4
3
2
2
/
1
,
2
/
1
2
,
min
11
10
21
20
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
−
−
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
γ
γ
α
α
since the minimum of these ratios correspond
to
11
10
γ
γ
,
non-basic
variables
can
be
removed. Thus we introduce a free variable,
u
1
as an additional non-basic
variable,
defined by
1
1
1
1
4
1
2
2
1
2
1
2
1
x
x
x
f
u
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
=
∂
∂
=
Note that now the basis has two basic
variables
x
2
and
x
1
(just entered). That is, we
have
(
)
(
)
2
1
1
,
,
x
x
x
and
u
s
x
B
NB
=
=
.
Expressing the basic
x
B
in terms of non-basic
x
NB
, we have,
1
1
4
1
u
x
−
=
(
)
.
2
1
2
1
8
15
4
2
1
1
3
1
2
s
u
x
x
x
and
−
+
=
−
−
=
The objective function, expressing in terms
of
x
NB
is,
.
2
3
16
97
4
1
2
1
2
1
8
15
3
4
1
2
2
1
2
1
1
1
u
s
u
s
u
u
f
−
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
−
⎟
⎠
⎞
⎜
⎝
⎛
−
+
+
⎟
⎠
⎞
⎜
⎝
⎛
−
=
Now,
2
3
0
0
−
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=
=
u
NB
x
s
F
;
0
2
1
1
0
0
=
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
u
u
f
u
NB
x
δ
δ
since
0
1
≤
∂
∂
u
f
for all
x
j
in
x
NB
and
,
0
=
∂
∂
u
f
the current solution is optimal. Hence the
optimal basic feasible solution to the given
problem is:
16
97
*
,
8
15
,
4
1
2
1
=
=
Z
x
x
Similarly we can find that by Wolfe’s
algorithm the optimal point is at (1/4, 15/8).
which was introduced by [14].
Thus for the optimal solution for the given
QP problem is
(
)
2
1
2
1
2
1
2
2
3
1
1 5
1
2
3
4
8
4
9 7
1
1 5
,
,
1 6
4
8
M a x Z
x
x
x
a t
x
x
∗
∗
=
+
−
⎛
⎞
=
⋅
+
⋅
−
⎜
⎟
⎝
⎠
⎛
⎞
=
=
⎜
⎟
⎝
⎠
Therefore the solution obtained by graphical
solution method, Kuhn-Tucker conditions,
Beale’s method and Wolf’s algorithm are
same. The computational cost is that by the
graphical solution method using MATLAB
Programming it will take very short time to
determine the plan of action and the solution
obtained
by
graphical
method
is
more
effective
than
any
other
methods
we
considered.