96
2 1
1
2 ,
,
k k
k k
k k
k
x g
s c
u H
c u
H ε
+ ≤
−
+ +
.
5.7 Defuzzification and solution of defuzzificated problem
In this section, we present the modified subgradient method Gasimov [57] and use it for solving the defuzzificated problems 5.55-5.57 for nonconvex constrained
problems and can be applied for solving a large class of such problems. Notice that, the constraints in problem 5.55-5.57 is generally not convex. These
problems may be solved either by the fuzzy decisive set method, which is presented by Sakawa and Yana [165], or by the linearization method of Kettani and Oral [83].
5.7.1 A modified subgradient method to fuzzy linear programming
For applying the subgradient method Gasimov [57] to the problem 5.54-5.57, we first formulate it with equality constraints by using slack variables
y
and
i
y
,
2 ,
1 +
+ =
m mT
i
. Then, we can be written as
λ max
, 5.99
, ,
2 1
2 1
= +
+ −
− =
∑
=
y z
X c
z z
y X
g
n j
j j
λ λ
, 5.100
= ,
, λ
y X
g
i
∑
+ =
= +
−
mT n
j i
i j
ij
y b
X a
1
ˆ
λ
,
2 ,
1 +
+ =
m mT
i
, 5.101
≥
j
X
,
, ≥
i
y y
,
1 ≤
≤
λ
,
mT n
j +
= ,1
,
2 ,
1 +
+ =
m mT
i
. 5.102 where
,...,
n
y y
y =
For this problem we define the set S as
} 1
, ,
, ,
{ ≤
≤ ≥
≥ =
λ λ
y X
y X
S
. Since
min max
λ λ
− −
=
and
,...,
2 +
+
=
m mT
g g
g
the augmented Lagrangian associated with the problem 5.99-5.102 can be written in the form
97
. 2
1 ˆ
2 1
2 1
2 1
2 2
1 ˆ
2 2
1 2
1 ,
,
1 1
∑ +
+ =
⎟ ⎠
⎞ ⎜
⎝ ⎛
+ −
∑ −
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ ∑
= +
− −
− ⎥
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎢ ⎣
⎡ ∑
+ +
= ⎟
⎠ ⎞
⎜ ⎝
⎛ +
− ∑
+ ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ +
+ ∑
= −
− +
− =
+ =
+ =
m mT
i y
b X
a u
y z
n j
j X
j c
z z
m mT
i y
b X
a y
z n
j j
X j
c z
z c
c u
X L
i i
j mT
n j
ij i
i i
j mT
n j
ij
λ λ
μ λ
λ λ
The modified subgradient method may be applied to the problem 5.99-5.102 in the following way:
Initialization Step. Choose a vector
, ,...,
,
1 1
2 1
1 1
c u
u u
m mT
+ +
with
1
≥ c
. Let
1 =
k
, and go to main step.
Main Step. Step 1 . Given
, ,...,
,
2 1
k k
m mT
k k
c u
u u
+ +
; solve the following subproblem :
, 2
1 1
ˆ 2
1 2
1 2
1 2
2 1
1 ˆ
2 2
1 2
1 min
∑ +
+ =
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ −
∑ +
= −
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ ∑
= +
− −
− ⎥
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎢ ⎣
⎡ ∑
+ +
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ +
− ∑
+ =
+ ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ +
+ ∑
= −
− +
− m
mT i
i y
i b
j X
mT n
j ij
a i
u y
z n
j j
X j
c z
z u
m mT
i i
y i
b mT
n j
j X
ij a
y z
n j
j X
j c
z z
c λ
λ λ
λ λ
. ,
, S
y X
∈ λ
Let
, ,
k k
k
y X
λ
be a solution. If
, ,
=
k k
k
y X
g λ
, then stop;
, ,...,
,
2 1
k k
m mT
k k
c u
u u
+ +
is a solution to dual problem,
, ,
k k
k
y X
λ
is a solution to problem 5.54-5.57. Otherwise, go to Step 2.
Step 2 . Let
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
+ +
− −
− =
∑
= +
2 1
2 1
1
y z
X c
z z
h u
u
n j
j j
k k
k
λ
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
+ −
− =
∑
+ =
+ mT
n j
i i
j ij
k k
i k
i
y b
X a
h u
u
1 1
ˆ
λ
,
2 ,
1 +
+ =
m mT
i ,
,
1 k
k k
k k
k k
y X
g h
c c
λ ε
+ +
=
+
98
where
k
h
and
k
ε
are positive scalar stepsizes and
k k
h ε
, replace k by k + 1; and repeat Step 1.
5.7.2 Fuzzy decisive set method
For a fixed value of
λ
, the problem 5.54-5.57 is a linear programming problem. Thus obtaining the optimal solution
λ
to the problem 5.54-5.57 is equivalent to determining the maximum value of
λ
so that the feasible set is nonempty. Bellow is presented the algorithm Gasimov [57], of this method for the problem 5.54-
5.57.
Algorithm Step 1. Set
λ
= 1 and test whether a feasible set satisfying the constraints of the problem 5.54-5.57 exists or not using phase one of the simplex method. If a feasible set exists,
set
λ
= 1: Otherwise, set
=
L
λ
and
1 =
R
λ
and go to the next step.
Step 2. For the value of
2
R L
λ λ
λ +
=
; update the value of
L
λ
and
R
λ
using the bisection method as follows :
λ λ =
L
if feasible set is nonempty for
λ
;
λ λ =
R
if feasible set is empty for
λ
. Hence, for each
λ
, we test whether a feasible set of the problem 5.54-5.57 exists or not using phase one of the Simplex method and determine the maximum value
λ
satisfying the constraints of the problem 5.54-5.57.
Example 5.1.
Solve the optimization problem, see Gasimov [40]
, ,
6 1
~ 3
~ 4
2 ~
1 ~
3 2
max
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ +
x x
x x
x x
x x
5.103
which take fuzzy parameters as
2 ,
3 3
~ ,
3 ,
2 2
~ ,
1 ,
1 1
~ L
L L
= =
=
and
3 ,
1 1
~ L
=
as used by Shaocheng [171]. That is,
99 ⎥⎦
⎤ ⎢⎣
⎡ =
1 3
2 1
ij
a
,
⎥⎦ ⎤
⎢⎣ ⎡
= 3
2 3
1
ij
d ⇒
⎥⎦ ⎤
⎢⎣ ⎡
= +
4 5
5 2
ij ij
d a
For solving this problem we must solve the folowing two subproblems:
, ,
6 3
4 2
3 2
max
2 1
2 1
2 1
2 1
1
≥ ≤
+ ≤
+ +
=
x x
x x
x x
x x
z
and
, ,
6 4
5 4
5 2
3 2
max
2 1
2 1
2 1
2 1
2
≥ ≤
+ ≤
+ +
=
x x
x x
x x
x x
z
Optimal solutions of these subproblems are
2 .
1 6
. 1
8 .
6
2 1
1
= =
=
x x
z
and
, 82
. 47
. 06
. 3
2 1
2
= =
=
x x
z
respectively. By using these optimal values, problem 5.103 can be reduced to the following equivalent non-linear programming problem:
, 1
3 2
3 6
3 2
4 06
. 3
8 .
6 06
. 3
3 2
max
2 1
2 1
2 1
2 1
2 1
2 1
≥ ≤
≤ ≥
+ −
− ≥
+ −
− ≥
− −
+
x x
x x
x x
x x
x x
x x
λ λ
λ λ
λ
that is
100 ,
1 6
3 1
2 3
4 3
2 1
74 .
3 06
. 3
3 2
max
2 1
2 1
2 1
2 1
≥ ≤
≤ ≤
+ +
+ ≤
+ +
+ +
≥ +
x x
x x
x x
x x
λ λ
λ λ
λ λ
λ
5.104
Lets solve problem 5.104 by using the fuzzy decisive set method. For
1 =
λ
, the problem can be written as
, 6
4 5
4 5
2 8
. 6
3 2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
=
L
λ
and
R
λ
=1; the new value of
2 1
2 1
= +
= λ
is tried. For
2 1
= λ
, the problem can be written as
, ,
6 2
5 4
4 2
7 2
3 9294
. 4
3 2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
=
L
λ
and
2 1
=
R
λ
, the new value of
4 1
2 2
1 =
+ =
λ
is tried. For
4 1
= λ
, the problem can be written as
101 ,
, 6
4 7
2 7
4 4
11 4
5 9941
. 3
3 2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
4 1
=
L
λ
and
2 1
=
R
λ
, the new value of
8 3
2 2
1 4
1 =
+ =
λ
is tried. For
8 3
= λ
, the problem can be written as
, ,
6 8
17 4
15 4
8 25
8 11
4618 .
4 3
2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is nonempty, by taking
8 3
=
L
λ
and
2 1
=
R
λ
, the new value of
16 7
2 2
1 8
3 =
+ =
λ
is tried. For
4375 .
16 7 =
= λ
, the problem can be written as
, ,
6 16
37 8
31 4
16 53
16 23
6956 .
4 3
2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is nonempty, by taking
8 3
=
L
λ
and
16 7
=
R
λ
, the new value
102
of
32 13
2 16
7 8
3 =
+ =
λ
is tried. For
40625 .
32 13 =
= λ
, the problem can be written as
, ,
6 32
71 32
122 4
32 103
32 45
5787 .
4 3
2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
8 3
=
L
λ
and
32 13
=
R
λ
, the new value of
64 25
2 32
13 8
3 =
+ =
λ
is tried. For
390625 .
64 25 =
= λ
, the problem can be written as
, ,
6 64
139 64
242 4
64 203
64 89
5202 .
4 3
2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
64 25
=
L
λ
and
32 13
=
R
λ
, the new value of
128 51
2 32
13 64
25 =
+ =
λ
is tried. For
398475 .
128 51 =
= λ
, the problem can be written as
103 ,
, 6
128 281
128 486
4 128
409 128
179 5494
. 4
3 2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
64 25
=
L
λ
and
128 51
=
R
λ
, the new value of
256 101
2 128
51 64
25 =
+ =
λ
is tried. The following value of
λ
are obtained in the next thirteen iterations:
256 101
=
λ
= 0.39453125
512 1203
=
λ
= 0.396484325
1024 407
=
λ
= 0.396972656
4096 1627
=
λ
= 0.397216796
8192 3255
=
λ
= 0.397338867
16384 6511
=
λ
= 0.397399902
32768 13021
=
λ
= 0.397369384
65536 26043
=
λ
= 0.397384643
131072 52085
=
λ
= 0.397377014
262144 104169
=
λ
= 0.3973731
524288 208337
=
λ
= 0.3973733
1048576 416675
= λ
= 0.3973723
Consequently, we obtain the optimal value of
λ
at the twenty first iteration by using the fuzzy decisive set method.
Now, lets solve the same problem by using the modified subgradient method. Before solving the problem, we first formulate it in the form
min max
λ λ
− −
=
104 ,
, ,
, 1
6 3
1 2
3 4
3 2
1 06
. 3
3 2
74 .
3
2 1
2 1
2 2
1 1
2 1
2 1
≥ ≥
≤ ≤
= +
− +
+ +
= +
− +
+ +
= +
+ −
−
p p
p x
x p
x x
p x
x p
x x
λ λ
λ λ
λ λ
where
1
, p p
and
2
p
are slack variables. The augmented Lagrangian function for this problem is
. 6
3 1
2 3
4 3
2 1
06 .
3 3
2 74
. 3
] 6
3 1
2 3
4 3
2 1
06 .
3 3
2 74
. 3
[ ,
,
2 2
1 2
1 2
1 1
2 1
2 1
2 2
2 1
2 1
2 1
2 2
1
p x
x u
p x
x u
p x
x u
p x
x p
x x
p x
x c
c u
x L
+ −
+ +
+ −
+ −
+ +
+ −
+ +
− −
− +
− +
+ +
+ +
− +
+ +
+ +
+ −
− +
− =
λ λ
λ λ
λ λ
λ λ
λ λ
λ
Let the initial vector is
, ,
,
1 1
2 1
1 1
c u
u u
= 0; 0; 0; 0 and lets solve the following subproblem
. 2
. 1
74 .
6 .
1 82
. 1
, ,
min
2 1
≤ ≤
≤ ≤
≤ ≤
x x
x L
λ
The optimal solutions of subproblem are obtained as
. 1
, ,
2 ,
, 8
. 4
, ,
1 1
1 1
1 3
1 1
1 2
1 1
1 1
2 1
− =
− =
= =
= =
λ λ
λ λ
p x
g p
x g
p x
g x
x
Since
, ,
1 1
1
≠ λ
p x
g
, we calculate the new values of Lagrangemultipliers
, ,
2 2
2 2
1 2
c u
u u
by using Step 2 of the modified subgradient method. The solutions of the second iteration are obtained as
105
6 2
2 2
3 6
2 2
2 2
6 2
2 2
1 2
1
10 31
. 2
, ,
10 8
. 3
, ,
10 9
, ,
3973723 .
75147 .
1475877 ,
1
− −
−
× =
× −
= ×
= =
= =
λ λ
λ λ
p x
g p
x g
p x
g x
x
Since
x g
is quite small, by Theorem 5.4
75147 .
, 1475877
. 1
2 1
= =
x x
and
= λ
3973723 .
are optimal solutions to the problem 5.100. This means that, the vector
,
2 1
x x
is a solution to the problem 5.99 which has the best membership grade
λ
. Note that, the optimal value of
λ
found at the second iteration of the modified subgradient method is approximately equal to the optimal value of
λ
calculated at the twenty first iteration of the fuzzy decisive set method.
Example 5.2.
Solve the optimization problem, see Gasimov [40]
, ,
4 ~
3 ~
2 ~
3 ~
2 ~
1 ~
max
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ +
x x
x x
x x
x x
5.105
which take fuzzy parameters as
2 ,
3 3
~ ,
2 ,
3 3
~ ,
2 ,
2 2
~ ,
1 ,
2 2
~ ,
1 ,
1 1
~ L
b L
L L
L =
= =
= =
=
and
3 ,
4 4
~
2
L b
= =
as used by Shaocheng [171]. That is,
⎥⎦ ⎤
⎢⎣ ⎡
= 3
2 2
1
ij
a
,
⎥⎦ ⎤
⎢⎣ ⎡
= 2
2 1
1
ij
d ⇒
⎥⎦ ⎤
⎢⎣ ⎡
= +
5 4
3 2
ij ij
d a
⎥⎦ ⎤
⎢⎣ ⎡
= 4
3
i
b
,
⎥⎦ ⎤
⎢⎣ ⎡
= 3
2
i
p ⇒
⎥⎦ ⎤
⎢⎣ ⎡
= +
7 5
i i
p b
.
To solving this problem, first, we must solve the folowing two subproblems:
106 ,
, 4
5 4
3 3
2 max
2 1
2 1
2 1
2 1
1
≥ ≤
+ ≤
+ +
=
x x
x x
x x
x x
z
and
, ,
7 3
2 5
2 max
2 1
2 1
2 1
2 1
2
≥ ≤
+ ≤
+ +
=
x x
x x
x x
x x
z
Optimal solutions of these subproblems are
1 1
2 1
1
= =
=
x x
z
and
, 5
. 3
5 .
3
2 1
2
= =
=
x x
z
respectively. By using these optimal values, problem 5.105 can be reduced to the following equivalent non-linear programming problem:
, 1
2 2
3 2
4 2
3 1
5 .
3 1
max
2 1
2 1
2 1
2 1
2 1
2 1
≥ ≤
≤ ≥
+ −
− ≥
+ −
− ≥
− −
+
x x
x x
x x
x x
x x
x x
λ λ
λ λ
λ
that is
, 1
6 3
1 2
3 4
3 2
1 74
. 3
06 .
3 3
2 max
2 1
2 1
2 1
2 1
≥ ≤
≤ ≤
+ +
+ ≤
+ +
+ +
≥ +
x x
x x
x x
x x
λ λ
λ λ
λ λ
λ
5.106
Lets solve problem 5.106 by using the fuzzy decisive set method.
107
For
1 =
λ
, the problem can be written as
, 1
5 4
1 3
2 5
. 3
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
=
L
λ
and
R
λ
=1; the new value of
2 1
2 1
= +
= λ
is tried. For
2 1
= λ
, the problem can be written as
, ,
2 5
4 3
2 2
5 2
3 25
. 2
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
=
L
λ
and
2 1
=
R
λ
, the new value of
4 1
2 2
1 =
+ =
λ
is tried. For
4 1
= λ
, the problem can be written as
, ,
4 13
2 7
2 5
2 5
4 9
4 5
625 .
1
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is empty, by taking
4 1
=
L
λ
and
4 1
=
R
λ
, the new value of
8 1
2 4
1 =
+ =
λ
is tried.
108
For
8 1
= λ
, the problem can be written as
, ,
8 29
4 13
4 9
8 22
8 17
8 9
3125 .
1
2 1
2 1
2 1
2 1
≥ ≤
+ ≤
+ ≥
+
x x
x x
x x
x x
and since the feasible set is nonempty, by taking
8 1
=
L
λ
and
4 1
=
R
λ
, the new value of
16 3
2 4
1 8
1 =
+ =
λ
is tried. The following value of
λ
are obtained in the next thirteen iterations:
16 3
=
λ
= 0.1875
32 5
=
λ
= 0.15625
64 11
=
λ
= 0.171875
128 23
=
λ
= 0.1796875
256 47
=
λ
= 0.18359375
512 93
=
λ
= 0.181640625
1024 187
=
λ
= 0.182617187
2048 375
=
λ
= 0.183349609
4096 751
=
λ
= 0.183166503
8192 1501
=
λ
= 0.397377014
16384 3001
=
λ
= 0.183166503
32768 6003
=
λ
= 0.183197021
65536 12007
=
λ
= 0.18321228
131072 24015
=
λ
= 0.183219909
62144 48029
=
λ
= 0.183216095
524288 96057
=
λ
= 0.183214187
1048576 192115
=
λ
= 0.183215141
2097152 383231
=
λ
= 0.183215618
109
4194304 786463
=
λ
= 0.183215856
8388608 1536927
=
λ
= 0.183215975
16777216 3073853
= λ
=0.183215916
Hence,, we obtain the optimal value of
λ
at the twenty fifth iteration by using the fuzzy decisive set method.
By using the modified subgradient we solve the same problem method. Before solving the problem, we first formulate it in the form
min max
λ λ
− −
=
, ,
, ,
1 4
2 3
2 2
3 2
1 1
5 .
2
2 1
2 1
2 2
1 1
2 1
2 1
≥ ≥
≤ ≤
= +
− +
+ +
= +
− +
+ +
= +
+ −
−
p p
p x
x p
x x
p x
x p
x x
λ λ
λ λ
λ λ
where
1
, p p
and
2
p
are slack variables. The augmented Lagrangian function for this problem is
. 4
2 3
2 2
3 2
1 1
5 .
2 ]
4 2
3 2
2 3
2 1
1 5
, 2
[ ,
,
2 2
1 2
1 2
1 1
2 1
2 1
2 2
2 1
2 1
2 1
2 2
1
p x
x u
p x
x u
p x
x u
p x
x p
x x
p x
x c
c u
x L
+ −
+ +
+ −
+ −
+ +
+ −
+ +
− −
− +
− +
+ +
+ +
− +
+ +
+ +
+ −
− +
− =
λ λ
λ λ
λ λ
λ λ
λ λ
λ
Let the initial vector is
, ,
,
1 1
2 1
1 1
c u
u u
= 0; 0; 0; 0 and lets solve the following subproblem
. 5
. 3
1 1
, ,
min
2 1
≤ ≤
≤ ≤
≤ ≤
x x
x L
λ
The optimal solutions of subproblem are obtained as
110 3
, ,
1 ,
, 5
. 2
, ,
1 1
1 1
1 3
1 1
1 2
1 1
1 1
2 1
= =
= =
= =
λ λ
λ λ
p x
g p
x g
p x
g x
x
Since
, ,
1 1
1
≠ λ
p x
g
, we calculate the new values of Lagrangemultipliers
, ,
2 2
2 2
1 2
c u
u u
by using Step 2 of the modified subgradient method. The solutions of the second iteration are obtained as
8 2
2 2
3 8
2 2
2 2
7 2
2 2
1 8
2 1
10 83
. 7
, ,
10 2
. 8
, ,
10 28
. 3
, ,
1832159 .
10 8
. 7
45804 ,
1
− −
− −
× −
= ×
= ×
= =
× =
=
λ λ
λ λ
p x
g p
x g
p x
g x
x
Since
x g
is quite small, by Theorem 5.4,
10 8
. 7
, 45804
. 1
8 2
1
≅ ×
= =
x x
and
= λ
0.1832159 are optimal solutions to the problem 5.106. This means that, the vector
,
2 1
x x
is a solution to the problem 5.105 which has the best membership grade
λ
. Note that, the optimal value of
λ
found at the second iteration of the modified subgradient method is approximately equal to the optimal value of
λ
calculated at the twenty first iteration of the fuzzy decisive set method.
5.8 Portfolio problem with fuzzy multi-objective
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