Transcript 幻灯片 1 - Shandong University

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


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
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 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 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







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.
……
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
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

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

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 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 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 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 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 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 characteristics

Sign restrictions
If a decision variable xi can only assume
nonnegative values, we add the sign
restriction xi≥0
If a decision variable xi allowed to assume both
positive and negative values, we say that xi
is unrestricted in sign.
For the Giaprtto problem,
x1≥0 and x2≥0
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.

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.

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

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

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 problems
constraint 1 is satisfied
2 x1+x2-100 = 0
100
80
x2
60
40
20
0
0
20
40
60
x1
80
100
The graphical solution of two-variable
linear programming problems
constraint 2 is satisfied
x1+x2-80 = 0
100
80
x2
60
40
20
0
0
20
40
60
x1
80
100
The graphical solution of two-variable
linear programming problems
constraint 3 is satisfied
100
80
x2
60
40
20
0
0
20
40
60
x1
80
100
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.
100
80
x2
60
40
20
0
0
20
40
60
x1
80
100
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.
3 x+2 y-180 = 0
100
80
60
y

40
z=60
20
z=100
z=180
0
0
20
40
60
x
80
100
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.
A
A
E
B A
B
C
Convex sets
B
A
D
Nonconvex sets
B
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.

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

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
max z=
2x2
1/40x1+1/60x2≤1
1/50x1+1/50x2≤1
x1,x2≥0
60
50
2
40
x
s.t.
3x1+
30
Z=100
20
Z=60
10
0
0
10
Z=120
20
30
x1
40
50
60
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
60
50
x
2
40
30
20
10
0
0
10
20
30
x1
40
50
60
Unbounded LP
max z=2x1 - x2
s.t.
x1 - x2≤1
2x1+ x2≥6
x1,x2≥0
10
8
z=4
z=6
x
2
6
4
2
0
0
2
4
6
x1
8
10