non linear programming problems

Non-linear Programming Problem Problem

Classical Optimization

• Classical optimization theory uses differential calculus to determine points of maxima and minima for unconstrained and constrained functions.

• This chapter develops necessary and sufficient conditions for determining maxima and minima of unconstrained and constrained functions.

Quadratic forms

Consider the function

xxaxxaxaxaxaXf

1,12112

2111)(

−−++

⋯⋯

The above function is called the quadratic or quadratic form in n-variables.

The above function can be written in the form XTAX, where X = [x1, x2, …, xn]T be a n-vector and A = (aij) be a n × n symmetric matrix.

nnnn xxaxxa 1,12112 −−++⋯

Quadratic formsThen the quadratic form

(or the matrix A (symmetric matrix)) is called• Positive semi definite if XTAX ≥ 0 for all X ≠ 0.

1( ) 2T

ii i ij i ji i j n

Q X X A X a x a x x≤ ≤ ≤

= = +∑ ∑∑

• Positive semi definite if XTAX ≥ 0 for all X ≠ 0.• Positive definite if XTAX > 0 for all X ≠ 0.• Negative semi definite if XT A X ≤ 0 for all X ≠ 0.• Negative definite if XTA X < 0 for all X ≠ 0.The above quantities of the quadratic form depend on the symmetric matrix A, therefore we have

1. Matrix minor testA necessary and sufficient condition for A (any square matrix ) to be :Positive definite (positive semi definite) is that all the nprincipal minors of A are > 0 ( ≥ 0).

Negative definite (negative semi definite) if kth principal

Negative definite (negative semi definite) if kth principalminor of A has the sign of (-1)k, k = 1, 2, …,n(kth principal minor of A is zero or has the sign of (-1)k,k=1,2,…,n )

Indefinite, if none of the above cases happen

2. Eigenvalue Test

Since matrix A is a real symmetric matrix in XTAX, it follows that its eigenvalues ( λi) are real. Then XTAX is

Positive definite (positive semi definite) if λi > 0(λi ≥ 0) i =1, 2, …,n.

Negative definite (negative semi definite) if λi < 0 ,(λi ≤ 0) i=1,2,…,n

Indefinite, if A has both positive and negative eigenvalues.

Examples : Decide the definiteness of the following quadratic functions:

21 53 xxx ++

32312123

21 4247107 xxxxxxxxx +−+−−−

323121321 4247107 xxxxxxxxx +−+−−−

21 xx −

Local maximum/Local minimum

Let f (X) = f (x1, x2,…,xn) be a real-valued function of the n variables

x1, x2, …, xn (we assume that f (X) is at least twice differentiable ).

A point X0 is said to be a local maximum of f (X) if there exists an

ε > 0 such thatε > 0 such that

A point X0 is said to be a local minimum of f (X) if there exists an

ε > 0 such that

0 0( ) ( ) for all jf X h f X h ε+ ≤ ≤"

0 0 0 0 0 01 2 0 1 2 0 1 1 2 2( , ,.., ) , ( , ,.., ) and ( , ,.., ) .T T T

n n n nh h h h X x x x X h x h x h x h= = + = + + +""

0 0( ) ( ) for all jf X h f X h ε+ ≥ ≤"

Absolute maximum/Absolute minimum

X0 is called an absolute maximum or global maximum of f (X) if

X0 is called an absolute minimum or global minimum of f (X) if

0( ) ( ) .f X f X X≤ ∀

X0 is called an absolute minimum or global minimum of f (X) if

0( ) ( ) .f X f X X≥ ∀

⋯++∇=−+ HhhhXfXfhXf T

21)()()( 000

Taylor Series expansion for functions of several variables

TheoremA necessary condition for X0 to be an optimum

point of f (X) is that

(that is all the first order partial derivatives arezero at X0.)

0)( 0 =∇ Xf

Definition: A point X0 for which is calleda stationary point of f (X) (potential candidate forlocal maximum or local minimum).

0)( 0 =∇ Xf

Let X0 be a stationary point of f (X). A sufficient conditionfor X0 to be a

local minimum of f (X) is that the Hessian matrixH(X0) is positive definite;

Theorem

H(X0) is positive definite;

local maximum of f (X) is that the Hessian matrixH(X0) is negative definite.

Hessian matrix

21 1 2 1

22 1 2 2

. .( ) .

f f fx x x x xf f f

x x x x xH x

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 2 1 2 2

( ) ...

x x x x xH x

f f fx x x x x

∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂

Example 1

Examine the following function for extreme points

f (x , x ,x ) = x +2 x +x x –x 2 – x 2 – x 2f (x1, x2,x3) = x1 +2 x3 +x2 x3 –x12 – x 22 – x3

Definition:

A function f (X)=f (x1,x2,…xn) of n variables is said to beconvex if for each pair of points X, Y on the graph, the linesegment joining these two points lies entirely above or onthe graph.

the graph.

i.e. f((1-λ) X + λY) ≤ (1-λ) f (X)+ λ f (Y)

for all 0 ≤ λ ≤ 1.

f is said to be strictly convex if for each pair of points X,Y on the graph,

f ((1-λ) X + λ Y) < (1-λ) f (X)+ λ f (Y)

for all λ such that 0 < λ < 1.

f is called concave (strictly concave) if – f is convex (strictly convex).

2 0d fdx

Convexity test for function of one variable

convex if

A function of one variable f(x) is

2 0d fdx

concave if

quantity convex Strictly convex

concave Strictly concave

f f - (f )2 ≥ 0 > 0 ≥ 0 > 0

Convexity test for functions of 2 variables

fxx fyy - (fxy)2 ≥ 0 > 0 ≥ 0 > 0

fxx ≥ 0 > 0 ≤ 0 < 0

fyy ≥ 0 > 0 ≤ 0 < 0

Results(1)The Sum of two convex functions is convex.

(2) Let f(X) = XTAX. Then f(X) is convex if XTAX is positive semi-definite.is positive semi-definite.

Example: Show that the following function is convex f (x1, x2) = -2 x1 x2 +x1

2 + 2x22

Constraint Optimization ProblemsProblems

Constrained optimization ProblemsKarush–Kuhn–Tucker (KKT) conditions:

Consider the problem maximize z = f(X) = f (x1, x2,…, xn)subject to g(X) ≤ 0 ⇒ [g (X) ≤ 0

subject to g(X) ≤ 0 ⇒ [g1(X) ≤ 0g2(X) ≤ 0

.gm(X) ≤ 0]

(the non-negativity restrictions, if any, are included in the above).

We define the Lagrangian function

L(X, S, λ) = f(X) – λ [g(X) + s2]

where s, s12, s2

2,..,sm2 ,are the non negative slack variables

added to g1(X) ≤ 0 ,…. gm(X) ≤ 0 to make them into equalities. Therefore

L(X, S, λ) = f(X) – [λ1{g1(X) + s12} + λ2{g2(X) +s2

+… +λm{gm(x) + sm2}]

KKT conditions …• KKT necessary conditions for optimality are given

( ) ( ) 0L f X g XXL

= ∇ − ∇ =∂∂

2 0, 1, 2, ... ,

( ( ) ) 0

L S i mSL g X S

∂= − = =

∂= − + =

0)()( =∇−∇ XgXf λ

The KKT necessary conditions for maximization problem are :

λ≥0

KKT conditions …

=∇−∇

i=1,2…m

These conditions apply to the minimization case as well,except that λ≤ 0.

KKT conditions …

gggf m 0..21 =∂

−−∂

−∂

λλλ

In scalar notation, this is given by

λi ≥ 0 i=1,2,….m

,...,2,1

∂−−

∂−

∂λλλ

i=1,2,….m

IMPORTANT: The KKT condition can be satisfied at a local minimum (or max.), a global minimum (or max.) as well as at a saddle point.

We can use the KKT condition to characterize all the stationary pointsof the problem, and then perform some additional testing to determine

of the problem, and then perform some additional testing to determinethe optimal solutions of the problem.

Sufficiency of the KKT conditions:

Sense of optimization

maximization

Required conditions

Objective function Solution space

Concave Convex set

maximization

minimization

ConcaveConvex

Convex set

Example: Use the KKT conditions to find the optimal solution for the following problem:

maximize f(x1, x2) = x1+ 2x2 – x23

subject to x1 + x2 ≤ 1

x1 ≥ 0

x2 ≥ 0

Solution: Here there are three constraints ,

g1(x1,x2) = x1+x2 - 1 ≤ 0

g2(x1,x2) = -x1 ≤ 0

g3(x1,x2) = -x2 ≤ 0The KKT necessary conditions for maximization problem are :

problem are :

λi ≥ 0

=∇−∇

i=1,2…m

Hence the KKT conditions become

∂−

λλλ

∂−

λλλ

λ1g1(x1,x2) = 0

λ g (x ,x ) = 0 (note: f is concave, g are

λ2g2(x1,x2) = 0 (note: f is concave, gi are

λ3g3(x1,x2) = 0 convex, maximization problem

g1(x1,x2) ≤ 0 these KKT conditions are

g2(x1,x2) ≤ 0 sufficient at the optimum point)

g3(x1,x2) ≤ 0 and λ1≥0, λ2≥0, λ3≥0

i.e. 1 – λ1 + λ2 = 0 (1)

2 – 3x22 – λ1 + λ3 = 0 (2)

λ1(x1 + x2 – 1) = 0 (3)

λ2 x1 = 0 (4)

λ3 x2 = 0 (5)

x1 + x2 – 1 ≤ 0 (6)

x1 ≥ 0 (7)

x2 ≥ 0 (8) λ1 ≥ 0 (9)

λ2 ≥ 0 (10) and λ3 ≥ 0 (11)

(1) gives λ1 = 1 + λ2 ≥ 1 >0 (using 10)

Hence (3) gives x1 + x2 = 1 (12)

Thus both x1, x2 cannot be zero, therefore let x1>0, then (4)gives λ2 = 0. therefore λ1 = 1

if now x2 = 0, then (2) gives 2 – 0 – 1 + λ3 = 0

=> λ < 0 not possible

=> λ3 < 0 not possible

Therefore x2 > 0, hence (5) gives λ3 = 0 and then (2) gives

x22 = 1/3 => x2 =1/√3.

And so x1 = 1- 1/√3. Therefore

Max f = 1 - 1/√3 + 2/√3 – 1/3√3 = 1 + 2/3√3.

Example: Use the KKT conditions to derive an optimal solution for the following problem:

minimize f (x1, x2) = x12+ x2

subject to x12 + x2

2 ≤ 9

x + x ≤ 1

x1 + x2 ≤ 1

Solution: Here there are two constraints,

g1(x1,x2) = x12+x2

2 - 9 ≤ 0

g2(x1,x2) = x1 + x2 -1 ≤ 0

Thus the KKT conditions are:

1 20, 0λ λ≤ ≤ as it is a minimization problem

2x1 - 2λ1x1 - λ2 = 0

1 - 2λ1x2 - λ2 = 02 2( 9) 0x xλ + − =

( 9) 0( 1) 0x xx x

+ − =

2 21 2

x xx x+ ≤

Now (from 2) gives 01 =λ 12 =λ Not possible.

Hence 01 ≠λ and so 922

21 =+ xx

Assume . So (1st equation of ) (2) gives02 =λ

0)1(2 11 =−λx Since we get x1= 001 ≤λ

From (5), we get 32 ±=x

2nd equation of (2) says (with ) x2 = -3,01 <λ 02 =λ

Thus the optimal solution is: The optimal value is :

1 2 1 210, 3, , 06

x x λ λ= =− =− = 3z = −

Maximize f(x) = 20x1 + 10 x2

Subject to x12 + x2

2 ≤ 1

x1 + 2x2 ≤ 2

Use the KKT conditions to find an optimal solution of the following problem:

x1≥ 0, x2 ≥ 0

max f occurs at x1 = 2/√5, x2 = 1/√5

Quadratic Programming

Quadratic Programming …

A quadratic programming problem is a non-linear programming problem of the form

Maximize subject to , 0

Tz CX X DXAX b X= +

≤ ≥

[ ] [ ] [ ], ,..., , , ,..., , , ,...,T TX x x x b b b b C c c c= = =where [ ] [ ] [ ]1 2 1 2 1 2, ,..., , , ,..., , , ,...,T Tn n nX x x x b b b b C c c c= = =

aaaaaa

dddddd

! Assume that the n × n matrix D is symmetric and negative-definite.

! This means that the objective function is strictly

! This means that the objective function is strictly concave.

! Since the constraints are linear, the feasible region is a convex set.

In scalar notation, the quadratic programming problem reads:

1 1 1Maximize 2

j j jj j ij i jj j i j n

z c x d x d x x= = ≤ < ≤

= + +∑ ∑ ∑ ∑

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

subject to . . .. . .

. . ., , . . . , 0

m m mn n m

a x a x a x ba x a x a x b

a x a x a x bx x x

+ + + ≤

Wolfe’s Method to solve a Quadratic Programming Problem:

! The solution to this problem is based on the KKTconditions. Since the objective function is strictlyconcave and the solution space is convex, theKKT conditions are also sufficient for optimum.

KKT conditions are also sufficient for optimum.

! Since there are m + n constraints, we have m + nLagrange multipliers; the first m of them aredenoted by λ1, λ2 , …, λm and the last n of themare denoted by µ1, µ2 , …, µn.

Wolfe’s Method…

The KKT (necessary) conditions are:

0,...,,,,...,,.1 2121 ≥nm µµµλλλ

njaxdc j

iiijj ,...,2,1,02.2

11==+−+ ∑∑

13. 0, 1, 2,. . .,

0 , 1, 2,. . .,

i ij j ij

a x b i m

− = =

14. , 1, 2,. . ., and 0, 1,2,. . .,

ij j i jj

a x b i m x j n=

≤ = ≥ =∑

Wolfe’s Method…

Denoting the (non-negative) slack variable for the ith constraint

by si, the 3rd condition can be written in an equivalent form

jjij bxa ≤∑

(Referred to as " Restricted Basis " conditions).

3. 0, 1, 2, . . .,0 , 1, 2,. . .,

s i mx j n

Wolfe’s Method…

Also condition(s) (2) can be rewritten as:

caxd jj

,...,2,1

=−+− ∑∑==

and condition(s) (4) can be rewritten as:

14. , 1, 2, . . .,

0 , 1, 2,. . .,

ij j i ij

a x s b i m

x j n=

nj ,...,2,1=

Wolfe’s Method…

Thus we have to find a solution of the following m + nlinear equations in the 2n + m unknowns jijx µλ ,,

njcaxd jj

iiij ,...,2,1,2

11==−+− ∑∑

1, 1 , 2 , . . . ,

i j j i ij

a x s b i m=

+ = =∑0 , 1 , 2 , . . . , ,

0 , 1 , 2 , . . . ,i i

s i mx j n

0 , 0 , 1 , 2 , . . . , ,0 , 0 , 1 , 2 , . . . ,

s i mx j n

≥ ≥ =

Wolfe’s Method…

Since we are interested only in a " feasible solution“of the above system of linear equations, we usePhase-I method to find such a feasible solution. Bythe sufficiency of the KKT conditions, it will be

the sufficiency of the KKT conditions, it will beautomatically the optimum solution of the givenquadratic programming problem.

Example-1:

2 21 2 1 2Maximize 8 4z x x x x= + − −

1 2subject to 2x x+ ≤

subject to 2, 0.

x xx x+ ≤

Denoting the Lagrange multipliers by λ1, µ1, and µ2, the KKT conditions are:

0,,.1 211 ≥µµλ

1 1 12. 8 2 04 2 0

− − + =

4 2 0i.e. 2 8

− − + =

+ − =

1 2 1 1 1 1 1 2 23. 2, 0x x s s x xλ µ µ+ + = = = =

All variables ≥ 0.

Introducing artificial variables R1, R2, we thus have toMinimizesubject to the constraints

21 RRr +=

=+−+

All variables ≥ 0 (We solve by " Modified Simplex " Algorithm).

=+−+

SxxRx µλ

0221111 === xxS µµλ

Basic r x1 x2 λ1 µ1 µ2 R1 R2 s1 Sol

r 1 0 0 0 0 0 -1 -1 0 0 2 2 2 -1 -1 0 0 0 12

R1 0 2 0 1 -1 0 1 0 0 8 R2 0 0 2 1 0 -1 0 1 0 4 S1 0 1 1 0 0 0 0 0 1 2 r 1 0 0 2 -1 -1 0 0 -2 8

r 1 0 0 2 -1 -1 0 0 -2 8 R1 0 0 -2 1 -1 0 1 0 -2 4 R2 0 0 2 1 0 -1 0 1 0 4 x1 0 1 1 0 0 0 0 0 1 2 r 1 0 4 0 1 -1 -2 0 2 0 λ 1 0 0 -2 1 -1 0 1 0 -2 4 R2 0 0 4 0 1 -1 -1 1 2 0 x1 0 1 1 0 0 0 0 0 1 2

Thus we have got the feasible solution

x1 = 2, x2 = 0, λ1 = 4, µ1 = 0, µ2 = 0

and the optimal value is: z = 12

Example-2

Maximize 21 1 2 38 2z x x x x= − + +

subject to

3 2 12, , 0

x x xx x x+ + ≤

Denoting the Lagrange multipliers by λ1,µ1, µ2, and µ3, the KKT conditions are:

1 1 12. 8 2 02 3 0

x λ µ

− − + =

− + =

1 1 2 31. , , , 0λ µ µ µ ≥

Example-2: Solution

2 3 01 2 3 0. . 2 8

3 22 1

− + =

+ − =

All variables ≥ 0.

1 2 3 1

1 1 2 2 3 3

3. 3 2 12,0

x x x SSx x xλ

µ µ µ

+ + + =

Solving this by " Modified Simplex Algorithm ", the optimal solution is:

and the optimal z =

1 2 311 25, , 03 9

x x x= = =

Basic r x1 x2 x3 λ1 µ1 µ2 µ3 R1 R2 R3 S1 Sol

r 1 0 0 0 0 0 0 0 -1 -1 -1 0 02 0 0 6 -1 -1 -1 0 0 0 0 11

R1 0 2 0 0 1 -1 0 0 1 0 0 0 8 R2 0 0 0 0 3 0 -1 0 0 1 0 0 2

R3 0 0 0 0 2 0 0 -1 0 0 1 0 1

S1 0 1 3 2 0 0 0 0 0 0 0 1 12

Since λ1 S1 = 0 and S1 is in the basis, λ1 cannot enter.

So we allow x1 to enter the basis and of course by minimum ratio test R1 leaves the basis.

r 1 0 0 0 5 0 -1 -1 -1 0 0 0 3

x1 0 1 0 0 1/2 -1/2 0 0 1/2 0 0 0 4 R2 0 0 0 0 3 0 -1 0 0 1 0 0 2

R3 0 0 0 0 2 0 0 -1 0 0 1 0 1

S1 0 0 3 2 -1/2 1/2 0 0 -1/2 0 0 1 8

Since λ1 S1 = 0 and S1 is in the basis, λ1 cannot enter.

So we allow x2 to enter the basis and of course by minimum ratio test S1 leaves the basis.

r 1 0 0 0 5 0 -1 -1 -1 0 0 0 3

x1 0 1 0 0 1/2 -1/2 0 0 1/2 0 0 0 4 R2 0 0 0 0 3 0 -1 0 0 1 0 0 2

R3 0 0 0 0 2 0 0 -1 0 0 1 0 1

x2 0 0 1 2/3 -1/6 1/6 0 0 -1/6 0 0 1/3 8/3

As S1 is not in the basis, now λ1 enters the basis .

And by minimum ratio test R3 leaves the basis.

r 1 0 0 0 0 0 -1 3/2 -1 0 -5/2 0 1/2

x1 0 1 0 0 0 -1/2 0 1/4 1/2 0 -1/4 0 15/4 R2 0 0 0 0 0 0 -1 3/2 0 1 -3/2 0 1/2λ1 0 0 0 0 1 0 0 -1/2 0 0 1/2 0 1/2

x2 0 0 1 2/3 0 1/6 0 -1/12 -1/6 0 1/12 1/3 11/4

Now µ3 enters the basis .

And by minimum ratio test R2 leaves the basis.

r 1 0 0 0 0 0 0 0 -1 -1 -1 0 0

x1 0 1 0 0 0 -1/2 1/6 0 1/2 -1/6 0 0 11/3

µ3 0 0 0 0 0 0 -2/3 1 0 2/3 -1 0 1/3λ1 0 0 0 0 1 0 -1/3 0 0 1/3 0 0 2/3

x2 0 0 1 2/3 0 1/6 -1/18 0 -1/6 1/18 0 1/3 25/9

This is the end of Phase I.

Thus the optimal solution is:

1 2 311 25, , 03 9

x x x= = =

Thus the optimal value z is: 193

2121 34236 xxxxxxz −−−+=Maximize

subject to 1 2 1x x+ ≤

Example-3

subject to 1 2

12 3 4

x xx x

Denoting the Lagrange multipliers by λ1, λ2, µ1, and µ2, the KKT conditions are:

1 2 1 21. , , , 0λ λ µ µ ≥

λ λ µ− − − − + =

1 2 1 2 1

1 2 1 2 2

1 2 1 2 1

1 2 1 2 2

2. 6 4 4 2 03 4 6 3 0i.e. 4 4 2 6

4 6 3 3

x xx xx xx x

λ λ µ

− − − − + =

+ + + − =

1 1 2 2

3. 1,2 3 4,

x x Sx x SS Sx x

and all variables ≥ 0.

1 1 2 2

Solving this by " Modified Simplex Algorithm ", the optimal solution is:

x1 = 1, x2 = 0 and the optimal z = 4.

Basic r x1 x2 λ1 λ2 µ1 µ2 R1 R2 S1 S2 Sol

r 1 0 0 0 0 0 0 -1 -1 0 0 0 8 10 2 5 -1 -1 0 0 0 0 9

R1 0 4 4 1 2 -1 0 1 0 0 0 6 R2 0 4 6 1 3 0 -1 0 1 0 0 3

S1 0 1 1 0 0 0 0 0 0 1 0 1

r 1 4/3 0 1/3 0 -1 2/3 0 -5/3 0 0 4

x2 0 2/3 1 1/6 1/2 0 -1/6 0 1/6 0 0 1/2 S1 0 1/3 0 -1/6 –1/2 0 1/6 0 -1/6 1 0 1/2

S2 0 0 0 -1/2 -3/2 0 1/2 0 -1/2 0 1 5/2

S2 0 2 3 0 0 0 0 0 0 0 1 4

R1 0 4/3 0 1/3 0 -1 2/3 1 -2/3 0 0 4

r 1 0 -2 0 -1 -1 1 0 -2 0 0 3

R1 0 4 4 1 2 -1 0 1 0 0 0 6

x1 0 1 3/2 1/4 3/4 0 -1/4 0 1/4 0 0 3/4

S1 0 0 -1/2 -1/4–3/4 0 1/4 0 -1/4 1 0 1/4

r 1 0 0 1 2 -1 0 0 -1 -4 0 2

x1 0 1 1 0 0 0 0 0 0 1 0 1µ2 0 0 -2 -1 –3 0 1 0 -1 4 0 1S2 0 0 1 0 0 0 0 0 0 -2 1 2

S2 0 0 0 -1/2 -3/2 0 1/2 0 -1/2 0 1 5/2

R1 0 0 0 1 2 -1 0 1 0 -4 0 2

r 1 0 0 0 0 0 0 -1 -1 0 0 0

λ1 0 0 0 1 2 -1 0 1 0 -4 0 2 x1 0 1 1 0 0 0 0 0 0 1 0 1

µ2 0 0 -2 0 –1 -1 1 1 -1 0 0 3

S2 0 0 1 0 0 0 0 0 0 -2 1 2

Thus the optimum solution is:

x1 = 1, x2 = 0, λ1 = 2, λ2 = 0, µ1 = 0, µ2 = 3

And the optimal value is: z = 4

Solve the following quadratic programming problem

Maximize 2221

2121 518205020 xxxxxxz −+−+=

subject to 6≤+ xx

subject to

0,1846

xxxxxx

Using Excel solver, the opt solution is: x1=2, x2=4Max z = 224

Remark:

If the problem is a minimization problem, say, Minimize z,

we convert it into a maximization problem, Maximize -z.

non linear programming problems

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