03_lp2

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Lecture 3 – Classic LPExamples

Topics

•Employee scheduling problem

•Energy distribution problem

•Feed mix problem

•Cutting stock problem

•Regression analysis

•Model Transformations

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Macrosoft has a 24-hour-a-day, 7-days-a-week tollfree hotline that is being set up to answer questionsregarding a new product. The following tablesummarizes the number of full-time equivalent

employees (FTEs) that must be on duty in each time block.

Interval Time FTEs1 0-4 15

2 4-8 103 8-12 404 12-16 705 16-20 40

6 20-0 35

Employee Scheduling

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•Macrosoft may hire both full-time and part-time

employees. The former work 8-hour shifts and thelatter 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.

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

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

Formulate an LP to determine how to staff

the hotline at minimum cost.

Constraints for Employee Scheduling

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Decision Variables

xt =# of full-time employees that begin the day at the

start of intervalt and work for 8 hours

yt = # of part-time employees that are assigned intervalt

min 121.6(x1 +• • • +x6) +51.8(y1 +• • • +y6)

s.t. 5

6 y1

≥ 1556 y2 ≥ 1056 y3 ≥ 4056 y4 ≥ 70

56 y5 ≥ 4056 y6 ≥ 35

(8× 15.20) (4× 12.95)

All shifts must

be covered

PT employee is 5/6 FT employee

x1 + x

6+

x1 + x2 +

x2 + x3 +

x3 + x4 +

x4 + x5 +

x5 + x6 +

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x1 +x6 ≥ 23

≥ 23

...

≥ 23

xt≥ 0, yt≥ 0 t =1,2,…,6

At least 2/3

workers must

be full time

More constraints:

Noe!ativity

(x6 +x1 +y1)

x1 +x2 (x1 +x2 +y2)

(x5 +x6 +y6)x5 +x6

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• An agricultural mill produces a different feed for cattle,

sheep, and chickens by mixing the following rawingredients: corn, limestone, soybeans, and fish meal.

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

quantities:

Ingredient, i Vitamins Protein Calcium Crude Fat

Corn 8 10 6 8Limestone 6 5 10 6Soybeans 10 12 6 6Fish Meal 4 18 6 9

"et aik # $uatity of utriet k per k! of i!rediet i

Nutrient, k

Feed Mix Problem

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

Demand (kg) Cattle Sheep Chicken10,000 6,000 8,000

• There are limited availabilities of the raw ingredients.

Supply (kg)Corn Limestone Soybeans Fish Meal

6,000 10,000 4,000 5,000

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

Vitamins Protein Calcium Crude fat

min max min max min max min maxCattle 6 -- 6 -- 7 -- 4 8Sheep 6 -- 6 -- 6 -- 4 8Chicken 4 6 6 -- 6 -- 4 8

si

d j

The above values represent bounds: l jk

andu jk

Constraints

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

Corn Limestone Soybeans Fish meal

cost/kg,ci 20¢ 12¢ 24¢ 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 }k∈ K nutrients { vitamins, protein, calcium, crude fat }

Costs and Notation

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Data

d j demand for product j (kg)si supply of ingredienti (kg)l jk lower bound on number of nutrients of typek

per kg of product jupper bound on number of nutrients of typek

per kg of product jcost per kg of ingredientiaik number of nutrientsk per kg of ingredienti

Decision Variables

xij amount (kg) of ingredienti used in producingproduct j

u jk

ci

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min ∑ ∑ cixij

s.t. ∑

∑

xij≥ 0

xij=d j

xij ≤si

∀ j∈ J

i∈ I j∈ J

i∈ I

∀ i∈ I j∈ J

∑ aikxij ≤u jkd ji∈ I

∀ j∈ J, k∈K

∑ aikxij≥l jkd ji∈ I

∀ j∈ J, k∈K

∀ i∈ I, j∈ J

LP Formulation of Feed Mix Problem

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Raw Materials Qualities Blendedcommodities

corn, limestone,soybeans, fish meal

protein, vitamins,calcium, crude fat

feed

butane, catalyticreformate,

heavy naphtha

octane, volatility, vapor pressure

gasoline

pig iron,ferro-silicon,carbide, various

alloys

carbon,manganese,chrome content

metals

≥ 2 raw ingredients ≥ 1 quality ≥ 1 commodity

Generalization of feed Mix Problem GivesBlending Problems

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•Three special orders for rolls of paper have been placedat a paper mill. The orders are to be cut from standardrolls of 10′ and 20′ widths.

Order Width Length

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

Trim-Loss or Cutting Stock problem

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Problem: What is trim-loss?

Decision variables:x j = length of roll cut using

pattern, j = 1, 2, … ?

% ′

&' ′

()

5)5′

2' ′

5''')

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10′ roll 20′ roll

x1

5′ 2 0 0 4 2 2 1 0 07′ 0 1 0 0 1 0 2 1 09′ 0 0 1 0 0 1 0 1 2

Trim loss 0 3 1 0 3 1 1 4 2

x2 x3 x4 x5 x6 x7 x8 x9

min z = 10(x1+x2+x3) + 20(x4+x5+x6+x7+x8+x9)

s.t. 2x1 + 4x4 + 2x5 + 2x6 +x7 ≥ 10,000

x2 +x5 + 2x7 +x8 ≥ 30,000

x3 + x6 +x8 + 2x9 ≥ 20,000

x j≥ '* j # &* 2*+*(

Patterns Possible

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Minimize Trim Loss + Overproduction

min z = 3x2 + x3 + 3x5 +x6 +x7+ 4x8

+ 5y1+ 7y2+ 9y3

s.t. 2x1 + 4x4 + 2x5 + 2x6 +x7 –y1 = 10,000

x2 + x5 + 2x7 +x8 –y2 = 30,000

x3 + x6 + x8 + 2x9 – y3 = 20,000

x j≥ '* j = 1,…,9;yi ≥ 0,i = 1, 2, 3

where yi is overproductio of width i

+ 2x9

Alternative Formulation

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Minimizing Piecewise LinearConvex Functions

•Definition of convexity

•Examples of objective functions

1. f(x) = maxk=1,…, p(ckx +dk)

2. f(x) =Σ j=1,nc j|x j|,c j> 0 for all j

3. f(x) = separable, piecewise linear, convex

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Definition of a Convex/Concave Function

•A function f :ℜn → ℜ is calledconvex if foreveryx andy ∈ ℜn, and everyλ ∈[0,1], wehave

f(λ x + (1 –λ )y) ≤λ f(x) + (1 –λ ) f(y)

•A function f :ℜn → ℜ is calledconcave if foreveryx andy ∈ ℜn, and everyλ ∈[0,1], wehave

f(λ x + (1 –λ )y) ≥λ f(x) + (1 –λ ) f(y)

•If f(x) is convex, then – f(x) is concave

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Minimizing the Maximum of Several Affine Functions

Problem:min maxk=1,…, p(ckx +dk)s.t. Ax ≥b

Transformed problem:minz

s.t.z ≥ckx +dk,k = 1,…, p

Ax ≥b

x

f(x) = max

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Problems Involving Absolute Values: Minimizing theL1-Norm

Problem: minΣ j=1,nc j|x j|,c j> 0 for all j

s.t. Ax ≥b

Transformation 1:

minΣ j=1,nc jz j

s.t. Ax ≥b

z j ≥ x j, j = 1,…,n

z j ≥ –x j, j = 1,…,n

Transformation 2:

minΣ j=1,nc j(x j+ +x j

-)

s.t. Ax + – Ax - ≥b

x + ≥ 0,x - ≥ 0

wherex j =x j+ – x j- for all j

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Data Fitting Example

• Problem: We are given p data points of the form(ak,bk),k = 1,…, p, whereak ∈ ℜn andbk ∈ ℜ,and wish to build a model that predicts the valueof the variableb from knowledge of the vectora.

• Assume a linear model:b =ax +x0, where (x,x0) isa parameter vector to be determined.

• Error: Given a particular values of (x,x0), theresidual (predictive error) at thek th data point is

defined by |akx + x0 –bk|.

• Objective: Find values of (x,x0) that best explainthe available data; i.e., minimize the error.

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Data Fitting Example (cont’d)

• Model 1: Minimize thelargest residual

min maxk |akx + x0 –bk| Transformed model 1:

minz

s.t.z ≥akx + x0 –bk ,k = 1,…, p

z ≥ – akx – x0+ bk,k = 1,…, p

• Model 2: Minimize thesum of residuals

minΣk=1, p |akx + x0 –bk|

Transformed model 2:

min Σk=1, p zk

s.t.zk≥akx + x0 –bk ,k = 1,…, p

zk ≥ –ak

x

– x0+ bk ,k = 1,…, p

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Data (a,b) = { (1,2) , (3,4) , (4,7) }

We want to “fit” a linear functionb =ax +x0 to these data

points; i.e., we have to choose optimal values forx andx0.

%6

5,32

&

& 2 3 , 5

b

a

•

•

•

Constrained Regression

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Objective: Find parametersx andx0 that minimize the

maximum absolute deviation between the dataak and the

fitted linebk =akx +x0.

bk and

bk

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

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

Decision variables x = slope of line known to be positivex0 = b-intercept positive or negative

observed

value

Predicted

value

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Letz = max { |bk − bk|:k = 1, 2, 3 }

Optimization model:

min z

wherebk =akx +x0

Objective function:

s.t. z ≥ |bk − bk|,k = 1, 2, 3

min max { |bk − bk|:k = 1, 2, 3 }

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Note: 2≥|x| iff 2≥ x and 2≥ −x

Thusz≥|bk − bk| is equivalent to

z ≥ akx +x0 − bk and z ≥ − akx –x0 +bk

Convert absolute value terms to linear terms:

Nonlinear constraints:

b1

− b1

= 1x +x0

– 2

b2 − b2 = 3x +x0 – 4

b3− b3 = 4x +x0 – 7

z≥

z≥

z≥

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Lettingx0 =x0+ − x0-,x0

+ ≥ 0,x0- ≥ 0, we finally get …

min zs.t. x +x0+− x0

-− z ≤ 2

≤ − 2

≤ 4

≤ − 4

≤ 7

≤ − 7

x,x0+,x0-,z ≥ 0

− x − x0++ x0-− z

3x +x0+− x0-− z

− 3x − x0++ x0-− z

4x +x0+− x0-− z

− 4x − x0++ x0-− z

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Separable Piecewise Linear Functions

• Model: min f (x) = f1(x1) + f2(x2) +. . . + f p(x p)

• For eachx j we are givenr break points:

0 <a j1 <a j2 <. . . <a jr < ∞

• Letc jt be the slope in the intervala j,t-1≤ x j ≤ a jt fort =1,…,r+1, wherea j0= 0 anda j,r+1 = ∞

• Lety jt be the portion ofx j lying in thetth interval,t = 1,…,r+1

x ja j1 a j2 a jr

f j(x j)

a j,r-10

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Transformation for f j(x j)

•Letx j =y j1 +y j2 +. . . +y j,r+1

•Model:

minc j1y j1 +c j2y j2 +. . .+c j,r+1y j,r+1 + f1(x1) +. . .

s.t.0≤ y j1 ≤ a j1

0≤ y j2 ≤ a j2 – a j1. . .

0≤ y jr ≤ a jr – a j,r-1

0≤ y j,r+1

and for everyt, ify jt > 0, then eachy jk is equal to its

upper bounda jk – a j,k-1, for allk <t.

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

Plants

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.

2

&

3

2

&

-emad

sectors

-emad re$uiremets

, .

% .

6 .

Energy Generation Problem (with piecewise linear objective)

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Formulate an LP that, when solved, will yield optimal power

generation and distribution levels.

Decision Variables

= power generated at plant 1 at operating level 1

″ ″ ″ ″ 1 ″ ″ ″ 2

x21 ″ ″ ″ ″ 2 ″ ″ ″ 1

x22 ″ ″ ″ ″ 2 ″ ″ ″ 2

= power sent from plant1 to demand sector 1

″ ″ ″ ″ 1 ″ ″ ″ 2

″ ″ ″ ″ 1 ″ ″ ″

3″ ″ ″ ″ 2 ″ ″ ″ 1

″ ″ ″ ″ 2 ″ ″ ″ 2

″ ″ ″ ″ 2 ″ ″ ″ 3

Problem Statement and Notation

w11w12

w13w21

w22w23

x11

x12

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Formulation

min 10x11+ 25x12 + 8x21 + 28x22

s.t. w11 + w12 + w13w21 + w22 + w23

w11 + w21 = 4w12 + w22 = 7

w13 + w23 = 60≤ x11≤6, 0≤ x12 ≤ 4

0≤ x21≤5, 0≤ x22 ≤ 6

w11,w12,w13,w21,w22,w32 ≥ 0

Note that we ca model the oliear operati! costs as a "P oly because

the efficiecies have the ri!ht kid of structure0 1 particular* the plat is

less efficiet more costly at hi!her operati! levels0 Thus the "P solutio

will automatically select level & first0

= x11+x12= x21+x22

Flow balace

-emad

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The above formulation can be generalized for any

number of plants, demand sectors, and generationlevels.

Indices/Setsi∈ I plants

demand sectorsgeneration levels

Data

Cik = unit generation cost ($/MW) for planti at levelk

uik= upper bound (MW) for planti at levelk

d j = demand (MW) in sector j

Decision Variables

xik = power (MW) generated at planti at levelk

wij = power (MW) sent from planti to sector j

j∈ J

k∈ K

General Formulation of PowerDistribution Problem

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min ∑ ∑

s.t. ∑ wij= ∑

∑

cikxik

xik

0 ≤ xik

≤uik

∀ i

∈ I,k

∈ K

0 ≤ wij ∀ i∈ I, j∈ J

wij= d j

k∈ K i∈ I

k∈ K j∈ J

∀ i∈ I

∀ j∈ J i∈ I

General Network Formulation

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Model Transformations

•Direction of optimization:

Minimize {c1 x1 + c2 x2 + … + cn xn}

⇔ Maximize {–c1 x1 – c2 x2 – … – cn xn}

4 restricted variables

x j = y

1 j – y

2 j where y

1 j ≥ 0, y

2 j≥ 0

4 7ostat term i ob8ective fuctio i!ore

4 No9ero lower bouds o variables

x j > l

j replace with x

j = y

j+ l

j where y

j ≥ 0

4 Nopositive variable

x j ≤ 0 replace with x j = – y j where y j ≥ 0

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What You Should Know About LP Problems

•How to formulate various types ofproblems.

•Difference between continuous andinteger variables.

•How to find solutions.

•How to transform variables andfunctions into the standard form.

03_lp2

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