Nonlinear ProgrammingI

Li Xiaolei

Introductory concepts

A general nonlinear programming problem (NLP)

can be expressed as follows:

mnm

n

n

n

borxxxg

borxxxg

borxxxgts

xxxfzor

) ,,( ),,,(

(1) ) ,,( ),,,(

) ,,( ),,,( ..

),,,( min)( max

21

2212

1211

21

objective function

constraints

Introductory concepts

DEFINATION

the feasible region for NLP(1) is the set of points

(x1,x2,…,xn) that satisfy the m constraints in (1). A

point in the feasible region is a feasible point, and

a point that is not in the feasible region is an

infeasible point.

Introductory concepts

DEFINATION

For a maximization problem, any point x* in the

feasible region for which f(x*) ≥f(x) holds for all

points x in the feasible region is an optimal

solution to the NLP. For a minimization problem,

x is the optimal solution if f(x*) ≤ f(x) for all

feasible x.

Examples of NLPs EXAMPLE 1

It costs a company c dollars per unit to

manufacture a product. If the company

charges p dollars per unit for the product,

customers demand D(p) units. To maximize

profits, what price should the firm charge?

Solution The firm’s decision variable is p.

Since the firm’s profit is (p-c)D(p), the firm

wants to solve the following unconstrained

maximization problem: max (p-c)D(p).

Examples of NLPs

EXAMPLE 2

If K units of capital and L units of labor are used,

a company can produce KL units of a

manufactured good. Capital can be purchased at

$4/unit and labor can be purchased at $1/unit. A

total of $8 is available to purchase capital and

labor. How can the firm maximized the quantity of

the good that can be manufactured?

Solution

Let K=units of capital purchased and L=units of

labor purchased. Then K and L must satisfy

4K+L≤8,K≥0, and L ≥ 0. Thus, the firm wants to

solve the following constrained maximization

problem:

0,

84 ..

max

LK

LKts

KLz

Differences Between NLPs and LPs

The feasible region for any LP is a convex set. If

an LP has an optimal solution, there is an

extreme point of the feasible region that is

optimal.

Even if the feasible region for an NLP is a convex

set, the optimal solution need not be an extreme

point of the NLP’s feasible region.

Figure to EXAMPLE 2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

k0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

1

2

3

4

5

6

7

8

9

k

l

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

k

lk l=2 = 0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

7

8

9

k

lk l=4 = 0

KL=1

KL=2

KL=4

4K+L=8

D

Thus, the optimal solution to the example is

z=4,k=1,l=4(point D).

Point D is not an extreme point of the NLP’s

feasible region.

In fact, the optimal solution for an NLP may not

be on the boundary of the feasible region.

Local Extremum

DEFINITION 1

For any NLP (maximization), a feasible point

x=(x1,x2,…,xn) is a local maximum if for sufficiently

small ε, any feasible point x’=(x1’,x2’,…,xn’) having

|xi-xi’|<ε (i=1,2,…,n) satisfies f(x)≥f(x’). Analogously,

for a minimization problem, a point x is a local

minimum if f(x)≤f(x’) holds for all feasible x’ that

are close to x.

For a NLP, local maximum maybe not an optimal solution.

z

x 0

A

B

C

z=f(x)

Convex and Concave Function

Let f(x1,x2,…,xn) be a function that is defined for all points (x1,x2,…,xn) in a convex set S.

DEFINITION 2 A function f(x1,x2,…,xn) is a convex function on a

convex set S if for any x’∈S and x’’ ∈S f(cx’+(1-c)x’’) ≤cf(x’)+(1-c)f(x’’) holds for 0≤c≤1.

A=(x’,f(x’)) D=(x”,f(x”))C=(cx’+(1-c)x”,cf(x’)+(1-c)f(x”))B=(cx’+(1-c)x”,f(cx’+(1-c)x”))From figure: f(cx’+(1-c)x”) ≤cf(x’)+(1-c)f(x”)

x

y

A

DC

x’ x”cx’+(1-c)x”

y=f(x)

B

DEFINITION 3

A function f(x1,x2,…,xn) is a concave function on a

convex set S if for any x’∈S and x’’ ∈S

f(cx’+(1-c)x’’) ≥cf(x’)+(1-c)f(x’’)

holds for 0≤c≤1.

From definition 2 and 3, we see that f(x1,x2,…,xn) is

a convex function if and only if - f(x1,x2,…,xn) is a

concave function, and conversely.

A=(x’,f(x’)) D=(x”,f(x”))C=(cx’+(1-c)x”,cf(x’)+(1-c)f(x”))B=(cx’+(1-c)x”,f(cx’+(1-c)x”))From figure: f(cx’+(1-c)x”) ≥cf(x’)+(1-c)f(x”)

x

y

A

DC

x’ x”cx’+(1-c)x”

y=f(x)

B

x

y

AC

x’ x”cx’+(1-c)x”

y=f(x)

B

f(x) is not a convex or a concave function.

A linear function of the form f(x)=ax+b is both a

convex and a concave function.

f(cx’+(1-c)x”)=a[cx’+(1-c)x”]+b

=c(ax’+b)+(1-c)(ax”+b)

=cf(x’)+(1-c)f(x”)

Theorem 1 Consider NLP(1) and assume it is a maximization

problem. Suppose the feasible region S for NLP (1) is a convex set. If f(x) is concave on S, then any local maximum for NLP(1) is an optimal solution to this NLP.

Theorem 1’ Consider NLP(1) and assume it is a minimization

problem. Suppose the feasible region S for NLP(1) is a convex set. If f(x) is convex on S, then any local minimum for NLP(1) is an optimal solution to this NLP.

Theorems 1 and 1’ demonstrate that if we are

maximizing a concave function (or minimizing a

convex function )over a convex feasible region S,

then any local maximum (or local minimum) will

solve NLP(1), we will repeatedly apply theorem 1

and 1’.

We now explain how to determine if a function

f(x) of a single variable is convex or concave.

Theorem 2

Suppose f’’(x) exists for all x in a convex set S.

Then f(x) is a convex function on S if and only if

f’’(x) ≥0 for all x in S.

Since f(x) is convex if and only if - f(x) is

concave, Theorem 2’ must also be true.

Theorem 2’

Suppose f’’(x) exists for all x in a convex set S.

Then f(x) is a concave function on S if and only if

f’’(x) ≤0 for all x in S.

Example 1. show that f(x) = is a convex function on

S = R1.

2. show that f(x) = is a concave function on

S =(0,∞).

Solution

1. f’’(x)=2≥0, so f(x) is convex on S.

2. f’’(x)=- /4≤0,so f(x) is a concave function

on S.

2x

x21

23x

How can we determine whether a function f(x1,x2,

…,xn) of n variables is convex or concave on a set

S R∈ n?

We assume that f(x1,x2,…,xn) has continuous

second-order partial derivatives.

We require three definitions.

Definition

The Hessian of f(x1,x2,…,xn) is the n×n matrix

whose ijth entry is

We let H(x1,x2,…,xn) denote the value of the

Hessian at (x1,x2,…,xn) . For example ,if

f(x1,x2) = , then

ji xx

f

2

2221

31 2 xxxx

22

26 1xH(x1,x2)=

Definition

An ith principal minor of an n×n matrix is the determinant of any i×i matrix obtained by deleting n–i rows and the corresponding n–i columns of the matrix.

Thus ,for the matrix ,the first principal minors are -2 and -4, and the second principal minor is -2(-4)-(-1)(-1)=7. for any matrix, the first principal minors are just the diagonal entries of the matrix.

41

12

Definition

The kth leading principal minor of an n×n matrix is

the determinant of the k×k matrix obtained by

deleting the last n-k rows and columns of the

matrix.

We let Hk(x1,x2,…,xn) be the kth leading principal

minor of the H matrix evaluated at the point (x1,x2,

…,xn) . Thus, if f(x1,x2) =

,then H1(x1,x2)=6, and H2(x1,x2)= 6x1(2)-2(2)=12x1-4

2221

31 2 xxxx

Theorem 3

Suppose has continuous

second-order partial derivatives for each

point = S. Then

is a convex function on S iff for each S,

all principal minors of H are nonnegative.

Example : Show that =

is a convex function on S= .

),...,,( 21 nxxxf

x ),...,,( 21 nxxx ),...,,( 21 nxxxf

x

),( 21 xxf 2221

21 2 xxxx

2R

Solution

The first principal minors are the diagonal entries

(both equal 2≥0). The second principal minor is

2(2)-2(2)=0 ≥0. Since for all point, all principal

minors of H are nonnegative, theorem 3 shows

that f(x1,x2) is a convex function on R2.

22

22),( 21 xxH

Theorem 3’

Suppose has continuous

second-order partial derivatives for each

point S. Then

is a concave function on S iff for each

and =1,2,…n, all nonzero principal

minors have the same sign as .

),...,,( 21 nxxxf

),...,,( 21 nxxxx ),...,,( 21 nxxxf

sxk

k)1(

Solving NLPs with one variable In this section, we explain how to solve the

NLP :

max (or min )

s.t. (4)

(If b=∞,the feasible region for NLP (4) is ,

and if a=-∞, the feasible region for (4) is .)

)(xf

bax ,

ax bx

Solving NLPs with one variable To find the optimal solution to (4), we find

all local maxima (or minima). A point that

is a local maximum or minimum for (4) is

called a local extremum. Then the optimal

solution to (4) is the local maximum

(minimum) having the largest (smallest)

value of . Of cause if a=-∞ or b=∞, (4)

may have no optimal solution .

)(xf

NLPs with no solution

y=f(x)

y

x b Max f(x)

s.t. x∈ (-∞ ,b ]

y=f(x)

y

x a

Min f(x)

s.t. x∈[a,∞)

Extremum candidates

Case 1 Points where a<x<b, and f’(x)=0

Case 2 Points where f’(x) does not exist.

Case 3 Endpoints a and b of the interval [a,b].

Case 1. points where and =0 Theorem 4: If =0, and <0, then is a

local maximum. If =0 and >0,then

is a local minimum.

Theorem 5:If =0, and

1 If the first non-vanishing (nonzero) derivative at

is an odd-order derivative( and so

on), then is not a local maximum or a local

minimum.

bxa )(' xf

)('' 0xf 0x)(' 0xf)(' 0xf )('' 0xf 0x

)(' 0xf

0x)()( 0

)5(0

3 xfxf ，）（0x

2 If the first non-vanishing derivative

at is positive and is an even-order

derivative, then is a local minimum.

3 If the first non-vanishing derivative at

is negative and is an even-order

derivative , then is a local maximum.

0x0x

0x

0x

Figure 14

a b

x1 x0 x2

y

x

y=f(x)

f ’(x0)>0

f(x1)<f(x0)

f(x2)>f(x0)

x0 not a local extremum

a b

x1 x0 x2

y

x

y=f(x)

f ’(x0)<0

f(x1)>f(x0)

f(x2)<f(x0)

x0 not a local extremum

Figure 14

a b

x0

y

x

y=f(x)

f ’(x0)=0

for x<x0,f ’(x)>0

for x>x0 f ’(x)<0

x0 is a local maximum

a b x0

y

x

y=f(x)

f ’(x0)=0

for x<x0,f ’(x)<0

for x>x0 f ’(x)>0

x0 is a local minimum

Figure 14

a b

x0

y

x

y=f(x)

f ’(x0)=0 but x0 is neither

a local maximum nor a

local minimum

Case 2. Point Where does not exist

If does not have a derivative at ,

may be a local maximum, a local minimum,

or neither (see Figure 15). In this case, we

determine whether is a local maximum

or a local minimum by checking values of

at points < and > near

The four possible cases that can occur that

are summarized in Table 7.

)(' xf

)(xf 0x 0x

0x

)(xf 1x 0x 2x 0x0x

Figure 15

x1 x2 x0

y

x

y=f(x)

x0 is not a local extremum

x1 x2 x0

y

x

y=f(x)

x0 is not a local extremum

(a) (b)

Figure 15

x1 x2 x0

y

x

y=f(x)

x0 is a local maximum

x1 x2 x0

y

x

y=f(x)

x0 is a local minimum

(c) (d)

Table 7 If does not exist

Relationship x0 Figure

Not local extremum 15a

Not localextremum 15b

local maximum 15c

local minimum 15d

)()();()( 2010 xfxfxfxf

)()();()( 2010 xfxfxfxf

)()();()( 2010 xfxfxfxf

)()();()( 2010 xfxfxfxf

)(' xf

Case 3.Endpoints a and b of [a,b]

figure 16

a b

y

x

y=f(x)

f ’(a)>0

a is a local minimum

a b

y

x

y=f(x)

f ’(a)<0

a is a local maximum

Case 3.Endpoints a and b of [a,b]

figure 16

a b

y

x

y=f(x)

f ’(b)>0

b is a local maximum

a b

y

x

y=f(x)

f ’(b)<0

b is a local minimum

Example 15

It costs a monopolist $5/unit to produce a

product. If he produces x units of the product,

each can be sold for 10-x dollars(0≤x≤10). To

maximize profit, how much should the

monopolist produce?

Solution

Let P(x) be the monopolist’s profit if he produces x

unites. Then

P(x)=x(10-x)-5x=5x-x2 (0≤x≤10)

Thus,the monopolist wants to solve the following

NLP:

max P(x)

s.t. 0≤x≤10

Classify all extremum candidates: Case 1 P’(x)=5-2x, so P’(2.5)=0. since P’’(x)=-2,

x=2.5 is a local maximum yielding a profit of P(2.5)=6.25.

Case 2 P’(x) exists for all points in [0,10],so there are no Case 2 candidates.

Case 3 a=0 has P’(0)=5>0,so a=0 is a local minimum; b=10 has P’(10)=-15<0, so b=10 is a local minimum.

Thus, x=2.5 is the only local maximum.

Example 16

Let

Find

63 )4(3)(

30 )1(2)(2

2

xforxxf

xforxxf

60 ..

)(max

xts

xf

Solution

Case 1 For 0≤x<3,f’(x)=-2(x-1) and f’’(x)=-2. For 3<x ≤6, f’(x)=2(x-4) and f’’(x)=2. Thus, f’(1)=f’(4)=0. Since f’’(1)<0, x=1 is a local maximum. Since f’’(4)>0, x=4 is a local minimum.

Case 2 We see that f(x) has no derivative at x=3(for x slightly less than 3, f’(x)is near -4, and for x slightly bigger than 3, f’(x)is near -2). Since f(2.9)=-1.61, f(3)=-2,and f(3.1)=-2.19, x=3 is not a local extremum.

Case 3 Since f’(0)=2>0, x=0 is a local minimum. Since f’(6)=4>0, x=6 is a maximum.

Thus, on [0,6], f(x) has a local maximum for x=1 and x=6. Since f(1)=2 and f(6)=1, we find that the optimal solution to the NLP occurs for x=1.