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Lecture 3 – Classic LP Examples. Topics Employee scheduling problem Energy distribution problem Feed mix problem Cutting stock problem Regression analysis Model Transformations. Examples of LP Formulations. 1. Employee Scheduling.

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Lecture 3 classic lp examples l.jpg

Lecture 3 – Classic LP Examples

Topics

  • Employee scheduling problem

  • Energy distribution problem

  • Feed mix problem

  • Cutting stock problem

  • Regression analysis

  • Model Transformations


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Examples of LP Formulations

1. Employee Scheduling

Macrosoft has a 24-hour-a-day, 7-days-a-week toll free hotline that is being set up to answer questions regarding a new product. The following table summarizes the number of full-time equivalent employees (FTEs) that must be on duty in each time block.

FTEs

Interval

Time

1

0-4

15

2

4-8

10

3

8-12

40

4

12-16

70

5

16-20

40

6

20-0

35


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Constraints for Employee Scheduling

  • Macrosoft may hire both full-time and part-time employees. The former work 8-hour shifts and the latter work 4-hour shifts; their respective hourly wages are $15.20 and $12.95. Employees may start work only at the beginning of 1 of the 6 intervals.

  • At least two-thirds of the employees working at any one time must be full-time employees.

  • Part-time employees can only answer 5 calls in the time a full-time employee can answer 6 calls. (i.e., a part-time employee is only 5/6 of a full-time employee.)

Formulate an LP to determine how to staff the hotline at minimum cost.


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Decision Variables

xt =# of full-time employees that begin the day at the start of interval t and work for 8 hours

yt = # of part-time employees that are assigned interval t

(4  12.95)

(8  15.20)

min

121.6(x1 + • • • + x6) +

51.8(y1 + • • • + y6)

5

³

s.t.

y1

15

x1 + x6 +

6

5

³

y2

10

x1 + x2 +

6

All shifts must be covered

5

³

y3

40

x2 + x3 +

6

5

³

y4

70

x3 + x4 +

6

5

³

y5

40

x4 + x5 +

6

5

³

y6

35

x5 + x6 +

6

PT employee is 5/6 FT employee


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More constraints:

2

x1 + x6

³

(x6 + x1 + y1)

3

At least 2/3 workers must be full time

2

³

(x1 + x2 + y2)

x1 + x2

3

.

.

.

2

³

x5 + x6

(x5 + x6 + y6)

3

Nonnegativity

xt ³ 0, yt ³ 0 t =1,2,…,6


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2. Energy Generation Problem (with piecewise linear objective)

Austin Municipal Power and Light (AMPL) would like to determine optimal operating levels for their electric generators and associated distribution patterns that will satisfy customer demand. Consider the following prototype system

Demand requirements

4 MW

1

1

Demand

sectors

Plants

7 MW

2

2

3

6 MW

The two plants (generators) have the following (nonlinear) efficiencies:

Plant 1

[ 0, 6 MW]

[ 6MW, 10MW]

Unit cost ($/MW)

$10

$25

Plant 2

[ 0, 5 MW]

[5MW, 11MW]

Unit cost ($/MW)

$8

$28

For plant #1, e.g., if you generate at a rate of 8MW (per sec), then the cost ($) is = ($10/MW)(6MW) + ($25/MW)(2MW) = $110.


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Problem Statement and Notation

Formulate an LP that, when solved, will yield optimal power generation and distribution levels.

Decision Variables

x

= power generated at plant

1 at operating level 1

11

1

x

²

2

²

²

²

²

²

²

12

2

1

x

²

²

²

²

²

²

²

21

2

x

2

²

²

²

²

²

²

²

22

y

1 to demand sector 1

= power sent from plant

11

2

y

1 ²²²

²

²

²

²

12

y

3

²

1

²

²

²

²

²

²

13

y

1

²

2

²

²

²

²

²

²

21

y

2

²

2

²

²

²

²

²

²

22

y

3

²

2

²

²

²

²

²

²

23


Slide8 l.jpg

Formulation

min

10x11 + 25x12 + 8x21 + 28x22

s.t.

y

y

y

= x11 + x12

+

+

11

12

13

y

y

y

= x21 + x22

+

+

21

22

23

y

y

+

= 4

11

21

y

y

+

= 7

12

22

y

y

= 6

+

13

23

0 £ x11 £ 6, 0 £x12£ 4

0 £ x21 £ 5, 0 £x22£ 6

y11, y12, y13, y21, y22, y32 0

Note that we can model the nonlinear operating costs as an LP only because the efficiencies have the right kind of structure. In particular, the plant is less efficient (more costly) at higher operating levels. Thus the LP solution will automatically select level 1 first.


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General Formulation of Power Distribution Problem

The above formulation can be generalized for any number of plants, demand sectors, and generation levels.

Indices/Sets

plants

i Î I

j Î J

demand sectors

generation levels

k Î K

Data

unit generation cost ($/MW) for plant i at level k

Cik

=

upper bound (MW) for plant i at level k

uik

=

=

dj

demand (MW) in sector j

Decision Variables

xik

= power (MW) generated at plant i at level k

yij

= power (MW) sent from plant i to sector j


Slide10 l.jpg

General Network Formulation

å

å

cikxik

min

iÎI

kÎK

xik

å

å

s.t.

yij

=

" i Î I

jÎJ

kÎK

å

yij = dj

" j Î J

iÎI

  • £xik£ uik " i Î I, k Î K

    0 £yij" i Î I, j Î J


Slide11 l.jpg

3. Feed Mix Problem

  • An agricultural mill produces a different feed for cattle,sheep, and chickens by mixing the following raw ingredients: corn, limestone, soybeans, and fish meal.

  • These ingredients contain the following nutrients: vitamins, protein, calcium, and crude fat in the following quantities:

Nutrient, k

Vitamins

Protein

Calcium

Crude Fat

Ingredient, i

Corn

8

10

6

8

Limestone

6

5

10

6

Soybeans

10

12

6

6

Fish Meal

4

18

6

9

Let aik= quantity of nutrient k per kg of ingredient i


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Constraints

  • The mill has (firm) contracts for the following demands.

Demand (kg)

Cattle

Sheep

Chicken

dj

10,000

6,000

8,000

  • There are limited availabilities of the raw ingredients.

Supply (kg)

Corn

Limestone

Soybeans

Fish Meal

si

6,000

10,000

4,0

00

5,000

  • The different feeds have “quality” bounds per kilogram.

Vitamins

Crude fat

Protein

Calcium

min max

min max

min max

min max

Cattle

6 --

6 --

7 --

4 8

Sheep

6 --

6 --

6 --

4 8

6 --

Chicken

4 6

6 --

4 8

The above values represent bounds:Ljk& Ujk


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Costs and Notation

  • Cost per kg of the raw ingredients is as follows:

Soybeans

Fish meal

Limestone

Corn

cost/kg, ci

24¢

12¢

20¢

12¢

Formulate problem as a linear program whose solution yields desired feed production levels at minimum cost.

Indices/sets

i  I

ingredients { corn, limestone, soybeans, fish meal }

j  J

products { cattle, sheep, chicken feeds }

nutrients { vitamins, protein, calcium, crude fat }

k  K


Slide14 l.jpg

Data

dj

demand for product j (kg)

si

supply of ingredient i (kg)

Ljk

lower bound on number of nutrients of type k per kg of product j

upper bound on number of nutrients of type k per kg of product j

Ujk

ci

cost per kg of ingredient i

aik

number of nutrients k per kg of ingredient i

Decision Variables

xij

amount (kg) of ingredient i used in producing product j


Slide15 l.jpg

å

aikxij£ Ujkdj

"j  J, kK

iÎI

å

aikxij ³ Ljk dj

"j  J, kK

iÎI

LP Formulation

å

å

cixij

min

jÎJ

iÎI

å

xij = dj

"j  J

s.t.

iÎI

xij£ si

å

"i  I

jÎJ

"i  I, j  J

xij³ 0


Slide16 l.jpg

Generalization of feed Mix Problem Gives

Blending Problems

Blended

commodities

Raw Materials

Qualities

corn, limestone,

protein, vitamins,

feed

soybeans, fish meal

calcium, crude fat

butane, catalytic

octane, volatility,

gasoline

reformate,

vapor pressure

heavy naphtha

pig iron,

carbon,

metals

ferro-silicon,

manganese,

carbide, various

chrome content

alloys

³

³

³

2 raw ingredients

1 quality

1 commodity


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4. Trim-Loss or Cutting Stock problem

  • Three special orders for rolls of paper have been placed at a paper mill. The orders are to be cut from standard rolls of 10¢ and 20¢ widths.

Length

Width

Order

1

¢

5

10,000¢

2

¢

7

30,000¢

3

¢

9

20,000¢

  • Assumption: Lengthwise strips can be taped together

  • Goal: Throw away as little as possible


Slide18 l.jpg

Problem: What is trim-loss?

¢

20

¢

10

5000'

5'

¢

5

9'

¢

7

Decision variables:xj= length of roll cut using pattern, j = 1, 2, … ?


Slide19 l.jpg

Patterns possible

¢

¢

10

roll

20

roll

x3

x9

x2

x8

x1

x5

x4

x7

x6

2

0

0

4

2

2

1

0

0

0

1

0

0

1

0

2

1

0

0

0

1

0

0

1

0

1

2

Trim loss

0

3

1

0

3

1

1

4

2

min

z =

10(x

+x

+x

) + 20(x

+x

+x

+x

+x

+x

)

7

9

1

2

3

4

5

6

8

³

s.t.

2x

+

4x

+

2x

+ 2x

+ x

10,000

1

7

4

5

6

³

+ x

x

+

x

+

2x

30,000

8

7

2

5

³

+ x

x

+

x

+ 2x

20,000

8

3

9

6

xj³ 0, j = 1, 2,…,9


Slide20 l.jpg

Alternative Formulation

+ 2x9

min

z =

3x2

+

x3

+

3x5

+ x6

+ x7

+

4x8

+ 5y1 + 7y2 + 9y3

4

s.t.

2x1

+

x4

+

2x5

+ 2x6

+ x7 – y1

=

10,000

x2

+

x5

+

2x7

+ x8

– y2

=

30,000

x3

+

x6

+

x8

+ 2x9

– y3

=

20,000

xj³ 0, j = 1,…,9; yi³ 0, i = 1, 2, 3

where yi is overproduction of width i


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5. Constrained Regression

Data (x,y) = { (1,2) , (3,4) , (4,7) }

y

·

7

6

5

·

4

3

·

2

1

x

1 2 3 4 5

We want to “fit” a linear function y = ax + b to these data points; i.e., we have to choose optimal values for a and b.


Slide22 l.jpg

Objective: Find parameters a and b that minimize the

maximum absolute deviation between the data yi and the fitted line yi = axi + b.

Ù

Ù

yi

yi

and

Ù

Predicted

value

observed

value

In addition, we’re going to impose a priori knowledge that the

slope of the line must be positive. (We don’t know about the intercept.)

known to be positive

a = slope of line

Decision variables

tive

positive or nega

b = y-intercept

b = b+-b-, b+ 0, b- 0


Slide23 l.jpg

Objective function:

Ù

min max { |yi-yi| : i = 1, 2, 3 }

Ù

where yi = axi + b

Ù

Let w = max { |yi-yi| : i = 1, 2, 3 }

Optimization model:

min w

Ù

s.t. w³ |yi-yi|, i = 1, 2, 3


Slide24 l.jpg

Nonlinear constraints:

Ù

w ³

-

 1a + b – 2 

y1

=

y1

Ù

-

y2

=

y2

w ³

 3a + b – 4 

Ù

w ³

-

=

y3

y3

 4a + b – 7 

Convert absolute value terms to linear terms:

Note: 2 ³|x| iff 2 ³x and 2 ³ -x

Ù

Thus w ³

-

is equivalent to

y

y

i

i

w³axi + b-yi and w³ -axi – b + yi


Slide25 l.jpg

so finally …

min w

-

s.t.

+

-

-

£

a + b

b

w

2

-

+

-

-

£

-

a-b

+ b

w

2

-

+

-

-

£

3a + b

b

w

4

-

-

+

-

-

£

-

3a

b

+ b

w

4

-

+

-

-

£

4a + b

b

w

7

-

+

-

+

-

£

-

-

4a

b

b

w

7

a, b+, b-, w³ 0


Model transformations l.jpg

Model Transformations

  • Direction of optimization:

    Minimize {c1x1 + c2x2 + … + cnxn}

    ÛMaximize {–c1x1 – c2x2 – … – cnxn}

  • Constant term in objective function  ignore

  • Nonzero lower bounds on variables:

  • xj>lj replace with xj= yj + lj whereyj 0

  • Nonpositive variable:

  • xj≤ 0 replace with xj= –yj where yj 0

  • Unrestricted variables:

  • xj = y1j– y2j where y1j 0, y2j  0


What you should know about lp problems l.jpg

What You Should Know About LP Problems

  • How to formulate various types of problems.

  • Difference between continuous and integer variables.

  • How to find solutions.

  • How to transform variables and functions into the standard form.


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