metode simplex

LINEAR LINEAR PROGRAMMINGPROGRAMMING

SIMPLEX METHODSIMPLEX METHOD

OutlineOutline

IntroductionIntroduction TerminologyTerminology Solving LP ProblemsSolving LP Problems

Problem StatementProblem Statement Algebric FormulationAlgebric Formulation Graphical FormulationGraphical Formulation

Simplex MethodSimplex Method

IntroductionIntroduction A A Linear ProgrammingLinear Programming problem is a special case of a problem is a special case of a

Mathematical ProgrammingMathematical Programming problem. problem.

LLinear programming involves identifying an inear programming involves identifying an extremeextreme (i.e., (i.e., minimum or maximum) point of a function minimum or maximum) point of a function f(x1, x2, ..., xn)f(x1, x2, ..., xn), , which furthermore satisfies a set of constraints, which furthermore satisfies a set of constraints, g(x1, g(x1, x2,...xn)≤ bix2,...xn)≤ bi..

LLinear programming is widely used in industry, governmental inear programming is widely used in industry, governmental organizations, ecological sciences, transportation and organizations, ecological sciences, transportation and business organizations to minimize objectives functions, business organizations to minimize objectives functions, which can be production costs, numbers of employees to hire, which can be production costs, numbers of employees to hire, or quantity of pollutants released, given a set of constraints or quantity of pollutants released, given a set of constraints such as availability of workers, of machines, or labors time. such as availability of workers, of machines, or labors time.

General Form Of LP General Form Of LP ProblemsProblems

MAX (or MIN):MAX (or MIN): cc11XX11 + c + c22XX22 + … + c + … + cnnXXnn

Subject to:Subject to: a11*x1+ a12*x2+…..+ a1n*xn ≤a11*x1+ a12*x2+…..+ a1n*xn ≤ b1b1

a21*x1+ a22*x2+..…+ a2n*xn ≤ a21*x1+ a22*x2+..…+ a2n*xn ≤

b2 b2 am1*x1+ am1*x1+

am2*x2+..…+ amn*xn ≤ bm,am2*x2+..…+ amn*xn ≤ bm,

and and

x1≥ 0, x2 ≥ 0, ....., xn ≥ 0.x1≥ 0, x2 ≥ 0, ....., xn ≥ 0.

Type of LP ProblemsType of LP Problems

There are five common types of There are five common types of decisions in which LP may play a roledecisions in which LP may play a role Product mixProduct mix Ingredient mixIngredient mix TransportationTransportation Production planProduction plan AssignmentAssignment

TerminologyTerminology Objective FunctionObjective Function: :

An output function that you want to maximize or minimize.An output function that you want to maximize or minimize. E.g. Maximize profit, yield, contributionE.g. Maximize profit, yield, contribution E.g. Minimize cost, space used, time taken, or riskE.g. Minimize cost, space used, time taken, or risk

Decision VariablesDecision Variables: : A set of input variables whose values you can change.A set of input variables whose values you can change.

E.g. The number of units of a product we should produce.E.g. The number of units of a product we should produce. E.g. The number of items to store in a warehouse.E.g. The number of items to store in a warehouse. E.g. The amount of material we should buy.E.g. The amount of material we should buy.

ConstraintsConstraints: : A set of restrictions on the permissible values (or A set of restrictions on the permissible values (or combinations of values) of the input variables.combinations of values) of the input variables.

E.g. Cannot produce less than 0 units of a product.E.g. Cannot produce less than 0 units of a product. E.g. Cannot use more than 1,000 yards of material.E.g. Cannot use more than 1,000 yards of material.

Terminology for SolutionsTerminology for Solutions FFeasible solutioneasible solution:: AA solution for which all the solution for which all the

constraints are satisfied. It is possible for a problem to constraints are satisfied. It is possible for a problem to have no feasible solutions. Given that there are feasible have no feasible solutions. Given that there are feasible solutions, the goal of the linear programming is to find solutions, the goal of the linear programming is to find which one is the best, as measured by the value of the which one is the best, as measured by the value of the objective function in the model. objective function in the model.

OOptimal solutionptimal solution:: AA feasible solution that has the most feasible solution that has the most favorable value of the objective function. Most favorable favorable value of the objective function. Most favorable value means the largest or the smallest value, depending value means the largest or the smallest value, depending upon whether the objective is maximization or upon whether the objective is maximization or minimization. minimization.

one optimal solutionone optimal solution multiple optimal solumultiple optimal soluttionsions no optimal solutions no optimal solutions

Solving LP ProblemsSolving LP Problems Graphical Solution ApproachGraphical Solution Approach - used mainly - used mainly

as a teaching toolas a teaching tool..

Simplex MethodSimplex Method - most common analytic - most common analytic tooltool..

Transportation MethodTransportation Method - one of the earliest - one of the earliest methodsmethods..

Assignment MethodAssignment Method - occasionally used in - occasionally used in OOM (Operations Management).M (Operations Management).

Steps in Formulating a Steps in Formulating a Linear Programming (LP) ModelLinear Programming (LP) Model

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1.1. Understand the problem.Understand the problem.

2.2. Identify the Identify the decision variablesdecision variables..

3.3. State the State the objective functionobjective function as a linear as a linear combination of the decision variables.combination of the decision variables.

4.4. State the State the constraintsconstraints:: upper or lower upper or lower boundsbounds on the decision variables, on the decision variables,

including non-negativity constraints if applicable.including non-negativity constraints if applicable. linear linear combinationscombinations of the decision variables. of the decision variables.

Problem StatementProblem StatementSleevelessSleeveless and and SleeveSleeve Example Example

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ReebokReebok Sports manufactures two types of Sports manufactures two types of t-shirtst-shirts: : sleeveless with logosleeveless with logo and and sleevesleeve. . How many How many sleevelessessleevelesses and how many and how many sleveesslevees should be produced per week, to maximize profitsshould be produced per week, to maximize profits, , given the following constraints…given the following constraints…

The (profit) contribution per The (profit) contribution per sleevelesssleeveless is $3.00, is $3.00, compared to $4.50 per compared to $4.50 per sleevesleeve..

SleeveSleeve use 0.5 yards of material; use 0.5 yards of material; sleevelessleeveless use 0.4 s use 0.4 yards. 300 yards of material are available.yards. 300 yards of material are available.

It requires 1 hour to manufacture one sIt requires 1 hour to manufacture one sleevelessleeveless and 2 hours for oneand 2 hours for one sleevee sleevee. 900 labors hours are . 900 labors hours are available.available.

There is unlimited demand for There is unlimited demand for sleevelesssleeveless but total but total demand for demand for sleevesleeve is 375 units per week. is 375 units per week.

Each Each sleevelesssleeveless uses 1 insignia logo and 600 uses 1 insignia logo and 600 insignia logos are in stock.insignia logos are in stock.

Algebraic Algebraic FormulationFormulation

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The Algebraic LP FormulationThe Algebraic LP Formulation

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Variables: Variables: number of number of SleevelessesSleevelesses, number of , number of SleevesSleeves

Objective function:Objective function: maximizemaximize ( $3.00 x ( $3.00 x SleevelessSleeveless ) + ( $4.50 x ) + ( $4.50 x SleeveSleeve ) )

Constraints:Constraints: Material: ( Material: ( 00.4 x .4 x SleevelessSleeveless ) + ( ) + ( 00.5 x .5 x SleeveSleeve ) <= 300 ) <= 300

yardsyards Labor: ( 1 x Labor: ( 1 x SleevelessSleeveless ) + ( 2 x ) + ( 2 x SleeveSleeve ) <= 900 ) <= 900

hrshrs Demand: ( 0 x Demand: ( 0 x SleevelessSleeveless ) + ( 1 x ) + ( 1 x SleeveSleeve ) <= 375 ) <= 375

unitsunits Logos: ( 1 x Logos: ( 1 x SleevelessSleeveless ) + ( 0 x ) + ( 0 x SleeveSleeve ) <= 600 ) <= 600

Non-Negativity:Non-Negativity: SleevelessSleeveless >= 0 >= 0 SleeveSleeve >= 0 >= 0

Graphical Graphical FormulationFormulation

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The Graphical LP FormulationThe Graphical LP Formulation

Material ConstraintMaterial Constraint

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evele

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evele

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SleeveSleeve

0.4 SleevelessSleeveless + 0.5 SleeveSleeve = 300 Yards

Place your Place your Decision VariablesDecision Variables on on the the axesaxes of the graph. of the graph.

• x = x = SleeveSleeve • y = y = SleevelessSleeveless• So, the line is:So, the line is:

0.4y + 0.5x = 3000.4y + 0.5x = 300

The Equation of a LineThe Equation of a Line

750 It is useful when the problem have It is useful when the problem have two variables.two variables.


Material ConstraintMaterial Constraint

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0.4 SleevelessSleeveless + 0.5 SleeveSleeve = 300 Yards

0.40.4 SleevelessSleeveless + 0.5 + 0.5 SleeveSleeve 300 yards happens to be an upper 300 yards happens to be an upper bound constraint. bound constraint.

To determine which side of the line to To determine which side of the line to shade, substitute the point shade, substitute the point (0,0)(0,0) into into the equation:the equation: 0.4(0.4(00) + 0.5() + 0.5(00) ) 300 300Since the inequality holds, (0,0) must Since the inequality holds, (0,0) must be within the feasible region, so shade be within the feasible region, so shade the side of the line where (0,0) lies.the side of the line where (0,0) lies.

(0,0)

The Area Under a LineThe Area Under a Line


Labor ConstraintLabor Constraint

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( 1 SleevelessSleeveless ) + ( 2 SleeveSleeve) = 900 HoursLabor


Demand ConstraintDemand Constraint

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Demand

( 0 SleevelessSleeveless ) + ( 1 SleeveSleeve ) = 375


Logos ConstraintLogos Constraint

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( 1 SleevelessSleeveless ) + ( 0 SleeveSleeve ) <= 600 logos


Feasible RegionFeasible Region

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Feasible Region


Objective Function – Isoprofit LinesObjective Function – Isoprofit Lines

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$600

( $3 SleevelessSleeveless ) + ( $4.5 SleeveSleeve ) = $600



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( $3 SleevelessSleeveless ) + ( $4.5 SleeveSleeve ) = $1800


Optimal SolutionOptimal Solution

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Isoprofit line tangent to point D. $3.00 Sleeveless + $4.50 Sleeve = $2400

Sle

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D

Notice that looking for the optimal solution Notice that looking for the optimal solution is a form of is a form of searchsearch process. process.



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$3,00 $4,50 TotalSleeveless Sleeve Profit

A 300 200 $1.800,00B 0 375 $1.687,50C 150 375 $2.137,50D 500 200 $2.400,00E 600 120 $2.340,00F 600 0 $1.800,00

F E

C

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D Optimum

Maximization vs MinimizationMaximization vs Minimization

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The position of your optimal point differs, depending upon The position of your optimal point differs, depending upon whether the objective is whether the objective is maximizationmaximization or or minimizationminimization..

X

Y

Isoprofit line

MaximizationMaximization

Optimum (Max)

X

Y

Isocost line

MinimizationMinimization

Optimum (Min)

Fundemental Theorem Of LPFundemental Theorem Of LP

If the optimal value of theIf the optimal value of the objective objective function in a linearfunction in a linear programming problem programming problem exists, exists, then that value mustthen that value must occur at one occur at one (or more) of the corner points of the (or more) of the corner points of the feasible regionfeasible region..

Existence of SolutionExistence of Solution

((A) If the feasible region for a linear programming A) If the feasible region for a linear programming problem is bounded, then both the maximum value and problem is bounded, then both the maximum value and the minimum value of the objective function always exist.the minimum value of the objective function always exist.

(B) If the feasible region is unbounded, and the (B) If the feasible region is unbounded, and the coefficients of the objective function are positive, then the coefficients of the objective function are positive, then the minimum value of the objective function exists, but the minimum value of the objective function exists, but the maximum value does not.maximum value does not.

(C) If the feasible region is empty (that is, there are no (C) If the feasible region is empty (that is, there are no points that satisfy all the constraints), the both the points that satisfy all the constraints), the both the maximum value and the minimum value of the objective maximum value and the minimum value of the objective function do not exist.function do not exist.

Simplex Simplex MethodMethod

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Real LP ProblemsReal LP Problems

Real-world LP problems often involve:Real-world LP problems often involve: Hundreds or thousands of constraintsHundreds or thousands of constraints Large quantities of dataLarge quantities of data Many products and/or servicesMany products and/or services Many time periodsMany time periods Numerous decision alternativesNumerous decision alternatives … … and other complicationsand other complications

Simplex MethodSimplex Method The simplex algorithm, which was discovered in1947 by The simplex algorithm, which was discovered in1947 by

George Dantzig, is a simple, straightforward method for George Dantzig, is a simple, straightforward method for solving linear programming problems.solving linear programming problems.

It has proved to be remarkably efficient method that is used It has proved to be remarkably efficient method that is used to solve huge problems on today’s computers. to solve huge problems on today’s computers.

Simplex method starts with a feasible solution and tests Simplex method starts with a feasible solution and tests whether or not it is optimum. If not, the method proceeds a whether or not it is optimum. If not, the method proceeds a better solution.better solution.

In an algebric procedure, it is much more convenient to In an algebric procedure, it is much more convenient to deal with equations than with inequality relationships. deal with equations than with inequality relationships. Therefore, the first step in the setting up the simplex Therefore, the first step in the setting up the simplex method is to convert inequality constraints into equality method is to convert inequality constraints into equality constraints. This conversion can be succeeded by constraints. This conversion can be succeeded by introducing introducing slack variablesslack variables. .

The Simplex AlgorithmThe Simplex Algorithm

Objective Function has no maximum.

All coefficients in the first row are positive

TableTable Formulation Formulation

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LP FormulationLP Formulation

Maximize Z=3xMaximize Z=3x11+4.5x+4.5x2 2 Objective FunctionObjective Function

where xwhere x11=sleeveless, x=sleeveless, x22=sleeve=sleeve

0.4 x0.4 x11+0.5x+0.5x22≤300 Material ≤300 Material

xx11+2x+2x22≤900 Labor≤900 Labor

xx22≤375 Demand≤375 Demand

xx11≤600 Logo≤600 Logo

xx11≥0, x≥0, x22≥0 Nonnegativity≥0 Nonnegativity

Constraints

LP Formulation 2LP Formulation 2

Converting inequality constraints into equality Converting inequality constraints into equality constraints by defining constraints by defining slack variablesslack variables..

0.4 x0.4 x11+0.5x+0.5x22++xx33 =300 Material =300 Material xx11+2x+2x22++xx44 =900 Labor=900 Labor xx22++xx55 =375 Demand=375 Demand xx11++xx66 =600 Logo=600 Logo The objective function can be defined as;The objective function can be defined as; Z-3xZ-3x11-4.5x-4.5x22-0-0xx33-0-0xx44-0-0xx55-0-0xx66=0=0instead ofinstead of

Z=Z=3x3x11+4.5x+4.5x22

Z x1 x2 x3 x4 x5 x6

Z 0 1 -3 -4,5 0 0 0 0 0

x3 1 0 0,4 0,5 1 0 0 0 300

x4 2 0 1 2 0 1 0 0 900

x5 3 0 0 1 0 0 1 0 375

x6 4 0 1 0 0 0 0 1 600

Basic variable

Eq. No.

Coefficient of Right Side

Row 0 [ -3 -4.5 0 0 0 0 : 0] -(-4.5) [ 0 1 0 0 1 0 : 375]---------------------------------------------------------- GAUSSIAN ELIMINATIONNew Row [ -3 0 0 0 4.5 0 : 1687.5]

Pivot column Pivot row

1.1. Step Determine the Step Determine the entering basicentering basic variable. Having the largest absolute value in Eq. 0 variable. Having the largest absolute value in Eq. 02.2. Step Determine the leaving basic variable;Step Determine the leaving basic variable; a) Picking out each coefficient in the column that is strictly positive,a) Picking out each coefficient in the column that is strictly positive, b) Dividing each of them into “right side” for the same row,b) Dividing each of them into “right side” for the same row, c) Identifying the the equation that has the smallest ratio.c) Identifying the the equation that has the smallest ratio.

1. Iteration1. Iteration

300/0.5=600900/2=450

375/1=375


Z x1 x2 x3 x4 x5 x6

Z 0 1 -3 0 0 0 4,5 0 1687,5

x3 1 0 0,4 0 1 0 -0,5 0 112,5

x4 2 0 1 0 0 1 -2 0 150

x2 3 0 0 1 0 0 1 0 375

x6 4 0 1 0 0 0 0 1 600

Basic variable

Eq. No.

Coefficient ofBasic

variable

Z x1 x2 x3 x4 x5 x6

Z 0 1 -3 0 0 0 4,5 0 1687,5

x3 1 0 0,4 0 1 0 -0,5 0 112,5

x4 2 0 1 0 0 1 -2 0 150

x2 3 0 0 1 0 0 1 0 375

x6 4 0 1 0 0 0 0 1 600

Basic variable

Eq. No.

Coefficient ofBasic

variable

x5

x2


Z x1 x2 x3 x4 x5 x6

Z 0 1 0 0 0 3 -1,5 0 2137,5

x3 1 0 0 0 1 -0,4 0,3 0 52,5

x1 2 0 1 0 0 1 -2 0 150

x2 3 0 0 1 0 0 1 0 375

x6 4 0 0 0 0 -1 2 1 450

Basic variable

Eq. No.

Coefficient ofBasic

variable

Z x1 x2 x3 x4 x5 x6

Z 0 1 0 0 0 3 -1,5 0 2137,5

x3 1 0 0 0 1 -0,4 0,3 0 52,5

x1 2 0 1 0 0 1 -2 0 150

x2 3 0 0 1 0 0 1 0 375

x6 4 0 0 0 0 -1 2 1 450

Basic variable

Eq. No.

Coefficient ofBasic

variable

x4

x1

Z x1 x2 x3 x4 x5 x6

Z 0 1 0 0 5 1 0 0 2400

x5 1 0 0 0 10/3 -4/3 1 0 175

x1 2 0 1 0 20/3 5/3 0 0 500

x2 3 0 0 1 -10/3 4/3 -1 0 200

x6 4 0 0 0 -20/3 5/3 0 1 100

Basic variable

Eq. No.

Coefficient ofBasic

variable

Depending on the optimality test, we found that solution is optimal becauseDepending on the optimality test, we found that solution is optimal becausenone of the coefficients in Eq. 0 are negative, so the algorithm is finished. none of the coefficients in Eq. 0 are negative, so the algorithm is finished.

Maximum Profit is $2400 when xMaximum Profit is $2400 when x11=500 and x=500 and x22=200.=200.

Reebok Sports must produce 500 sleeveless and 200 slevees per week to Reebok Sports must produce 500 sleeveless and 200 slevees per week to maximize the profit. At this condition the profit will be $2400....!maximize the profit. At this condition the profit will be $2400....!


Thank you......Thank you......

metode simplex

Documents

linear programming lp

linear combinations

set of constraints

set of input variables

riskdecision variables

output function

function fx1

problem statementsleeveless