lems
5.3.1
Canonical dual problem
The integer constraints in the problem (Pbqp) is treated as equality ones, x2i = 1, i =
in Chapter 3, we can easily get the canonical dual problem. Let
G( ) = Q + diag( ), and
Sa ={ 2 Rn| det(G( )) 6= 0}.
The canonical dual function is formulated as
⇧d( ) = 1
2f
TG( ) 1f 1
2e
T , (5.22)
and the canonical dual problem is defined as
(Pbqpd ) ext{⇧d( )| 2 Sa}. (5.23)
Notice that if f = 0, the function ⇧d( ) is equal to the objective function in
the problem (5.12), which is thus same to the problem of maximizing ⇧d over the
positive semidefinite region defined by
Sc+ ={ 2 Sa| G( ) ⌫ 0}.
5.3.2
Global optimality conditions
The following theorem characterizes the primal-dual relation.
Theorem 30 (Complementary-dual principle) If ¯ 2 Sa is a critical point of
⇧d( ), the vector ¯ x = G( ¯ ) 1f (5.24) is a feasible solution of (Pbqp). Let ¯x2 X and ¯ = ¯x f x (Q ¯¯ x). (5.25)
If ¯ 2 Sa, then ¯ is a critical point of ⇧d( ).
For both statements, we have
⇧( ¯x) = ⇧d( ¯ ). (5.26) Proof: From the assumption of ¯ being a critical point and the definition of ¯x, we have r ⇧d( ¯ ) = ⇢ 1 2f TG( ¯ ) 1I iG( ¯ ) 1f n i=1 1 2e = 1 2x¯ x¯ 1 2e = 0,
where Ii denotes an all-zero matrix except for the entry (i, i), which is equal to one.
It is verified that ¯x is a feasible solution of the primal problem (Pbqp). The definition
of ¯x in equation (5.24) also implies that ¯x and ¯ satisfy the equilibrium equation G( ¯ ) ¯x f = 0, (5.27)
from which we have
¯
x ¯ = f Q ¯x. (5.28) The equality (5.28) is further equivalent to
¯
x x¯ ¯ = ¯x f x (Q ¯¯ x). Thus, we have
¯ = ¯x f x (Q ¯¯ x) = diag( ¯x)f diag( ¯x)Q ¯x. (5.29)
Now we are ready to prove the equation (5.26):
⇧d( ¯ ) = 1 2f TG( ¯ ) 1f 1 2e T¯ = 1 2f Tx¯ 1 2e T (diag( ¯x)f diag( ¯x)Q ¯x) = 1 2x¯ TQ ¯x fTx = ⇧( ¯¯ x).
Let ¯ be defined by the equation (5.25). From the equivalence between (5.27) and (5.29), it can be easily proved that ¯ is a critical point if ¯ 2 Sa.
The theorem is proved. 2
Besides the positive semidefinite regionS+
c , we also introduce the negative semidef-
inite region:
Sc ={ 2 Sa| G( ) 0}.
First,S+
c and Sc are convex sets. Second, from the expression of Hessian of the dual
function ⇧d( ), we notice that ⇧d( ¯ ) is concave on S+
c and convex on Sc . Hence,
any critical point in S+
c is a maximizer of ⇧d over Sc+, and any critical point in Sc
is a minimizer of ⇧d over S
c . We have the following result.
Theorem 31 For any given matrix Q 2 Sn and vector f 2 Rn, suppose ¯ is a
critical point of the dual function ⇧d( ) and ¯x = G( ¯ ) 1f . 1. If ¯ 2 S+
c , then ¯x is a global minimizer of ⇧(x) on X ; we have
⇧( ¯x) = min
x2X⇧(x) = max2S+ c
2. If ¯ 2 Sc , then ¯x is a global maximizer of ⇧(x) on X ; we have ⇧( ¯x) = max
x2X ⇧(x) = min2Sc
⇧d( ) = ⇧d( ¯ ). (5.31)
Proof: The last equalities in (5.30) and (5.30) are obvious, since ⇧d( ) is concave
onS+
c and convex on Sc . For the minimizer, the weak duality in (3.30) shows that
we always have
⇧d( ¯ ) = max
2Sc+
⇧d( ) min
x2X⇧(x).
Thus, by Theorem 30, we have
⇧( ¯x) = ⇧d( ¯ ) = min x2X⇧(x),
and it is proved that ¯x is a minimizer of ⇧(x) on X .
The second part of the theorem can be similarly proved by applying the fact that the maximization of ⇧(x) is equivalent to the minimization of ⇧(x) since there are finite feasible solutions inX . 2 The equation (5.30) shows that the critical point ¯ is the maximizer of the dual function ⇧d( ¯ ) over S+
c . On the other hand, if the maximizer is a critical point of
⇧d( ¯ ), we can claim that the corresponding ¯x defined by the equation (5.24) is a
global optimal solution of the primal problem. Follow the discussion in Section 3.3.4, the maximizer can be found by solving the following SDP problem:
max ,⌧ ⌧ (5.32) s.t. ✓ 2G( ) f fT 1 2e T ⌧ ◆ ⌫ 0
Let ( ¯ , ¯⌧ ) be a maximizer. Then, if G( ¯ ) 0, ¯ must be a critical point of ⇧d and ¯
x must be a global solution. While if det(G( ¯ )) = 0, ¯ may not be a critical point and ⇧d( ¯ ) is only a lower bound for the primal problem. It shows that the integer
problem can be converted into a convex optimization problem, which can be solved efficiently by well-developed convex optimization methods.
Corollary 32 Suppose that ¯ is a maximizer of the problem (5.32) and ¯x = G( ¯ ) 1f .
If G( ¯ ) 0, then ¯x is a global optimal solution of the problem (Pbqp).
5.3.3
Existence and uniqueness
The following result gives a criterion of existence and uniqueness of a critical point inS+
c .
Theorem 33 If, for any 0 with det(G( 0)) = 0 and G( 0)⌫ 0 and any 2 Sc+,
we have
lim
t!0+⇧
d(
0+ t ) = 1, (5.33)
then the canonical dual problem (Pd
Proof: The assumption in the equation (5.33), plus the fact that ⇧d( ) approaches
to minus infinity as any entry of increases infinitely, implies that the function ⇧d( ) is coercive on the convex set S+
c . Since ⇧d( ) is concave over Sc+, it has at
least one maximizer, which must be a critical point. We use ¯ to denote one of the maximizers. The uniqueness results from the fact that the Hessian of ⇧d( ) is
negative definite at the critical point ¯ . 2
5.3.4
Examples
Example 1 Let Q = 2 3 3 1 , and f = ✓ 1 2 ◆ .In this case, the dual function has four critical points,
¯1 = (4, 6), ¯2 = (6, 2), ¯3 = (0, 0), and ¯4 = ( 2, 4),
with function values
⇧d( ¯1) = 5.5 < ⇧d( ¯2) = 3.5 < ⇧d( ¯3) = 1.5 < ⇧d( ¯4) = 4.5.
The corresponding solutions of the primal problem are
¯
x1 = ( 1, 1), ¯x2 = (1, 1), ¯x3 = (1, 1), and ¯x4 = ( 1, 1).
By checking the eigenvalues of Q+diag( ), we find Q+diag( 1)⌫ 0, Q+diag( 4)
0, and Q+diag( 2) and Q+diag( 3) are indefinite. Thus, Theorem 30 and Theorem
31 are demonstrated. Example 2 Let Q = 2 4 922 1409 16 1 6 80 3 5 , and f = 0 @ 26 1 1 A . The dual function has eight critical points:
¯1 = (12, 128, 73), ¯2 = (10, 119, 72), ¯3 = (11, 134, 7),
¯4 = (11, 125, 8), ¯5 = (19, 21, 78), ¯6 = (3, 12, 79),
¯7 = (20, 15, 2). ¯8 = (2, 6, 1)
The corresponding primal solutions are ¯ x1 = (0, 1, 1), x¯2 = (1, 1, 1), x¯3 = (0, 1, 0), ¯ x4 = (1, 1, 0), x¯5 = (1, 0, 1), x¯6 = (0, 0, 1), ¯ x7 = (1, 0, 0). x¯8 = (0, 0, 0)
with function values
⇧( ¯x1) = 97, ⇧( ¯x2) = 96, ⇧( ¯x3) = 64, ⇧( ¯x4) = 64,
⇧( ¯x5) = 47, ⇧( ¯x6) = 39, ⇧( ¯x7) = 9, ⇧( ¯x8) = 0.
It can be verified that Q + diag( ¯1) is positive definite, Q + diag( ¯8) is negative
definite and all Q + diag( ¯i) for i = 2, . . . , 7 are indefinite. The function values show
that ¯x1 is the minimizer and ¯x8 is the maximizer. Thus, Theorem 30 and Theorem
31 are explained.