part 4 nonlinear programming 4.5 quadratic programming (qp)

Part 4 Nonlinear Programming

4.5 Quadratic Programming (QP)

Introduction

Quadratic programming is the name given to the procedure that minimizes a quadratic function of n variables subject to m linear inequality and/or equality constraints.

A number of practical optimization problems can be naturally posed as QP problems, such as constrained least squares, optimal control of linear system with quadratic cost functions, and the solution of liear algebraic equations.

Standard QP Problems

1

2

1min

2

. .

T Tf

s t

x c x x Qx

g x Ax b 0

g x x 0

Kuhn-Tucker Conditions

T T

T

f

μ g λ h 0

g 0

μ g 0

μ 0

1 1 2 2

1 2

T T

T T

T T T T

f

f

μ g λ h 0

μ g μ g 0

c x Q μ A μ 0

1

2

1

2

or and

gg 0

g

g Ax b 0

Ax b s s 0

g x 0

i

1 2 2 1

21

1

2

0 g =0 1, 2,

0

Ti

T T T T T

T T

i

μ g =

μ g = μ s + μ x = μ x + s μ

x= μ s

μ

μμ 0

μ

2 1

2 1

2

1

let

-

-

T T T T

T

T

μ x Q μ A c

Ax b s

μ Qx A μ c

s Ax b

xμ cQ Aw = z = M = q =

μs bA 0

Find and such that

0

,

T

w z

w = Mz + q

w z

w z 0

Complementary Problem

Definitions

A nonnegative solution , to the system of the

equation is called feasible solution to the

complementary problem.

A feasible solution , to the complementary

problem that also satisfies the c

w z

w = Mz + q

w z

ondition 0 is

called a complementary solution.

Note that 0 0 1,2, , .

Thus, , is referred to as the complementary pair.

T

Ti i

i i

w z i m n

w z

w z

w z

Basic Ideas of Complementary Pivot Method - 1

If (which may or may not be true),

then there exists an obvious complementary

solution, i.e.,

(trivial)

q 0

w = q

z = 0

Basic Ideas of Complementary Pivot Method - 2

0 0

0 0

If otherwise, rearrange the complementary problem

formulation

- or

Introduce a new term to both sides of the equation

1

1where, , -min , and

1

ii

z z

z z q

ww Mz = q I -M = q

z

w - Mz + e = w - Mz = q e

w w + e e

if .

w 0 z = 0

Almost Complementary Solution

0

0 0

Note that, if , then

complementary solution!

z

z

w Mz e q

w 0

z 0

z = 0

w = w 0

Example

2 21 1 1 2 2

1 2 1 2

min 6 2 2 2

. . 2 , 0

6 4 2; ; 1 1 ; 2

0 2 4

4 2 1 6

2 4 1 ; 0

1 1 0 2

T

f x x x x x

s t x x x x

x

c Q A b

cQ -AM q

-bA 0

Initial Tableau

w1’ w2’ w3’ z1 z2 z3 z0 q

w1’ 1 0 0 -4 2 -1 -1 -6

w2’ 0 1 0 2 -4 -1 -1 0

w3’ 0 0 1 1 1 0 -1 2

Step 1

To determine the initial almost elementary solution, the variable z0 is brought into the basis, replacing the basic variable with the most negative value.

Step 1

w1’ w2’ w3’ z1 z2 z3 z0 q

z0 -1 0 0 4 -2 1 1 6

w2’ -1 1 0 6 -6 0 0 6

w3’ -1 0 1 5 -1 1 0 8

Step-2 Principles

In essence, the complementary pivot algorithm proceeds to find a sequence of almost complementary solutions until z0 becomes zero. To do this, the basis changes must be done in such a way that

(a) The complementary relation between between the variables

must be maintained, i.e., 0 for 1, 2, , .

(b) The basic solution remains nonnegative.

i iw z i m n

Step-2 Procedure

• To satisfy (a), the nonbasic variable that enters the basis in the next tableau is always the complement of the basic variable that just left the basis in the last tableau. (Complementary Rule)

• To satisfy (b), minimum ratio test is used to determine which basic variable leaves the basis.

Step 2.1

w1’ w2’ w3’ z1 z2 z3 z0 q

z0 -1/3 -2/3 0 0 2 1 1 2

z1 -1/6 1/6 0 1 -1 0 0 1

w3’ -1/6 -5/6 1 0 4 1 0 3

Step 2.2

w1’ w2’ w3’ z1 z2 z3 z0 q

z0 -1/4 -1/4 -1/2 0 0 1/2 1 1/2

z1 -1/5 -1/24 1/4 1 0 1/4 0 7/4

z2 -1/24 -5/24 1/4 0 1 1/4 0 3/4

Step 2.3

w1’ w2’ w3’ z1 z2 z3 z0 q

z3 -1/2 -1/2 -1 0 0 1 2 1

z1 -1/12 -1/12 1/2 1 0 0 -1/2 3/2

z2 -1/12 -1/12 1/2 0 1 0 -1/2 1/2

Termination Criteria

1. z0 leaves the basis, or

2. The minimum ratio test fail, since all coefficients in the pivot column are nonpositive. Therefore, no solution.

*1 1

* *2 2

*3 1

3

21 11

2 21

z x

z x f

z

x