Linear Programming Solution Techniques: Graphical and Computer Methods
Published byModified over 4 years ago
Presentation on theme: "Linear Programming Solution Techniques: Graphical and Computer Methods"— Presentation transcript:
1 Linear Programming Solution Techniques: Graphical and Computer Methods
2 Learning ObjectivesUnderstand 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 IntroductionManagement 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 orProvide services - such as package delivery, health services, or investment decisions.To solve problems of resource allocation one may use mathematical programming.
4 Linear ProgrammingLinear 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 constraintsLP 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 minimizedConstraints - 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 FormulationProcess of translating problem scenario into simple LP model framework with set of mathematical relationships.SolutionMathematical relationships resulting from formulation process are solved to identify optimal solution.Interpretation and What-if AnalysisProblem solver or analyst works with manager tointerpret 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 = 10This equation is not linear: 2A2 + 5B3 + 3AB = 10LP 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 produceMaximize overall profit given its limited resources.
11 Maximization Model Examples: Beaver Creek ExampleFlair Furniture ExampleGalaxy Industries Example
12 Beaver Creek Maximization Problem (1 of 18) Problem DefinitionBeaver Creek Maximization Problem (1 of 18)Product mix problem - Beaver Creek Pottery CompanyHow 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 DefinitionBeaver Creek Example (2 of 18)Resource hrs of labor per dayAvailability: 120 lbs of clayDecision x1 = number of bowls to produce per dayVariables: x2 = number of mugs to produce per dayObjective Maximize Z = $40x1 + $50x2Function: Where Z = profit per dayResource x1 + 2x2 40 hours of laborConstraints: 4x1 + 3x2 120 pounds of clayNon-Negativity x1 0; x2 0Constraints:
14 Beaver Creek Example (3 of 18) Problem DefinitionBeaver Creek Example (3 of 18)Complete Linear Programming Model:Maximize Z = $40x1 + $50x2subject to: x1 + 2x2 404x1 + 3x2 120x1, x2 0
15 Feasible Solutions Beaver Creek Example (4 of 18) A feasible solution does not violate any of the constraints:Example x1 = 5 bowlsx2 = 10 mugsZ = $40x1 + $50x2 = $700Labor constraint check:1(5) + 2(10) = 25 < 40 hours, within constraintClay 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 bowlsx2 = 20 mugsZ = $1400Labor 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 Models 102030405060Coordinates for graphical analysis
21 Coordinates for Graphical Analysis Graphical Representation of ConstraintsCoordinate Axes - Beaver Creek Example (6 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Coordinates for Graphical Analysis
22 Graph of Labor Constraint Graphical Representation of ConstraintsBeaver Creek Example- Labor Constraint (7 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Graph of Labor Constraint
23 Graphical Representation of Constraints Beaver Creek Example-Labor Constraint Area(8 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Labor Constraint Area
24 Graphical Representation of Constraints Beaver Creek Example-Clay Constraint Area(9 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Clay Constraint Area
25 Graph of Both Model Constraints Graphical Representation of ConstraintsBeaver Creek Example-Both Constraints (10 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Graph of Both Model Constraints
26 Feasible Solution Area Beaver Creek Example (11 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Feasible Solution Area
27 Graphical Solution: Isoprofit Line Solution Method Optimal solution is the point in feasible region that produces highest profitThere 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 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Set Objective Function = 800Objective Function Line for Z = $800
29 Alternative Objective Function Lines Graphical Solution (Isoprofit Line Method)Alternative Objective Function Solution LinesBeaver Creek Example (13 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Alternative Objective Function Lines
30 Identification of Optimal Solution Graphical Solution (Isoprofit Line Method)Optimal SolutionBeaver Creek Example (14 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Identification of Optimal Solution
31 Optimal Solution Coordinates Graphical Solution (Isoprofit Line Method)Optimal Solution CoordinatesBeaver Creek Example (15 of 18)Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Optimal Solution Coordinates
32 Corner Point PropertyIt 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 + $50x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Solution at All Corner Points
34 Optimal Solution with Z = 70x1 + 20x2 Optimal Solution for a New Objective FunctionBeaver Creek Example (17 of 18)Maximize Z = $70x1 + $20x2subject to: 1x1 + 2x2 404x1 + 3x2 120x1, x2 0Optimal Solution with Z = 70x1 + 20x2
35 Slack VariablesStandard 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 ModelBeaver Creek Example (18 of 18)Max Z = 40x1 + 50x2 + s1 +s2subject to:1x1 + 2x2 + s1 = 404x1 + 3x2 + s2 = 120x1, x2, s1, s2 0Where:x1 = number of bowlsx2 = number of mugss1, s2 are slack variablesSolution 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 < 240There are 100 painting hours available. 2T + 1C 100The marketing specified chairs limit constraint.C 60The 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 + 0s3Subject 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 constraint4T + 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 Area2T + 1C 100Any 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 AreaFeasible 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 = ZSelect 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 = 100If one solves these two equations with two unknowns for Tand 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) = $0Point 2 (T = 0, C = 60)profit = $7(0) + $5(60) = $300Point 3 (T = 15, C = 60)profit = $7(15) + $5(60) = $405Point 4 (T = 30, C = 40)profit = $7(30) + $5(40) = $410Point 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 to1200 pounds of special plastic.40 hours of production time per week.
57 Problem Definition The Galaxy Industries Example (2 of 9) Marketing requirementTotal production cannot exceed 800 dozens.Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 450.Technological inputSpace Rays requires 2 pounds of plastic and3 minutes of labor per dozen.Zappers requires 1 pound of plastic and4 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 profitper dozen).The current production plan consists of:Space Rays = 550 dozensZapper = 100 dozensProfit = 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 VariablesGalaxy 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 to2X1 + 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)X2The plastic constraint:2X1+X2<=12001200The Plastic constraintTotal production constraint:X1+X2<=800Infeasible600Production mix constraint:X1-X2<=450Production Time3X1+4X2<=2400FeasibleX1Boundary points.600800Interior points.There are three types of feasible pointsExtreme points.
64 Isoprofit Line Solution Method Galaxy Industries Example (7 of 9) We now demonstrate the search for an optimal solutionStart at some arbitrary profit, say profit = $2,000...X21200Then increase the profit, if possible......and continue until it becomes infeasibleProfit =$5040Profit = $2,3,4,800Recall the feasible Region600X1400600800
65 Isoprofit Line Solution Method Galaxy Industries Example (8 of 9) 1200X2Let’s take a closer look atthe optimal point800Infeasible600FeasibleregionFeasibleregionX1400600800
66 Optimal solution Galaxy Industries Example (9 of 9) Space Rays = 480 dozensZappers = 240 dozensProfit = $5040This 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 ProblemHoliday Meal Turkey Ranch ExampleNavy 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 DefinitionFertilizer 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 DefinitionFertilizer Mix Example (2 of 7)Decision Variables: x1 = bags of Super-Grox2 = bags of Crop-QuickThe Objective Function:Minimize Z = $6x1 + 3x2Model 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 ConstraintsFertilizer Mix Example (3 of 7)Minimize Z = $6x1 + $3x2subject to: 2x1 + 4x2 164x2 + 3x2 24x1, x2 0Graph of Both Model Constraints
72 Feasible Solution Area Fertilizer Mix Example (4 of 7)Minimize Z = $6x1 + $3x2subject to: 2x1 + 4x2 164x2 + 3x2 24x1, x2 0Feasible Solution Area
73 Optimum Solution Point Optimal Solution PointFertilizer Mix Example (5 of 7)Minimize Z = $6x1 + $3x2subject to: 2x1 + 4x2 164x2 + 3x2 24x1, x2 0Optimum Solution Point
74 Fertilizer Mix Example (6 of 7) Surplus VariablesFertilizer 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 SolutionFertilizer Mix Example (7 of 7)Minimize Z = $6x1 + $3x2 + 0s1 + 0s2subject to: 2x1 + 4x2 – s1 = 164x2 + 3x2 – s2 = 24x1, x2, s1, s2 0Graph 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, and0.5 units of iron.One pound of Brand B contains:10 units of protein,3 units of vitamins, and0 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 + 3BSubject 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+0s3subject to constraints5A + 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 1½4A + 3B 485A + 10B 90Nonnegativity 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 MethodHoliday Meal Chicken Ranch Example (9 of 10)Solving for corner point 2 with two equations with values 8.4 for A and4.8 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 centsPoint 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 problemMix two sea ration products: Texfoods, Calration.Minimize the total cost of the mix.Meet the minimum requirements ofVitamin A, Vitamin D, and Iron.
87 Navy Sea Rations Example (2 of 4) Complete ModelNavy Sea Rations Example (2 of 4)Decision variablesX1 (X2) The number of portions of Texfoods (Calration) product used in a serving.The ModelMinimize 0.60X X2Subject to20X X2 ≥ Vitamin A25X X2 ≥ Vitamin D50X X2 ≥ IronX1, X ≥ 0
88 Navy Sea Rations Example (3 of 4) Graphical SolutionNavy Sea Rations Example (3 of 4)5The Iron constraintFeasible Region4Vitamin “D” constraint2Vitamin “A” constraint245
89 Summary of the optimal solution Navy Sea Rations Example (4 of 4) Texfood product = 1.5 portionsCalration product = 2.5 portionsCost =$ 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 MethodDetermine 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 MethodSelect 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 problemIdentify 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 solutionsInfeasible solutionsUnbounded solutions
95 Redundancy: A redundant constraint is constraint that does not affect feasible region in any way. Maximize Profit= 2X + 3Ysubject to:X + Y 202X + Y 30X 25X, Y 0
96 Infeasibility: A condition that arises when an LP problem has no solution that satisfies all of its constraints.X + 2Y 62X + Y 8X 7
97 Unboundedness: Sometimes an LP model will not have a finite solution Maximize profit= $3X + $5Ysubject to:X 5Y 10X + 2Y 10X, 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.
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.