Dual Gauge Programs, with Applications to

Dual Gauge Programs, with Applications to Quadratic Programming and The Minimum-Norm Problem by Robert M. Freund Working Paper #1726-85

Revised August 1986

Abstract A gauge function f(.) positively homogeneous and

is a nonnegative convex function that satisfies f(O)=O.

are specific instances of a gauge function.

Norms and pseudonorms This paper presents a

gauge duality theory for a gauge program, which minimizing the value of a gauge function f(.)

is

is the problem of

over a convex set.

The gauge dual program is also a gauge program, unlike the standard Lagrange dual.

We present sufficient conditions on f(-)

the existence of optimal

solutions to the gauge

dual, with no duality gap.

qualifications on the constraints.

The

gauge dual

programs,

program is

than the standard dual,

Keywords:

and

independent of any

to the minimum lp norm problem.

shown to provide a smaller duality gap

in a certain sense discussed

in the text.

Gauge function, norm, quadratic program, Lagrange dual,

Running Header: , , l 11,

its

The theory is applied to a class

duality.

_ll,

program and

These sufficient conditions are

relatively weak and are easy to verify, and are

of convex quadratic

that ensure

Dual Gauge Programs.

Introduction A gauge function

function f(.):RnRu{(+

=}

is a nonnegative

convex

that is positively homogeneous and satisfies

f(O)=O.

and pseudonorms are specific instances of a gauge function. program

is

defined as an optimization

P:

f(.)

is

the form

quadratic

to Mx > b

a gauge

programming fall

function.

Many problems

into this category,

programming,

in mathematical

including strictly convex

linear programming,

problem on a polyhedron (see Luenberger

and the minimum norm

[9]).

Duality for programs

similar to P have been studied by Eisenberg [2], recently been

generalized by Gwinner

instances where explicit Lagrange programs

A gauge

minimize f(x) subject

where

problem of

Norms

like

P can be stated,

[7].

whose work has most

Glassey [6]

has examined

duals of convex homogeneous

without

reference to primal

variables. This

paper presents a gauge duality theory for gauge programs

that contrasts, but particular,

is related to,

the gauge dual D of P

the Lagrange dual

of P.

In

is also a gauge program, unlike

its

Lagrange dual.

are

feasible values of the primal and dual objective functions,

z.v > D.

The gauge duality theory states that if

z and v then

1, with equality only if z and v are optimal values of P and

This inequality is analagous to the weak duality relationship z

> v for the Lagrange dual.

We present sufficient

that ensure that optimal

solutions

and that z-v=1

solutions.

for these

1

conditions on f(.)

to the dual gauge programs exist These sufficient conditions are

III

relatively weak and are

easy to verify.

They are independent of any

qualification on the constraints of the gauge program, unlike the Slater condition In the case applicable to than the dual

for Lagrange duality, for example. of quadratic

a class

programming, the theory developed

of quadratic

programs that

class of strictly convex quadratic

is equivalent

program different

is slightly broader

programs.

The gauge

(by a monotone transformation) to a quadratic from the

Lagrange dual.

Another application of the gauge duality theory is problem of minimizing the The gauge dual certain

is

is shown

lp norm of a vector over a polyhedron.

to provide a smaller duality gap

sense discussed in the text)

hence provides a better for feasible values of

to the

than the standard dual,

lower bound on the primal the dual,

(in a

than does

and

objective value,

the standard dual

program. A final application of the gauge duality theory programming, where the gauge dual

is different

is to linear

(but equivalent to)

the standard linear programming dual. In order 1 reviews

to lay the groundwork for the ensuing theory,

basic polarity properties

of gauge functions.

presents the gauge duality theory, which theorem, and necessary conditions valid.

Section

Section 2

includes a weak duality

for a strong duality theorem to

be

The duality theory of Section 2 is generalized to gauge

programs with nonlinear constraints

in Section 3.

Section 2 is applied to selected mathematical Section 4. programming,

This

section first

The theory of

programming problem in

discusses convex quadratic

followed by a duality analysis of the minimum

problem, for which strictly convex quadratic programming

2

lp norm

is a

special case.

The discussion shows that the duality gap for the

gauge dual is in a certain sense smaller than that of the standard dual.

Section 4 concludes with an analysis of linear programming in

the context of gauge duality theory.

3

III

1.

Preliminaries Let R denote

the set

numbers and

real

positively homogeneous

f(ax)

(i.e.,

= af(x)

and pseudonorms are gauge functions.

A

let R=Ru{(+}.

convex, nonnegative,

is

f(-)

f(.):Rn-R is called a gauge if

function

Norms

of

for a > 0),

and f(O)=O.

A gauge function need

not be symmetric and can take on the value +,

unlike a pseudonorm

or a norm. An example

of a gauge function

f(x)=f(xl,

x2,

( Note

that f(x)

that

f(x)

= 0

x3)

=

all x =

if 2x1-x2-x3 =

1 (Xl-x3),(xl-x2)12

(+

otherwise. the plane

is finite only on for

is

In

(a,a,a).

{xER 3 12xl-x 2 -x

this example,

3

f(.)

= 0),

and

is

symmetric. For any convex set C c by C

=

{yeRnlyTx

1 for all

containing the origin,

the polar of C,

Rn,

xC}.

is a gauge

function,

and

is a closed convex set

C

(see Rockafellar

[12],

p.

If

121).

if C is defined by

C = then f(-)

is defined

= C if and only if C is a closed

and C

convex set containing the origin f(.)

denoted C,

f(x)

{(xRnl

(1)

< 1)

can be represented by f(x)

where by convention, closed function and only if

(i.e.,

the origin, function,

({

Furthermore, f(.)

level sets of f(.)

Also,

the function

(2)

> OxEuC).

= +=O.

inf

all of the

C is a closed set.

that contains closed gauge

we denote

= inf

are closed)

for any closed convex set C f(.)

defined by

called the closed gauge function

corresponding to C.

4

is a

(2)

is a

if

For any gauge function =

fo(y)

f(.),

inf

Then f(.)

is a closed gauge.

(ylfo(y)

1),


olyTx

< vf(x) for all x).

If C =

1},

(xlf(x)

whereby f(.) is closed,

if f(.)

is

polar function f(.) by

closed,

since

then fOO(x) =

C

then C

(3) =

is closed.

f(x),

because C ° O

=

C.

The following summarizes the above statements: Remark

1 (see Rockafellar

f(.)ofO(.)

[12],

p.

129).

induces a one-to-one symmetric correspondence in the

of all closed gauges on R n .

class

The polarity operation

the origin are polar to functions are polar

Two closed convex sets containing

each other

if and only if

their gauge

X]

to each other.

We also have: Remark

2.

If f(.)

is a closed gauge function,

then f(.)

and f(.)

can be written as: f(x) =

sup yTx yECO

and

In the case when f(x) IX{{q,

=

fO(y)

that yTx < Holder

where 1/p +

=

fO(y)

xHIp,

1/q =

the lp norm,

1

p


defined by x E Rn and w

defined by f(x,w)

= g(x)

and

fo(y,z) where y E Rn and z

go(y)

if

z =

0

+*

if

z

0

=

Rm.

6

is

E

R,

a gauge

then

2.

Dual Gauge Programs Consider

the following nonlinear program:

P:

to

subject where f(.)

is

z = f(x)

minimize

Mx > b The Lagrange dual of P

a closed gauge function.

is

formulated as

{f(x) -

supremum {infimum >

XT(Mx-b))),

x

to

which can be simplified

-

supremum {bTX X > O

f*(MTX))},

or LD:

maximize

v = bTX

subject to

fo(MTX)
0 Dual

programs for classes of programs

developed by Eisenberg how to construct problems.

All

[2]

explicit duals

[7].

Glassey

[6] has

shown

for such

problem for which

is convex and positively homogeneous

need not be nonnegative,

to be finite-valued).

P have been

(with no primal variables)

three authors work with a primal

the objective function f(.) (f(.)

and Gwinner

that include

as in our

primal, but

is restricted

When applied to a gauge program P, however,

the dual programs that each author develops is the program LD above. function with the polar

If we exchange the objective constraint in LD, D:

we obtain a dual gauge program: minimize subject

v =

fo(MTX)

to bTX =

1

X > 0

7

gauge

III

Together, Note

the pair

that both the primal

problems,

and

to primal appears

(P) and dual

programs,

(D) are minimization

their objective functions

are polar gauge

functions.

variables X are restricted to be nonnegative and correspond

The dual

side

P and D constitute dual gauge

constraints.

in the dual

(RHS) of

in

The constraint matrix M in

the primal

the objective function, and the right-hand

the primal appears

in the equality constraint in the

dual. The definition of other

types

constraint

of of

the gauge dual program D can be extended to

linear constraints P is Mix

= ()

bi,

in the then in

ith variable Xi to be unrestricted

standard way. the dual D,

in sign

(less

If

the

ith

we require

than or equal

the to

zero.) Note

that the dual of the dual

write the dual

in

to:

y-MTX = 0 1

X > 0 = f(y).

s with constraints

To see this,

v = g(y,X)

bTX =

where g(y,X)

the primal.

the format:

minimize subject

is

Then

(i),

(i)

(x)

(ii)

(t)

(iii)

(s).

if we associate the variables x,

(ii),

(iii),

and

is: minimize

z = gO(x,bt-Mx+s)

subject to:

t = 1 s

>

0.

8

t,

and

the dual of this program

Thus

fOO(x) =

gO(x,w) =

However,

f(x) when w=O, and gO(x,w) =

+.

O.

if w

the last program can be written as z = f(x)

minimize

Mx-s =

subject to:

b

s >O is precisely the primal

which

P.

The vector of variables x if x

said to be feasible i.e.

satisfy the linear constraints,

(X)

otherwise x

(X)

= +),

(fO(MTX)

If x

infeasible.

is

then x

feasible and f(x) If P

is

(X)




Mx > b

is feasible but

(X)

for P (D)

is a strongly feasible program.

We have the following preliminary duality result for dual

gauge

programs. Theorem 1.

Let z

respectively.

and v*

respectively,

then zv >

v*

=

+~.

is essentially infeasible,

i.e.,

z

=

+o.

= O,

then D is essentially

(iii)

If v*

= O,

then P

(iv)

If x and X are feasible solutions values z and v, optimal

respectively,

solutions of P and D,

If x and X are

and

(iii)

f(x)

= zv,

and zv =

1, then x and X are

respectively.

by Remark 3.

then we have

This result shows

follow by contradiction from

9

1.

for P and D with objective

strongly feasible for P and D,

1 = XTb < XTMx < fo(MTX) (ii)

1, and hence z*v* > i.e.,

If z*

and

(D),

infeasible,

(ii)

(i),

and

strongly feasible for P and D, with objective

values z and v,

PROOF:

(P)

Then:

If x and X are

(i)

be optimal values of

(i).

(iv)

follows

III

from z.

(i)

is achieved by

bound on z*, and

by noting that 1/v is a lower

[XI (i)

Assertion value

of

corresponds to the standard duality result that the less

is

the max program

Assertions

equals min, then both are optimal.

f(x)

>

0 for any x,

the min

of

if max

(iii)

and

is

Because f(.)

= 0,

the program P

the same way that a standard

in

limit,

program would have a value of -, i.e.,

If z*

whereby z* > O.

has achieved its absolute lower

infeasible,

(ii)

in the standard theory.

correspond to unbounded cases a gauge,

to the value

corresponds to the result that

(iv)

program, and assertion

than or equal

and hence the dual program is

= +.

v*

In order to prove a strong duality theorem for the dual gauge programs their the

optimal values,

it is necessary

A convex set S c Rn satisfies

A gauge

both C and C C =

(xERnlf(x)

the projection property if all

S are closed convex sets,

transformation A:Rn4Rm, set.

to impose some qualifications on

We proceed as follows.

function f(.).

projections of

and that P and D both attain

that z*v*=l

P and D that asserts

(z

RmIz=Ax for

E

function f(.)

satisfies

i.e.,

if for any is

some xES)

a closed convex

the projection qualification

satisfy

the projection

property, where C is

< 1).

For notational

convenience,

assumed to be related by relations

linear

(1) and

(2)

f(-)

if

given by

and C will

be

of the previous section,

for the remainder of this paper. f(-)

satisfies the projection

f(.)

satisfies the projection

Note that by definition, qualification if and only if qualification.

If

C and C

are convex and compact,

the projection qualification, and hence projection qualification

for

1


1.

sets UC and X are convex and have no points in common.

There thus exists a hyperplane satisfies the closed.

if

{uIU=yTx for some yECO}.

by the projection qualification.

polyhedron such that UC n X =

PROOF:

= sup

= yTx for some yECO}

a closed interval,

qualification and

= yTx for some yCO.

finite, then f(x) We

However,

projection

that separates them.

qualification, all

Furthermore, X is polyhedron.

Because

projections

f(.)

of uC are

Consequently, according to

remark 5 in the appendix, there exists a hyperplane that separates from X and does that yTx must be

not meet UC.

< a for all xEUC,

Therefore, there exists

and yTx

a

>

[X]

We can now prove:

11 aa ·--

for all xEX.

positive, whereby by scaling we can presume

completing the proof.

-Ls(B

is a

__

UC

(y,a)E(Rn,R) such Since x=O EUC, a

it is equal

to

1,

11

Theorem 2. and

let

Assume that f(.)

z* and v*

satisfies

be the optimal

the projection qualification,

values of P and D,

respectively.

Then: (i)

If P and D are both

strongly feasible, then

z*v* =1;

and

the optimal values of P and D are achieved for some x*

(ii)

and

*.

P is

essentially infeasible and v*=O

v*=O;

is

(i.e.

z*=*)

if and only if

achieved for some feasible X*

if P is

infeasible. (iii)

D is essentially z*=O;

and

z*=O

infeasible

(i.e.

v*=o)

if and only if

is achieved for some feasible x*

if D is

infeasible. Let X = {xERnlMx > b}.

PROOF:

(i)

then 0

< z*

< +=

x

f(x)

=

E

X,

Lemma

and 0

/v*.

1, there

< +.

< v*

Assume

exists yRn

so does

convex set, since yTx y=MTX,

XTb

fO(MTX)

=

i.e.

(1/v*)C,

1.

x.

and


0 with

Thus P achieves

= f(x*)

Regarding

= l/v*.

its minimum at some (ii),

For the

the

"if"

"only if"

Then for any finite

, whereby there exists a vector yRn as described in

12

Now

and

a contradiction.

assume that P is essentially infeasible. UC

(-a,d],

shows that D achieves

establishing

the statement has been

or

is feasible for D,

its minimum at some feasible point x*, A parallel argument

(1/v*)C n X=¢,

therefore (yTxlxE(l/v*)C) is a closed

satisfy Mx > b,

Now X = X/XTb

for some

But since C satisfies the projection

a closed interval

(1/XTb) fO(y)

Then

that satisfies yTx < 1 for all

> 1 for all x that >

We must now show that

the contrary.

and yTx > 1 for all feasible property,

If P and D are strongly feasible,

part o f part, > 0,

Lemma

Let X and X be constucted as above.

1.

for D and f(MTX) < If

1/U,

P is infeasible,

exists X > 0 with MTX This completes The proof Theorem

= 0 and

(iii)

1/u for any U > 0,

i.e.

v*=O.

then by a theorem of the alternative, there

the proof of of


0,

qualification is

then UC

shown

stemming from an that

to guarantee the

(i) of Theorem 2 asserts

qualification is sufficient optimum z* where 0

< z*

The proof


x/2) and

_I

The

open separation.

The rather stronger condition that the dual

qualification is not sufficient

can

of Theorem

for a gauge program to attain

(ii)

to be sufficient in this case.

and

"open separation"

can be openly separated from X.

sufficient

Although part

z*=O.

need not be differentiable,

(see the appendix), which states

for some

2 indicates,

the feasible region

the feasible region of P.

essentially on arguments

and indirectly

of Theorem

a constraint qualification is not needed, because is polyhedral.

6]

_____

(2).

an Let Thus

its

III

we have

It

0

xO,

X2=0

+o=

x 1 70,

X2=0

+00

X2

=

f(xl,x2)

that both C and C°

the feasible

Let

< 0 {(Y1,Y2)ER2ly2

=

to derive C

is straightforward

verify

> O

X2

x2/(2x 2 )

satisfy the projection qualification.

region X of P be defined by X =

-2

= X1

which

= 1/(2x2),

/(2x2)

However,

z*=0.

Thus

spirit of

follows.

z*

and v*

(rel

there

to

(xl,x 2 )

for

condition for strong duality in P and D in is given below.

We

define X =

proceed as {xERnJAx > b),


so

so X'

is

raylike. We now must show that set that does not contain

if X is a nonempty closed, the origin,

then X"=X.

yTx > 1 for all yX', whereby xX", and Then there

exists an element z of X"

Because X is closed and (z)

strictly separates with the property then by and so

convex, from X,

Let xeX.

so X c X".

that

Suppose XX".

(y,a) E (Rn,

for all xX, and yTz < a.

yTz > 1=a, a contradiction.

This

Thus a < O.

Also yTz < a

Because OX and X is nonempty, X'

yTx

> 0 for every xX, (y + ey)Tx

Therefore

(y + ey)

(y + ey)Tz > yTz < a

is

for every e >

1 for all e > 0,

.

and so yTz

> 0,

Because

for every xX.

This

in

turn implies

that

> O,

which contradicts [X]

3 implies that the

dual of NGD

closed, and X is a closed,

contain the origin.

yX',

is nonempty.

< O.

Lemma f(-)

EX'

> 0,

for all xX.

yTx > 1 for every xX. > 1 for any

R),

Because X is raylike,

xX and all e > 1, and hence yTx > 0

Then,

If a

being the case,

yT(ex) > a for all

Let y be any element of X'.

Then

there exists a hyperplane that

rescaling we can assume that a=1.

< O.

raylike

is not contained in X.

and so there exists

that yTx > a

convex,

Of

course,

convex,

if X does

17

is precisely NGP whenever raylike

set that does not

contain the origin,

then

li

z*=O,

X'-$,

and v*=,

primal and dual,

where

z* and v*

are the optimal values of the

respectively.

The following result is analogous to Theorem 1: Theorem 1A (Weak Duality).

Let

NGP and NGD,

Then

(i)

respectively.

z* and v*

be optimal

values for

If x and y are strongly feasible for NGP and NGD, with objective values z and v, hence

respectively, the zv > 1, and

z*v* > 1.

(ii)

If z*=O,

then NGD is essentially infeasible,

(iii)

If v*=O,

then NGP

(iv)

If x and y are strongly feasible

is

essentially infeasible,

objective values

z and v,

then x and y are

optimal

i.e.

v*=+".

i.e.

z*=+o

for NGP and NGD with

respectively,

and zv =

1,

solutions of NGP and NGD,

respectively. PROOF:

If x and y are strongly feasible

1 < yTx < fo(y) f(x) = (ii),

(iii),

and

(iv)

zv,

by Remark 3.

follow from

To see how to obtain the from NGP and NGD, X =

let

a raylike

satisfying bTX

Y = (yERnly=MTX for some X > 0

(i),

The

(xeRnjMx > b}

> be for some Furthermore, > 1,

satisfying bTX

and

and

> 1}. X'

=

and define =

1),

and note that X'

linearly constrained gauge program P

is equivalent to the program P:

Define X =

> 1) = (ERnMx

raylike and contains Y.

result shows

[X]

set that contains X.

(yERnly=MTX for some X > 0

is

This

then

linearly constrained problems P and D

P be as given.

(xeRnlxEeX for some

Then X is

(i).

for NGP and NGD,

minimize f(x) subject xEX

18

because even though the feasible region of P contains the feasible region of P, every point xEX has at least as large an objective function value as a corresponding point xX.

The nonlinear gauge dual

of P is D:

minimize fO(MT\) subject to bTX > 1 >0

However, because f(.) is

homogeneous, we can restrict our .attention

to those X > 0 for which bTX = 1, obtaining the equivalent program: minimize fo(MTX)

D:

to bTX = 1

subject

>

0

linear gauge dual.

which is the

We have the following strong duality result for the nonlinear Let E = {xERnlf(x)
0,

then we can presume a = 1 by rescaling if necessary.

Then yX'

19

III

and v*

f(y) =


0, which Similarly,

0 for all XEX.

xz*C, and hence for all xC.

properly separate X from z*C.

that a > 0,

for NGD.

for every xX, there exists

turn means yTx =

that yTx =

l/z*.

is optimal

E z*C,

Thus yT(6x

X.

is attained

is attained in the primal,

If z* so

(z*/f(xO))xO

(1-6)x E

This in

< 0.

Thus H does not

xO

Because

6 > 1 such that implies yTx

C,

rel int

1 and y

=

Then ytxO > 0 because xO

= 0, because OEz*C.

because x

Also,

sup yTx < a/z* XEZ*C

(/z*)

since vz* > 1, we have z*v* =

However, If a

sup yTx = xEC

in

This contradiction the dual and

then the above proof

z*v* =

1.

is still valid,

long as z*C and X can be properly separated by a hyperplane H.

z*C and X cannot be properly separated, appendix,

there

exists XE

(rel

int

X)n(rel

int z*C and OEz*C, there exists

xE rel

whereby x

(z*/6)C, and so f(x)

then by Theorem 6 of

can be properly separated, and so v*

6 > 1 such that

is attained in

the

Because

int z*C).

a contradiction.

< z*,

If

6x + (1-6)OEz*C, Thus z*C and X

the dual,

and

z*v*=1.

A parallel primal.

argument

establishes that z*

is attained in

the

[X]

The nonlinear gauge duality theory parallels the gauge duality theory for linear constraints. there

It

is only natural

then to examine if

is a parallel duality construction that extends the Lagrange-

type dual LD to handle nonlinear constraints. and does not

contain the origin,

is closed,

NGP can be written as

f(x)

minimize subject

If X

to

yTx

> g(y)

20

for all y E cone X',

convex,

where cone X' g(y)

= max

=

{yeRnly = aw for some a > 0 and wX') and

(a > Oy = aw for

The above program

some wX').

suitable format so that Gwinner's dual

[7]

is in a

can be constructed, which

is: g(y)

maximize

GwD:

subject

yTx < f(x) for all x

to

y E cone X' Because be

if f(y)

yTx < f(x) for all x if and only


0 and

is feasible for NLD.

dual nonlinear gauge programs NGP and NGD make up a neat

theory in terms of

polar

explicit dual variables constraints

functions and antipolar sets.

However,

the

y do not directly correspond to primal

(though they do indirectly),

mention of constraints per se

and there is no formal

in either the primal

special

case of the general nonlinear theory is

theory,

instances

of which will be seen

or the dual.

One

of course the linear

in the next

section.

An open

issue regarding nonlinear gauge duality is under what circumstances can the

nonlinear gauge dual program be written

a finite number of

constraints whose

for every primal constraint?

21

variables

explicitly in terms of include a multiplier

li

4.

Applications In this section we explore a number of mathematical

models that correspond to a gauge program, programming,

problems

involving the

Many of the applications will minimize x subject where f(-) that of P.

However,

including convex quadratic

lp norm,

involve programs

and

linear programming.

of the form:

f(Nx+d)

to

is a gauge.

programming

Ax > b

Note that this format does not conform to

it

is equivalent to:

g(x,s)

minimize x,s subject to

where g(x,s)

= f(s).

Ax -Nx + Is

Furthermore,

> b = d

f(-)

qualification if and only if g(.,.)

satisfies

does.

the projection

The gauge dual

of this

program is:

minimize

g(ATX-NTU,

)

X,U

subject to

+ dT

bTX

X> But gO(y,t)

= f(t) when y=O,

U

=

and gO(y,t)

last program becomes minimize subject to

fO(u)

subject

ATX-NTU bTX+dTU

to

X>

22

1

0

0

= 0 = 1

= +

for y #

O;

thus this

A.

Convex Quadratic Programming The standard convex quadratic program QP:

minimize subject

(1/2)xTQx

is given by

+ qTx

Ax > b

to

where Q is a symmetric positive semi-definite matrix.

into the form Q = MTM for some square

the matrix Q can be factored matrix M. lies can

If M

is nonsingular

in the row space of

M,

(i.e.,

subject which

is positive definite),

then q = MTs for some vector sERn,

or

if q

and QP

(1/2)xTMTMx + sTMx to

Ax > b

is equivalent to the gauge GP:

minimize subject

where f(.)

= #'U 2 ,

program

lMx+sU 2 to

Ax > b

and so f(.)

satisfies

Note that QP and GP are equivalent identical

and their objective

transformation.

the projection qualification.

in that their

constraints are

functions differ by a strictly monotone

In examining the duality properties of QP and GP,

first study the case where Q is

the case when Q is positive Q is

Q

be written as minimize

will

Furthermore,

positive definite,

followed by

semi-definite.

Positive Definite When Q is positive definite the Lagrange dual bTX -

maximize X

of QP

is

(1/2)(XTA-qT)Q-1(ATX-q) (LQP),

subject

to

see Dorn [1].

X

> 0

The guage dual

11(MT)-IATX1{

minimize

of GP,

on the other hand,

is

2

(GD), subject

to

(bT+qTQ-lAT)X X > 0

= 1

23

we

II'

to

is equivalent

which

the quadratic program

minimize

XTAQ-1ATX

subject to

(bT+qTQ-1AT)X = 1 X >O

(D).

Note that both LQP and GD are strictly convex quadratic programs. constraints of LQP consist of the nonnegativity conditions

X > 0, whereas

the single equality constraint (bT+qTQ-1AT)X = 1.

GD also includes The

The

following theorem shows

the relationship between the gauge dual

programs GD or GD and the Lagrange dual LQP: Theorem 3. (i)

is a solution to

If

*

a)

t*

0

dual

LQP,

b) (ii)

is positive definite,

If Q

t*

GD and

t* = X*TAQ-1ATX*,

if and only if X = X*/t* solves

t

0 if and

only if

\*

then

the Lagrange

infeasible.

= 0 if and only if QP is

is a solution to LQP and t = bTX

If a)

then

+ qTQ-1ATX, then

= X/t solves

the guage

dual

GD, b)

t = 0 if and only if QP has a solution x to Ax > b satisfying Mx + s = O,

i.e.,

if and only if GD is

infeasible. PROOF:

The proof follows

Tucker conditions

The

for GD and LQP.

direct substitution. Q

from an examination of the Karush-Kuhntransformations follow from

[X]

is Positive Semi-Definite We now turn our attention to the broader case,

semi-definite and q lies some vector sRn.

in the row space

In this case,

of M,

whence q = MTs for

the Lagrange dual

24

when Q is positive

of QP

is

bTX -

maximize

(1/2)xTMTMx

',x

(LQP'),

subject

to

as in Dorn [1].

ATX -

MTMx = MTs X> O

The gauge dual of GP is

minimize X,u

null 2

subject to

-ATX + MTu = bTX

+ sTU

=

0

(GD'),

1

X > 0

which is equivalent to the quadratic program minimize

uTu

X, u subject

-ATX + MTU = 0

to

bTX

+

sTU

=

(GD')

1

X > 0

Analogous to Theorem 3, Theorem 4 demonstrates the relationship between the two different dual quadratic programs LQP' and GD'. Theorem 4.

If Q is positive semi-definite, Q = MTM and q = MTs for

some s ERn,

then

(i)

(X*,u*)

constitute an optimal solution to the gauge dual

GD' if and only if there exists x*, t* such that

(ii)

(a)

-ATX*+MTU*=O

(b)

bTX*+sTu*=l

(c)

X*

(d)

Ax* > bt*

(e)

U*=Mx*+st*

(f)

X*TAx*=X*Tbt*

> 0

t*=O if and only if QP is infeasible. only if

t$*O if and

=X*/t*, x=x*/t* constitute a solution to the

Lagrange dual LQP'.

25

III

(iii)

(X,x) constitute an optimal solution to the Lagrange dual LQP'

if and only if there exists z such that

(a)

X > 0

(b)

AT\-MTM=MTs

(c)

Az > b

(d)

MTMx=MTMz

(e)

XTAz=XTb

XTb+sTs+sTMz=o if and only if QP has a solution Ax > b,

(iv)

Mx+s=O, i.e.

if and only if GD' is infeasible.

XTb+sTs+sTMz~o if and only if (X*,U*) solves the gauge dual t*

D'

with multipliers x*,

t*

given by:

= 1/(XTb+sTs+sTMz)

X* = Xt* U*

=

x*

= zt*.

The conditions

(Mz+s)t* [X]

(i) and (iii)

of this theorem are simply the Karush-

Kuhn-Tucker conditions, and the transformations in

(ii)

and (iv)

follow from direct substitution. B.

Programs with the l If f(x)

=

xllp,

Norm

1 < p
d

(GPp)

26

f(y) =

if

The gauge dual

of GPp, derived using P and D, nBTXq

minimize

subject

is

to

CTX

=

0-

dTX

=

1

(GDq)

X > 0 Because f(x) results

of Theorem

Note

=

Hxlp satisfies the projection qualification, 2 are valid

solution s),

which is a derived equivalent of

i-I

instance of GPp, by setting

C

M

is a more general

When pl, p,

, d =

s

program than QP.

(l/p)

p

w

X,Z

The Lagrange

and p = 2.

the program GPp is equivalent to

minimize

subject

the

(when Q is positive semi-definite and q=MTs has a

can be cast as an

B

Thus GPp

for GPp and GDq.

that the program GP,

quadratic program QP

the

(LPp) to

dual of

Bw+Cz > this

d

program

is

maximize XTd-(1/q)UBTn q q subject

to

= 0

CT

(LDq)

X> 0 The gauge dual GDq and the Lagrange dual LDq bear a relationship that generalizes the case of quadratic 4.

This

programming in Theorems 3 and

relationship is demonstrated below.

notation xP,

where xERn,

In the theorem,

denotes the vector whose

(sign xj)(Ixjlp).

27

the

jth component

is

If pl and p,

Theorem 5.

If X*

(i)

t*

then

is an optimal

solution

to the gauge

dual GDq and

= \*T(B)(BTX*)q-1, then

(a)

t'*O if

and only if

=X*/t*

solves the Lagrange

dual LDq. (b)

t*=O if

and only if GPp

If X is an optimal

(ii)

(a)

XTd~O

is infeasible.

solution to

the Lagrange dual LDq, then

if and only if X*=X/XTd solves the gauge

dual GDq. (b)

XTd=o

if and only if GPp has an optimal

with value The proof of

this

0,

Cz > d has a solution.

i.e.

theorem follows

solution [X]

from examining the Karush-Kuhn-

Tucker conditions and substituting in the transformations as given. Although

the programs GPp and LPp are equivalent

objective functions differ by a monotone dual GDq of GPp will yield a better the

optimal

To see this, LPp, and

let

let

(w,z)

and

in GPp and LPp,

X = X/(dTX)

h =

dTX -

values of

X and X in

the gauge

lower bound on

the Lagrange dual

LDq for LPp. to GPp and

be the corresponding objective values of =

wllp,

' =

(l/p)

P Uwlp

Let X >

.

(w,z)

0 be any

LDq with a positive objective value, and hence

is a feasible solution to GDq. (l/q)JJBTx\

larger)

be any strongly feasible solution

namely

feasible solution to

transformation),

(i.e.,

solution to GPp than will

(their

p

p

Let g =

be the corresponding objective

BTXUq, and function

the programs GDq and LDq, respectively.

Then

1/g and h each represent a lower bound on the optimal primal objective values for GPp and LPp, and the values numbers

greater than or equal

the respective dual pairs

g and

/h are

to one that measure the duality gap in

of programs,

28

as a ratio

of the primal

objective

function value

to the dual objective

function value.

We

have: Lemma This

3.

Under

the above assumptions,

lemma states

ratio)