optimization
DESCRIPTION
Optimization. Assoc. Prof. Dr. Pelin Gündeş [email protected]. Optimization. Basic Information Instructor: Assoc. Professor Pelin Gundes (http://atlas.cc.itu.edu.tr/~gundes/) E-mail: [email protected] Office Hours: TBD by email appointment - PowerPoint PPT PresentationTRANSCRIPT
2
Optimization
Basic Information• Instructor: Assoc. Professor Pelin Gundes
(http://atlas.cc.itu.edu.tr/~gundes/)• E-mail: [email protected]• Office Hours: TBD by email appointment• Website:
http://atlas.cc.itu.edu.tr/~gundes/teaching/Optimization.htm
• Lecture Time: Wednesday 13:00 - 16:00• Lecture Venue: M 2180
3
Optimization literatureTextbooks:
1. Nocedal J. and Wright S.J., Numerical Optimization, Springer Series in Operations Research, Springer, 636 pp, 1999.
2. Spall J.C., Introduction to Stochastic Search and Optimization, Estimation, Simulation and Control, Wiley, 595 pp, 2003.
3. Chong E.K.P. and Zak S.H., An Introduction to Optimization, Second Edition, John Wiley & Sons, New York, 476 pp, 2001.
4. Rao S.S., Engineering Optimization - Theory and Practice, John Wiley & Sons, New York, 903 pp, 1996.
5. Gill P.E., Murray W. and Wright M.H., Practical Optimization, Elsevier, 401 pp., 2004.
6. Goldberg D.E., Genetic Algorithms in Search, Optimization and Machine Learning, Addison Wesley, Reading, Mass., 1989.
7. S. Boyd and L. Vandenberghe, Convex Optimization, Cambridge University Press, 2004.(available at http://www.stanford.edu/~boyd/cvxbook/)
4
Optimization literatureJournals:
1. Engineering Optimization
2. ASME Journal of Mechnical Design
3. AIAA Journal
4. ASCE Journal of Structural Engineering
5. Computers and Structures
6. International Journal for Numerical Methods in Engineering
7. Structural Optimization
8. Journal of Optimization Theory and Applications
9. Computers and Operations Research
10. Operations Research and Management Science
5
Optimization
Course Schedule:
1. Introduction to Optimization
2. Classical Optimization Techniques
3. Linear programming and the Simplex method
4. Nonlinear programming-One Dimensional Minimization Methods
5. Nonlinear programming-Unconstrained Optimization Techniques
6. Nonlinear programming-Constrained Optimization Techniques
7. Global Optimization Methods-Genetic algorithms
8. Global Optimization Methods-Simulated Annealing
9. Global Optimization Methods- Coupled Local Minimizers
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Optimization
Course Prerequisite:
• Familiarity with MATLAB, if you are not familiar with MATLAB, please visit
http://www.ece.ust.hk/~palomar/courses/ELEC692Q/lecture%2006%20-%20cvx/matlab_crashcourse.pdf
http://www.ece.ust.hk/~palomar/courses/ELEC692Q/lecture%2006%20-%20cvx/official_getting_started.pdf
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Optimization
• 70% attendance is required!
• Grading:
Homeworks: 15%
Mid-term projects: 40%
Final Project: 45%
8
Optimization
• There will also be lab sessions for MATLAB exercises!
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1. Introduction1. Introduction
• Optimization is the act of obtaining the best result under given circumstances.
• Optimization can be defined as the process of finding the conditions that give the maximum or minimum of a function.
• The optimum seeking methods are also known as mathematical programming techniques and are generally studied as a part of operations research.
• Operations research is a branch of mathematics concerned with the application of scientific methods and techniques to decision making problems and with establishing the best or optimal solutions.
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1. Introduction1. Introduction• Operations research (in the UK) or operational research (OR)
(in the US) or yöneylem araştırması (in Turkish) is an interdisciplinary branch of mathematics which uses methods like:
– mathematical modeling– statistics– algorithms to arrive at optimal or good decisions in complex
problems which are concerned with optimizing the maxima (profit, faster assembly line, greater crop yield, higher bandwidth, etc) or minima (cost loss, lowering of risk, etc) of some objective function.
• The eventual intention behind using operations research is to elicit a best possible solution to a problem mathematically, which improves or optimizes the performance of the system.
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1. Introduction1. Introduction
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1. Introduction
Historical development
• Isaac Newton (1642-1727) (The development of differential calculus methods of optimization)
• Joseph-Louis Lagrange (1736-1813) (Calculus of variations, minimization of functionals, method of optimization for constrained problems)
• Augustin-Louis Cauchy (1789-1857) (Solution by direct substitution, steepest descent method for unconstrained optimization)
13
1. Introduction
Historical development
• Leonhard Euler (1707-1783)
(Calculus of variations, minimization of
functionals)
• Gottfried Leibnitz (1646-1716)
(Differential calculus methods
of optimization)
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1. Introduction
Historical development
• George Bernard Dantzig (1914-2005)
(Linear programming and Simplex method (1947))
• Richard Bellman (1920-1984)
(Principle of optimality in dynamic
programming problems)
• Harold William Kuhn (1925-)
(Necessary and sufficient conditions for the optimal solution of programming problems, game theory)
15
1. Introduction
Historical development
• Albert William Tucker (1905-1995) (Necessary and sufficient conditions for the optimal solution of programming problems, nonlinear programming, game theory: his PhD student was John Nash)
• Von Neumann (1903-1957) (game theory)
16
1. Introduction1. Introduction
• Mathematical optimization problem:
• f0 : Rn R: objective function• x=(x1,…..,xn): design variables (unknowns of the problem,
they must be linearly independent)• gi : Rn R: (i=1,…,m): inequality constraints
• The problem is a constrained optimization problem
mibxg
xf
ii ,....,1,)( subject to
)( minimize 0
17
1. Introduction1. Introduction
• If a point x* corresponds to the minimum value of the function f (x), the same point also corresponds to the maximum value of the negative of the function, -f (x). Thus optimization can be taken to mean minimization since the maximum of a function can be found by seeking the minimum of the negative of the same function.
18
1. Introduction
Constraints
• Behaviour constraints: Constraints that represent limitations on the behaviour or performance of the system are termed behaviour or functional constraints.
• Side constraints: Constraints that represent physical limitations on design variables such as manufacturing limitations.
19
1. Introduction
Constraint Surface• For illustration purposes, consider an optimization problem with only
inequality constraints gj (X) 0. The set of values of X that satisfy the equation gj (X) =0 forms a hypersurface in the design space and is called a constraint surface.
20
1. Introduction
Constraint Surface• Note that this is a (n-1) dimensional subspace, where n is the
number of design variables. The constraint surface divides the design space into two regions: one in which gj (X) 0and the other in which gj (X) 0.
21
1. Introduction
Constraint Surface• Thus the points lying on the hypersurface will satisfy the constraint
gj (X) critically whereas the points lying in the region where gj (X) >0 are infeasible or unacceptable, and the points lying in the region where gj (X) < 0 are feasible or acceptable.
22
1. Introduction
Constraint Surface• In the below figure, a hypothetical two dimensional design space is
depicted where the infeasible region is indicated by hatched lines. A design point that lies on one or more than one constraint surface is called a bound point, and the associated constraint is called an active constraint.
23
1. Introduction
Constraint Surface• Design points that do not lie on any constraint surface are known as
free points.
24
1. Introduction
Constraint Surface
Depending on whether a particular design point belongs to the acceptable or unacceptable regions, it can be identified as one of the following four types:
• Free and acceptable point
• Free and unacceptable point
• Bound and acceptable point
• Bound and unacceptable point
25
1. Introduction
• The conventional design procedures aim at finding an acceptable or adequate design which merely satisfies the functional and other requirements of the problem.
• In general, there will be more than one acceptable design, and the purpose of optimization is to choose the best one of the many acceptable designs available.
• Thus a criterion has to be chosen for comparing the different alternative acceptable designs and for selecting the best one.
• The criterion with respect to which the design is optimized, when expressed as a function of the design variables, is known as the objective function.
26
1. Introduction
• In civil engineering, the objective is usually taken as the minimization of the cost.
• In mechanical engineering, the maximization of the mechanical efficiency is the obvious choice of an objective function.
• In aerospace structural design problems, the objective function for minimization is generally taken as weight.
• In some situations, there may be more than one criterion to be satisfied simultaneously. An optimization problem involving multiple objective functions is known as a multiobjective programming problem.
27
1. Introduction
• With multiple objectives there arises a possibility of conflict, and one simple way to handle the problem is to construct an overall objective function as a linear combination of the conflicting multiple objective functions.
• Thus, if f1 (X) and f2 (X) denote two objective functions, construct a new (overall) objective function for optimization as:
where 1 and 2 are constants whose values indicate the relative importance of one objective function to the other.
)()()( 2211 XXX fff
28
1. Introduction
• The locus of all points satisfying f (X) = c = constant forms a hypersurface in the design space, and for each value of c there corresponds a different member of a family of surfaces. These surfaces, called objective function surfaces, are shown in a hypothetical two-dimensional design space in the figure below.
29
1. Introduction
• Once the objective function surfaces are drawn along with the constraint surfaces, the optimum point can be determined without much difficulty.
• But the main problem is that as the number of design variables exceeds two or three, the constraint and objective function surfaces become complex even for visualization and the problem has to be solved purely as a mathematical problem.
30
Example
Example:
Design a uniform column of tubular section to carry a compressive load P=2500 kg f
for minimum cost. The column is made up of a material that has a yield stress of 500
kgf/cm2, modulus of elasticity (E) of 0.85e6 kgf/cm2, and density () of 0.0025 kgf/cm3.
The length of the column is 250 cm. The stress induced in this column should be less
than the buckling stress as well as the yield stress. The mean diameter of the column
is restricted to lie between 2 and 14 cm, and columns with thicknesses outside the
range 0.2 to 0.8 cm are not available in the market. The cost of the column includes
material and construction costs and can be taken as 5W + 2d, where W is the weight
in kilograms force and d is the mean diameter of the column in centimeters.
31
Example
Example:
The design variables are the mean diameter (d) and tube thickness (t):
The objective function to be minimized is given by:
t
d
x
x
2
1X
121 282.92525)( xxxddtldWf X
32
Example
• The behaviour constraints can be expressed as:
stress induced ≤ yield stress
stress induced ≤ buckling stress
• The induced stress is given by:
21i
2500stress induced
xxdt
P
33
Example
• The buckling stress for a pin connected column is given by:
where I is the second moment of area of the cross section of the column given by:
dtl
EI
2
2
b area sectionalcross
load bucklingEuler stress buckling
)(8
)(8
)()()()()()(64
))()((64
)(64
22
2121
22
22
2244
xxxxtddt
tdtdtdtdtdtd
ddddddddI ioioioio
34
Example
• Thus, the behaviour constraints can be restated as:
• The side constraints are given by:
0)250(8
))(1085.0(2500)(
05002500
)(
2
22
21
62
212
211
xx
xxg
xxg
X
X
8.02.0
142
t
d
35
Example
• The side constraints can be expressed in standard form as:
08.0)(
02.0)(
014)(
02)(
26
25
14
13
xg
xg
xg
xg
X
X
X
X
36
Example
• For a graphical solution, the constraint surfaces are to be plotted in a two dimensional design space where the two axes represent the two design variables x1 and x2. To plot the first constraint surface, we have:
• Thus the curve x1x2=1.593 represents the constraint surface g1(X)=0. This curve can be plotted by finding several points on the curve. The points on the curve can be found by giving a series of values to x1 and finding the corresponding values of x2
that satisfy the relation x1x2=1.593 as shown in the Table below:
593.105002500
)( 2121
1 xxxx
g
X
x1 2 4 6 8 10 12 14
x2 0.7965 0.3983 0.2655 0.199 0.1593 0.1328 0.114
37
Example
• The infeasible region represented by g1(X)>0 or x1x2< 1.593 is shown by hatched lines. These points are plotted and a curve P1Q1 passing through all these points is drawn as shown:
38
Example
• Similarly the second constraint g2(X) < 0 can be expressed as:
• The points lying on the constraint surface g2 (X)=0 can be obtained as follows (These points are plotted as Curve P2Q2:
3.47)( 22
2121 xxxx
x1 2 4 6 8 10 12 14
x2 2.41 0.716 0.219 0.0926 0.0473 0.0274 0.0172
39
Example
• The plotting of side constraints is simple since they represent straight lines.
• After plotting all the six constraints, the feasible region is determined as the bounded area ABCDEA
3.47)( 22
2121 xxxx
40
Example
• Next, the contours of the objective function are to be plotted before finding the optimum point. For this, we plot the curves given by:
for a series of values of c. By giving different values to c, the contours of f can be plotted with the help of the following points.
constant
282.9)( 121
cxxxf X
41
Example
• For
• For
• For
• For
0.50282.9)( 121 xxxf X
x2 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x1 16.77 12.62 10.10 8.44 7.24 6.33 5.64
0.40282.9)( 121 xxxf X
x2 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x1 13.40 10.10 8.08 6.75 5.79 5.06 4.51
C)point corner he through t(passing 58.31282.9)( 121 xxxf X
x2 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x1 10.57 7.96 6.38 5.33 4.57 4 3.56
B)point corner he through t(passing 53.26282.9)( 121 xxxf X
x2 0.1 0.2 0.3 0.4 0.5 0.6 0.7
x1 8.88 6.69 5.36 4.48 3.84 3.36 2.99
42
Example
• These contours are shown in the
figure below and it can be seen
that the objective function can not
be reduced below a value of 26.53
(corresponding to point B) without
violating some of the constraints.
Thus, the optimum solution is
given by point B with d*=x1*=5.44
cm and t*=x2*=0.293 cm with
fmin=26.53.
43
ExamplesExamples
Design of civil engineering structures• variables: width and height of member cross-sections
• constraints: limit stresses, maximum and minimum dimensions
• objective: minimum cost or minimum weight
Analysis of statistical data and building empirical models from measurements
• variables: model parameters
• Constraints: physical upper and lower bounds for model parameters
• Objective: prediction error
44
Classification of optimization problemsClassification of optimization problems
Classification based on:
• Constraints– Constrained optimization problem
– Unconstrained optimization problem
• Nature of the design variables– Static optimization problems
– Dynamic optimization problems
45
Classification of optimization problemsClassification of optimization problems
Classification based on:
• Physical structure of the problem– Optimal control problems
– Non-optimal control problems
• Nature of the equations involved– Nonlinear programming problem
– Geometric programming problem
– Quadratic programming problem
– Linear programming problem
46
Classification of optimization problemsClassification of optimization problems
Classification based on:
• Permissable values of the design variables– Integer programming problems
– Real valued programming problems
• Deterministic nature of the variables– Stochastic programming problem
– Deterministic programming problem
47
Classification of optimization problemsClassification of optimization problems
Classification based on:
• Separability of the functions– Separable programming problems
– Non-separable programming problems
• Number of the objective functions– Single objective programming problem
– Multiobjective programming problem
48
Geometric Programming
• A geometric programming problem (GMP) is one in which the objective function and constraints are expressed as posynomials in X.
49
50
Quadratic Programming Problem
• A quadratic programming problem is a nonlinear programming problem with a quadratic objective function and linear constraints. It is usually formulated as follows:
subject to
where c, qi,Qij, aij, and bj are constants.
ji
n
jij
n
ii
n
ii xxQxqcF
111
)(X
nix
mjbxa
i
ji
n
iij
,,2,1,0
,,2,1,1
51
Optimal Control Problem
• An optimal control (OC) problem is a mathematical programming problem involving a number of stages, where each stage evolves from the preceding stage in a prescribed manner.
• It is usually described by two types of variables: the control (design) and the state variables. The control variables define the system and govern the evolution of the system from one stage to the next, and the state variables describe the behaviour or status of the system in any stage.
52
Optimal Control Problem
• The problem is to find a set of control or design variables such that the total objective function (also known as the performance index) over all stages is minimized subject to a set of constraints on the control and state variables.
• An OC problem can be stated as follows:
Find X which minimizes
subject to the constraints
where xi is the ith control variable, yi is the ith control variable, and fi is the contribution of the ith stage to the total objective function; gj, hk and qi are functions of xj, yk and xi and yi, respectively, and l is the total number of stages.
),( )(l
1iiii yxff
X
lky
ljx
liyyyxq
kk
jj
iiiii
,,2,1,0)(
,,2,1,0)(
,,2,1,),( 1
h
g
53
Integer Programming Problem
• If some or all of the design variables x1,x2,..,xn of an optimization problem are restricted to take on only integer (or discrete) values, the problem is called an integer programming problem.
• If all the design variables are permitted to take any real value, the optimization problem is called a real-valued programming problem.
54
Stochastic Programming Problem
• A stochastic programming problem is an optimization problem in which some or all of the parameters (design variables and/or preassigned parameters) are probabilistic (nondeterministic or stochastic).
• In other words, stochastic programming deals with the solution of the optimization problems in which some of the variables are described by probability distributions.
55
Separable Programming Problem
• A function f (x) is said to be separable if it can be expressed as the sum of n single variable functions, f1(x1), f2(x2),….,fn(xn), that is,
• A separable programming problem is one in which the objective function and the constraints are separable and can be expressed in standard form as:
Find X which minimizes
subject to
where bj is constant.
i
n
ii xff
1
)(X
)()(1
i
n
ii xff
X
mjbxgg ji
n
iijj ,,2,1,)()(
1
X
56
Multiobjective Programming Problem
• A multiobjective programming problem can be stated as follows:
Find X which minimizes f1 (X), f2 (X),…., fk (X)
subject to
where f1 , f2,…., fk denote the objective functions to be minimized simultaneously.
mjg j ,...,2,1,0)( X
57
Review of mathematics
Concepts from linear algebra:Concepts from linear algebra: Positive definiteness Positive definiteness
• Test 1: A matrix A will be positive definite if all its eigenvalues are positive; that is, all the values of that satisfy the determinental equation
should be positive. Similarly, the matrix A will be negative definite if its eigenvalues are negative.
0 I A
58
Review of mathematicsReview of mathematics
Positive definitenessPositive definiteness• Test 2: Another test that can be used to find the positive definiteness of a matrix A of
order n involves evaluation of the determinants
• The matrix A will be positive definite if and only if all the values A1, A2, A3,An are positive
• The matrix A will be negative definite if and only if the sign of Aj is (-1)j for j=1,2,,n • If some of the Aj are positive and the remaining Aj are zero, the matrix A will be positive
semidefinite
333231
232221
131211
3
2221
12112
11
aaa
aaa
aaa
A
aa
aaA
aA
nnnnn
n
n
n
aaaa
aaaa
aaaa
aaaa
A
321
3333231
2232221
1131211
3
59
Review of mathematicsReview of mathematics
Negative definitenessNegative definiteness
• Equivalently, a matrix is negative-definite if all its eigenvalues are negative
• It is positive-semidefinite if all its eigenvalues are all greater than or equal to zero
• It is negative-semidefinite if all its eigenvalues are all less than or equal to zero
60
Concepts from linear algebra:Concepts from linear algebra:
Nonsingular matrix: The determinant of the matrix is not zero.
Rank: The rank of a matrix A is the order of the largest nonsingular square submatrix of A, that is, the largest submatrix with a determinant other than zero.
Review of mathematicsReview of mathematics
61
Review of mathematicsReview of mathematics
Solutions of a linear problem Minimize f(x)=cTxSubject to g(x): Ax=bSide constraints: x ≥0• The existence of a solution to this problem depends on the
rows of A.
• If the rows of A are linearly independent, then there is a unique solution to the system of equations.
• If det(A) is zero, that is, matrix A is singular, there are either no solutions or infinite solutions.
62
Review of mathematicsReview of mathematics
Suppose
The new matrix A* is called the augmented matrix- the columns of b are added to A. According to the theorems of linear algebra:
• If the augmented matrix A* and the matrix of coefficients A have the same rank r which is less than the number of design variables n: (r < n), then there are many solutions.
• If the augmented matrix A* and the matrix of coefficients A do not have the same rank, a solution does not exist.
• If the augmented matrix A* and the matrix of coefficients A have the same rank r=n, where the number of constraints is equal to the number of design variables, then there is a unique solution.
1.50.511-
3111 A*
1.5
3b A
5.011
111
63
Review of mathematicsReview of mathematics
In the example
The largest square submatrix is a 2 x 2 matrix (since m = 2 and m < n). Taking the submatrix which includes the first two columns of A, the determinant has a value of 2 and therefore is nonsingular. Thus the rank of A is 2 (r = 2). The same columns appear in A* making its rank also 2. Since
r < n, infinitely many solutions exist.
1.50.511-
3111 A*
1.5
3b A
5.011
111
64
Review of mathematicsReview of mathematics
In the example
One way to determine the solutions is to assign ( n-r) variables arbitrary values and use them to determine values for the remaining r variables. The value n-r is often identified as the degree of freedom for the system of equations.
In this example, the degree of freedom is 1 (i.e., 3-2). For instance x3 can be assigned a value of 1 in which case x1=0.5 and x2=1.5
1.50.511-
3111 A*
1.5
3b A
5.011
111
65
HomeworkHomework What is the solution of the system given below?
Hint: Determine the rank of the matrix of the coefficients and the augmented matrix.
12:
1:
2:
213
212
211
xxg
xxg
xxg
66
2. Classical optimization techniques2. Classical optimization techniques Single variable optimization
• Useful in finding the optimum solutions of continuous and differentiable functions
• These methods are analytical and make use of the techniques of differential calculus in locating the optimum points.
• Since some of the practical problems involve objective functions that are not continuous and/or differentiable, the classical optimization techniques have limited scope in practical applications.
67
2. Classicial optimization techniques2. Classicial optimization techniques Single variable optimization
• A function of one variable f (x) has a relative or local minimum at x = x* if f (x*) ≤ f (x*+h) for all sufficiently small positive and negative values of h
• A point x* is called a relative or local maximum if f (x*) ≥ f (x*+h) for all values of h sufficiently close to zero.
Local minimum
Global minima
Local minima
68
2. Classicial optimization techniques2. Classicial optimization techniques Single variable optimization
• A function f (x) is said to have a global or absolute minimum at x* if f (x*) ≤ f (x) for all x, and not just for all x close to x*, in the domain over which f (x) is defined.
• Similarly, a point x* will be a global maximum of f (x) if f (x*) ≥ f (x) for all x in the domain.
69
Necessary conditionNecessary condition
• If a function f (x) is defined in the interval a ≤ x ≤ b and has a relative minimum at x = x*, where a < x* < b, and if the derivative df (x) / dx = f’(x) exists as a finite number at x = x*, then f’ (x*)=0
• The theorem does not say that the function necessarily will have a minimum or maximum at every point where the derivative is zero. e.g. f’ (x)=0 at x= 0 for the function shown in figure. However, this point is neither a minimum nor a maximum. In general, a point x* at which f’(x*)=0 is called a stationary point.
70
Necessary conditionNecessary condition
• The theorem does not say what happens if a minimum or a maximum occurs at a point x* where the derivative fails to exist. For example, in the figure
depending on whether h approaches zero through positive or negative values, respectively. Unless the numbers or are
equal, the derivative f’ (x*) does not exist. If f’ (x*) does not exist, the theorem is not applicable.
(negative) mor (positive) *)()*(
lim -
0
m
h
xfhxfh
FIGURE 2.2 SAYFA 67
m m
71
Sufficient conditionSufficient condition
• Let f’(x*)=f’’(x*)=…=f (n-1)(x*)=0, but f(n)(x*) ≠ 0. Then f(x*) is – A minimum value of f (x) if f (n)(x*) > 0 and n is even
– A maximum value of f (x) if f (n)(x*) < 0 and n is even
– Neither a minimum nor a maximum if n is odd
72
Example
Determine the maximum and minimum values of the function:
Solution: Since f’(x)=60(x4-3x3+2x2)=60x2(x-1)(x-2),
f’(x)=0 at x=0,x=1, and x=2.
The second derivative is:
At x=1, f’’(x)=-60 and hence x=1 is a relative maximum. Therefore,
fmax= f (x=1) = 12
At x=2, f’’(x)=240 and hence x=2 is a relative minimum. Therefore,
fmin= f (x=2) = -11
5404512)( 345 xxxxf
)494(60)( 23 xxxxf
73
Example
Solution cont’d:
At x=0, f’’(x)=0 and hence we must investigate the next derivative.
Since at x=0, x=0 is neither a maximum nor a minimum, and it is an inflection point.
0at 240)41812(60)( 2 xxxxf
0)( xf
74
Multivariable optimization with no Multivariable optimization with no constraintsconstraints
• Definition: rth Differential of f
If all partial derivatives of the function f through order r ≥ 1 exist and are continuous at a point X*, the polynomial
is called the rth differential of f at X*.
kji
r
kj
n
ki
n
j
n
i
r
xxx
fhhhfd
*)(
*)(111
XX
r summations
75
Multivariable optimization with no Multivariable optimization with no constraintsconstraints
• Example: rth Differential of f
when r = 2 and n = 3, we have
kji
r
kj
n
ki
n
j
n
i
r
xxx
fhhhfd
*)(
*)(111
XX
r summations
)(2)(2)(2
)()()(
)(*)*,*,()(
31
2
3132
2
3221
2
21
23
2232
2
2222
1
22
1
23
1
3
1321
22
X*X*X*
X*X*X*
X*X*
xx
fhh
xx
fhh
xx
fhh
x
fh
x
fh
x
fh
xx
fhhxxxfdfd
jij
ji
i
76
Multivariable optimization with no Multivariable optimization with no constraintsconstraints
• Definition: rth Differential of f
The Taylor series expansion of a function f (X*) about a point X* is given by:
where the last term, called the remainder is given by:
)h*,()*(!
1
)*(!3
1)*(
!2
1)*()*()( 32
XX
XXXXX
NN Rfd
N
fdfddfff
*X-Xh
hX*,hX*,
and 10 where
)()!1(
1)( 1
fdN
R NN
77
Example
Find the second order Taylor’s series approximation of the function
about the point
Solution: The second order Taylor’s series approximation of the function f about point X* is given by
313
22321 ),,( xexxxxxxf
2-
0
1
X*
2
0
1
!2
1
2
0
1
2
0
1
)( 2 fddfff X
78
Example cont’d
where
2
2
0
1
ef
23
2113
2233221
33
22
11
2
0
1
])2([
2
0
1
2
0
1
2
0
1
2
0
1
33
ehehexhxhxxheh
x
fh
x
fh
x
fhdf
xx
79
Example cont’d
where
231
23
222
31232211233
22
21
31
2
3132
2
3221
2
21
23
22
322
22
221
22
1
23
1
3
1
2
2-4
2
0
1
)](2)2(2)0(2)()2()0([
2
0
1
)222
(
2
0
1
2
0
1
33
ehhheh
ehhxhhhhexhxhh
xx
fhh
xx
fhh
xx
fhh
x
fh
x
fh
x
fh
xx
fhhfd
xx
jiji
ji
80
Example cont’d
Thus, the Taylor’s series approximation is given by:
Where h1=x1-1, h2=x2, and h3=x3+2
)2(-4!2
1)()( 2
3123
22231
22 ehhhehhheef X
81
Multivariable optimization with no Multivariable optimization with no constraintsconstraints
• Necessary condition If f(X) has an extreme point (maximum or minimum) at X=X*
and if the first partial derivatives of f (X) exist at X*, then
• Sufficient condition A sufficient condition for a stationary point X* to be an
extreme point is that the matrix of second partial derivatives (Hessian matrix) of f (X*) evaluated at X* is – Positive definite when X* is a relative minimum point– Negative definite when X* is a relative maximum point
0*)(*)(*)(21
XXXnx
f
x
f
x
f
82
ExampleExample
Figure shows two frictionless rigid bodies (carts) A and B connected by three linear elastic springs having spring constants k1, k2, and k3. The springs are at their natural positions when the applied force P is zero. Find the displacements x1 and x2 under the force P by using the principle of minimum potential energy.
83
ExampleExample
Solution: According to the principle of minimum potential energy, the system will be in equilibrium under the load P if the potential energy is a minimum. The potential energy of the system is given by:
Potential energy (U)
= Strain energy of springs-work done by external forces
The necessary condition for the minimum of U are
2221
2123
212 ]
2
1)(
2
1
2
1[ Pxxkxxkxk
0)(
0)(
211232
123121
Pxkxxkx
U
xxkxkx
U
323121
322
323121
31
)(*
*
kkkkkk
kkPx
kkkkkk
Pkx
84
ExampleExample
Solution cont’d: The sufficiency conditions for the minimum at (x1*,x2*) can also be verified by testing the positive definiteness of the Hessian matrix of U. The Hessian matrix of U evaluated at (x1*,x2*) is:
The determinants of the square submatrices of J are
Since the spring constants are always positive. Thus the matrix J is positive definite and hence (x1*,x2*) corresponds to the minimum of potential energy.
313
332
*)*,(
22
2
21
2
21
2
21
2
*)*,(
21
21 kkk
kkk
x
U
xx
U
xx
U
x
U
xx
xxJ
0
0
323121313
3322
32321
kkkkkkkkk
kkkJ
kkkkJ
85
Semi-definite caseSemi-definite case
The sufficient conditions for the case when the Hessian matrix of the given function is semidefinite:
• In case of a function of a single variable, the higher order derivatives in the Taylor’s series expansion are investigated
86
Semi-definite caseSemi-definite case
The sufficient conditions for a function of several variables for the case when the Hessian matrix of the given function is semidefinite:
• Let the partial derivatives of f of all orders up to the order k ≥ 2 be continuous in the neighborhood of a stationary point X*, and
so that dk f |X=X* is the first nonvanishing higher-order differential of f at X*. • If k is even:
– X* is a relative minimum if dk f |X=X* is positive definite– X* is a relative maximum if dk f |X=X* is negative definite– If dk f |X=X* is semidefinite, no general conclusions can be drawn
• If k is odd, X* is not an extreme point of f(X*)
0|
110|
*XX
*XX
fd
krfdk
r
87
Saddle pointSaddle point
• In the case of a function of two variables f (x,y), the Hessian matrix may be neither positive nor negative definite at a point (x*,y*) at which
In such a case, the point (x*,y*) is called a saddle point.• The characteristic of a saddle point is that it corresponds
to a relative minimum or maximum of f (x,y) wrt one variable, say, x (the other variable being fixed at y=y* ) and a relative maximum or minimum of f (x,y) wrt the second variable y (the other variable being fixed at x*).
0
y
f
x
f
88
Saddle pointSaddle point
Example: Consider the function f (x,y)=x2-y2. For this function:
These first derivatives are zero at x* = 0 and y* = 0. The Hessian matrix of f at (x*,y*) is given by:
Since this matrix is neither positive definite nor negative definite, the point ( x*=0, y*=0) is a saddle point.
yy
fx
x
f2 and 2
20
02
J
89
Saddle pointSaddle point
Example cont’d: It can be seen from the figure that f (x, y*) = f (x, 0) has a relative minimum and f
(x*, y) = f (0, y) has a relative maximum at the saddle point (x*, y*).
90
ExampleExample
Find the extreme points of the function
Solution: The necessary conditions for the existence of an extreme point are:
These equations are satisfied at the points: (0,0), (0,-8/3), (-4/3,0), and (-4/3,-8/3)
642),( 22
21
32
3121 xxxxxxf
0)83(83
0)43(43
22222
2
11121
1
xxxxx
f
xxxxx
f
91
ExampleExample
Solution cont’d: To find the nature of these extreme points, we have to use the sufficiency conditions. The second order partial derivatives of f are given by:
The Hessian matrix of f is given by:
0 864621
2
222
2
121
2
xx
fx
x
fx
x
f
860
046
2
1
x
xJ
92
ExampleExample
Solution cont’d:
If J1=|6x1+4| and , the values of J1 and J2 and
the nature of the extreme point are as given in the next slide:
860
046
2
1
x
xJ
860
046
2
1
x
x2J
93
Example
Point X Value of J1 Value of J2 Nature of J Nature of X
f (X)
(0,0) +4 +32 Positive definite
Relative minimum
6
(0,-8/3) +4 -32 Indefinite Saddle point
418/27
(-4/3,0) -4 -32 Indefinite Saddle point
194/27
(-4/3,-8/3) -4 +32 Negative definite
Relative
maximum
50/3
94
Multivariable optimization with equality Multivariable optimization with equality constraintsconstraints
• Problem statement: Minimize f = f (X) subject to gj(X)=0, j=1,2,…..,m where
Here m is less than or equal to n, otherwise the problem becomes overdefined and, in general, there will be no solution.
• Solution:– Solution by direct substitution– Solution by the method of constrained variation– Solution by the method of Lagrange multipliers
nx
x
x
2
1
X
95
Solution by direct substitutionSolution by direct substitution
For a problem with n variables and m equality constraints:
• Solve the m equality constraints and express any set of m variables in terms of the remaining n-m variables
• Substitute these expressions into the original objective function, the result is a new objective function involving only n-m variables
• The new objective function is not subjected to any constraint, and hence its optimum can be found by using the unconstrained optimization techniques.
96
Solution by direct substitutionSolution by direct substitution
• Simple in theory
• Not convenient from a practical point of view as the constraint equations will be nonlinear for most of the problems
• Suitable only for simple problems
97
ExampleExample
Find the dimensions of a box of largest volume that can be inscribed in a sphere of unit radius
Solution: Let the origin of the Cartesian coordinate system x1, x2, x3 be at the center of the sphere and the sides of the box be 2x1, 2x2, and 2x3. The volume of the box is given by:
Since the corners of the box lie on the surface of the sphere of unit radius, x1, x2 and x3 have to satisfy the constraint
321321 8),,( xxxxxxf
123
22
21 xxx
98
ExampleExample
This problem has three design variables and one equality constraint. Hence the equality constraint can be used to eliminate any one of the design variables from the objective function. If we choose to eliminate x3:
Thus, the objective function becomes:
f(x1,x2)=8x1x2(1-x12-x2
2)1/2
which can be maximized as an unconstrained function in two variables.
2/122
213 )1( xxx
99
ExampleExample
The necessary conditions for the maximum of f give:
which can be simplified as:
From which it follows that x1*=x2*=1/3 and hence x3*= 1/3
0])1(
)1[(8
0])1(
)1[(8
2/122
21
222/12
2211
2
2/122
21
212/12
2212
1
xx
xxxx
x
f
xx
xxxx
x
f
021
02122
21
22
21
xx
xx
100
ExampleExample
This solution gives the maximum volume of the box as:
To find whether the solution found corresponds to a maximum or minimum, we apply the sufficiency conditions to f (x1,x2) of the equation f (x1,x2)=8x1x2(1-x1
2-x22)1/2. The
second order partial derivatives of f at (x1*,x2*) are given by:
*)*,(at 3
32
)1(2)1(1
8
)1(
8
21
2/122
2112/12
221
31
22
21
22/12
221
2121
2
xx
xxxxx
x
xx
x
xx
xx
x
f
33
8max f
101
ExampleExample
The second order partial derivatives of f at (x1*,x2*) are given by:
*)*,(at 3
32
)1(2)1(1
8
)1(
8
21
2/122
2122/12
221
32
22
21
12/12
221
2122
2
xx
xxxxx
x
xx
x
xx
xx
x
f
*)*,(at 3
16
])1(
)1[(1
8
)1(
8)1(8
21
2/122
21
222/12
2212
221
21
2/122
21
222/12
221
21
2
xx
xx
xxx
xx
x
xx
xxx
xx
f
102
ExampleExample
Since
the Hessian matrix of f is negative definite at (x1*,x2*). Hence the point (x1*,x2*) corresponds to the maximum of f.
0 and 02
21
2
22
2
21
2
22
2
xx
f
x
f
x
f
x
f
103
Solution by constrained variationSolution by constrained variation
• Minimize f (x1,x2) subject to g(x1,x2)=0
• A necessary condition for f to have a minimum at some point (x1*,x2*) is that the total derivative of f (x1,x2) wrt x1 must be zero at (x1*,x2*)
• Since g(x1*,x2*)=0 at the minimum point, any variations dx1 and dx2 taken about the point (x1*,x2*) are called admissable variations provided that the new point lies on the constraint:
022
11
dxx
fdx
x
fdf
0),( 2*21
*1 dxxdxxg
104
Solution by constrained variationSolution by constrained variation
• Taylor’s series expansion of the function about the point (x1*,x2*):
• Since g(x1*, x2*)=0
• Assuming
• Substituting the above equation into
),(at0 *2
*12
21
1
xxdxx
gdx
x
gdg
0),(),(),(),( 2*2
*1
21
*2
*1
1
*2
*12
*21
*1
dxxxx
gdxxx
x
gxxgdxxdxxg
02
x
g
1*2
*1
2
12 ),(
/
/dxxx
xg
xgdx
022
11
dxx
fdx
x
fdf
0)/
/( 1),(
22
1
1
*2
*1
dxx
f
xg
xg
x
fdf
xx
105
• The expression on the left hand side is called the constrained variation of f
• Since dx1 can be chosen arbitrarily:
• This equation represents a necessary condition in order to have (x1*,x2*) as an extreme point (minimum or maximum)
Solution by constrained variationSolution by constrained variation
0)/
/( 1),(
22
1
1
*2
*1
dxx
f
xg
xg
x
fdf
xx
0)(),(
1221
*2
*1
xxx
g
x
f
x
g
x
f
106
A beam of uniform rectangular cross section is to be cut from a log having a circular cross secion of diameter 2 a. The beam has to be used as a cantilever beam (the length is fixed) to carry a concentrated load at the free end. Find the dimensions of the beam that correspond to the maximum tensile (bending) stress carrying capacity.
Example Example
107
Solution: From elementary strength of materials, we know that the tensile stress induced in a rectangular beam at any fiber located at a distance y from the neutral axis is given by
where M is the bending moment acting and I is the moment of inertia
of the cross-section about the x axis. If the width and the depth of the rectangular beam shown in the figure are 2x and 2y, respectively, the maximum tensile stress induced is given by:
Example Example
I
M
y
23
max 4
3
)2)(2(121 xy
M
yx
Myy
I
M
108
solution cont’d: Thus for any specified bending moment, the beam is said to have maximum tensile stress carrying capacity if the maximum induced stress (max) is a minimum. Hence we need to minimize k/xy2 or maximize Kxy2, where k=3M/4 and K=1/k, subject to the constraint
This problem has two variables and one constraint; hence the equation
can be applied for finding the optimum solution.
Example Example
222 ayx
0)(),(
1221
*2
*1
xxx
g
x
f
x
g
x
f
109
Solution: Since
we have:
Equation gives:
Example Example
222
21
ayxg
ykxf
31
22
2
ykxy
f
ykxx
f
yy
g
xx
g
2
2
0)(),(
1221
*2
*1
xxx
g
x
f
x
g
x
f
)**,(at0)2(2)2( 3122 yxxykxyykx
110
Solution: that is
Thus the beam of maximum tensile stress carrying capacity has a depth of 2 times its breadth. The optimum values of x and y can be obtained from the above equation and
as:
Example Example
*2* xy
222 ayxg
32* and
3*
ay
ax
111
Necessary conditions for a general problem
• The procedure described can be generalized to a problem with n variables and m constraints.
• In this case, each constraint equation gj(x)=0, j=1,2,..,m gives rise to a linear equation in the variations dxi, i=1,2,…,n.
• Thus, there will be in all m linear equations in n variations. Hence any m variations can be expressed in terms of the remaining n-m variations.
• These expressions can be used to express the differential of the objective function, df, in terms of the n-m independent variations.
• By letting the coefficients of the independent variations vanish in the equation df = 0, one obtains the necessary conditions for the cnstrained optimum of the given function.
Solution by constrained variationSolution by constrained variation
112
Necessary conditions for a general problem
• These conditions can be expressed as:
• It is to be noted that the variations of the first m variables (dx1, dx2,.., dxm) have been expressed in terms of the variations of the remaining n-m variables (dxm+1, dxm+2,.., dxn) in deriving the above equation.
Solution by constrained variationSolution by constrained variation
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
113
Necessary conditions for a general problem
• This implies that the following relation is satisfied:
• The n-m equations given by the below equation represent the necessary conditions for the extremum of f(X) under the m equality constraints, gj(X) = 0, j=1,2,…,m.
Solution by constrained variationSolution by constrained variation
0,,,
,,,
21
21
m
m
xxx
gggJ
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
114
Minimizesubject to
Solution: This problem can be solved by applying the necessary conditions given by
ExampleExample)(
2
1)( 2
423
22
21 yyyyf Y
015652)(
010532)(
43212
43211
yyyyg
yyyyg
Y
Y
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
115
Solution cont’d: Since n = 4 and m = 2, we have to select two variables as independent variables. First we show that any arbitrary set of variables can not be chosen as independent variables since the remaining (dependent) variables have to satisfy the condition of
In terms of the notation of our equations, let us take the independent variables as x3=y3 and x4=y4 so that x1=y1 and x2=y2. Then the Jacobian becomes:
and hence the necessary conditions can not be applied.
ExampleExample
0,,,
,,,
21
21
m
m
xxx
gggJ
021
21
,
,
2
2
1
2
2
1
1
1
21
21
y
g
y
g
y
g
y
g
xx
ggJ
116
Solution cont’d: Next, let us take the independent variables as x3=y2
and x4=y4 so that x1=y1 and x2=y3. Then the Jacobian becomes:
and hence the necessary conditions of
can be applied.
ExampleExample
0251
31
,
,
3
2
1
2
3
1
1
1
21
21
y
g
y
g
y
g
y
g
xx
ggJ
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
117
Solution cont’d: The equation
give for k = m+1=3
ExampleExample
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
042)22()610()35(
512
312 12312
312
3
2
1
2
2
2
3
1
1
1
2
1
312
2
2
1
2
3
2
2
1
1
1
3
1
213
yyyyy
yyy
y
g
y
g
y
g
y
g
y
g
y
g
y
f
y
f
y
f
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
118
Solution cont’d: For k = m+2=4
From the two previous equations, the necessary conditions for the minimum or the maximum of f is obtained as:
ExampleExample
072)65()1825()35(
516
315 314314
314
3
2
1
2
4
2
3
1
1
1
4
1
314
2
2
1
2
4
2
2
1
1
1
4
1
214
yyyyyy
yyy
y
g
y
g
y
g
y
g
y
g
y
g
y
f
y
f
y
f
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
24143
21
2
7272
2
1
yyyyy
yy
119
Solution cont’d: When the equations
are substituted, the equations
take the form:
ExampleExample
015652)(
010532)(
43212
43211
yyyyg
yyyyg
Y
Y
24143
21
2
7272
2
1
yyyyy
yy
151615
10118
42
42
yy
yy
120
Solution cont’d: from which the desired optimum solution can be obtained as:
ExampleExample
37
30*
74
155*
37
5*
74
5*
4
3
2
1
y
y
y
y
121
Sufficiency conditions for a general problem
• By eliminating the first m variables, using the m equality constraints, the objective function can be made to depend only on the remaining variables xm+1, xm+2, …,xn. Then the Taylor’s series expansion of f , in terms of these variables, about the extreme point X* gives:
where is used to denote the partial derivative of f wrt xi (holding all the other variables xm+1, xm+2, …,xi-1, xi+1, xi+2,…,xn constant) when x1, x2, …,xm are allowed to change so that the constraints gj(X*+dX)=0, j=1,2,…,m are satisfied; the second derivative is used to denote a similar meaning.
Solution by constrained variationSolution by constrained variation
ji
gji
n
mj
n
mii
g
n
mi i
dxdxxx
fdx
x
f)fdf
2
111 2
1()( *XX*X
gixf )/(
gji xxf )/( 2
122
Example
Consider the problem of minimizing
Subject to the only constraint
Since n=3 and m=1 in this problem, one can think of any of the m variables, say x1, to be dependent and the remaining n-m variables, namely x2 and x3, to be independent.
Here the constrained partial derivative means the rate of change of f with respect to x2 (holding the other independent variable x3 constant) and at the same time allowing x1 to change about X* so as to satisfy the constraint g1(X)=0
Solution by constrained variationSolution by constrained variation
),,()( 321 xxxff X
08)( 23
22
211 xxxg X
gxf )/( 2
123
Example
In the present case, this means that dx1 has to be chosen to satisfy the relation
since g1(X*)=0 at the optimum point and dx3= 0 (x3 is held constant.)
Solution by constrained variationSolution by constrained variation
022
isthat
0)()()()()(
2*21
*1
33
12
2
11
1
111
dxxdxx
dxx
gdx
x
gdx
x
ggdg X*X*X*X*X*X
124
Example
Notice that (df/dxi)g has to be zero for i=m+1, m+2,...,n since the dxi appearing in the equation
are all independent. Thus, the necessary conditions for the existence of constrained optimum at X* can also be expressed as:
Solution by constrained variationSolution by constrained variation
ji
gji
n
mj
n
mii
g
n
mi i
dxdxxx
fdx
x
f)fdf
2
111 2
1()( *XX*X
nmmix
f
gi
,...,2,1,0
125
Example
It can be shown that the equations
are nothing bu the equation
Solution by constrained variationSolution by constrained variation
nmmix
f
gi
,...,2,1,0
nmmk
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
g
x
f
x
f
x
f
x
f
xxxxx
gggfJ
m
mmm
k
m
mk
mk
mk
mk
m ,,2,1where0,,,,,
,,,,
21
2
2
2
1
22
1
2
1
1
11
21
321
21
126
• A sufficient condition for X* to be a constrained relative minimum (maximum) is that the quadratic form Q defined by
is positive (negative) for all nonvanishing variations dxi and the matrix
has to be positive (negative) definite to have Q positive (negative) for all choices of dxi
Sufficiency conditions for a general problem
ji
g
n
mj ji
n
mi
dxdxxx
fQ
1
2
1
gngmngmn
gnmgmmgm
x
f
xx
f
xx
f
xx
f
xx
f
x
f
2
2
2
2
1
2
1
2
21
2
21
2
127
• The computation of the constrained derivatives in the sufficiency condition is difficult and may be prohibitive for problems with more than three constraints
• Simple in theory
• Difficult to apply since the necessary conditions involve evaluation of determinants of order m+1
Solution by constrained variationSolution by constrained variation
128
Problem with two variables and one constraint:
Minimize f (x1,x2)
Subject to g(x1,x2)=0
For this problem, the necessary condition was found to be:
By defining a quantity , called the Lagrange multiplier as:
Solution by Lagrange multipliersSolution by Lagrange multipliers
0)/
/(
),(12
2
1
*2
*1
xxx
g
xg
xf
x
f
*)*,(2
2
21
/
/
xxxg
xf
129
Problem with two variables and one constraint:
Necessary conditions for the point (x1*,x2*) to be an extreme point The problem can be rewritten as:
In addition, the constraint equation has to be satisfied at the extreme point:
Solution by Lagrange multipliersSolution by Lagrange multipliers
0)(),(
11
*2
*1
xxx
g
x
f
0)(),(
22
*2
*1
xxx
g
x
f
0),(),(21 *
2*1
xx
xxg
130
Problem with two variables and one constraint:
• The derivation of the necessary conditions by the method of Lagrange multipliers requires that at least one of the partial derivatives of g(x1,x2) be nonzero at an extreme point.
• The necessary conditions are more commonly generated by constructing a function L,known as the Lagrange function, as
Solution by Lagrange multipliersSolution by Lagrange multipliers
0),(),(),,( 212121 xxgxxfxxL
131
Problem with two variables and one constraint:
• By treating L as a function of the three variables x1, x2 and , the necessary conditions for its extremum are given by:
Solution by Lagrange multipliersSolution by Lagrange multipliers
0),(),,(
0),(),(),,(
0),(),(),,(
2121
212
212
212
211
211
211
xxgxxL
xxx
gxx
x
fxx
x
L
xxx
gxx
x
fxx
x
L
132
Example: Find the solution using the Lagrange multiplier method.
SolutionThe Lagrange function is
ExampleExample
0
022
02
),( of minimum for the conditionsnecessary The
)(),(),(),,(
222
31
22
22221
ayxL
yykxy
L
xykxx
L
yxf
ayxykxyxgyxfyxL
0),(
subject to
),( Minimize
222
21
ayxyxg
ykxyxf
133
Solution cont’dwhich yield:
ExampleExample
3
a2* and
3
ax*
:assolution optimum thegives 0
withalongrelation This obtained. becan *)2/1(*relation which thefrom
22
222
423
y
ayxL
yx
xy
k
yx
k
134
Necessary conditions for a general problem:
Minimize f(X)
subject to
gj (X)= 0, j=1, 2,….,m
The Lagrange function, L, in this case is defined by introducing one Lagrange multiplier j for each constraint gj(X) as
Solution by Lagrange multipliersSolution by Lagrange multipliers
)()()()(),,,,,,,( 22112121 XXXX mmmn gggfxxxL
135
By treating L as a function of the n+m unknowns, x1, x2,…,xn,1, 2,…, m, the necessary conditions for the extremum of L, which also corresponds to the solution of the original problem are given by:
The above equations represent n+m equations in terms of the n+m unknowns, xi and j
Solution by Lagrange multipliersSolution by Lagrange multipliers
mjgL
nix
g
x
f
x
L
jj
i
jm
jj
ii
,,2,1,0)(
,,2,1,01
X
136
The solution:
The vector X* corresponds to the relative constrained minimum of f(X) (sufficient conditions are to be verified) while the vector * provides the sensitivity information.
Solution by Lagrange multipliersSolution by Lagrange multipliers
*
*2
*1
*
*2
*1
*and
mnx
x
x
X*
137
Sufficient Condition
A sufficient condition for f(X) to have a constrained relative minimum at X* is that the quadratic Q defined by
evaluated at X=X* must be positive definite for all values of dX for which the constraints are satisfied. If
is negative for all choices of the admissable variations dxi, X* will be a constrained maximum of f(X)
Solution by Lagrange multipliersSolution by Lagrange multipliers
ji
n
j ji
n
i
dxdxxx
LQ
1
2
1
ji
n
j ji
n
i
dxdxxx
LQ )(
1
2
1
*X*,
138
A necessary condition for the quadratic form Q to be positive (negative) definite for all admissable variations dX is that each root of the polynomial zi, defined by the following determinantal equation, be positive (negative):
• The determinantal equation, on expansion, leads to an (n-m)th-order polynomial in z. If some of the roots of this polynomial are positive while the others are negative, the point X* is not an extreme point.
Solution by Lagrange multipliersSolution by Lagrange multipliers
139
Find the dimensions of a cylindirical tin (with top and bottom) made up of sheet metal to maximize its volume such that the total surface area is equal to A0=24.
Solution
If x1 and x2 denote the radius of the base and length of the tin, respectively, the problem can be stated as:
Maximize f (x1,x2) = x12x2
subject to
Example 1Example 1
2422 02121 Axxx
140
Solution
Maximize f (x1,x2) = x12x2
subject to
The Lagrange function is:
and the necessary conditions for the maximum of f give:
Example 1Example 1
2422 02121 Axxx
)22(),,( 02121
2121 AxxxxxxL
022
02
0242
02121
121
2
21211
AxxxL
xxx
L
xxxxx
L
141
Solution
that is,
The above equations give the desired solution as:
Example 1Example 1
121
21
2
1
2x
xx
xx
21 2
1xx
2/1
0
2/1
0*2
2/1
0*1 24
* and,3
2,
6
AA
xA
x
142
SolutionThis gives the maximum value of f as
If A0 = 24, the optimum solution becomes
To see that this solution really corresponds to the maximum of f, we apply the sufficiency condition of equation
Example 1Example 12/13
0
54*
A
f
16* and,1*,4,2 *2
*1 fxx
143
SolutionIn this case:
Example 1Example 1
222 **1)(
21
2
12
xxx
LL
*X*,
442 **2)(2
1
2
11
xx
LL
*X*,
0)(2
2
2
22
*X*,x
LL
1624 *2
*1
)(1
111
xxx
gg
*X*,
42 *1
)(2
112
xx
gg
*X*,
144
SolutionThus, equation
becomes
Example 1Example 1
0
0416
402
1624
z
z
145
Solutionthat is,
This gives
Since the value of z is negative, the point (x1*,x2*) corresponds to the maximum of f.
Example 1Example 1
0192272 32 z
17
12z
146
Find the maximum of the function f (X) = 2x1+x2+10 subject to g (X)=x1
2+2x22 = 3 using the Lagrange multiplier method. Also find the
effect of changing the right-hand side of the constraint on the optimum value of f.
Solution The Lagrange function is given by:
The necessary conditions for the solution of the problem are:
Example 2Example 2
)23(102)( 22121 xxxxL X,
221
22
1
23
041
02
xxL
xx
L
x
L
147
Solution The solution of the equation is:
The application of the sufficiency condition yields:
Example 2Example 2
2*
13.0
97.2*2
*1
x
xX*
0
052.01
52.080
10
041
440
10
2
2
z
z
x
xz
z
0
01211
122221
111211
gg
gzLL
gLzL
148
Solution
Hence X* will be a maximum of f with f* = f (X*)=16.07
Example 2Example 2
2972.6
082704.0
z
zz
149
Minimize f (X)
subject to
gj (X) ≤ 0, j=1, 2,…,m The inequality constraints can be transformed to equality constraints by adding nonnegative slack variables, yj
2, as
gj (X) + yj2 = 0, j = 1,2,…,m
where the values of the slack variables are yet unknown.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
150
Minimize f(X) subject to
Gj(X,Y)= gj (X) + yj2=0, j=1, 2,…,m
where is the vector of slack variables
This problem can be solved by the method of Lagrange multipliers. For this, the Lagrange function L is constructed as:
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
my
y
y
2
1
Y
smultiplier Lagrange of vector theiswhere
)()()(
2
1
1
m
j
m
jjGfL
YX,XY,X,
151
The stationary points of the Lagrange function can be found by solving the following equations (necessary conditions):
(n+2m) equations (n+2m) unknowns
The solution gives the optimum solution vector X*, the Lagrange multiplier vector, *, and the slack variable vector, Y*.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
mjyy
L
mjygGL
nix
g
x
f
x
L
jjj
jjjj
i
jm
jj
ii
,,2,1,02)(
,,2,1,,0(()(
,,2,1,0)()()(
2
1
Y,X,
X)Y)X,Y,X,
XXY,X,
152
Equation
ensure that the constraints
are satisfied, while the equation
implies that either j=0 or yj=0
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
mjygGL
jjjj
,,2,1,,0(()( 2
X)Y)X,Y,X,
mjg j ,,2,1,0)( X
mjyy
Ljj
j
,,2,1,02)( Y,X,
153
• If j=0, it means that the jth constraint is inactive and hence can be ignored.
• On the other hand, if yj= 0, it means that the constraint is active (gj = 0) at the optimum point.
• Consider the division of the constraints into two subsets, J1 and J2, where J1 + J2 represent the total set of constraints.
• Let the set J1 indicate the indices of those constraints that are active at the optimum point and J2 include the indices of all the inactive constraints.
• Those constraints that are satisfied with an equality sign, gj= 0, at the optimum point are called the active constraints, while those that are satisfied with a strict inequality sign, gj< 0 are termed inactive constraints.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
154
• Thus for j J1, yj = 0 (constraints are active), for j J2, j=0 (constraints are inactive), and the equation
can be simplified as:
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
nix
g
x
f
i
jm
Jjj
i
,,2,1,01
nix
g
x
f
x
L
i
jm
jj
ii
,,2,1,0)()()(1
XXY,X,
155
Similarly, the equation
can be written as:
The equations (1) and (2) represent n+p+(m-p)=n+m equations in the n+m unknowns xi (i=1,2,…,n), j (j J1), and yj (j J2), where p denotes the number of active constraints.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
22
1
,0)(
,0)(
Jjyg
Jjg
jj
j
X
X
mjygGL
jjjj
,,2,1,,0(()( 2
X)Y)X,Y,X,
nix
g
x
f
i
jm
Jjj
i
,,2,1,01
(1)
(2)
156
Assuming that the first p constraints are active, the equation
can be expressed as:
These equations can be collectively written as
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
nix
g
x
g
x
g
x
f
i
pp
iii
,,2,1,,22
11
nix
g
x
f
i
jm
Jjj
i
,,2,1,01
ly.respective ,constraintjth theandfunction objective theof gradients theare and where
2211
j
pp
gf
gggf
157
Equation
indicates that the negative of the gradient of the objective function can be expressed as a linear combination of the gradients of the active constraints at the optimum point.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
/
/
/
and
/
/
/
2
1
2
1
nj
j
j
j
n xg
xg
xg
g
xf
xf
xf
f
pp gggf 2211
158
• A vector S is called a feasible direction from a point X if at least a small step can be taken along S that does not immediately leave the feasible region.
• Thus for problems with sufficiently smooth constraint surfaces, vector S satisfying the relation
can be called a feasible direction.
Multivariable optimization with Multivariable optimization with inequality constraints-Feasible regioninequality constraints-Feasible region
0 jT gS
159
• On the other hand, if the constraint is either linear or concave, any vector satisfying the relation
can be called a feasible region.
• The geometric interpretation of a feasible direction is that the vector S makes an obtuse angle with all the constraint normals.
Multivariable optimization with Multivariable optimization with inequality constraints-Feasible regioninequality constraints-Feasible region
0 jT gS
160
Multivariable optimization with Multivariable optimization with inequality constraints-Feasible regioninequality constraints-Feasible region
161
• Further we can show that in the case of a minimization problem, the j values (j J1), have to be positive. For simplicity of illustration, suppose that only two constraints (p=2) are active at the optimum point.
• Then the equation
reduces to
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
pp gggf 2211
2211 ggf
162
• Let S be a feasible direction at the optimum point. By premultiplying both sides of the equation
by ST, we obtain:
where the superscript T denotes the transpose. Since S is a feasible direction, it should satisfy the relations:
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
2T
21T
1T ggf SSS
2211 ggf
0
0
2
1
gS
gST
T
163
• Thus if, 1 > 0 and 2 > 0 the quantity STf is always positive.
• As f indicates the gradient direction, along which the value of the function increases at the maximum rate, STf represents the component of the increment of f along the direction S.
• If STf > 0, the function value increases, the function value increases as we move along the direction S.
• Hence if 1 and 2 are positive, we will not be able to find any direction in the feasible domain along which the function value can be decreased further.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
164
• Since the point at which the equation
is valid is assumed to be optimum, 1 and 2 have to be positive.
• This reasoning can be extended to cases where there are more than two constraints active. By proceeding in a similar manner, one can show that the j values have to be negative for a maximization problem.
Multivariable optimization with Multivariable optimization with inequality constraintsinequality constraints
0
0
2
1
gS
gST
T
165
Kuhn-Tucker Conditions
• The conditions to be satisfied at a constrained minimum point, X*, of the problem can be expressed as:
• These conditions are in general not sufficient to ensure a relative minimum.
• There is only a class of problems, called convex programming problems, for which the Kuhn-Tucker conditions are necessary and sufficient for a global minimum.
1,0
,,2,1,01
Jj
nix
g
x
f
j
i
j
Jjj
i
166
Kuhn-Tucker Conditions
• Those constraints that are satisfied with an equality sign, gj=0, at the optimum point are called the active constraints. If the set of active constraints is not known, the Kuhn-Tucker conditions can be stated as follows:
mj
mjg
mjg
nix
g
x
f
j
j
jj
i
jm
jj
i
,,2,1,0
,,2,1,0
,,2,1,0
,,2,1,01
167
Kuhn-Tucker Conditions
• Note that if the problem is one of maximization or if the constraints are of the type gj ≥ 0, the j have to be nonpositive in the equations below :
• On the other hand, if the problem is one of maximization with constraints in the form gj ≥ 0, the j have to be nonnegtaive in the above equations.
mj
mjg
mjg
nix
g
x
f
j
j
jj
i
jm
jj
i
,,2,1,0
,,2,1,0
,,2,1,0
,,2,1,01
168
Constraint Qualification
• When the optimization problem is stated as: Minimize f(X) subject to gj (X) ≤ 0, j=1, 2,…,m hk(X) = 0, k=1,2,….,p the Kuhn-Tucker conditions become
ly.respective ,0 and 0
sconstraint with theassociated smultiplier Lagrange thedenote and where
,,2,1,0
,,2,1,0
,,2,1,0
,,2,1,0
0
j
11
kj
k
j
k
j
jj
k
p
kkj
m
jj
hg
mj
pkh
mjg
mjg
hgf
169
Constraint Qualification• Although we found that the Kuhn-Tucker conditions represent the necessary
conditions of optimality, the following theorem gives the precise conditions of optimality:
• Theorem: Let X* be a feasible solution to the problem of
Minimize f(X)
subject to
gj (X) ≤ 0, j=1, 2,…,m
hk(X) = 0, k=1,2,….,p
If gj(X*), j J1 and hk(X*), k=1,2,…,p are linearly independent, there exist * and * such that (X*, *, *) satisfy the equations below:
mj
pkh
mjg
mjg
hgf
j
k
j
jj
k
p
kkj
m
jj
,,2,1,0
,,2,1,0
,,2,1,0
,,2,1,0
011
170
Example 1
Consider the problem:
Minimize f(x1,x2)=(x1-1)2 +x22
subject to
g1 (x1,x2) =x13-2x2≤ 0
g2 (x1,x2) =x13+2x2≤ 0
Determine whether the constraint qualification and the Kuhn-Tucker conditions are satisfied at the optimum point.
171
Example 1
Solution: The feasible region and the contours of the objective function are shown in the figure below. It can be seen that the optimum solution is at (0,0).
172
Example 1
Solution cont’d: Since g1 and g2 are both active at the optimum point (0,0), their gradients can be computed as:
It is clear that g1(X*) and g2(X*) are not linearly independent. Hence the constraint qualification is not satisfied at the optimum point.
2
0
2
3*)X(
2
0
2
3*)X(
)0,0(
21
2
)0,0(
21
1
xg
xg
173
Example 1 Solution cont’d: Noting that:
The Kuhn Tucker conditions can be written using the equations
as:
Since equation (E4) is not satisfied and equation (E5) can be satisfied for negative values of 1= 2 also, the Kuhn-Tucker conditions are not satisfied at the optimum point.
0
2
2
)1(2*)X(
)0,0(2
1
x
xf
1,0
,,2,1,01
Jj
nix
g
x
f
j
i
j
Jjj
i
(E4) 0
(E3) 0
(E2) 0)2()2(0
(E1) 0)0()0(2
2
1
21
21
174
Example 2
A manufacturing firm producing small refrigerators has entered into a contract to supply 50 refrigerators at the end of the first month, 50 at the end of the second month, and 50 at the end of the third. The cost of producing x refrigerators in any month is given by $(x2+1000). The firm can produce more refrigerators in any month and carry them to a subsequent month. However, it costs $20 per unit for any refrigerator carried over from one month to the next. Assuming that there is no initial inventory, determine the number of refrigerators to be produced in each month to minimize the total cost.
175
Example 2
Solution:
Let x1, x2, x3 represent the number of refrigerators produced in the first, second and third month respectively. The total cost to be minimized is given by:
total cost= production cost + holding cost
2123
22
21
21123
22
21321
2040
)100(20)50(20)1000()1000()1000(),,(
xxxxx
xxxxxxxxxf
176
Example 2
0150),,(
0100),,(
050),,(
3213213
213212
13211
xxxxxxg
xxxxxg
xxxxg
Solution cont’d:
The constraints can be stated as:
The first Kuhn Tucker condition is given by:
(E3)02
(E2)0202
(E1)0402
isthat
3,2,1,0
33
322
3211
33
22
11
x
x
x
ix
g
x
g
x
g
x
f
iiii
177
Example 2
Solution cont’d:
The second Kuhn Tucker condition is given by :
(E6)0)150(
(E5)0)100(
(E4)0)50(
isthat
3,2,1,0
3213
212
11
xxx
xx
x
jg jj
178
Example 2
Solution cont’d:
The third Kuhn Tucker condition is given by :
(E9)0)150(
(E8)0)100(
(E7)0)50(
is,that
3,2,10
321
21
1
xxx
xx
x
jg j
179
Example 2
Solution cont’d:
The fourth Kuhn Tucker condition is given by :
(E12)0
(E11)0
(E10)0
is,that
3,2,10
3
2
1
jj
180
Example 2
Solution cont’d:
The solution of Eqs.(E1) to (E12) can be found in several ways. We proceed to solve these equations by first noting that either 1=0 or x1=50 according to (E4).
Using this information, we investigate the following cases to identify the optimum solution of the problem:
• Case I: 1=0
• Case II: x1=50
181
Example 2
Solution cont’d:
• Case I: 1=0
Equations (E1) to (E3) give:
22
20x
(E13) 22
10
2
321
322
33
x
x
182
Example 2
Solution cont’d:
• Case I: 1=0
Substituting Equations (E13) into Eqs. (E5) and (E6) give:
(E14) 0) 2
3180(
0)130(
323
322
183
Example 2
Solution cont’d:
Case I: 1=0
The four possible solutions of Eqs. (E14) are:
1. 2=0, -180- 2-3/2 3=0. These equations along with Eqs. (E13) yield the solution:
2=0, 3=-120, x1=40, x2=50, x3=60
This solution satisfies Eqs.(E10) to (E12) but violates Eqs.(E7) and (E8) and hence can not be optimum.
184
Example 2
Solution cont’d:
Case I: 1=0
The second possible solution of Eqs. (E14) is:
2. 3=0, -130- 2-3=0. The solution of these equations leads to:
2=-130, 3=0, x1=45, x2=55, x3=0
This solution satisfies Eqs.(E10) to (E12) but violates Eqs.(E7) and (E9) and hence can not be optimum.
185
Example 2
Solution cont’d:
Case I: 1=0
The third possible solution of Eqs. (E14) is:
3. 2=0, 3=0. Equations (E13) give:
x1=-20, x2=-10, x3=0
This solution satisfies Eqs.(E10) to (E12) but violates the constraints Eqs.(E7) and (E9) and hence can not be optimum.
186
Example 2
Solution cont’d:
Case I: 1=0
The third possible solution of Eqs. (E14) is:
4. -130- 2-3=0, -180- 2-3/2 3=0. The solutions of these equations and Equations (E13) give:
2 =-30, 3 =-100, x1=45, x2=55, x3=50
This solution satisfies Eqs.(E10) to (E12) but violates the constraint Eq.(E7) and hence can not be optimum.
187
Example 2
23211
32322
33
2120240
(E15) 2220220
2
xx
xxx
x
Solution cont’d:
Case II: x1=50. In this case, Eqs. (E1) to (E3) give:
Substitution of Eqs.(E15) in Eqs
give:
(E16) 0)150)(2(
0)100)(2220(
3213
2132
xxxx
xxxx
(E6)0)150(
(E5)0)100(
3213
212
xxx
xx
188
Example 2
Solution cont’d:
Case II: x1=50. Once again, there are four possible solutions to Eq.(E16) as indicated below:
1. -20 - 2x2 + 2x3 = 0, x1 + x2 + x3 -150 = 0: The solution of these equations yields:
x1 = 50, x2 = 45, x3 = 55
This solution can be seen to violate Eq.(E8) which says:
(E8)0)100( 21 xx
189
Example 2
Solution cont’d:
Case II: x1=50. Once again, there are four possible solutions to Eq.(E16) as indicated below:
2. -20 - 2x2 + 2x3 = 0, -2x3 = 0: The solution of these equations yields:
x1 = 50, x2 = -10, x3 = 0
This solution can be seen to violate Eqs.(E8) and (E9) which say:
(E9)0)150(
(E8)0)100(
321
21
xxx
xx
190
Example 2
Solution cont’d:
Case II: x1=50. Once again, there are four possible solutions to Eq.(E16) as indicated below:
3. x1 + x2 -100 = 0, -2x3 = 0: The solution of these equations yields:
x1 = 50, x2 = 50, x3 = 0
This solution can be seen to violate Eq. (E9) which say:
(E9)0)150( 321 xxx
191
Example 2
Solution cont’d:
Case II: x1=50. Once again, there are four possible solutions to Eq.(E16) as indicated below:
4. x1 + x2 -100 = 0, x1 + x2 + x3 -150 = 0 : The solution of these equations yields:
x1 = 50, x2 = 50, x3 = 50
This solution can be seen to satisfy all the constraint Eqs.(E7-E9) which say:
(E9)0)150(
(E8)0)100(
(E7)0)50(
321
21
1
xxx
xx
x
192
Example 2
Solution cont’d:
Case II: x1=50.
The values of 1 , 2 , and 3 corresponding to this solution can be obtained from
as:
23211
32322
33
2120240
(E15) 2220220
2
xx
xxx
x
100,20,20 321
193
Example 2
Solution cont’d:
Case II: x1=50.
Since these values of i satisfy the requirements:
this solution can be identified as the optimum solution. Thus
100,20,20 321
(E12)0
(E11)0
(E10)0
3
2
1
50*,50*,50* 321 xxx
194
Convex functions• A function f(X) is said to be convex if for any pair of points
that is, if the segment joining the two points lies entirely above or on the graph of f(X).
• A convex function is always bending upward and hence it is apparent that the local minimum of a convex function is also a global minimum
)()1()()1(
1,0 , all and
and
1212
)2(
)2(2
)2(1
2
)1(
)1(2
)1(1
1
XXXX
XX
fff
x
x
x
x
x
x
nn
195
Convex functions
• A function f(x) is convex if for any two points x and y, we have
• A function f(X) is convex if the Hessian matrix
is positive semidefinite.
• Any local minimum of a convex function f(X) is a global minimum
x)(x)(yff(x)f(y) T
ji xxf /)(2 XH(X)
196
Concave function
• A function f(X) is called a concave function if for any two points X1 and X2, and for all 0 1,
that is, if the line segment joining the two points lies entirely below or on the graph of f(X).
• It can be seen that a concave function bends downward and hence the local maximum will also be its global maximum.
• It can be seen that the negative of a convex function is a concave function.
)()1()()1( 1212 XXXX fff
197
Concave function
• Convex and concave functions in one variable
198
Concave function
• Convex and concave functions in two variables
199
Example
Determine whether the following function is convex or concave.
f(x) = ex
Solution:
convex.strictly is f(x) Hence x.of valuesreal allfor 0)(2
2
xedx
fdxH
200
Example
Determine whether the following function is convex or concave.
f(x) = -8x2
Solution:
concave.strictly is f(x) Hence x.of valuesreal allfor 016)(2
2
dx
fdxH
201
Example
Determine whether the following function is convex or concave.
f(x1,x2) = 2x13-6x2
2
Solution:
Here
Hence H(X) will be negative semidefinite and f(X) is concave for x1 ≤ 0
120
012)( 1
22
2
21
2
21
2
21
2
x
x
f
xx
f
xx
f
x
f
H X
0for 0
0for 012
1
1121
2
x
xxx
f
0for 0
0for 0144
1
11
x
xxH(X)
202
Example
Determine whether the following function is convex or concave.
Solution:
15236534),,( 21312123
22
21321 xxxxxxxxxxxxf
1001
066
168
)(
23
2
32
2
31
2
32
2
22
2
21
2
31
2
21
2
21
2
x
f
xx
f
xx
f
xx
f
x
f
xx
f
xx
f
xx
f
x
f
H X
203
Example
Determine whether the following function is convex or concave.
Solution cont’d:
Here the principal minors are given by:
and hence the matrix H(X) is positive definite for all real values of x1, x2, x3. Therefore f(X) is strictly convex function.
15236534),,( 21312123
22
21321 xxxxxxxxxxxxf
0114
1001
066
168
01266
68
088
204
Convex programming problem
• When the optimization problem is stated as:
Minimize f (X) subject to gj (X) ≤ 0, j = 1, 2,…,m it is called a convex programming problem if the objective function f
(X), and the constraint functions, gj (X) are convex.• Supposing that f (X) and gj(X), j=1,2,…,m are convex functions, the
Lagrange function can be written as:
2
1
)()()( jj
m
Jj ygfL
XXλY,X,
205
Convex programming problem
• If j ≥ 0, then jgj(X) is convex, and since jyj=0 from
L(X,Y,) will be a convex function
• A necessary condition for f (X) to be a relative minimum at X* is that
L(X,Y,) have a stationary point at X*. However, if L(X,Y,) is a convex
function, its derivative vanishes only at one point, which must be an
absolute minimum of the function f (X). Thus the Kuhn-Tucker conditions
are both necessary and sufficient for an absolute minimum of f (X) at X*.
2
1
)()()( jj
m
Jj ygfL
XXλY,X,
mjyy
Ljj
j
,,2,1,02)( Y,X,
206
Convex programming problem
• If the given optimization problem is known to be a convex programming
problem, there will be no relative minima or saddle points, and hence the
exteme point found by applying the Kuhn-Tucke conditions is guaranteed
to be an absolute minimum of f (X). However, it is often very difficult to
ascertain whether the objective and constraint functions involved in a
practical engineering problem are convex.
207
Linear Programming I:Simplex method
• Linear programming is an optimization method applicable for the solution of problems in which the objective function and the constraints appear as linear functions of the decision variables.
• Simplex method is the most efficient and popular method for solving general linear programming problems.
• At least four Nobel prizes were awarded for contributions related to linear programming (e.g. In 1975, Kantorovich of the former Soviet Union and T.C. Koopmans of USA were awarded for application of LP to the economic problem of allocating resources).
208
Linear Programming I:Simplex method-Applications
• Petroleum refineries– choice of buying crude oil from several different sources with
differing compositions and at differing prices– manufacturing different products such as aviation fuel, diesel fuel,
and gasoline, in varying quantities– Constraints due to the restrictions on the quantity of the crude oil
from a particular source, the capacity of the refinery to produce a particular product
– A mix of the purchased crude oil and the manufactured products is sought that gives the maximum profit
• Optimal production plan in a manufacturing firm– Pay overtime rates to achieve higher production during periods of
higher demand• The routing of aircraft and ships can also be decided using
LP
209
Standard Form of a Linear Programming Problem
• Scalar form
ariablesdecision v theare and
constants,known are ),,2,1;,,2,1( and , where
0
0
0
sconstraint thesubject to
),,,( Minimize
2
1
2211
22222121
11212111
221121
j
ijjj
n
mnmnmm
nn
nn
nnn
x
njmiabc
x
x
x
bxaxaxa
bxaxaxa
bxaxaxa
xcxcxcxxxf
210
Standard Form of a Linear Programming Problem
• Matrix form
mnmmm
n
n
nmn
aaaa
aaaa
aaaa
c
c
c
b
b
b
x
x
x
f
321
2232221
1131211
2
1
2
1
2
1
,,
where
0
sconstraint thesubject to
)( Minimize
a
cbX
X
baX
XcTX
211
Characteristic of a Linear Programming Problem
• The objective function is of the minimization type
• All the constraints are of the equality type
• All the decision variables are nonnegative
• The number of the variables in the problem is n. This includes the slack and surplus variables.
• The number of constraints is m (m < n).
212
Characteristic of a Linear Programming Problem
• The number of basic variables is m (same as the number of constraints).
• The number of nonbasic variables is n-m.
• The column of the right hand side b is positive and greater than or equal to zero.
• The calculations are organized in a table.
• Only the values of the coefficients are necessary for the calculations. The table therefore contains only coefficient values, the matrix A in previous discussions. These are the coefficients in the constraint equations.
213
Characteristic of a Linear Programming Problem
• The objective function is the last row in the table. The constraint coefficients are written first.
• Row operations consist of adding (subtracting)a definite multiple of the pivot row from other rows of the table.
214
Transformation of LP Problems into Standard Form
• The maximization of a function f(x1,x2,…,xn ) is equivalent to the minimization of the negative of the same function. For example, the objective function
Consequently, the objective function can be stated in the minimization form in any linear programming problem.
nn
nn
xcxcxcff
xcxcxcf
2211
2211
maximize
toequivalent is
minimize
215
Transformation of LP Problems into Standard Form
• A variable may be unrestricted in sign in some problems. In such cases, an unrestricted variable (which can take a positive, negative or zero value) can be written as the difference of two nonnegative variables.
• Thus if xj is unrestricted in sign, it can be written as xj=xj'-xj", where
• It can be seen that xj will be negative, zero or positive, depending on whether xj" is greater than, equal to, or less than xj
’
.
0" and 0' jj xx
216
Transformation of LP Problems into Standard Form
If a constraint appears in the form of a “less than or equal to” type of inequality as:
it can be converted into the equality form by adding a
nonnegative slack variable xn+1 as follows:
knknkk bxaxaxa 2211
knnknkk bxxaxaxa 12211
217
Transformation of LP Problems into Standard Form
If a constraint appears in the form of a “greater than or equal to” type of inequality as:
it can be converted into the equality form by subtracting a variable as:
where xn+1 is a nonnegative variable known as a surplus variable.
knknkk bxaxaxa 2211
knnknkk bxxaxaxa 12211
218
Geometry of LP Problems
Example: A manufacturing firm produces two machine parts using lathes, milling machines, and grinding machines. The different machining times required for each part, the machining times available for different machines, and the profit on each machine part are given in the following table. Determine the number of parts I and II to be manufactured per week to maximize the profit.
Type of machine Machine time required (min)
Machine Part I
Machine time required (min)
Machine Part II
Maximum time available per week (min)
Lathes 10 5 2500
Milling machines 4 10 2000
Grinding machines 1 1.5 450
Profit per unit $50 $100
219
Geometry of LP Problems
Solution: Let the machine parts I and II manufactured per week be denoted by x and y, respectively. The constraints due to the maximum time limitations on the various machines are given by:
Since the variables x and y can not take negative values, we have
220
Geometry of LP Problems
Solution: The total profit is given by:
Thus the problem is to determine the nonnegative values of x and y that satisfy the constraints stated in Eqs.(E1) to (E3) and maximize the objective function given by (E5). The inequalities (E1) to (E4) can be plotted in the xy plane and the feasible region identified as shown in the figure. Our objective is to find at least one point out of the infinite points in the shaded region in figure which maximizes the profit function (E5).
221
Geometry of LP Problems
Solution: The contours of the objective function, f, are defined by the linear equation:
As k is varied, the objective function line is moved parallel to itself. The maximum value of f is the largest k whose objective f, unction line has at least one point in common with the feasible region). Such a point can be identified as point G in the figure. The optimum solution corresponds to a value of x*=187.5, y*=125, and a profit of $21,875.00.
222
Geometry of LP Problems
Solution: In some cases, the optimum solution may not be unique. For example, if the profit rates for the machine parts I and II are $40 and $100 instead of $50 and $100 respectively, the contours of the profit function will be parallel to side CG of the feasible region as shown in the figure. In this case, line P”Q” which coincides with the boundary line CG will correspond to the maximum feasible profit. Thus there is no unique optimal solution to the problem and any point between C and G online P”Q” can be taken as optimum solution with a profit value of $ 20000.
223
Geometry of LP Problems
Solution: There are three other possibilities. In some problems. In some prolems, the feasible region may not be a closed convex polygon. In such a case, it may happen that the profit level can be increased to an infinitely large value without leaving the feasible region as shown in the figure. In this case, the solution of the linear programming problem is said to be unbounded.
224
Geometry of LP problems
• On the other extreme, the constraint set may be empty in some problems. This could be due to the inconsistency of the constraints; or, sometimes, even though the constraints may be consistent, no point satisfying the constraints may also satisfy the nonnegativity restrictions.
• The last possible case is when the feasible region consists of a single point. This can occur only if the number of constraints is at least equal to the number of variables. A problem of this kind is of no interest to us since there is only one feasible point and there is nothing to be optimized.
225
Solution of LP Problems
A linear programming problem may have
• A unique and finite optimum solution
• An infinite number of optimal solutions
• An unbounded solution
• No solution
• A unique feasible point
226
Solution of LP Problems
A linear programming problem may have a unique and finite optimum solution:
The condition necessary for this to occur is:
• The objective function and the constraints have dissimilar slopes
• The feasible region is bounded/closed.
227
Solution of LP Problems
A linear programming problem may have infinite solution.
The condition necessary for this to occur is:
• The objective function must be parallel to one of the constraints.
228
Solution of LP Problems
A linear programming problem may have no solution.
The condition necessary for this to occur is:
• There is no point that is feasible. There is no solution to the problem.
229
Solution of LP Problems
A linear programming problem may have unbounded solution.
• In this case, the feasible region is not bounded.
• The presence of an unbounded solution also suggests that the formulation of the problem may be lacking. Additional meaningful constraints can be accomodated to define the solution.
230
Geometrical characteristics of the graphical solution of LP Problems
• The feasible region is a convex polygon*
• The optimum value occurs at an extreme point or vertex of the feasible region
*A convex polygon consists of a set of points having the property that the line segment joining any two points in the set is entirely in the convex set. In problems having more than two decision variables, the feasible region is called a convex polyhedron
231
Definitions
• Point in n-Dimensional Space A point X in an n-dimensional space is characterized by an ordered set of n values or coordinates (x1,x2,…,xn). The coordinates of X are also called the components of X.
• Line segment in n-Dimensions (L) If the coordinates of two points A and B are given by xj(1) and
xj(2) (j=1,2,…,n), the line segment (L) joining these points is the collection of points X() whose
coordinates are given by xj= xj(1)+(1- ) xj
(2), j=1,2,…,n, with 0 ≤ ≤ 1. Thus,
In one dimension, it is easy to see that the definition is in accordance with our experience
(2)(1) XXXX )1( L
10],[)( )1()2()2( xxxx
0
A
x(1)
x()
B
x(2)
10,)1()( )2()1( xxx
232
Definitions
• Hyperplane In n-dimensional space, the set of points whose coordinates satisfy a linear equation
is called a hyperplane. A hyperplane, H, is represented as H(a,b) = {X| aT X = b}
A hyperplane has n-1 dimensions in an n-dimensional space. For example, in three dimensional space it is a plane, and in two dimensional space it is a line.
bxaxa nn XaT11
233
Definitions
• Hyperplane continued: The set of points whose coordinates satisfy a linear inequality like
is called a closed half space, closed due to the inclusion of an equality sign in the inequality above. A hyperplane partitions the n-dimensional space (En) into two closed half-spaces, so that
bxaxa nn 11
bH
bH
Xa
XaT
T
X
X
234
Definitions
• Convex set: A convex set is a collection of points such that if X(1) and X(2) are any two points in the collection, the line segment joining them is also in the collection. A convex set, S, can be defined mathematically as follows:
A set containing only one point is always considered to be convex. Some examples of convex sets in two dimensions are shown shaded.
10 ,)1(
where
then ,, If
(2)(1)
)2((1)
XXX
X SS XX
235
Definitions
Nonconvex sets:
• The sets depicted by the shaded region on the figure below are not convex.
• The L-shaped region, for example, is not a convex set because it is possible to find two points a and b in the set such that not all points on the line joining them belong to the set.
236
Definitions
Convex Polyhedron and Convex Polytope:
• A convex polyhedron is a set of points common to one or more half spaces• A convex polyhedron that is bounded is called a convex polytope
237
Definitions
Vertex or Extreme Point:
• This is a point in the convex set that does not lie on a line segment joining two other points of the set. e.g. – Every point on the circumference of a circle– Each corner point of a polygon
Feasible Solution:
• In an LP problem, any solution that satisfies the constraints
is called a feasible solution
0X
baX
238
Definitions
Basic Solution:
• A basic solution is one in which n-m variables are set equal to zero. A basic solution can be obtained by setting n-m variables to zero and solving the constraint
simultaneously. In general, for n design variables and m constraints, the number of basic solutions is given by the combination
Basis:
• The collection of variables not set equal to zero to obtain the basic solution is called the basis.
baX
)!(!
!
mnm
n
m
n
239
DefinitionsBasic Feasible Solution:
• This is a basic solution that satisfies the nonnegativity conditions of
Nondegenerate Feasible Solution:
• This is a basic feasible solution that has got exactly m positive xi
Optimal Solution:
• A feasible solution that optimizes (minimizes) the objective function is called an optimal solution
Optimal Basic Solution:
• This is a basic feasible solution for which the objective function is optimal
0X
240
TheoremsTheorem 1:
• The intersection of any number of convex sets is also convex
Theorem 2:
• The feasible region of a linear programming problem is convex
241
TheoremsTheorem 3:
• Any local minimum solution is global for a linear programming problem
242
TheoremsTheorem 4:
• Every basic feasible solution is an extreme point of the convex set of feasible solutions
Theorem 5:
• Let S be a closed, bounded convex polyhedron with Xie, i=1 to p, as the set of its extreme points. Then any
vector X Є S can be written as:
Theorem 6:
• Let S be a closed convex polyhedron. Then the minimum of a linear function over S is attained at an extreme point of S.
1,01
1
p
iii
ei
p
ii
XX
243
Solution of a system of linear simultaneous equations
• Consider the following system of n equations in n unknowns:
• Assuming that this set of equations possesses a unique solution, a method of solving the system consists of reducing the equations to a form known as canonical form
)(E
)(E
)(E
)(E
2211
333232131
222222121
111212111
nnnnnnn
nn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
bxaxaxa
244
Solution of a system of linear simultaneous equations
• The equation:
will not be altered under the following elementary operations:– Any equation Er is replaced by the equation kEr , where k is a nonzero constant– Any equation Er is replaced by the equation Er+ kEs, where Es is any other equation of the system
)(E
)(E
)(E
)(E
2211
333232131
222222121
111212111
nnnnnnn
nn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
bxaxaxa
245
Solution of a system of linear simultaneous equations
Pivot operations• By making use of the elementary operations, the system can be reduced to a convenient equivalent form as
follows. Let us select some variable xi and try to eliminate it from all the equations except the jth one (for which aji is nonzero). This can be accomplished by dividing the jth equation by aji and subtracting aki times the result from each of the other equations, k=1,2,...,j-1,j+1,...,n. The resulting system of equations can be written as:
where the primes indicate that the a’ij and b’j are changed from the original system. This procedure of eliminating a particular variable from all but one equations is called a pivot operation.
0
0
1
0
0
0
11,11,2211
1,111,111,122,111,1
,11,11,2211
1,111,111,122,111,1
2211,211,2222121
1111,111,1212111
nnnniiniiinnn
jnnjiijiiijjj
jnnjniijiiijjj
jnniiijiiijjj
nniiiii
nniiiii
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
246
Solution of a system of linear simultaneous equations
• The system of equations
produced by the pivot operation have exactly the same solution as the original set
2211
33232131
22222121
11212111
nnnnnn
nn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
bxaxaxa
0
0
1
0
0
0
11,11,2211
1,111,111,122,111,1
,11,11,2211
1,111,111,122,111,1
2211,211,2222121
1111,111,1212111
nnnniiniiinnn
jnnjiijiiijjj
jnnjniijiiijjj
jnniiijiiijjj
nniiiii
nniiiii
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
bxaxaxxaxaxa
247
Solution of linear simultaneous equations - Canonical form
• Next time, if we perform a new pivot operation by eliminating xs, s≠i, in all the equations except the tth equation, t≠j, the zeros or the 1 in the ith column will not be disturbed. The pivotal operations can be repeated by using a different variable and equation each time until the system of the equations is reduced to the form:
This system of equations is said to be in canonical form and has been obtained after carrying out n pivot operations.
1000
0100
0010
0001
321
3321
2321
1321
nn
n
n
n
bxxxx
bxxxx
bxxxx
bxxxx
248
Solution of linear simultaneous equations - Canonical form
• From the canonical form, the solution vector can be directly obtained as:
• Thus the solution above is the desired solution of
nibx ii ,,2,1,
2211
33232131
22222121
11212111
nnnnnn
nn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
bxaxaxa
249
Pivotal reduction of a general system of equations
• Instead of a square system, let us consider a system of m equations in n variables with n ≥ m. This system of equations is assumed to be consistent so that it will have at least one solution
• The solution vector X that satisfy the above equation are not evident from the equations. However, it is possible to reduce this system to an equivalent canonical system from which at least one solution can readily be deduced.
2211
22222121
11212111
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
250
Pivotal reduction of a general system of equations
• If pivotal operations with respect to any set of m variables, say, x1, x2,….,xm, are carried, the resulting set of equations can be written as follows: Canonical system with pivotal variables x1, x2,….,xm
• One special solution that can always be deduced is:
• This solution is called a basic solution since the solution vector contains no more than m nonzero terms
iables var
tindependen variables
Constants or Nonpivotal Pivotal
100
010
001
11,21
2211,221
1111,121
mnmnmmmm
nnmmm
nnmmm
bxaxaxxx
bxaxaxxx
bxaxaxxx
nmmi
mibi
,,2,1,0
,,2,1,
251
Pivotal reduction of a general system of equations
• One special solution that can always be deduced is:
• The pivotal variables xi, i=1,2,…,m are called the basic variables and the other variables xi, i=m+1,m+2,…,n are called the nonbasic variables.
• This is not the only solution, but it is the one most readily deduced from
nmmi
mibi
,,2,1,0
,,2,1,
iables var
tindependen variables
Constants or Nonpivotal Pivotal
100
010
001
11,21
2211,221
1111,121
mnmnmmmm
nnmmm
nnmmm
bxaxaxxx
bxaxaxxx
bxaxaxxx
252
Pivotal reduction of a general system of equations cont’d
If all bi", i=1,2,…,m in the solution given by
are nonnegative, it satisfies Eqs
and hence it can be called a basic feasible solution.
nmmi
mibi
,,2,1,0
,,2,1,
0X
baX
253
Motivation of the simplex method
• Given a system in canonical form corresponding to a basic solution, we have seen how to move to a neighboring basic solution by a pivot operation
• Thus one way to find the optimal solution of the given LP problem is to generate all the basic solutions and pick the one that is feasible and corresponds to the optimal value of the objective function
• This can be done because the optimal solution, if one exists, always occurs at an extreme point or vertex of the feasible domain.
254
Motivation of the simplex method
• If there are m equality constraints in n variables with n m, a basic solution can be obtained by setting any of the n-m variables equal to zero.
• The number of basic solutions to be inspected is thus equal to the number of ways in which m variables can be selected from a set of n variables, that is,
• e.g. If n = 10 and m=5, we have 252 basic solutions
If n = 20 and m=10, we have 184756 basic solutions
!)!(
!
mmn
n
m
n
255
Motivation of the simplex method (contd)
• Usually we do not have to inspect all these basic solutions since many of them will be infeasible
• However, for large values of n and m , this is still a very large number to inspect one by one
• Hence, what we really need is a computational schemethat examines a sequence of basic feasible solutions, each of which corresponds to a lower value of f until a minimum is reached
• The simplex method of Dantzig is a powerful scheme for obtaining a basic feasible solution.
256
Motivation of the simplex method (contd)
• The simplex method of Dantzig is a powerful scheme for obtaining a basic feasible solution.
• If the solution is not optimal, the method provides for finding a neighboring basic feasible solution that has a lower or equal value of f .
• The process is repeated until, in a finite number of steps, an optimum is found.
257
Motivation of the simplex method (contd)
• The first step involved in the simplex method is to construct an auxiliary problem by introducing certain variables known as artificial variables into the standard form of the linear programming problem.
• The primary aim of adding artificial variables is to bring the resulting auxiliary problem into a canonical form from which its basic feasible solution can be obtained immediately.
• Starting from this canonical form, the optimal solution of the original linear programming problem is sought in two phases.
258
Motivation of the simplex method (contd)
• The first phase is intended to find a basic feasible solution to the original linear programming problem.– It consists of a sequence of pivot operations that produces a succession of
different canonical forms from which the optimal solutiobn of the auxiliary problem can be found.
• The second phase is intended to find the optimal solution of the original LP problem.– It consists of a second sequence of pivot operations that enables us to move
from one basic feasible solution to the next of the original LP problem.
• In this process, the optimal solution of the problem, if one exists, will be identified. The sequence of different canonical forms that is necessary in both phases of the simplex method is generated according to the simplex algorithm.
259
Simplex Algorithm
• The starting point of the simplex algorithm is always a set of equations, which includes the objective function along with the equality constraints of the problem in canonical form.
• Thus, the objective of the simplex algorithm is to find the vector X 0 that minimizes the function f (X) and satisfies the equations:
where aij, cj, bj, and fo are constants. Notice that (-f) is treated as a basic variable in the canonical form
onmnmmm
mnmnmmmm
nnmmm
nnmmm
fxcxcfxxx
bxaxaxxx
bxaxaxxx
bxaxaxxx
1121
11,21
2211,221
1111,121
000
100
010
001
260
Simplex Algorithm
• The basic solution which can be deduced is:
• In Phase I of the simplex method, the basic solution corresponding to the canonical form obtained after the introduction of the artificial variables will be feasible for the auxiliary problem
• Phase II of the simplex method starts with a basic feasible solution of the original linear programming problem. Hence the initial canonical form at the start of the simplex algorithm will always be a basic feasible solution.
nmmix
ff
mibx
i
n
ii
,,2,1,0
,,2,1,
0
261
Simplex Algorithm
• We know from Theorem
‘Let S be a closed convex polyhedron. Then the minimum of a linear function over S is attained at an extreme point of S.’
that the optimal solution of a linear programming problem lies at one of the basic feasible solutions.
• Since the simplex algorithm is intended to move from one basic feasible solution to the other through pivotal operations, before moving to the next basic feasible solution, we have to make sure that the present basic feasible solution is not the optimal solution.
• By merely glancing at the number cj", j=1,2,…,n, we can tell whether or not the present basic feasible solution is optimal.
262
Simplex Algorithm
Identifying an optimal point:• Theorem: A basic feasible solution is an optimal solution with a
minimum objective function value of f0" if all the cost coefficients cj" ,
j=m+1, m+2, …,n, in the equations
are nonnegative.
onmnmmm
mnmnmmmm
nnmmm
nnmmm
fxcxcfxxx
bxaxaxxx
bxaxaxxx
bxaxaxxx
1121
11,21
2211,221
1111,121
000
100
010
001
263
Improving a nonoptimal basic feasible solution
• From the last row of the Equation
we can write the objective function as:
for the solution given by
onmnmmm
mnmnmmmm
nnmmm
nnmmm
fxcxcfxxx
bxaxaxxx
bxaxaxxx
bxaxaxxx
1121
11,21
2211,221
1111,121
000
100
010
001
0
110
f
xcxcff j
n
mjji
m
ii
nmmix
ff
mibx
i
n
ii
,,2,1,0
,,2,1,
0
264
Improving a nonoptimal basic feasible solution
• If at least one cj" is negative, the value of f can be reduced by making the corresponding xj" > 0.
• In other words, the nonbasic variable xj, for which the cost coefficient is cjn
is negative, is to be made a basic variable in order to reduce the value of the objective function.
• At the same time, due to the pivotal operation, one of the current basic variables will become nonbasic and hence the values of the new basic variables are to be adjusted in order to bring the value of f less than f0" .
265
Improving a nonoptimal basic feasible solution
• If there are more than one cj" < 0, the index s of the nonbasic variable xs which is to be made basic is chosen such that
• Although this may not lead to the greatest possible decrease in f (since it may not be possible to increase xs very far), this is intuitively at least a good rule for choosing the variable to become basic.
• It is the one generally used in practice because it is simple and it usually leads to fewer iterations than just choosing any cj" < 0.
• If there is a tie in applying the above equation, (i.e., if more than one cj" has the same minimum value), we select one of them arbitrarily as cs"
0 minimum js cc
266
Improving a nonoptimal basic feasible solution
• Having decided on the variable xs to become basic, we increase it from zero holding all other nonbasic variables zero and observe the effect on the current basic variables. From Eq.
we can obtain
onmnmmm
mnmnmmmm
nnmmm
nnmmm
fxcxcfxxx
bxaxaxxx
bxaxaxxx
bxaxaxxx
1121
11,21
2211,221
1111,121
000
100
010
001
0,
0,
0,
0,
0
2222
1111
sss
msmsmm
ss
ss
cxcff
bxabx
bxabx
bxabx
267
Improving a nonoptimal basic feasible solution
• Since cs" < 0, the equation
suggests that the value of xs should be made as large as possible in order to reduce the value of f as much as possible.
• However, in the process of increasing the value of xs, some of the variables xi (i=1,2,...,m) in the Equation
may become negative.
0,
0,
0
sss
msmsmm
cxcff
bxabx
0,
0,
2222
1111
bxabx
bxabx
ss
ss
268
Improving a nonoptimal basic feasible solution
• If all the coefficients ais" 0, then xs can be made infinitely large without making any xi < 0, i=1,2,...,m
• In such a case, the minimum value of f is minus infinity and the linear programming problem is said to have an unbounded solution.
• If at least one ais" is positive, the maximum value that xs can take without making xi negative is bi" / ais"
• If there are more than one ais" > 0, the largest value xs* that xs can take is
given by the minimum of the ratios bi" / ais" for which ais" > 0. Thus,
is
i
rs
rs a
b
a
bx * minimum
0 ais"
269
Improving a nonoptimal basic feasible solution
• The choice of r in the case of a tie, assuming that all b"i > 0 , is arbitrary.
• If for any b"i for which a"is > 0 is zero in the Eq.
xs can not be increased by any amount. Such a solution is called a degenerate solution.
0,
0,
2222
1111
bxabx
bxabx
ss
ss
270
Improving a nonoptimal basic feasible solution
• In the case of a nondegenerate basic feasible solution, a new basic feasible solution can be constructed with a lower value of the objective function.
• By substituting the value of xs* given by
into
is
i
rs
rs a
b
a
bx * minimum
0 ais"
0,
0,
0,
0,
0
2222
1111
sss
msmsmm
ss
ss
cxcff
bxabx
bxabx
bxabx
271
Improving a nonoptimal basic feasible solution (contd)
• We obtain
which can readily be seen to be a feasible solution different from the previous one.
0*
0
*
and ,,2,1,0
0
and ,,2,1,0
*
fxcff
sjnmmjx
x
rimixabx
xx
ss
j
r
sisii
ss
272
Improving a nonoptimal basic feasible solution (contd)
• Since a“rs > 0 in Equation
a single pivot operation on the element a“rs in the system of equations
will lead to a new canonical form from which the basic feasible solution of the equation
can easily be deduced.
is
i
rs
rs a
b
a
bx * minimum
0 ais"
onmnmmm
mnmnmmmm
nnmmm
nnmmm
fxcxcfxxx
bxaxaxxx
bxaxaxxx
bxaxaxxx
1121
11,21
2211,221
1111,121
000
100
010
001
sjnmmjx
x
rimixabx
xx
j
r
sisii
ss
and ,,2,1,0
0
and ,,2,1,0
**
273
Improving a nonoptimal basic feasible solution (contd)
• Also, the equation
shows that the basic feasible solution corresponds to a lower objective function value compared to that of
• This basic feasible solution can be again be tested for optimality by seeing whether all c"i > 0 in the new canonical form.
• If the solution is not optimal, the entire procedure of moving to another basic feasible solution from the present one has to be repeated.
nmmix
ff
mibx
i
n
ii
,,2,1,0
,,2,1,
0
0*
0 fxcff ss
274
Improving a nonoptimal basic feasible solution (contd)
• In the simplex algorithm, this procedure is repeated in an iterative manner until the algorithm finds either
1. A class of feasible solutions for which f - or
2. An optimal basic feasible solution with all c"i 0, i=1,2,...,n
• Since there are only a finite number of ways to choose a set of m basic variables out of n variables, the iterative process of the simplex algorithm will terminate in a finite number of cycles.
275
Example
Maximize F = x1 + 2x2 + x3
subject to
3,2,1,0
64
652
22
321
321
321
ix
xxx
xxx
xxx
i
276
Example
Solution: We first change the sign of the objective function to convert it to a minimization problem and the signs of the inequalities where necessary so as to obtain nonnegative avlues for bi (to see whether an initial basic feasible solution can be obtained readily. The resulting problem can be stated as:
Minimize f = -x1 - 2x2 - x3
subject to
3,2,1,0
64
652
22
321
321
321
ix
xxx
xxx
xxx
i
277
Example
Solution cont’d: By introducing the slack variables x4 ≥ 0, x5 ≥ 0, and x6 ≥ 0, the system of equations can be stated in canonical form as:
where x4, x5 , x6 and -f can be treated as basic variables.
02
(E1) 6 4
652
2 2
321
6321
5321
4321
fxxx
xxxx
xxxx
xxxx
278
Example
Solution cont’d: The basic solution corresponding to Eqs. (E1) is given by:
which can be seen to be feasible.
Since the cost coefficients corresponding to nonbasic variables in Eqs. (E1) are negative (c1” = -1, c2” = -2, c3” = -1), the present solution given by Eqs.(E2) is not optimum.
0
(E2) ) variables(nonbasic 0
) variables(basic 6 ,6,2
321
654
f
xxx
xxx
279
Example
Solution cont’d: To improve the present basic feasible solution, we first decide the variable xs to be brought into the basis as:
Thus x2 enters the next basic set. To obtain the new canonical form, we select the pivot element ars” such that
2)0min( 2 ccc js
)"
"(min
0"is
i
ars
r
a
b
a
bis
280
Example
Solution cont’d: In the present case, s = 2 and a12” and a32” are ≥ 0. Since b1”/ a12” =2/1 and b3”/ a32” =6/1 , xr=x1. By pivoting on a12” , the new system of equations can be obtained as:
02303
(E3) 4202
8404
2 2
4321
64321
54321
4321
fxxxx
xxxxx
xxxxx
xxxx
281
Example
Solution cont’d: The basic feasible solution corresponding to this canonical system is given by:
Since c3” = -3, the present solution is not optimum. As cs”=min(ci”<0) = c3” , xs=x3 enters the next basis.
4
(E4) ) variables(nonbasic0
) variables(basic4,8,2
431
652
f
xxx
xxx
282
Example
Solution cont’d: To find the pivot element ars” , we find the ratios bi”/ ais” > 0. In the Equation below only a23 ” and a23 ” > 0
Hence:
02303
(E3) 4202
8404
2 2
4321
64321
54321
4321
fxxxx
xxxxx
xxxxx
xxxx
2
4 and
4
8
33
3
23
2
a
b
a
b
283
Example
Solution cont’d: Since both of these ratios are the same, we arbitrarily select a23 ” as the pivot element. Pivoting on a23 ” gives the following canonical system of equations:
The basic feasible solution corresponding to this canonical form is:
104
3
4
11006
(E5) 02
1
2
3000
24
1
4
1101
4 4
1
4
503
54321
654321
54321
54321
fxxxxx
xxxxxx
xxxxx
xxxxx
10
(E6) ) variables(basic 0
) variables(basic 02,,4
541
632
f
xxx
xxx
284
Example
Variables
Basic variables
x1 x2 x3 x4 x5 x6 -f bi” b1”/ ais” for
ais”>0
x4 2 1 -1 1 0 0 0 2 2
x5 2 -1 5 0 1 0 0 6
x6 4 1 1 0 0 1 0 6 6
-f -1 -2 -1 0 0 0 1 0
Most negative ci (x2 enters the next basis)
Smaller one (x4 drops from the next basis)
1
285
ExampleResult of pivoting
Variables
Basic variables
x1 x2 x3 x4 x5 x6 -f bi” b1”/ ais” for
ais”>0
x2 2 1 -1 1 0 0 0 2
x5 4 0 4 1 1 0 0 8 2
x6 2 0 2 -1 0 1 0 4 2
-f 3 0 -3 2 0 0 1 4
Most negative ci (x3 enters the next basis)
Select this arbitrarily. (x5 drops from the next basis)
4
286
ExampleResult of pivoting
Variables
Basic variables
x1 x2 x3 x4 x5 x6 -f bi”
x2 3 1 0 5/4 1/4 0 0 4
x3 1 0 1 1/4 1/4 0 0 2
x6 0 0 0 -3/2 -1/2 1 0 0
-f 6 0 0 11/4 3/4 0 1 10
All ci are ≥ 0 and hence the present solution is optimum
287
Example 2Unbounded solution
Minimize f = -3x1- 2x2
subject to
0,0
623
1
21
21
21
xx
xx
xx
288
Example 2Unbounded solution
Solution: Introducing the slack variables x3 ≥ 0 and x4 ≥ 0, the given system of equations can be written in canonical form as:
The basic feasible solution corresponding to this canonical form is given by:
023
(E1) 623
1
21
421
321
fxx
xxx
xxx
0
(E2) ) variables(nonbasic0
) variables(basic6,1
21
43
f
xx
xx
289
Example 2Unbounded solution
Solution: Introducing the slack variables x3 ≥ 0 and x4 ≥ 0, the given system of equations can be written in canonical form as:
The basic feasible solution corresponding to this canonical form is given by:
023
(E1) 623
1
21
421
321
fxx
xxx
xxx
0
(E2) ) variables(nonbasic0
) variables(basic6,1
21
43
f
xx
xx
290
Example 2Unbounded solution
Solution: Since the cost coefficients corresponding to the nonbasic variables are negative, the solution given by Eq.( E2) is not optimum. Hence the Simplex procedure is applied to the canonical system of Eqs.(E1) starting from the solution Eqs.(E2). The computations are done in tableau form as shown in the next slides.
291
Variables
Basic variables
x1 x2 x3 x4 -f bi” b1”/ ais” for
ais”>0
x3 1 -1 1 0 0 1 1
x4 3 -2 0 1 0 6 2
-f -3 -2 0 0 1 0
Example 2Unbounded solution
Smaller value (x3 leaves the basis)
Most negative ci (x1 enters the next basis)
1
292
Variables
Basic variables
x1 x2 x3 x4 -f bi” b1”/ ais” for
ais”>0
x1 1 -1 1 0 0 1
x4 0 1 -3 1 0 3 3
-f 0 -5 3 0 1 3
Example 2Results of pivoting
Smaller value (x4 leaves the basis)
Most negative ci (x2 enters the next basis)
1
293
Variables
Basic variables
x1 x2 x3 x4 -f bi” b1”/ ais” for
ais”>0
x1 1 0 -2 1 0 4
x2 0 1 -3 1 0 3
-f 0 0 -12 5 1 18
Example 2Results of pivoting
Both ais” are negative (i.e. No variable leaves the basis
Most negative ci (x3 enters the next basis)
294
Example 2Unbounded solution
• At this stage, we notice that x3 has the most negative cost coefficient and hence it should be brought into the next basis.
• However, since all the coefficients ai3” are negative, the value of f can be decreased indefinitely without violating any of the constraints if we bring x3 into the basis. Hence the problem has no bounded solution.
• In general, if all the coefficients of the entering variable xs (ais”) have negative or zero values at any iteration, we can conclude that the problem has an unbounded solution.
295
Example 3Infinite number of solutions
Minimize f = -40x1- 100x2
subject to
0,0
90032
2000104
2500510
21
21
21
21
xx
xx
xx
xx
296
Example 3Infinite number of solutions
Solution: By adding the slack variables x3 ≥ 0, x4 ≥ 0 and x5 ≥ 0, the equations can be written in canonical form as follows:
010040
90032
2000104
2500510
21
521
421
321
fxx
xxx
xxx
xxx
297
Example 3Infinite number of solutions
Smaller value (x4 leaves the basis)
Most negative ci (x2 enters the next basis)
10
Variables
Basic variables
x1 x2 x3 x4 x5 -f bi” b1”/ ais” for
ais”>0
x3 10 5 1 0 0 0 2500 500
x4 4 0 1 0 0 2000 200
x5 2 3 0 0 1 0 900 300
-f -40 -100 0 0 0 1 0
298
Example 3Infinite number of solutions
Variables
Basic variables
x1 x2 x3 x4 x5 -f bi”
x3 8 0 1 -1/2 0 0 1500
x2 4/10 1 0 -1/10 0 0 200
x5 8/10 0 0 -3/10 1 0 300
-f 0 0 0 10 0 1 20,000
299
Example 3Infinite number of solutions
• Since all ci” ≥ 0, the present solution is optimum. The optimum values are given by:
• It can be observed from the last row of the preceding tableau that the cost coefficient corresponding to the nonbasic variable x1 (c1”) is zero. This is an indication that an alternative solution exists.
• Here x1 can be brought into the basis and the resulting new solution will also be a basic feasible solution. For example, introducing x1 into the basis in place of x3 (i.e., by pivoting on a13”), we obtain the new canonical system of equations as shown in the following tableau:
000,20
) variables(nonbasic0
) variables(basic300,1500,200
min
41
532
f
xx
xxx
300
Example 3: Infinite number of solutions
Variables
Basic variables
x1 x2 x3 x4 x5 -f bi”
x1 1 0 1/8 -1/16 0 0 1500/8
x2 0 1 -1/20 1/8 0 0 125
x5 0 0 -1/10 -1/4 1 0 150
-f 0 0 0 10 0 1 20,000
301
Example 3: Infinite number of solutions
• The solution corresponding to this canonical form is given by:
• Thus, the value of f has not changed compared to the preceding value since x1 has a zero cost coefficient in the last row of the preceding tableau. Thus, the two basic (optimal) feasible solutions are:
000,20
) variables(nonbasic0
) variables(basic150,125,8/1500
min
43
521
f
xx
xxx
150
0
0
125
8/1500
and
300
0
1500
200
0
21 XX
302
Example 3: Infinite number of solutions
• An infinite number of nonbasic (optimal) feasible solutions can be obtained by taking any weighted average of the two solutions as:
• It can be verified that the solution X* will always give the same value of -20,000 for f for all 0 ≤ ≤ 1.
10
150150
0
1500
751258
1500)1(
150)1(300
0
1500
125)1(2008
1500)1(
*
*
*
*
*
)1(
5
4
3
2
1
x
x
x
x
x
X*
XXX* 21
303
Two Phases of the Simplex Method
• This problem is to find nonnegative values for the variables x1, x2,….,xn that satisfy the equations
and minimize the objective function given by
2211
22222121
11212111
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
fxcxcxc nn 2211
304
Two Phases of the Simplex Method
• The general problems encountered in solving this problem are:
– An initial feasible canonical form may not be readily available. This is the case when the linear programming problem does not have slack variables for some of the equations or when the slack variables have negative coefficients.
– The problem may have redundancies and/or inconsistencies, and may not be solvable in nonnegative numbers.
• The two-phase simplex method can be used to solve the problem.
305
Two Phases of the Simplex Method
• Phase I of the simplex method uses the simplex algorithm itself to find whether the linear programming problem has a feasible solution.If a feasible solution exists, it provides a basic feasible solution in canonical form ready to initiate Phase II of the method.
• Phas II in turn, uses the simplex algorithm to find whether the problem has a bounded optimum. If a bounded optimum exists, it finds the basic feasible solution which is optimal.
306
Two Phases of the Simplex Method
The simplex method is described in the following steps:1. Arrange the original system of the equations below so that
all the constant terms bi are positive or zero by changing, where necessary, the signs on both sides of the equations.
2211
22222121
11212111
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
307
Two Phases of the Simplex Method
2. Introduce to this system a set of artificial variables y1, y2, …. ym (which serve as basic variables in Phase I), where each yi ≥ 0 , so that it becomes
The objective function can be written as:
0
2211
222222121
111212111
i
mmnmnmm
nn
nn
b
byxaxaxa
byxaxaxa
byxaxaxa
0)(2211 fxcxcxc nn
308
Two Phases of the Simplex Method
3. Phase I of the method: Define a quantity w as the sum of the artificial variables:
and use the simplex algorithm to find xi ≥ 0 (i=1,2,…,n), yi ≥ 0 (i=1,2,…,m) which minimize w and satisfy the following equations:
myyyw 21
0)(2211 fxcxcxc nn
0
2211
222222121
111212111
i
mmnmnmm
nn
nn
b
byxaxaxa
byxaxaxa
byxaxaxa
309
Two Phases of the Simplex Method
3. Phase I of the method cont’d: Consequently consider the array:
0)(y
0)(
21
2211
2211
222222121
111212111
-wyy
-fxcxcxc
byxaxaxa
byxaxaxa
byxaxaxa
m
nn
mmnmnmm
nn
nn
310
Two Phases of the Simplex Method3. Phase I of the method cont’d: This array is not in canonical form;
however, it can be rewritten as a canonical system with basic variables y1, y2, …. ym, -f, and –w by subtracting the sum of the first m equations from the last to obtain the new system which provides the initial basic solution that is necessary for starting Phase I:
)...(
,...,2,1),(
where
-)(
0)(
210
21
02211
2211
2211
222222121
111212111
m
miiii
nn
nn
mmnmnmm
nn
nn
bbbw
niaaad
w-wxdxdxd
-fxcxcxc
byxaxaxa
byxaxaxa
byxaxaxa
311
Two Phases of the Simplex Method
4. w is called the infeasibility form and has the property that if as a result of Phase I, with a minimum of w>0, no feasible solution exists for the original linear programming problem stated in the equations:
the procedure is terminated.
2211
22222121
11212111
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
0)(2211 fxcxcxc nn
312
Two Phases of the Simplex Method
4. On the other hand, if the minimum of w=0, the resulting array will be in canonical form and hence initiate Phase II by eliminating the w equation as well as the columns corresponding to each of the artificial variables y1, y2,..,ym
from the array.
5. Apply the simplex algorithm to the adjusted canonical system at the end of Phase I to obtain a solution, if a finite exists which optimizes the value of f.
313
Example
• Minimize
5 to1,0
23
02433
:sconstraint thesubject to
232
54321
54321
54321
ix
xxxxx
xxxxx
xxxxxf
i
314
Example
Solution:
Step I: As the constants on the right hand side of the constraints, are already nonnegative, the application of step I is unnecessary.
Step II: Introducing the artificial variables y1 ≥ 0, and y2 ≥ 0, the equations can be written as follows:
0 232
(E1) 2 3
0 2433
54321
254321
154321
fxxxxx
yxxxxx
yxxxxx
315
Example
Solution:
Step III: By defining the infeasibility form w as
0
(E2) 0 232
2 3
0 2433
21
54321
254321
154321
wyy
fxxxxx
yxxxxx
yxxxxx
21 yyw
316
Example
Solution:
Step III cont’d: This array can be rewritten as a canonical system with basic variables as y1, y2,-f,-w by subtracting the sum of the first two equations of (E2) from the last equation of E2. Thus the last equation of (E2) becomes:
Since this canonical system (first three equations of (E2) and (E3), provides an initial basic feasible solution, Phase I of the simplex method can be started. The Phase I computations are shown in tableau form.
(E3) 2055-24 54321 --wxxxxx
317
Example
Solution: Step III cont’d: Since this canonical system (first three equations of (E2) and (E3) provides an initial basic feasible solution, Phase I of the simplex method can be started.
Basic variables
Admissable variables Artificial variables Value of
bi”/ ais”>0
x1 x2 x3 x4 x5 y1 y2 bi”
y1 3 -3 4 2 -1 1 0 0 0
y2 1 1 1 3 1 0 1 2 2/3
-f 2 3 2 -1 1 0 0 0
-w -4 2 -5 -5 0 0 0 -2
Most negative
Smaller value (y1
drops from next basis)