Management decisions in many organizations involve trying ...aa2015_topic-05_linear... · Solution: Graphical and Computer ... Computers play an important rolein the solution of LP

Profit Maximization / Cost Minimization Solution: Graphical and Computer

Methods

1

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 or Provide services - such as package delivery, health

services, or investment decisions.To solve problems of resource allocation one may use

mathematical programming. 2

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

3

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 to be 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.

LP Model Components and Formutaion

4

DEVELOPMENT OF A LP MODEL LP applied extensively to problems areas - medical, transportation, operations, financial, marketing, accounting, human resources, and agriculture.

Development and solution of all LP models can be examined in a four step process:

(1) identification of the problem as solvable by LP(2) formulation of the mathematical model.(3) solution. (4) interpretation. 5

Basic Steps of Development Model

Formulation– Process of translating problem scenario into simple LP model

framework with a 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 the manager to

• interpret results and implications of problem solution.• investigate changes in input parameters and model

variables and impact on problem solution results. 6

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

7

Maximization model examples

BEAVER Creek example Flair Furniture example GALAXY Industries example

8

Resource Requirements

Product Labor (hr/unit)

Clay (lb/unit)

Profit ($/unit)

Bowl 1 4 40

Mug 2 3 50

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:

9

PROBLEM DEFINITION: BEAVER CREEK EXAMPLE (2 of 18)

Resource 40 hrs of labor per dayAvailability: 120 lbs of clay

Decision Variables x1 = number of bowls to produce per dayx2 = number of mugs to produce per day

Objective Maximize Z = $40x1 + $50x2Function: Where Z = profit per day

Resource 1x1 + 2x2 40 hours of laborConstraints: 4x1 + 3x2 120 pounds of clay

Non-Negativity x1 0; x2 0 Constraints:

10

PROBLEM DEFINITION:BEAVER CREEK EXAMPLE (3 of 18)

Solve:

Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40

4x1 + 3x2 120x1, x2 0

11

A feasible solution does not violate any of the constraints:

Example x1 = 5 bowlsx2 = 10 mugsZ = $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

FEASIBLE SOLUTIONS:BEAVER CREEK EXAMPLE (4 of 18)

12

An infeasible solution violates at least one of the constraints:Example x1 = 10 bowls

x2 = 20 mugsZ = $1400

Labor constraint check:1(10) + 2(20) = 50 > 40 hours, violates the constraint

INFEASIBLE SOLUTIONS:BEAVER CREEK EXAMPLE (5 of 18)

13

The set of all points that satisfy all the constraints of the model is called

a

FEASIBLE REGION

14

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.

GRAPHICAL SOLUTION OF LINEAR PROGRAMMING MODELS

15

Primary advantage of two-variable LP models (such as BeaverCreek problem) is their solution can be graphically illustrated usingtwo-dimensional graph.

GRAPHICAL REPRESENTATION OF CONSTRAINTS:COORDINATE AXES-BEAVER CREEK EXAMPLE (6 of 18)


4x1 + 3x2 120x1, x2 0

Coordinates for Graphical Analysis

16

GRAPHICAL REPRESENTATION OF CONSTRAINTS:-BEAVER CREEK EXAMPLE-LABOR CONSTRAINT (7 of 18)


4x1 + 3x2 120x1, x2 0

Graph of Labor Constraint 17

GRAPHICAL REPRESENTATION OF CONSTRAINTS:-BEAVER CREEK EXAMPLE-LABOR CONSTRAINT AREA(8 of 18)


4x1 + 3x2 120x1, x2 0

Labor Constraint Area18

GRAPHICAL REPRESENTATION OF CONSTRAINTS:BEAVER CREEK EXAMPLE-CLAY CONSTRAINT AREA(9 of 18)


4x1 + 3x2 120x1, x2 0

Clay Constraint Area 19


4x1 + 3x2 120x1, x2 0

Graph of Both Model Constraints

GRAPHICAL REPRESENTATION OF CONSTRAINTS:BEAVER CREEK EXAMPLE- BOTH CONSTRAINTS (10 of 18)

20

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, oneyielding 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. 21

ISOPROFIT LINE METHOD:BEAVER CREEK EXAMPLE (12 of 18)


4x1 + 3x2 120x1, x2 0

Objective Function Line for Z = $800

Set Objective Function = 800

22

ISOPROFIT LINE METHOD -ALTERNATIVE OBJECTIVE FUNCTION SOLUTION LINES:BEAVER CREEK EXAMPLE (13 of 18)


4x1 + 3x2 120x1, x2 0

Alternative Objective Function Lines23

ISOPROFIT LINE METHOD-OPTIMAL SOLUTION):BEAVER CREEK EXAMPLE (14 of 18)


4x1 + 3x2 120x1, x2 0

Identification of the Optimal Solution24

ISOPROFIT LINE METHOD - OPTIMAL SOLUTION COORDINATES:BEAVER CREEK EXAMPLE (15 of 18)


4x1 + 3x2 120x1, x2 0

Optimal Solution Coordinates25

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.

26

GRAPHICAL SOLUTION - CORNER POINT SOLUTION METHOD :BEAVER CREEK EXAMPLE(16 of 18)


4x1 + 3x2 120x1, x2 0

Solution at All Corner Points27

OPTIMAL SOLUTION FOR A NEW OBJECTIVE FUNCTION:BEAVER CREEK EXAMPLE (17 of 18)


4x1 + 3x2 120x1, x2 0

Optimal Solution with Z = 70x1 + 20x228

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.

SLACK VARIABLES

29

STANDARD FORM OF LINEAR PROGRAMMING MODEL:BEAVER CREEK EXAMPLE (18 of 18)

Max Z = 40x1 + 50x2 + s1 +s2subject to:1x1 + 2x2 + s1 = 40

4x1 + 3x2 + s2 = 120x1, x2, s1, s2 0

Where:x1 = number of bowlsx2 = number of mugss1, s2 are slack variables

Solution Points A, B, and C with Slack30

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 the best possible combination of tables and chairs to manufacture

in order to attain maximum profit.31

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.

32

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)

33

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.

34

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)

35

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)

36

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 (chairs limit constraint)T 0 (non-negativity constraint on tables)C 0 (non-negativity constraint on chairs) s1 s2 s3 0 (non-negativity constraints on slacks)

37

GRAPHICAL REPRESENTATION OF CONSTRAINTS:FLAIR FURNITURE EXAMPLE (8 of 19)

4T + 3C 240

Carpentry time constaint

38


Carpentry Time Constraint(feasible area)

Carpentry time and the feasible region

Any point below line satisfies constraint.

39


Painting Time Constraint and the Feasible Area

2T + 1C 100

Any point on line satisfies equation:2T + 1C = 100(30,40) yields 100. 40


Chair Limit Constraint and the Feasible Solution Area

Feasible solution area is constrained by three limiting lines 41

ISOPROFIT LINE SOLUTION METHOD:FLAIR FURNITURE EXAMPLE (12 of 19)

Let objective function (that is, $7T + $5C) guide one towards an optimal point in the 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?

42

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 = 30.43


One can check for higher values of Z to find an optimal solution.

210 is not the highest possible value.

44


Isoprofit lines ($210, $280, $350) are all parallel.

45


Optimal Solution:Corner Point 4: T=30 (tables) and C=40 (chairs) with $410 profit 46

ISOPROFIT LINE SOLUTION METHOD:FLAIR FURNITURE EXAMPLE (17 of 19) Optimal solution occurs at the maximum point in the feasible

region. It occurs at the 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. 47

CORNER POINT SOLUTION METHOD:FLAIR FURNITURE EXAMPLE (18 of 19)

From the figure one knows that the feasible region for Flair’s problem has five corner points, namely, 1, 2, 3, 4, and 5, respectively.

To find the point yieldingthe maximum profit, one finds coordinates of each corner point and computes profit level at each point.

48

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 .

49

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.

50

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.


51

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 dozensZapper = 100 dozensProfit = 4900 dollars per week


52

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

53

• 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

DECISION VARIABLES:GALAXY INDUSTRIES EXAMPLE (4 of 9)

54

Max. 8X1 + 5X2 (Weekly profit)subject to2X1 + 1X2 < = 1200 (Plastic)3X1 + 4X2 < = 2400 (Production Time)X1 + X2 < = 800 (Total production)X1 - X2 < = 450 (Mix)Xj> = 0, j = 1,2 (Nonnegativity)

BUILDING THE COMPLETE MATHEMATICAL MODEL: GALAXY INDUSTRIES EXAMPLE (5 of 9)

55

1200

600

The Plastic constraint

Feasible

The plastic constraint: 2X1+X2<=1200

X2

Infeasible

Production Time3X1+4X2<=2400

Total production constraint:X1+X2<=800

600

800

Production mix constraint:X1-X2<=450

• There are three types of feasible pointsInterior points.

Boundary points.

Extreme points.

X1

GRAPHICAL REPRESENTATION OF CONSTRAINTS:GALAXY INDISTRIES EXAMPLE (6 of 9)

56

600

800

1200

400 600 800

X2

X1

We now demonstrate the search for an optimal solutionStart at some arbitrary profit, say profit = $2,000...

Profit = $ 000

2,

Then increase the profit, if possible...

3,4,...and continue until it becomes infeasibleProfit =$5040

ISOPROFIT LINE SOLUTION METHOD:GALAXY INDUSTRIES EXAMPLE (7 of 9)

57

600

800

1200

400 600 800

X2

X1

Let’s take a closer look at the optimal point

FeasibleregionFeasibleregion

Infeasible

ISOPROFIT LINE SOLUTION METHOD:GALAXY INDUSTRIES EXAMPLE (8 of 9)

58

OPTIMAL SOLUTION:GALAXY INDUSTRIES EXAMPLE (9 of 9)

Space Rays = 480 dozensZappers = 240 dozensProfit = $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).59

MINIMIZATION MODEL EXAMPLES: FERTILIZER MIX PROBLEM HOLIDAY MEAL CHICKEN RANCH

EXAMPLE NAVY SEA RATIONS EXAMPLE

60

A MINIMIZATION LP PROBLEMMany 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.

61

PROBLEM DEFINITION:FERTILIZER MIX EXAMPLE (1 of 7)

Chemical Contribution

Brand Nitrogen (lb/bag)

Phosphate (lb/bag)

Super-gro 2 4

Crop-quick 4 3

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 the following data ?

62

PROBLEM DEFINITION:FERTILIZER MIX EXAMPLE (2 of 7)

Decision Variables:x1 = bags of Super-Grox2 = 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)

63

GRAPHICAL REPRESENTATION OF CONSTRAINTS:FERTILIZER MIX EXAMPLE (3 of 7)

Minimize Z = $6x1 + $3x2subject to: 2x1 + 4x2 16

4x2 + 3x2 24x1, x2 0

Graph of Both Model Constraints64

FEASIBLE SOLUTION AREA:FERTILIZER MIX EXAMPLE (4 of 7)


4x2 + 3x2 24x1, x2 0

Feasible Solution Area65

OPTIMAL SOLUTION POINT:FERTILIZER MIX EXAMPLE (5 of 7)


4x2 + 3x2 24x1, x2 0

Optimum Solution Point

66

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 surplus variables in the farmer problem

constraints:2x1 + 4x2 - s1 = 16 (nitrogen)4x1 + 3x2 - s2 = 24 (phosphate)

67

Minimize Z = $6x1 + $3x2 + 0s1+ 0s2subject to: 2x1 + 4x2 – s1 = 16

4x2 + 3x2 – s2 = 24x1, x2, s1, s2 0

Graph of Fertilizer Example

GRAPHICAL SOLUTION:FERTILIZER MIX EXAMPLE (7 of 7)

68

PROBLEM DEFINITION:HOLIDAY MEAL CHICKEN RANCH (HMCR) EXAMPLE (1 of 10)

Buy two brands of feed for good, low-cost diet for chickens. 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 vitamin, and 0 units of iron. 69

PROBLEM DEFINITION:HMCR EXAMPLE (2 of 10)

Brand A feed costs ranch $0.02 per pound, while Brand Bfeed costs $0.03 per pound.

Ranch owner would like lowest-cost diet that meetsminimum monthly intake requirements for each nutritional ingredient.

70

PROBLEM DEFINITION:HMCR EXAMPLE (3 of 10)

71

BUILDING THE COMPLETE MATHEMATICAL MODEL:HMCR EXAMPLE (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.

72

BUILDING THE STANDARD LP MODEL:HMCR 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)

73

GRAPHICAL REPRESENTATION OF CONSTRAINTS:HMCR EXAMPLE (6 of 10)

Drawing Constraints:

5A + 10B 90

4A + 3B 48

½A 1½

Nonnegativity ConstraintA 0, B 0

74

GRAPHICAL SOLUTION METHOD-ISOCOST LINE METHOD: HMCR EXAMPLE (7 of 10)

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

75

ISOCOST LINE METHOD:HMCR EXAMPLE (8 of 10)

• Isocost line is moved parallel to 54-cent solution line toward lower left origin.

• Last point to touch the isocost line while still in contact with the feasible region, is corner point 2.

76

• Solving for corner point 2 with two equations produces 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

ISOCOST LINE METHOD:HMCR EXAMPLE (9 of 10)

77

CORNER POINT SOLUTION METHOD:HMCR 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

78

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.

79

• Decision variables:– X1 (X2) -- The number of portions of Texfoods

(Calration) product used in a serving.

• The Model:Minimize 0.60X1 + 0.50X2Subject to

20X1 + 50X2 ≥ 100 Vitamin A25X1 + 25X2 ≥ 100 Vitamin D50X1 + 10X2 ≥ 100 Iron

X1, X2 ≥ 0

COMPLETE MODEL: NAVY SEA RATIONS EXAMPLE (2 of 4)

80

GRAPHICAL SOLUTION:NAVY SEA RATIONS EXAMPLE (3 of 4)

5

4

2

2 4 5

Feasible Region

Vitamin “D” constraint

Vitamin “A” constraint

The Iron constraint

81

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.

82

SUMMARY OF THE GRAPHICAL SOLUTION METHODS (1 of 3)

1. Plot the model constraints accepting them as equalities,

2. Considering the inequalities of the constraints identify thefeasible solution region, that is, the area that satisfies all constraints simultaneously.

3. Select one of two following graphical solution techniquesand proceed to solve problem.

- Isoprofit or Isocost Method.

- Corner Point Method

83


Corner Point Method Determine the coordinates of each of the corner points of

the feasible region by solving simultaneous equations at each point.

Compute the 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), or lowest cost (minimization).

84


Isoprofit or Isocost Method Select an arbitrary value for profit or cost, and plot an isoprofit

/ isocost line to reveal its slope. Maintain the 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 the optimal solution as coordinates of the point that is touched by the highest possible isoprofit line or lowest possible isocost line (by solving the simultaneous equations)

Read optimal coordinates and compute the optimal profit or cost.

85

SPECIAL SITUATIONS IN SOLVING LP PROBLEMS

(IRREGULAR TYPES OF LP PROBLEMS)

86

For some linear programming models, the general rules do not apply.

Special types of problems include those with:Redundancy Infeasible solutionsUnbounded solutionsMultiple optimal solutions

IRREGULAR TYPES OF LINEAR PROGRAMMING PROBLEMS

87

Redundancy: A redundant constraint is a constraint that does not affect the feasible region in any way.

Maximize Profit = 2X + 3Ysubject to:X + Y 202X + Y 30X 25X, Y 0

88

Infeasibility: A condition that arises when an LPproblem has no solution that satisfies all of itsconstraints.

X + 2Y 6

2X + Y 8

X 7

89

Unboundedness: Sometimes an LP model will nothave a finite solution

Maximize profit

= $3X + $5Y

subject to:

X 5

Y 10

X + 2Y 10

X, Y 090

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 the problem

which lies in the direction in which isoprofit (or

isocost) line is located.

– In other words, when they have the same slope. 91

EXAMPLE: MULTIPLE OPTIMAL SOLUTIONS

Maximize profit =$3x + $2ySubject to:

6X + 4Y 24X 3X, Y 0

92

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 the line between corner points 1 and 2 provides an optimal X and Y combination. 93

SETTING UP AND SOLVING LP PROBLEMS USING EXCEL’S SOLVER

Using solver to solve Flair Furniture problem:Recall decision variables T ( Tables ) and C ( Chairs ) in Flair Furniture problem:

Maximize profit = $7T + $5CSubject to constraints4T + 3C 240 (carpentry constraint)2T + 1C 100 (painting constraint)C 60 (chairs limit constraint)T, C 0 (non-negativity)

94

SOLVER SPREADSHEET SETUP

95

LP EXCEL AND SOLVER PARTS

Changing cells• Solver refers to decision variables as changing cells. • In Flair Furniture example, there are two decision

variables cells B5 and C5 to represent number of tables to make (T) and number of chairs to make (C), respectively.

96

LP EXCEL AND SOLVER PARTSChanging Cells• In each of excel layouts, for clarity, changing cells

(decision variables) have been shaded yellow.

Changing Cells ( B5 and C5 )

97

LP EXCEL AND SOLVER PARTSTarget CellObjective function, referred to as target cell by solver,

= SUMPRODUCT(B6:C6,$B$5:$C$5) This is equivalent to =B6*B5+C6*C5

Target Cell

98

LP EXCEL AND SOLVER PARTSConstraints

Each constraint has three parts -

(1) A left hand side (LHS) part consisting of every term

to the left of the equality or inequality sign.

(2) A right hand side (RHS) part consisting of all terms to

the right of the equality or inequality sign.

(3) An equality or inequality sign. 99

LP EXCEL AND SOLVER PARTSConstraintsEach constraint has three parts1. A left hand side (LHS) part. 2. A right hand side (RHS) part. 3. Equality or inequality sign.

1 23

100

ENTERING INFORMATION IN SOLVERInvoke Solver by clicking Tools|Solver • Specify Target Cell (D6)• Specify Changing Cells (highlight B5, C5)

Flair Furniture

T CTables Chairs

Number Of UnitsProfit 7 5 0 <-ObjectiveConstraints:Carpentry Hours 4 3 0 <= 240Painting Hours 2 1 0 <= 100Chairs Limit 1 0 <= 60

LHS Sign RHS

101

CONSTRAINTSSpecifying Constraints

• Use "Add" constraints to enter relevant cell references for LHS and RHS.

• Either add constraints one at a time or add blocks of constraints having same sign (<=, >=, or =) at the same time.

• Since all constraints have same <= sign one chose to highlight all LHS D8:D10 on left and F8:F10 on right with <= sign.

102

CONSTRAINTSSpecifying Constraints

103

SOLVER OPTIONS• Click on Options

button to get Solver Options window

• One must check boxes titled– Assume Linear

Model – Assume Non-

Negative

104

SOLVING MODELWhen Solve button is clicked, Solver executes model and results appear as shown.

Solver Results window also indicates the availability of three reports –

- Answer. - Sensitivity. - Limits.

105

SOLUTIONOptimal solution indicated that one should make 30 Tables

and 40 chairs with an optimal profit of $ 410.

Flair Furniture

T CTables Chairs

Number Of Units 30 40Profit 7 5 410 <-ObjectiveConstraints:Carpentry Hours 4 3 240 <= 240Painting Hours 2 1 100 <= 100Chairs Limit 1 40 <= 60

LHS Sign RHS

106

POSSIBLE MESSAGES IN RESULTS WINDOW

107

FLAIR FURNITURE SOLVER ANSWER REPORT

108

Documents

Management decisions in many organizations involve trying ...aa2015_topic-05_linear... · Solution: Graphical and Computer ... Computers play an important rolein the solution of LP