Linear Programming Model 2 MBA PPT

7/31/2019 Linear Programming Model 2 MBA PPT

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Linear Programming

Module 2


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Introduction To Linear Programming

Today many of the resources needed asinputs to operations are in limited supply.

Operations managers must understand theimpact of this situation on meeting their

objectives. Linear programming (LP) is one way that

operations managers can determine howbest to allocate their scarce resources.


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A model consisting of linear relationships

representing a firms objective and resourceconstraints

Linear Programming (LP)

LP is a mathematical modeling technique used todetermine a level of operational activity in order toachieve an objective, subject to restrictions called

constraints


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Linear Programming (LP) in OM

There are five common types of decisions inwhich LP may play a role

Product mix

Production plan

Ingredient mix

Transportation

Assignment


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LP Model Formulation Decision variables

Depends on the type of the LP problem

These variables are controllable

The variables can be the quantities of the resources to be

allocated or the number of units to be produced or sold The decision maker has to determine the value of these

variables

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 Problems on Product Mix

Objective

To select the mix of products or services thatresults in maximum profits for the planning period

Decision Variables

How much to produce and market of each product

or service for the planning period Constraints

Maximum amount of each product or servicedemanded; Minimum amount of product orservice policy will allow; Maximum amount ofresources available


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LP Problems on Production Plan Objective

To select the mix of products or services thatresults in maximum profits for the planning period

Decision Variables

How much to produce on straight-time labor andovertime labor during each month of the year

ConstraintsAmount of products demanded in each month;Maximum labor and machine capacity available ineach month; Maximum inventory space available

in each month


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Recognizing LP ProblemsCharacteristics of LP Problems

A well-defined single objective must be

stated.There must be alternative courses of action.

The total achievement of the objective mustbe constrained by scarce resources or other

restraints.

The objective and each of the constraintsmust be expressed as linear mathematical

functions.


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Steps in Formulating LP Problems

1. Define the objective. (min or max)

2. Define the decision variables. (positive, binary)3. Write the mathematical function for the objective.

4. Write a 1- or 2-word description of each constraint.

5. Write the right-hand side (RHS) of each constraint.

6. Write for each constraint.

7. Write the decision variables on LHS of each constraint.

8. Write the coefficient for each decision variable in each

constraint.


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Cycle Trends is introducing two new lightweight

bicycle frames, the Deluxe and the Professional, to be

made from aluminum and steel alloys. The anticipatedunit profits are $10 for the Deluxe and $15 for the

Professional.

The number of pounds of each alloy needed per

frame is summarized on the next slide. A supplierdelivers 100 pounds of the aluminum alloy and 80

pounds of the steel alloy weekly. How many Deluxe

and Professional frames should Cycle Trends produce

each week?

Example: LP Formulation


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Aluminum Alloy Steel Alloy

Deluxe 2 3Professional 2 4

Pounds of each alloy needed per frame



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Define the objective

Maximize total weekly profit

Define the decision variables

x1 = number of Deluxe frames produced weekly

x2 = number of Professional frames producedweekly

Write the mathematical objective function

Max Z = 10x1 + 15x2


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Example: LP FormulationWrite a one- or two-word description of each

constraint

Aluminum availableSteel available

Write the right-hand side of each constraint

100

80

Write for each constraint

< 100

< 80


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Write all the decision variables on the left-hand side of each constraintx1 x2 < 100x1 x2 < 80

Write the coefficient for each decision in eachconstraint

+ 2x1 + 4x2 < 100

+ 3x1 + 2x2 < 80


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LP in Final Form

Max Z = 10x1 + 15x2Subject To

2x1 + 4x2 < 100 ( aluminum constraint)

3x1 + 2x2 < 80 ( steel constraint)x1 , x2 > 0 (non-negativity constraints)


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LP Formulation: Example

Maximize Z= 40 x1 + 50 x2

Subject to

x1 + 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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19

Montana Wood Products manufacturers two-high quality products, tables and chairs. Its profit is

Rs15 per chair and Rs 21 per table. Weekly production

is constrained by available labor and wood. Each chair

requires 4 labor hours and 8 board feet of wood whileeach table requires 3 labor hours and 12 board feet of

wood. Available wood is 2400 board feet and available

labor is 920 hours. Management also requires at least

40 tables and at least 4 chairs be produced for every

table produced. To maximize profits, how many chairs

and tables should be produced?



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20


Define the objective Maximize total weekly profit


x1 = number of chairs produced weekly x2 = number of tables produced weekly


Max Z = 15x1

+ 21x2


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21


Write a one- or two-word description of each constraint

Labor hours available

Board feet available

At least 40 tables

At least 4 chairs for every table


920 2400

40

4 to 1 ratio

Write for each constraint

< 920 < 2400

> 40

4 to 1


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22


Write all the decision variables on the left-hand side of eachconstraint

x1 x2 < 920

x1 x2 < 2400

x2 > 40

4 to 1 ratio x1 / x2 4/1

Write the coefficient for each decision in each constraint

+ 4x1 + 3x2 < 920 + 8x1 + 12x2 < 2400

x2 > 40

x1 4 x2


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24

The Sureset Concrete Company producesconcrete. Two ingredients in concrete are sand (costs

Rs 6 per ton) and gravel (costs Rs 8 per ton). Sand and

gravel together must make up exactly 75% of the

weight of the concrete. Also, no more than 40% of theconcrete can be sand and at least 30% of the concrete

be gravel. Each day 2000 tons of concrete are

produced. To minimize costs, how many tons of gravel

and sand should be purchased each day?



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25


Define the objective Minimize daily costs


x1 = tons of sand purchased x2 = tons of gravel purchased


Min Z = 6x1

+ 8x2


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26


Write a one- or two-word description of each constraint 75% must be sand and gravel

No more than 40% must be sand

At least 30% must be gravel


.75(2000)

.40(2000)

.30(2000)Write for each constraint

= 1500

< 800

> 600


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27


Write all the decision variables on the left-hand sideof each constraint

x1 x2 = 1500

x1

< 800

x2 > 600

Write the coefficient for each decision in each constraint

+ x1 + x2 = 1500

+ x1 < 800

x2 > 600


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LP in Final Form Min Z = 6x1 + 8x2

Subject To

x1 + x2 = 1500 ( mix constraint) x1 < 800 ( mix constraint)

x2 > 600 ( mix constraint )

x1

, x2

> 0 (non-negativity constraints)


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29

LP Problem

Galaxy Ind. produces two water guns, the Space Rayand the Zapper. Galaxy earns a profit of Rs3 for every

Space Ray and Rs2 for every Zapper. Space Rays and

Zappers require 2 and 4 production minutes per unit,

respectively. Also, Space Rays and Zappers require .5and .3 pounds of plastic, respectively. Given

constraints of 40 production hours, 1200 pounds of

plastic, Space Ray production cant exceed Zapper

production by more than 450 units; formulate theproblem such that Galaxy maximizes profit.


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30

R = # of Space Rays to produceZ = # of Zappers to produce

Max Z = 3.00R + 2.00Z

ST

2R + 4Z 2400 cant exceed available hours (40*60)

.5R + .3Z 1200 cant exceed available plastic

R - S 450 Space Rays cant exceed Zappers by more

than 450

R, S 0 non-negativity constraint

LP Model


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Graphical Solution Method

1. Plot model constraint on a set of coordinatesin a plane

2. Identify the feasible solution space on thegraph where all constraints are satisfied

simultaneously

3. Plot objective function to find the point onboundary of this space that maximizes (or

minimizes) value of objective function


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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 = 404x1 + 3x2 = 120

4x1 + 8x2 = 160-4x1 - 3x2 = -120

5x2 = 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

248


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Extreme Corner Points

x1

= 224 bowls

x2 = 8 mugs

Z= 1,360 x1 = 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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4x1 + 3x2 120 lb

x1 + 2x2 40 hr

40

30

20

10

0

B

|10

|20

|30

|40

x1

x2

C

A

Z= 70x1 + 20x2Optimal point:x1 = 30 bowls

x2 =0 mugs

Z= $2,100

Objective Function


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Minimization Problem

Two brands of fertilizer available - Super-gro,Crop-quick.

Field requires at least 16 pounds of nitrogen and 24

pounds of phosphate.

Super-gro costs $6 per bag, Crop-quick $3 per bag.Problem : How much of each brand to purchase to

minimize total cost of fertilizer given following data ?


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Minimize Z= $6x1 + $3x2

subject to 2x1 + 4x2 16 lb of nitrogen

4x1 + 3x2 24 lb of phosphate

x1, x2 0

CHEMICAL CONTRIBUTION

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

Gro-plus 2 4Crop-fast 4 3


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Graphical method

Complete model formulation:

minimize Z = $6x1 + 3x2

subject to

2x1 + 4x2 16 lb of

nitrogen

4x1 + 3x2 24 lb of

phosphatex1, x2 0


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A Minimization Model Example

Feasible Solution Area

minimize Z = $6x1 + 3x2

subject to

2x1 + 4x2 16 lb of nitrogen


x1, x2 0


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A Minimization Model Example

Optimal Solution Point

minimize Z = $6x1 + 3x2subject to

2x1 + 4x2 16 lb of nitrogen


x1, x2

0


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Irregular Types of Linear Programming

Problems

For some linear programming models, the general

rules do not apply. Special types of problems include those with:

1. Multiple optimal solutions

2. Infeasible solutions

3. Unbounded solutions


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Multiple Optimal Solutions

Objective function is parallel to a

constraint line:

maximize Z=$40x1 + 30x2

subject to

1x1 + 2x2 40 hours of labor

4x2

+ 3x2

120 pounds of clay

x1, x2 0

where x1 = number of bowls

x2 = number of mugs


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An Infeasible Problem

Every possible solution violates

at least one constraint:

maximize Z = 5x1 + 3x2

subject to

4x1 + 2x2

8x1 4

x2 6

x1, x2 0


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An Unbounded Problem

Value of objective function

increases indefinitely:

maximize Z = 4x1 + 2x2

subject to

x1 4

x2 2

x1, x2 0

Linear Programming Model 2 MBA PPT

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