linear programming model formulation and graphical solution mba ppt

7/31/2019 Linear Programming Model Formulation and Graphical Solution MBA PPT

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Linear ProgrammingModel Formulation and

Graphical Solution

By Babasab Patil

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Part I: Linear Programming

Model Formulation and Graphical Solution

Model Formulation

A Maximization Model Example

Graphical Solutions of Linear Programming Models

A Minimization Model Example

Irregular Types of Linear Programming Models

Characteristics of Linear Programming Problems


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Linear Programming - An Overview

Objectives of business firms frequently include maximizing profit or

minimizing costs.

Linear programming is an analysis technique in which linear algebraicrelationships represent a firms decisions given a business objective

and resource constraints.

Steps in application:

1- Identify problem as solvable by linear programming.

2- Formulate a mathematical model of the unstructured problem.

3- Solve the model.


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Model Components and Formulation

Decision variables: mathematical symbols representing

levels of activity of a firm.

Objective function: a linear mathematical relationship

describing an objective of the firm, in terms of decision

variables, that is maximized or minimized

Constraints: restrictions placed on the firm by the

operating environment stated in linear relationships of the

decision variables.

Parameters: numerical coefficients and constants used in

the objective function and constraint equations.


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A Maximization Model Example (1 of 2)

Problem Definition

Product mix problem - Beaver Creek Pottery Company

How many bowls and mugs should be produced to

maximize profits given labor and materials constraints?

Product resource requirements and unit profit:

Resource Requirements

Product Labor(hr/unit)

Clay(lb/unit)

Profit($/unit)

Bowl 1 4 40Mug 2 3 50


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A Maximization Model Example (2 of 2)Resource availability:

40 hours of labor per day

120 pounds of clay

Decision Variables:

x1=number of bowls to produce/day

x2= number of mugs to produce/day

Objective function

maximize Z = $40x1 + 50x2

where Z= profit per day

Resource Constraints:

1x1 + 2x2 40 hours of labor

4x1 + 3x2 120 pounds of clay

Non-negativity Constraints:

x10; x2 0

Complete Linear Programming Model:

maximize Z=$40x1 + 50x2

subject to1x1 + 2x2 40

4x2 + 3x2 120

x1, x2 0


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Feasible/Infeasible Solutions

A feasible solution does not violate any of the constraints:

Example x1= 5 bowls

x2= 10 mugs

Z = $40 x1 + 50x2= $700

Labor constraint check:

1(5) + 2(10) = 25 < 40 hours, within constraint

Clay constraint check:4(5) + 3(10) = 70 < 120 pounds, within constraint

An infeasible solution violates at least one of the constraints:

Example x1 = 10 bowls

x2 = 20 mugsZ = $1400

Labor constraint check:

1(10) + 2(20) = 50 > 40 hours, violates constraint


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Graphical Solution of Linear Programming Models

Graphical solution is limited to linear programming models

containing only two decision variables. (Can be used with

three variables but only with great difficulty.)

Graphical methods provide visualization of how a solution

for a linear programming problem is obtained.


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Graphical Solution of a Maximization Model

Coordinate Axes

maximize Z=$40x1

+ 50x2

subject to



x1, x2 0

Coordinates for graphical analysis


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Labor Constraint

maximize Z=$40x1 + 50x2subject to



x1, x2 0

Graph of the labor constraint line


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Labor Constraint Area

maximize Z=$40x1

+ 50x2

subject to



x1, x2 0

The labor constraint area


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Clay Constraint Area

maximize Z=$40x1

+ 50x2

subject to



x1, x20

The constraint area for clay


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Both Constraints


subject to



x1, x2 0

Graph of both model Constraints


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Feasible Solution Area


subject to



x1, x2 0

The feasible solution area constraints


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Objective Function = $800

Z= $800 = $40x1 + 50x2

subject to1x1 + 2x2 40 hours of labor


x1, x2 0

Objective function line for Z 5 $800


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Alternative Objective Functions

Z=$800, $1200, $1600 = $40x1 + 50x2

subject to1x1 + 2x2 40 hours of labor


x1, x2 0

Alternative objective function lines for profits, Z, of $800, $1,200, and $1,600


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Optimal Solution

Z= $800 =$40x1 + 50x2

subject to



x1, x2 0

Identification of optimal solution point


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Optimal Solution Coordinates


subject to

1x1 + 2x2 40 hours of labor4x2 + 3x2 120 pounds of clay

x1, x2 0

Optimal solution coordinates


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Corner Point Solutions


subject to


4x2 + 3x2 120 pounds of clayx1, x2 0

Solutions at all corner points


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Optimal Solution for New Objective Function


subject to


4x2 + 3x2

120 pounds of clayx1, x2 0

The optimal solution with Z 5 70x1 1 20x2


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Slack Variables

Standard form requires that all constraints be in the form ofequations.

A slack variable is added to a constraint to convert it to

an equation (=).

A slack variable represents unused resources.

A slack variable contributes nothing to the objectivefunction value.


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Complete Linear Programming Model in Standard Form

maximize Z=$40x1 + 50x2 + 0s1 + 0s2

subject to

1x1 + 2x2 + s1 = 40

4x2 + 3x2 + s2 = 120

x1,x2,s1,s2 = 0

where x1 = number of bowlsx2 = number of mugs

s1, s2 are slack variables

Solutions at points A, B, and C with slack


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Problem Definition

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 ?

Chemical Contribution

BrandNitrogen(lb/bag)

Phosphate(lb/bag)

Super-gro 2 4

Crop-quick 4 3


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A Minimization Model Example Model Construction

Decision variables

x1 = bags of Super-gro

x2 = bags of Crop-quick

The objective function:

minimize Z = $6x1 + 3x2

where $6x1 = cost of bags of Super-gro

3x2 = cost of bags of Crop-quick

Model constraints:

2x1 + 4x2 16 lb (nitrogen constraint)

4x1 + 3x2 24 lb (phosphate constraint)

x1, x2 0 (nonnegativity constraint)


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Feasible Solution Area


subject to

2x1 + 4x2 16 lb of nitrogen

4x1 + 3x2 24 lb of phosphate

x1, x2 0

Feasible solution area



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Optimal Solution Point


subject to

2x1 + 4x2 16 lb of nitrogen

4x1 + 3x2 24 lb of phosphate

x1, x2 0

The optimal solution point


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Surplus Variables

A surplus variable is subtracted from a constraint to convert it to anequation (=).

A surplus variable represents an excess above a constraint requirement

level.

Surplus variables contribute nothing to the calculated value of the

objective function.

Subtracting slack variables in the farmer problem constraints:2x1 + 4x2 - s1 = 16 (nitrogen)

4x1 + 3x2 - s2 = 24 (phosphate)



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Graphical Solutions

minimize Z = $6x1 + 3x2 + 0s1 + 0s2

subject to2x1 + 4x2 - s1 = 16

4x1 + 3x2 - s2 = 24

x1, x2, s1, s2 = 0

Graph of the fertilizer example


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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

Multiple Optimal Solutions


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Multiple Optimal Solutions

Objective function is parallel to a

constraint line:


subject to

1x1 + 2x2 40 hours of labor4x2 + 3x2 120 pounds of clay

x1, x2 0

where x1 = number of bowls

x2 = number of mugs

Graph of the Beaver Creek Pottery Company example with multiple optimal solutions

An Infeasible Problem


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An Infeasible Problem

Every possible solution violates

at least one constraint:

maximize Z = 5x1 + 3x2

subject to

4x1 + 2x2 8

x1 4

x2 6

x1, x2 0

Graph of an infeasible problem

An Unbounded Problem


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An Unbounded Problem

Value of objective function

increases indefinitely:


subject to

x1 4

x2 2

x1, x2 0

An unbounded problem


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Characteristics of Linear Programming Problems

A linear programming problem requires a decision - a choice amongst

alternative courses of action.

The decision is represented in the model by decision variables.

The problem encompasses a goal, expressed as an objective function,that the decision maker wants to achieve.

Constraints exist that limit the extent of achievement of the objective.

The objective and constraints must be definable by linear mathematical

functional relationships.


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QP5013

LINEAR PRORAMMING 35

Properties of Linear Programming Models

Proportionality - The rate of change (slope) of the

objective function and constraint equations is constant.

Additivity - Terms in the objective function and constraint

equations must be additive.

Divisability -Decision variables can take on any fractional

value and are therefore continuous as opposed to integer

in nature.

Certainty - Values of all the model parameters are assumed

to be known with certainty (non-probabilistic).


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Example Problem No. 1

Problem Statement

- Hot dog mixture in 1000-pound batches.

- Two ingredients, chicken ($3/lb) and beef ($5/lb),

- Recipe requirements:

at least 500 pounds of chicken

at least 200 pounds of beef.

- Ratio of chicken to beef must be at least 2 to 1.

- Determine optimal mixture of ingredients that will minimize costs.


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Example Problem No. 1

Solution

Step 1: Identify decision variables.

x1 = lb of chicken

x2 = lb of beef

Step 2: Formulate the objective function.


where Z = cost per 1,000-lb batch

$3x1 = cost of chicken

5x2 = cost of beef


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Example Problem No.2


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Solve the following model graphically:


subject to

x1 + 2x2 10

6x1 + 6x2 36

x1

4

x1,x2 0

Step 1: Plot the constraint s as equations:

The constraint equations



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p


subject to

x1 + 2x2 1

6x1 + 6x2 36

x1 4

x1,x2 0

Step 2: Determine the feasiblesolution area:

The feasible solution space and extreme points



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subject to

x1 + 2x2 10

6x1 + 6x2 36

x1 4

x1,x2 0

Steps 3 and 4:

Determine the solution points

and optimal solution.

Optimal solution point


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Part II: Linear Programming

Modeling Examples

A Product Mix Example

A Diet Example

An Investment Example A Marketing Example

A Transportation Example

A Blend Example A Multiperiod Scheduling Example

A Data Envelopment Analysis Example


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Product Mix Example

Problem Definition

- Four-product T-shirt/sweatshirt manufacturing company.

- Must complete production within 72 hours

- Truck capacity = 1,200 standard sized boxes.

- Standard size box holds12 T-shirts.

- One-dozen sweatshirts box is three times size of standard box.

- $25,000 available for a production run.

- 500 dozen blank T-shirts and sweatshirts in stock.

- How many dozens (boxes) of each type of shirt to produce?


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Product Mix Example

Data

ProcessingTime (hr)Per dozen

Cost($)

per dozen

Profit($)

per dozen

Sweatshirt - F 0.10 36 90

SweatshirtB/F

0.25 48 125

T-shirt - F 0.08 25 45

T-shirt - B/F 0.21 35 65

Product Mix Example


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Product Mix Example

Model Construction

Decision variables:

x1 = sweatshirts, front printingx2 = sweatshirts, back and front printing

x3 = T-shirts, front printing

x4 = T-shirts, back and front printing

Objective function:

maximize Z = $90x1 + 125x2 + 45x3 + 65x4

Model constraints:

0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72 hr

3x1 + 3x2 + x3 + x4 1,200 boxes

$36x1 + 48x2 + 25x3 + 35x4 $25,000

x1 + x2 500 dozen sweatshirts

x3 + x4 500 dozen T-shirts

Product Mix Example


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Computer Solution with QM for Windows

maximize Z = $90x1 + 125x2 + 45x3 + 65x4

subject to:

0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72

3x1 + 3x2 + x3 + x4 1,200 boxes

$36x1 + 48x2 + 25x3 + 35x4 $25,000

x1 + x2 500 dozed sweatshirts


x1, x2, x3, x4 0

Product Mix Example


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Computer Solution with QM for Windows (continued)

maximize Z = $90x1 + 125x2 + 45x3 + 65x4

subject to:

0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72

3x1 + 3x2 + x3 + x4 1,200 boxes

$36x1 + 48x2 + 25x3 + 35x4 $25,000

x1 + x2 500 dozed sweatshirts


x1, x2, x3, x4 0

Diet Example


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Diet Example

Data and Problem Definition

Breakfast to include at least 420 calaries, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12

grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.

BreakfastFood Calories Fat(g) Cholesterol(mg) Iron(mg) Calcium(mg) Protein(g) Fiber(g) Cost($)

1. Bran cereal (cup)2. Dry cereal (cup)3. Oatmeal (cup)4. Oat bran (cup)5. Egg6. Bacon (slice)

7. Orange8. Milk-2% (cup)9. Orange juice (cup)10. Wheat toast (slice)

90110100

907535

65100120

65

022253

0401

0000

2708

01200

642310

1001

2048128

300

522503

26

345672

1913

523400

1003

0.180.220.100.120.100.09

0.400.160.500.07


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Diet Example

Model Construction: Decision Variables

x1 = cups of bran cereal

x2 = cups of dry cereal

x3 = cups of oatmeal

x4 = cups of oat bran

x5 = eggs

x6 = slices of bacon

x7 = oranges

x8 = cups of milk

x9 = cups of orange juice

x10 = slices of wheat toast

Diet Example


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Diet Example

Model Summary

minimize Z =0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6+ 0.40x7 + 0.16x8 + 0.50x9

0.07x10

subject to

90x1 + 110x2 + 100x3 + 90x4 + 75x5 + 35x6 + 65x7 + 100x8 + 120x9 + 65x10 420

2x2 + 2x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10 20

270x5 + 8x6 + 12x8 30

6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10 5

20x1 + 48x2 + 12x3 + 8x4+ 30x5 + 52x7 + 250x8 + 3x9 + 26x10 400

3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7+ 9x8+ x9 + 3x10 20

5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10 12

xi 0

An Investment Example


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An Investment Example

Model Summary

maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4

subject tox1 14,000

x2 - x1 - x3- x4 0

x2 + x3 21,000

-1.2x1 + x2 + x3 - 1.2 x4 0

x1 + x2 + x3 + x4 = 70,000

x1, x2, x3, x4 0

where

x1 = amount invested in municipal bonds ($)

x2 = amount invested in certificates of deposit ($)

x3 = amount invested in treasury bills ($)

x4 = amount invested in growth stock fund($)

A Marketing Example


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A Marketing Example

Data and Problem Definition

- Budget limit $100,000

- Television time for four commercials

- Radio time for 10 commercials- Newspaper space for 7 ads

- Resources for no more than 15 commercials and/or ads.

Exposure

(people/ad orcommercial)

Cost

Televisioncommercial

20,000 $15,000

Radio commercial 12,000 6,000

Newspaper ad 9,000 4,000

Transportation Example


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Transportation Example

Problem Definition and Data

Warehouse supply of televisions sets: Retail store demand for television sets:

1- Cinncinnati 300 A. - New York 150

2- Atlanta 200 B. - Dallas 250

3- Pittsburgh 200 C. - Detroit 200

total 700 total 600

To StoreFromWarehouse

A B C

1 $16 $18 $11

2 14 12 13

3 13 15 17

A Blend Example


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Determine the optimal mix of the three components in each grade of motor oil that will maximize

profit. Company wants to produce at least 3,000 barrels of each grade of motor oil.

ComponentMaximum Barrels

Available/dayCost/barrel

1 4,500 $12

2 2,700 10

3 3,500 14

Grade Component Specifications Selling Price ($/bbl)

Super At least 50% of 1

Not more than 30% of 2

$23

Premium At least 40% of 1Not more than 25% of 3

20

Extra At least 60% of 1At least 10% of 2

18

A Blend Example


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Decision Variables and Model Summary

Decision variables: The quantity of each of the three components used in each grade of

gasoline (9 decision variables); xij = barrels of component i used in motor oil grade j per day,

where i = 1, 2, 3 and j = s (super), p(premium), and e(extra).Model Summary: maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e

subject to

x1s + x1p + x1e 4,500

x2s + x2p + x2e 2,700

x3s + x3p + x3e 3,500

0.50x1s - 0.50x2s - 0.50x3s 0

0.70x2s - 0.30x1s - 0.30x3s 0

0.60x1p - 0.40x2p - 0.40x3p 0

0.75x3p - 0.25x1p - 0.25x2p 0

0.40x1e- 0.60x2e- - 0.60x3e 0

0.90x2e - 0.10x1e - 0.10x3e

0x1s + x2s + x3s 3,000

x1p+ x2p + x3p 3,000

x1e+ x2e + x3e 3,000

xij 0


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A Multiperiod Scheduling Example


Production capacity : 160 computers per weekAdditional 50 computers with overtime

Assembly costs: $190/comp. regular time; $260/comp. overtime

Inventory cost: $10/comp. per week

Order schedule: Week Computer Orders1 105

2 170

3 230

4 180

5 150

6 250

A M lti i d S h d li E l


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A Multiperiod Scheduling Example


Decision variables:

rj = regular production of computers per week j (j = 1, 2, 3, 4, 5, 6)

oj = overtime production of computers per week j (j = 1, 2, 3, 4, 5, 6)

ij = extra computers carried over as inventory in week j (j = 1, 2, 3, 4, 5)

Model summary:

minimize Z = $190(r1 + r2 + r3 + r4 + r5 + r6) + $260(o1 + o2 + o3 + o4 + o5 +o6) + 10(i1, + i2 + i3 + i4 + i5)subject to rj 160 (j = 1, 2, 3, 4, 5, 6)

oj 150 (j = 1, 2, 3, 4, 5, 6)

r1 + o1 - i1 105

r2 + o2 + i1 - i2 170

r3 + o3 + i2 - i3

230r4 + o4 + i3 - i4 180

r5 + o5 + i4 - i5 150

r6 + o6 + i5 250

rj, oj, ij 0

A Data Envelopment Analysis (DEA) Example


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Problem Definition and DataDEA compares a number of service units of the same type based on their inputs (resources) and

outputs. The result indicates if a particular unit is less productive, or efficient, than other units.

Elementary school comparison:input 1 = teacher to student ratio output 1 = average reading SOL score

input 2 = supplementary $/student output 2 = average math SOL score

input 3 = parent education level output 3 = average history SOL score

Inputs Outputs

School 1 2 3 1 2 3

Alton .06 $260 11.3 86 75 71

Beeks .05 320 10.5 82 72 67

Carey .08 340 12.0 81 79 80

Delancey .06 460 13.1 81 73 69



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Decision variables:

xi = a price per unit of each output where i = 1, 2, 3

yi = a price per unit of each input where i = 1, 2, 3

Model summary:

maximize Z = 81x1 + 73x2 + 69x3

subject to

.06 y1 + 460y2 + 13.1y3 = 1

86x1 + 75x2 + 71x3.06y1 + 260y2 + 11.3y3

82x1 + 72x2 + 67x3 .05y1 + 320y2 + 10.5y3

81x1 + 79x2 + 80x3 .08y1 + 340y2 + 12.0y3

81x1 + 73x2 + 69x3 .06y1 + 460y2 + 13.1y3

xi, yi 0

E l P bl S l i


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Example Problem Solution

Problem Statement and Data

- Canned catfood, Meow Chow; dogfood, Bow Chow.

- Ingredients/week: 600lb horse meat; 800 lb fish; 1000 lb cereal.

- Recipe requirement: Meow Chow at least half fish; Bow Chow at least

half horse meat.

- 2,250 sixteen-ounce cans available each week.

-Profit /can: Meow Chow $0.80; Bow Chow$0.96.

- How many cans of Bow Chow and Meow Chow should be produced

each week in order to maximize profit?



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Model Formulation

Step 1: Define the Decision Variables

xij

= ounces of ingredient i in pet food j per week, where i = h (horse meat), f (fish) and

c (cereal), and j = m (Meow chow) and b (Bow Chow).

Step 2: Formulate the Objective Function

maximize Z = $0.05(xhm + xfm + xcm) + 0.06(xhb + xfb + xcb)

Step 3: Formulate the Model Constraints

Amount of each ingredient available each week:

xhm + xhb 9,600 ounces of horse meat

xfm + xfb 12,800 ounces of fish

xcm + xcb 16,000 ounces of cereal additive

Recipe requirements:

Meow Chow xfm/(xhm + xfm + xcm) 1/2, or, - xhm + xfm- xcm 0

Bow Chow xhb/(xhb + xfb + xcb) 1/2, or, xhb- xfb - xcb 0

Can content constraint: xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces

Example Problem SolutionModel Summary and Solution with QM for Windows


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Model Summary and Solution with QM for Windows

Step 4: Model Summary

maximize Z = $0.05 xhm + 0.05 xfm + 0.05 xcm + 0.06 xhb + 0.06 xfb + 0.06 xsubject to xhm + xhb 9,600 ounces of horse meat

xfm + xfb 12,800 ounces of fishxcm + xcb 16,000 ounces of cereal additive

- xhm + xfm- xcm 0

xhb- xfb - xcb 0

xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces

xij 0


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Chap 2 - No 36 & 38; page 62;

Chap 3No 8, 9, 10, 17, 19; pg 92;

Chap 4 - 20, 21; pg 146 Group assignment: Case Problem Summer

Sports Camp at State University

Please try these questions & we will discuss it

during our class.

Additional Exercises

linear programming model formulation and graphical solution mba ppt

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