4.8 Quadratic Programming 229
Noting that
∑n
r= 1
x(r^1 )yr(^2 )=^13 ( 10834 )+^13 ( 10837 )+^13 ( 10837 )=^13
Eq. (4.71) can be used to find
{x(i^2 )} =
xi(^1 )y(i^2 )
∑^3
r= 1
x(r^1 )yr(^2 )
= 3
34
324
37
324
37
324
=
34
108
37
108
37
108
Set the new iteration number ask=k+ 1 =2 and go to step 2. The procedure
is to be continued until convergence is achieved.
Notes:
1.AlthoughX(^2 )=Y(^2 )in this example, they need not be, in general, equal to
one another.
2.The value off atX(^2 )is
f(X(^2 ))= 2 ( 10834 )+ 10837 − 10837 =^1727 < f (X(^1 ))=^1827
4.8 Quadratic Programming
Aquadratic programming problem can be stated as
Minimizef (X)=CTX+^12 XTDX (4.72)
subject to
AX≤B (4.73)
X≥ 0 (4.74)
where
X=
x 1
x 2
..
.
xn
, C=
c 1
c 2
..
.
cn
, B=
b 1
b 2
..
.
bm
,
D=
d 11 d 12 · · ·d 1 n
d 21 d 22 · · ·d 2 n
..
.
dn 1 dn 2 · · ·dnn
, and A=
a 11 a 12 · · · a 1 n
a 21 a 22 · · · a 2 n
..
.
am 1 am 2 · · ·amn
In Eq. (4.72) the term XTDX / 2 represents the quadratic part of the objective
function withDbeing a symmetric positive-definite matrix. IfD= 0 , the problem
