56:171 Operations Research

1
56171 Operations Research
Instructor Prof. Yong Chen TA Qingyu Yang
M/W/F 1230 - 120
Fall 2005
2
Introduction

• INSTRUCTORYong Chen
• - Background
• - Availability
• TEXT
• - Text Book
• - Lecture Notes
• COURSE
• - Web site, Email list
• - Computing
• - Attendance Policy
• - Prerequisites
• - Grading Policy
• TA
• Lab/Recitation

3
Background

• B. E. in computer science, Tsinghua University,
  China, 1998
• Ph. D., 2003, Industrial and Operations
  Engineering, University of Michigan
• Assistant professor, Dept. of Mechanical and
  Industrial Engineering, 2003now
• Research area Quality and reliability
  engineering Sensor system design and analysis

4
Prerequisites

• Basic linear algebra and calculus knowledge
• Basic probability and statistics knowledge
• Basic computing skills

5
Review Questions
How many Yes do you get?

• Do you know how to solve a system of linear
  equations using Gaussian elimination?
• Can you draw a line on x-y plane given its
  equation?
• Do you know the meaning of convex and concave
  functions?
• Can you calculate the first and second
  derivatives of a given function?
• Do you know what is conditional probability?
• Do you know what is probability density function?
• Do you know the definition of expectation and
  variance of a random variable?
• Do you know how to use Excel to do some simple
  calculation?

6
Grading Policy

• Attendance/participation 5
• Homework 20
• Exam 1 25
• Exam 2 25
• Exam 3 25

• - Grade for attendance is based on random quizzes
  in lectures and labs
• - Homework should be submitted in-class on the
  due date
• - No late homework is acceptable
• but one HW grade is not counted in the final
  grading.

7
Course Coverage

• Deterministic
• Linear programming
• Transportation models
• Assignment models
• Integer programming
• Nonlinear programming
• Network models
• PERT/CPM

• Stochastic
• Markov chains
• Queueing theory

8
What is Operations Research?

• From military research on (military) operations
• Today, operations research means a scientific
  approach to decision making, which seeks to
  determine how best to design and operate a
  system, usually under conditions requiring the
  allocation of scarce resources.

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Decision-Making

• We all make decisions, all of the time
• Day-to-day choices
• Business decisions
• Public policy decisions
• Feasibility vs. optimality decisions
• Easy vs. difficult

10
Course Selection Example

• You have to decide your choice of courses for all
  the semesters that you are here
• Your choices are restricted by a number of rules
• You have to achieve a minimum number of credit
  hours
• You have to take each of the required courses at
  some point during the program

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Course Selection Example, Cont.

• You can take a course only if you have taken its
  pre-reqs
• You can not take two courses with time conflict
• You can not take more than a maximum number of
  credits each semester
• You have to take enough GEC courses from
  specified categories
• You have to satisfy the EFA requirements
• You can not take classes before 9 because you
  hate getting up early

12
Course Selection Example, Cont.

• The rules limit the possible choices you have,
  but there are still a large number of choices!
• How can you pick the best set of courses for
  yourself?

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Course Selection Example, Cont.

• You need to decide your choice of courses
• There are a large number of choices which satisfy
  the given rules
• You need to find the courses that maximize some
  measure of your performance, e.g., expected GPA
• This is an optimization problem

14
Another Example The Diet Problem

• Given a collection of foods (e.g. milk,
  chocolate, orange juice, pizza), determine how
  much of each food to eat in a given day
• Goal is to minimize cost or calories, or maximize
  satisfaction
• Have to satisfy rules limiting our choices

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The Diet Problem, Cont.

• An example
• Minimize the cost of my diet, subject to
  satisfying minimum requirements of protein, and
  maximum limits on calories and fat

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The Diet Problem, Cont.
Decisions Milk Chocolate Orange juice Pizza

• Objective
• Minimize cost

• Rules
• Minimum protein requirement
• Maximum calories limit
• Maximum fat limit

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The Diet Problem, Cont.

• Stigler 1945
• Posed problem
• Solved heuristically
• Dantzig 1963, 1990
• Solved optimally in 1947 using simplex method
• Appears in many Operations Research texts
• New journal articles still appearing

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The Diet Problem, cont.

• IDEAL DIET
• 1.31 cups wheat flour 1.32 cups rolled oats
• 16 oz. milk 3.86 tbsp peanut butter
• 7.28 tbsp lard 0.0108 oz. beef
• 1.77 bananas 0.0824 oranges
• 0.707 cups cabbage 0.314 carrots
• 0.387 potatoes 0.53 cups pork and beans

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Optimization Problem

• Optimization problem
• Decisions
• Means of comparing decisions
• Rules governing interactions between decisions

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Mathematical Models

• Once an optimization problem is defined in words,
  we need to find an appropriate mathematical model
  to analyze/solve it
• A mathematical model captures the essence of the
  problem
• An idealized version/approximation of the problem

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Formulation / Model

• Formulation or model mathematical
  representation of an optimization problem
• Decision variables
• Objective function
• Constraints
• Parameters (Constants)

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The Diet ProblemDecision Variables

• Decision variables a number of variables whose
  values are to be determined
• Define the decision variables as
• xm Gallons of milk consumed daily
• xc Bars of chocolate consumed daily
• xo Gallons of orange juice consumed daily
• xp Pizzas consumed daily

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The Diet ProblemObjective Function

• Objective minimize daily cost
• Let ci , for i m, c, o, p be the current cost
  per unit for item i
• Objective function cmxm ccxc coxo cpxp
• Objective function the measure of performance
  expressed as a function of the decision variables

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The Diet ProblemConstraints

• Constraints Restrictions on the values that can
  be assigned to the decision variables
• We may want to limit our diet so that
• total fat content in the diet does not exceed
  some limit
• total calories do not exceed some limit
• total protein intake is at least some minimum
  amount
• We need the data for the fat, calorie, and
  protein value per unit of each item and the
  limits we want to meet

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The Diet ProblemConstraints, Cont.

• Suppose fi , wi , pi , for i m, , p are the
  values of fat, calories, and proteins per unit of
  item i, respectively
• Suppose F, W, and P are the daily limits on fat,
  calories, and protein, respectively
• We have all the data to write our constraints

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The Diet ProblemConstraints, Cont.

• We want
• Daily fat intake fmxm fcxc foxo fpxp ? F
• Daily calorie intake wmxm wcxc woxo wpxp ?
  W
• Daily protein intake pmxm pcxc poxo ppxp ?
  P
• And, none of the intake amounts for each food is
  negative

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The Diet ProblemParameters

• Parameters the constants used to specify the
  objective function and the constraints
• For the diet problem, ci, fi, wi, pi, F, W, and P
  are parameters
• In practice, it is difficult to determine the
  parameters exactly
• Sensitivity analysis is needed

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The Diet ProblemComplete Model

• Minimize cmxm ccxc coxo cpxp
• Subject to
• fmxm fcxc foxo fpxp ? F
• wmxm wcxc woxo wpxp ? W
• pmxm pcxc poxo ppxp ? P
• xm ? 0, xc ? 0, xo ? 0, xp ? 0

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Other Applications

• Optimization has been used in a number of serious
  applications yielding huge profits
• A good place to find information about these
  applications is the Interfaces journal
• Available electronically

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Applications from Interfaces

• Recovering from major airline disruptions (Horner
  2002)
• Reducing travel costs and player fatigue in NBA
  (Bean and Birge 1980)
• Assigning managers to construction projects
  (LeBlanc et al. 2000)
• Managing consumer credit delinquency (Makuch et
  al. 1992)
• Manufacturing of beer cans (Katok and Ott 2000)

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More Applications

• Scheduling prototype vehicle testing at Ford
  (Chelst et al. 2001)
• Portfolio construction (Bertsimas et al. 1999)
• Scheduling police patrol officers (Taylor and
  Huxley 1989)
• Planning closure and realignment of army bases
  (Dell 1999)
• Assigning restoration capacity in a
  telecommunication network (Ambs et al. 2000)
• Crew scheduling for airlines (Butchers 2000)

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Ch. 3 Introduction to Linear Programming

• The Linear Programming Model
• Examples
• Assumptions of Linear Programming
• Graphic Solutions of LP
• Excel Solver
• Suggested Readings 3.1-3.6 of HL

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Start of Linear Programming

• George Dantzig (1914 2005),
• Father of Linear Programming
• Junior Statistician U.S. Bureau of Labor
  Statistics (1937-39)
• Head of USAF Combat Analysis Branch (1941-46)
• PhD Mathematics, Cal Berkeley (1946)
• Invented Simplex method for solving linear
  programs (1947)
• Medal of Science, 1975, for his work in LP

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Linear Programs

• Programming means planning
• Linear programs (LPs) have
• Linear objective function
• Linear constraints
• Continuous variables
• Typically very easy to solve, even when quite
  large

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Why Linear Programs?

• Many real-world problems can be modeled as LPs
• Many other problems can be approximated as LPs
• LP solution techniques provide the foundation for
  solution methods for many other structures of
  Mathematical Programs (MPs)

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Function Types

• Linear functions have the general form
• f(x1, x2, , xn) c1x1 c2x2 cnxn where c1,
  c2, , cn are constants
• Linear functions are simple

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Function Types (Cont.)

• Examples of non-linear functions
• Polynomial f(x, y,z) x2 y2 z2
• Cross terms f(x, y, z) xy
• Exponential f(x) ex
• Maximum f(x, y, z) max x, y, z
• Absolute f(x) x
• More complex

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Linear Constraints

• Linear constraints are of three types
• ?
• pmxm pcxc poxo ppxp ? P, must have a
  minimum quantity of proteins
• ?
• fmxm fcxc foxo fpxp ? F, at most F units of
  fat in the diet
• P. 46 of textbookradiation therapy example

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General Form of Linear Constraints

• Any linear constraint has the following general
  form

a linear function of decision variables ? ? a constant
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Not linear Constraints

• Examples
• x2 y2 ? 1
• xy y 2z ? 1
• (x y z) is integer
• Again, more complex

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Example 1 Wyndor Glass

• Wyndor makes doors and windows. They have three
  plants. A batch of doors requires 1 hour at Plant
  1 plus 3 hours at Plant 3. A batch of windows
  requires 2 hours at Plant 2 plus 2 hours at Plant
  3. Plant 1 is available for 4 hours per week,
  Plant 2 for 12 hours per week, and Plant 3 for 18
  hours per week. The profit per door batch is
  3000 and the profit per window batch is 5000.
  What should Wyndor manufacture to maximize
  profits?

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Wyndor GlassResource Consumption
Production Time per Batch, hours Production Time per Batch, hours
Product Product Production time
Plant Door Window Available hours per week
1 1 0 4
2 0 2 12
3 3 2 18
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Wyndor Glass (Cont.)

• What are our decisions?
• How many doors should we make?
• How many windows should we make?
• What is our goal?
• Maximize profits
• What are our rules?
• Dont exceed capacity at plant 1
• Dont exceed capacity at plant 2
• Dont exceed capacity at plant 3

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Wyndor GlassLP Model

• What are our decisionsDecision Variables
• x1 gt 0 batches of doors per week
• x2 gt 0 batches of windows per week
• What is our goalObjective Function
• Maximize 3000 x1 5000 x2
• What are our rulesConstraints
• (Plant 1) x1 ? 4
• (Plant 2) 2x2 ? 12
• (Plant 3) 3x1 2x2 ? 18
• (Non-negativity) x1 ? 0, x2 ? 0

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Example 2 Lego Chair and Table
Make tables and chairs to maximize profits.
Profit 16 for each Table, 10 for each
Chair Each table uses 2 large blocks and 2 small
blocks Each chair uses 1 large block and 2 small
blocks You are limited by the availability of
material. You only have 6 large blocks and 8
small blocks. How many tables and chairs should
you make to maximize profit?
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How to Make the Chair and Table?
47

Model

• Decision variables
• t of tables
• c of chairs
• Objective function
• Maximize 16t10c
• Constraints

2t c lt 6
6 (large legos)
8 (small legos)
2t 2c lt 8
48
Optimal Solution of Lego Game
of tables
2
of chairs
2
Total profit
52
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Product Mix

• There are n different products that I can
  produce. Each of these products consumes certain
  quantity of m different resources. Every unit of
  a product i, for i 1,, n uses aij units of
  resource j, for j 1,, m. I can make a profit
  of pi per unit of item i. The amount of resource
  j available is uj. What should I do to maximize
  my profit?

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Connection

• Product mix is a general version of Wyndor Glass
  example and Lego chair and table example
• Wyndor glass n 2, m 3
• Lego chair and table n2, m2
• Many business resource allocation problems take
  this form

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Product Mix General Model

• Decision variables
• xi Amount of item i produced daily
• Objective function
• maximize profit p1x1 p2x2 pnxn
• Constraints
• (Resource Availability)
• a1j x1 a2j x2 anj xn ? uj for j
  1,, m
• (Non-negativity)
• xi ? 0 for i 1,, n

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Assumption of LPs

• When we write a problem as a linear program, we
  are making a few assumptions about the underlying
  process
• Proportionality The contribution of a decision
  variable to the objective function or any one of
  the constraints is proportional to its value,
  e.g.,
• The daily fat in-take from the pizza is
  proportional to the amount of pizza eaten daily
• of small blocks used is proportional to the
  number of chairs made
• The total profit is proportional to the number of
  chairs or tables made

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Assumptions (Cont.)

• Additivity The total contribution to the
  objective and left hand side of each constraint
  is the sum of individual contributions of each
  activity
• The total cost of daily food is the sum of costs
  from individual food items
• The total profit is the sum of profits from chair
  and table
• Divisibility Each decision can take any real
  value
• Daily amount of milk consumption
• Amount invested in a one-year CD at the beginning
  of year 1

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Assumptions (Cont.)

• Big assumption Certainty
• The value of each parameter needed in the linear
  programming is known with certainty
• This assumption is almost never satisfied
• Sensitivity analysis can help to find out the
  robustness of our optimal solutions to
  uncertainty of data

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LP ModelA Standard Form

• Max c1x1 c2x2 cnxn
• subject to
• (functional constraints)
• a11x1 a12x2 a1nxn ? b1
• a21x1 a22x2 a2nxn ? b2
• am1x1 am2x2 amnxn ? bm
• (nonnegativity constraints)
• x1 ? 0, x2 ? 0, , xn ? 0
• The number of variables is n and the number of
  constraints is mn

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Notation

• We can use some notation to write linear programs
  compactly
• We can write c1x1 c2x2 cnxn as
• We can write each constraint aj1x1 aj2x2
  ajnxn ? bj as

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Standard Form

• Using the previous notation, the standard form of
  LP can be written as
• max
• st

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Some Terminology

• Any specification of values for the decision
  variables (x1,, xn) is called a solution.
• A feasible solution is a solution for which all
  the constraints are satisfied.
• An infeasible solution is a solution for which at
  least one constraint is violated.
• A feasible solution is called an optimal solution
  if there is no other feasible solution with
  objective function value better than it
• The optimal value of a problem is the objective
  value of an optimal solution to the problem

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Example

• Min 2x1 3x2
• subject to x1 5x2 3
• 3x1 2
• x1 ? 0, x2 ? 0
• In this case, x1 2/3 and x2 (3-2/3)/5 7/15
  is a feasible solution for the problem
• In fact, it is the only solution to the problem
• So it is also the optimal solution to the problem
  as well
• Optimal value of the problem is
  2(2/3)3(7/15)2.73

60
Wyndor Glass Co. Problem
max Z3x1 5x2 (in K) s.t. x1 ?
4 (1) 2x2 ? 12 (2) 3x1 2x2 ?
18 (3) x1 ? 0, x2 ? 0

• x1 0, x2 0 is a feasible solution with Z0
• Is it optimal?
• x1 0, x2 4 is a feasible solution with Z20,
  better than (0, 0)
• Is this the optimal solution?

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Feasible Region
The feasible region is the collection of all
feasible solutions
x2
x1
(0,0)
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Moving Isovalue Line

• Now that we have the feasible region, how do we
  find the best solution?
• (0,4) has value 20, all solutions with value 20
  lie on the isovalue line 3x1 5x2 20

x2
x1
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Finding the Optimal Point
The optimal point occurs at the intersection of
these two lines
2x212 Plant 2 3x12x2 18 Plant 3
x1 2 x2 6
Optimal value Z3x15x236Maximal profit is 36K
64
Graphical Solutions

• Min -2x - y
• St x y lt 3
• x lt 2
• y lt 2
• x, y gt 0

-2x - y -1
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Graphical Solutions

• Min -2x - y
• St x y lt 3
• x lt 2
• y lt 2
• x, y gt 0

66
Graphical Solutions

• Min -2x - y
• St x y lt 3
• x lt 2
• y lt 2
• x, y gt 0

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Graphical Solutions

• Min -2x - y
• St x y lt 3
• x lt 2
• y lt 2
• x, y gt 0

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Solving LPs Graphically

• Procedures
• Identify feasible region
• Plot an isovalue line corresponding to a feasible
  solution
• Move line in improving direction and find the
  last isovalue line touching the feasible region
• Any point(s) on the intersection of the last
  isovalue line and feasible region are optimal
  solutions

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Finding Optimal Solutions

• Min 3x1 x2
• s.t. x1 x2 lt 6
• x1 lt 4
• x2 lt 4
• x1, x2 gt 0

Which point is optimal?
70
Finding Optimal Solutions

• Min -2x1 x2
• s.t. x1 x2 lt 6
• x1 lt 4
• x2 lt 4
• x1, x2 gt 0

Which point is optimal?
71
Finding Optimal Solutions

• Max x1 x2
• s.t. x1 x2 lt 6
• x1 lt 4
• x2 lt 4
• x1, x2 gt 0

Which point is optimal?
72
Graphical LP Solutions

• Works well for 2 decision variables
• Possible for 3 decision variables
• Impossible for 4 variables
• Other solution approaches necessary
• Good to illustrate concepts, aid in conceptual
  understanding

73
Example

• 3.2-2(a) of HL The colored area in the
  following graph represents the feasible region of
  a LP problem whose objective function is to be
  maximized. Label the following statement as True
  or False and give an example of an objective
  function that illustrates your answer.
• (a) If (3,3) produces a larger value of the
  objective function than (0, 2) and (6, 3), then
  (3,3) must be an optimal solution.

x2
x1
74
Property of Optimal Solution

• In all cases there is a corner point of the
  feasible solution region that is an optimal
  solution
• Is this a coincidence?
• NO!
• Important result Whenever a linear program has
  an optimal solution and it has a corner, there is
  always an optimal solution on one of the corners
  of the feasible region
• Note that the statement does not say that a
  linear program always has an optimal solution, it
  does not say that all optimal solutions have to
  be on the corners, in fact it does not presume
  that there will be corner points!

75
Pathological Cases
What are the possibilities that a linear program
does not have an optimal solution at a corner
point?

• Infeasibility There is no feasible solution to
  the linear program, e.g.,
• Min x1
• s.t. x1 ? -1
• x1 ? 0
• There is no real value that is simultaneously
  less than 1 and greater than 0

76
Infeasible LPs
77
Pathological Cases

• Unboundedness The linear program is feasible but
  the optimal value is not finite, e.g.,

• max x1 x2
• s.t.
• x1 ? 3
• x2 ? 4
• x1, x2 ? 0

78
Unbounded LPs
79
Pathological Cases

• No corner points The feasible region has no
  corner points

min x1 s.t. x1 ? 0 x2 is free
Any solution with x1 0 is optimal!
This case can never happen for LPs in a standard
form
80
Pathological Cases

• Are there other possibilities when a linear
  program may not have an optimal solution at a
  corner point?
• NO!
• Any linear program falls under one of the four
  cases (i) infeasible, (ii) unbounded, (iii) no
  corner point, (iv) has an optimal solution at a
  corner point
• There can be multiple optimal solutions to a
  linear program

81
Multiple Optima
What if the objective function of the Wyndor
problem was 3x1 2x2 instead of 3x1 5x2?
The optimal value of 18 is achieved by all the
solutions on the line segment joining (2,6) and
(4,3)!
82
Multiple Optima
83
Summary

• A linear program always satisfies one of the four
  cases
• Pathological cases
• It is infeasible
• It is unbounded
• It has no corner points but it has optimal
  solutions
• Normal case
• It has an optimal solution at one of the corner
  point feasible solution (among possibly many
  others)

84
Summary

• We shall always transform a linear program into
  one of the standard forms before solving it
• We dont need to worry about the third
  pathological possibility
• We only need to worry about infeasibility and
  unboundedness
• When our LP (in standard form) is not infeasible
  or unbounded, there is a corner point feasible
  solution which is an optimal solution

85
Introduction to Excel Solver
86
Loading Solver

• Standard with every version of Excel
• Insert MS Office or Excel master CD
• Click on Add/Delete components
• Open Add-In tools
• Click on Solver or add all
• Click OK
• Solver should now appear in the Tools menu

87
56171 Operations Research
Ch. 4 Solving Linear Programs The Simplex Method
88
Outline

• The Simplex Method
• Simplex Method for Standard Form
• Theory of the Simplex Method
• Simplex Method for other LP problems
• Suggested Reading 4.1, 4.2, 4.4, 4.5, 4.6

89
Review of Wyndor Glass Example
Product Mix LP. Wyndor makes doors and windows.
They have three plants. A batch of doors requires
1 hour at Plant 1 plus 3 hours at Plant 3. A
batch of windows requires 2 hours at Plant 2 plus
2 hours at Plant 3. Plant 1 is available for 4
hours per week, Plant 2 for 12 hours per week,
and Plant 3 for 18 hours per week. The profit per
door batch is 3000 and the profit per window
batch is 5000. What should Wyndor manufacture to
maximize profits?
Max Z 3x1 5x2 profits (in thousands of
) s.t. 1x1 ? 4 Plant 1 2x2
? 12 Plant 2 3x1 2x2 ? 18 Plant 3 x1, x2
? 0 non-negativity
90
Standard Augmented Forms
Standard Form with Nonnegative RHS
Max Z 3x1 5x2 s.t. 1x1 ? 4
2x2 ? 12 3x1 2x2 ? 18 x1, x2 ? 0
Augmented Form
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
91
Geometric Representation of Simplex Method
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
x2
(0,9)
(2,6)
(0,6)
(4,3)
x1
(6, 0)
(4, 0)
(0,0)
92
Geometric Representation of Simplex Method
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
x2
(0,9)
(2,6)
(0,6)
x1 0 x2 0 s1 4 s2 12 s3 18 Z 0
(4,3)
x1
(6, 0)
(4, 0)
(0,0)
93
Geometric Representation of Simplex Method
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
x2
(0,9)
(2,6)
(0,6)
x1 0 x2 6 s1 4 s2 0 s3 6 Z 30
(4,3)
x1
(6, 0)
(4, 0)
(0,0)
94
Geometric Representation of Simplex Method
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
x2
(0,9)
(2,6)
(0,6)
x1 2 x2 6 s1 2 s2 0 s3 0 Z 36
(4,3)
x1
(6, 0)
(4, 0)
(0,0)
95
Algebraic Representation
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0

• 3 equations in 5 unknowns
• Multiple solutions
• Guided search to move to optimal solution
• Simplex Method

96
Tabular Representation
Max Z 3x1 5x2 s.t. 1x1 s1
4 2x2 s2 12 3x1 2x2 s3
18 x1 , x2 , s1 , s2 , s3 ? 0
Initial Simplex Method Tableau
x1 x2 s1 s2 s3 RHS
Z -3 -5 0 0 0 0
s1 1 0 1 0 0 4
s2 0 2 0 1 0 12
s3 3 2 0 0 1 18
97
Simplex Method (Tabular Form)
x1 x2 s1 s2 s3 RHS
Z -3 -5 0 0 0 0
s1 1 0 1 0 0 4
s2 0 2 0 1 0 12
s3 3 2 0 0 1 18

• s1, s2 and s3 in this tableau represent basic
  variables
• x1 and x2 are non-basic variables
• Basic solutions are obtained by setting the
  non-basic variables to zero and solve the basic
  variables
• Basic solutions represent corner points
• Systematically change basic solution to improve
  objective function
• while maintaining feasibility!

98
Basic Variables in Simplex Tableaux
x1 x2 s1 s2 s3 RHS
Z -3 -5 0 0 0 0
s1 1 0 1 0 0 4
s2 0 2 0 1 0 12
s3 3 2 0 0 1 18

• Any column corresponding to a basic variable in
  Simplex tableaux should have one and only one
  nonzero element, which is equal to 1
• The collection of the basic variables is called
  the basis

99
Each Iteration of Simplex Method

• Optimality Test the current basic solution is
  optimal if and only if every coefficient in row 0
  is nonnegative (?0)
• Determine the entering basic variable by
  selecting the variable with the most negative
  coefficient in row 0.
• Determine the leaving basic variable by applying
  the minimum ratio test
• Pick each coefficient in pivot column that is gt0
• Divide each of them into the RHS
• Identify the row with smallest ratio
• Solve the new basic solution by using elementary
  row operations (multiply a row by a constant or
  add/subtract a multiple of the pivot row to/from
  another row)

100
More Simplex Examples
Max Z 2x1 - x2 x3 s.t. 3x1 x2 x3
? 60 x1 - x2 2x3 ? 10 x1 x2 -
x3 ? 20 x1 , x2 , x3 ? 0
x1 x2 x3 s1 s2 s3 RHS
Z -2 1 -1 0 0 0 0
s1 3 1 1 1 0 0 60
s2 1 -1 2 0 1 0 10
s3 1 1 -1 0 0 1 20
101
More Simplex Examples
Max Z 60x1 30x2 20x3 s.t. 8x1 6x2
x3 ? 48 4x1 2x2 1.5x3 ? 20 2x1
1.5x20.5x3 ? 8 x2 ? 5 x1 , x2 , x3 ?
0
x1 x2 x3 s1 s2 s3 s4 RHS
Z -60 -30 -20 0 0 0 0 0
s1 8 6 1 1 0 0 0 48
s2 4 2 1.5 0 1 0 0 20
s3 2 1.5 0.5 0 0 1 0 8
s4 0 1 0 0 0 0 1 5
102
Theory of the Simplex Method
For any LP with feasible solutions and a bounded
feasible region

• If there is exactly one optimal solution, then it
  must be a corner-point feasible (CPF) solution
• If there are multiple optimal solutions, then at
  least two must be adjacent CPF solutions
• There are a finite number of CPF solutions
• A CPF solution is optimal if there are no other
  adjacent CPF solutions that are better

103
Corner Point Feasible Solutions
X
X
104
Alternate Optima
105
Finite Number of CPF Solutions
An upper bound of the number of CPF solutions
Example m50 constraints, n100 decision
variables
Greater than thenumber of atoms in Universe!
106
Adjacent CPF Solutions
x2
X
larger Z
smaller Z
x1
107
Solving Other Types of Linear Programs
108
Finding a Feasible Solution?
Constraints in ? Form Minimization Problems
Equality Constraints
Not Feasible!
109
Equality Constraints
Max Z 2x1 3x2 s.t. 1x1 2x2 ? 4 x1
x2 3 x1 , x2 ? 0
Note x1 x2 0 is not feasible
How to achieve feasibility?
110
Big M Method
Strategy Start with artificial variables, then
remove artificial variables from the basic
variables using penalty M
Max Z 2x1 3x2 s.t. 1x1 2x2 ? 4 x1
x2 3 x1 , x2 ? 0
Max Z 2x1 3x2 - M a1 s.t. 1x1 2x2
s1 4 x1 x2 a1
3 x1 , x2 , s1, a1 ? 0
Note x1 x2 0 is now feasible
Add artificial variable a1
111
Big M Simplex Tableaux
Max Z 2x1 3x2 M a1 0 s.t. 1x1
2x2 s1 4 x1 x2
a1 3 x1 , x2 , s1, a1 ? 0
An extra Nonzero Element need To remove!
x1 x2 s1 a1 RHS
Z -2 -3 0 M 0
s1 1 2 1 0 4
a1 1 1 0 1 3
112
Big M Simplex Tableau
Initial Solution x1 0 x2 0 s1 4 a1 3
113
? Constraints
Max Z 2x1 5x2 3x3 s.t. 1x1 2x2 x3 ?
20 2x1 4x2 x3 50 x1 , x2 , x3 ? 0
Subtract surplus variableto create equality
1x1 2x2 x3 s1 20
Add artificial variablefor equality
1x1 2x2 x3 s1 a1 20
artificial variables?
114
? Big M
Max Z 2x1 5x2 3x3 Ma1 Ma2
s.t. 1x1 2x2 x3 s1 a1
20 2x1 4x2 x3 a2
50 x1 , x2 , x3 , s1 , a1 , a2 ? 0
Tableau with revised row 0
115
Variables Allowed to be Negative
xj allowed to be any value ( or )
Substitute xj xj xj
xj , xj ? 0
116
Negative RHSs
0.4x1 0.3x2 ? 10
Is exactly equivalent to
Multiply by 1
0.4x1 0.3x2 ? 10
Change sign of the inequality and use
corresponding methods for the new constraint
117
Minimization Problems
Min Z 0.4x1 0.3x2
Is exactly equivalent to
Multiply by 1
Max -Z 0.4x1 0.3x2
118
Summary of All Types of LP
Situations Solutions Initial Basic Variables
? constraint Slack variable Slack variable
Equality constraint Artificial variable Big M Method Artificial variable
? constraint Surplus variables artificial variables Big M Artificial variable (NOT surplus var.)
Variables allowed to be negative Change of variables
Negative RHS Multiply both sides by -1
Minimization problems Multiply objective function by -1
Combination of cases use Big M method last
119
LP Solution Problems

• Unbounded Solutions
• No Feasible Solutions

120
Unbounded Solutions
x1 x2 s1 s3 RHS
Z -3 -5 0 0 0
s1 1 0 1 0 4
s3 3 -2 0 1 6

• No coefficient in pivot column is positive
• No leaving basic variable
• Can bring in unlimited x2
• Z increases without limit!
• LP is unbounded

121
Unbounded Solutions
unbounded
x2
x1
(4, 0)
(0,0)
122
Example

• Breadco Bakeries bake two kinds of bread french
  and sourdough. Each loaf of french bread can be
  sold for 36cents, and each loaf of sourdough
  bread for 30cents. A loaf of french bread
  requires 1 yeast packet and 6 oz of flour
  sourdough requires 1 yeast packet and 5 oz of
  flour. At present Breadco has 5 yeast packets
  and 10 oz of flour. Additional yeast packets can
  be purchased at 3 cents each, and additional
  flour at 4 cents/oz. Formulate and solve an LP
  that can be used to maximize Breadcos profits.

123
No Feasible Solutions
Min Z 2x1 3x2 s.t. ½x1 ¼x2 ? 4 x1 3x2
? 36 x1 x2 10 x1 , x2 ? 0
x1 x2 s1 s2 a1 a2 RHS
-Z 2M-1 0 0 M 0 4M-3 -30-6M
s1 1/4 0 1 0 0 -1/4 3/2
a1 -2 0 0 -1 1 -3 6
x2 1 1 0 0 0 1 10
124
Introduction to Interior Point Solution Approach

• Starts at inner feasible point
• Moves through interior of feasible region
• Always improves objective function
• Longer computer time per iteration
• Less iterations for large problems
• Faster than Simplex method for huge problems

x2
(0,9)
(2,6)
(0,6)
(4,3)
x1
(6, 0)
(0,0)
(4, 0)
125
Sensitivity and Duality Analysis in LP
126
Outline

• Binding vs. Slack Constraints
• LP Sensitivity
• Ranges of optimality
• Shadow prices
• LP Duality
• Suggested Readings 4.7, parts of 6.1, 6.2, 6.6

127
Binding vs. Slack Constraints
128
Binding vs. Slack Constraints
BindingConstraints
SlackConstraint
129
Ranges of Optimality

• Objective function (OF) coefficients
• Right-hand sides (RHS)

130
Review of Wyndor Glass Example
Product Mix LP. Wyndor makes doors and windows.
They have three plants. A batch of doors requires
1 hour at Plant 1 plus 3 hours at Plant 3. A
batch of windows requires 2 hours at Plant 2 plus
2 hours at Plant 3. Plant 1 is available for 4
hours per week, Plant 2 for 12 hours per week,
and Plant 3 for 18 hours per week. The profit per
door batch is 3000 and the profit per window
batch is 5000. What should Wyndor manufacture to
maximize profits?
Max Z 3x1 5x2 profits (in thousands of
) s.t. 1x1 ? 4 Plant 1 2x2
? 12 Plant 2 3x1 2x2 ? 18 Plant 3 x1, x2
? 0 non-negativity
131
Wyndor Glass Example
How much can c1 or c2 changewithout changing the
optimalsolution?
Max Z c1x1 c2x2 profits (in thousands of
) s.t. 1x1 ? 4 Plant 1 2x2
? 12 Plant 2 3x1 2x2 ? 18 Plant 3 x1, x2
? 0 non-negativity
132
Allowable Range to Stay Optimal

• Allowable range to stay optimal is the range of
  values for cj over which the current optimal
  solution remains optimal, assuming no change in
  the other coefficients.

133
Graphical Sensitivity Analysis
x2
(0,9)
(2,6)
(0,6)
(4,3)
x1
(6, 0)
(4, 0)
(0,0)
Allowable range to stay optimal for c1 0 ? c1 ?
7.5
134
Graphical Sensitivity Analysis
x2
(0,9)
(2,6)
(0,6)
(4,3)
x1
(6, 0)
(0,0)
(4, 0)
135
Ranges of Optimality

• Objective function coefficients
• Right-hand sides

136
Wyndor Glass Example
How much can the RHSschange so that the current
optimal CPF solution is still feasible?
Max Z 3x1 5x2 profits (in thousands of
) s.t. 1x1 ? b1 Plant 1
2x2 ? b2 Plant 2 3x1 2x2 ? b3 Plant 3 x1,
x2 ? 0 non-negativity
137
Allowable Range to Stay Feasible

• Allowable range to stay feasible is the range of
  values for bi over which the current optimal CPF
  solution remains feasible, assuming no change in
  the other right-hand sides.

138
Graphical Sensitivity Analysis
Max Z 3x1 5x2 s.t. 1x1 ? 4
2x2 ? 12 3x1 2x2 ? 18 x1, x2 ? 0
x2
(0,9)
(0,6)
x1
(6, 0)
(4, 0)
(0,0)
139
Graphical Sensitivity Analysis
x2
(0,9)
(0,6)
x1
(6, 0)
(4, 0)
(0,0)
140
Graphical Sensitivity Analysis
x2
(0,9)
(0,6)
X
x1
(6, 0)
(4, 0)
(0,0)
Allowable range to stay feasible for b1 is 2 ? b1
141
Graphical Sensitivity Analysis
Max Z 3x1 5x2 s.t. 1x1 ? 4
2x2 ? 12 3x1 2x2 ? 18 x1, x2 ? 0
x2
(0,9)
(0,6)
x1
(6, 0)
(4, 0)
(0,0)
142
Graphical Sensitivity Analysis
x2
(0,9)
(0,6)
X
x1
(6, 0)
(4, 0)
(0,0)
143
Graphical Sensitivity Analysis
Max Z 3x1 5x2 s.t. 1x1 ? 4
2x2 ? 12 3x1 2x2 ? b3 x1, x2 ? 0
x2
(0,9)
(0,6)
X
x1
(6, 0)
(4, 0)
(0,0)
Allowable range to stay feasible for b3 is 12 ?
b3 ?24
144
Shadow Prices
145
Wyndor Glass Example
How much is additionalRHS resource worth to us?
Max Z 3x1 5x2 profits (in thousands of
) s.t. 1x1 ? 4 Plant 1 2x2
? 12 Plant 2 3x1 2x2 ? 18 Plant 3 x1, x2
? 0 non-negativity
146
Graphical Solution
Max Z 3x1 5x2 (Z36) s.t. 1x1 ?
4 2x2 ? 12 3x1 2x2 ? 18 x1, x2
? 0
x2
(0,9)
(2,6)
(0,6)
(4,3)
x1
(6, 0)
(0,0)
(4, 0)
147
Resource Shadow Price

• The shadow price of resource i (bi) is the rate
  at which Z could be improved by slightly
  increasing the amount of this resource
• Shadow price 0 for any slack constraint.

148
Algebraic Solution

• The simplex method identifies the shadow price by
  the coefficients of the slack variables in row 0
  of the final simplex tableau.

Final Simplex Tableau
x1 x2 s1 s2 s3 RHS
Z 0 0 0 3/2 1 36
s1 0 0 1 1/3 -1/3 2
x2 0 1 0 1/2 0 6
x1 1 0 0 -1/3 1/3 2
149
LP Duality
150
Dual LP Example
Primal Problem
Dual Problem
Max Z 3x1 5x2 s.t. 1x1 0x2 ? 4 0x1 2x2
? 12 3x1 2x2 ? 18 x1 , x2 ? 0
Min W 4y1 12y2 18y3 s.t. 1y1 0y2 3y3 ?
3 0y1 2y2 2y3 ? 5 y1 , y2, y3 ? 0
y1 0 y2 3/2 y3 1 W 36
x1 2 x2 6 Z 36
151
Dual Interpretation
MaximizeProfits
Minimize Total Implicit Price
Value
Resources
Shadow Prices
Quantities
Owner of resources
Buyer of resources
152
Weak Duality

• If x is a feasible solution for the primal
  problem and y a feasible solution for the dual,
  then Z ? W
• That is, the value of the dual is an upper bound
  on the primal problem

153
Strong Duality

• If Z is the optimal value for the primal problem
  and W the optimal value for the dual, then Z
  W
• That is, the optimal value of the primal and the
  optimal value of the dual are equal

154
Example

• Consider the following problem
• Max Z2x17x24x3
• s.t.
• x1 2x2 x3 ? 10
• 3x1 3x2 2x3 ? 10
• and
• x1?0, x2 ?0, x3 ?0.
• Construct the dual problem for this primal
  problem.
• Use the dual problem to demonstrate that the
  optimal value of Z for the primal problem cannot
  exceed 25.

155
Example

• Prove that for any linear programming problem in
  standard form if the primal problem has an
  unbounded feasible region that permits increasing
  Z indefinitely, then the dual problem has no
  feasible solutions.

156
Example

• Consider the following LP
• Max Z-x15x2
• s.t.
• x1 2x2 ? 0.5
• -x1 3x2 ? 0.5
• and
• x1?0, x2 ?0.
• What is the missed value at the following Row 0
  of the optimal tableau?

x1 x2 s1 s2 RHS
Z 0 0 0.4 1.4 ?
157
Why do we care?

• Dual problem may be easier to solve
• Number of constraints affects computation far
  more than number of variables for simplex method
• Primal has 1000 constraints, 100 variables
• Dual has 100 constraints, 1000 variables
• Dual is easier to solve
• Evaluate a primal solution using dual feasibility
• Sensitivity Analysis
• Weak duality often used in solving integer
  programs (branch and bound)
• Economic interpretation and insight

158
Chapter 8
Transportation and Assignment Problems
159
Previous LP Applications

• Product Mix
• Investment
• Diet/Nutrition

160
Two Additional Important LP Applications

• Transportation Problems
• Shipping Planning
• Production Scheduling
• Water Distribution
• Assignment Problems
• Suggested readings 8.18.3

161
Transportation Problem
162
Transportation Problem Example
High Plains Electronics manufactures MP3 players
at three separate overseas plants. It ships
finished products to three US distribution
centers. Shipping costs (per unit) are shown
below, as are the production capacities of each
plant and the demand from each distribution
center (units per week)
Supply Demand
700
163
Transportation Problem Defined

• Minimize total shipping cost (or maximize
  profit) of products from source to destinations
• Supply equals demand
• Distribution quantities often have integer values
• Integrality assured all basic solutions
  (including optimal solution) have integer value
  when supply/demand are integer
• Linear costs assumed

164
General LP Form
Optimize some objective
Supply constraints
Demand constraints
Non-negativity constraints
165
General LP Form
166
Prohibited Routes

• What if some routes are infeasible or prohibited?
• Create allocation cost that is so large that it
  will be quickly forced leaving from the basis
  Big M
• Example

167
Supply ? Demand

• What if supply does not equal demand?
• Feasible solutions exist only if total supply
  equals total demand

• Create dummy source or destination
• Example

168
Check Processing Example (Dummy Destination)
A bank has two sites at which checks are
processed. Site 1 can process 10,000 checks per
day, and site 2 can process 6000 checks per day.
The bank processes 3 types of checks vendor,
salary, and personal. The processing cost per
check depends on the site (see table). Each day,
5000 checks of each type must be processed.
Formulate a transportation problem to minimize
the daily cost of processing checks.
Site Vendor checks Salary checks Personal checks
1 5 cents 4 cents 2 cents
2 3 cents 4 cents 5 cents
169
Production Scheduling Example
The NORTHERN AIRPLANE COMPANY builds commercial
airplanes. The production of the jet engines must
be scheduled for the next 4 months. The unit
storage cost is 0.015 million dollars per month.
The demand, supply capacity, and production costs
are listed in the table. The production manager
wants a schedule for the number of engines to be
produced in each of the 4 months to minimize the
total production and storage costs.
Month Scheduled Installation Maximum Production Unit Cost of Production Unit Cost of Storage
1 10 25 1.08 0.015
2 15 35 1.11 0.015
3 25 30 1.10 0.015
4 20 10 1.13
170
Water Distribution Example (Dummy Source)
Two reservoirs are available to supply the water
needs of three cities. Each reservoir can supply
50 million gallons of water per day. Each city
would like to receive 40 million gallons per day.
For each million gallons per day of unmet
demand, there is a penalty. At city 1, the
penalty is 20 at city 2, the penalty is 22
and at city 3, the penalty is 23. The cost of
transporting 1 million gallons of water from each
reservoir to each city are shown in the table.
Formulate a transportation problem that can be
used to minimize the sum of shortage and
transport costs.
City 1 City 2 City 3
Reservoir 1 7 8 10
Reservoir 2 9 7 8
171
Transportation Problem Solutions

• Transportation problem is a special type of
  linear programming problem. So it can be solved
  by simplex method
• A streamlined procedure (the transportation
  simplex method) is available to achieve
  tremendous computational savings by exploiting
  the special structure of transportation problems.

172
Assignment Problem
173
Assignment Problem Example
MACHINECO has 4 machines and 4 jobs to be
completed. Each machine must be assigned to
complete one job. The time required to set up
each machine for completing each job is shown in
the table. MACHINECO wants to minimize the total
setup time needed to complete the 4 jobs.
Job 1 Job 2 Job 3 Job 4
Machine 1 14 5 8 7
Machine 2 2 12 6 5
Machine 3 7 8 3 9
Machine 4 2 4 6 10
174
General LP Form
Optimize some objective
Resource constraints
Task constraints
If i is not assigned to j
If i is assigned to j
175
Assignment Problem

• The assignment problem is a special type of
  transportation problem
• Assign discrete resources (people, machines,
  vehicles, ) to tasks (jobs, routes, )
• Number of resources and tasks are equal
• Each resource is assigned to one task
• Each task is assigned to one resource
• Binary solutions assured
• Linear costs assumed
• Highly degenerate (many zero basic variables)

176
Assignment Problem Extensions

• resources unequal to tasks
• Use dummy resources or tasks
• Prohibited assignments
• Use large allocation cost Big M

177
Another Assignment Problem Example
A company has decided to assign 3 new machines to
4 locations. The estimated cost in dollars per
hour of materials handling involving each of the
machines is given in the table for the respective
locations. Location 2 is not suitable for
machine 2.
Location 1 Location 2 Location 3 Location 4
Machine 1 13 16 12 11
Machine 2 15 -- 13 20
Machine 3 5 7 10 6
178
Assignment Solutions

• Could try complete enumeration
• n! assignments possible
• n 10 means 10!3.6 million possible assignments
• Linear Programming solution
• Small problem can be solved by the general
  simplex method
• Applying transportation simplex method is a
  relatively fast way
• Specialized algorithms available, which are in
  preference to the transportation simplex method

179
Ch. 12
Integer Programming
180
Outline

• Integer Programs
• General Integer Models
• 0-1 (Binary) Models
• Mixed Integer Models
• IP Examples

• IP Solution Techniques
• Branch and Bound
• Application articles
• Suggested Readings 11.1-11.7

181
Integer Programming

• An integer programming problem (IP) is an LP in
  which some or all of the variables are required
  to be integers.
• IP is generally much more difficult to solve than
  LP due to
• Exponential growth of number of solutions
• Property of simplex method is invalid for IP

182
An IP Example
Max Z 33x1 12x2 s.t. x1 2x2 ? 4 5x1
2x2 ? 16 2x1 x2 ? 4 x1, x2 ? 0 and
integer
Optimal LP Solution x1 8/3 x2 4/3 ZLP 104
Optimal IP Solution x1 2 x2 3 ZIP 102
Feasible and optimal
Rounded LP Solution x1 2 x2 1 ZRLP 78
Not feasible, superoptimal
Feasible, Not optimal
183
Bounds on IPs

• For a max (min) IP
• The LP obtained by omitting all integer
  constraints on variables is called the LP
  relaxation of the IP
• The optimal value of the LP relaxation is an
  upper-bound (lower-bound) to the IP
• If all variables in the optimal solution of LP
  relaxation are integers, then the optimal
  solutions of LP and IP are the same

184
General IP Example
Pawtucket University is planning to buy new
copier machines for its library. Two different
models are considered Model A and B. Model A
can handle 20,000 copies a day, and costs 6,000.
Model B can handle 10,000 copies a day, but
costs only 4,000. At least six copiers are
needed. And at least one of them is model A.
The copiers need to be able to handle a capacity
of at least 75,000 copies with minimum cost.
185
Binary IP Problems

• General IP problems important,but not compelling
• Integer Programming of greater importance is to
  model yes-no decisions
• BIP models

186
Binary IP Example
The CALIFORNIA MFG. COMPANY is considering
expansion by building a new factory in either Los
Angeles or San Francisco, or both cities. It
also is considering buiding at most one new
warehouse, but the choice of location is
restricted to a city where a new factory is being
built. The net present value and the capital
required for each alternative is in the table.
The total capital available is 10 million. The
objective is to maximize the total net present
value.
Net Present Value Capital Required
Factory in LA 9 million 6 million
Factory in SF 5 million 3 million
Warehouse in LA 6 million 5 million
Warehouse in SF 4 million 2 million
187
Mixed Integer Programs

• MIP
• Both integer and continuous variables

max z 3x1 2x2 s.t. x1 x2? 6 x1, x2 ? 0, x1
integer
188
Other Integer Programming Examples
189
Set-Covering Problem
There are six cities (cities 1-6) in Kilroy
County. The county must determine where to build
fire stations. The county wants to build the
minimum number of fire stations needed to ensure
that at least one fire station is within 15
minutes of each city. The times required to
drive between the cities are shown in the table.
Formulate a BIP to tell Kilroy how many fire
stations should be built and where they should be
located.
1 2 3 4 5 6
1 0 10 20 30 30 20
2 10 0 25 35 20 10
3 20 25 0 15 30 20
4 30 35 15 0 15 25
5 30 20 30 15 0 14
6 20 10 20 25 14 0
190
Either-Or Constraints
Dorian Auto is considering manufacturing three
types of autos compact, midsize, and large. The
resources required for, and the profits yielded
by, each type of car are shown in the table. At
present, 6000 tons of steel and 60,000 hours of
labor are available. For production of a type of
car to be economically feasible, at least 1000
cars of that type must be produced. Formulate an
IP to maximize Dorians profit.
Compact Midsize Large
Steel required (tons) 1.5 3 5
Labor Required (hrs) 30 25 40
Profit (k) 2 3 4
191
Summary of IP Formulation

• Mutually exclusive alternatives
• At most one of x1, x2, , xn can be equal to 1
  x1x2xn?1
• Exactly one of x1, x2, , xn must be equal to 1
  x1x2xn1
• Contingent decisions
• x1 can be equal to 1 only if x2 is equal to 1
  x1?x2
• Set-Covering Problem
• For member i of set 1, let Si be the acceptable
  members of set 2 used to cover member i
• Either-or constraint
• Use auxiliary binary variable and big M

192
Solving Integer Programs
193
BIP Branch and Bound Example
The CALIFORNIA MFG. COMPANY is considering
expansion by buiding a new factory in either Los
Angeles or San Francisco, or both cities. It
also is considering buiding at most one new
warehouse, but the choice of location is
restricted to a city where a new factory is being
built. The net present value and the capital
required for each alternative is in the table.
The total capital available is 10 million. The
objective is to maximize the total net present
value.
Net Present Value Capital Required
Factory in LA 9 million 6 million
Factory in SF 5 million 3 million
Warehouse in LA 6 million 5 million
Warehouse in SF 4 million 2 million
194
BB Solution
Solution tree
1 All , B16, X(5/6,1,0,1)
x1 0
x1 1
Z -?
2 B9, X(0,1,0,1)
3 B16, X(1, 0.8, 0, 0.8)
x2 0
x2 1
Z 9
Fathomed
5 B16, X(1,1,0,0.5)
4 B13, X(1, 0, 0.8, 0)
x3 0
x3 1
Fathomed
7 InfeasibleFathomed
6 B16 X(1,1,0, 0.5)
x4 0
x4 1
9 InfeasibleFathomed
8 X(1,1,0,0) Fathomed
Z 14
195
Summary of BIP Branch and Bound (Maximization
Prob.)

• For minimization problem first convert to a
  maximization problem by multiplying -1 to the
  objective fn (remember to change the sign back
  for the optimal solution)
• Initialization
• Set initial incumbent (Z) -?
• Apply bounding, fathoming, and optimality test to
  the whole problem
• Iteration
• Branching branch remaining subproblems created
  most recently (break ties by selecting larger
  bound ) by fixing next variable at 0 and 1.
  Choose the first one in the natural ordering to
  be the branching variable.
• Bounding the optimal Z value of LP relaxation
  gives the bound. Rounding down bound if all obj.
  fn. coefficients are integers.
• Fathoming Apply the three fathoming tests.
• Optimality test Stop when there are no remaining
  subproblems the current incumbent is optimal.

196
Summary of Fathoming Tests

• Test 1 Its bound?Z
• Test 2 Its LP relaxation has no feasible
  solution
• Test 3 The optimal solution for its LP
  relaxation is integer. If this solution is
  better than the incumbent, it becomes the new
  incumbent, and test 1 is reapplied to all
  unfathomed subproblems.

197
Another BIP BranchBound Example
Max Z 4x1 9x26x3 s.t. 5x18x26x3 ?
12 xi 0 or 1 (i1,2,3)
198
Review of Mixed Integer Programs

• MIP
• Both integer and continuous variables

max z 3x1 2x2 s.t. x1 x2? 6 x1, x2 ? 0, x1
integer
199
MIP Branch-And-Bound

• Changes from BIP branch-and-bound
• Select branching variable only from the
  integer-restricted variables that have a
  noninteger value in the optimal solution for the
  current LP relaxation
• Branching based on
• xj?xj and xj?xj1
• xjbranching variable, xjgreatest integer
  ?xj
• Never rounding down Z for MIP (due to possible
  non-integer decision variables)