LP - I Lecture

7/30/2019 LP - I Lecture

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Linear Programming(An Optimization Technique)

General Optimization Problem I nvolves :

1. Decision Variables

2. Objective Criterion (Max/Min)

3. Constraints: (i) Equality (ii) Inequality

L inear Programming(LP) : It is one of the important Optimization

Techniques. Most Versatile, powerful & useful technique. Developed in 1947 by George B. Dantzig during

World War II to solve military problems, whileworking with US AirForce.


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(1)xcMin.)Zor(Max.Optimizen

1j

jj

The word Linear in LP indicates -relationship among variables both in objectivefunction and constraints.

while Programming means - SystematicMathematical technique.

General Linear Programming Problem (LPP)

with n variables and m constraints can be statedas follows:

)2(...2,1;),,(1

mibxa i

n

j

jij

)3(...2,1;0 njxj


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Types of Variables Decision Variables : xj are decision variables

Slack Variables

mibwxa in

j

ijij ...2,1;1

+

bxan

jijij

1

wi are slack variables.


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Artificial Variables :Required for solution methodology

Surplus Variables

bxa

n

jijij

1

mibsxa in

jijij ...2,1;

1

-

si are surplus variables


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Components of LPP :

The three important components of LPP

are as follows:

(a) Decision Variables

(b) Objective Function - Max. or Min.

(c) Constraints-limitations.


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1. Certainty:cj, aij, bi are known & constants.

2. Divisibility: Values of decision variable

can be integer or fractional.

3. Additivity: Total contribution = Sum of

contribution of all variables4. Linearity:Relationship among variables both

in objective function & constraints.

Assumptions of LP


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Formulation of LPP :

Steps to be followed.

1.To decide (define) decision variables.

2. To formulate objective function.

3. To formulate constraints.


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CASE - 1 Data for LP Formulation.

Factory: Two products P1 & P2

Profit/piece of P1 = Rs.3/-

Profit/piece of P2 = Rs.5/-

Time Constraint:

Time required/piece of P1 = 3 hrs.

Time required/piece of P2 = 2 hrs.

Max. time available = 18 hrs.

Material Constraint:

Max. Material available for P1 = 4 pieces.

Max. Material available for P2 = 6 pieces.


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CASE - 1 Formulation of LPP :

Let decision variable x1 = no of pieces of P1

x2 = no of pieces of P2

Max. Z = 3x1 + 5x2 Objective Equation

ConstraintTimeIxx )(182213 -+

ConstraintMaterialIIx )(41 -

ConditionsnegativityNonxx - 02,1

ConstraintMaterialIIIx )(62 -


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CASE - 2

If Time Constraint is modified to:

Time available is not less than 18 hours.







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CASE - 3

If Time Constraint is modified to:

Time available is exactly 18 hours.







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1. Feasible Solution: Solution with values of

decision variables(xj) which satisfy allConstrints & Non-negativity condition.

2. Basic Solution: For a set of m equations

in n variables (n>m), basic solution is a

solution obtained by setting (n-m)

variables equal to zero and solving for

remaining m equations in m variables.

Types of Solution

No. of possible Basic solutions = nCm

= n!/(m!.(n-m)!)


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(i) Basic Variables : Having values > 0.

(ii) Non-basic Variables :Having values = 0.

3. Basic Feasible Solution : It is a BasicSolution which satisfies (3).

4. Non-degenerate/degenerate B.F.S

5. Optimal Basic Feasible Solution : It is a

Basic Feasible Solution which optimizes

objective function.

6. Unbounded Solution :

7. Infeasible Solution :

8. Unique/Alternative Optimal Solutions :


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Methods to solve LPP :

1.Graphical Method - for only two variables.2. Simplex Method - Universal method.

3. Assignment Method - Special method.4. Transportation Method - Special method.

Note :Methods (2),(3) and (4) are iterative

methods.


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(1) Graphical Method

( for only two variables)


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3

6

9

2 4 6 x1

F.R.

Opt. Pt. (2,6)

Z=0.

Z=15.

Zopt = 36.

III

I

II

02,1

)(62)(41

)(182213/

2513

--

-+

+

xx

IIIxIIx

Ixxts

xxZMaxGraphical Method

Case 1.


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3

6

9

2 4 6 x1

F.R.

Opt. Pt.(4,6)

Z=15.

Zopt = 42

III

I

II02,1

)(62

)(41

)(182213/

2513

-

-

-+

+

xxIIIx

IIx

Ixxts


Case 2.


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3

6

9

2 4 6 x1

F.L.

Opt. Pt.(2,6)

Z=15.

Zopt = 36

III

I

II

02,1

)(62)(41

)(182213/

2513

-

-

-+

+

xx

IIIxIIx

Ixxts


Case 3.


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3

6

9

2 4 6 x1

F.R.

III

I

II

Graphical Method

Special Case.

Alternative OptimalSolutions


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3

6

9

2 4 6 x1

F.R.(Unbounded)

Opt. Pt.

(Min.)

III

II

I

GraphicalMethod

Special Case.

Z line

Opt. Sol. (Max)

(Unbounded)


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3

6

9

2 4 6 x1

F.R.Not

possible

InfeasibleSolution

GraphicalMethod

Special Case.


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4

5

4 5 x1

x2

20-2-3

2

-1

-1

-2

-3

-4

III

III

F.R.

Z=6

Opt. Pt. (0.5,

4.5)

Zopt = 12.5

02,1

)(521

)(3213

)(4212/

2312

-+

-+-

--

+-

xx

IIIxx

IIxx

Ixxts

xxZMax

GraphicalMethod

Special Case.


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Graphical Methodtodiscuss typical constraints

to identify regions of

constraints.

Typical Case ?62213 +l ttH


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4 5 x1

x2

0-2-3 -1

-1

-3

-2

1

2

3

21 3

Typical Case ?62213 -+ xxplottoHow

62213 -+ xx

),Point(xxWhen 022102 --

)Point(xxWhen 3,03201 --

Typical Case ?62213 l ttH


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2

3

4 5 x1

x2

10-2-3

1

-1

-1

-2

-3

2 3

Typical Case ?62213 - xxplottoHow

62213 - xx

),Point(xxWhen 022102

)Point(xxWhen 3,03201 --

Typical Case ?021 +l ttH


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4 5 x1

x2

0-2-3 -1

-1

-3

-2

1

2

3

21 3

Typical Case ?021 + xxplottoHow

021 +xx

21 xx - )Point(xxWhen 1,11211 --arrowFor

0102 xx


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Recapitulate

What is LPP ? Types of variables.

Components of LPP.

Assumptions of LP. Formulation of LPP.

Types of solutions.

Graphical Method.

Unique & Alternative Optimal Solution,

Unbounded FR, Infeasible solution.


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(2) Simplex Method

(Universal method)


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Simplex Method to solve

Max. Problem with All constraints.


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02,1

)(62

)(41

)(182213/

2513

-

-

-+

+

xx

IIIx

IIx

Ixxts

xxZMax

Standard Form:

Max Z = 3x1+5x2+0w1+0w2+0w3

3x1+2x2+w1+0w2+0w3 = 18

x1+0x2+0w1+w2+0w3 = 4

0x1+x2+0w1+0w2+w3 = 6

Simplex Method

Case 1.

To

prepare initial Tableau:


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ci xi bi x1 x2 w1 w2 w3

Ij 3 5 0 0 0

Ij = (cj -Ej) = Cj- (

aij.ci)

cJ 3 5 0 0 0

Toprepare initial Tableau:

Tableau - I

0 w1 18 3 2 1 0 0

0 w2 4 1 0 0 1 0

0 w3 6 0 1 0 0 1

Z = bi

.ci

= 0

Ej 0 0 0 0 0

Tableau I


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Ij Z = 0 3 5 0 0 0

cj 3 5 0 0 0

Tableau - I

0 w1 18 3 2 1 0 0

0 w2 4 1 0 0 1 0

0 w3 6 0 1 0 0 1

Ratio18/2 = 9

4/0 =

6/1 = 6

Key Column Max +ve Ij Key Row Min positive ratio.


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How to get next tableau ?

Leaving variable : w3

Entering variable : x2

If the key element is 1, then key row remain same inthe new simplex tableau

If the key element is other than 1, then divide eachelement in the key row (including b value) by the keyelement to find new values of that row

Make the other elements in the key column equal to0 by performning elementary row operations withnew key row obtained as above.

T bl I


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ci xi bi x1 x2 w1 w2 w30 w1 18 3 2 1 0 0

0 w2 4 1 0 0 1 0

0 w3 6 0 1 0 0 1

Ij Z = 0 -3 -5 0 0 0

Ratio18/2 = 9

4/0 =

6/1 = 6

cj 3 5 0 0 0Tableau - I

18 18 - (6*2)/1 = 6

R1 2R3 gives

2 2-2(1) = 0 , 3 3- 2(0) = 3 and 1 12(0) = 1

T bl I I


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0 w1 6 3 0 1 0 -2

0 w2 4 1 0 0 1 0

Ij Z = 30 3 0 0 0 -5

Ratio6/3 = 2

4/1 = 4

6/0 =

cj 3 5 0 0 0

Key Column Max +ve Ij

Key Row Min positive ratio.

Tableau - I I

5 x2 6 0 1 0 0 1


Tableau I I I


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3 x1 2 1 0 1/3 0 -2/3

Ij Z = 36 0 0 - 1 0 - 3

cj 3 5 0 0 0Tableau - I I I

5 x2 6 0 1 0 0 1

This is the final Tableau.

The Optimal Solution is x1 = 2, x2 = 6

giving Z = 36


0 w2 2 0 0 -1/3 1 2/3


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Interpretation of

Simplex Method through

Graphical method.

I nterpretation of Simplex Method


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3

6

9

2 4 6 x1

F.R.

Opt. Pt. (2,6)

Z=15.

Zopt = 36.

III

I

II

02,1

)(62

)(41

)(182213/

2513

-

-

-+

+

xx

IIIx

IIx

Ixxts

xxZMax

I nterpretation of Simplex Method

Through Graphical Method

T1

T3

T2

GATE 2002


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GATE - 2002(1) A furniture manufacture produces Chairs & Tables.

The wood working department is capable of producing

200 chairs or 100 tables or any proportionatecombinations of these per week. The weekly demand

for chairs and tables is limited to 150 and 80 units

respectively.The profit from a chair is Rs. 100 and that

from a table is Rs. 300.

Set up the problem as a Linear Program.

Determine optimal product mix and optimal valueof objective function.

If a profit of each table drops to Rs. 200 per unit,

what is the product mix and profit ?

(a)

(b)

(c)

GATE 2002


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GATE - 2002

MaxZ = 100x1+ 300x2

S/t1

100

2

200

1+

xx

x1 150

x2 80

x1,x2 0

Formulation(a)

(b) Optimal Solution


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(40, 80) Z = 28000

Z = 15000

100

200

150

50

80

x2

x1

(b) Optimal Solution

(c) Multiple Optimal Solutions


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(40, 80) Z = 20000

Z = 15000

100

200

150

50

80

x2

x1

75 Opt. Line

(c) Multiple Optimal Solutions

GATE 2003


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GATE - 2003(1) A manufacture produces two types of products, 1 and

2, at production levels of x1 and x2 respectively. The

profit is given 2x1 + 5x2. The production constraints

are :

The maximum profit which can meet the constraint is

(A) (D)(C)(B)29 38 44 75

x1 + 3x2 40

3x1 + x2 24

x1 + x2 10

x1 0, x2 0


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40

x2

x1

10

10

20

Z = 50

Opt. Pt. (0,10) Z = 50

Ans. = (C) 44

25

GATE 2000


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GATE - 2000

Max Z = 4x1+ 6x2 + x3

S/t

x1, x2, x3 0

2x1+ x2 + 3x3 5

Solve :

If x2 2 is added then what will be the

solution ?

GATE 2000


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GATE - 2000

Solution :

x1 = 0, x2 = 5, x3 = 0 giving z = 30

Through Simplex Method, in

two iterations the solution ofbasic problem is :

If x2 2 is added then the solution through

Simplex Method, in three iterations will be :

x1

= 3/2, x2

= 2, x3

= 0 giving z = 18

GATE - 2008


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GATE 2008Max Z = 4x1

+ 6x2

x1, x2 0

3x1+ 2x2 6

2x1+ 3x2 6

Q.1 After introducing slack variables w1 and w2, the initial

feasible solution is represented by the tableau below.

ci xi bi x1 x2 w1 w2

Ij 0 -4 -6 0 0

cj 4 6 0 0

0 w1 6 3 2 1 0

0 w2 6 2 3 0 1

GATE - 2008


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GATE 2008

After some Simplex interactions, the following tableau is obtained.

ci xi bi x1 x2 w1 w2

Ij 12 0 0 0 2

cj 4 6 0 0

0 w1

2 5/3 0 1 -1/3

6 x2 2 2/3 1 0 1/3

LP - I Lecture

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