1 Decision Science BMGT 825 Fall 2006, UNK Professor: Ron Konecny Ph.D. University of Nebraska at Kearney

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Decision ScienceBMGT 825

Fall 2006, UNK

Professor: Ron Konecny Ph.D.

University of Nebraska at Kearney

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Table of Contents

Class Syllabus Homework Assignments Daily Discussion Topics GP/LP Basics Sample Problems

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Contact Information

Office Hours:249W West CenterPhone 865-8366email: [email protected] hours: M-TH 2:45-3:45

Appointments can be arranged for times outside of normal office hours. Students are encouraged to seek extra assistance if needed. Please do not delay visiting during office hours if you have questions on the material. Please call ahead for an appointment outside of normal office hours.

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Required Course Materials

1. An Introduction to Management Science, 11th ed., 2005, Anderson, Sweeney, Williams

2. The Goal, 3rd revised edition, 2004, Goldratt and Cox, North River Press 3. Decision Science 2005 Computer software (Provided by instructor –free-)4. Access to Windows NT/ME/XP platform computer with internet access and

Excel 2002 or newer.5. The Department of Management is a member of the Microsoft Academic

Alliance. This permits all MBA students taking BMGT 825 to receive free software. The list is extensive. You may download and use full versions of any/all of these software packages beyond the end of this class. You will receive an email from the MSDNAA (Microsoft Developer Network Academic Alliance) giving you a user name and a password.

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Course Description

Course Description:Recent developments relating to business application of linear programming, simplex method, transportation method, post-optimality analysis, game theory, utility theory, PERT-CPM, queuing theory, dynamic programming, Markov chains, Decision tree analysis, time series, analysis and forecasting.

Course Objectives:

To introduce students to some quantitative methods and techniques of management science. To cultivate their skill in the application of those methods and techniques. To encourage students to apply the learned tools in business applications. To present state-of-the-art modeling techniques.

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Course Evaluation

Course Performance Evaluation300 points : 3 examinations: 100 points each100 points : homework submission 50 points : Notebook & short papers450 points total

Grade assignments will correspond to standard UNK policy,Notebooks will be graded on completeness, organization, presentation, and neatness.

A detailed listing of homework assignments and daily discussion topics is contained on the Decision Science Software CD.

Other supporting information for the class may be found at http://Platteriver.unk.edu/BMGT825

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Course Evaluation - continued

Course Outline: Test Schedule• Exam 1: Due September 20 - Chapters 1 - 6, 8, Introduction to

Management Science, linear programming, sensitivity analysis, integer programming, and applications.

• Exam 2: Due November 8 - Chapters 7, 15 & handouts, transportation & assignment methods, multi-objective programming, integer multi-objective programming, set notation.

• Final: December 13 - Chapters 4, 10, 13 data envelopment analysis, program evaluation and review technique, critical path method, simulation.

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Graphical & Algebraic solutions Chapter 2, Problem 21 Chapter 2, Problem 34 Chapter 2, Problem 50- Include the these three hand solved problems in the notebook.

Reading Assignment

•Chapter 1, pg. 1-17

•Chapter 2, all sections

•The Goal, chapters 1-20

Homework Assignment, due 8/30

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Linear Programming Chapter 2, Problem 21 – use software Chapter 2, Problem 34 – use software Chapter 2, Problem 50 – use software Chapter 4, Problem 1 Chapter 4, Problem 2 (modified a and c) Chapter 4, Problem 3 Chapter 4, Problem 8

Reading Assignment

•Chapter 3

•Chapter 4, sections 1-3

•The Goal – Chapters 21-40

All homework problems Include the following printouts•model sheet•output sheet•variable sheet

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Linear Programming Chapter 4, Problem 14 Chapter 4, Problem 16 Chapter 4, Problem 17 Margaret Black Farm - on Platteriver website

Multi-period Problems Antelope Endowment Fund (AEF)

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Exam #1 is due. Electronic submission due by 5:59pm on the http://platteriver.unk.edu/bmgt825 web server.

Academic Dishonesty. The maintenance of academic honesty and integrity is a vital concern of the University community. Any student found guilty of academic dishonesty shall be subject to both academic and disciplinary sanctions. Academic dishonesty includes, but is not limited to, the following: 1. Cheating. Copying or attempting to copy from an academic test or examination of another student; using or attempting to use unauthorized materials, information, notes, study aids or other devices for any academic test, examination or exercise; engaging or attempting to engage the assistance of another individual in misrepresenting the academic performance of a student; or communicating information in an unauthorized manner to another person for an academic test, examination or exercise. 2. Fabrication and Falsification. Falsifying or fabricating any information or citation in any academic exercise, work, speech, test or examination. Falsification is the alteration of information, while fabrication is the invention or counterfeiting of information. 3. Plagiarism. Presenting the work of another as one's own (i.e., without proper acknowledgment of the source) and submitting examinations, theses, reports, speeches, drawings, laboratory notes or other academic work in whole or in part as one's own when such work has been prepared by another person or copied from another person. 4. Abuse of Academic Materials. Destroying, defacing, stealing, or making inaccessible library or other academic resource material. 5. Complicity in Academic Dishonesty. Helping or attempting to help another student to commit an act of academic dishonesty. 6. Falsifying Grade Reports. Changing or destroying grades, scores, or markings on an examination or in an instructor's records. 7. Misrepresentation to Avoid Academic Work. Misrepresentation by fabricating an otherwise justifiable excuse such as illness, injury, accident, etc., in order to avoid or delay timely submission of academic work or to avoid or delay the taking of a test or examination. 8. Other. Academic units and members of the faculty may prescribe and give student prior notice of additional standards of conduct for academic honesty in a particular course, and violation of any such standard of conduct shall constitute misconduct under this Code of Conduct and the University Disciplinary Procedures.

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Transportation & Assignment Problems – without set notation Chapter 7, Problem 2 (transportation) Chapter 7, Problem 20 (assignment) Chapter 7, Problem 26 (transshipment)

Reading Assignment


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- with set notation Chapter 7, Problem 2 (transportation) Chapter 7, Problem 20 (assignment) Chapter 7, Problem 26 (transshipment) Chapter 7, Problem 29 (transshipment) Chapter 4, Problem 8 (cyclical set)

Reading Assignment


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Set Notation Problems Loper Dairy Bank Location Problem Section 8.3

– The bank location problem is on page 396. You do not need to solve as an integer problem. You must use set notation. Hint: Variable X is only one dimensional.

Production Scheduling, Section 4.3Production Scheduling Hints (pages 173 – 179)1.You will need to create a variable name set for March, April, May, and June. March has the previous inventory. Remember, the For Every and Sum statements may contain a condition such as •For Every t, t>1 : •The above statement type of statement will permit you to look back to the previous month such as•For Every t, t>1 : INVENTORY.Comp802B.(t-1) + …

2. You will need to set the ending inventory for March for both the Component332A and the Component802B individually. For example:•Inventory.Comp802B.March = 500•Inventory.Comp322A.March = 200

3. You can construct the Machine and Labor capacity constraints on page 176 with one “For Every” statement. Treat the storage capacity separately.4. You will need variables for production, ending inventory, and two variables for inventory change. •One variable (the book uses I1 ) for month to month increases •One variable (the book used D1) for month to month decreases (see page 177).

5. The objective function (see page 175) has four summation statements.6. The answer on page 178 shows two-dimensional variables and one-dimensional variables.

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All problems must use set notation

Production and Inventory Application, Section 7.4 (page 374) Blending Problem, Section 4.4 (page 183) Multi-Commodity Transportation

Mixed Integer Linear Programs Chapter 8, problem 13 , a and b American Settlement Problem

Reading Assignment

•Chapter 8

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Set Notation Problems Queens on a chess board (bonus) or Sudoku (bonus)

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Multi-criteria Decision Problems (or Goal Programming) Chapter 15, Problem 2 Chapter 15, Problem 3 Chapter 15, Problem 4 Chapter 15, Problem 5 Chapter 15, Problem 7 Kearney City Council

Reading Assignment

•Chapter 15, sec 1, 2

•Chapter 4, sec 5

•“Data Envelopment Analysis: Partial Survey and Applications for Management Accounting,” Callen, Jeffrey L, Journal of Management Accounting Research, Vol 3, Fall 1991

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Exam #2 is due. Class will meet at 5:59 on 11/9/2005

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Project Scheduling – Microsoft Project

- Critical Path Method Chapter 10, Problem 6 Chapter 10, Problem 7 Chapter 10, Problem 14 (PERT) Chapter 10, Problem 20 Chapter 10, Problem 22

Reading Assignment

• Chapter 10

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PERT, by simulation Chapter 10, Problem 14

– Use Excel, with Normal Distribution

– Use Excel, with Beta Distribution

Reading Assignment

•Chapter 10, review

•Handout material

Notebooks are Due at the end of class.In class final is next week.

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Discussion Topics (8/23, Week 1)

August 23– Introduction to Management Science– Graphic solution

• Loper Dairy

• Millworks Plywood

– Algebraic Solution– RHS and objective coefficient ranging

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Discussion Topics (8/30 Week 2)

August 30– Discuss homework

• shadow prices• RHS and objective coefficient ranging

– Features of DS software– Introduction to Decision Science Software

• installing• Editing, Saving, Printing Simplex method

– Discuss problem types• Production mix• Ingredient mix• Production Scheduling

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September 6– Discuss The Goal and homework problems

– Identity equations• Efficiency

• Profit, revenue, and cost functions

– Multi-period investment problems

– Natural resource problems

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September 13– Discuss The Goal and homework problems

Distribute Exam #1• covers material in first 4 weeks of class

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September 20– Transportation and assignment examples

Discuss Exam #1

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September 27– Discuss transportation homework– Summation notation exercise– Introduction to set notation

• name sets• variable sets• coefficient sets• ordered and cyclical sets

– summation notation– ‘for every’ notation– Software utilization of set notation

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October 4– Discuss homework– Coefficient tables– Applications

• Bank location example

• Old Farms

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October 11– Discuss homework– Generalized LP forms using set notation– Pattern usage with set notation– Advanced set referencing

• Queens on a chess board• Knights on a chess board• subsets

– Mixed Integer Linear Programming• Discrete variables• Fixed costs• Multiple choice and Mutually exclusive constraints

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October 18

– Homework problems

– Introduction Goal Programming (multiple-objective)• Kearney City Council

– Integer Goal Programming

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October 25

– Homework problems

– Integer & multiple-objective programming using sets

– Introduction to Data Envelopment Analysis• Callen reading

• facets, production frontier, relative efficiency, scale efficiency, most productive scale size

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November 2– Discuss homework– Introduction to Networking models

• Project Network Graph (digraph),

• Critical Path Method – CPM

– Distribute Exam 2

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November 8– Networking models Continued

• Critical Path Method (CPM)

• CPM with crashing

• Activity schedule

• LP/GP solutions

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November 15– Other Networking models

• Program Evaluation and Review Technique(PERT)

• simulation

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November 29– Evaluation of PERT & CPM

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December 6– Homework notebooks due– Advances in Management Science

• Genetic Algorithms

• Expert Systems

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December 14– Final Exam, in class

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An Introduction to Linear Programming

Linear Programming Problem Problem Formulation A Maximization Problem Graphical Solution Procedure

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Requirements of a Linear Programming Problem All linear programming (LP) problems seek to maximize

or minimize some quantity, such as profit or cost. This is called Optimization of the Objective Function.

The quantity of the objective is limited by a system of constrains.– Land, labor, capital, prices, contracts, limited resources

There must exist multiple alternatives. However, some of these alternatives may give rather poor results.

The objective function and constraints in an LP must be expressed in terms of linear equations or inequalities

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Problem Formulation

Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

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Guidelines for Model Formulation

Understand the problem thoroughly. Write a verbal description of the objective. Write a verbal description of each constraint. Define the decision variables. Write the objective in terms of the decision

variables. Write the constraints in terms of the decision

variables.

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Linear Programming using the Decision Science software

Equation label double click on the box to create or change a label.

Duplication

By dragging the square dot on the right bottom a cell you my duplicate a line numerous times.

Equation EditingYou may edit an equation by placing the cursor on a cell and pressing the F2 key. If you do not want to keep the change just press the ESC key before leaving the cell.

Creating, Saving, and Printing a file is performed from the menu bar.

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Goal Programming using the Decision Science software

Documentation

By starting a line with a double quotation a line is considered strictly as documentation. Documentation lines appear in green.

Adding PrioritiesYou may add an extra priority by placing the cursor in the priority list and pressing Edit - Insert Row or ALT-E-I.

You may also remove a priority by pressing Edit Remove Row

To create a new Goal Program/Multi-criteria Decision Model select File-New-Goal program from the menu bar.

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Sample Problems

Loper Dairy (linear program) Millworks Plywood (linear program) Antelope Development Fund (set notation) Old Farms (set notation) Queens on a chess board (set notation) Knights on a chess board (set notation) Kearney City Council (goal program) Shortest Route - map (transshipment, set) Multi-commodity Shipping (3D set) American Settlement Problem (set) Suncoast Office Supplies (goal program) Hospital Evaluation (Data Envelopment Analysis)

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Loper DairyAt Loper Dairy specialty yogurt and cheese are produced and

sold nationally. The production of one case of yogurt requires 2 machine

hours and 1 labor hour. Profit for selling one case of yogurt is $8.00.

The production of one case of cheese requires 2 machine hours and 2 labor hours. Profit for selling one case of cheese is $12.00

There are only 120 machine hours and 80 labor hours available.

Profits equal revenue minus costs.

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Loper Dairy

Cost Availability machine hours per machine hours perCase of Yogurt Case of Cheeselabor hours per labor hours perCase of Yogurt Case of Cheese

RevenueProfit

Resource Requirements

$16/machine hour

$10/labor hour

120 machine hours

80 labor hours

Yogurt Cheese

$50/Case of Yogurt $64/Case of Cheese


Machine

Labor

2 2

1 2

Yogurt Revenue $50

Machine Cost -$32

Labor Cost -$10

$8Yogurt Profit

Cheese Revenue $64

Machine Cost -$32

Labor Cost -$20

$12Cheese Profit

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Loper Dairy – LP model•Identify the decision variables

•What can change in the model?•Labor hours? Machine hours? Cases of yogurt produced? Cases of cheese produced?• Y - Cases of yogurt produced • C - Cases of cheese produced.

•Write the objective function•What do we want to do?

Max Profit = 8 Y + 12 C

cheese of cases C cheese of cases

$ + 12of yogurt cases Y

yogurtof case$

Max Profit($)= 8

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Loper Dairy – LP model•Write the constraints

120 (machine hours)2 Y + 2 C


Machine hours + 2of yogurt cases Y

yogurtof caseMachine hours

2 120

machine hours

80 (labor hours)1 Y + 2 C


Labor hours + 2of yogurt cases Y

yogurtof caseLabor hours

1 80 labor hours

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Loper Dairy – Graphical Solution

20 40 60 80

20

40

60

80

00

Cheese

Yog

urt ≤ 120 (machine hours)2 Y + 2 C

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20 40 60 80

20

40

60

80

00

Cheese

Yog

urt

≤ 80 (labor hours)1 Y + 2 C

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20 40 60 80

20

40

60

80

00

Cheese

Yog

urt


≤ 120 (machine hours)2 Y + 2 C

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20 40 60 80

20

40

60

80

00

Cheese

Yog

urt


≤ 120 (machine hours)2 Y + 2 C

Feasible Region

Infeasible

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Loper Dairy – Corner Point Principle

20 40 60 80

20

40

60

80

00

Cheese

Yog

urt

1 Y + 2 C ≤ 80 (labor hours)

2 Y + 2 C ≤ 120 (machine hours)

Feasible Region

Max Profit= 8 Y + 12 CSubject to:

( C, Y ) Profit

(0, 0)

(40, 0)

(0, 60)

(20, 40) $0$480$480$560

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120

104

656

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Loper Dairy – Sensitivity Analysis

Y=16.0C=44.0

2.0 2.0

1.0 2.0

Y+ C

Y+ C

Y+ C

Feasible Region

Profit=

Max Profit =Subject to

Cheese

Yog

urt

20 40 60 800

20

40

60

80

0

(machine hours)

(labor hours)

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Loper Dairy - Linear Program

The objective function multiplies the profit for each item sold by the number of items sold.

The variable “Yogurt” represents the number of cases of yogurt produced and sold. The variable “Cheese” represents the number of cases of cheese produced and sold.

The “LaborHours” constraint illustrates that when one case of yogurt is produced then one labor hours are used. Similarly, when one case of cheese is produced then two labor hours are used.

Examining the “Variables” table is useful in making sure the model describes the problem correctly. The “count” column shows how many times a variable is used in the model. If the count equals one for a variable then most likely there is a spelling or logic error.

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Loper Dairy - simplex tableau

Initial Simplex Tableau - Selecting the “Tableau” sheet activates the simplex presentation. The cursor is placed on the pivot element (the intersection of the pivot column and pivot row)

The “Final Tableau” can be viewed by selecting “final” under the “Step” menu option or by clicking on the “Tableau” sheet after running solving the problem using “Run Local Solution.” The cursor will appear over the word “Basis” when the solution is optimal.

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Loper Dairy - identity variable and equations

Determining profit for an individual product can be quite tedious. The table at right shows how the profit for one case of yogurt is determined. If you were to determine the profit for selling an automobile with thousands of parts the calculation would be nearly impossible, especially if the purchase prices changed often.

Revenue $50 per case- Machine Cost ($32) 2 hours at $16/hour- Labor Cost ($10) 1 hour at $10/hour

Profit $8 per case

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Loper Dairy - Decreasing Cost

machine hours per machine hours perCase of Yogurt Case of Cheeselabor hours per labor hours perCase of Yogurt Case of Cheese

Revenue

Machine

Labor

2 2

1 2

Resource Requirements


Yogurt Cheese

At Loper Dairy specialty yogurt and cheese are produced and sold nationally. As a member of the production management team you are responsible for determining the proper allocation of the resources of machinery (capital) and labor to produce a mix of yogurt and cheese that results in the maximum profit.

Labor is $10/hour for the first 80 hours and $15/hour for the next 20 hours (overtime). There are three possible contracts for the machine time. Contract #1 has a machine costs of $20/hour. Contract #2 has a machine cost of $15/hour plus a $200 start fee. Contract #3 has a machine cost of $10/hour plus a $600 start fee. There are only 120 hours of machine time available for any contract.

Production resource requirements are listed in the table below.

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Competing Technologies

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Millworks Plywood

At Millworks Plywood has a contract to produce 800 sheets of Grade A plywood and 600 sheets of Grade B plywood. Two separate production lines are available to produce plywood.

The first production line, Alpha Line, can produce 10 sheets of grade A plywood and 10 sheets of grade B plywood in one hour at a machine cost of $5.00 per hour.

The second production line, Beta Line, can produce 20 sheets of grade A plywood and 10 sheets of grade B plywood in one hour as a machine cost of $7.00 per hour.

Minimize the cost to Millworks Plywood in meeting the contract.

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Hartman Company, Problem 4.2

Modify the information in the problem to reflect the changes below.

Labor-HoursDepartment Prod 1 Prod 2 Available Cost/Hour

A 1 0.35 100 12B 0.3 0.2 36 15C 0.2 0.5 50 8

Revenue 48.10 26.20

Product(hours/unit)

Do not use the Profit contribution/unit row as presented in the text problem.

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Antelope Endowment Fund

The Antelope Endowment Fund has five million dollars to invest. The AEF gives $400,000 in scholarships to university students at the beginning of each year. The AEF can invest the available funds in common stock, treasury bills (T-bills), and local bank certificate of deposits (CD’s).

For every dollar invested in common stock a profit of $.10 is expected after one year and T-bills are expected to earn $.14 on the dollar at the end of the second year. The T-bills should be held for two years to avoid excess transactions costs. CD’s must be held for 3 three years for a return of $.25 for every dollar invested.

As Chief Executive Officer you plan to maximize the total value of the fund at the end of your term. Since your term expires at the end of four years, you plan to sell all assets at the end of the forth year for the next CEO to invest at their discretion. As a risk averse organization, the AEF board desires to hold at least 30% of all monies invest in T-Bills and at least 25% in CD’s. Further, investments in stocks in the third year must be at least 10% above investments in stocks in the second year. Investments in stocks in the fourth year should be less than 80% of the monies invested in stock in the third year. Finally, AEF must hold $200,000 in cash on hand for emergency purposes. Maximize the AEF assets.

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AEF - Investment PossibilitiesYear 1 Year 2 Year 3 Year 4

Stock1

Stock2

Stock3

Stock4

TBill1

TBill2

TBill3

CD1

CD2

Cash1

Cash2

Cash3

Cash4

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Hart Manufacturing, Prob 8.11

Modify the information in the problem to reflect the changes below.

Cost/Department Prod 1 Prod 2 Prod 3 Available Hour

A 1.50 3.00 2.00 450 2.00B 2.00 1.00 2.50 350 4.00C 0.25 0.25 0.25 50 8.00

Revenue 38.00 40.00 46.00

Product (hours/unit)

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Old Farms

Mark is a crop consultant for Old Farms near Loop River. The farm raises three crops; corn, alfalfa, and soybeans. For simplicity assume that it is possible to raise dry-land crops or irrigated crops in any field. The Old Farm may sell as much of corn, alfalfa, or soybeans that they can raise (there is no sales limit). However, when the crop is sold may greatly effect the farm's profit. Futures prices for the spring are higher than the expected price for harvest time. It is possible to store a limited amount corn, alfalfa, and soybeans for the spring. The consultant has recommended that you store no more than 50% of your alfalfa yield for spring sale. Water limitations differ from field to field. There is, however, an overall limitation of 4200 acre-ft of water that Old Farms can use for the entire growing season.

Sale Price Sale Price(harvest) (spring)

Corn $2.00/bushel $2.50/bushelAlfalfa $42.00/ton $51.00/tonSoybeans $4.50/bushel $5.25/bushel

•What is the best production plan for each section? •What is the value of one more acre-foot of water? •What is the marginal value of the storage capacity of each crop?

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Storage CapacitySince Old Farm stores alfalfa in round bails there is no limitation regarding storage. Storage capacity for corn and soybeans combined is 150,000 bushels.

Dry-land yields and irrigated yields vary from one parcel of land to another. There is no quality difference between dry-land and irrigated crops. The following table represents the potential yields by crop in each field.

Field\Crop Corn Alfalfa Soybeans Corn Alfalfa SoybeansSoutheast 110 bu/acre 1.0 tons/acre 35 bu/acre 180 bu/acre 1.6 tons/acre 47 bu/acre North 110 bu/acre 0.9 tons/acre 38 bu/acre 200 bu/acre 1.5 tons/acre 51 bu/acre Northwest 90 bu/acre 1.5 tons/acre 39 bu/acre 210 bu/acre 1.5 tons/acre 53 bu/acre West 105 bu/acre 1.1 tons/acre 30 bu/acre 190 bu/acre 1.4 tons/acre 41 bu/acre Southwest 95 bu/acre 1.2 tons/acre 27 bu/acre 160 bu/acre 1.5 tons/acre 40 bu/acre

Expected Dry Land Yields Expected Irrigated Yields

Water requirement for each crop also differs by parcel. The following table gives the water requirement for each crop planted in each field. The NRD (Natural Resource District) limits the amount of water allocated to each field

Water Limit Field SizeField\Crop Corn Alfalfa Soybeans Acre-ft/fieldSoutheast 1.5 acre-ft 2.3 acre-ft 0.8 acre-ft 1500 acre-ft 1920 acresNorth 1.4 acre-ft 0.0 acre-ft 0.7 acre-ft 1700 acre-ft 2240 acresNorthwest 1.2 acre-ft 2.1 acre-ft 0.8 acre-ft 1300 acre-ft 820 acresWest 1.6 acre-ft 2.6 acre-ft 0.9 acre-ft 800 acre-ft 1280 acresSouthwest 1.6 acre-ft 0.0 acre-ft 0.8 acre-ft 200 acre-ft 640 acres

Water requirements by Crop

Old Farms - continued

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Old Farms Linear Program (set)

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Old Farms - Coefficient Tables

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Queens on a Chess Board

The objective is to place as many queens on a chess board as possible with one stipulation. At most one queen my lay on any row, any column, and any diagonal.

The purpose of this exercise is to demonstrate advanced features in variable indexing.

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diagA.r2: Q.r2.c1 + Q.r1.c2 <= 1 diagA.r3: Q.r3.c1 + Q.r2.c2 + Q.r1.c3 <= 1 diagA.r4: Q.r4.c1 + Q.r3.c2 + Q.r2.c3 + Q.r1.c4 <= 1 diagA.r5: Q.r5.c1 + Q.r4.c2 + Q.r3.c3 + Q.r2.c4 + Q.r1.c5 <= 1 diagA.r6: Q.r6.c1 + Q.r5.c2 + Q.r4.c3 + Q.r3.c4 + Q.r2.c5 + Q.r1.c6 <= 1 diagA.r7: Q.r7.c1 + Q.r6.c2 + Q.r5.c3 + Q.r4.c4 + Q.r3.c5 + Q.r2.c6 + Q.r1.c7 <= 1 diagA.r8: Q.r8.c1 + Q.r7.c2 + Q.r6.c3 + Q.r5.c4 + Q.r4.c5 + Q.r3.c6 + Q.r2.c7 + Q.r1.c8 <= 1

diagB.r1: Q.r1.c1 + Q.r2.c2 + Q.r3.c3 + Q.r4.c4 + Q.r5.c5 + Q.r6.c6 + Q.r7.c7 + Q.r8.c8 <= 1 diagB.r2: Q.r2.c1 + Q.r3.c2 + Q.r4.c3 + Q.r5.c4 + Q.r6.c5 + Q.r7.c6 + Q.r8.c7 <= 1 diagB.r3: Q.r3.c1 + Q.r4.c2 + Q.r5.c3 + Q.r6.c4 + Q.r7.c5 + Q.r8.c6 <= 1 diagB.r4: Q.r4.c1 + Q.r5.c2 + Q.r6.c3 + Q.r7.c4 + Q.r8.c5 <= 1 diagB.r5: Q.r5.c1 + Q.r6.c2 + Q.r7.c3 + Q.r8.c4 <= 1 diagB.r6: Q.r6.c1 + Q.r7.c2 + Q.r8.c3 <= 1 diagB.r7: Q.r7.c1 + Q.r8.c2 <= 1

diagC.c2: Q.r8.c2 + Q.r7.c3 + Q.r6.c4 + Q.r5.c5 + Q.r4.c6 + Q.r3.c7 + Q.r2.c8 <= 1 diagC.c3: Q.r8.c3 + Q.r7.c4 + Q.r6.c5 + Q.r5.c6 + Q.r4.c7 + Q.r3.c8 <= 1 diagC.c4: Q.r8.c4 + Q.r7.c5 + Q.r6.c6 + Q.r5.c7 + Q.r4.c8 <= 1 diagC.c5: Q.r8.c5 + Q.r7.c6 + Q.r6.c7 + Q.r5.c8 <= 1 diagC.c6: Q.r8.c6 + Q.r7.c7 + Q.r6.c8 <= 1 diagC.c7: Q.r8.c7 + Q.r7.c8 <= 1

diagD.c2: Q.r1.c2 + Q.r2.c3 + Q.r3.c4 + Q.r4.c5 + Q.r5.c6 + Q.r6.c7 + Q.r7.c8 <= 1 diagD.c3: Q.r1.c3 + Q.r2.c4 + Q.r3.c5 + Q.r4.c6 + Q.r5.c7 + Q.r6.c8 <= 1 diagD.c4: Q.r1.c4 + Q.r2.c5 + Q.r3.c6 + Q.r4.c7 + Q.r5.c8 <= 1 diagD.c5: Q.r1.c5 + Q.r2.c6 + Q.r3.c7 + Q.r4.c8 <= 1 diagD.c6: Q.r1.c6 + Q.r2.c7 + Q.r3.c8 <= 1 diagD.c7: Q.r1.c7 + Q.r2.c8 <= 1

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Queens - output table

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Knights on a Chess Board

The objective is to place as many knights on a chess board as possible without any knight jeopardizing any other knight.

The purpose of this exercise is to demonstrate advanced features in variable indexing and output table formatting.

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Spreadsheet Formatting

Formatting CellsFormatting spreadsheet cell font size, font color, background color, boarders, alignment, and numeric presentation is performed by double clicking on the right mouse button. The “Workbook Designer” appears permitting Excel type format changes. When done editing press alt-F4 or click on the (x) in the top right corner to return to the Decision Science screen.

Equations in Cells

DS permits Excel type equations in any of the coefficient tables. Equations may reference cells in other tables. In the example at left, an “if” statement places a “K” in a cell if there is a “1” in the corresponding cell in the “Knight” table.

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Multi-commodity TransportationThe purpose of this problem is to demonstrate the use of three-dimensional indexing. Multi-commodity transportation problems can become extremely large with hundreds of thousands of constraints and variables. However, with the use of set notation such problems are manageable.

Suppose that a steel firm ship produces Bands, Plates, and Coils at three different foundries; Gary, Cleveland, and Pittsburgh. These item are shipped to Framingham, Detroit, Lansing, Windsor, St. Louis, Fremont, and Lafayette. Supply from each steel firm and demand for each city are listed below.

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Multi-commodity Transportation- Costs

Transportation costs (variable costs) of shipping products from the supplier to the destination are given in the following tables. There is also a limitation that no more that 625 units (all items combined) are permitted to ship along any one route.

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Multi-commodity Transportation- Model

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American Settlement Problem Your niece asked you to help construct an early American settlement for a science project from

her oversized box of Legos. Her box contains 20 big brown pieces, 30 small brown pieces, 22 white pieces, 15 red pieces, 40 tan pieces, 20 body pieces, 4 blue hats, 4 green hats, 4 black hats, and 20 rods.

In the community there are nine types of people: farmers, carpenters, forgers (includes all smiths and metal workers), clergy, teachers, fishermen, grain millers (includes bakers), soldiers (includes hunters), and administrators. Each member of the community offers unique physical, mental, and spiritual insights.

In the town you may build eleven types of structures: churches, schools, houses, fences, boat piers, boats, wagons, granaries, market, bakery shops (serves also as a grain mill), and carpentry shops (serves also as a lumber mill). You may also construct four types of animals: horses, mules, sheep, and chickens.

Community Requirements If a bakery is built then there must be at least one miller. If a carpentry shop is built then there

must be at least one carpenter. If a church is built then there must be at least one clergy. At most, only one church can be built. The number of fishermen must be greater than or equal to the number of boats. The number of boats must be greater than or equal to the number of piers. The number of schools must be less than or equal to the number of teachers. The community only needs one school. The number of farmers must be greater than or equal to the number of wagons. A fence protects the animals from predators and from running away, therefore, the number of fences built must be less than or equal to the number of animals. Each person unit represents one family, which must have housing. One house can hold one or two families.

In order for the community to survive it must have 30 organization points, 40 contentment points, and 20 food points.

Maximize the number of people in the community.

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American Settlement Problem

Contentment Organization FoodPeopleadministrators 1 4 -2carpenters 2 1 -3clergy 3 2 -2farmers 0 -1 4fishermen -2 0 2forgers 1 0 -3millers 0 0 1soldiers -1 1 -2teachers 2 2 -1Structureschurch 10 2bakery 1 2boat 1carpentry shop 2fence 1granary 2house 1 1market 3 2pier 1 1school 5 1wagons 2 1Animalschickens 2horses 1mules 2sheep 1

Persons in the community produce food, contentment, and organization for the community. Negative values represent the creation of discontentment and disorganization.

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American Settlement Problem

Big Small Blue Green BlackBrown Brown White Red Tan Body Hat Hat Hat rods

Peopleadministrators 1 1carpenters 1 1 1 2clergy 1 1 1farmers 1 1 3fishermen 1 1 1 2forgers 1 1 1millers 1 1 1soldiers 1 1 1teachers 1 1 1Structureschurch 4 3 2 1 1bakery 2 2 2boat 1 1 1 2carpentry shop 2 1 1fence 1granary 2 1 2house 1 1 1 1market 3 3 1 1 1 1pier 2 2school 3 2 2 2wagons 1 1 2Animalschickens 2 1horses 1 1 3mules 2 1 2

sheep 3 1

Construction Requirements - Construction of each community member, animal, and building requires a different combination of Legos pieces. The table below lists the strengths of each member and the Lego piece requirement.

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Kearney City CouncilA proposal to develop 80 acres of land was presented to the Kearney City Council. Negotiations with the developers of Shipwreck Point has lead to the following goals:

Priority 1: Build at least 500 family units.

Priority 2: Add at least 60 million dollars to the property tax base.

Priority 3: The amount services financed by the city must remain under $250,000.

Priority 4: Provide space for at least a 5 acre park

Priority 5: At least 40% of the family units must live in single family dwellings

Housing Statistics applying to this project are given in the following table

Single DeluxeFamily Condo Apartment

Land Usage per dwelling .25 acre .8 acre .75 acreFamily Units per dwelling 1 4 6Tax Base per dwelling 200,000$ 640,000$ 600,000$ Required Services per Dwelling 1,500$ 1,200$ 2,500$

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Kearney City Council - Variable List

integer

integer

integer

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Shortest Route using capacitated transshipment

Major Cities•Kearney, NE•Albuquerque, NM•Cheyenne, WY•Dallas, TX•Denver, CO•Des Moines, IA•Kansas City, MO•Minneapolis, MN•Oklahoma City, OK•Omaha, NE

Find the shortest route between any two cities

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Shortest Route Map - model

The miles.i.j<>’null’ statement permits exclusion of routes. Traditionally, route prohibitions are created by listing the miles between the two prohibited routes as a very large value. If mileage in the MILES table from i to j is not listed then the route is considered prohibited. Any given mileage, even 0, is considered valid and the route is included. The miles.i.j in constraints StartCity and StopCity is not used directly in the constraint equation but is used to limit the routes possible.

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Shortest Route - tablesDetermine the shortest route from Cheyenne, WY to Dallas, TX.

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Shortest Route - tablesDetermine the shortest route from Cheyenne, WY to Dallas, TX.

The optimal route is stored in the ROUTES table as well as the OUTPUT table.

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Suncoast Office Supplies See section 15.2 in Anderson, Sweeney, and Williams

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Suncoast Office Supplies - Variable List

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Hospital Evaluation - DEA See section 4.5 in Anderson, Sweeney, Williams

The DEA variable identification table

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Hospital Evaluation - continued

The DEA data input table. This is found on the ‘Model’ spreadsheet

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Hospital Evaluation - output

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Summation Notation - one dimensional

The use of summation notation greatly simplifies the addition of a range of values.

The following summation shows a one dimension summation of the variable X.

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Summation Notation - 2 dimensional

Client1 Client2 Client3

Terry X1,1 X1,2 X1,3

Carle X2,1 X2,2 X2,3

McClymondsX3,1 X3,2 X3,3

Variable NamesClient1 Client2 Client3

Terry 10 15 9Carle 9 18 5McClymonds 6 14 3

Completion Times

See table 7.3 in text

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‘For Every’ Notation

Client1 Client2 Client3

Terry X1,1 X1,2 X1,3

Carle X2,1 X2,2 X2,3

McClymondsX3,1 X3,2 X3,3

Variable NamesClient1 Client2 Client3

Terry 10 15 9Carle 9 18 5McClymonds 6 14 3

Completion Times

This example continues from the ‘Summation Notation’ . Example from ASW text, figure 7.4

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F-117 Stealth Fighter

B-2 Bomber

F-22 Raptor

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Survey

Name of your favorite cartoon character: _______________

1. How many DVDs have you rented this month?

2. How many parking tickets from UNK have you received this year?

3. Chocolate chip recipe: _______________________

_______________________

_______________________

_______________________

Documents

1 Decision Science BMGT 825 Fall 2006, UNK Professor: Ron Konecny Ph.D. University of Nebraska at Kearney