1 Additional examples LP Let : X 1, X 2, X 3, ………, X n = decision variables Z = Objective function or linear function Requirement: Maximization of the

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Additional examplesAdditional examplesLPLP

Additional examplesAdditional examplesLPLP

Let: X1, X2, X3, ………, Xn = decision variables

Z = Objective function or linear function

Requirement: Maximization of the linear function Z.Z = c1X1 + c2X2 + c3X3 + ………+ cnXn …..Eq (1)

subject to the following constraints:

…..Eq (2)

Formulating LP ProblemsFormulating LP Problems

X-podsX-pods BlueBerrysBlueBerrys Available HoursAvailable HoursDepartmentDepartment (X(X11)) (X(X22)) This WeekThis Week

Hours Required Hours Required to Produce 1 Unitto Produce 1 Unit

ElectronicElectronic 44 33 240240

AssemblyAssembly 22 11 100100

Profit per unitProfit per unit $7$7 $5$5

Decision Variables:Decision Variables:XX11 = number of X-pods to be produced= number of X-pods to be produced

XX22 = number of BlueBerrys to be produced= number of BlueBerrys to be produced

Table B.1Table B.1


Objective Function:Objective Function:

Maximize Profit = $7XMaximize Profit = $7X11 + $5X + $5X22

There are three types of constraints Upper limits where the amount used is ≤ the

amount of a resource Lower limits where the amount used is ≥ the

amount of the resource Equalities where the amount used is = the

amount of the resource


Second Constraint:Second Constraint:

2X2X11 + 1X + 1X22 ≤ 100 (hours of assembly time) ≤ 100 (hours of assembly time)

AssemblyAssemblytime availabletime available

AssemblyAssemblytime usedtime used is ≤is ≤

First Constraint:First Constraint:

4X4X11 + 3X + 3X22 ≤ 240 (hours of electronic time) ≤ 240 (hours of electronic time)

ElectronicElectronictime availabletime available

ElectronicElectronictime usedtime used is ≤is ≤

Graphical SolutionGraphical Solution

Can be used when there are two Can be used when there are two decision variablesdecision variables

1.1. Plot the constraint equations at their Plot the constraint equations at their limits by converting each equation to an limits by converting each equation to an equalityequality

2.2. Identify the feasible solution space Identify the feasible solution space

3.3. Create an iso-profit line based on the Create an iso-profit line based on the objective functionobjective function

4.4. Move this line outwards until the optimal Move this line outwards until the optimal point is identifiedpoint is identified


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Number of X-podsNumber of X-pods

XX11

XX22

Assembly (constraint B)Assembly (constraint B)

Electronics (constraint A)Electronics (constraint A)Feasible region

Figure B.3Figure B.3


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XX11

XX22

Assembly (constraint B)Assembly (constraint B)

Electronics (constraint A)Electronics (constraint A)Feasible region


Iso-Profit Line Solution Method

Choose a possible value for the objective function

$210 = 7X1 + 5X2

Solve for the axis intercepts of the function and plot the line

X2 = 42 X1 = 30


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XX11

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(0, 42)

(30, 0)(30, 0)

$210 = $7X$210 = $7X11 + $5X + $5X22


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XX11

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$210 = $7X$210 = $7X11 + $5X + $5X22

$350 = $7X$350 = $7X11 + $5X + $5X22

$420 = $7X$420 = $7X11 + $5X + $5X22

$280 = $7X$280 = $7X11 + $5X + $5X22


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$410 = $7X$410 = $7X11 + $5X + $5X22

Maximum profit lineMaximum profit line

Optimal solution pointOptimal solution point(X(X11 = 30, X = 30, X22 = 40) = 40)

Corner-Point MethodCorner-Point Method

Figure B.7Figure B.7 1

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Corner-Point MethodCorner-Point Method The optimal value will always be at a corner

point

Find the objective function value at each corner point and choose the one with the highest profit

Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350


point


Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Solve for the intersection of two constraints

2X1 + 1X2 ≤ 100 (assembly time)4X1 + 3X2 ≤ 240 (electronics time)

4X1 + 3X2 = 240

- 4X1 - 2X2 = -200

+ 1X2 = 40

4X1 + 3(40) = 240

4X1 + 120 = 240

X1 = 30


point


Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410

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The Galaxy Industries Production Problem – The Galaxy Industries Production Problem – A Prototype ExampleA Prototype Example

• Galaxy manufactures two toy doll models:– Space Ray. – Zapper.

• Resources are limited to– 1000 pounds of special plastic.– 40 hours of production time per week.

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• Marketing requirement– Total production cannot exceed 700 dozens.

– Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 350.

• 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.

The Galaxy Industries Production Problem – The Galaxy Industries Production Problem – A Prototype ExampleA Prototype Example

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• The 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), while remaining within the marketing guidelines.

• The current production plan consists of:

Space Rays = 450 dozenZapper = 100 dozenProfit = $4100 per week

The Galaxy Industries Production Problem –The Galaxy Industries Production Problem – A Prototype Example A Prototype Example

8(450) + 5(100)

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

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A linear programming model can provide an insight and an intelligent solution to this problem.

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• Decisions variables::

– X1 = Weekly production level of Space Rays (in dozens)

– X2 = Weekly production level of Zappers (in dozens).

• Objective Function:

– Weekly profit, to be maximized

The Galaxy Linear Programming ModelThe Galaxy Linear Programming Model

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Max 8X1 + 5X2 (Weekly profit)subject to2X1 + 1X2 1000 (Plastic)

3X1 + 4X2 2400 (Production Time)

X1 + X2 700 (Total production)

X1 - X2 350 (Mix)

Xj> = 0, j = 1,2 (Nonnegativity)

The Galaxy Linear Programming ModelThe Galaxy Linear Programming Model

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Using a graphical presentation

we can represent all the constraints,

the objective function, and the three

types of feasible points.

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The non-negativity constraints

X2

X1

Graphical Analysis – the Feasible RegionGraphical Analysis – the Feasible Region

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1000

500

Feasible

X2

Infeasible

Production Time3X1+4X2 2400

Total production constraint: X1+X2 700 (redundant)

500

700

The Plastic constraint2X1+X2 1000

X1

700


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1000

500

Feasible

X2

Infeasible

Production Time3X1+4X22400

Total production constraint: X1+X2 700 (redundant)

500

700

Production mix constraint:X1-X2 350

The Plastic constraint2X1+X2 1000

X1

700


• There are three types of feasible pointsInterior points. Boundary points. Extreme points.

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Solving Graphically for an Solving Graphically for an Optimal SolutionOptimal Solution

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The search for an optimal solutionThe search for an optimal solution

Start at some arbitrary profit, say profit = $2,000...Then increase the profit, if possible...

...and continue until it becomes infeasible

Profit =$4360

500

700

1000

500

X2

X1

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Summary of the optimal solution Summary of the optimal solution

Space Rays = 320 dozen Zappers = 360 dozen Profit = $4360

– This solution utilizes all the plastic and all the production hours.

– Total production is only 680 (not 700).

– Space Rays production exceeds Zappers production by only 40

dozens.

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– If a linear programming problem has an optimal solution, an extreme point is optimal.

Extreme points and optimal solutionsExtreme points and optimal solutions

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• For multiple optimal solutions to exist, the objective function must be parallel to one of the constraints

Multiple optimal solutionsMultiple optimal solutions

•Any weighted average of optimal solutions is also an optimal solution.

Copyright 2006 John Wiley & Sons, Inc. Supplement 13-32

Minimization ProblemMinimization Problem

CHEMICAL CONTRIBUTIONCHEMICAL CONTRIBUTION

BrandBrand Nitrogen (lb/bag)Nitrogen (lb/bag) Phosphate (lb/bag)Phosphate (lb/bag)

Gro-plusGro-plus 22 44

Crop-fastCrop-fast 44 33

Minimize Minimize ZZ = $6x = $6x11 + $3x + $3x22

subject tosubject to

22xx11 ++ 44xx22 16 lb of nitrogen 16 lb of nitrogen

44xx11 ++ 33xx22 24 lb of phosphate 24 lb of phosphate

xx11, , xx22 0 0

Copyright 2006 John Wiley & Sons, Inc. Supplement 13-33

14 14 –

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0 0 –|22

|44

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|1414 xx11

xx22

A

B

C


x1 = 0 bags of Gro-plusx2 = 8 bags of Crop-fastZ = $24

Z = 6x1 + 3x2

Dual problem (2 vars primal)Dual problem (2 vars primal)

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Documents

1 Additional examples LP Let : X 1, X 2, X 3, ………, X n = decision variables Z = Objective function or linear function Requirement: Maximization of the