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Linear Programming

Graphical method

Submitted by

Priyanka shrivastavaISBE -B

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Agenda

Formulation Of LPP

Graphical method-1. Maximization case

2. Minimization case

Feasible region.

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LP Model Formulation

Decision variables mathematical symbols representing levels of activity of an operation

Objective function a linear relationship reflecting the objective of an operation

most frequent objective of business firms is to maximize profit

most frequent objective of individual operational units (such as a

production or packaging department) is to minimize cost

Constraint a linear relationship representing a restriction on decision making

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LP Model Formulation

Max/min z = c1x1 + c2x2 + ... + cnxn

subject to: a11x1 + a12x2 + ... + a1nxn(, =, ) b1a21x1 + a22x2 + ... + a2nxn(, =, ) b2

:

am1x1 + am2x2 + ... + amnxn(, =, ) bm

xj = decision variables

bi = constraint levels

cj = objective function coefficients

aij = constraint coefficients 3

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Graphical Method

Steps -

1. Identify the problem

- The decision variables

- The objective function

- The constraints restrictions

2. Draw a graph & Identify the feasible region.

3. Obtain the point on the feasible region.

4. Interpret the result.

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LP Model: Example

Labor Clay Revenue

PRODUCT (hr/unit) (lb/unit) ($/unit)Bowl 1 4 40

Mug 2 3 50

There are 40 hours of labor and 120 pounds of clay available

each day

Decision variables

x1 = number of bowls to produce

x2 = number of mugs to produce

RESOURCE REQUIREMENTS

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LP Formulation: Example

Maximize Z= $40 x1 + 50 x2

Subject tox1 + 2x2 40 hr (labor constraint)

4x1 + 3x2 120 lb (clay constraint)

x1 , x2 0

Solution is x1 = 24 bowls x2 = 8 mugs

Revenue = $1,360

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Graphical Solution: Example

4 x1 + 3 x2 120 lb

x1 + 2 x2 40 hr

Area common toboth constraints

50

40

30

20

10

0 |10

|60

|50

|20

|30

|40 x1

x2

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Computing Optimal Values

x1 + 2x2 = 40

4x1 + 3x2 = 120

4x1 + 8x2 = 160

-4x1 - 3x2 = -1205x2 = 40

x2 = 8

x1 + 2(8) = 40

x1 = 24

4 x1 + 3 x2 120 lb

x1 + 2 x2 40 hr

40

30

20

10

0 |10

|20

|30

|40

x1

x2

Z= $50(24) + $50(8) = $1,360

24

8

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Extreme Corner Points

Optimalsolution-

x1 = 24 bowls

x2 = 8 mugs

Z= $1,360x1 = 30 bowls

x2 = 0 mugs

Z= $1,200

x1 = 0 bowls

x2 = 20 mugs

Z= $1,000

A

B

C|20

|30

|40

|10 x1

x2

40

30

20

10

0

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Minimization Problem

CHEMICAL CONTRIBUTION

Brand Nitrogen (lb/bag) Phosphate (lb/bag)

Gro-plus 2 4Crop-fast 4 3

MinimizeZ= $6x1 + $3x2

subject to

2x1 + 4x2 16 lb of nitrogen

4x1 + 3x2 24 lb of phosphate

x1,x2 0 10

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12

10

8

6

4

2

0 |2

|4

|6

|8

|10

|12

|14

x1

x2

A

BC

Graphical Solution

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

Z = 6x1 + 3x2

X1= 8 bags

X2=0 bags

Z= 48$

X1=1.8 bags

X2=4.4 bagsZ=31.8$

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The Graphical Analysis of Linear

Programming

The set of all points that satisfy all

the constraints of the model is calleda

FEASIBLE REGION

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Max z= 8X1 + 5X2 (Weekly profit)

subject to

2X1 + 1X2 1000 (Plastic)

3X1 + 4X2 2400 (Production Time)

X1 + X2

700 (Total production)X1 - X2 350 (Mix)

Xj> = 0, j = 1,2 (Non negativity)

The Linear Programming Model-

Example

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The non-negativity constraints

X2

X1

Graphical Analysisthe Feasible Region

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1000

500

Feasible

X2

Infeasible

ProductionTime3X1+4X22400

Total production constraint:

X1+X2 700 (redundant)500

700

Production mixconstraint:

X1-X2 350

The Plastic constraint

2X1+X2 1000

X1

700

Graphical Analysisthe FeasibleRegion

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The 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 infeasibleProfit =$4360

500

700

1000

500

X2

X1

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Infeasibility: Occurs when a model has no feasible

point.

Unboundness: Occurs when the objective can become

infinitely large (max), or infinitely small (min).

Models Without Unique Optimal

Solutions

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1

No point, simultaneously,

lies both above line and

below lines and

1

2 32

3

Infeasible Model

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Unbounded solution

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