1 introduction to linear programming chapter 3 from l. winston copyright (c) 2004 brooks/cole, a...

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Introduction to Linear Programming

Chapter 3 from L. Winston

Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.

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Example 1: Giapetto’s Woodcarving

Giapetto’s, Inc., manufactures wooden soldiers and trains. Each soldier built:

• Sell for $27 and uses $10 worth of raw materials.

• Increase Giapetto’s variable labor/overhead costs by $14.

• Requires 2 hours of finishing labor.

• Requires 1 hour of carpentry labor.

Each train built:• Sell for $21 and used $9 worth of raw materials.

• Increases Giapetto’s variable labor/overhead costs by $10.

• Requires 1 hour of finishing labor.

• Requires 1 hour of carpentry labor.

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Ex. 1 - continued

Each week Giapetto can obtain: All needed raw material. Only 100 finishing hours. Only 80 carpentry hours.

Demand for the trains is unlimited.

At most 40 soldiers are bought each week.

Giapetto wants to maximize weekly profit (revenues – costs).

Formulate a mathematical model of Giapetto’s situation that can be used maximize weekly profit.

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Example 1: Solution

The Giapetto solution model incorporates the characteristics shared by all linear programming problems. Decision variables should completely describe the decisions to

be made.• x1 = number of soldiers produced each week.

• x2 = number of trains produced each week.

The decision maker wants to maximize (usually revenue or profit) or minimize (usually costs) some function of the decision variables. This function to maximized or minimized is called the objective function.

• For the Giapetto problem, fixed costs do not depend upon the the values of x1 or x2.

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Ex. 1 - Solution continued

Giapetto’s weekly profit can be expressed in terms of the decision variables x1 and x2:

Weekly profit =

weekly revenue – weekly raw material costs – weekly variable costs = 3x1 + 2x2

Thus, Giapetto’s objective is to choose x1 and x2 to maximize weekly profit. The variable z denotes the objective function value of any LP.

Giapetto’s objective function is: Maximize z = 3x1 + 2x2

The coefficient of an objective function variable is called an objective function coefficient.

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As x1 and x2 increase, Giapetto’s objective function grows larger.

For Giapetto, the values of x1 and x2 are limited by the following three restrictions (often called constraints): Each week, no more than 100 hours of finishing time may be

used. (2 x1 + x2 ≤ 100).

Each week, no more than 80 hours of carpentry time may be used. (x1 + x2 ≤ 80).

Because of limited demand, at most 40 soldiers should be produced. (x1 ≤ 40).

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The coefficients of the decision variables in the constraints are called the technological coefficients. The number on the right-hand side of each constraint is called the constraint’s right-hand side (or rhs).

To complete the formulation of a linear programming problem, the following question must be answered for each decision variable. Can the decision variable only assume nonnegative values, or is

the decision variable allowed to assume both positive and negative values?

• If the decision variable can assume only nonnegative values, the sign restriction xi ≥ 0 is added.

• If the variable can assume both positive and negative values, the decision variable xi is unrestricted in sign (often abbreviated urs).

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For the Giapetto problem model, combining the sign restrictions x1≥0 and x2≥0 with the objective function and constraints yields the following optimization model:

Max z = 3x1 + 2x2 (objective function)

Subject to (s.t.)

2 x1 + x2 ≤ 100 (finishing constraint)

x1 + x2 ≤ 80 (carpentry constraint)

x1 ≤ 40 (constraint on demand for soldiers)

x1 ≥ 0 (sign restriction)


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Linearity

A function f(x1, x2, …, xn) of x1, x2, …, xn is a linear function if and only if for some set of constants, c1, c2, …, cn, f(x1, x2, …, xn) = c1x1 + c2x2 + … + cnxn.

For any linear function f(x1, x2, …, xn) and any number b, the inequalities f(x1, x2, …, xn) ≤ b and f(x1, x2, …, xn) ≥ b are linear inequalities.

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A linear program

A linear programming problem (LP) is an optimization problem for which we do the following: Attempt to maximize (or minimize) a linear function (called the

objective function) of the decision variables. The values of the decision variables must satisfy a set of

constraints. Each constraint must be a linear equation or inequality.

A sign restriction is associated with each variable. For any variable xi, the sign restriction specifies either that xi must be nonnegative (xi ≥ 0) or that xi may be unrestricted in sign.

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Linear objective functions

The fact that the objective function for an LP must be a linear function of the decision variables has two implications:

1. The contribution of the objective function from each decision variable is proportional to the value of the decision variable. For example, the contribution to the objective function for 4 soldiers is exactly four times the contribution of 1 soldier.

2. The contribution to the objective function for any variable is independent of the other decision variables. For example, no matter what the value of x2, the manufacture of x1 soldiers will always contribute 3x1 dollars to the objective function.

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Linear constraints

Analogously, the fact that each LP constraint must be a linear inequality or linear equation has two implications:

1. The contribution of each variable to the left-hand side of each constraint is proportional to the value of the variable. For example, it takes exactly 3 times as many finishing hours to manufacture 3 soldiers as it does 1 soldier.

2. The contribution of a variable to the left-hand side of each constraint is independent of the values of the variable. For example, no matter what the value of x1, the manufacture of x2 trains uses x2 finishing hours and x2 carpentry hours.

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Other features of LP

The first item in each list is called the proportionality assumption of linear programming.

The second item in each list is called the additivity assumption of linear programming.

The divisibility assumption requires that each decision variable be permitted to assume fractional values.

The certainty assumption is that each parameter (objective function coefficients, right-hand side, and technological coefficients) are known with certainty.

The feasible region of an LP is the set of all points satisfying all the LP’s constraints and sign restrictions.

For a maximization problem, an optimal solution to an LP is a point in the feasible region with the largest objective function value.

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X1

1 2 3 4

1

2

3

4

X2

-1

-1

X1

1 2 3 4

1

2

3

4

X2

-1

-1

Graphical solution to a two-variable LP problem

Any LP with only two variables can be solved graphically. The variables are always labeled x1 and x2 and the coordinate

axes the x1 and x2 axes.

Satisfies 2x1 + 3x2 ≥ 6

Satisfies 2x1 + 3x2 ≤ 6

2x1 + 3x2 = 6

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Example 1 cont’d

Since the Giapetto LP has two variables, it may be solved graphically.

The feasible region is the set of all points satisfying the constraints:2 x1 + x2 ≤100 (finishing constraint)

x1 + x2 ≤ 80 (carpentry constraint)

x1 ≤ 40 (demand constraint)



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Example 1 cont’d

The set of points satisfying the Giapetto LP is bounded by the five sided polygon DGFEH. Any point on or in the interior of this polygon (the shade area) is in the feasible region.

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Example 1 cont’d

Having identified the feasible region for the Giapetto LP, a search can begin for the optimal solution which will be the point in the feasible region with the largest z-value.

To find the optimal solution, graph a line on which the points have the same z-value. In a max problem, such a line is called an isoprofit line while in a min problem, this is called the isocost line. The figure shows the isoprofit lines for z = 60, z = 100, and z = 180.

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Binding constraints

A constraint is binding if the left-hand and right-hand side of the constraint are equal when the optimal values of the decision variables are substituted into the constraint. In the Giapetto LP, the finishing and carpentry constraints are

binding.

A constraint is nonbinding if the left-hand side and the right-hand side of the constraint are unequal when the optimal values of the decision variables are substituted into the constraint. In the Giapetto LP, the demand constraint for wooden soldiers is

nonbinding since at the optimal solution (x1 = 20), x1 < 40.

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Convex sets of points

A set of points S is a convex set if the line segment joining any two pairs of points in S is wholly contained in S.

For any convex set S, a point p in S is an extreme point if each line segment that lies completely in S and contains the point P has P as an endpoint of the line segment.

Extreme points are sometimes called corner points, because if the set S is a polygon, the extreme points will be the vertices, or corners, of the polygon. The feasible region for the Giapetto LP will be a convex set.

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Linear programming solver

By our own hand (e.g. using Simplex method)

Using commercial solvers CPLEX, LINDO, …

Using MS Excel

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Using Excel solver to obtain LP solution

Setting & data input

1

3

2

Max z = 3x1 + 2x2

s.t. 2 x1 + x2 ≤ 100

x1 + x2 ≤ 80

x1 ≤ 40

x1, x2 ≥ 0

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Using Excel solver – continued

Input formula B3 =C3*C2+D3*D2 B4 =C4*C2+D4*D2 B5 =C5*C2+D5*D2 B6 =C6*C2+D6*D2

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5

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B3 =SUMPRODUCT(C$2:D$2,C3:D3)B3 =SUMPRODUCT(C$2:D$2,C4:D4)B3 =SUMPRODUCT(C$2:D$2,C5:D5)B3 =SUMPRODUCT(C$2:D$2,C6:D6)

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Setting linear and nonnegative options

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8

9

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Obtain solution

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Example 2 : Dorian Auto

Dorian Auto manufactures luxury cars and trucks.

The company believes that its most likely customers are high-income women and men.

To reach these groups, Dorian Auto has embarked on an ambitious TV advertising campaign and will purchase 1-mimute commercial spots on two type of programs: comedy shows and football games.

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Ex. 2: continued

Each comedy commercial is seen by 7 million high income women and 2 million high-income men and costs $50,000.

Each football game is seen by 2 million high-income women and 12 million high-income men and costs $100,000.

Dorian Auto would like for commercials to be seen by at least 28 million high-income women and 24 million high-income men.

Use LP to determine how Dorian Auto can meet its advertising requirements at minimum cost.

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Example 2: Solution

Dorian must decide how many comedy and football ads should be purchased, so the decision variables are: x1 = number of 1-minute comedy ads.

x2 = number of 1-minute football ads.

Dorian wants to minimize total advertising cost.

Dorian’s objective functions is: min z = 50x1 + 100x2

Dorian faces the following constraints: Commercials must reach at least 28 million high-income women.

(7x1 + 2x2 ≥ 28).

Commercials must reach at least 24 million high-income men. (2x1 + 12x2 ≥ 24).

The sign restrictions are necessary, so x1, x2 ≥ 0.

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Ex. 2 – Solution continued

Like the Giapetto LP, The Dorian LP has a convex feasible region.

The feasible region for the Dorian problem, however, contains points for which the value of at least one variable can assume arbitrarily large values.

Such a feasible region is called an unbounded feasible region.

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To solve this LP graphically begin by graphing the feasible region.

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Since Dorian wants to minimize total advertising costs, the optimal solution to the problem is the point in the feasible region with the smallest z value.

An isocost line with the smallest z value passes through point E and is the optimal solution at x1 = 3.4 and x2 = 1.4.

Both the high-income women and high-income men constraints are satisfied, both constraints are binding.

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Some problems

Does the Dorian model meet the four assumptions of linear programming? The proportionality assumption is violated because at a certain

point advertising yields diminishing returns. Even though the additivity assumption was used in writing: (Total

viewers) = (Comedy viewer ads) + (Football ad viewers) many of the same people might view both ads, double-counting of such people would occur thereby violating the assumption.

The divisibility assumption is violated if only 1-minute commercials are available. Dorian is unable to purchase 3.6 comedy and 1.4 football commercials.

The certainty assumption is also violated because there is no way to know with certainty how many viewers are added by each type of commercial.

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3.3 Special Cases

The Giapetto and Dorian LPs each had a unique optimal solution.

Some types of LPs do not have unique solutions. Some LPs have an infinite number of solutions (alternative or

multiple optimal solutions). Some LPs have no feasible solutions (infeasible LPs). Some LPs are unbounded: There are points in the feasible

region with arbitrarily large (in a maximization problem) z-values.

The technique of goal programming is often used to choose among alternative optimal solutions.

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No optimal solutions

It is possible for an LP’s feasible region to be empty, resulting in an infeasible LP.

Because the optimal solution to an LP is the best point in the feasible region, an infeasible LP has no optimal solution.

For a maximization problem, an unbounded LP occurs if it is possible to find points in the feasible region with arbitrarily large z-values, which corresponds to a decision maker earning arbitrarily large revenues or profits.

For a minimization problem, an LP is unbounded if there are points in the feasible region with arbitrarily small z-values.

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Example 6: Diet Problem

My diet requires that all the food I get come from one of the four “basic food groups”.

At present, the following four foods are available for consumption: brownies, chocolate ice cream, cola and pineapple cheesecake.

Each brownie costs 50¢, each scoop of ice cream costs 20¢, each bottle of cola costs 30¢, and each piece of pineapple cheesecake costs 80¢.

Each day, I must ingest at least 500 calories, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat.

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Ex. 6 - continued

The nutritional content per unit of each food is given in the following table.

Formulate a linear programming model that can be used to satisfy my daily nutritional requirements at minimum cost.

Type of food Calories Chocolate Sugar Fat Brownie 400 3 2 2

Chocolate ice cream

200 2 2 4

Cola 150 0 4 1 Pineapple

cheesecake 500 0 4 5

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Example 6: Solution

As always, begin by determining the decisions that must be made by the decision maker: how much of each type of food should be eaten daily.

Thus we define the decision variables: x1 = number of brownies eaten daily.

x2 = number of scoops of chocolate ice cream eaten daily.

x3 = bottles of cola drunk daily.

x4 = pieces of pineapple cheesecake eaten daily.

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My objective is to minimize the cost of my diet.

The total cost of any diet may be determined from the following relation:(total cost of diet) = (cost of brownies) + (cost of ice cream) +(cost of cola) + (cost of cheesecake).

The decision variables must satisfy the following four constraints: Daily calorie intake must be at least 500 calories. Daily chocolate intake must be at least 6 oz. Daily sugar intake must be at least 10 oz. Daily fat intake must be at least 8 oz.

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Express each constraint in terms of the decision variables.

The optimal solution to this LP is found by combining the objective function, constraints, and the sign restrictions.

The chocolate and sugar constraints are binding, but the calories and fat constraints are nonbinding.

A version of the diet problem with a more realistic list of foods and nutritional requirements was one of the first LPs to be solved by computer.

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3.5 A Work-Scheduling Problem

Many applications of linear programming involve determining the minimum-cost method for satisfying workforce requirements.

The results may be different solved manually vs using CPLEX, LINGO or Excel Solver.

One type of work scheduling problem is a static scheduling problem.

In reality, demands change over time, workers take vacations in the summer, and so on, so the post office does not face the same situation each week. This is a dynamic scheduling problem.

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Scheduling postal workers

Each postal worker works for 5 consecutive days, followed by 2 days off, repeated weekly.

Minimize the number of postal workers (for the time being, we will permit fractional workers on each day.)

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Formulating as an LP

Select the decision variables Let x1 be the number of workers who start working on Monday,

and work till Friday. Let x2 be the number of workers who start on Tuesday …

Let x3, x4, …, x7 be defined similarly.

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Minimize z = x1 + x2 + x3 + x4 + x5 + x6 + x7

subject to

x1 + x4 + x5 + x6 + x7 17

x1 + x2 + x5 + x6 + x7 13

x1 + x2 + x3 + x6 + x7 15

x1 + x2 + x3 + x4 + x7 19

x1 + x2 + x3 + x4 + x5 14

x2 + x3 + x4 + x5 + x6 16

x3 + x4 + x5 + x6 + x7 11

xj 0 for j = 1 to 7

The linear program

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On the selection of decision variables

Optimal solution x1,…,x7 = 4/3, 10/3, 2, 22/3, 0, 10/3, 5.

Obviously this solution doesn’t work! We will see that the optimal integer programming solution is

actually x1,…,x7 = 4, 4, 2, 6, 0, 4, and 3.

• Compare two optimal solutions!

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Modeling issues

In reality this problem actually requires a dynamic analysis rather than the current static one.

When the scale of problems like this one gets very large, LP ideas become rather inefficient. Heuristic solutions may therefore be a suitable solution technique

The model does not allow for part-time workers or overtime hours, and alternative objective functions may include maximizing the number of weekend days off.

We may need to obtain the target workloads using ideas from other subjects like queueing theory to be able to estimate the number of workers required each day.

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(x1 + x2)/(x1 + x2 + x3 + x4 + x5 + x6 + x7) .3

(x1 + x2 ) .3 x1 + .3 x2 + .3 x3 + .3 x4 + .3 x5 + .3 x6 + .3 x7

-.7 x1 - .7 x2 + .3 x3 + .3 x4 + .3 x5 + .3 x6 + .3 x7 0

A ratio constraint:

Suppose that we need to ensure that at least 30% of the workers have Sunday off.

How do we model this?

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3.8 Blending Problems

Situations in which various inputs must be blended in some desired proportion to produce goods for sale are often amenable to linear programming analysis.

Such problems are called blending problems.

Some examples of how linear programming has been used to solve blending problems. Blending various types of crude oils to produce different types of

gasoline and other outputs. Blending various chemicals to produce other chemicals. Blending various types of metal alloys to produce various types

of steels.

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Example 12 Oil Blending

Sunco Oil manufactures three types of gasoline.

Each type is produced by blending three types of crude oil.

The price of gasoline and the cost of crude oil is given in the following table

Gas Sale price Crude Cost price

1 70 1 45 2 60 2 35 3 50 3 25

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Ex 12 cont’d

It costs $4 to transfer each barrel of crude into gasoline.

At most 5,000 barrels of each type of crude can be purchased each day.

At most 14,000 barrels of gas can be made each day.

Sunco must produce at least 3000, 2000, and 1000 barrels of types 1, 2, and 3 gas to satisfy customer demand.

This demand will increase if money is spent on advertising. Each $1 of advertising increases demand by 10 barrels of any type of gas.

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Ex 12 cont’d

The three types of gas and three types of crude all differ in octane and sulfur content.

The tables below show how much sulfur and octane the crude oil has, and the amounts that the gasoline must have after production.

Crude 1 2 3 Octane rating 12 6 8 Sulfur content 0.5% 2.0% 3.0% Gasoline 1 2 3 Minimum octane 10 8 6 Maximum Sulfur 1% 2% 1%

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Ex 12 Solution

Sunco Oil must decide how much to spend on advertising each type of gas and how it will produce each type of gas from the three types of crude oil.

Let’s use ai to mean the advertising budget for each gas, and xij to represent the number of barrels of crude i used to produce gas j.

i,j=1,2,3.

Summing xij over i or j, gives the total amount of crude purchased or gas sold.

We also need to assume the production outputs cannot be stored for later sale. That is, everything produced must be sold.

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Ex 12 Solution cont’d

Sunco Oil’s objective is to maximize profits using the function: Max z = (70-45-4)x11 + (60-45-4)x12 + (50-45-4)x13 +(70-35-4)x21

+ (60-35-4)x22 + (50-35-4)x23 + (70-25-4)x31 + (60-25-4)x32 + (50-25-4)x33 – a1 – a2 – a3

Reduction of the coefficients can be done now to simplify the expression.

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Demand constraints come from the information given previously. Note the increased possible demand from advertising, and that the constraints are equalities. x11 + x21 + x31 – 10a1 = 3000

x12 + x22 + x32 – 10a2 = 2000

x13 + x23 + x33 – 10a3 = 1000

Supply constraints: x11 + x12 + x13 5000

x21 + x22 + x23 5000

x31 + x32 + x33 5000

Production constraint: i j xij 14000

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Average octane constraints: (12x11 + 6x21 + 8x31)/ixi1 10

(12x12 + 6x22 + 8x32)/ixi2 8

(12x13 + 6x23 + 8x33)/ixi3 6

Note that these constraints have greater than or equal to limitations.

Rearranging gives: 2x11 – 4x21 – 2x31 0

4x12 – 2x22 0

6x13 + 2x33 0 (Redundant)

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Sulfur content constraints: (0.005x11 + 0.02x21 + 0.03x) /ixi1 0.01

(0.005x12 + 0.02x22 + 0.03x32) /ixi2 0.02

(0.005x13 + 0.02x23 + 0.03x33) /ixi3 0.01

Note that sulfur is a poison to the environment.

Rearranging gives: -0.005x11 + 0.01x21 + 0.02x31 0

-0.015x12 + 0.01x32 0

-0.005x13 + 0.01x23 + 0.02x33 0

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Modeling issues

In reality we would want to run this model daily to account for changes in the cost of crude oil and the market price obtainable for the finished product.

We may also want to allow for some diminishing return for the advertising budgets.

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3.9 Production Process Models

An LP model can be formulated to represent a simple production process.

The key step is to determine how the outputs from a later stage of the process are related to the outputs from an earlier stage.

A common mistake in these type of LP models is with different units of measurement in constraints.

If there are doubts about a constraint, then make sure that all terms in the constraint have the same units.

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Ex 13 Brute production process

Rylon Corp manufactures Brute and Chanelle perfumes.

Raw material costs $3 per pound (lb).

Processing each lb of raw material takes 1 hour of laboratory time and yields 3oz of regular Brute and 4oz of regular Chanelle.

Regular Brute sells for $7/oz and regular Chanelle for $6/oz.

Regular Brute and Chanelle can be processed more to obtain luxury Brute and Chanelle products which sells for $18/oz and $14/oz respectively.

1oz of luxury Brute can be made from 1oz of regular Brute with 3 hours of lab time and costs $4 for processing.

1oz of regular Chanelle can be converted into 1oz of luxury Chanelle using 2 hours of lab time and costs $4 for processing.

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Ex 13 cont’d

Rylon has 6000 hours of lab time available each year.

A maximum of 4000 pounds of raw material is purchasable.

Assume all other costs for the lab are fixed, and can be ignored.

Also assume no stockpiling.

Formulate this scenario as a LP to maximize profit.

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Ex 13 Solution

Decision variables need to be chosen carefully. x1 for annual regular Brute sales (oz)

x2 for luxury Brute,

x3 for regular Chanelle,

x4 for luxury Chanelle, and

x5 for the number of pounds of raw material to be purchased annually.

The objective function is therefore: max z = 7x1 + 14x2 + 6x3 + 10x4 – 3x5.

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Raw material constraint: x5 4000

Laboratory constraint: 3x2 + 2x4 + x5 6000

We need to add constraints for the fact that raw material purchased is converted into the four perfumes. x1 + x2 = 3x5 and

x3 + x4 = 4x5

Rearranging gives: x1 + x2 – 3x5 = 0 and

x3 + x4 – 4x5 = 0.

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Up until now all the LP examples have been static, or one-period, models.

Linear programming can also be used to determine optimal decisions in multiperiod, or dynamic, models.

Dynamic models arise when the decision maker makes decisions at more than one point in time.

In a dynamic model, decisions made during the current period influence decisions made during future periods.

Multiperiod Decision Problems: Inventory Model

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Example 14: Sailco Inventory

Sailco Corporation must determine how many sailboats should be produced during each of the next four quarters.

Demand is known for the next four quarters. Sailco must meet demands on time.

At the beginning of the first quarter Sailco has an inventory of 10 boats.

Sailco must decide at the beginning of each quarter how many boats should be produced during each quarter.

Sailco can produce 40 boats (cost=$400) per quarter with regular labor hours.

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Ex. 14 - continued

They can produce more if their workers work overtime. These boats cost $450 to produce.

At the end of each quarter a carrying or holding cost of $20 per boat is incurred.

Use linear programming to determine a production schedule to minimize the sum of production and inventory costs during the next four quarters.

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Example 14: Solution

For each quarter Sailco must determine the number of sailboats that should be produced by regular-time and by overtime labor.

The decision variables are: xt = number of sailboats produced by regular time labor during

quarter t yt = number of boats produced by overtime labor during quarter t

it = number of sailboats on hand at the end of quarter t

Sailco’s objective function is: min z = 400x1 + 400x2 + 400x3 + 400x4 + 450y1 + 450y2

+ 450y3 + 450y4 + 20i1 +20i2 +20i3+20i4

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Sailco’s constraints: The four constraints xt 40 ensure that each periods regular

time production does not exceed 40. Inventory constraints for each time period can now be added.

Let dt be the demand during quarter t,

• it = it-1 + (xt + yt) – dt (t = 1,2,3,4)

Note that quarter t’s demand will be met on time if and only if (iff) it ≥ 0

The sign restriction ensures that each quarter’s demand will be met on time.

All it, xt, and yt 0

Note that any production plan that had yt 0 and xt 40 could not be optimal.

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3.11 Multiperiod Financial Models

Linear programming can be used to model multiperiod cash management problems.

The key is to determine the relations of cash on hand during different periods.

One real world example is to use LP to determine a bond model that maximizes profit from bond purchases and sales subject to constraints that minimize the firm’s risk exposure.

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Ex 15 Finco investment problem

Finco must determine the investment strategy for the next three years.

At time 0, $100,000 is available for investment.

There are five investments – A to E.

The table shows the cash flows for each investment.

Investment A B C D E Time 0 -1 0 -1 -1 0 Time 1 0.5 -1 1.2 0 0 Time 2 1 0.5 0 0 -1 Time 3 0 1 0 1.9 1.5

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Ex 15 cont’d

To ensure that Finco has a diversified portfolio, at most $75,000 can be placed in any investment.

In addition to the five investments, Finco can earn 8% on any cash not placed in these investments using the money markets.

Assume that any cash inflow can be re-invested in another investment immediately.

No borrowing is allowed.

Formulate the LP that will maximize the cash on hand at the end of time 3.

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Ex 15 Solution

As well as the five investment decision variables, Finco must decide how much money will be placed on the money markets.

Let A to E mean the amount of $ invested in these investments and St is the $ invested in the money market at time t.

Finco must maximize the inflows of cash at time 3, givingmax z = B + 1.9D + 1.5E +1.08S2

Note that some investments do not finish at time 3 and are left out of the objective function.

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We need a constraint for the cash flows at each time period (0 to 2). A + C + D + S0 = 100000

0.5A – B + 1.2C + 1.08S0 – S1 = 0

A + 0.5B – E +1.08S1 – S2 = 0

A, B, C, D, E 75000 A, B, C, D, E, S0, S1, S2 0

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3.12 Multiperiod Work Scheduling

Have seen that LP can be used to schedule employees in a static environment.

LP can also be used to schedule employee training when a firm faces a demand that changes over time.

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Example 16 Multiperiod work scheduling

CSL is a chain of computer stores that must respond to growth in demand for its services.

In the next five months CSL will need 6000, 7000, 8000, 9500, and 11000 hours of skilled repair time.

At the beginning 50 skilled technicians work for CSL, and each one can work up to 160 hours per month.

New technicians can be hired but need training. This takes one month. Training requires 50 hours of supervision from skilled technicians.

Each skilled technician costs $2000 per month, while a trainee costs $1000 when in training.

At the end of each month 5% of the skilled technicians leave CSL and go to Plum Computers.

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Ex 16 Solution

First define the decision variables.

Let xt be the number of trainees and yt the number of skilled technicians at the start of month t.

Total labor cost is therefore:1000txt + 2000tyt

Obviously, we wish to minimize this quantity. We know that y1=50, but we need to know how to find the other yt.

This is found using a generalization of:y2 = 0.95y1 + x1

This can be converted to give the constraints:xt-1 + 0.95yt-1 – yt = 0

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In each month the skilled technicians either do the regular work or train the new staff.

The constraint for month t will be: 160yt – 50xt dt

where dt is 6000, 7000, 8000, 9500, and 1000 for the next five months.

Note that the values of yt actually need to be integer for this formulation to give the optimal solution exactly, but the formulation still works quite nicely.

75

Summary

Remember that any formulation must have an objective function, decision variables, and constraints.

The most difficult part of the problem is the definition of the decision variables. Getting these wrong will stuff it all up. Getting them right is an art – there is not always one way to generate a set of decision variables.

Keep things as simple as you need to.

Make sure all word constraints are turned into inequality or equality constraints.

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Homework #1

A company supplies goods to three customers, who each require 30 units. The company has two warehouses. Warehouse 1 has 40 units available, and warehouse 2 has 30 units available. The cost of shipping 1 unit from warehouse to customer are shown in the table below. There is a penalty for each unmet customer unit of demand: With customer 1, a penalty cost of $90 incurred; with customer 2, $80, and with customer 3, $110. Formulate an LP to minimize the sum of shortage and shipping costs, and obtain the optimal shipping plan.

To

From Customer 1 Customer 2 Customer 3

Warehouse 1 $15 $35 $25

Warehouse 2 $10 $50 $40

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Homework #2

You own a wheat warehouse with a capacity of 20,000 bushels. At the beginning of month 1, you have 6,000 bushels of wheat. Each month, wheat can be bought and sold at the price per 1,000 bushels given in the table on the right.

The sequence of events during each month is as follows:

a. You observe our initial stock of wheat

b. You can sell any amount of wheat up to your initial stock at the current month’s selling price.

c. You can buy (at the current month’s buying price) as much wheat as you want, subject to the warehouse size limitation.

Your goal is to formulate an LP that can be used to determine how to maximize the profit earned over the next 10 months, and to get the optimal solution.

MonthSelling

Price ($)Purchasing

Price ($)

1

2

3

4

5

6

7

8

9

10

3

6

7

1

4

5

5

1

3

2

8

8

2

3

4

3

3

2

5

5

1 introduction to linear programming chapter 3 from l. winston copyright (c) 2004 brooks/cole, a...

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