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OutlineRoadmap for the NPP segment:

1 A cookbook example

2 Preliminaries: role of convexity

3 Existence of a solution

4 Necessary conditions for a solution: inequality constraints

5 The constraint qualification

6 The Lagrangian approach

7 Interpretation of the Lagrange Multipliers

8 Constraints that are binding vs satisfied with equality

9 Necessary conditions for a solution: eq and ineq constraints

10 Sufficiency conditions

1 Bordered Hessians2 Pseudo-concavity

1 The relationship between quasi- and pseudo-concavity

11 The basic sufficiency theorem

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Cookbook NPP Example I

maxx

u(x) = 0.5n

i=1

(xi xi)2 s.t.

n

i=1

pixi y, x0 (1)

L(x,) = u(x) +(yn

i=1

pixi) = n

i=1

(xi xi)2

2+(y

n

i=1

pixi)

Differentiate Lw.r.t. each element ofxand also.

L(x,)

xi = u(x)

xi pi= (xi xi)pi 0 (2a)

L(x,)

= (ypixi) 0 (2b)

We also have thecomplementary slackness conditions:

L(x,)

xixi = 0 (2c)

L(x,)

= 0 (2d)

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Cookbook NPP Example II

1 Diet constraint is binding iffy=ni=1 pixi>y. Assume its binding.

2 From (2a), we get x1x1

p1

= x2x2p2

, i.e., that

p1(x2 x2) = p2(x1x1). (3)

3 Letting=p1x2 p2x1, rearrange(3)to getp1x2p2x1= .

4

Combine this expr with(2b)

p1x2 p2x1 = (4a)

p2x2 + p1x1 = y (4b)

Solve forx1 : eliminatex2 by subtractingp2 (4a)fromp1 (4b)

x1 (p22+ p

21) = p1yp2 = x

1 =

p1yp2

p22+ p21

(4c)

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Cookbook NPP Example III

Solve forx2 : eliminatex1 by addingp1 (4a) top2 (4b)

x2 (p22+ p

21) = p2y+ p1 = x

2 =

p2y+ p1

p22+ p21

(4d)

5 Check for a contradiction. No contradiction iffxi 0.

6 Finally, solve for. From(2b), we have(x1 x1)p1= 0, so that

= x1x

1

p1

= (p22+ p

21)x1 (p1yp2)

p1(p2

2+ p2

1)

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Existence

Preliminaries:

Convexity, quasi-ness and constrained optimizationThe canonical form of the nonlinear programming problem (NPP)

maximizef(x)subject tog(x)b (5)

Conditions for existence of a solution the NPP:Theorem: Iff :A R is continuous andstrictly quasi-concave, andAiscompact, nonempty and convex, then fattains a unique maximum onA.

Cant directly apply this theorem to (5)

constraint set isnt necessarily convex, non-empty, compact

Theorem: Iff :RnR is continuous and strictly quasi-concave andgj : RnR isquasi-convexfor eachj, thenif fattains a local maximumonA={x Rn :j,gj(x)bj}, this max is the unique global maximumoff onA.

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Convexity and quasi-ness properties

x1x1

x2

x2

FOC doesnt imply max FOC is local max but not soln

Constraint set is not convex Upper contour set of ob jective not convex

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Separating Hyperplanes

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Want to be able to separate by a hyperplane

upper contour set of objective function

constraint set

Role of quasi-ness properties of functions

A function isquasi-concave(quasi-convex) if all of its upper (lower)

contour sets are convex

A function isstrictly quasi-concave (convex)if

all of its upper (lower) contour sets are strictly convex sets (no flat edges)all of its level sets have empty interior

To make everything work nicely1 Require your objective function to be strictly quasi-concave

2 Define your feasible set as the intersection of lower contour sets ofquasi-convexfunctions,

e.g.,gj(x)bj, forj=1,...m.intersection of convex sets is convex

your feasible set is convex

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KKT necessary conditions for a soln: inequality constraints

The KKT conditions in Mantra format:

(Except for a bizarre exception) a necessary condition forx to solve aconstrained maximization problem is that the gradient vector of the

objective function atx belongs to the nonnegative cone defined by the

gradient vectors of the constraints that are satisfied with equality at x.

The KKT conditions in math format:

Ifxsolves the maximization problemand the constraint qualification holdsatxthen there exists a vector Rm+ s.t.

f(x) =Jg(x)

Moreover,has the property: gj(x)

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The KKT and the mantra

x1

x2

a necessary condition ... is that the gradient vector of the objective

function at x belongs to the nonnegative cone defined by the gradient

vectors of the constraints satisfied with equality at x

3 dotted curves represent level sets of 3differentobjective functions

in each case, dashed black arrows (gradients of objectives) belong to

appropriate non-negative cone, defined by solid arrows

redarrows:s of constraintsnotsatisfied with=at point being examined

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Necessary conditions for a Maximum

In this picture the mantra fails

fdoesnt belong to cone defined byg1 andg2

dashed line perpular tofpoints into interior of constraint setangle betweendx andf is acute

directiondxincreasesf& strictly decreases BOTH constraints

angles betweendx andgi are obtuse

tail of red arrow cant solve the NPPg1

g2

f

dx

(dx,f)< 90

(dx,g2) > 90

KTGradients

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Important issues

The role of the constraint qualification (CQ):

1 it ensures that the linearized version of the constraint set is, locally, a good

approximation of the true nonlinear constraint set.2

the CQ will be satisfied atx if the gradients of the constraints that aresatisfied with equality atx form a linear independent set

Interpretation of the Lagrangian

Constraint satisfied with equality vs binding constraints constraint.

The KKT and equality constraints

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The Constraint Qualification

g1(x)g

2

(x

)

f(x)

constraint set

linearized constraint set

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The Constraint Qualification with quasi-convex constraints

g1(x)g2(x)

f(x)

constraint set

linearized constraint set

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KKT and the Lagrangian Approach

The Lagrangian function for f :RnR,g: RnRm andb Rm.

L(x,) = f(x) +(bg(x)) = f(x) +m

j=1

j(bjgj(x))

The first order conditions for an extremum of Lon RnRm+ are:(x,)s.t.

fori=1,...n,L(x,)/xi = 0 (6)

forj=1,...m,L(x,)/j 0 (7)

jL(x,)/j = 0. (8)

Compare KKT with (6) f(x)T =

TJg(x)

f(x)/xi =

m

j=1

j gj(x)/xi

(7) says forj=1,...m,(bjgj(x))0 or(bg(x))0.

(8) complementary slackness:(bjgj(x))>0 implies j=0;

Note no nonnegativity constraints: we treat these likeanyotherconstraints() November 14, 2013 15 / 24

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Lagrangian as the shadow value of the constraint

Theorem: At a solution(x,)to the NPP,L(x,) =f(x).

Fact: j is the shadow value of the jth constraint:

M(b) = f(x(b)) = L(x(b),(b)) = f(x) + (bg(x)).

dM(b)

dbj=

dL(x(b),(b))

dbj=

n

i=1

fi(x(b))

m

k=1

k(b)gki(x(b))

dxidbj

+m

k=1

dk

dbj(bkg

k(x(b))) + j(b)

For eachi, the term in curly brackets is zero, by the KT conditions.

For eachk,

If(bk gk(x(b))) =0, then dkdbk (bk gk(x(b)))is zero.

If(bk gk(x(b)))

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Interpretation of the Lagrangian: One inequality constraint

Gradient of objective and constraint are collinear:i,f(x)

xi= g(x

)xi

.i.e., is the ratio of the norms of the two gradients

in both figures, dashed line represents relaxation of constraint byb=1.the arrows representf(x)andg(x).

in one panel|| f(x)||small;||g(x)||big;in the other,|| f(x)||big;||g(x)||small

which is which (picture only tells you the answer forg)?

f(x) or g(x)? f(x) or g(x)?

Figure 1. Interpretation of the Lagrangian

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Interpretation of the Lagrangian: One inequality constraint

Gradient of objective and constraint are collinearis the ratio of the norms of the two gradients

left panel:|| f(x)||small;||g(x)||big; interpretationright panel:|| f(x)||big;||g(x)||small; interpretation

is the shadow value of the constraint: bang for a buck

g(x) =b + 1 (g given dx)

g(x) =b

g(x)

f(x)

f() = f(x)f() = f(x) + small f

= ||f(x)||

||g(x)|| is small

g(x) =b + 1 (g given dx)

g(x) = b

f(x)

g(x)

f() = f(x)f() = f(x) + BIG f

= ||f(x)||

||g(x)|| is big

Figure 1. Interpretation of the Lagrangian

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Interpretation of the Lagrangian: Two inequality constraints

Gradient of objective and one constraint are nearly collinear

1 is relatively large

relatively large gain fromb1 =.2 is relatively small

relatively small gain fromb2 =.

g1

(x) b1g1(x) b1+

g2(x) b2

g2(x) b2+

g1(x)

g2(x)

f(x)

Figure 1. Two multipliers, one big, one small

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Constraint satisfied with equality vs binding

Informal defn: a constraint isbindingif the maximized value of objective

increases when the constraint is slightly relaxed.

g1(x) b1

g2(x) b2

g1(x)

g2(x)

f(x)

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R l h fi i i hif NE

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Relax the first constraint: optimum shifts NE

g1(x) b1

g1(x) b1 +

g2(x) b2

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R l th d t i t ti d t

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Relax the second constraint: optimum doesnt move

g1(x) b1

g2(x) b2

g2(x) b2 +

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KKT ith lit t i t d fi d i liti

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KKT with equality constraints defined as inequalities

KKT with equality constraints

technically, dont absolutelyneedto introduce them

g(x) =biffg(x)bandg(x) bcan represent an equality constraint as two inequality constraints

apply the mantra as usual

butthere are high costs of doing this

you lose uniqueness of the Lagrangian multiplier

full rank sufficient condition for the CQ is neversatisfied

Lower contour set of g corresponding to b

Upper contour set of g corresponding to b

x1

x2

g(x) = b

x1

x2

x3

g(x)

f(x)

f(x)

g(x)

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KKT necessar conditions eq alit & ineq alit constraints

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KKT necessary conditions: equality & inequality constraints

The KKT conditions in full generality

f :RnR,g: RnRm,h: RnR are twice continuously diffable

b Rm andc R.

The canonical form of the nonlinear programming problem (NPP)

maximizef(x)subject tog(x)b and h(x) =c

The KKT conditions:Ifxsolves the maximization problemand the constraint qualification holdsatxthen there exist vectors Rm+ and R

s.t.

f(x) = Jg(x) + Jh(x)

Moreover,has the property: gj(x)