Optimization methods
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Optimization Methods. LI Xiao-lei [email protected] http://www.csce.sdu.edu.cn ftp://202.194.201.155:112/upload. Optimization Tree. Introduction to Operations Research. Encyclopedia of Mathematics Optimization Theory See Operations Research

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Optimization methods

Optimization Methods

LI Xiao-lei

[email protected]

http://www.csce.sdu.edu.cn

ftp://202.194.201.155:112/upload


Optimization tree

Optimization Tree


Introduction to operations research

Introduction to Operations Research

  • Encyclopedia of Mathematics

    Optimization Theory

    See Operations Research

  • During World War II, British military leaders asked scientists and engineers to analyze several military problems. The application of mathematics and the scientific method to military operations was called operations research.

  • Today, the term oprations 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.


The methodology of operations research

The Methodology of operations research

Seven-step procedure

  • Step1. Formulate the problem

    • Specify the organization’s objectives and the parts of the system that must be studied before the problem can be solved.

  • Step2. Observe the system

    • The analyst collects data to estimate the values of parameters that affect the organization’s problem.


The methodology of operations research1

The Methodology of operations research

  • Step3. Formulate a methematical model of the problem

    • The analyst develops a mathematical model of the problem.

  • Step4. Verify the model and use the model for prediction

    • The analyst now tries to determine if the mathematical model is an accurate representation of reality.

  • Step5. Select a suitable alternative

    • Given a model and a set of alternatives, the analyst now choose the alternative that best meets the organization’s objectives.


The methodology of operations research2

The Methodology of operations research

  • Step6. Present the results and conclusions of the study to the organization

    • The analyst presents the model and recommendations from step5 to the to the decision making individual or group. Let the organization choose the one that best meets its needs.

  • Step7. Implement and evaluate recommendations

    • If the organization has accepted the study, the analyst aids in implementing the recommendations.

    • The system must be constantly monitored to ensure that the recommendations are enabling the organization to meet its objectives.


Successful applications of operations research

Successful applications of operations research

  • Police patrol officer scheduling in San Francisco.

  • Reducing fuel cost in electric power industry.

  • Designing an ingot mold stripping facility at Betnlehem Steel.

  • Gasoline blending at Texaco.

  • Scheduling trucks at north american van lines.

  • Inventory management at Blue Bell.

  • … …


Optimization methods

Reference:

OPERATIONS RESEARCH:Mathematical Programming(THIRD EDITION)

WAYNE L. WINSTON, 2003

More:

Anything about Operations Research, Management Science, Decision Sciences, Mathematical Programming, Logistics Research.


Introduction to linear programming

Introduction to Linear Programming

  • LP is atool for solving optimization problems. In 1947,George Dantzig developed the simplex algorithm for solving LP problem, since then, LP has been used to solve optimization problems in industries as diverse as banking, education, petroleum, and trucking.

  • In a survey of Fortune 500 firms,85% of those responding said that they had used LP.


Linear programming problem

Linear Programming Problem

  • Example 1

    Giapetto’s Woodcarving, Inc., manufactures two types of wooden toys: soldiers and trains.

    Requires two types of skilled labor: carpentry and finishing.

    A Soldier sells for $27 and uses $10 worth of raw materials, each costs variable labor and overhead by $14. requires 2 hours of finishing labor and 1 hour of carpentry labor.

    A train sells for $21 and uses $9 worth of raw materials, each cost variable labor and overhead by $10. requires 1 hour of finishing labor and 1 hour of carpentry labor.

    Each week, Giapetto can obtain all the needed raw material but only 100finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers are bought.

    Formulate a mathematical model to maximize Giapetto’s weekly profit (revenues -costs).


Solution explore characteristics

Solution – explore characteristics

  • Decision Variables

    Giapetto must decide how many soldiers and trains should be manufactured each week. With this end, we define

    x1=number of soldiers produced each week.

    x2=number of trains produced each week.

  • Objective Function

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


Solution explore characteristics1

Solution – explore characteristics

Weekly revenues

=weekly revenues from soldiers + weekly revenues from trains

=(dollars/soldier) (soldiers/week) + (dollars/train) (trains/week)

=27x1+21x2

Also,

Weekly raw material costs=10x1+9x2

Other weekly variable costs=14x1+10x2

Then Giapetto wants to maximize,

(27x1+21x2)-(10x1+9x2)-(14x1+10x2)= 3x1+2x2


Solution explore characteristics2

Solution – explore characteristics

We use the variable z to denote the objective function value of LP. Giapetto’s objective function is,

Maximize z=3x1+2x2

The coefficient of a variable in the objective function is called the objective function coefficient of the variable.


Solution explore characteristics3

Solution – explore characteristics

  • Constraints

    The value of x1 and x2 are limited by the following three restrictions.

    Constraint 1 Each week, no more than 100 hours of finishing time may be used.

    Constraint 2 Each week, no more than 80 hours of carpentry time may be used.

    Constraint 3 Because of limited demand, at most 40 soldiers should be produced each week.


Solution explore characteristics4

Solution – explore characteristics

Constraint 1

Total finishing hrs./Week

=(finishing hrs./soldier) (soldiers made/week) +(finishing hrs./train) (trains made/week)

=2x1+1x2=2x1+x2

Constraint 1 may be expressed by,

2x1+x2≤100

Note: For a constraint to be reasonable, all terms in the constraint must have the same units.


Solution explore characteristics5

Solution – explore characteristics

  • Constraint 2

    Total carpentry hrs./week

    =(carpentry hrs./soldier) (soldiers/week) + (carpentry hrs./train) (trains/week)

    =1x1+1x2=x1+x2

    The constraint 2 may be written as,

    x1+x2≤80

  • Constraint 3

    x1≤40


Solution explore characteristics6

Solution – explore characteristics

  • Sign restrictions

    If a decision variable xican only assume nonnegative values, we add the sign restrictionxi≥0

    If a decision variable xiallowed to assume both positive and negative values, we say that xiis unrestricted in sign.

    For the Giaprtto problem,

    x1≥0 and x2≥0


Solution explore characteristics7

Solution – explore characteristics

  • Optimization model

    Max z=3x1+2x2

    s.t. 2x1 + x2≤100

    x1 + x2≤80

    x1 ≤40

    x1 ≥0

    x2≥0

    s.t. (subject to) means that the values of the decision variables must satisfy all the constraints and all the sign restrictions.


Optimization methods

  • DEFINATION 1

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

  • DEFINATION 2

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


Optimization methods

  • DEFINATION 3

    A linear programming problem (LP) is an optimization problem for which we do the following:

    • We attempt to maximize (or minimize) a linear function of the decision variables. The function that is to be maximized or minimized is called the objective function.

    • The values of the decision variables must satisfy a set of constraints. Each constraint must be a linear equation or linear inequality.

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


Feasible region and optimal solution

Feasible Region and Optimal Solution

  • DEFINATION 4

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

  • DEFINATION 5

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


The graphical solution of two variable linear programming problems

The graphical solution of two-variable linear programming problems

  • Finding the feasible solution

    The feasible region for the Giapetto problem is the set of all points (x1.x2) satisfying all the inequalities,

    • 2x1 + x2≤100 (Constraints)

    • x1 + x2≤80

    • x1≤40

    • x1≥0 (Sign restrictions)

    • x2≥0


The graphical solution of two variable linear programming problems1

The graphical solution of two-variable linear programming problems

  • constraint 1 is satisfied


The graphical solution of two variable linear programming problems2

The graphical solution of two-variable linear programming problems

  • constraint 2 is satisfied


The graphical solution of two variable linear programming problems3

The graphical solution of two-variable linear programming problems

  • constraint 3 is satisfied


The graphical solution of two variable linear programming problems4

The graphical solution of two-variable linear programming problems

  • Any point on the polygon or interior is the feasible region; Any other point fails to satisfy at least one of the inequalities.


The graphical solution of two variable linear programming problems5

The graphical solution of two-variable linear programming problems

  • Finding the optimal solution

    The optimal solution will be the point in the feasible region with the largest value of objective function z=3x1+2x2.


Convex sets extreme points and lp

Convex Sets, Extreme Points, and LP

  • DEFINITION

    A set of points S is a convex set if the line segment joining any pair of points in S is wholly constrained in S.

E

B

A

A

B

A

B

A

B

C

D

Convex sets

Nonconvex sets


Convex sets extreme points and lp1

Convex Sets, Extreme Points, and LP

  • DEFINATION

  • For any convex set S, a point P in S is an extreme point (corner point) if each line segment that lies completely in S and constrains the point P has P as an endpoint of the line segment.


Optimization methods

Convex Sets, Extreme Points, and LP

  • The feasible region for the Giapetto problem is a convex set. It can be shown that the optimal solution is an extreme points.

    Note:

    Any LP that has an optimal solution has an extreme point that is optimal. This reduces the set of points yielding an optimal solution from the entire feasible region (infinite) to the set of extreme points (finite).


Special cases

Special Cases

  • Three type of LPs that do not have unique optimal solutions.

    • Some LPs have an infinite number of optimal solutions (alternative or multiple optimal solutions)

    • Some LPs have no feasible solutions (infeasible LPs)

    • Some LPs are unbounded: there are points in the feasible region with arbitrarily large (in a max problem) z-values.


Alternative or multiple optimal solutions

Alternative or multiple optimal solutions

max z= 3x1+ 2x2

s.t. 1/40x1+1/60x2≤1

1/50x1+1/50x2≤1

x1,x2≥0

Z=100

Z=60

Z=120


Infeasible lp

Infeasible LP

max z= 3x1+ 2x2

s.t. 1/40x1+1/60x2≤1

1/50x1+1/50x2≤1

x1 ≥30

x2≥20

x1,x2≥0


Unbounded lp

Unbounded LP

max z=2x1 - x2

s.t. x1 - x2≤1

2x1+ x2≥6

x1,x2≥0


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