Linear Programming (LP) - Gunadarma

Linear Programming(LP)

The Simplex Method

A. Primal Simplex Method

Overall Idea of Simplex Method

Simplex Method translates the geometricdefinition of the extreme point into analgebraic definition

Initial Step: all the constraints are put in astandard form

In standard form: all the constraints areexpressed as equations by augmentingslack and surplus variables as necessary

Overall Idea

The conversion of inequality to equationnormally results in a set of simultaneousequations in which the number of variablesexceeds the number of equations

The Equation might yield an infinitenumber of solution points

Extreme Points ↔ Basic Solutions

Overall Idea

Linear Algebra Theory:A basic solution is obtained

by setting to zero as many variable as differencebetween the total number of variables and thetotal number of equation

solving for the remaining variable

resulting in a unique solution

Standard LP Form

To develop a general solution method, LPproblem should be put in a commonformat

Development of the SimplexMethod

Two types of Simplex Method

Primal Simplex Method

Dual Simplex Method

Basically, the difference between the twomethod lies on the method of choosing theinitial setup

Standard LP Form

The Properties of Standard LP Form

All constraints are equations (Primal SimplexMethod requires a non-negative right-handside)

All the variables are non-negative

The Objective Function may be maximizationor minimization

Standard LP Form:Constraints

A constraint of type () is converted toequation by adding a slack variable to theleft side of the constraint

For example:

x1 + 2x2 6

x1 + 2x2 + s1 = 6, s1 0

Standard LP Form:Constraints

A constraint of type () is converted toequation by subtracting a surplus variableform the left side of the constraint

For example:

3x1 + 2x2 – 3x3 5

3x1 + 2x2 – 3x3 – s2 = 5, s2 0

Standard LP FormConstraints

The right side of an equation can alwaysbe made non-negative by multiplying bothsides by –1

For example:

2x1 + 3x2 – 7x3 = –5

–2x1 – 3x2 +7x3 = +5

Standard LP FormConstraints

The direction of an inequality is reversedwhen both sides are multiplied by –1

For example:

2x1 – x2 –5 = –2x1 + x2 5

Standard LP FormVariables

Unrestricted variable xi can be expressedin terms of two non-negative variables

xi = xi’ – xi”, where xi’, xi” 0

xi’ > 0, xi” = 0, and vice versa

Slack ↔Surplus

Standard LP FormObjective Function

Maximization or Minimization

Maximization = Minimization of thenegative of the function

For example:

Maximize z = 5x1 + 2x2 + 3X3 equivalent toMinimize (–z) = – 5x1 – 2x2 – 3X3

Standard LP FormExample

Write the following LP Model in standardform:

Minimize z = 2x1 + 3X2

Subject to (with constraints): x1 + x2 = 10

–2 x1 +3 x2 –5

7 x1 – 2 x2 6

x1 unrestricted

x2 0

Basic Solution

For m: number of equations And n: number of unknowns (variables) Basic solution is determined

by setting n – m (number of) variables equal to zero The n – m variables are called the non-basic

variables solving the m remaining variables The m remaining variables are called the basic

variable resulting a unique solution

Basic SolutionExample

2x1 + x2 + 4x3 + x4 = 2

x1 + 2x2 + 2x3 + x4 = 3

m = 4

n = 2

Basic solution is associated with m – n

(= 4 – 2 = 2) zero variables

Number of possible basic solution = mCn=

4C2 =4!/2!2! = 6

Basic SolutionExample

2x1 + x2 + 4x3 + x4 = 2

x1 + 2x2 + 2x3 + x4 = 3

Set x2 = 0 and x4 = 0

2x1 + 4x3 = 2

x1 + 2x3 = 3inconsistent

Basic SolutionExample 2x1 + x2 + 4x3 + x4 = 2

x1 + 2x2 + 2x3 + x4 = 3

Set x3 = 0 and x4 = 0 (non-basic variables)

x1 and x2 are basic variables

2x1 + x2 = 2

x1 + 2x2 = 3

x1 = 1/3

x2 = 4/3

Basic feasible solution: x1 = 1/3, x2 = 4/3, x3 = 0 and x4 = 0

Basic SolutionExample 2x1 + x2 + 4x3 + x4 = 2

x1 + 2x2 + 2x3 + x4 = 3

Set x1 = 0 and x2 = 0

x3 and x4

4x3 + x4 = 2

2x3 + x4 = 3

x3 = – 1/2

x2 = 4/3infeasible

Basic Feasible Solution

A basic solution is said to be feasible if allits solution values are non-negative

Else: infeasible basic solution

Primal and Dual Simplex

All iterations in Primal Simplex Methodare always associated to feasible basicsolutions only

Primal Simplex Method deals with feasibleextreme points only

Primal and Dual Simplex

Iterations in Dual Simplex Method endonly if the last iteration is infeasible

Both methods yield feasible basic solutionas stipulated by the non-negativitycondition of the LP model

Primal Simplex MethodThe Reddy Mikks Company XE: tons produced daily of exterior paint

XI: tons produced daily of interior paint

Objective Function to be satisfy:

Maximize z = 3XE + 2XI

Subject to these constraints:

XE + 2XI 6

2XE + XI 8

– XE + XI 1

XI 2

XE, XI 0

Primal Simplex MethodThe Reddy Mikks Company

Set the constraints to equations

xE + 2xI 6 xE + 2xI + sI = 6

2xE + xI 8 2xE + xI + s2 = 8

– xE + xI 1 – xE + xI + s3 = 1

xI 2 xI + s4= 2

xI, xE, sI, s2, s3, s4 0

m = 4

n = 2


Basic Solution

m = 4

n = 2 (xE and xI)

If xE and xI = 0 then

xE + 2xI + sI = 6 sI = 6

2xE + xI + s2 = 8 s2 = 8

– xE + xI + s3 = 1 s3 = 1

xI + s4= 2 s4= 2

BasicFeasibleSolution

Primal Simplex MethodThe Reddy Mikks Company Convert the Objective Function

Maximize z = 3xE + 2xI

z – 3xE – 2xI = 0

Include the slack variables

Maximize

z – 3xE – 2xI + 0sI + 0s2 + 0s3 + 0s4 = 0

Primal Simplex MethodThe Reddy Mikks Company Maximize

z – 3xE – 2xI + 0sI + 0s2 + 0s3 + 0s4 = 0

Later (in the next sub-chapters), the slack andthe substitute variables are written as commonvariables

Maximize

z – 3xE – 2xI + 0x1 + 0x2 + 0x3 + 0x4 = 0

Primal Simplex MethodThe Reddy Mikks Company Incorporate the objective function with the

basic feasible solution

xE and xI : entering variables

sI, s2, s3 ,and s4 : leaving variables

Start the iterations with the values of sI, s2,s3 ,and s4 = zero

Final Result of the iterations the values ofxE and xI in z = zero

Primal Simplex MethodEasiness (a commentary)

Each equation has a slack variable

The right hand of all constraints are non-negative

Entering Variables inMaximization and Minimization

Optimality Condition

The entering variable in Maximization isthe non-basic variable with the mostnegative coefficient in the z-equation

The optimum of Maximization is reachedwhen all the non-basic coefficients arenon-negative (or zero)

A Tie may be broken arbitrarily

Entering Variables inMaximization and Minimization

Optimality Condition

The entering variable in Minimization is thenon-basic variable with the most positivecoefficient in the z-equation

The optimum of Minimization is reachedwhen all the non-basic coefficients arenon-positive (Zero)


Leaving Variables inMaximization and Minimization

Feasibility Condition

For both Maximization and Minimization,the leaving variable is the current basicvariable having the smallest intercept(minimum ratio with strictly positivedenominator) in the direction of the enteringvariable


Primal Simplex Method:The Formal Iterative Steps

Step 0: Using the standard form with allnon-negative right hand sides, determinea starting feasible solution

Step 1: Select an entering variable fromamong the current non-basic variablesusing the optimality condition

Primal Simplex Method:The Formal Iterative Steps

Step 2: Select the leaving variable fromthe current basic variables using thefeasibility condition

Step 3: Determine the value of the newbasic variables by making the enteringand the leaving variable non basic

Stop if the optimum solution is achieved;else Go to Step 1

Primal Simplex MethodThe Reddy Mikks Company Incorporate the objective function with the

basic feasible solution

xE and xI : entering variables

sI, s2, s3 ,and s4 : leaving variables

Start the iterations with the values of sI, s2,s3 ,and s4 = zero

Final Result of the iterations the values ofxE and xI in z = zero


Maximize

z – 3XE – 2XI + 0sI + 0s2 + 0s3 + 0s4 = 0

xE + 2xI + sI = 6 Raw Material A

2xE + xI + s2 = 8 Raw Material B

– xE + xI + s3 = 1 Demand x1 #1

xI + s4= 2 Demand x1 #2

Put all the equation in the Table (Tableau)

Use Gauss-Jordan Method (Swapping)

Primal Simplex Method:The Tableau

21000100s4

101001-10s3

80010120s2

60001210s1

00000-2-31z

Inter-

cept

solutions4s3s2s1xIxEzBasic

This column can be omitted

-2100010s4

-101001-1s3

48001012s2

66000121s1

00000-2-3z

Inter-

cept

solutions4s3s2s1xIxEBasic

Entering Column

Pivot

Equa-tion(PE)

Pivot ElementLeaving Variable

Gauss – Jordan

1. Pivot Equation

New Pivot Equation (NPE)= Old PivotEquation : Pivot Element

2. All other equations, including z

New Equation = (Old Equation) – (itsentering column coefficient NPE)

s4

s3

4001/201/21xE

s1

z


Pivot Element = 2

NPE

4001/201/21NPE

00000-2-3Z

(OLD)

Coefficient xE for z = -3

-1200-3/20-3/2-3-3NPE

-1200-3/20-3/2-33NPE

12003/20-1/20Z

(NEW)

z

s4

s3

4001/201/21xE

s1

12003/20-1/20z


4001/201/21NPE

Coefficient xE for s1 = 1

4001/201/211NPE

s1

6000121s1 (OLD)

4001/201/211NPE

200-1/213/20s1

(NEW)

4

2

12

solution

s4

s3

001/201/21xE

00-1/213/20s1

003/20-1/20z

s4s3s2s1xIxEBasic

4001/201/21NPE

Coefficient xE for S3 = -1

-400-1/20-1/2-1-1NPE

s3

5011/203/20s3

(NEW)

101001-1s3(OLD)

-400-1/20-1/2-1-1NPE

5

4

2

12

solution

s4

011/203/20s3

001/201/21xE

00-1/213/20s1

003/20-1/20z

s4s3s2s1xIxEBasic

4001/201/21NPE

Coefficient xE for S4 = 0

00000000NPE

s4

2010010s4

(NEW)

2100010s4

00000000NPE

2

5

4

2

12

solution

010010s4

011/203/20s3

001/201/21xE

00-1/213/20s1

003/20-1/20z

s4s3s2s1xIxEBasic

Result of Iteration 1

22010010s4

10/35011/203/20s3

84001/201/21XE

4/3200-1/213/20s1

12003/20-1/20z

Inter-

cept


Entering Column

Pivot

Equa-tion(PE)

Pivot ElementLeaving Variable

s4

s3

xE

4/300-1/32/310xi

z


Pivot Element = 3/2

NPE

4/300-1/32/310NPE

12003/20-1/20Z

(OLD)

Coefficient xI for z = -1/2

-2/3001/6-1/3-1/20-1/2NPE

12 2/3004/31/300Z

(NEW)

z

-2/3001/6-1/3-1/20-1/2NPE

s4

s3

xE

4/300-1/32/310xi

12 2/3004/31/300z


4/300-1/32/310NPE

4001/201/21xE

(OLD)

Coefficient xi for x3 = 1/2

-2/300-1/61/31/201/2NPE

10/3002/3-1/301xE

(NEW)

xE

-2/300-1/61/31/201/2NPE

s4

s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 +2/3004/31/300z


4/300-1/32/310NPE

5011/203/20s3

(OLD)

Coefficient xi for s3 = 3/2

200-1/213/203/2NPE

3011-100s3

(NEW)

s3

200-1/213/203/2NPE

s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


4/300-1/32/310NPE

2010010s4

(OLD)

Coefficient xi for s4 = 1

4/300-1/32/3101NPE

2/3011/3-2/300s4

(NEW)

s4

4/300-1/32/3101NPE

2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


Result of Iteration 2 = END

Interpretation of the SimplexTableau

Information from the tableau Optimum Solution

Status of Resources

Dual Prices (Unit worth of resources) and ReducedCost

Sensitivity of the optimum solution to the changes ofin availability of resources

marginal profit/cost (objective function coefficients)

The usage of the resources by the model activities

Interpretation of the SimplexTableau: Optimum Solution

2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


Interpretation of the SimplexTableau: Optimum Solution

The Reddy Mikks Company

Optimum Solution x1 = 4/3

x2 = 10/3

Objective Function:

Maximize z = 3XE + 2XI

z = 3(4/3) + 2(10/3) = 12 2/3

Interpretation of the SimplexTableau: Status of Resources

abundants > 0

scarces = 0

Status of ResourceSlack Variable


2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z



abundantLimit of demand forinterior paint

s4 = 2/3

scarceRaw material Bs2 = 0

scarceRaw material As1 = 0

s3 = 3

SlackVariable

abundantLimit of excess ofinterior over exteriorpaint

Status ofResource

Resource


Which of the scarce resources should begiven priority in the allocation of additionalfunds to improve profit mostadvantageously?

Compare the dual price of the scarceresources

Interpretation of the SimplexTableau: Dual Price

2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


Dual Price

y4 = 0Limit of demandfor interior paint

s4 = 0

y2 =4/3 thousanddollars per ton ofmaterial B

Raw material Bs2 = 4/3

y1 = 1/3 thousanddollars per ton ofmaterial A

Raw material As1 = 1/3

s3 = 0

SlackVariable

y3 = 0Limit of excess ofinterior overexterior paint

Dual Price function(y)

Resource

Interpretation of the SimplexTableau: Maximum Change inResource Availability

Maximum Change in Resource Availabilityof Resource i = Di

Two Cases:

Di > 0

Di < 0

4/326xi

12 2/3120z

2

(Optimum)10

10/348xE

Right-Side Element (Solution) inIteration

Equation

2/322s4

351s3

4/326xi (raw material A)

12 2/3120z

2

(Optimum)10

10/348xE (raw material B)

Right-Side Element(Solution) in Iteration

Equation

2/322s4 (Demand xI #2)

351s3 (Demand xI #1)

4/3 + ?2 + D16 + D11 (raw material A)

12 2/3 + ?120z

2

(Optimum)10

10/3 + ?482 (raw material B)

Right-Side Element(Solution) in Iteration

Equation

2/3 + ?224 (Demand xI #2)

3 + ?513 (Demand xI #1)

2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


Coefficients for D1

Right-Side Element (Solution)in Iteration

Equation

4/3 + 2/3 D12/31 (raw material A)

12 2/3 + 1/3 D11/3z

2

(Optimum)s1

10/3 – 1/3 D1-1/32 (raw material B)

2/3 – 2/3D1-2/34 (Demand xI #2)

3 - 1D1-13 (Demand xI #1)

D1 -2D1 1Overall

SatisfiedD1 12/3 – 2/3D1 0

Case

D1 < 0D1 > 0

D1 3

D1 10

Satisfied

-2 D1 1

3 - 1D1 0

10/3 – 1/3 D1 0

4/3 + 2/3 D1 0

Satisfied

Satisfied

D1 -2

-2 D1 1

Any change outside this range (i.e.decreasing raw material A by more than 2tons or increasing raw material A by morethan 1 ton) will lead to infeasibility and a newset of basic variables (Chapter of SensitivityAnalysis)

Maximum Change in ResourceAvailability

-2 D1 1 feasible solution

Any change outside this range (i.e. decreasingraw material A by more than 2 tons or increasingraw material A by more than 1 ton) will lead toinfeasibility and a new set of basic variables(See Chapter of Sensitivity Analysis)

Max and Min of raw material A with dual price y1

= 1/3 when Max = 6 + 1 = 7 and Min = 6 – 2 = 4

Interpretation of the SimplexTableau: Maximum Change inMarginal Profit/Cost Changing the coefficients in z-row Case 1: in accordance with basic variables

Case 2: in accordance with non-basicvariables

Maximum change in marginal profit/cost =di

For exampled1 : Maximum change profit in

accordance with xE

Interpretation of the SimplexTableau: Maximum Change inMarginal Profit/Cost

2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3004/31/300z


2/3011/3-2/300s4

3011-100s3

10/3002/3-1/301xE

4/300-1/32/310xi

12 2/3+10/3 d1004/3+2/3 d11/3 - 1/3d100z


Case 1 : Basic Variables

Objective Function: z = 3xE – 2xI

Objective Function: z = 3xE – 2xI

Coefficient xE = cE =31/3 – 1/3 d1 0 d1 1

4/3 + 2/3 d1 0 d1 –2

– 2 d1 1 3 – 2 cE 1 + 1 1 cE 4

1 cE 4

Current optimum remains unchanged for the rangeof cE [1. 4]

The value of z will change according to theexpression: 12 2/3+10/3 d1, – 2 d1 1

Case 2: Non - Basic Variables

Changes in their original objectivecoefficients can only affect only their z-equation coefficient and nothing else

Corresponding column is not pivoted

In general, the change di of the originalobjective of a non-basic variable alwaysresults in decreasing the objectivecoefficient in the optimum tableau by thesame amount


Objective Function


cI = 2 cI = 2 + d2

Coefficient xI

1/2 – d2 0 d2 1/2

cI 2 + 1/2

cI 5/3

2100010s4

5011/203/20s3

4001/201/21xE

200-1/213/20s1

20005/201/20z


Maximize z = 5xE + 2xI OptimumSolution

ReducedCost

Interpretation of the SimplexTableau: Reduced Cost

Reduced Cost = the optimum objectivecoefficients of the non-basic variables

Reduced Cost = net rate of decrease inthe optimum objective value resulting fromincreasing of the associated non-basicvariable


Reduced Cost = cost of resources toproduce per unit (xi) input – its revenueper unit output

Reduced Cost > 0 cost > revenue

No economic advantage in producing theoutput

Reduced cost < 0 cost < revenuecandidate for becoming positive in theoptimum solution


2100010s4

5011/203/20s3

4001/201/21xE

200-1/213/20s1

20005/201/20z



OptimumSolution

ReducedCost


Optimum objective function

z + 1/2 xI + 5/2 s1 = 20

z = 20 – 1/2 xI – 5/2 s1


Reduced Cost = the optimum objectivecoefficients of the non-basic variables

Reduced Cost = net rate of decrease inthe optimum objective value resulting fromincreasing of the associated non-basicvariable


Reduced Cost = cost of resources toproduce per unit (xi) input – its revenueper unit output

Reduced Cost > 0 cost > revenue

No economic advantage in producing theoutput

Reduced cost < 0 cost < revenuecandidate for becoming positive in theoptimum solution


2100010s4

5011/203/20s3

4001/201/21xE

200-1/213/20s1

20005/201/20z



OptimumSolution

ReducedCost


Optimum objective function

z + 1/2 xI + 5/2 s1 = 20

z = 20 – 1/2 xI – 5/2 s1

Interpretation of the SimplexTableau: Reduced Cost Unused economic activity = non-basic variable

can become economically viable in two ways:(Logically)

#1 by decreasing its per unit use of the resource

#2 by increasing its per unit revenue (throughprice increase)

Combination of the two ways

#1 is more viable than #2 since #1 is inaccordance with efficiency

while #2 is related to market condition in whichother factors influence

The End

This is the end of Chapter 3A

Linear Programming (LP) - Gunadarma

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