = 4x1 + 2x2
subject to: x1  4
x2  2
x1, x2  0
Figure 6.20
Graph of an Unbounded Problem
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Characteristics of Linear Programming Problems
A linear programming problem requires a decision—a
choice amongst alternative courses of action.
The decision is represented in the model by decision
variables, of which any mix represents a choice.
The problem encompasses a goal, expressed as an
objective function, that the decision maker wants to
achieve (optimize = minimize or maximize, as appropriate).
Constraints exist that limit the extent of achievement of the
objective.
The objective and constraints must be definable by linear
mathematical functional relationships.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Properties 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).
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Problem Statement
Example Problem No. 1 (1 of 4)
Hot dog mixture in 1000-pound batches.
Two ingredients, chicken ($3/lb) and beef ($5/lb).
Recipe requirements:
at least 500 pounds of chicken
at least 200 pounds of beef
Ratio of chicken to beef must be at least 2 to 1.
Determine optimal mixture of ingredients that will minimize
costs.
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Solution
Example Problem No. 1 (2 of 4)
Step 1:
Identify decision variables.
x1 = lb of chicken
x2 = lb of beef
Step 2:
Formulate the objective function.
Minimize Z = $3x1 + $5x2
where Z = cost per 1,000-lb batch
$3x1 = cost of chicken
$5x2 = cost of beef
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Solution
Example Problem No. 1 (3 of 4)
Step 3:
Establish Model Constraints
x1 + x2 = 1,000 lb
x1  500 lb of chicken
x2  200 lb of beef
x1/x2  2/1, or x1  2 x2, or x1 - 2x2  0
x1, x2  0
The Model: Minimize Z = $3x1 + 5x2
subject to: x1 + x2 = 1,000 lb
x1  500
x2  200
x1 - 2x2  0
x1,x2  0
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Solution
Example Problem No. 1 (3 of 4)
Step 4:
Determine the corners:
x2  200 lb of beef
A
x1 + x2 = 1,000 lb
B
x1 - 2x2  0
A: x2=200 & x1+x2=1,000
 (x1,x2)=(800,200)
 Z(A) = $3x1+$5x2=$3,400
B: x1=2x2 & x1+x2=1,000
 (x1,x2)=(667,333)
 Z(B) = $3x1+$5x2=$3,667
B
.
.A
B is the optimal choice!
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Example Problem No. 2 (1 of 3)
Solve the following model
graphically:
Maximize Z = 4x1 + 5x2
subject to:
x1 + 2x2  10
6x1 + 6x2  36
x1  4
x1, x2  0
Step 1: Plot the constraints
as equations
Figure 6.21
Constraint Equations
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Example Problem No. 2 (2 of 3)
Maximize Z = 4x1 + 5x2
subject to:
x1 + 2x2  10
6x1 + 6x2  36
x1  4
x1, x2  0
Step 2: Determine the
feasible solution space
Figure 6.22
Feasible Solution Space and Extreme Points
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Example Problem No. 2 (3 of 3)
Maximize Z = 4x1 + 5x2
subject to:
x1 + 2x2  10
6x1 + 6x2  36
x1  4
x1, x2  0
Step 3: Determine the corners;
Step 4: Evaluate the objective
function and identify the
optimal solution.
Figure 6.22
Optimal Solution Point
Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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Chapter 6 - Linear Programming: Model Formulation and Graphical Solution
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