1. Linear Programming Solution Techniques: Graphical and Computer Methods

2. 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.

3. 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.

4. 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

5. 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.

6. 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.

7. 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.

8. 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

9. Basic Assumptions of a LP Model
Conditions of certainty exist.
Proportionality in objective function and constraints (1 unit – 3 hours, 3 units- 9 hours).
Additivity (total of all activities equals sum of individual activities).
Divisibility assumption that solutions need not necessarily be in whole numbers (integers); ie.decision variables can take on any fractional value.

10. 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.

11. Maximization Model Examples:
Beaver Creek Example
Flair Furniture Example
Galaxy Industries Example

12. Beaver Creek Maximization Problem (1 of 18)
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:

13. Beaver Creek Example (2 of 18)
Problem Definition
Beaver Creek Example (2 of 18)
Resource hrs of labor per day
Availability: 120 lbs of clay
Decision x1 = number of bowls to produce per day
Variables: x2 = number of mugs to produce per day
Objective Maximize Z = $40x1 + $50x2
Function: Where Z = profit per day
Resource x1 + 2x2  40 hours of labor
Constraints: 4x1 + 3x2  120 pounds of clay
Non-Negativity x1  0; x2  0
Constraints:

14. Beaver Creek Example (3 of 18)
Problem Definition
Beaver Creek Example (3 of 18)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0

15. 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

16. 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

17. The set of all points that satisfy all the constraints of the model is called
FEASIBLE REGION

18. 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.

19. 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.

20. Graphical Representation of LP Models10 20 30 40 50 60
Coordinates for graphical analysis

21. Coordinates for Graphical Analysis
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

22. Graph of Labor Constraint
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

23. 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

24. 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

25. Graph of Both Model Constraints
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

26. 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

27. 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.

28. Set Objective Function = 800
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

29. Alternative Objective Function Lines
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

30. Identification of Optimal Solution
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

31. Optimal Solution Coordinates
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

32. 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.

33. Solution at All Corner Points
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

34. Optimal Solution with Z = 70x1 + 20x2
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

35. 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.

36. Solution Points A, B, and C with Slack
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

37. 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.

38. 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.

39. 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)

40. 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.

41. 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)

42. Building the Complete Mathematical Model Flair Furniture Example (6 of 19)
Maximize profit = $7T + $5C (objective function)
Subject to constraints -
4T + 3C  (carpentry constraint)
2T + 1C  (painting constraint)
C  (chairs limit constraint)
T  (non-negativity constraint on tables)
C  (non-negativity constraint on chairs)

43. 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 = (painting constraint)
C + s3 = (chairs limit constraint)
T  (non-negativity constraint on tables)
C  (non-negativity constraint on chairs)
s1 s2 s3  0 (non-negativity constraints on slacks)

44. Carpentry time constraint
Graphical Representation of Constraints Flair Furniture Example (8 of 19)
Carpentry time constraint
4T + 3C  240

45. Carpentry Time Constraint (feasible area)
Graphical Representation of Constraints Flair Furniture Example (9 of 19)
Carpentry Time Constraint (feasible area)
Any point below line satisfies constraint.

46. 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.

47. 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

48. 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?

49. 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.

50. 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.

51. Isoprofit Line Solution Method Flair Furniture Example (15 of 19)
Isoprofit lines ($210, $280, $350) are all parallel.

52. 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

53. 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 = 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.

54. 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.

55. 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 .

56. 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.

57. 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.

58. 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

59. Management is seeking a production schedule that will increase the company’s profit.

60. Galaxy Industries Example (4 of 9)
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

61. Building the Complete Mathematical Model Galaxy Industries Example (5 of 9)
Max 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = (Plastic)
3X1 + 4X2 < = (Production Time)
X1 + X2 < = (Total production)
X X2 < = (Mix)
Xj> = 0, j = 1,2 (Nonnegativity)

62. Graphical Representation of Constraints
Galaxy Industries Example (6 of 9) X2
The plastic constraint:
2X1+X2<=1200
1200
The Plastic constraint
Total production constraint:
X1+X2<=800
Infeasible
600
Production mix constraint:
X1-X2<=450
Production Time
3X1+4X2<=2400
Feasible X1
Boundary points.
600
800
Interior points.
There are three types of feasible points
Extreme points.

63. Solving Graphically for an
Optimal Solution

64. Isoprofit Line Solution Method Galaxy Industries Example (7 of 9)
We now demonstrate the search for an optimal solution
Start at some arbitrary profit, say profit = $2,000... X2
1200
Then increase the profit, if possible...
...and continue until it becomes infeasible
Profit =$5040
Profit = $ 2, 3, 4,
800
Recall the feasible Region
600 X1
400
600
800

65. Isoprofit Line Solution Method Galaxy Industries Example (8 of 9)
1200 X2
Let’s take a closer look at
the optimal point
800
Infeasible
600
Feasible
region
Feasible
region X1
400
600
800

66. 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).

67. Minimization Model Examples
Fertilizer Mix Problem
Holiday Meal Turkey Ranch Example
Navy Sea Rations Example

68. 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.

69. Fertilizer Mix Example (1 of 7)
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 ?

70. Fertilizer Mix Example (2 of 7)
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)

71. Graph of Both Model Constraints
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

72. 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

73. Optimum Solution Point
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

74. Fertilizer Mix Example (6 of 7)
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)

75. Graph of Fertilizer Example
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

76. 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.

77. 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.

78. Problem Definition Holiday Meal Chicken Ranch Example (3 of 10)

79. Building the Complete Mathematical Model Holiday Meal Chicken Ranch Problem (4 of 10)
Minimize cost (in cents) = 2A + 3B
Subject to:
5A + 10B  90 (protein constraint)
4A + 3B  48 (vitamin constraint)
½A  1½ (iron constraint)
A  0, B  0 (nonnegativity constraint)
Where:
A denotes number of pounds of Brand A feed, and B denote number of pounds of Brand B feed.

80. 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 = (vitamin constraint)
½A - s3 = 1½ (iron constraint)
A, B, s1,s2 s3  0 (nonnegativity)

81. 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

82. One can start by drawing a 54-cent cost line 2A + 3B. = 54
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

83. 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.

84. Holiday Meal Chicken Ranch Example (9 of 10)
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

85. 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 = cents

86. 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.

87. Navy Sea Rations Example (2 of 4)
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 0.60X X2
Subject to
20X X2 ≥ Vitamin A
25X X2 ≥ Vitamin D
50X X2 ≥ Iron
X1, X ≥ 0

88. Navy Sea Rations Example (3 of 4)
Graphical Solution
Navy Sea Rations Example (3 of 4) 5
The Iron constraint
Feasible Region 4
Vitamin “D” constraint 2
Vitamin “A” constraint 2 4 5

89. 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.

90. Summary of Graphical Solution Methods (1 of 3)
Plot the model constraints accepting them as equalities,
Considering the inequalities of the constraints identify the feasible solution region, that is, area that satisfies all constraints simultaneously.
Select one of two following graphical solution techniques and proceed to solve problem.
Isoprofit or Isocost Method.
Corner Point Method

91. 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). 

92. 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.

93. Special Situations in Solving LP Problems
(Irregular Types of LP Problems)

94. 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

95. 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

96. 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

97. 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

98. 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.

99. Example: Multiple Optimal Solutions
Maximize profit =
$3x + $2y
Subject to:
6X + 4Y  24
X  3
X, Y  0

100. 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.