mcgraw-hill/irwin © the mcgraw-hill companies, inc., 2003 9.1 chapter 7 (binary integer...

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003

9.1

Chapter 7 (Binary Integer Programming)Including two CD-rom supplements


9.2

Assumptions of linear programming

Proportionality: the gross margin and resource requirements per unit of activity are assumed to be constant regardless of the level of the activity use

Additivity: no interaction effects between activities

Homogeneity: all units of the same resource or activity are identical

Continuity: resources can be used and activities produced in any fractional units

Deterministic coefficients: all coefficients in the model are known with certainty

Optimization: One proper objective function to be maximized or minimized

Finiteness: only a finite number of activities and constraints is considered


9.3

Types of Integer Programming Problems

• Pure integer programming problems are those where all the decision variables must be integers.

• Mixed integer programming problems only require some of the variables (the “integer variables”) to have integer values so the divisibility assumption holds for the rest (the “continuous variables”).

• Binary variables are variables whose only possible values are 0 and 1.

• Binary integer programming (BIP) problems are those where all the decision variables restricted to integer values are further restricted to be binary variables.

– Such problems can be further characterized as either pure BIP problems or mixed BIP problems, depending on whether all the decision variables or only some of them are binary variables.


9.4

The TBA Airlines Problem

• TBA Airlines is a small regional company that specializes in short flights in small airplanes.

• The company has been doing well and has decided to expand its operations.

• The basic issue facing management is whether to purchase more small airplanes to add some new short flights, or start moving into the national market by purchasing some large airplanes, or both.

Question: How many airplanes of each type should be purchased to maximize their total net annual profit?


9.5

Data for the TBA Airlines Problem

SmallAirplane

LargeAirplane

CapitalAvailable

Net annual profit per airplane $1 million $5 million

Purchase cost per airplane 5 million 50 million $100 million

Maximum purchase quantity 2 —


9.6

Linear Programming Formulation

Let S = Number of small airplanes to purchase

L = Number of large airplanes to purchase

Maximize Profit = S + 5L ($millions)

subject to

Capital Available: 5S + 50L ≤ 100 ($millions)

Max Small Planes: S ≤ 2

and

S ≥ 0, L ≥ 0.


9.7

Graphical Method for Linear Programming


9.8

Violates Divisibility Assumption of LP

• Divisibility Assumption of Linear Programming: Decision variables in a linear programming model are allowed to have any values, including fractional values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values.

• Since the number of airplanes purchased by TBA must have an integer value, the divisibility assumption is violated.


9.9

Integer Programming Formulation

Let S = Number of small airplanes to purchase

L = Number of large airplanes to purchase

Maximize Profit = S + 5L ($millions)

subject to

Capital Available: 5S + 50L ≤ 100 ($millions)

Max Small Planes: S ≤ 2

and

S ≥ 0, L ≥ 0

S, L are integers.


9.10

Graphical Method for Integer Programming

• When an integer programming problem has just two decision variables, its optimal solution can be found by applying the graphical method for linear programming with just one change at the end.

• We begin as usual by graphing the feasible region for the LP relaxation, determining the slope of the objective function lines, and moving a straight edge with this slope through this feasible region in the direction of improving values of the objective function.

• However, rather than stopping at the last instant the straight edge passes through this feasible region, we now stop at the last instant the straight edge passes through an integer point that lies within this feasible region.

• This integer point is the optimal solution.


9.11

Graphical Method for Integer Programming


9.12

Spreadsheet Model

3456789

1011121314

B C D E F GSmall Airplane Large Airplane

Unit Profit ($millions) 1 5

Capital CapitalSpent Available

Capital ($millions) 5 50 100 <= 100

Total ProfitSmall Airplane Large Airplane ($millions)

Units Produced 0 2 10<=

Maximum Small Airplanes 2

Capital Per Unit Produced


9.13

The default Tolerance field on the Solver Options dialog (relevant only for ILP models) is 5%. This means that the Solver ILP optimization procedure is continued only until the ILP solution OV is within 5% of the ILP’s optimum OV.

A higher Tolerance speeds up Solver at the risk of a reported solution further from the true ILP optimum.Setting Tolerance to 0% forces Solver to find the ILP optimum but with much longer solution times.


9.14

Applications of Binary Variables

• Since binary variables only provide two choices, they are ideally suited to be the decision variables when dealing with yes-or-no decisions.

• Examples:– Should we undertake a particular fixed project?

– Should we make a particular fixed investment?

– Should we locate a facility in a particular site?


9.15

California Manufacturing Company

• The California Manufacturing Company is a diversified company with several factories and warehouses throughout California, but none yet in Los Angeles or San Francisco.

• A basic issue is whether to build a new factory in Los Angeles or San Francisco, or perhaps even both.

• Management is also considering building at most one new warehouse, but will restrict the choice to a city where a new factory is being built.

Question: Should the California Manufacturing Company expand with factories and/or warehouses in Los Angeles and/or San Francisco?


9.16

Data for California Manufacturing

DecisionNumber

Yes-or-NoQuestion

DecisionVariable

Net PresentValue

(Millions)

CapitalRequired(Millions)

1 Build a factory in Los Angeles? x1 $8 $6

2 Build a factory in San Francisco? x2 5 3

3 Build a warehouse in Los Angeles? x3 6 5

4 Build a warehouse in San Francisco? x4 4 2

Capital Available: $10 million


9.17

Binary Decision Variables

DecisionNumber

DecisionVariable

PossibleValue

Interpretationof a Value of 1

Interpretationof a Value of 0

1 x1 0 or 1Build a factory inLos Angeles

Do not buildthis factory

2 x2 0 or 1Build a factory inSan Francisco

Do not buildthis factory

3 x3 0 or 1Build a warehouse inLos Angeles

Do not buildthis warehouse

4 x4 0 or 1Build a warehouse inSan Francisco

Do not buildthis warehouse


9.18

Algebraic Formulation

Let x1 = 1 if build a factory in L.A.; 0 otherwisex2 = 1 if build a factory in S.F.; 0 otherwisex3 = 1 if build a warehouse in Los Angeles; 0 otherwisex4 = 1 if build a warehouse in San Francisco; 0 otherwise

Maximize NPV = 8x1 + 5x2 + 6x3 + 4x4 ($millions)subject to

Capital Spent: 6x1 + 3x2 + 5x3 + 2x4 ≤ 10 ($millions)Max 1 Warehouse: x3 + x4 ≤ 1Warehouse only if Factory: x3 ≤ x1

x4 ≤ x2

andx1, x2, x3, x4 are binary variables.


9.19

Spreadsheet Model

3456789

1011121314151617181920

B C D E F GNPV ($millions) LA SF

Warehouse 6 4

Factory 8 5

Capital Required($millions) LA SFWarehouse 5 2 Capital Capital

Spent AvailableFactory 6 3 9 <= 10

Total MaximumBuild? LA SF Warehouses Warehouses

Warehouse 0 0 0 <= 1<= <=

Factory 1 1

Total NPV ($millions) 13


9.20

Sensitivity Analysis with Solver Table

2324252627282930313233343536

B C D E F GCapital Available Warehouse Warehouse Factory Factory Total NPV

($millions) in LA? in SF? in LA? in SF? ($millions)0 0 1 1 13

5 0 1 0 1 96 0 1 0 1 97 0 1 0 1 98 0 1 0 1 99 0 0 1 1 1310 0 0 1 1 1311 0 1 1 1 1712 0 1 1 1 1713 0 1 1 1 1714 1 0 1 1 1915 1 0 1 1 19


9.21

Management’s Conclusion

• Management’s initial tentative decision had been to make $10 million of capital available.

• With this much capital, the best plan would be to build a factory in both Los Angeles and San Francisco, but no warehouses.

• An advantage of this plan is that it only uses $9 million of this capital, which frees up $1 million for other projects.

• A heavy penalty (a reduction of $4 million in total net present value) would be paid if the capital made available were to be reduced below $9 million.

• Increasing the capital made available by $1 million (to $11 million) would enable a substantial ($4 million) increase in the total net present value. Management decides to do this.

• With this much capital available, the best plan is to build a factory in both cities and a warehouse in San Francisco.


9.22

Some Other Applications

• Investment Analysis– Should we make a certain fixed investment?

– Examples: Turkish Petroleum Refineries (1990), South African National Defense Force (1997), Grantham, Mayo, Van Otterloo and Company (1999)

• Site Selection– Should a certain site be selected for the location of a new facility?

– Example: AT&T (1990)

• Designing a Production and Distribution Network– Should a certain plant remain open? Should a certain site be selected for a new

plant? Should a distribution center remain open? Should a certain site be selected for a new distribution center? Should a certain distribution center be assigned to serve a certain market area?

– Examples: Ault Foods (1994), Digital Equipment Corporation (1995)

All references available for download at www.mhhe.com/hillier2e/articles


9.23

Some Other Applications

• Dispatching Shipments– Should a certain route be selected for a truck? Should a certain size truck be used?

Should a certain time period for departure be used?– Examples: Quality Stores (1987), Air Products and Chemicals, Inc. (1983),

Reynolds Metals Co. (1991), Sears, Roebuck and Company (1999)

• Scheduling Interrelated Activities– Should a certain activity begin in a certain time period?– Examples: Texas Stadium (1983), China (1995)

• Scheduling Asset Divestitures– Should a certain asset be sold in a certain time period?– Example: Homart Development (1987)

• Airline Applications:– Should a certain type of airplane be assigned to a certain flight leg? Should a certain

sequence of flight legs be assigned to a crew?– Examples: American Airlines (1989, 1991), Air New Zealand (2001)

All references available for download at www.mhhe.com/hillier2e/articles


9.24

Wyndor with Setup Costs (Variation 1)

Suppose that two changes are made to the original Wyndor problem:

1. For each product, producing any units requires a substantial one-time setup cost for setting up the production facilities.

2. The production runs for these products will be ended after one week, so D and W in the original model now represent the total number of doors and windows produced, respectively, rather than production rates. Therefore, these two variables need to be restricted to integer values.


9.25

Graphical Solution to Original Wyndor Problem

0 2 4 6 8

8

6

4

2

Production ratefor windows

Production rate for doors

FeasibleRegion

(2, 6)

Optimal solution

10

P = 3,600 = 300 D + 500 W

W

D


9.26

Net Profit for Wyndor Problem with Setup Costs

Net Profit ($)

Number ofUnits Produced Doors Windows

0 0(300) – 0 = 0 0 (500) – 0 = 0

1 1(300) – 700 = –400 1(500) – 1,300 = –800

2 2(300) – 700 = –100 2(500) – 1,300 = –300

3 3(300) – 700 = 200 3(500) – 1,300 = 200

4 4(300) – 700 = 500 4(500) – 1,300 = 700

5 Not feasible 5(500) – 1,300 = 1,200

6 Not feasible 6(500) – 1,300 = 1,700


9.27

Feasible Solutions for Wyndor with Setup Costs


9.28


Let D = Number of doors to produce,W = Number of windows to produce,y1 = 1 if perform setup to produce doors; 0 otherwise,y2 = 1 if perform setup to produce windows; 0 otherwise .

Maximize P = 300D + 500W – 700y1 – 1,300y2

subject toOriginal Constraints:

Plant 1: D ≤ 4Plant 2: 2W ≤ 12Plant 3: 3D + 2W ≤ 18

Produce only if Setup:Doors: D ≤ 99y1

Windows: W ≤ 99y2

andD ≥ 0, W ≥ 0, y1 and y2 are binary.


9.29

Spreadsheet Model

3456789

1011121314151617

B C D E F G HDoors Windows

Unit Profit $300 $500Setup Cost $700 $1,300

Hours HoursUsed Available

Plant 1 1 0 0 <= 4Plant 2 0 2 12 <= 12Plant 3 3 2 12 <= 18

Doors WindowsUnits Produced 0 6

<= <= Production Profit $3,000Only If Setup 0 99 - Total Setup Cost $1,300

Setup? 0 1 Total Profit $1,700

Hours Used Per Unit Produced


9.30

Wyndor with Mutually Exclusive Products(Variation 2)

Suppose that now the only change from the original Wyndor problem is:

• The two potential new products (doors and windows) would compete for the same customers. Therefore, management has decided not to produce both of them together.

– At most one can be chosen for production, so either D = 0 or W = 0, or both.


9.31

Feasible Solution forWyndor with Mutually Exclusive Products


9.32


Let D = Number of doors to produce,W = Number of windows to produce,y1 = 1 if produce doors; 0 otherwise,y2 = 1 if produce windows; 0 otherwise.

Maximize P = 300D + 500Wsubject to

Original Constraints:Plant 1: D ≤ 4Plant 2: 2W ≤ 12Plant 3: 3D + 2W ≤ 18

Auxiliary variables must =1 if produce any:Doors: D ≤ 99y1

Windows: W ≤ 99y2

Mutually Exclusive: y1 + y2 ≤ 1and

D ≥ 0, W ≥ 0, y1 and y2 are binary.


9.33

Spreadsheet Model

3456789

10111213141516171819

B C D E F GDoors Windows

Unit Profit $300 $500

Hours HoursUsed Available

Plant 1 1 0 0 <= 4Plant 2 0 2 12 <= 12Plant 3 3 2 12 <= 18

Doors WindowsUnits Produced 0 6

<= <= Total MaximumOnly If Produce 0 99 Produced To Produce

Produce? 0 1 1 <= 1

Total Profit$3,000



9.34

Wyndor with Either-Or Constraints(Variation 3)

Suppose that now the only change from the original Wyndor problem is:

• The company has just opened a new plant (plant 4) that is similar to plant 3, so the new plant can perform the same operations as plant 3 to help produce the two new products (doors and windows).

• However, management wants just one of the plants to be chosen to work on these new products. The plant chosen should be the one that provides the most profitable product mix.


9.35

Data for Wyndor with Either-Or Constraints(Variation 3)

Production Time Used forEach Unit Produced (Hours)

Production TimeAvailable

per Week (Hours)Plant Doors Windows

1 1 0 4

2 0 2 12

3 3 2 18

4 2 4 28

Unit Profit $300 $500


9.36

Graphical Solution with Plant 3 or Plant 4


9.37


Let D = Number of doors to produce,

W = Number of windows to produce,

y = 1 if plant 4 is used; 0 if plant 3 is used

Maximize P = 300D + 500W

subject to

Plant 1: D ≤ 4

Plant 2: 2W ≤ 12

Plant 3: 3D + 2W ≤ 18 + 99y

Plant 4: 2D + 4W ≤ 28 + 99(1 – y)

and

D ≥ 0, W ≥ 0, y1 and y2 are binary.


9.38

Spreadsheet Model

3456789

10111213141516

B C D E F G HDoors Windows

Unit Profit $300 $500Modified

Hours Hours HoursUsed Available Available

Plant 1 1 0 4 <= 4 4Plant 2 0 2 10 <= 12 12Plant 3 3 2 22 <= 117 18Plant 4 2 4 28 <= 28 28

Doors WindowsUnits Produced 4 5 Total Profit $3,700

Which Plant to Use? (0=Plant 3, 1=Plant 4) 1



9.39

Good Products Company Production Planning

• The Research and Development Division of the Good Products Company has developed three possible new products.

• To avoid undue diversification of the company’s product line, management has imposed the following restriction:

– From the three possible new products, at most two should be chosen to be produced.

• Each of these products can be produced in either of two plants. For administrative reasons, management has imposed the following restriction:

– Just one of the two plants should be chosen to be the sole producer of the two new products.


9.40

Data for the Good Products Company

Production Time Used for EachUnit Produced (Hours)

Production TimeAvailable perWeek (Hours)Plant Product 1 Product 2 Product 3

1 3 4 2 30

2 4 6 2 40

Unit Profit 5 7 3 ($thousands)

Sales potential 7 5 9 (units per week)


9.41


Let xi = Number of units of product i to produce per week (i = 1, 2, 3),yi = 1 if product i is produced; 0 otherwise (i = 1, 2, 3),y4 = 1 if plant 2 is used; 0 if plant 1 is used

Maximize Profit = 5x1 + 7x2 + 3x3 ($thousands)subject to

Auxiliary variables must =1 if produce any & Max Sales:Product 1: x1 ≤ 7y1

Product 2: x2 ≤ 5y2

Product 3: x3 ≤ 9y3

Either plant 1 (y4 = 0) or plant 2 (y4 = 1):Plant 1: 3x1 + 4x2 + 2x3 + 99y4 ≤ 30Plant 2: 4x1 + 6x2 + 2x3 + 99(1 – y4) ≤ 40

andxi ≥ 0 (i = 1, 2, 3), yi are binary (i = 1, 2, 3, 4).


9.42

Spreadsheet Model

3456789

101112131415161718192021

B C D E F G H IProduct 1 Product 2 Product 3

Unit Profit ($thousands) 5 7 3Modified

Hours Hours HoursHours Used Per Unit Produced Used Available Available

Plant 1 3 4 2 34.5 <= 129 30Plant 2 4 6 2 40 <= 40 40

Product 1 Product 2 Product 3Units Produced 5.5 0 9

<= <= <=Only If Produce 7 0 9Maximum Sales 7 5 9 Total Maximum

Produced To ProduceProduce? 1 0 1 2 <= 2

Total Profit($thousands)

Which Plant to Use? (0=Plant 1, 1=Plant 2) 1 54.5


9.43

Supersuds Corporation Marketing Plan

• The Supersuds Corporation is developing its marketing plan for next year’s new products.

• For three of these products, the decision has been made to purchase a total of five TV spots for commercials on national television networks.

• Each spot will feature a single product.

Question: How should the five spots be allocated to these three products?


9.44

Data for the Supersuds Corp. Problem

Profit (Millions)

Number of TV Spots Product 1 Product 2 Product 3

0 $0 $0 $0

1 1 0 –1

2 3 2 2

3 3 3 4


9.45


Let yij = 1 if there are j TV spots for product i; 0 otherwise (i = 1, 2, 3; j = 1, 2, 3)

Maximize Profit = y11 + 3y12 + 3y13 + 2y22 + 3y23 – y31 + 2y32 + 4y33 ($millions)

subject to

Mutually Exclusive:

Product 1: y11 + y12 + y13 ≤ 1

Product 2: y21 + y22 + y23 ≤ 1

Product 3: y31 + y32 + y33 ≤ 1

Total available spots: y11 + 2y12 + 3y13 + y21 + 2y22 + 3y23 + y31 + 2y32 + 3y33 ≤ 5

and

yij are binary (i = 1, 2, 3; j = 1, 2, 3).


9.46

Spreadsheet Model

3456789

101112131415161718

B C D E F G H IProfit

($millions) Product 1 Product 2 Product 3Number 1 1 0 -1

of 2 3 2 2Spots 3 3 3 4

Solution Product 1 Product 2 Product 3 TotalNumber 1 0 0 0 Profit

of 2 1 0 0 ($millions)Spots 3 0 0 1 7

Total 1 0 1<= <= <=

Max Of One 1 1 1 Total RequiredSpots Spots

Number of Spots 2 0 3 5 = 5


9.47

Southwestern Airways Crew Scheduling

• Southwestern Airways needs to assign crews to cover all its upcoming flights.

• We will focus on assigning 3 crews based in San Francisco (SFO) to 11 flights.

Question: How should the 3 crews be assigned 3 sequences of flights so that every one of the 11 flights is covered?


9.48

Southwestern Airways Flights

Seat tl e (SEA)

San Francisco (SFO)

Los Angel es (LAX)

Denver (DEN)

Chicago ORD)


9.49

Data for the Southwestern Airways Problem

Feasible Sequence of Flights

Flights 1 2 3 4 5 6 7 8 9 10 11 12

1. SFO–LAX 1 1 1 1

2. SFO–DEN 1 1 1 1

3. SFO–SEA 1 1 1 1

4. LAX–ORD 2 2 3 2 3

5. LAX–SFO 2 3 5 5

6. ORD–DEN 3 3 4

7. ORD–SEA 3 3 3 3 4

8. DEN–SFO 2 4 4 5

9. DEN–ORD 2 2 2

10. SEA–SFO 2 4 4 5

11. SEA–LAX 2 2 4 4 2

Cost, $1,000s 2 3 4 6 7 5 7 8 9 9 8 9


9.50


Let xj = 1 if flight sequence j is assigned to a crew; 0 otherwise. (j = 1, 2, … , 12).

Minimize Cost = 2x1 + 3x2 + 4x3 + 6x4 + 7x5 + 5x6 + 7x7 + 8x8 + 9x9 + 9x10 + 8x11 + 9x12

(in $thousands)

subject to

Flight 1 covered: x1 + x4 + x7 + x10 ≥ 1

Flight 2 covered: x2 + x5 + x8 + x11 ≥ 1

: :

Flight 11 covered: x6 + x9 + x10 + x11 + x12 ≥ 1

Three Crews: x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 ≤ 3

and

xj are binary (j = 1, 2, … , 12).


9.51

Spreadsheet Model

3456789

101112131415161718192021222324

B C D E F G H I J K L M N O P QFlight Sequence

1 2 3 4 5 6 7 8 9 10 11 12Cost ($thousands) 2 3 4 6 7 5 7 8 9 9 8 9 At

LeastIncludes Segment? Total One

SFO-LAX 1 0 0 1 0 0 1 0 0 1 0 0 1 >= 1SFO-DEN 0 1 0 0 1 0 0 1 0 0 1 0 1 >= 1SFO-SEA 0 0 1 0 0 1 0 0 1 0 0 1 1 >= 1LAX-ORD 0 0 0 1 0 0 1 0 1 1 0 1 1 >= 1LAX-SFO 1 0 0 0 0 1 0 0 0 1 1 0 1 >= 1ORD-DEN 0 0 0 1 1 0 0 0 1 0 0 0 1 >= 1ORD-SEA 0 0 0 0 0 0 1 1 0 1 1 1 1 >= 1DEN-SFO 0 1 0 1 1 0 0 0 1 0 0 0 1 >= 1DEN-ORD 0 0 0 0 1 0 0 1 0 0 1 0 1 >= 1SEA-SFO 0 0 1 0 0 0 1 1 0 0 0 1 1 >= 1SEA-LAX 0 0 0 0 0 1 0 0 1 1 1 1 1 >= 1

Total Number1 2 3 4 5 6 7 8 9 10 11 12 Sequences of Crews

Fly Sequence? 0 0 1 1 0 0 0 0 0 0 1 0 3 <= 3

Total Cost ($thousands) 18


9.52

Integer Programming

• When are “non-integer” solutions okay?– Solution is naturally divisible

• e.g., $, pounds, hours

– Solution represents a rate

• e.g., units per week

– Solution only for planning purposes

• When is rounding okay?– When numbers are large

• e.g., rounding 114.286 to 114 is probably okay.

• When is rounding not okay?– When numbers are small

• e.g., rounding 2.6 to 2 or 3 may be a problem.

– Binary variables

• yes-or-no decisions


9.53

The Challenges of Rounding

• Rounded Solution may not be feasible.

• Rounded solution may not be close to optimal.

• There can be many rounded solutions.

– Example: Consider a problem with 30 variables that are non-integer in the LP-solution. How many possible rounded solutions are there?

1 2 3 4 5

1

2

3

4

5

x1

x2


9.54

How Integer Programs are Solved

1 2 3 4 5

1

2

3

4

5

x1

x2


9.55

How Integer Programs are Solved

1 2 3 4 5

1

2

3

4

5

x1

x2


9.56


• Making “yes-or-no” type decisions– Build a factory?

– Manufacture a product?

– Do a project?

– Assign a person to a task?

• Set-covering problems– Make a set of assignments that “cover” a set of requirements.

• Fixed costs– If a product is produced, must incur a fixed setup cost.

– If a warehouse is operated, must incur a fixed cost.


9.57

Example #1 (Capital Budgeting)

• Norwood Development is considering the potential of four different development projects.

• Each project would be completed in at most three years.

• The required cash outflow for each project is given in the table below, along with the net present value of each project to Norwood, and the cash that is available each year.

Cash Outflow Required ($million)

CashAvailable($million)Project 1 Project 2 Project 3 Project 4

Year 1 9 7 6 11 28

Year 2 6 4 3 0 13

Year 3 6 0 4 0 10

NPV 30 16 22 14

Question: Which projects should be undertaken?


9.58


Let yi = 1 if project i is undertaken; 0 otherwise (i = 1, 2, 3, 4).

Maximize NPV = 30y1 + 16y2 + 22y3 + 14y4

subject to

Year 1: 9y1 + 7y2 + 6y3 + 11y4 ≤ 28 ($million)

Year 2 (cumulative): 15y1 + 11y2 + 9y3 + 11y4 ≤ 41 ($million)

Year 3 (cumulative): 21y1 + 11y2 + 13y3 + 11y4 ≤ 51 ($million)

and

yi are binary (i = 1, 2, 3, 4).


9.59

Spreadsheet Solution

12345678910111213

A B C D E F G H I

Norwood Development Capital Budgeting

Project 1 Project 2 Project 3 Project 4NPV ($million) 30 16 22 14

Cumulative CumulativeOutflow Available

Year 1 9 7 6 11 22 <= 28Year 2 15 11 9 11 35 <= 41Year 3 21 11 13 11 45 <= 51

Total NPVProject 1 Project 2 Project 3 Project 4 ($million)

Undertake? 1 1 1 0 68

Cumulative Outflow Required ($million)


9.60

Additional Considerations(Logic and Dependency Constraints)

• At least one of projects 1, 2, or 3

• Project 2 can’t be done unless project 3 is done

• Either project 3 or project 4, but not both

• No more than two projects total

Question: What constraints would need to be added for each of these additional considerations?


9.61

Example #2 (Set Covering Problem)

• The Washington State legislature is trying to decide on locations at which to base search-and-rescue teams.

• The teams are expensive, so they would like as few as possible.

• Response time is critical, so they would like every county to either have a team located in that county or in an adjacent county.

Question: Where should search-and-rescue teams be located?


9.62

The Counties of Washington State

1

2

3

4

7

6

9

10

11

12

8

5

13

14

15

16

17

18

19

20

2122

2325

24

26 27 28

29 30

31 32

33

3435

36

37

1. Clallum2. J efferson3. Grays Harbor4. Pacific5. Wahkiakum6. Kitsap7. Mason8. Thurston9. Whatcom10. Skagit11. Snohomish12. King13. Pierce14. Lewis15. Cowlitz16. Clark17. Skamania18. Okanogan

19. Chelan20. Douglas21. Kittitas22. Grant23. Yakima24. Klickitat25. Benton26. Ferry27. Stevens28. Pend Oreille29. Lincoln30. Spokane31. Adams32. Whitman33. Franklin34. Walla Walla35. Columbia36. Garfield37. Asotin

Counties


9.63


Let yi = 1 if a team is located in county i; 0 otherwise (i = 1, 2, … , 37).

Minimize Number of Teams = y1 + y2 + … + y37

subject to

County 1 covered: y1 + y2 ≥ 1

County 2 covered: y1 + y2 + y3 + y6 + y7 ≥ 1

County 3 covered: y2 + y3 + y4 + y7 + y8 + y14 ≥ 1

:

County 37 covered: y32 + y36 + y37 ≥ 1

and

yi are binary (i = 1, 2, … , 37).


9.64


123456789101112131415161718192021222324

A B C D E F G H I J K L M N

Search & Rescue Location

# Teams # TeamsCounty Team? Nearby County Team? Nearby

1 Clallam 0 1 >= 1 19 Chelan 0 2 >= 12 Jefferson 1 1 >= 1 20 Douglas 0 1 >= 13 Grays Harbor 0 2 >= 1 21 Kittitas 1 1 >= 14 Pacific 0 1 >= 1 22 Grant 0 1 >= 15 Wahkiakum 0 1 >= 1 23 Yakima 0 3 >= 16 Kitsap 0 1 >= 1 24 Klickitat 0 1 >= 17 Mason 0 1 >= 1 25 Benton 0 1 >= 18 Thurston 0 1 >= 1 26 Ferry 0 1 >= 19 Whatcom 0 1 >= 1 27 Stevens 1 1 >= 110 Skagit 1 1 >= 1 28 Pend Oreille 0 1 >= 111 Snohomish 0 1 >= 1 29 Lincoln 0 1 >= 112 King 0 1 >= 1 30 Spokane 0 1 >= 113 Pierce 0 2 >= 1 31 Adams 0 1 >= 114 Lewis 1 2 >= 1 32 Whitman 0 2 >= 115 Cowlitz 0 2 >= 1 33 Franklin 1 1 >= 116 Clark 0 1 >= 1 34 Walla Walla 0 1 >= 117 Skamania 1 2 >= 1 35 Columbia 0 1 >= 118 Okanogan 0 1 >= 1 36 Garfield 1 1 >= 1

37 Asotin 0 1 >= 1Total Teams: 8


9.65

Example #3 (Fixed Costs)

• Woodridge Pewter Company is a manufacturer of three pewter products: platters, bowls, and pitchers.

• The manufacture of each product requires Woodridge to have the appropriate machinery and molds available. The machinery and molds for each product can be rented at the following rates: for the platters, $400/week; for the bowls, $250/week; for the pitcher, $300/week.

• Each product requires the amounts of labor and pewter given in the table below. The sales price and variable cost are also given in the table.

LaborHours

Pewter(pounds)

SalesPrice

VariableCost

Platter 3 5 $100 $60

Bowl 1 4 85 50

Pitcher 4 3 75 40

Available 130 240

Question: Which products should be produced, and in what quantity?


9.66


Let x1 = Number of platters produced,x2 = Number of bowls produced,x3 = Number of pitchers produced,yi = 1 if lease machine and mold for product i; 0 otherwise (i = 1, 2, 3).

Maximize Profit = ($100–$60)x1 + ($85–$50)x2 + ($75–$40)x3 – $400y1 – $250y2 – $300y3

subject toLabor: 3x1 + x2 + 4x3 ≤ 130 hoursPewter: 5x1 + 4x2 + 3x3 ≤ 240 poundsAllow production only if machines and molds are purchased:

x1 ≤ 99y1

x2 ≤ 99y2

x3 ≤ 99y3

andxi ≥ 0, and yi are binary (i = 1, 2, 3).


9.67


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A B C D E F G H

Woodridge Pewter Company

Platters Bowls PitchersSales Price $100 $85 $75

Variable Cost $60 $50 $40Fixed Cost $400 $250 $300

Constraint Usage (per unit produced) Total AvailableLabor (hrs.) 3 1 4 60 <= 130

Pewter (lbs.) 5 4 3 240 <= 240

Lease Equipment? 0 1 0Revenue $5,100

Production Quantity 0 60 0 Variable Cost $3,000<= <= <= Fixed Cost $250

Produce only if Lease 0 99 0 Profit $1,850


9.68


• Making “yes-or-no” type decisions– Build a factory?

– Manufacture a product?

– Do a project?

– Assign a person to a task?

• Fixed costs– If a product is produced, must incur a fixed setup cost.

– If a warehouse is operated, must incur a fixed cost.

• Either-or constraints– Production must either be 0 or ≥ 100.

• Subset of constraints– meet 3 out of 4 constraints.


9.69

Capital Budgeting with Contingency Constraints(Yes-or-No Decisions)

• A company is planning their capital budget over the next several years.

• There are 10 potential projects they are considering pursuing.

• They have calculated the expected net present value of each project, along with the cash outflow that would be required over the next five years.

• Also, suppose there are the following contingency constraints:– at least one of project 1, 2 or 3 must be done,

– project 4 and project 5 cannot both be done,

– project 7 can only be done if project 6 is done.

Question: Which projects should they pursue?


9.70

Data for Capital Budgeting Problem

Cash Outflow Required ($million)

CashAvailable($million)

Project

1 2 3 4 5 6 7 8 9 10

Year 1 1 4 0 4 4 3 2 8 2 6 25

Year 2 2 2 2 2 2 4 2 3 3 6 25

Year 3 3 2 5 2 4 2 3 4 8 2 25

Year 4 4 4 5 4 5 3 1 2 1 1 25

Year 5 1 1 0 6 5 5 5 1 1 2 25

NPV 20 25 22 30 42 25 18 35 28 33 ($million)


9.71


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A B C D E F G H I J K L M N O

Capital Budgeting with Contingency Constraints

Project Project Project Project Project Project Project Project Project Project1 2 3 4 5 6 7 8 9 10

NPV ($million) 20 25 22 30 42 25 18 35 28 33Cumulative Cumulative

Cumulative Cash Outflow Required ($million) Total Outflow AvailableYear 1 1 4 0 4 4 3 2 8 2 6 22 <= 25Year 2 3 6 2 6 6 6 4 11 5 12 44 <= 50Year 3 6 8 7 8 10 8 7 15 13 14 73 <= 75Year 4 10 12 12 12 15 11 8 17 14 15 97 <= 100Year 5 11 13 12 18 20 16 13 18 15 17 117 <= 125

Project Project Project Project Project Project Project Project Project Project Total NPV1 2 3 4 5 6 7 8 9 10 ($million)

Undertake? 1 1 1 0 1 1 1 0 1 1 213

Contingency ConstraintsProject 1,2,3 3 >= 1Project 4,5 1 <= 1Project 7 1 <= 1 Project 6


9.72

Electrical Generator Startup Planning (Fixed Costs)

• An electrical utility company owns five generators.

• To generate electricity, a generator must be started up, and associated with this is a fixed startup cost.

• All of the generators are shut off at the end of each day.

Generator

A B C D E

Fixed Startup Cost $2,450 $1,600 $1,000 $1,250 $2,200

Variable Cost (per MW) $3 $4 $6 $5 $4

Capacity (MW) 2,000 2,800 4,300 2,100 2,000

Question: Which generators should be started up to meet the total capacity needed for the day (6000 MW)?


9.73


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A B C D E F G H I J

Electrical Utility Generator Startup Planning

Generator A Generator B Generator C Generator D Generator EFixed Startup Cost $2,450 $1,600 $1,000 $1,250 $2,200Cost per Megawatt $3 $4 $6 $5 $4Max Capacity (MW) 2,000 2,800 4,300 2,100 2,000

Startup? 1 1 0 1 0Total MW MW Needed

MW Generated 2,100 3,000 0 900 0 6000 >= 6,000<= <= <= <= <=

Capacity 2,000 2,800 0 2,100 0

Fixed Cost $5,300Variable Cost $22,800

Total Cost $28,100


9.74

Quality Furniture (Either-Or Constraints)

• Reconsider the Quality Furniture Problem:– The Quality Furniture Corporation produces benches and picnic tables. The firm

has a limited supply of two resources: labor and wood. 1,600 labor hours are available during the next production period. The firm also has a stock of 9,000 pounds of wood available. Each bench requires 3 labor hours and 12 pounds of wood. Each table requires 6 labor hours and 38 pounds of wood. The profit margin on each bench is $8 and on each table is $18.

• Now suppose that they would not produce any fewer than 200 units of either product (i.e., either produce 0 or at least 200).

Question: What product mix will maximize their total profit?


9.75


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A B C D E F G

Quality Furniture (with either-or constraints)

Benches TablesProfit $8.00 $18.00

Min Production (if any) 200 200

Resources ResourcesUsed Available

Labor 3 6 1600 <= 1,600Wood 12 38 6400 <= 9,000

Produce? 1 0

Min Production 200 0<= <= Total Profit

Production Quantities 533.33 0 $4,266.67<= <=

Max Production 533 0Max Possible 533 237

Use of Resources


9.76

Meeting a Subset of Constraints

• Consider a linear programming model with the following constraints, and suppose that meeting 3 out of 4 of these is good enough

– 12x1 + 24x2 + 18x3 ≥ 2,400

– 15x1 + 32x2 + 12x3 ≥ 1,800

– 20x1 + 15x2 + 20x3 ≤ 2,000

– 18x1 + 21x2 + 15x3 ≤ 1,600


9.77

Meeting a Subset of Constraints

Let yi = 1 if constraint i is enforced; 0 otherwise.

Constraints:

y1 + y2 + y3 + y4 ≥ 3

12x1 + 24x2 + 18x3 ≥ 2,400y1

15x1 + 32x2 + 12x3 ≥ 1,800y2

20x1 + 15x2 + 20x3 ≤ 2,000 + M (1 – y3)

18x1 + 21x2 + 15x3 ≤ 1,600 + M (1 – y4)

where M is a large number.


9.78

Facility Location

• Consider a company that operates 5 plants and 3 warehouses that serve customers in 4 different regions.

• To lower costs, they are considering streamlining by closing one or more plants and warehouses.

• Associated with each plant are fixed costs, shipping costs, and production costs. Each plant has a limited capacity.

• Associated with each warehouse are fixed costs and shipping costs. Each warehouse has a limited capacity.

Questions:Which plants should they keep open?

Which warehouses should they keep open?

How should they divide production among the open plants?

How much should be shipped from each plant to each warehouse, and from each warehouse to each customer?


9.79

Data for Facility Location Problem

FixedCost

(per month)

(Shipping + Production) Cost(per unit)

Capacity(units per

month)WH #1 WH #2 WH #3

Plant 1 $42,000 $650 $750 $850 400

Plant 2 50,000 500 350 550 300

Plant 3 45,000 450 450 350 300

Plant 4 50,000 400 500 600 350

Plant 5 47,000 550 450 350 375

Fixed Cost(per month)

Shipping Cost (per unit)

Capacity(per month)Cust. 1 Cust. 2 Cust. 3 Cust. 4

WH #1 $45,000 $25 $65 $70 $35 600

WH #2 25,000 50 25 40 60 400

WH #3 65,000 60 20 40 45 900

Demand: 250 225 200 275


9.80


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A B C D E F G H I J K L M

Plant to WarehouseShipping + Production FixedCost Warehouse 1 Warehouse 2 Warehouse 3 Cost Capacity

Plant 1 $650 $750 $850 $42,000 400Plant 2 $500 $350 $550 $50,000 300Plant 3 $450 $450 $350 $45,000 300Plant 4 $400 $500 $600 $50,000 350Plant 5 $550 $450 $350 $47,000 375

Shipment Total ActualQuantities Warehouse 1 Warehouse 2 Warehouse 3 Shipped Capacity Open? Total Costs

Plant 1 0 0 0 0 <= 0 0 Shipping Cost (P-->W) $332,500Plant 2 0 300 0 300 <= 300 1 Shipping Cost (W-->C) $37,375Plant 3 0 0 275 275 <= 300 1 Fixed Cost (P) $142,000Plant 4 0 0 0 0 <= 0 0 Fixed Cost (W) $90,000Plant 5 0 0 375 375 <= 375 1 Total Cost $601,875

Total Shipped 0 300 650

Warehouse to CustomerShipping FixedCost Customer 1 Customer 2 Customer 3 Customer 4 Cost Capacity

Warehouse 1 $25 $65 $70 $35 $45,000 600Warehouse 2 $50 $25 $40 $60 $25,000 400Warehouse 3 $60 $20 $40 $45 $65,000 900

Shipment Shipped Shipped ActualQuantities Customer 1 Customer 2 Customer 3 Customer 4 Out In Capacity Open?

Warehouse 1 0 0 0 0 0 <= 0 <= 0 0Warehouse 2 250 0 50 0 300 <= 300 <= 400 1Warehouse 3 0 225 150 275 650 <= 650 <= 900 1Total Shipped 250 225 200 275

>= >= >= >=Needed 250 225 200 275

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