Intro Management Science 472.21 2 Fall 2011 Bruce Duggan Providence University College

Intro Management Science

472.212

Fall 2011Bruce Duggan

Providence University College

This Week

Review Cases from ch 1

Linear Programming ch 2 formulas & graphs

Case 1: Clean Clothes Corner

A. Current volume?

she’s just breaking even

v =cf

p-cv

v =$1,700.00

$1.10 - $0.25

Case 1: Clean Clothes Corner

B. Increase needed to break even?

v =cf

p-cv

v =$16,200.00/12$1.10 - $0.25

Case 1: Clean Clothes Corner

C. Monthly profit?

Z = vp - cf - vcv

Z = 4,300.00 $1.10

- ($1,700.00 + $1,350.00)

- 4,300 $0.25

Case 1: Clean Clothes Corner

D. If lower price?

BE?

Z? Z = vp - cf - vcv

v =cf

p-cv

Case 1: Clean Clothes Corner

E. Which is the better choice?

Z with new equipment?

Z without new equipment?

Case 2: Ocobee

Which option is better?

make the rafts yourself?

buy them from North Carolina?

ch 2: Linear Programming

George Dantzighttp://forum.stanford.edu/blog/?p=27

Linear Programming

Jargon Linear programming

• l.p.

• “figuring stuff out with basic algebra”

Model formulation• Stating our problem in words/math/graphs

Sensitivity analysis• “What happens if…?”

Linear Programming

Jargon Why is there jargon?

handout

Applications

Kellogg pg 35

Nutrition Coordinating Center pg 46

Soquimich pg 51

Example: Maximization

The St. Adolphe Historical Museum We have a group of older volunteers

• The St. Adolphe Craft League

They’ve offered to make toothpick tchochkes to sell at the gift shop

• Red River ox carts

• the first church in St. Adolphe

We can sell everything

they make

St. Adolphe Craft League

They want to know: How many ox carts? How many churches?

Goal To make the most profit possible for the

museum

St. Adolphe Craft League

Resource availability 40 hrs of labor 120 boxes of toothpicks

Decision variables x1 = number of ox carts to make

x2 = number of churches to make

St. Adolphe Craft League

Product resource requirements and unit profit:

41

32

Product

cart

church

Profit ($/unit)

40

50

Material (boxes/unit)

Labour (hr/unit)

Resource Requirements

St. Adolphe Craft League

Objective function Maximize Z = $40x1 + $50x2

Resource constraints 1x1 + 2x2 40 hours of labor

4x1 + 3x2 120 boxes of toothpicks

Non-Negativity constraints x1 0; x2 0

Maximize Z = $40x1 + $50x2

subject to:

1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

Problem definition Complete linear programming model

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

Model formulation: l.p.

no computers yet

St. Adolphe Craft League

words

math graphs

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

x2

0 10 20 30 40

10

20

30

40

x1

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

x2

0 10 20 30 40

10

20

30

40

x1

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

x2

0 10 20 30 40

10

20

30

40

x1

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

x2

0 10 20 30 40

10

20

30

40

x1

St. Adolphe Craft League

x2

0 10 20 30 40

10

20

30

40

x1

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

x1 = 0 ox cartsx2 = 20 churchesZ = $1,000

x1 = 30 ox cartsx2 = 0 churchesZ = $1,200

x1 = 24 ox cartsx2 = 8 churchesZ = $1,360

Linear Programming

lp has 2 main tools maximization

• most profit

minimization• least cost

Z means profit

Z means cost

Example: Minimization

Friesen Farms section of land needs at least

• 16 lb nitrogen

• 24 lb phosphate

2 brands of fertilizer available• DeSallaberry Superior

• Carmen Crop

Goal• Meet fertilizer needs at minimum cost

Problem• How much of each brand should you buy?

words

math

graphs

Friesen Farms

Chemical Contributions

ProductNitrogen (lb/bag)

Phosphate (lb/bag)

Cost ($/bag)

DeSallaberry Superior

Carmen Crop

words

math

graphs

Friesen Farms

Chemical Contributions

ProductNitrogen (lb/bag)

Phosphate (lb/bag)

Cost ($/bag)

DeSallaberry Superior

2 4 $6

Carmen Crop 4 3 $3

words

math

graphs

Friesen Farms

Objective function Minimize Z = $6x1 + $3x2

Decision variables x1 = bags of DeSallaberry to buy

x2 = bags of Carmen to buy

words

math

graphs

Friesen Farms

Objective function Minimize Z = $6x1 + $3x2

Model constraints 2x1 + 4x2 16 (lb) nitrogen constraint

4x1 + 3x2 24 (lb) phosphate constraint

x1, x2 0 non-negativity constraint

words

math

graphs

Min Z = $6x1 + $3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

Friesen Farms

Model formulation: l.p.

words

math

graphs

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

Friesen Farms

x2

0 2 4 6 8

2

4

6

8

x1

words

math

graphs

Friesen Farms

x2

0 2 4 6 8

2

4

6

8

x1

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

Friesen Farms

x2

0 2 4 6 8

2

4

6

8

x1

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

x2

0 2 4 6 8

2

4

6

8

x1

Friesen Farms

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

x1 = 0 bags of DeSallaberry x2 = 8 bags of CarmenZ = $24

x1 = 5 DeSallaberryx2 = 2 CarmenZ = $36

x1 = 8 DeSallaberry x2 = 0 CarmenZ = $48

x2

0 2 4 6 8

2

4

6

8

x1

Friesen Farms

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

x2

0 2 4 6 8

2

4

6

8

x1

Friesen Farms

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

x2

0 2 4 6 8

2

4

6

8

x1

Friesen Farms

Min Z = $6x1 + 3x2

s.t. 2x1 + 4x2 ≥ 16

4x1 + 3x2 ≥ 24

x1, x2 0

Surplus Variableswhat’s left over - don’t contribute to - “slack”

x1 = 0 bags of DeSallaberry x2 = 8 bags of Carmens1 = 16 lb of nitrogens2 = 0 lb of phosphateZ = $2400

x1 = 4.8 DeSallaberryx2 = 1.6 Carmens1 = 0 nitrogens2 = 0 phosphateZ = $3360

x1 = 8 DeSallaberry x2 = 0 Carmens1 = 0 nitrogens2 = 8 phosphateZ = $4800

On computer

much easier to do

goals up to now the idea the formulas

l.p.

usual characteristics & limitations clear goal choice amongst alternatives “certainty”

• non-probabilistic

constraints exist

relationships• linear

• slope constant

additivity divisibility for graphical solution

• 2 variables

Assignment

ch 2 problems in group

• 2

• 38

yourself• 1

• 16

Max Z = $40x1 + $50x2

s.t. 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

St. Adolphe Craft League

Model formulation: l.p.

Next Week

review ch 2 problems

ch 3 on the computer sensitivity analysis