ba 452 lesson b.5 binary indicator variables 11readingsreadings chapter 7 integer linear programming

BA 452 Lesson B.5 Binary Indicator Variables 11

Readings

Readings

Chapter 7Integer Linear Programming


Overview

Overview


Overview

Machine Use Indicators are binary variables where 0 indicates no use and 1 indicates positive use. They are part of a simple linear model of the fixed cost of machine use.

Resource Allocation Problems with Fixed Costs trade off the advantage of using a variety of inputs to conform to fixed resources with the positive fixed cost of using each input.

Product Mix Problems with Fixed Costs trade off the advantage of producing a variety of goods to conform to fixed resources with the positive fixed cost of producing each good.

Make or Buy Decisions with Fixed Costs trade off the lower unit cost of producing a good yourself with the positive fixed cost of production.

Relational Constraints such as “either project i or project j is completed” or “both project i and project j are completed” can be written as linear constraints in binary indicator variables.

Capital Budgeting Problems maximize the net present value or net return from a selection of projects that each require a fixed amount of capital. Relational constraints are often imposed.


Tool Summary Use binary variables to indicate whether an activity, such as a

production run, is undertaken. Write a multiple-choice constraint: The sum of two or more binary

variables equals 1, so any feasible solution choose one variable to equal 1.

Write a mutually-exclusive constraint: The sum of two or more binary variables is at most 1, so any feasible solution chooses at most one variable to equal 1. All variables could equal 0.

Write a conditional constraint: An inequality constraint so that one binary variable cannot equal unless certain other binary variables also equal 1.

Write a corequisite constraint: An equality constraint of binary variables, so are either both 0 or both 1.

Tool Summary


Machine Indicators

Machine Indicators


Overview

Machine Use Indicators are binary variables where 0 indicates no use and 1 indicates positive use. They are part of a simple linear model of the fixed cost of machine use.

For example, consider a standard Resource Allocation Problem with Machines, where the maximum machine resource available is 20. Suppose fixed setup costs of using a machine restrict machine use X so “either X = 0 or 6 < X < 20”. But “either X = 0 or 6 < X < 20” is not a linear constraint. To make “either X = 0 or 6 < X < 20” a linear constraint, introduce binary variable Y and replace the standard resource constraint X < 20 with “6Y < X < 20Y”. The linear constraints “6Y < X < 20Y” can be satisfied in one of two ways. On the one hand, if Y = 0, then the constraints read “6(0) < X < 20(0)” and so are satisfied only by X = 0. On the one hand, if Y = 1, then the constraints read “6(1) < X < 20(1)” and so are satisfied only by 6 < X < 20.

Machine Indicators


Question: Ji-Yang Plastic makes plastic parts in Taiwan used in automobiles and computers. One of its major contracts involves the production of plastic printer cases for a computer company’s portable printers. The printer cases can be produced on two injection molding machines. The M-100 machine has a production capacity of 25 printer cases per hour, and the M-200 machine has a production capacity of 40 printer cases per hour. Both machines use the same chemical to produce the printer cases; the M-100 uses 40 pounds of raw material per hour, and the M-200 uses 50 pounds per hour.

Machine Indicators


The computer company asked Ji-Yang to produce as many of the cases as possible during the upcoming week; it will pay $18 for each case. However, next week is a regularly scheduled vacation period for most of Ji-Yang’s production employees. During this time, annual maintenance is performed on all equipment. Because of the downtime for maintenance, the M-100 is only available for at most 15 hours, and the M-200 for at most 10 hours.

Machine Indicators


The supplier of the chemical used in the production process informed Ji-Yang that a maximum of 1000 pounds of the chemical material will be available for next week’s production; the cost for this raw material is $6 per pound. In addition to the raw material cost, Ji-Yang estimates that the hourly cost of operating the M-100 and the M-200 are $50 and $75, respectively.

Machine Indicators


However, because of the high setup cost on both machines, management requires that, if a machine is used at all, it must be used for at least 5 hours.

(That last constraint makes the fixed cost of producing with a machine equal to the cost of using the machine for 5 hours.)

Machine Indicators


Answer: Define decision variables:

• M1 = number of hours spent on the M-100 machine• M2 = number of hours spent on the M-200 machine

Total revenue = 18(25) M1 + 18(40) M2 = 450 M1 + 720 M2

Total cost = 6(40) M1 + 6(50) M2 + 50 M1 + 75 M2 = 290 M1 + 375 M2

Machine Indicators


Mixed binary formulation Define objective:

Maximize (profit = revenue-cost) 160 M1 + 345 M2 Define constraints:

• M1 < 15 (M-100 maximum)• M2 < 10 (M-200 maximum)• 40M1 + 50 M2 < 1000 (Raw Material)• M1 > 5 U1 (M-100 minimum, if M1 > 0)• M2 > 5 U2 (M-200 minimum, if M2 > 0)• 15U1 > M1 (so U1 = 1 if M1 > 0)• 10U2 > M2 (so U2 = 1 if M2 > 0)• M1, M2 > 0 and U1, U2 binary (0 or 1)

Machine Indicators


Mixed binary formulation

Machine Indicators


Mixed binary solution

Use M1 for 12.5 hours, and M2 for 10 hours.

Machine Indicators


Resource Allocation with Fixed Cost



Overview

Resource Allocation Problems with Fixed Cost trade off the advantage of using a variety of inputs to conform to fixed resources with the positive fixed cost of using each input. The simplest way to model those fixed costs in a linear program is with binary (0 or 1) variables used as indicators, where 0 values indicate no production and 1 indicates positive production.

For example, consider a standard Resource Allocation Problem, where for one of the inputs, the maximum resource available is 10, and the cost of using X is 3X. Now suppose there is an added fixed setup costs of 15. To model those fixed costs in a linear program, introduce binary variable Y, replace the resource supply constraint X < 10 with “X < 10Y”, and add the cost term 15Y so total cost becomes 3X + 15Y. The constraint “X < 10Y” can be satisfied in one of two ways. On the one hand, if X = 0, then the constraint reads “0 < 10Y” and so can be satisfied by Y = 0, and so total cost = 3X. On the one hand, if X > 0, then the constraint “X < 10Y” can only be satisfied by Y = 1, and so total cost = 3X + 15.



Question: A businessman is considering opening a small specialized trucking firm. To make the firm profitable, it is estimated that it must have a daily trucking capacity of at least 12 tons. Two types of trucks are appropriate for the specialized operation. Their characteristics and costs are summarized in the table below. Note that each large truck require 2 drivers for long-haul trips. There are 8 potential drivers available and there are facilities for at most 4 trucks.



The businessman's objective is to minimize the total cost outlay for trucks.

Capacity Drivers Truck Cost (tons) Needed

Small $10,000 2 1Large $30,000 4 2

Formulate and graphically solve a linear programming model for this problem. How many of each type of truck should be used? Reformulate the problem if the trucking firm incurs a fixed setup cost of $2 if it uses any positive quantity of small trucks, and a fixed setup cost of $4 if it uses any positive quantity of large trucks?



Answer:Let S = the number of small trucks usedLet L = the number of large trucks used

Min 10000S + 30000Ls.t. 2S + 4L > 12 (capacity constraint)

S + 2L < 8 (driver constraint) S + L < 4 (facility constraint) S, L > 0

A graph of the feasible set and iso-value lines shows the optimal solution occurs where the first and third constraints bind (the second constraint is redundant). Solving the binding form of those two constraints yields the optimal solution: S = 2, L = 2.



Question: How would the formulation change if the trucking firm incurs a fixed setup cost of $2 if it uses any positive quantity of small trucks, and a fixed setup cost of $4 if it uses any positive quantity of large trucks?



Answer: Let SS = 1 if Small Trucks are used; 0 if notLet SY = 1 if Large Trucks are used; 0 if notObjective: Min 10000S + 30000L + 2SS + 4SYInput Constraints:

2S + 4L > 12 (capacity constraint) S + 2L < 8 (driver constraint) S + L < 4 (facility constraint) S, L > 0

Setup Constraints (given input constraints imply S < 4, L < 4):

S < 4SSL < 4SL



Product Mix with Fixed Cost



Overview

Product Mix Problems with Fixed Cost trade off the advantage of producing a variety of goods to conform to fixed resources with the positive fixed cost of producing each good. The simplest way to model those fixed costs in a linear program is with binary (0 or 1) variables used as indicators, where 0 values indicate no production and 1 indicates positive production.

For example, consider a standard Product Mix Problem, where for one of the outputs, the maximum production demanded is 10, and the cost of producing X is 3X. Now suppose there is an added fixed setup costs of 15. To model those fixed costs in a linear program, introduce binary variable Y, replace the demand constraint X < 10 with “X < 10Y”, and add the cost term 15Y so total cost becomes 3X + 15Y. The constraint “X < 10Y” can be satisfied in one of two ways. On the one hand, if X = 0, then the constraint reads “0 < 10Y” and so can be satisfied by Y = 0, and so total cost = 3X. On the one hand, if X > 0, then the constraint “X < 10Y” can only be satisfied by Y = 1, and so total cost = 3X + 15.



Question: Iron Elegance seeks to maximize profit by making two products from steel. It just received this month's allocation of 19 pounds of steel. It takes 2 pounds of steel to make a unit of product 1, and 3 pounds of steel to make a unit of product 2. The physical plant has the capacity to make at most 6 units of product 1, and at most 8 units of total product (product 1 plus product 2).

Product 1 sells for price 7, has marginal cost 2, and fixed cost 3. Product 2 sells for price 8, has marginal cost 1, and fixed cost 2.

Formulate a linear program to maximize profit.



Answer: A mixed integer linear program can be set up to solve this problem. Binary variables are used to indicate whether or not we setup to produce the two products. Let x1 and x2 denote this month's production level of

product 1 and product 2. y1 = 1 if x1 > 0; y1 = 0 if not

y2 = 1 if x2 > 0; y2 = 0 if not

Make or Buy with Fixed Cost


The total monthly profit =

(profit per unit of product 1) x (monthly production of product 1)

+ (profit per unit of product 2) x (monthly production of product 2)

= (7-2)x1 + (8-1)x2 - 3y1 - 2y2

Maximize total monthly profit: Max 5x1 + 7x2 - 3y1 - 2y2



Here is a mathematical formulation of constraints. The total amount of steel used during monthly production =

(steel used per unit of product 1) x (monthly production of product 1)

+ (steel used per unit of product 2) x (monthly production of product 2)

= 2x1 + 3x2

That quantity must be less than or equal to the allocated 19 pounds of steel (the inequality < in the constraint below assumes excess steel can be freely disposed; if disposal is impossible, then use equality =) :

2x1 + 3x2 < 19 The constraint that the physical plant has the capacity to make at

most 6 units of product 1 is formulated

x1 < 6 The constraint that the physical plant has the capacity to make at

most 8 units of total product (product 1 plus product 2) is

x1 + x2 < 8



Given resource constraint 2x1 + 3x2 < 19 implies x1 < 19/2 and x2 < 19/3, setup indicators

x1 < (19/2) y1

x2 < (19/3) y2



Adding the non-negativity and binary constraints complete the formulation.

Max 5x1 + 7x2 - 3y1 - 2y2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1 < (19/2) y1

x2 < (19/3) y2

x1 > 0 and x2 > 0

y1 = 0 or 1

y2 = 0 or 1

Objective

function

Standard constraints

Non-negativity


Setup constraints

Binary


The Management Scientist can solve this mixed integer linear program of 2 binary variables Yi and 2 continuous variables Xi.



Overview

Make or Buy Decisions with Fixed Cost trade off the lower unit cost of producing a good yourself with the positive fixed cost of production.The simplest way to model those fixed costs in a linear program is with binary (0 or 1) variables used as indicators, where 0 values indicate no production and 1 indicates positive production.

For example, consider a standard Make or Buy Decision, where the maximum production demanded is 10, and the cost of producing X is 3X. Now suppose there is an added fixed setup costs of 15. To model those fixed costs in a linear program, introduce binary variable Y, replace the demand constraint X < 10 with “X < 10Y”, and add the cost term 15Y so total cost becomes 3X + 15Y. The constraint “X < 10Y” can be satisfied in one of two ways. On the one hand, if X = 0, then the constraint reads “0 < 10Y” and so can be satisfied by Y = 0, and so total cost = 3X. On the one hand, if X > 0, then the constraint “X < 10Y” can only be satisfied by Y = 1, and so total cost = 3X + 15.



Question: Sony produces remote controllers for both televisions and DVD players. Each controller consists of three subassemblies that are made by Sony --- a base, a cartridge, and a keypad. Both controllers use the same base subassembly, but different cartridge and keypad subassemblies.

Sony’s sales forecast is that 7000 TV controllers and 5000 DVD controllers will be needed to satisfy demand during the upcoming Christmas season. Because only 500 hours of in-house manufacturing time is available, Sony considers buying some, or all, of the subassemblies from outside suppliers.



If Sony makes a subassembly, it incurs a fixed setup cost as well as a variable manufacturing cost. For each subassembly, the following are the fixed setup costs, the manufacturing time, the manufacturing costs, and the purchase costs:

Subassembly Setup Cost ($) Manufacturing Time per Unit

(min.)

Manufacturing Cost per Unit

($)

Purchase Cost per Unit ($)

Base 1000 0.9 0.40 0.65

TV cartridge 1200 2.2 2.90 3.45

DVD cartridge 1900 3.0 3.15 3.70

TV keypad 1500 0.8 0.30 0.50

DVD keypad 1500 1.0 0.55 0.70



Formulate a linear program to determine the optimal production and purchase of each subassembly.



Answer: A mixed integer linear program can be set up to solve this problem. Binary variables are used to indicate whether or not we setup to produce the subassemblies. SB = 1 if bases are produced; 0 if not STVC = 1 if TV cartridges are produced; 0 if not SVCRC = 1 if VCR cartridges are produced; 0 if not STVP = 1 if TV keypads are produced; 0 if not SVCRP = 1 if VCR keypads are produced; 0 if not



Continuous variables are used for the number of units made or purchased. BM = Number of bases made BP = Number of bases purchased TVCM = Number of TV cartridges made TVCP = Number of TV cartridges purchased VCRCM = Number of VCR cartridges made VCRCP = Number of VCR cartridges purchased TVPM = Number of TV keypads made TVPP = Number of TV keypads purchased VCRPM = Number of VCR keypads made VCRPP = Number of VCR keypads purchased



Minimize cost

(manufacturing cost)

0.4BM+2.9TVCM+3.15VCRCM+0.3TVPM+0.55VCRPM

(purchase cost)

+0.65BP+3.45TVCP+3.7VCRCP+0.5TVPP+0.7VCRPP

(setup cost)

+1000SB+1200STVC+1900SVCRC+1500STVP+1500SVCRP



Demand constraints

1) BM+BP=12000

2) TVCM+TVCP=7000

3) VCRCM+VCRCP=5000

4) TVPM+TVPP=7000

5) VCRPM+VCRPP=5000



Manufacturing time constraint (in minutes)

6) 0.9BM+2.2TVCM+3VCRCM+0.8TVPM+1VCRPM < 30000

Setup indicators (given demand constraints)

7) BM-12000SB < 0

8) TVCM-7000STVC < 0

9) VCRCM-5000SVCRC < 0

10) TVPM-7000STVP < 0

11) VCRPM-5000SVCRP < 0



The Management Scientist can solve this mixed integer linear program of 5 binary variables and 10 continuous variables.



Relational Constraints



Overview

Relational Constraints such as “either project i or project j is completed” or “both project i and project j are completed” can be written as linear constraints in binary indicator variables.



Binary variables can allow mathematical formulations of some intricate verbal descriptions.

Let xi and xj represent binary variables designating whether projects i and j have been completed (1 means completed, 0 means not completed).

The constraint “either project i or project j is completed” is formulated xi + xj > 1. In particular, xi + xj > 1 if both projects are completed (both variables equal 1)

That is the standard use of “or” in quantitative or mathematical work. It is the inclusive “or”.

For another example, if you are told “your friend is either at the snack bar or on the tennis court”, then you are told the truth if your friend is on the tennis court.



The constraint “either project i or project j is completed, but not both” is formulated xi + xj = 1. In particular, xi + xj = 1 is false if both projects are completed (both variables equal 1).

The relation “xi + xj = 1” is sometimes called the exclusive “or”.

There is debate over whether the English word “or” is inclusive or exclusive.

Some say “you may have coffee or tea” is an example of an exclusive “or”. But it turns out that “or” is not a logical or at all because “you may have coffee or tea” is not considered true if you can only have coffee.

In this class, “or” will only mean the inclusive “or”.



SummaryLet xi be a binary variables that is 1 if project i is done; xj likewise.

xi + xj = 1 means “either project i or project j is done, but not both”

xi + xj < 1 means “project i and project j are not both done”

xi + xj < 1 means “project i and project j will not both be done”

xi + xj > 1 means “at least one of project i or project j is done”

xi < xj means “if project i is done, then project j is done”

xi > xj means “project i is done if project j is done”

xi < xj means “project i is done only if project j is done”

xi = xj means “project i is done if, and only if, project j is done”

xi < xj means “project i will not be done unless project j is done”

xi < xj means “project i will not be done unless project i and project j are done”



Question: Springer Verlag, a publisher of college textbooks, must decide which new books to adopt and publish next year. The books under consideration are described in the first column below, along with their expected three-year sales in the second column:

Book Sales John Susan Monica

Business Math 20 12 40 X

Finite Math 32 9 24 X

General Statistics 17 16 X 30

Mathematical Statistics 10 7 X 24

Business Statistics 25 8 X 16

Finance 18 X X 14



Three individuals in the company can be assigned to these projects, all of whom have varying amounts of time available. John has 60 days, Susan has 52 days, and Monica has 43 days. The days required by each person to complete each project are showing in the third, fourth, and fifth columns above. For example, if the business calculus book is published, it will require 30 days of John’s time and 40 days of Susan’s time. (X means 0 time.)



Springer Verlag will not publish more than two statistics books in a single year. And one of the math books must be published, but not both. And the business math will be published only if the finance is published. And the general statistics will be published if the finite math is published. And the mathematical statistics and the finance will not both be published.

Formulate a model to decide which books should be published to maximize sales.



Answer:

where 1 is the business calculus book, 2 is the finite math book, … .

Here is a binary programming model for maximizing projected sales (thousands of units) subject to the restrictions mentioned.


xi =1 if book i is scheduled for publication0 otherwise{Let



Max 20x1 + 32x2 + 17x3 + 10x4 + 25x5 + 18x6

s.t.

12x1 + 9x2 + 16x3 + 7x4 + 8x5 < 60 John

40x1 + 24x2 < 40 Susan

30x3 + 24x4 + 16x5 + 14x6 < 40 Monica

x3 + x4 + x5 < 2 No. of Stat Books

x1 + x2 = 1 Math Book

x1 - x6 < 0 Only if

x2 - x3 < 0 If

x4 + x6 < 1 Not both


Capital Budgeting

Capital Budgeting


Capital Budgeting

Overview

Capital Budgeting Problems maximize the net present value or net return from a selection of projects that each require a fixed amount of capital. Relational constraints are often imposed.


Question: Fry’s Electronics is planning to expand its sales operation by offering new electronic appliances. The company has identified seven new product lines it can carry.

Initial Floor Space Exp. Rate

Product Line Invest. (Sq.Ft.) of Return

1. TV/VCRs $ 6,000 125 8.1%2. TVs 12,000 150 9.0 3. Projection TVs 20,000 200 11.0 4. VCRs 14,000 40 10.2 5. DVD Players 15,000 40 10.5 6. Video Games 2,000 20 14.1 7. Home Computers 32,000 100 13.2

Capital Budgeting


Fry's has decided that they should not stock projection TVs unless they stock either TV/VCRs or TVs. Also, they will not stock both VCRs and DVD players, and they will stock video games if they stock TVs. Finally, the company wishes to introduce at least three new product lines.

If the company has $45,000 to invest and 420 sq. ft. of floor space available, formulate an integer linear program for Fry's to maximize its overall expected return.

Capital Budgeting


xj

Define the Decision Variables

= 1 if product line j is introduced;

= 0 otherwise.Product line 1 = TV/VCRsProduct line 2 = TVsProduct line 3 = Projection TVsProduct line 4 = VCRsProduct line 5 = DVD PlayersProduct line 6 = Video GamesProduct line 7 = Home Computers

Capital Budgeting


Invest. Space Return1. $ 6,000 125

8.1%2. 12,000 150 9.0 3. 20,000 200 11.0 4. 14,000 40 10.2 5. 15,000 40 10.5 6. 2,000 20 14.1 7. 32,000 100

13.2

Define the Objective Function

Maximize total expected return:

Max .081(6000)x1 + .09(12000)x2 + .11(20000)x3 + .102(14000)x4

+ .105(15000)x5 + .141(2000)x6 + .132(32000)x7

Capital Budgeting


Invest. Space Return1. $ 6,000 125

8.1%2. 12,000 150 9.0 3. 20,000 200 11.0 4. 14,000 40 10.2 5. 15,000 40 10.5 6. 2,000 20 14.1 7. 32,000 100

13.2

1. Constrain total investment by $45,000: 6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7 < 45

2. Constrain space by 420 square feet: 125x1 +150x2 +200x3 +40x4 + 40x5 + 20x6 + 100x7 < 420

3. Stock projection TVs only if stock TV/VCRs or TVs: x1 + x2 > x3 or x1 + x2 - x3 > 0

4. Do not stock both VCRs and DVD players: x4 + x5 < 1

5. Stock video games if they stock TV's: x2 - x6 < 0

6. Introduce at least 3 new lines:

x1 + x2 + x3 + x4 + x5 + x6 + x7 > 3

Capital Budgeting


Interpretation:

Introduce: TV/VCRs, Projection TVs, and DVD Players

Do Not Introduce: TVs, VCRs, Video Games, and Home Computers

Total Expected Return: $4,261

Surplus: $4,000 Investment 55 square feet of space

Capital Budgeting


BA 452 Quantitative Analysis

End of Lesson B.5