om2-1linear programming chapter 6 supplement linear programming

OM2-1 Linear Programming

Chapter 6 Supplement

Linear Programming


Linear Programming

• Linear Programming (LP) deals with the problems of allocating limited resources among competing activities in the best possible way (optimal)

• A linear program consist of a linear objective function and a set of linear constraints


Linear Programming Model

• Objective: the goal of an LP model is maximization or minimization

• Decision variables: amounts of either inputs or outputs

• Constraints: limitations that restrict the available alternatives

• Parameters: numerical values


Linear Programming Assumptions

• Linearity: the impact of decision variables is linear in constraints and objective function

• Divisibility: noninteger values of decision variables are acceptable

• Certainty: values of parameters are known and constant

• Nonnegativity: negative values of decision variables are unacceptable


Linear Programming Application Procedure

• Parameter Estimation

• Problem Formulation

• Optimal Solution Graphical Method

Simplex Method

Computer Solution

Other Methods

• Sensitivity Analysis


Linear Programming Application Areas

• Production

• Inventory

• Financial

• Marketing

• Distribution

• Sports

• Agriculture


Linear Programming: Some Definitions

• Solution: A solution is a set of values of the decision variables

• Feasible Solution: A feasible solution is a solution for which all the constraints are satisfied

• Optimal Solution: An optimal solution is a feasible solution which optimizes the objective function


Linear Programming: Types of Solutions

• Single Optimal Solution

• Multiple Optimal Solutions

• No Optimal Solution


Graphical Linear Programming

• Set up objective function and constraints in mathematical format

• Plot the constraints

• Identify the feasible solution space

• Plot the objective function

• Determine the optimum solution


Graphical Linear Programming

• Maximize Z = 4X1 + 5X2

Subject to

• X1 + 3X2 < 12(constraint 1)

• 4X1 + 3X2 < 24 (constraint 2)

• X1 > 0

• X2 > 0


Linear Programming Example

0

2

4

6

8

10

12

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

PlotConstraint 1X1 + 3X2 = 12



0

2

4

6

8

10

12

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

AddConstraint 24X1 + 3X2 = 24

Constraint 1X1 + 3X2 = 12

Solution space



0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Z x x

x x

x x

x x

4 5

3 12

4 3 24

0

1 2

1 2

1 2

1 2,

Z = 60

Z = 40

Z = 20

X1

X2


LP Formulation and Computer Solution: Problem 11. A manufacturing firm has discontinued production of a certain unprofitable product line. This created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products; call them products 1, 2, and 3. The available capacity on the machines that might limit output is summarized in the following table:

Machine type Available time (in machine hours per week) Milling machine 500 Lathe machine 350 Drilling machine 150

The number of machine hours required for each unit of the respective products is: Productivity coefficient (in machine hours per unit)

Machine type Product 1 Product 2 Product 3 Milling machine 9 3 5 Lathe machine 5 4 0 Drilling machine 3 0 2

The sales department indicates that the sales potential for products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week. The profit would be $30, $12, and $15, respectively, on products 1, 2, and 3. Formulate and solve the linear programming model for determining how much of each product the firm should produce to maximize profits.


Linear Programming Problem 1: Formulation

• Let Xi be the number of units of product type i to be produced per week, i = 1, 2, 3

• Maximize Z = 30X1 + 12X2 + 15X3

Subject to

• 9X1 + 3X2+ 5X3 < 500 (Milling)

• 5X1 + 4X2 < 350 (Lathe)

• 3X1 + 2X3 < 150 (Drill)

• X3 < 20 (Sales Potential)

• X1 > 0, X2 > 0, X3 > 0


Slack and Surplus• Binding constraint: a constraint that forms

the optimal corner point of the feasible solution space

• Slack: when the optimal values of decision variables are substituted into a less than or equal to constraint and the resulting value is less than the right side value

• Surplus: when the optimal values of decision variables are substituted into a greater than or equal to constraint and the resulting value exceeds the right side value


Linear Programming Problem 1: Solution Using LINGO Software

• Objective value: 1742.857Objective value: 1742.857

• Variable Value Reduced CostVariable Value Reduced Cost

• X1 26.19048 0.0000000X1 26.19048 0.0000000

• X2 54.76190 0.0000000X2 54.76190 0.0000000

• X3 20.00000 0.0000000X3 20.00000 0.0000000

• Row Slack or Surplus Dual PriceRow Slack or Surplus Dual Price

• PROFIT 1742.857 1.000000PROFIT 1742.857 1.000000

• MILLING 0.0000000 2.857143MILLING 0.0000000 2.857143

• LATHE 0.0000000 0.8571429LATHE 0.0000000 0.8571429

• DRILL 31.42857 0.0000000DRILL 31.42857 0.0000000

• SALESPOT 0.0000000 0.7142857SALESPOT 0.0000000 0.7142857


Sensitivity Analysis

• Range of optimality: the range of values for which the solution quantities of the decision variables remains the same

• Range of feasibility: the range of values for the fight-hand side of a constraint over which the shadow price (dual price) remains the same

• Shadow prices: negative values indicating how much a one-unit decrease in the original amount of a constraint would decrease the final value of the objective function


Linear Programming Problem 1: Solution Using LINGO Software• Ranges in which the basis is unchanged:Ranges in which the basis is unchanged:

• Objective Coefficient RangesObjective Coefficient Ranges

• Current Allowable AllowableCurrent Allowable Allowable

• Variable Coefficient Increase DecreaseVariable Coefficient Increase Decrease

• X1 30.00000 0.7500000 15.00000X1 30.00000 0.7500000 15.00000

• X2 12.00000 12.00000 0.6000000X2 12.00000 12.00000 0.6000000

• X3 15.000 INFINITY 0.7142857X3 15.000 INFINITY 0.7142857

• Righthand Side RangesRighthand Side Ranges

• Row Current Allowable AllowableRow Current Allowable Allowable

• RHS Increase DecreaseRHS Increase Decrease

• MILLING 500.0000 55.00000 137.5000MILLING 500.0000 55.00000 137.5000

• LATHE 350.0000 183.3333 73.33334LATHE 350.0000 183.3333 73.33334

• DRILL 150.0000 INFINITY 31.42857DRILL 150.0000 INFINITY 31.42857

• SALESPOT 20.00000 27.50000 20.00000SALESPOT 20.00000 27.50000 20.00000


Linear Programming Problem 1: Solution Using EXCEL (a)

LP Example 1

Variables Product 1 Product 2 Product 3 Total Profit26.19047619 54.76190476 20 1742.857143

Constraints RHSMilling 500 500Lathe 350 350Drill 118.5714286 150Sales Potential 20 20


Linear Programming Problem 1: Solution Using EXCEL Software (b)Microsoft Excel 8.0e Answer ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 11:21:09 AM

Target Cell (Max)Cell Name Original Value Final Value

$F$4 Total Profit 0 1742.857143

Adjustable CellsCell Name Original Value Final Value

$B$4 Product 1 0 26.19047619$C$4 Product 2 0 54.76190476$D$4 Product 3 0 20

ConstraintsCell Name Cell Value Formula Status Slack

$B$6 Milling Product 1 500 $B$6<=$C$6 Binding 0$B$7 Lathe Product 1 350 $B$7<=$C$7 Binding 0$B$8 Drill Product 1 118.5714286 $B$8<=$C$8 Not Binding 31.42857143$B$9 Sales Potential Product 1 20 $B$9<=$C$9 Binding 0$B$4 Product 1 26.19047619 $B$4>=0 Not Binding 26.19047619$C$4 Product 2 54.76190476 $C$4>=0 Not Binding 54.76190476$D$4 Product 3 20 $D$4>=0 Not Binding 20


Linear Programming Problem 1: Solution Using EXCEL Software (c)

Microsoft Excel 8.0e Sensitivity ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 11:21:19 AM

Adjustable CellsFinal Reduced

Cell Name Value Gradient$B$4 Product 1 26.19047619 0$C$4 Product 2 54.76190476 0$D$4 Product 3 20 0

ConstraintsFinal Lagrange

Cell Name Value Multiplier$B$6 Milling Product 1 500 2.857142857$B$7 Lathe Product 1 350 0.857142857$B$8 Drill Product 1 118.5714286 0$B$9 Sales Potential Product 1 20 0.714285714


Linear Programming Problem 1: Solution Using EXCEL Software (d)

Microsoft Excel 8.0e Limits ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 11:21:37 AM

TargetCell Name Value

$F$4 Total Profit 1742.857143

Adjustable Lower Target Upper TargetCell Name Value Limit Result Limit Result

$B$4 Product 1 26.19047619 0 957.1428571 26.19047619 1742.857143$C$4 Product 2 54.76190476 0 1085.714286 54.76190476 1742.857143$D$4 Product 3 20 -1.77636E-15 1442.857143 20 1742.857143


LP Formulation And Computer Solution: Problem 2

2. A farmer is raising lamas for the market, and she wishes to determine the quantities of the available types of feed that should be given to each lama to meet certain nutritional requirements at minimum cost. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with the daily nutritional requirements and feed costs:

Nutritional ingredient

Kilogram of corn

Kilogram of tankage

Kilogram of alfalfa

Minimum daily requirement

Carbohydrate 90 20 40 200 Protein 30 80 60 180 Vitamins 10 20 60 150 Cost(cents) 21 18 15

Formulate and solve the linear programming model for this problem.


Linear Programming Problem 2: Formulation

• Let X1 X2 X3 be the kilograms of corn, tankage, and alfalfa, respectively.

• Minimize Z = 21X1 + 18X2 + 15X3

Subject to

• 90X1 + 20X2+ 40X3 > 200 (Carbo)

• 30X1 + 80X2 + 60X3 > 180 (Protein)

• 10X1 + 20X2 + 60X3 > 150 (Vitamin)

• X1 > 0, X2 > 0, X3 > 0


Linear Programming Problem 2: Solution Using LINGO Software

• Objective value: 60.42857Objective value: 60.42857

• Variable Value Reduced CostVariable Value Reduced Cost

• X1 1.142857 0.0000000X1 1.142857 0.0000000

• X2 0.0000000 4.428571X2 0.0000000 4.428571

• X3 2.428571 0.0000000X3 2.428571 0.0000000

• Row Slack or Surplus Dual PriceRow Slack or Surplus Dual Price

• COST 60.42857 1.000000COST 60.42857 1.000000

• CARBOHY 0.0000000 -0.1928571CARBOHY 0.0000000 -0.1928571

• PROTEIN 0.0000000 -0.1214286PROTEIN 0.0000000 -0.1214286

• VITAMIN 7.142857 0.0000000VITAMIN 7.142857 0.0000000


Linear Programming Problem 2: Solution Using LINGO Software• Ranges in which the basis is unchanged:Ranges in which the basis is unchanged:

• Objective Coefficient RangesObjective Coefficient Ranges

• Current Allowable AllowableCurrent Allowable Allowable

• Variable Coefficient Increase DecreaseVariable Coefficient Increase Decrease

• X1 21.00000 12.75000 9.299998X1 21.00000 12.75000 9.299998

• X2 18.00000 INFINITY 4.428571X2 18.00000 INFINITY 4.428571

• X3 15.00000 2.818181 5.666667X3 15.00000 2.818181 5.666667

• Righthand Side RangesRighthand Side Ranges

• Row Current Allowable AllowableRow Current Allowable Allowable

• RHS Increase DecreaseRHS Increase Decrease

• CARBOHY 200.0000 25.00000 80.00000CARBOHY 200.0000 25.00000 80.00000

• PROTEIN 180.0000 120.0000 6.000000PROTEIN 180.0000 120.0000 6.000000

• VITAMIN 150.0000 7.142857 INFINITYVITAMIN 150.0000 7.142857 INFINITY


Linear Programming Problem 2: Solution Using EXCEL Software (a)

LP Example 2

Cost 21 18 15Variables Feed 1 (Corn) Feed 2 (Tankage) Feed 3 (Alfalfa) Total Cost

1.142857143 0 2.428571429 60.42857143

Requirements Achieved RHSCarbohydrates 90 20 40 200 200Protein 30 80 60 180 180Vitamins 10 20 60 157.1428571 150


Linear Programming Problem 2: Solution Using EXCEL Software (b)Microsoft Excel 8.0e Answer ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 2:34:19 PM

Target Cell (Min)Cell Name Original Value Final Value

$G$5 Total Cost 0 60.42857143

Adjustable CellsCell Name Original Value Final Value

$B$5 Feed 1 (Corn) 0 1.142857143$C$5 Feed 2 (Tankage) 0 0$D$5 Feed 3 (Alfalfa) 0 2.428571429

ConstraintsCell Name Cell Value Formula Status Slack

$F$8 Carbohydrates Achieved 200 $F$8>=$G$8 Binding 0$F$9 Protein Achieved 180 $F$9>=$G$9 Binding 0$F$10 Vitamins Achieved 157.1428571 $F$10>=$G$10 Not Binding 7.142857143$B$5 Feed 1 (Corn) 1.142857143 $B$5>=0 Not Binding 1.142857143$C$5 Feed 2 (Tankage) 0 $C$5>=0 Binding 0$D$5 Feed 3 (Alfalfa) 2.428571429 $D$5>=0 Not Binding 2.428571429


Linear Programming Problem 2: Solution Using EXCEL Software (c)Microsoft Excel 8.0e Sensitivity ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 2:34:36 PM

Adjustable CellsFinal Reduced

Cell Name Value Gradient$B$5 Feed 1 (Corn) 1.142857143 0$C$5 Feed 2 (Tankage) 0 4.428571429$D$5 Feed 3 (Alfalfa) 2.428571429 0

ConstraintsFinal Lagrange

Cell Name Value Multiplier$F$8 Carbohydrates Achieved 200 0.192857143$F$9 Protein Achieved 180 0.121428571$F$10 Vitamins Achieved 157.1428571 0


Linear Programming Problem 2: Solution Using EXCEL Software (d)

Microsoft Excel 8.0e Limits ReportWorksheet: [Book2]Sheet1Report Created: 2/14/2000 2:34:43 PM

TargetCell Name Value

$G$5 Total Cost 60.42857143

Adjustable Lower Target Upper TargetCell Name Value Limit Result Limit Result

$B$5 Feed 1 (Corn) 1.142857143 1.142857143 60.42857143 #N/A #N/A$C$5 Feed 2 (Tankage) 0 0 60.42857143 #N/A #N/A$D$5 Feed 3 (Alfalfa) 2.428571429 2.428571429 60.42857143 #N/A #N/A

om2-1linear programming chapter 6 supplement linear programming

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