chapter 2 lp

8/8/2019 Chapter 2 LP

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Lecture 2


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1. Write down the problem preferablyin a tabular form.

2. Formulate the objective andconstraint functions.

3. For maximization problem allconstraints should be

4. For minimization problem, all

constraints should be 5. If required, multiply the constraint

equation with -1

6. Standardize the problemIntroduce slack variables to converteach constraint to an equality.

Add artificial variables to constraints tohave a starting solution. Normally

required for constraints with sign


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11.Identify the key element onecommon toboth key columnandkey row. Non-basic variable

corresponding to Key columnwillreplace the Basic variablecorresponding to the Key Rowandprepare a Revised tableu

12.Perform row operations tocomplete the new solution:

Write down Replacement Rowbydividing each element of the Key Row

with Key ElementFor each rowother than the Key Row,find the values using the formula:New Row Element = Old Row Element

less Replacement Row value x Row

Element in Key Column.


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13.Check solution for optimality.

To be optimal, ij 0 for allvariables for maximization

problem.and ij 0 for minimizationproblem.

14. If optimal, stop. If not go tostep 9 to prepare a new tablefor improved solution


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No artificial variable present in thebasis

It is a feasible solution

It is a feasible solution and the ijvalues of all non-basic variables

are 0 for a max problem (or 0for a min problem)

The solution is optimal

It is an optimal solution and the ijvalues of all non-basic variablesare < 0 for a max problem (or > 0

for a min problem)The solution is optimal and unique


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It is an optimal solution and the ijvalue of one or more non-basicvariables is equal to zero

The solution is not unique: theproblem has multiple optimalsolutions

It is final in terms of the ij valuesand has an artificial variable in thebasis

It indicates infeasibility

It is non-optimal and the elements ofthe key column are all zero/negative

The problem has unboundedsolution


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1. Non-negative bi values

Multiply a constraint on both sidesby -1 if it involves a negative bivalue

2. Non-negative variables

Replace every unrestricted

variable by difference of two non-negative variables


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(continued)

Maximize Z = 8x1 + 5x2 +12x3Subject to 2x1 + 5x2 + 4x3 44

3x1 7x2 6x3 10

7x1 + 2x2 +4x3 = 54

x1,x2 0 x3: unrestricted insign

Should change as follows:

(i) Letx3 =x4 x5, wherex4,x5 0(ii) 3x1 7x2 6x3 10 be replacedas

3x1 + 7x2 + 6x3 10

Revised LPP:

Maximize Z = 8x1 + 5x2 +12x4 12x5Subject to 2x1 + 5x2 + 4x4 4x5 44

3x1 + 7x2 + 6x4 6x5 10

7x1

+ 2x2

+ 4x4

4x5

= 54

x1, x2, x4, x5 0


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Obtain a revised solution(steps)

1. Select a variable to enter toimprove the solution.

2. Select a variable to leave thesolution.

3. Perform row operations to

complete the new solution.4. Repeat above steps untiloptimality is achieved


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Example 3.2 dataMax Z = 5x1 + 10x2 + 8x3 Contribution

St 3x1 + 10x2 + 2x3 60 Fabrication Hrs

4x1

+ 4x2

+ 4x3

72 Finishing Hrs

2x1 + 4x2 + 4x3 100 Packaging Hrs

x1, x2, x3 0

Conditions for application of simplexmethod are both satisfied here

We need 3 slack variables to convertthe inequalities in to equations

The problem becomes:

Max Z = 5x1 + 10x2 + 8x3 + 0S1 + 0S2 + 0S3St 3x1 + 10x2 + 2x3 + S1 = 60

4x1 + 4x2 + 4x3 + S2 = 72

2x1 + 4x2 + 4x3 + S3 = 100

x1, x2, x3, S1, S2, S3 0


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Basis x1 x2 x3 S1 S2 S3 bi bi/aij

S1 0 3 5* 2 1 0 0 60 12

S2 0 4 4 4 0 1 0 72 18

S3 0 2 4 5 0 0 1 100 25

Cj 5 10 8 0 0 0

Sol 0 0 0 60 72 100 Z=0

j 5 10 8 0 0 0

Simplex Tableau 1

Includes variables that

yield identity matrix

Key element

Key rowKey column


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Basis x1 x2 x3 S1 S2 S3 bi bi/aij

x2 10 3/5 1 2/5 1/5 0 0 12 30

S2 0 8/5 0 12/5 -4/5 1 0 24 10

S3 0 -2/5 0 17/5 -4/5 0 1 52 2 0/17

Cj 5 10 8 0 0 0

Sol 0 12 0 0 24 52 120

j -1 0 4 -2 0 0

Simplex Tableau 2

To derive values in the next tableau,

Divide each element of key row by

key element: 3/5, 5/5, 2/5, 1/5 etc.

For other rows (for example row 2):

4 4 3/5 = 8/5

4 4 1 = 0

4 4 2/5 = 12/5

0 4 1/5 = -4/5 etc.


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Basis x1 x2 x3 S1 S2 S3 bi

x2 10 1/3 1 0 1/3 -1/6 0 8

x3 8 2/3 0 1 -1/3 5/12 0 10

S3 0 -8/3 0 0 1/3 -17/12 1 18

Cj 5 10 8 0 0 0

Sol 0 8 10 0 0 18 160

j -11/3 0 4 -2/3 -5/3 0

Simplex Tableau 3: Optimal Solution

Solution optimal as all j 0

Optimal Product mix: x1 = 0, x2 = 8 and x3 = 10

Maximum contribution = Rs 160

Capacity utilization: Fabrication and Finishing Full; Packaging 18 hours unutilized

No production ofx1 since every unit of it producedwould result in a loss of Rs 11/3

Production of a unit of x1 will result in losing 1/3

unit ofx2 and 2/3 unit ofx3 together with a releaseof 8/3 hours of Packaging time


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1. Standardise the problem

Introduce surplus and artificial variables

for all constraintsIntroduce artificial variables for all =constraints

In the objective function assign co-efficients to each of the artificial

variables [For max problems: assign Mand for min problems: assignM]

2. Obtain initial solution (withslack/artificial variables)

3. Test if solution is optimal (to beoptimal, ij 0 for all variables)

4. If optimal, stop and exit; else go to step5

5. If not, obtain a revised solution and goback to step 3


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Needed to be used to obtain initialsolution

Introduced where a constraint is of

or = type

May also be required in maximisationproblems

Expected to leave the basis one-by-one

in successive simplex iterations

Solution infeasible as long as one ormore artificial variables present in thebasis

Not expected to be present in the basisof the final solution

In case artificial variable present in basisof final solution, there is infeasibility


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Phase I

1. Standardise the problemFor each artificial variable, assignco-efficients in the objectivefunction

For max problem: assign 1

For min problem: assign 1

For every other variable, assign co-efficient of zero

2. Solve by Simplex Method. If theobjective function value for

optimal solution is zero, move toPhase II


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Phase II

1. Begin with Simplex Tableau ofPhase I final solution, eliminate allentries for artificial variables andreplace the zero co-efficients ofdecision variables by original co-efficient values

2. Solve the modified problem bySimplex Method


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A solution in which a basic variable hassolution value equal to zero isdegenerate solution

The basic variable which has solutionvalue of zero is called degeneratevariable

Whenever there is a tie in the

replacement ratios (bi/aij) to beselected, the next solution will be adegenerate solution

A degenerate solution may or may not

be optimalRevision ofa non-optimal solution leadsto no improvement (in terms of the Z-value) if the degenerate variable is the

outgoing variable


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Basis x1 x2 S1 S2 S3 bi bi/aij

S1 0 6 3 1 0 0 18 6

S2 0 3 1 0 1 0 8 8

S3 0 4 5 0 0 1 30 6

Cj 28 30 0 0 0

Sol 0 0 18 8 30

j 28 30 0 0 0

Simplex Tableau 1

Max Z = 28x1 + 30x2Stt 6x1 + 3x2 18

3x1 + x2 8

4x1 + 5x2 30

x1, x2 0With slack variables S1,S2 and S3, solution follows:

Next solutionwouldbe degenerate

Tie

Key column


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(continued)

Basis x1 x2 S1 S2 S3 bi bi/aij

S1 0 18/5 0 1 0 -3/5 0 0

S2 0 11/5 0 0 1 -1/5 2 10/11

X2 30 4/5 1 0 0 1/5 6 15/2

Cj 28 30 0 0 0

Sol 0 6 0 2 0 180

j 4 0 0 0 -6

Simplex Tableau 2: Non-optimal Solution

Basis x1 x2 S1 S2 S3 bi

x1

28 1 0 5/18 0 -1/6 0

S2 0 0 0 -11/18 1 1/6 2

x2 30 0 1 -2/9 0 1/3 6

Cj 28 30 0 0 0

Sol 0 6 0 2 180

j 0 0 -10/9 0 -16/3

Simplex Tableau 3: Optimal Solution

Degenerate outgoing variable


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(continued)

A tie in the minimum replacementratios in Tableau 1 results in the solutionin next tableau to be degenerate (even

if the outgoing variable was chosen tobe S1 instead of S3)

Since all values in the key column arepositive, the least ratio to be considered

is 0 and the outgoing variable (S1) isdegenerate

As a result, there is no change in theobjective function value: it remains at

180The solution in Tableau 3 is alsodegenerate

While the solution in Tableau 2 is non-

optimal, the solution in Tableau 3 isoptimal


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Mark the wrong statement:

1. An LPP with n variables and mconstraints gives nCm basic solutions.

2. A basic solution with all non-negative variables is called basicfeasible solution.

3. For any LPP, there is a uniqueextreme point of the feasible regioncorresponding to every basic

solution.

4. The basic feasible solution thatmaximises or minimises, as the case

may be, is called optimal solution.


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Mark the correct alternative.

A feasible solution is the one which

1. satisfies all constraints of theproblem.

2. is necessarily an optimal solution.

3. makes use ofall available resources.

4. yields more than one way to achievethe objective.


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Which of the following statementsis not true about application ofsimplex method?

1. The RHS of each constraint shouldbe greater-than-or-equal-to zero.

2. All decision variables of the problemshould be non-negative.

3. All constraints of the given problemshould be either e or u type.

4. All constraints should be convertedinto = type, with slack/surplusvariables which, like decisionvariables, are also non-negative.


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Choose the incorrect statement:

1. Simplex method is an iterativeprocess wherein variables aresubstituted until optimal solution isreached.

2. Successive simplex tableaus always

yield better solution in terms of theobjective function.

3. For a minimisation problem, theoptimal solution is reached when all

(j u 0, with no artificial variable inthe basis.

4. The key column indicates theincoming variable and the key row

represents outgoing variable.


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If(j = cj zj is equal to 24 7M forx1; 28 7M for x2; 45 4M for x3and 28 2M in respect of x4 in a

minimisation problem, then theentering variable would be:

1. x1

2. x2

3. x3

4. x4


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1. First Key element can never benegative.

2. Key element cannot be zero.

3. Key element has to be positive.

4. Key element can be negative, zero or

positive.


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The bi values and the elements inkey column are (12, 0, 18, 6)and (-2, 8, 1, 3) respectively.

Which of the rows will be selectedas the key row?

1. First

2. Second

3. Third

4. Fourth


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Which of these is not true?

1. All basic variables in an LPP have (j =0.

2. A slack variable cannot be there inthe basis ofoptimal solution.

3. Ifa problem has an optimal solution,the artificial variables, if any,introduced must be driven out ofthe basis one-by-one.

4. An outgoing variable in a simplextableau cannot re-enter the basis infollow-up iterations.


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Which of the following statementsis not true about artificialvariables?

1. They are introduced when initialsolution to a problem cannot beobtained for want of identity sub-matrix in the simplex tableau.

2. Each of such variables has to beassigned a very large negative co-efficient in the objective functionwhen it is of the maximisation type.

3. They help to obtain an initial feasiblesolution to the problem.

4. They bear no tangible relationshipwith the decision problem.


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1. Ifa non-basic variable in the optimal

solution to a problem has (j = 0, itindicates multiple optimal solutions.

2. Ifan artificial variable is found in thebasis of the final solution to an LPP,

it implies feasibility.3. A constraint with e sign does not

require artificial variable to beintroduced.

4. A slack variable is used to convert utype and a surplus variable is usedto convert e type of constraint intoequality.


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1. If there are no non-negativereplacement ratios in a simplex

tableau, then unboundness isindicated.

2. In a tableau, the key columnelements are 5, 0, 0, 7 while the

corresponding bi values are 20, 6, 8and 0. This indicates unboundedsolution.

3. A tie in the minimum replacement

ratio implies that the next solutionwould be degenerate.

4. It is possible for the initial solutionto an LP to be degenerate.


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Which of these is not correct?

1. The solution to a maximisation LPPis unique if (j values for all non-

basic variables are less than zero.2. If infeasibility is present, it is

detected in phase I of the two-phasemethod.

3. In two-phase method, the artificialvariables are each assigned a co-efficient of 1 in case of minimisation,and 1 in case of maximisation

problems.4. If the given problem has an optimal

solution, the artificial variables areall removed one by one in phase I ofthe two-phase method.


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If there is no non-negativereplacement ratio in a solutionwhich is sought to be improved,then the solution is indicated to be

1. infeasible

2. unbounded

3. degenerate

4. unique optimal


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Which of the f ollowing is notcorrect?

1.A

s and when a degeneratevariable is outgoing, no changewill take place in the objectivefunction value.

2. Degeneracy may be a temporaryphenomenon.

3. It is possible for the repetition ofsame tableaus in the course of

successive iterations whendegeneracy is encountered.

4. A degenerate solution cannot beoptimal.


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Degeneracy in LPP (1) renders thesolution infeasible, (2) leads tomultiple optimal solutions, (3)increases computations withoutaffecting the solution as long asthe outgoing variable happens to

be degenerate. Which of thesestatements is/are correct?

1. 1, 2 and 3

2. 1 and 3

3. 2 and 3

4. 3 only

chapter 2 lp

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