Transcript Managerial Decision Modeling with Spreadsheets

Linear Programming Solution
Techniques:
Graphical and Computer Methods
Learning Objectives
 Understand basic assumptions and properties of
linear programming (LP).
 Use graphical solution procedures for LP problems
with only two variables to understand how LP
problems are solved.
 Understand special situations such as redundancy,
infeasibility, unboundedness, and alternate optimal
solutions in LP problems.
 Understand how to set up LP problems on a
spreadsheet and solve them using Excel’s solver.
Introduction
 Management decisions in many organizations involve
trying to make most effective use of resources
(Machinery, labor, money, time, warehouse space, and
raw materials) in order to:
 Produce products - such as computers, automobiles, or
clothing or
 Provide services - such as package delivery, health services,
or investment decisions.
 To solve problems of resource allocation one may use
mathematical programming.
Linear Programming
Linear programming (LP) is the most common type of
mathematical programming.
LP seeks to maximize or minimize a linear objective
function subject to a set of linear constraints
LP assumes all relevant input data and parameters are
known with certainty (deterministic models).
Computers play an important role in the solution of LP
problems
LP 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.
Development of a LP Model

LP applied extensively to problems areas -
 medical, transportation, operations,
 financial, marketing, accounting,
 human resources, and agriculture.

Development of all LP models can be examined in three step
process:
(1) identification of the problem as solvable by LP
(2) formulation of the mathematical model.
(3) solution.
(4) interpretation.
Three Steps of Developing a LP Problem
Formulation
– Process of translating problem scenario into simple LP
model framework with set of mathematical relationships.
Solution
– Mathematical relationships resulting from formulation
process are solved to identify optimal solution.
Interpretation and What-if Analysis
– Problem solver or analyst works with manager to
• interpret results and implications of problem solution.
• investigate changes in input parameters and model
variables and impact on problem solution results.
Linear Equations and Inequalities
• This is a linear equation:
2A + 5B = 10
• This equation is not linear:
2A2 + 5B3 + 3AB = 10
• LP uses, in many cases, inequalities like:
A+BC
or A + B  C
Basic Assumptions of a LP Model
1.
Conditions of certainty exist.
2.
Proportionality in objective function and constraints
(1 unit – 3 hours, 3 units- 9 hours).
3.
Additivity (total of all activities equals sum of
individual activities).
4.
Divisibility assumption that solutions need not
necessarily be in whole numbers (integers);
ie.decision variables can take on any fractional value.
Formulating a LP Problem
• A common LP application is product mix problem.
– Two or more products are usually produced using limited
resources - such as personnel, machines, raw materials, and
so on.
• Profit firm seeks to maximize is based on profit
contribution per unit of each product.
• Firm would like to determine – How many units of each product it should produce
– Maximize overall profit given its limited resources.
Maximization Model Examples:
 Beaver Creek Example
 Flair Furniture Example
 Galaxy Industries Example
Problem Definition
Beaver Creek Maximization Problem (1 of 18)
• 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
Problem Definition
Beaver Creek Example (2 of 18)
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
Problem Definition
Beaver Creek Example (3 of 18)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to:
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Feasible Solutions
Beaver Creek Example (4 of 18)
• 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
Infeasible Solutions
Beaver Creek Example (5 of 18)
• 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
The set of all points that satisfy all the
constraints of the model is called
a
FEASIBLE REGION
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.
Primary advantage of two-variable LP models (such as
Beaver Creek problem) is their solution can be
graphically illustrated using two-dimensional graph.
Allows one to provide an intuitive explanation of how
more complex solution procedures work for larger LP
models.
Graphical Representation of LP Models
60
50
40
30
Coordinates
for graphical
analysis
20
10
0
10
20
30
40
50
60
Graphical Representation of Constraints
Coordinate Axes - Beaver Creek Example (6 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Coordinates for Graphical Analysis
Graphical Representation of Constraints
Beaver Creek Example- Labor Constraint (7 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Graph of Labor Constraint
Graphical Representation of Constraints
Beaver Creek Example-Labor Constraint Area
(8 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Labor Constraint Area
Graphical Representation of Constraints
Beaver Creek Example-Clay Constraint Area
(9 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Clay Constraint Area
Graphical Representation of Constraints
Beaver Creek Example-Both Constraints (10 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Graph of Both Model Constraints
Feasible Solution Area
Beaver Creek Example (11 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Feasible Solution Area
Graphical Solution:
Isoprofit Line Solution Method
 Optimal solution is the point in feasible region that produces
highest profit
 There are many possible solution points in region.
 How do we go about selecting the best one, one yielding
highest profit?
 Let objective function (that is, $$40x1 + $50x2) guide one
towards optimal point in feasible region.
 Plot line representing objective function on graph as a straight
line.
Graphical Solution (Isoprofit Line Method)
Beaver Creek Example (12 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Set Objective Function =
800
Objective Function Line for Z = $800
Graphical Solution (Isoprofit Line Method)
Alternative Objective Function Solution Lines
Beaver Creek Example (13 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Alternative Objective Function Lines
Graphical Solution (Isoprofit Line Method)
Optimal Solution
Beaver Creek Example (14 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Identification of Optimal Solution
Graphical Solution (Isoprofit Line Method)
Optimal Solution Coordinates
Beaver Creek Example (15 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Optimal Solution Coordinates
Corner Point Property
It is a very important property of Linear
Programming problems:
This property states optimal solution to LP problem will
always occur at a corner point.
Graphical Solution (Corner Point Solution
Method)
Beaver Creek Example (16 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Solution at All Corner Points
Optimal Solution for a New Objective Function
Beaver Creek Example (17 of 18)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Optimal Solution with Z = 70x1 + 20x2
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.
Standard Form of Linear Programming Model
Beaver Creek Example (18 of 18)
Max Z = 40x1 + 50x2 + s1 +s2
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
Solution Points A, B, and C with Slack
Problem Definition
Flair Furniture Maximization Example (1 of 19)
Company Data and Constraints  Flair Furniture Company produces tables and chairs.
 Each table requires: 4 hours of carpentry and 2 hours of painting.
 Each chair requires: 3 hours of carpentry and 1 hour of painting.
 Available production capacity: 240 hours of carpentry time and 100 hours
of painting time.
 Due to existing inventory of chairs, Flair is to make no more than 60 new
chairs.
 Each table sold results in $7 profit, while each chair produced yields $5
profit.
Flair Furniture’s problem:
 Determine best possible combination of tables and chairs to manufacture
in order to attain maximum profit.
Decision Variables
Flair Furniture Example (2 of 19)
Problem facing Flair is to determine how many
chairs and tables to produce to yield maximum
profit?
In Flair Furniture problem, there are two unknown
entities:
 T- number of tables to be produced.
 C- number of chairs to be produced.
Objective Function
Flair Furniture Example (3 of 19)
 Objective function states the goal of problem.
 What major objective is to be solved?
 Maximize profit!
 An LP model must have a single objective function.
In Flair’s problem, total profit may be expressed as:
Using decision variables T and C Maximize
$7 T + $5 C
($7 profit per table) x (number of tables produced) +
($5 profit per chair) x (number of chairs produced)
Constraints
Flair Furniture Example (4 of 19)
Denote conditions that prevent one from selecting
any specific subjective value for decision variables.
In Flair Furniture’s problem, there are three
restrictions on solution.
Restrictions 1 and 2 have to do with available
carpentry and painting times, respectively.
Restriction 3 is concerned with upper limit on the
number of chairs.
Constraints
Flair Furniture Example (5 of 19)
 There are 240 carpentry hours available.
 4T + 3C < 240
 There are 100 painting hours available.
 2T + 1C  100
 The marketing specified chairs limit constraint.
 C  60
 The non-negativity constraints.

T  0 (number of tables produced is  0)

C  0 (number of chairs produced is  0)
Building the Complete Mathematical Model
Flair Furniture Example (6 of 19)
Maximize profit = $7T + $5C
(objective function)
Subject to constraints -
4T + 3C  240
(carpentry constraint)
2T + 1C  100
(painting constraint)
C  60
(chairs limit constraint)
T  0
(non-negativity constraint on tables)
C  0 (non-negativity constraint on chairs)
Converting inequalities into equalities by using slack
Flair Furniture Example (7 of 19)
Maximize profit = $7T + $5C + 0s1 + 0s2 + 0s3
Subject to constraints 4T + 3C + s1 = 240
(carpentry constraint)
2T + 1C + s2 = 100
(painting constraint)
C + s3 = 60
T  0
(chairs limit constraint)
(non-negativity constraint on tables)
C  0 (non-negativity constraint on chairs)
s1 s2 s3  0 (non-negativity constraints on slacks)
Graphical Representation of Constraints
Flair Furniture Example (8 of 19)
Carpentry time constraint
4T + 3C  240
Graphical Representation of Constraints
Flair Furniture Example (9 of 19)
Carpentry Time Constraint (feasible area)
Any point below
line satisfies
constraint.
Graphical Representation of Constraints
Flair Furniture Example (10 of 19)
Painting Time Constraint and the Feasible Area
2T + 1C  100
Any point on line
satisfies equation:
2T + 1C = 100
(30,40) yields 100.
Any point below line
satisfies constraint.
Graphical Representation of Constraints
Flair Furniture Example (11 of 19)
Chair Limit Constraint and Feasible Solution Area
Feasible solution
area is contained
by three limiting
lines
Isoprofit Line Solution Method
Flair Furniture Example (12 of 19)
 Let objective function (that is, $7T + $5C) guide
one toward an optimal point in feasible region.
 Plot line representing objective function on graph.
 One does not know what $7T + $5C equals at an
optimal solution.
 Without knowing this value, how does one plot
relationship?
Isoprofit Line Solution Method
Flair Furniture Example (13 of 19)
 Write objective function: $7 T + $5 C = Z
 Select any arbitrary value for Z.
 For example, one may choose a profit ( Z ) of $210.

Z is written as: $7 T + $5 C = $210.
 To plot this profit line:
Set T = 0 and solve objective function for C.
 Let T = 0, then $7(0) + $5C = $210, or C = 42.

Set C = 0 and solve objective function for T.
 Let C = 0, then $7T + $5(0) = $210, or T = 30.
Isoprofit Line Solution Method
Flair Furniture Example (14 of 19)
One can check
for higher values
of Z to find an
optimal solution.
210 is not
highest possible
value.
Isoprofit Line Solution Method
Flair Furniture Example (15 of 19)
Isoprofit lines ($210,
$280, $350) are all
parallel.
Isoprofit Line Solution Method
Optimal Solution
Flair Furniture Example (16 of 19)
Optimal Solution:
Corner Point 4: T=30 (tables) and C=40 (chairs) with $410 profit
Isoprofit Line Solution Method
Optimal Solution
Flair Furniture Example (17 of 19)
 Optimal solution occurs at maximum point in the feasible
region.
 Occurs at intersection of carpentry and painting constraints:
- Carpentry constraint equation: 4T + 3C = 240
- Painting constraint equation: 2T + 1C = 100
If one solves these two equations with two unknowns for T
and C (for Corner Point 4), Optimal Solution is found:
T=30 (tables) and C=40 (chairs) with $410 profit.
Corner Point Solution Method
Flair Furniture Example (18 of 19)
From the figure one knows
feasible region for Flair’s
problem has five corner
points, namely, 1, 2, 3, 4,
and 5, respectively.
To find point yielding
maximum profit, one finds
coordinates of each corner
point and computes profit
level at each point.
Corner Point Solution Method
Flair Furniture Example (19 of 19)
• Point 1 (T = 0, C = 0)
profit = $7(0) + $5(0) = $0
• Point 2 (T = 0, C = 60)
profit = $7(0) + $5(60) = $300
• Point 3 (T = 15, C = 60)
profit = $7(15) + $5(60) = $405
• Point 4 (T = 30, C = 40)
profit = $7(30) + $5(40) = $410
• Point 5 (T = 50, C = 0)
profit = $7(50) + $5(0) = $350 .
Problem Definition
The Galaxy Industries Example (1 of 9)
• Galaxy manufactures two toy models:
– Space Ray.
– Zapper.
• Resources are limited to
– 1200 pounds of special plastic.
– 40 hours of production time per week.
Problem Definition
The Galaxy Industries Example (2 of 9)
Marketing requirement
– Total production cannot exceed 800 dozens.
– Number of dozens of Space Rays cannot exceed
number of dozens of Zappers by more than 450.
• Technological input
– Space Rays requires 2 pounds of plastic and
3 minutes of labor per dozen.
– Zappers requires 1 pound of plastic and
4 minutes of labor per dozen.
Problem Definition
The Galaxy Industries Example (3 of 9)
Current production plan calls for:
– Producing as much as possible of the more
profitable product, Space Ray ($8 profit per dozen).
– Use resources left over to produce Zappers ($5
profit
per dozen).
• The current production plan consists of:
Space Rays = 550 dozens
Zapper
= 100 dozens
Profit
= 4900 dollars per week
Management is seeking a production schedule
that will increase the company’s profit.
Decision Variables
Galaxy Industries Example (4 of 9)
• Decision variables:
– X1 = Production level of Space Rays (in dozens per
week).
– X2 = Production level of Zappers (in dozens per week).
• Objective Function:
- Weekly profit, to be maximized
Building the Complete Mathematical Model
Galaxy Industries Example (5 of 9)
Max 8X1 + 5X2
(Weekly profit)
subject to
2X1 + 1X2 < = 1200 (Plastic)
3X1 + 4X2 < = 2400 (Production Time)
X1 + X2 < = 800
(Total production)
X1 - X2 < = 450
(Mix)
Xj> = 0, j = 1,2
(Nonnegativity)
Graphical Representation of Constraints
Galaxy Industries
Example (6 of 9)
X2
1200
The plastic constraint:
The
Plastic constraint
2X1+X2<=1200
Total production constraint:
X1+X2<=800
Infeasible
600
Production
Feasible
Time
3X1+4X2<=2400
Production mix
constraint:
X1-X2<=450
Boundary points.
600
800
Interior points.
• There are three types of feasible points
Extreme points.
X1
Solving Graphically for an
Optimal Solution
Isoprofit Line Solution Method
Galaxy Industries Example (7 of 9)
We now demonstrate the search for an optimal solution
Start atX2
some arbitrary profit, say profit = $2,000...
1200
Then increase the profit, if possible...
...and continue until it becomes infeasible
800
Profit
4,
Profit
= $=$5040
2,
3,
000
600
X1
400
600
800
Isoprofit Line Solution Method
Galaxy Industries Example (8 of 9)
1200
X2
Let’s take a closer look at
the optimal point
Infeasible
800
600
Feasible
Feasible
region
region
X1
400
600
800
Optimal solution
Galaxy Industries Example (9 of 9)
Space Rays = 480 dozens
Zappers
= 240 dozens
Profit
= $5040
– This solution utilizes all the plastic and all the
production hours.
– Total production is only 720 (not 800).
– Space Rays production exceeds Zapper by only 240
dozens (not 450).
Minimization Model Examples
 Fertilizer Mix Problem
 Holiday Meal Turkey Ranch
Example
 Navy Sea Rations Example
A Minimization LP Problem
Many LP problems involve minimizing objective such as cost instead
of maximizing profit function.
Examples:
– Restaurant may wish to develop work schedule to meet staffing
needs while minimizing total number of employees.
– Manufacturer may seek to distribute its products from several
factories to its many regional warehouses in such a way as to
minimize total shipping costs.
– Hospital may want to provide its patients with a daily meal plan
that meets certain nutritional standards while minimizing food
purchase costs.
Problem Definition
Fertilizer Mix 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 ?
C h e m ica l C o n trib u tio n
N itro g e n
(lb /b a g )
P h o sp h a te
(lb /b a g )
S u p e r-g ro
2
4
C ro p -q u ick
4
3
B ra n d
Problem Definition
Fertilizer Mix Example (2 of 7)
Decision Variables:
x1 = bags of Super-Gro
x2 = bags of Crop-Quick
The Objective Function:
Minimize Z = $6x1 + 3x2
Model Constraints:
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
x1, x2  0 (non-negativity constraint)
Graphical Representation of Constraints
Fertilizer Mix Example (3 of 7)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
Graph of Both Model Constraints
Feasible Solution Area
Fertilizer Mix Example (4 of 7)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
Feasible Solution Area
Optimal Solution Point
Fertilizer Mix Example (5 of 7)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
Optimum Solution Point
Surplus Variables
Fertilizer Mix 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)
Graphical Solution
Fertilizer Mix Example (7 of 7)
Minimize Z = $6x1 + $3x2 + 0s1 + 0s2
subject to:
2x1 + 4x2 – s1 = 16
4x2 + 3x2 – s2 = 24
x1, x2, s1, s2  0
Graph of Fertilizer Example
Problem Definition
Holiday Meal Chicken Ranch Example (1 of 10)
 Buy two brands of feed for good, low-cost diet for
turkeys.
 Each feed may contain three nutritional ingredients
(protein, vitamin, and iron).
 One pound of Brand A contains:
 5 units of protein,
4 units of vitamin, and
0.5 units of iron.
 One pound of Brand B contains:
10 units of protein,
3 units of vitamins, and
0 units of iron.
Problem Definition
Holiday Meal Chicken Ranch Example (2 of 10)
Brand A feed costs ranch $0.02 per pound, while Brand
B feed costs $0.03 per pound.
• Ranch owner would like lowest-cost diet that meets
minimum monthly intake requirements for each
nutritional ingredient.
Problem Definition
Holiday Meal Chicken Ranch Example (3 of 10)
Building the Complete Mathematical Model
Holiday Meal Chicken Ranch Problem (4 of 10)
Minimize cost (in cents)
Subject to:
5A + 10B  90
4A + 3B  48
½A
 1½
A  0, B  0
= 2A + 3B
(protein constraint)
(vitamin constraint)
(iron constraint)
(nonnegativity constraint)
Where:
A denotes number of pounds of Brand A feed, and B
denote number of pounds of Brand B feed.
Building the Standard LP Model
Holiday Meal Chicken Ranch Example (5 of 10)
Minimize cost (in cents)=2A+3B+0s1+0s2+0s3
subject to constraints
5A + 10B - s1 = 90
(protein constraint)
4A + 3B - s2 = 48
(vitamin constraint)
½A - s3 = 1½
(iron constraint)
A, B, s1,s2 s3  0
(nonnegativity)
Graphical Representation of Constraints
Holiday Meal Chicken Ranch Problem (6 of 10)
Drawing Constraints:
½A  1½
4A + 3B  48
5A + 10B  90
Nonnegativity Constraint
A  0, B  0
Graphical Solution Method:Isocost Line Method
Holiday Meal Chicken Ranch Example (7 of 10)
One can start by
drawing a 54-cent
cost line
2A + 3B. = 54
Isocost Line Method
Holiday Meal Chicken Ranch Example (8 of 10)
•
•
Isocost line is moved parallel to 54-cent solution line toward lower left
origin.
Last point to touch isocost line while still in contact with feasible
region is corner point 2.
Isocost Line Method
Holiday Meal Chicken Ranch Example (9 of 10)
• Solving for corner point 2 with two equations with values 8.4 for A and
4.8 for B, minimum optimal cost solution is:
2A + 3B = (2)(8.4) + (3)(4.8) = 31.2
Corner Point Solution Method
Holiday Meal Chicken Ranch Example (10 of 10)
•
Point 1 - coordinates (A = 3, B = 12)
– cost of 2(3) + 3(12) = 42 cents.
•
Point 2 - coordinates (A = 8.4, b = 4.8)
– cost of 2(8.4) + 3(4.8) = 31.2 cents
•
Point 3 - coordinates (A = 18, B = 0)
– cost of (2)(18) + (3)(0) = 36 cents.
• Optimal minimal cost solution:
Corner Point 2, cost = 31.2 cents
Problem Definition
Navy Sea Rations Example (1 of 4)
• A cost minimization diet problem
– Mix two sea ration products: Texfoods, Calration.
– Minimize the total cost of the mix.
– Meet the minimum requirements of
Vitamin A, Vitamin D, and Iron.
Complete Model
Navy Sea Rations Example (2 of 4)
• Decision variables
– X1 (X2) --
The number of portions of Texfoods
(Calration) product used in a
serving.
• The Model
Minimize 0.60X1 + 0.50X2
Subject to
20X1 + 50X2
25X1 + 25X2
50X1 + 10X2
X1, X2
≥ 100 Vitamin A
≥ 100 Vitamin D
≥ 100 Iron
≥0
Graphical Solution
Navy Sea Rations Example (3 of 4)
5
4
The Iron constraint
Feasible Region
Vitamin “D” constraint
2
Vitamin “A” constraint
2
4
5
Summary of the optimal solution
Navy Sea Rations Example (4 of 4)
–
–
–
–
Texfood product = 1.5 portions
Calration product = 2.5 portions
Cost =$ 2.15 per serving.
The minimum requirement for Vitamin D and iron
are met with no surplus.
– The mixture provides 155% of the requirement for
Vitamine A.
Summary of Graphical Solution Methods (1 of 3)
1. Plot the model constraints accepting them as
equalities,
2. Considering the inequalities of the constraints
identify the feasible solution region, that is, area
that satisfies all constraints simultaneously.
3. Select one of two following graphical solution
techniques and proceed to solve problem.
- Isoprofit or Isocost Method.
- Corner Point Method
Summary of Graphical Solution Methods (2 of 3)
Corner Point Method
 Determine coordinates of each of corner points of
the feasible region by solving simultaneous
equations at each point.
 Compute profit or cost at each point by substituting
the values of coordinates into the objective function
and solving for results.
 Identify the optimal solution as a corner point with
highest profit (maximization problem), or lowest cost
(minimization).
Summary of Graphical Solution Methods(3 of 3)
Isoprofit or Isocost Method
 Select an arbitrary value for profit or cost, and plot an
isoprofit / isocost line to reveal its slope.
 Maintain same slope and move the line up or down until it
touches the feasible region at one point. While moving
the line up or down consider whether the problem is a
maximization or a minimization problem
 Identify optimal solution as coordinates of point touched
by highest possible isoprofit line or lowest possible
isocost line (by solving the simultaneous equations)
 Read optimal coordinates and compute optimal profit or
cost.
Special Situations in Solving LP Problems
(Irregular Types of LP Problems)
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
Redundancy: A redundant constraint is constraint that
does not affect feasible region in any way.
Maximize Profit
= 2X + 3Y
subject to:
X + Y  20
2X + Y  30
X  25
X, Y  0
Infeasibility: A condition that arises when an LP
problem has no solution that satisfies all of its
constraints.
X + 2Y  6
2X + Y  8
X  7
Unboundedness: Sometimes an LP model will
not have a finite solution
Maximize profit
= $3X + $5Y
subject to:
X  5
Y  10
X + 2Y  10
X, Y  0
Multiple Optimal Solutions
• An LP problem may have more than one optimal
solution.
– Graphically, when the isoprofit (or isocost) line
runs parallel to a constraint in problem which
lies in direction in which isoprofit (or isocost)
line is located.
– In other words, when they have same slope.
Example: Multiple Optimal Solutions
Maximize profit =
$3x + $2y
Subject to:
6X + 4Y  24
X  3
X, Y  0
Example: Multiple Optimal Solutions
 At profit level of $12, isoprofit line will rest directly on top of first
constraint line.
 This means that any point along line between corner points 1 and 2
provides an optimal X and Y combination.