Transcript Linear Programming Model

Department of Business Administration
FALL 2007-08
Management Science
by
Asst. Prof. Sami Fethi
© 2007 Pearson Education
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Outline: What You Will Learn . . .

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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Linear Programming: An Overview

Objectives of business decisions frequently involve
maximizing profit or minimizing costs.

Linear programming is an analytical 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.

Implementation
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Model Components

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 - this function is to be maximized or minimized.

Constraints – requirements or 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 constraints.
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Summary of Model Formulation Steps
Step 1 : Clearly define the decision variables
Step 2 : Construct the objective function
Step 3 : Formulate the constraints
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation
A Maximization Example (1 of 3)

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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation
A Maximization 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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation
A Maximization Example (3 of 3)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to:
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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) = 70 < 120 pounds, within constraint
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Graphical Solution of LP 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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CH 2:Linear Programming: Model Formulation and Graphical Solution
Coordinate Axes
Graphical Solution of Maximization Model (1 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure 2.2 Coordinates for Graphical Analysis
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
Labor const. 1(0) + 2x2 =40
x2 = 20
x1= 40
Figure 2.3 Graph of Labor Constraint
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Labor Constraint Area
Graphical Solution of Maximization Model (3 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Check Points:
1(10) + 2(10)  40
30  40 hr.
1(40) + 2(30)  40 hr.
100 not  40 hr.
Figure 2.4 Labor Constraint Area
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
Clay const. :
4(0) + 3x2 =120
x2 = 40
x1= 30
Figure 2.5 Clay Constraint Area
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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 2.6 Graph of Both Model Constraints
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
e.g. Point R ( not S an T) satisfies both
constraints, thus it is a feasible solution
Point. Point S satisfies the clay const.,
But exceeds the labor const., this is
İnfeasible. Point T satisfies neither
constraint so, it is also infeasible.
Figure 2.7 Feasible Solution Area
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Objective Function 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
•Say Z = $800, Objective Function is
$800 = $40x1 + $50x2
x1=20 and x2=16
•If you consider profits $ 1200 and $ 1600,
they may still provide a feasible solution.
•The former is outside the feasible solution
area but part of the line remains within the
feasible area. The latter is not possible
Figure 2.8 Objection Function Line for Z = $800
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
• 800 = $40x1 + $50x2
x1=20, x2= 16
• 1200 = $40x1 + $50x2
x1=30, x2= 24
•1600= $40x1 + $50x2
x1=40, x2= 32
Figure 2.9 Alternative Objective Function Lines
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
•The bottom line is that profit
increases as the objective line
moves away from the origion and
still touching a point in the feasible
solution area (e.g., Point B).
Figure 2.10 Identification of Optimal Solution
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Optimal Solution Coordinates CH 2:Linear Programming: Model Formulation and Graphical Solution
Graphical Solution of Maximization Model (10 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
To find out Optimal solution
coordinates, we use the both
constraints.
x1 = 40 -2x2
4(40 -2x2) + 3x2 =120
160-8x2 + 3x2 =120
-5x2 = -40
X2 = 8
x1 = 24
Z = $40 (24) + $50 (8)
Z = $ 1360 max daily profit possible
Figure 2.11 Optimal Solution Coordinates
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Extreme (Corner) Point Solutions
Graphical Solution of Maximization Model (11 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
 Extreme points: are corner points on the
Boundary of the feasible solution area.
To determine the variable solution values,
Constraint equations are solved at the
Optimal extreme point.
The optimal solution point is the last point
Objective function touches as it leaves
the feasible solution area.
Figure 2.12 Solutions at All Corner Points
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Optimal Solution for New Objective Function
Graphical Solution of Maximization Model (12 of 12)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
If the model constraints for labor
or clay change, obviously location
of the objective function is different
from the previous one.
The reason for this change is that
the new profit coefficients give the
linear objective function a new
slope.
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Figure 2.13 Optimal Solution with Z = 70x1 + 20x2 24
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Optimal Solution for New Objective Function
Slope and General Equation form
 The Slope can be determined by transforming the objective function into the
general equation for a straight line as such
y=a+bx
In this case:
Z = 70x1 + 20x2
20x2 =Z -70x1
X2 = (Z/20) –(7/2)x1
 This transformation identifies the slope of the new objective function as – 7/2
(downward) compared to the origional (-4/5).
This shows us that optimal solution will be at point C.
X2 = 0
x1= 30
Z = $70 (30) + $20 (8)
Z = $ 2100
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Slack Variables


Standard form requires that all constraints be in the form of
equations (equalities).
A slack variable is added to a  constraint (weak
inequality) to convert it to an equation (=).

A slack variable typically represents an unused resource.

A slack variable contributes nothing to the objective function
value.
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Linear Programming Model: Standard Form
Max Z = 40x1 + 50x2 + s1 + s2
subject to:1x1 + 2x2 + s1 = 40
4x2 + 3x2 + s2 = 120
x1, x2, s1, s2  0
Where:
x1 = number of bowls
x2 = number of mugs
s1, s2 are slack variables
Figure 2.14 Solution Points A, B, and C with Slack
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Linear Programming Model: Standard Form
Max Z = 40x1 + 50x2 + s1 + s2
subject to:1x1 + 2x2 + s1 = 40
4x2 + 3x2 + s2 = 120
x1, x2, s1, s2  0
Now, let us consider hypotheyical
solution of x1= 5 and x2 = 10 for the
system eq. Above:
1x1 + 2x2 + s1 = 40
1(5) + 2(10) + s1 = 40
s1 = 15 hrs. of labor
25 hrs. are not used
The amount of unused labor
4x1 + 3x2 + s2 = 120
4(5) + 3(10)+ s2 = 120
s2 = 70 lb. of clay
50 lb are not used
The amount of unused clay
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Linear Programming Model: Standard Form
Max Z = 40x1 + 50x2 + s1 + s2
subject to:1x1 + 2x2 + s1 = 40
4x1 + 3x2 + s2 = 120
x1, x2, s1, s2  0
Now, let us consider hypotheyical
solution of x1= 0 and x2 = 0 for the
system eq. Above:
1x1 + 2x2 + s1 = 40
1(0) + 2(0) + s1 = 40
s1 = 40 hrs. of labor
The amount of unused labor
4x1 + 3x2 + s2 = 120
4(0) + 3(0)+ s2 = 120
s2 = 120 lb. of clay
The amount of unused clay
No production takes place at the
origion so all the resources are unused
and slack variables contribute nothing to
profit or to the objective function value
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation
A Minimization 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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation
A Minimization 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)
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CH 2:Linear Programming: Model Formulation and Graphical Solution
LP Model Formulation and Constraint Graph
A Minimization Example (3 of 7)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
 for 1st eq., when x1=0
2(0) + 4x2  16 x2=4
When x2=0
2 (x1) + 4(0)  16 x1=8
for 2nd eq., when x1=0
4(0) + 3x2  24 x2=8
When x2=0
4(x1) + 3(0)  16 x1=6
Figure 2.16 Graph of Both Model Constraints
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Feasible Solution Area
A Minimization Example (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
The optimal solution of a
minimization problem is at the
extreme point closest to the
origin, in other words, zero being
the lowest cost possible
Figure 2.17 Feasible Solution Area
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Optimal Solution Point
A Minimization Example (5 of 7)
Minimize
subject to:
Z = $6x1 + $3x2
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
As the objective function edges
toward the origin, the last point
İt touches in the feasible solution
Area is A.
At point A
As x1 = 0 and x2=8
Z = $6(0) + $3(8)
Z = $ 24
Figure 2.18 Optimum Solution Point
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Surplus Variables
A Minimization Example (6 of 7)

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.

Subtracting slack variables in the farmer problem
constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Graphical Solutions
A Minimization 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
Now, let us consider hypotheyical solution of
x1= 0 and x2 = 10 for the system eq. Above:
2x1 + 4x2 – s1 = 16
2(0) + 4(10) – s1 = 16
s1 = 24 lb. of nitrogen
The extra amount of nitrogen
4x1 + 3x2 – s2 = 24
4(0) + 3(10) – s2 = 24
s2 = 6lb. of phosphate
The extra amount of phosphate
Surplus variable contribute nothing to the
overall cost of model or will not affect the
farmer’s cost.
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Figure 2.19 Graph of Fertilizer Example
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Multiple Optimal Solutions
Beaver Creek Pottery Example
Objective function is parallel to a
constraint line.
Maximize Z=$40x1 + 50x2 change to
Maximize Z=$40x1 + 30x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Where:
x1 = number of bowls
x2 = number of mugs
Now,Obj. Fuct is parallel to second
const.
Both line have the same slope 0f – 4/3
so it touches the whole line segment BC.
This means that every point along this
line is optimal or have same profit. The
endpoints of this line segment B and C
are referred to as the alternate optimal
solution
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Figure 2.20 Example with Multiple Optimal Solutions
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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
Point A satisfies only first const
whereas Point C satisfies only second
and third consts and point B satisfies
none of the
Consts.
The three consts do not overlap to
form a feasible solution area so there is
no solution to the problem.
Figure 2.21 Graph of an Infeasible Problem
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An Unbounded Problem CH 2:Linear Programming: Model Formulation and Graphical Solution
Value of increases indefinitely:
Maximize Z = 4x1 + 2x2
subject to: x1  4
x2  2
x1, x2  0
The objective function increases
without bound thus a solution is
Never reached.
Unlimited profits are not possible in
the real world.
This is like an unfeasible solution
typically reflects an error in definning
the problem or in formulating the
model.
Figure 2.22 Graph of an Unbounded Problem
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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.
The problem encompasses a goal, expressed as an
objective function, that the decision maker wants to
achieve.
Constraints exist that limit the extent of achievement of
the objective.
The objective and constraints must be definable by
linear mathematical functional relationships.



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CH 2:Linear Programming: Model Formulation and Graphical Solution
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).
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Problem Statement
Example Problem No. 1 (1 of 3)
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.
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Solution
Example Problem No. 1 (2 of 3)
Step 1:
Identify decision variables.
x1 = lb of chicken in mixture (1000 lb.)
x2 = lb of beef in mixture (1000 lb.)
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
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CH 2:Linear Programming: Model Formulation and Graphical Solution
Solution
Example Problem No. 1 (3 of 3)
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 - 2x2  0
x1, x2  0
The Model: Minimize Z = $3x1 + 5x2
subject to: x1 + x2 = 1,000 lb
x1  50
x2  200
x1 - 2x2  0
x1,x2  0
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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 2.23 Constraint Equations
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CH 2:Linear Programming: Model Formulation and Graphical Solution
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 2.24 Feasible Solution Space and Extreme Points
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© 2007/08, Sami Fethi, EMU, All Right Reserved.
CH 2:Linear Programming: Model Formulation and Graphical Solution
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 and 4: Determine
the solution points and
optimal solution
Figure 2.25 Optimal Solution Point
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© 2007/08, Sami Fethi, EMU, All Right Reserved.
CH 2:Linear Programming: Model Formulation and Graphical Solution
End of Chapter
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Operations Research
© 2007/08, Sami Fethi, EMU, All Right Reserved.