qp5013 – linear proramming1 part i: linear programming model formulation and graphical solution...

QP5013 – LINEAR PRORAMMING 1

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


Linear Programming - An Overview

• Objectives of business firms frequently include maximizing profit or minimizing costs.

• Linear programming is an analysis technique in which linear algebraic relationships represent a firm’s 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.


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.


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 40

Mug 2 3 50


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 to

1x1 + 2x2 40

4x2 + 3x2 120

x1, x2 0


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 mugs Z = $1400 Labor constraint check: 1(10) + 2(20) = 50 > 40 hours, violates constraint


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.


Graphical Solution of a Maximization Model Coordinate Axes


subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0

Coordinates for graphical analysis


Graphical Solution of a Maximization Model Labor Constraint



Graph of the labor constraint line


Graphical Solution of a Maximization Model Labor Constraint Area



The labor constraint area


Graphical Solution of a Maximization Model Clay Constraint Area



The constraint area for clay


Graphical Solution of a Maximization Model Both Constraints



Graph of both model Constraints


Graphical Solution of a Maximization Model Feasible Solution Area



The feasible solution area constraints


Graphical Solution of a Maximization Model Objective Function = $800

Z= $800 = $40x1 + 50x2


Objective function line for Z 5 $800


Graphical Solution of a Maximization Model Alternative Objective Functions

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


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


Graphical Solution of a Maximization Model Optimal Solution

Z= $800 =$40x1 + 50x2


Identification of optimal solution point


Graphical Solution of a Maximization Model Optimal Solution Coordinates



Optimal solution coordinates


Graphical Solution of a Maximization Model Corner Point Solutions



Solutions at all corner points


Graphical Solution of a Maximization Model Optimal Solution for New Objective Function



The optimal solution with Z 5 70x1 1 20x2


Slack Variables

• Standard form requires that all constraints be in the form of equations.

• 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 objective function value.


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 bowls x2 = number of mugs s1, s2 are slack variables

Solutions at points A, B, and C with slack


A Minimization Model Example 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

Brand

Nitrogen (lb/bag)

Phosphate (lb/bag)

Super-gro

2

4

Crop-quick

4

3


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)


A Minimization Model ExampleComplete Model Formulation and Constraint Graph

Complete model formulation:


subject to

2x1 + 4x2 16 lb of nitrogen

4x1 + 3x2 24 lb of phosphate

x1, x2 0


A Minimization Model ExampleFeasible Solution Area


subject to



x1, x2 0

Feasible solution area


A Minimization Model Example Optimal Solution Point


subject to



x1, x2 0

The optimal solution point


A Minimization Model Example Surplus Variables

• A surplus variable is subtracted from a constraint to convert it to an equation (=).

• 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)


A Minimization Model Example Graphical Solutions

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

subject to

2x1 + 4x2 - s1 = 16

4x1 + 3x2 - s2 = 24

x1, x2, s1, s2 = 0

Graph of the fertilizer example


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

Objective function is parallel to a

constraint line:


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

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


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

Value of objective function increases indefinitely:


subject to

x1 4

x2 2

x1, x2 0

An unbounded problem


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.


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).


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.


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


Example Problem No.1 Solution (continued)

Step 3: Establish Model Constraints

x1 + x2 = 1,000 lb

x1 500 lb of chicken

x2 200 lb of beef

x1/x2 2/1 or x1 - 2x2 0

x1,x2 0

The model: minimize Z = $3x1 + 5x2

subject to

x1 + x2 = 1,000 lb

x1 50

x2 200

x1 - 2x2 0

x1,x2 0


Example Problem No.2

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




subject to

x1 + 2x2 1

6x1 + 6x2 36

x1 4

x1,x2 0

Step 2: Determine the feasible solution area:

The feasible solution space and extreme points




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


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


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?


Product Mix Example Data

Processing Time (hr) Per dozen

Cost ($)

per dozen

Profit ($)

per dozen Sweatshirt - F

0.10

36

90

Sweatshirt – B/F

0.25

48

125

T-shirt - F

0.08

25

45

T-shirt - B/F

0.21

35

65


Product Mix Example Model Construction

Decision variables:

x1 = sweatshirts, front printing

x2 = 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 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 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 ExampleData 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.

Breakfast Food

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. Egg 6. Bacon (slice) 7. Orange 8. Milk-2% (cup) 9. Orange juice (cup) 10. Wheat toast (slice)

90 110 100

90 75 35 65

100 120

65

0 2 2 2 5 3 0 4 0 1

0 0 0 0

270 8 0

12 0 0

6 4 2 3 1 0 1 0 0 1

20 48 12

8 30

0 52

250 3

26

3 4 5 6 7 2 1 9 1 3

5 2 3 4 0 0 1 0 0 3

0.18 0.22 0.10 0.12 0.10 0.09 0.40 0.16 0.50 0.07


Diet ExampleModel 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 ExampleModel 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 Model Summary

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

subject to

x1 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 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 or commercial)

Cost

Television commercial

20,000 $15,000

Radio commercial 12,000 6,000

Newspaper ad 9,000 4,000


Transportation ExampleProblem 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 Store From Warehouse

A B C

1 $16 $18 $11

2 14 12 13

3 13 15 17


A Blend Example Problem Definition and Data

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.

Component

Maximum Barrels

Available/day

Cost/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 1

Not more than 25% of 3

20

Extra

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

18


A Blend Example 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 0

x1s + x2s + x3s 3,000

x1p+ x2p + x3p 3,000

x1e+ x2e + x3e 3,000

xij 0


A Multiperiod Scheduling Example Problem Definition and Data

Production capacity : 160 computers per week

Additional 50 computers with overtime

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

Inventory cost: $10/comp. per week

Order schedule: Week Computer Orders

1 105

2 170

3 230

4 180

5 150

6 250


A Multiperiod Scheduling Example Decision Variables and Model Summary

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 230

r4 + o4 + i3 - i4 180

r5 + o5 + i4 - i5 150

r6 + o6 + i5 250

rj, oj, ij 0


A Data Envelopment Analysis (DEA) Example Problem Definition and Data

DEA 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


A Data Envelopment Analysis (DEA) Example Decision Variables and Model Summary

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


Example Problem SolutionProblem 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?


Example Problem Solution 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 Solution 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 x subject to xhm + xhb 9,600 ounces of horse meat xfm + xfb 12,800 ounces of fish xcm + 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


• Chap 2 - No 36 & 38; page 62;

• Chap 3 – No 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

qp5013 – linear proramming1 part i: linear programming model formulation and graphical solution...

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