Kesimpulan Saran Aplikasi Relasi Peluang Bersyarat Fuzzy pada Sistem Informasi Fuzzy
Intan R, Mukaidono M . 2004. Fuzzy
Conditional Probability Relations and their Applications in Fuzzy Information
Systems. Knowledge and Information Systems 2004 6 : 345-365.
http:www.proquest.pqdweb.comfuzzy relation. [14 Februari 2008]
Klir GJ, Yuan B . 1995. Fuzzy Sets and
Fuzzy Logic : Theory and Applications. Prentice-Hall, New Jersey.
Pawlak Z . 1991. Rough Sets Theoretical
Aspects of Reasoning about Data. Kluwer
Kusumadewi, S
. 2002. Analisis Desain Sistem Fuzzy Menggunakan Tool Box
Matlab. Graha Ilmu, Yogyakarta.
Yager RR.
1990. Ordinal Measures of Specificity. Int J Gen Syst 17:57-72.
Zadeh LA
. 1970. Similarity Relations and Fuzzy Orderings. Inf Sci 32 :177-200.
LAMPIRAN
LAMPIRAN 1 Hasil Transformasi Tabel 3
Tabel 17 I U, A = {c
1
= ‘w
1
’, c
2
= ‘x
1
’, c
3
= ‘y
1
’, b
1
= ‘z
1
’}
U c
1
= ’w
1
’ c
2
= ’x
1
’ c
3
= ’ y
1
’ b
1
= ’ z
1
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
1 1 1 1 1
1 u
2
1 0 0 1 u
3
0 0 0 0 u
4
1 1 1 0 1
u
5
0 0 1 0 u
6
1 1 1 1 1
1 u
7
0 0 0 1 u
8
1 0 0 1 ∑ 5 3 4 5
3 2
1 2
3 1
1 1
2 3
m in , ,
, 2
, ,
m in , ,
3 c c c b
C B c c c
δ
= =
∑ ∑
1 2
3 1
1 1
m in , ,
, 2
, .
m in 5
c c c b B C
b
δ
= =
∑ ∑
Tabel 18 I U, A = {c
1
= ‘w
1
’, c
2
= ‘x
2
’, c
3
= ‘y
3
’, b
1
= ‘z
1
’}
U c
1
= ’w
1
’ c
2
= ’x
2
’ c
3
= ’ y
3
’ b
1
= ’ z
1
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
1 0 0 1 u
2
1 1 1 1 1
1 u
3
0 1 1 0 u
4
1 0 0 0 u
5
0 1 0 0 u
6
1 0 0 1 u
7
0 1 1 1 u
8
1 1 1 1 1
1 ∑ 5 5 4
5 2
2
1 2
3 1
2 1
2 3
m in ,
, ,
2 ,
1, m in
, ,
2 c c
c b C B
c c c
δ =
= =
∑ ∑
1 2
3 1
2 1
m in ,
, ,
2 ,
. m in
5 c c
c b B C
b δ
= =
∑ ∑
Tabel 19 I U, A = {c
1
= ‘w
2
’, c
2
= ‘x
2
’, c
3
= ‘y
3
’, b
1
= ‘z
2
’}
U c
1
= ’w
2
’ c
2
= ’x
2
’ c
3
= ’ y
3
’ b
1
= ’ z
2
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
0 0 0 0 u
2
0 1 1 0 u
3
1 1 1 1 1
1 u
4
0 0 0 1 u
5
1 1 0 1 u
6
0 0 0 0 u
7
1 1 1 0 1
u
8
0 1 1 0 ∑ 3 5 4 3
2 1
1 2
3 1
3 1
2 3
m in ,
, ,
1 ,
, m in
, ,
2 c c
c b C B
c c c
δ =
=
∑ ∑
1 2
3 1
3 1
m in ,
, ,
1 ,
. m in
3 c c
c b B C
b δ
= =
∑ ∑
Tabel 20 I U, A = {c
1
= ‘w
1
’, c
2
= ‘x
1
’, c
3
= ‘y
1
’, b
1
= ‘z
2
’}
U c
1
= ’w
1
’ c
2
= ’x
1
’ c
3
= ’ y
1
’ b
1
= ’ z
2
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
1 1 1 0 1
u
2
1 0 0 0 u
3
0 0 0 1 u
4
1 1 1 1 1
1 u
5
0 0 1 1 u
6
1 1 1 0 1
u
7
0 0 0 0 u
8
1 0 0 0 ∑ 5 3 4 3
3 1
1 2
3 1
4 1
2 3
m in ,
, ,
1 ,
, m in
, ,
3 c c
c b C B
c c c
δ =
=
∑ ∑
1 2
3 1
4 1
m in ,
, ,
1 ,
. m in
3 c c
c b B C
b δ
= =
∑ ∑
Tabel 21 I U, A = {c
1
= ‘w
2
’, c
2
= ‘x
2
’, c
3
= ‘y
1
’, b
1
= ‘z
2
’}
U c
1
= ’w
2
’ c
2
= ’x
2
’ c
3
= ’ y
1
’ b
1
= ’ z
2
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
0 0 1 0 u
2
0 1 0 0 u
3
1 1 0 1 u
4
0 0 1 1 u
5
1 1 1 1 1
1 u
6
0 0 1 0 u
7
1 1 0 0 u
8
0 1 0 0 ∑ 3 5 4 3
1 1
1 2
3 1
5 1
2 3
m in ,
, ,
1 ,
1, m in
, ,
1 c c
c b C B
c c c
δ =
= =
∑ ∑
1 2
3 1
5 1
m in ,
, ,
1 ,
. m in
3 c c
c b B C
b δ
= =
∑ ∑
Tabel 22 I U, A = {c
1
= ‘w
1
’, c
2
= ‘x
1
’, c
3
= ‘y
1
’, b
1
= ‘z
1
’}
U c
1
= ’w
1
’ c
2
= ’x
1
’ c
3
= ’ y
1
’ b
1
= ’ z
1
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
1 1 1 1 1
1 u
2
1 0 0 1 u
3
0 0 0 0 u
4
1 1 1 0 1
u
5
0 0 1 0 u
6
1 1 1 1 1
1 u
7
0 0 0 1 u
8
1 0 0 1 ∑ 5 3 4 5
3 2
1 2
3 1
6 1
2 3
m in ,
, ,
2 ,
, m in
, ,
3 c c
c b C B
c c c
δ =
=
∑ ∑
1 2
3 1
6 1
m in ,
, ,
2 ,
. m in
5 c c
c b B C
b δ
= =
∑ ∑
Tabel 23 I U, A = {c
1
= ‘w
2
’, c
2
= ‘x
2
’, c
3
= ‘y
3
’, b
1
= ‘z
1
’}
U c
1
= ’w
2
’ c
2
= ’x
2
’ c
3
= ’ y
3
’ b
1
= ’ z
1
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
0 0 0 1 u
2
0 1 1 1 u
3
1 1 1 0 1
u
4
0 0 0 0 u
5
1 1 0 0 u
6
0 0 0 1 u
7
1 1 1 1 1
1 u
8
0 1 1 1 ∑ 3 5 4 5
2 1
1 2
3 1
7 1
2 3
m in ,
, ,
1 ,
, m in
, ,
2 c c
c b C B
c c c
δ =
=
∑ ∑
1 2
3 1
7 1
m in ,
, ,
1 ,
. m in
5 c c
c b B C
b δ
= =
∑ ∑
Tabel 24 I U, A = {c
1
= ‘w
1
’, c
2
= ‘x
2
’, c
3
= ‘y
3
’, b
1
= ‘z
1
’}
U c
1
= ’w
1
’ c
2
= ’x
2
’ c
3
= ’ y
3
’ b
1
= ’ z
1
’ min c
1
, c
2
, c
3
min c
1
, c
2
, c
3,
b
1
u
1
1 0 0 1 u
2
1 1 1 1 1
1 u
3
0 1 1 0 u
4
1 0 0 0 u
5
0 1 0 0 u
6
1 0 0 1 u
7
0 1 1 1 u
8
1 1 1 1 1
1 ∑ 5 5 4 5
2 2
1 2
3 1
8 1
2 3
m in ,
, ,
2 ,
1, m in
, ,
2 c c
c b C B
c c c
δ =
= =
∑ ∑
1 2
3 1
8 1
m in ,
, ,
2 ,
. m in
5 c c
c b B C
b δ
= =
∑ ∑
LAMPIRAN 2 Perhitungan Tabel 5 untuk Contoh 4
Derajat kemiripan antara setiap himpunan fuzzy hasil dari fuzzy partition dengan data crisp yang ada pada Tabel 5 akan dihitung dengan menggunakan FCPR sebagai berikut :
R
E
he, N =
| |
| |
he N
N ∩
=
min0,1 1
= 0, R
E
he, ES =
| |
| |
he ES
ES ∩
=
min0,1 1
= 0, R
E
he, JHS =
| |
| |
he JHS
JHS ∩
=
min0,1 1
= 0, R
E
he, SHS =
| |
| |
he SHS
SHS ∩
=
min0.1,1 1
= 0.1, R
E
he, BA =
| |
| |
he BA
BA ∩
=
min0.8,1 1
= 0.8, R
E
he, MS =
| |
| |
he MS
MS ∩
=
min1,1 1
= 1, R
E
he, PhD =
| |
| |
he PhD
PhD ∩
=
min1,1 1
= 1, R
E
me, N =
| |
| |
me N
N ∩
=
min0,1 1
= 0, R
E
me, ES =
| |
| |
me ES
ES ∩
=
min0, 0.2 1
= 0.2, R
E
me, JHS =
| |
| |
me JHS
JHS ∩
=
min0.5,1 1
= 0.5,
R
E
me, SHS =
| |
| |
me SHS
SHS ∩
=
min0.9,1 1
= 0.9, R
E
me, BA =
| |
| |
me BA
BA ∩
=
min0.2,1 1
= 0.2, R
E
me, MS =
| |
| |
me MS
MS ∩
=
min0,1 1
= 0, R
E
me, PhD =
| |
| |
me PhD
PhD ∩
=
min0,1 1
= 0, R
E
le, N =
| |
| |
le N
N ∩
=
min1,1 1
= 1, R
E
le, ES =
| |
| |
le ES
ES ∩
=
min0.8,1 1
= 0.8, R
E
le, JHS =
| |
| |
le JHS
JHS ∩
=
min0.5,1 1
= 0.5, R
E
le, SHS =
| |
| |
le SHS
SHS ∩
=
min0,1 1
= 0, R
E
le, BA =
| |
| |
le BA
BA ∩
=
min0,1 1
= 0, R
E
le, MS =
| |
| |
le MS
MS ∩
=
min0,1 1
= 0, R
E
le, PhD =
| |
| |
le PhD
PhD ∩
=
min0,1 1
= 0, R
S
hs, 100 =
| 100 |
| 100 | hs
∩
=
min0,1 1
= 0, R
S
hs, 125 =
| 125 |
| 125 | hs
∩
=
min0,1 1
= 0, R
S
hs, 150 =
| 150 |
| 150 | hs
∩
=
min0,1 1
= 0, R
S
hs, 175 =
| 175 |
| 175 | hs
∩
=
min0,1 1
= 0, R
S
hs, 200 =
| 200 |
| 200 | hs
∩
=
min0,1 1
= 0, R
S
hs, 250 =
| 250 |
| 250 | hs
∩
=
min0,1 1
= 0, R
S
hs, 255 =
| 255 |
| 255 | hs
∩
=
min0.1,1 1
= 0.1, R
S
hs, 275 =
| 275 |
| 275 | hs
∩
=
min0.5,1 1
= 0.5, R
S
hs, 300 =
| 300 |
| 300 | hs
∩
=
min1,1 1
= 1, R
S
hs, 315 =
| 315 |
| 315 | hs
∩
=
min1,1 1
= 1, R
S
hs, 340 =
| 340 |
| 340 | hs
∩
=
min1,1 1
= 1, R
S
hs, 350 =
| 350 |
| 350 | hs
∩
=
min1,1 1
= 1, R
S
hs, 355 =
| 355 |
| 355 | hs
∩
=
min1,1 1
= 1, R
S
hs, 360 =
| 360 |
| 360 | hs
∩
=
min1,1 1
= 1, R
S
hs, 374 =
| 374 |
| 374 | hs
∩
=
min1,1 1
= 1, R
S
hs, 400 =
| 300 |
| 300 | hs
∩
=
min1,1 1
= 1, R
S
hs, 415 =
| 415 |
| 415 | hs
∩
=
min1,1 1
= 1, R
S
hs, 420 =
| 420 |
| 420 | hs
∩
=
min1,1 1
= 1, R
S
hs, 470 =
| 470 |
| 470 | hs
∩
=
min1,1 1
= 1, R
S
hs, 500 =
| 500 |
| 500 | hs
∩
=
min1,1 1
= 1, R
S
ms, 100 =
| 100 |
| 100 | ms
∩
=
min0,1 1
= 0, R
S
ms, 125 =
| 125 |
| 125 | ms
∩
=
min0.5,1 1
= 0.5, R
S
ms, 150 =
| 150 |
| 150 | ms
∩
=
min1,1 1
= 1, R
S
ms, 175 =
| 175 |
| 175 | ms
∩
=
min1,1 1
= 1, R
S
ms, 200 =
| 200 |
| 200 | ms
∩
=
min1,1 1
= 1, R
S
ms, 250 =
| 250 |
| 250 | ms
∩
=
min1,1 1
= 1, R
S
ms, 255 =
| 255 |
| 255 | ms
∩
=
min0.9,1 1
= 0.9, R
S
ms, 275 =
| 275 |
| 275 | ms
∩
=
min0.5,1 1
= 0.5, R
S
ms, 300 =
| 300 |
| 300 | ms
∩
=
min0,1 1
= 0, R
S
ms, 315 =
| 315 |
| 315 | ms
∩
=
min0,1 1
= 0, R
S
ms, 340 =
| 340 |
| 340 | ms
∩
=
min0,1 1
= 0, R
S
ms, 350 =
| 350 |
| 350 | ms
∩
=
min0,1 1
= 0, R
S
ms, 355 =
| 355 |
| 355 | ms
∩
=
min0,1 1
= 0,
R
S
ms, 360 =
| 360 |
| 360 | ms
∩
=
min0,1 1
= 0, R
S
ms, 374 =
| 374 |
| 374 | ms
∩
=
min0,1 1
= 0, R
S
ms, 400 =
| 300 |
| 300 | ms
∩
=
min0,1 1
= 0, R
S
ms, 415 =
| 415 |
| 415 | ms
∩
=
min0,1 1
= 0, R
S
ms, 420 =
| 420 |
| 420 | ms
∩
=
min0,1 1
= 0, R
S
ms, 470 =
| 470 |
| 470 | ms
∩
=
min0,1 1
= 0, R
S
ms, 500 =
| 500 |
| 500 | ms
∩
=
min0,1 1
= 0, R
S
ls, 100 =
| 100 |
| 100 | ls
∩
=
min1,1 1
= 1, R
S
ls, 125 =
| 125 |
| 125 | ls
∩
=
min0.5,1 1
= 0.5, R
S
ls, 150 =
| 150 |
| 150 | ls
∩
=
min0,1 1
= 0, R
S
ls, 175 =
| 175 |
| 175 | ls
∩
=
min0,1 1
= 0, R
S
ls, 200 =
| 200 |
| 200 | ls
∩
=
min0,1 1
= 0, R
S
ls, 250 =
| 250 |
| 250 | ls
∩
=
min0,1 1
= 0, R
S
ls, 255 =
| 255 |
| 255 | ls
∩
=
min0,1 1
= 0, R
S
ls, 275 =
| 275 |
| 275 | ls
∩
=
min0,1 1
= 0, R
S
ls, 300 =
| 300 |
| 300 | ls
∩
=
min0,1 1
= 0, R
S
ls, 315 =
| 315 |
| 315 | ls
∩
=
min0,1 1
= 0, R
S
ls, 340 =
| 340 |
| 340 | ls
∩
=
min0,1 1
= 0, R
S
ls, 350 =
| 350 |
| 350 | ls
∩
=
min0,1 1
= 0, R
S
ls, 355 =
| 355 |
| 355 | ls
∩
=
min0,1 1
= 0, R
S
ls, 360 =
| 360 |
| 360 | ls
∩
=
min0,1 1
= 0, R
S
ls, 374 =
| 374 |
| 374 | ls
∩
=
min0,1 1
= 0, R
S
ls, 400 =
| 300 |
| 300 | ls
∩
=
min0,1 1
= 0, R
S
ls, 415 =
| 415 |
| 415 | ls
∩
=
min0,1 1
= 0, R
S
ls, 420 =
| 420 |
| 420 | ls
∩
=
min0,1 1
= 0, R
S
ls, 470 =
| 470 |
| 470 | ls
∩
=
min0,1 1
= 0, R
S
ls, 500 =
| 500 |
| 500 | ls
∩
=
min0,1 1
= 0.
LAMPIRAN 3 Perhitungan Tabel 5
Tabel 25 Transformasi Tabel 5
U le me he ls ms hs min
le, ls min
le, ms min
le, hs min
me, ls
u
01
0 0 1 0 0 1
0 0 0 0 u
02
0 0.9 0.1 0 1 0
0 0 0 0 u
03
0 0 1 0 0 1
0 0 0 0 u
04
0.5 0.5 0 0 1 0
0 0.5 0 0 u
05
0.8 0.2 0 0.5 0.5 0
0.5 0.5 0 0.2 u
06
0 0.9 0.1 0 1 0
0 0 0 0 u
07
0 0 1 0 0 1
0 0 0 0 u
08
0 0.9 0.1 0 1 0
0 0 0 0 u
09
0 0 1 0 0 1
0 0 0 0 u
10
0 0.9 0.1 0 0.5 0.5
u
11
1 0 0 1 0 0
1 0 0 0 u
12
0 0.9 0.1 0 0 1
0 0 0 0 u
13
0 0.2 0.8 1
0 0 u
14
0 0.9 0.1 0 0 1 u
15
0 0.2 0.8 0 0 1 u
16
0 0.9 0.1 0 1 0 u
17
0 0.2 0.8 0 0 1 u
18
0 0 1 0 0 1 0 u
19
0.8 0.2 0 0.5 0.5 0 0.5 0.5
0.2 u
20
0 0.9 0.1 0 1 0 u
21
0 0.2 0.8 0 0 1 u
22
0 0.9 0.1 0 0.9 0.1 0 u
23
0 0.2 0.8 0 0 1 u
24
0 0.9 0.1 0 1 0 ∑ 3.1 10.9 10 2 9.4 12.6 2
1.5 0.4
Lanjutan Tabel 25
U min
me, ms min
me, hs min
he, ls min
he, ms min
he, hs min
ls, le min
ls, me min
ls, he
u
01
0 0 0 0 1 0 0 0 u
02
0.9 0 0 0.1 0 0 0 0 u
03
0 0 0 0 1 0 0 0 u
04
0.5 0 0 0 0 0 0 0 u
05
0.2 0 0 0 0 0.5 0.2 0
u
06
0.9 0 0 0.1 0 0 0 0 u
07
0 0 1
0 0 0 u
08
0.9 0 0 0.1 0 0 0 0 u
09
0 0 0 0 1 0 0 0 u
10
0.5 0.5 0.1
0.1 0 0 0
u
11
0 0 0 0 0 1 0 0 u
12
0 0.9 0 0 0.1 0 0 0 u
13
0 0.2 0 0 0.8 0 0 0 u
14
0 0.9 0.1
0 0 0 u
15
0 0.2 0 0 0.8 0 0 0 u
16
0.9 0 0 0.1 0 0 0 0 u
17
0 0.2 0.8
0 0 0 u
18
0 0 0 0 1 0 0 0 u
19
0.2 0 0 0 0 0.5 0.2 0
u
20
0.9 0 0 0.1 0 0 0 0 u
21
0 0.2 0 0 0.8 0 0 0 u
22
0.9 0.1 0 0.1 0.1 0 0 0 u
23
0 0.2 0 0 0.8 0 0 0 u
24
0.9 0 0 0.1 0 0 0 0 ∑ 7.7 3.4 0 0.8 9.4 2 0.4 0
Lanjutan Tabel 25
U min
ms, le min
ms, me min
ms, he min
hs, le min
hs, me min
hs, he
u
01
0 0 0 0 1
u
02
0 0.9 0.1
0 0 0 u
03
0 0 0 0 1
u
04
0.5 0.5 0 0 0 0 u
05
0.5 0.2 0 0 0 0 u
06
0 0.9 0.1
0 0 0 u
07
0 0 1
u
08
0 0.9 0.1
0 0 0 u
09
0 0 0 0 1
u
10
0 0.5 0.1
0.5 0.1
u
11
0 0 0 0 0
u
12
0 0 0 0.9
0.1 u
13
0 0 0 0.2
0.8 u
14
0 0 0.9
0.1 u
15
0 0 0 0.2
0.8 u
16
0 0.9 0.1
0 0 0 u
17
0 0 0.2
0.8 u
18
0 0 0 0 1
u
19
0.5 0.2 0 0 0 0 u
20
0 0.9 0.1
0 0 0 u
21
0 0 0 0.2
0.8 u
22
0 0.9 0.1
0.1 0.1
u
23
0 0 0 0.2
0.8 u
24
0 0.9 0.1
0 0 0 ∑ 1.5 7.7 0.8 0 3.4 9.4
ϕ
ℜ
{le}, {ls} =
m in , 2
1, m in
2 le ls
ls =
=
∑ ∑
ϕ
ℜ
{ls},{me} =
m in , 0.4
0 .04, m in
10 .9 ls m e
m e =
=
∑ ∑
ϕ
ℜ
{me},{ls} =
m in ,
0 .4 0 .2,
m in 2
m e ls ls
= =
∑ ∑
ϕ
ℜ
{ms},{me}=
m in ,
7.7 0.71,
m in 10 .9
m s m e m e
= =
∑ ∑
ϕ
ℜ
{he},{ls} =
m in ,
0, m in
2 he ls
ls =
=
∑ ∑
ϕ
ℜ
{hs},{me} =
m in ,
3.4 0.31,
m in 10.9
hs m e m e
= =
∑ ∑
ϕ
ℜ
{ls},{le}=
m in , 2
0.6 5, m in
3 .1 ls le
le =
=
∑ ∑
ϕ
ℜ
{le},{hs}=
m in , 0,
m in 12.6
le h s h s
= =
∑ ∑
ϕ
ℜ
{ms},{le}=
m in ,
1.5 0.4 8,
m in 3.1
m s le le
= =
∑ ∑
ϕ
ℜ
{me},{hs} =
m in ,
3 .4 0.27,
m in 12 .6
m e hs h s
= =
∑ ∑
ϕ
ℜ
{hs},{le} =
m in ,
0, m in
3 .1 hs le
le =
=
∑ ∑
ϕ
ℜ
{he},{hs}=
m in ,
9.4 0 .75,
m in 12 .6
h e hs hs
= =
∑ ∑
ϕ
ℜ
{le},{ms}=
m in , 1.5
0 .16, m in
9 .4 le m s
m s =
=
∑ ∑
ϕ
ℜ
{ls},{he} =
m in , 0,
m in 10
ls he he
= =
∑ ∑
ϕ
ℜ
{me},{ms}=
m in ,
7.7 0.8 2,
m in 9 .4
m e m s m s
= =
∑ ∑
ϕ
ℜ
{ms},{he} =
m in ,
0.8 0.0 8,
m in 10
m s he he
= =
∑ ∑
ϕ
ℜ
{he},{ms} =
m in ,
0.8 0 .09,
m in 9.4
h e m s m s
= =
∑ ∑
ϕ
ℜ
{hs},{he} =
m in ,
9 .4 0.94 .
m in 1 0
h s he he
= =
∑ ∑
LAMPIRAN 4 Hasil Transformasi Tabel 8 untuk Contoh 5
Tabel 26 Relasi RN=j,C=m,G=A
N = j C = m
G = A minN= j,G=A minN=j,C=m minN=j,C=m,G=A
1 1 1 1
1 1
1 0 0 1 0 0
0 1 0 0 0 1
0 0 1 ∑ = 3
∑ = 2 ∑ = 3
∑ = 1 ∑ = 1
∑ = 1 1.
0 .1
m in ,
1 ˆ G= | N=
, m in
3 G N
Q A
j N
ℜ
= =
∑ ∑
2.
0.1
min , ,
1 ˆ G=A | N=j, C=m
1. min ,
1 G N C
Q N C
ℜ
= = =
∑ ∑
LAMPIRAN 5 Hasil Transformasi Tabel 9 untuk Contoh 6
Tabel 27 Relasi RN, G
N = j N = p
G = A G = B
Min G=A,N=j
min G=B, N=j
min G=A, N=p
min G=B,N=p
1 0 1 0 1
1 0 0 1 1
1 0 0 1 1
0 1 0 1 1
0 1 1 0 1
0 1 1 0 1
∑ = 3 ∑ = 3 ∑ = 3
∑ = 3 ∑ = 1
∑ = 2 ∑ = 2
∑ = 1 1.
0 .1
m in =
, =
1 ˆ G= | N=
, m in
= 3
G A N
j Q
A j
N j
ℜ
= =
∑ ∑
2.
0.1
m in = ,
= 2
ˆ G= | N= ,
m in =
3 G
B N j
Q B
j N
j
ℜ
= =
∑ ∑
3.
0.1
m in = ,
= 1
ˆ G= | N= ,
m in =
3 G
B N p
Q B
p N
p
ℜ
= =
∑ ∑
4.
0.1
m in =
, =
2 ˆ G= | N =
, m in
= 3
G A N
p Q
A p
N p
ℜ
= =
∑ ∑
LAMPIRAN 6 Hasil Transformasi Tabel 10 untuk Contoh 6
Tabel 28 Relasi RC, G
C=m C =b
C =k G = A
G = B min
C=m,G=A min
C=m,G=B min
C=b,G=A 1 0 0 1 0
1 0 1 0 0 1
1 0 0 1 0 1
1 1 0 0 0 1
0 0 1 1 0 1
0 1 0 1 0 1
∑ = 2 ∑ = 2 ∑ = 2 ∑ = 3
∑ = 3 ∑ = 1
∑ = 2 ∑ = 2
Lanjutan Tabel 28
min C=’b’,G=’B’
min C=b,G=A
min C=b,G=B
0 0 0 0 0 0
0 0 0 1 0 1
0 1 0 0 1 0
∑ = 1 ∑ = 2
∑ = 1 1.
0.1
m in =
, =
1 ˆ = |
= ,
m in =
2 G
A C m
Q G
A C
m C
m
ℜ
= =
∑ ∑
2.
0 .1
m in =
, =
1 ˆ = |
= ,
m in =
2 G
A C b
Q G
A C
b C
b
ℜ
= =
∑ ∑
3.
0.1
m in =
, =
1 ˆ = |
= ,
m in =
2 G
A C k
Q G
A C
k C
k
ℜ
= =
∑ ∑
4.
0.1
m in = ,
= 1
ˆ = | =
, m in
= 2
G B C
m Q
G B
C m
C m
ℜ
= =
∑ ∑
5.
0.1
m in = ,
= 1
ˆ = | =
, m in
= 2
G B C
k Q
G B
C k
C k
ℜ
= =
∑ ∑
6.
0.1
m in = ,
= 1
ˆ = | =
. m in
= 2
G B C
b Q
G B
C b
C b
ℜ
= =
∑ ∑
LAMPIRAN 7 Hasil Transformasi Tabel 4 untuk Contoh 7 dan Contoh 8
Tabel 29 RU, he, hs
U he hs minhe,hs maxhe,hs
u
01
1 1 1
1 u
02
0.1 0 0 0.1
u
03
1 1 1
1 u
04
0 0 u
05
0 0 u
06
0.1 0 0 0.1
u
07
1 1 1
1 u
08
0.1 0 0 0.1
u
09
1 1 1
1 u
10
0.1 0.5 0.1
0.5 u
11
0 0 u
12
0.1 1 0.1
1 u
13
0.8 1
0.8 1
u
14
0.1 1 0.1
1 u
15
0.8 1 0.8
1 u
16
0.1 0 0.1
u
17
0.8 1 0.8
1 u
18
1 1 1
1 u
19
0 0 u
20
0.1 0 0.1
u
21
0.8 1 0.8
1 u
22
0.1 0.1 0.1
0.1 u
23
0.8 1 0.8
1 u
24
0.1 0 0.1
∑ 10 12.6 9.4 13.2
LAMPIRAN 8 Perhitungan dengan Menggunakan Persamaan 27 yaitu Peluang Query untuk Objek u
i
dengan Diberikan “he AND hs”
ˆ Q
ℜ
u
1
|he,hs =
1
min ,
, 1
0.106, min
, 9.4
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
2
|he,hs =
2
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
3
|he,hs =
3
min ,
, 1
0.106, min
, 9.4
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
4
|he,hs =
4
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
5
|he,hs =
5
min ,
, 0,
min ,
9.4 u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
6
|he,hs =
6
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
7
|he,hs =
7
min ,
, 1
0.106, min
, 9.4
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
8
|he,hs =
8
min ,
, 0,
min ,
9.4 u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
9
|he,hs =
9
min ,
, 1
0.106, 9.4
min ,
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
10
|he,hs =
10
min ,
, 0.1
0.0106, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
12
|he,hs =
12
min ,
, 0.1
0.0106, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
13
|he,hs =
13
min ,
, 0.8
0.085, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
14
|he,hs =
14
min ,
, 0.1
0.0106 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
15
|he,hs =
15
min ,
, 0.8
0.085, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
16
|he,hs =
16
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
17
|he,hs =
17
min ,
, 0.8
0.085, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
18
|he,hs =
18
min ,
, 1
0.106, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
19
|he,hs =
19
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
20
|he,hs =
20
min ,
, 0,
min ,
9.4 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
21
|he,hs =
21
min ,
, 0.8
0.085, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
22
|he,hs =
22
min ,
, 0.1
0.0106, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
23
|he,hs =
23
min ,
, 0.8
0.085, 9.4
min ,
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
24
|he,hs =
24
min ,
, 0.
min ,
9.4 u
he hs he hs
= =
∑ ∑
LAMPIRAN 9 Perhitungan dengan Menggunakan Persamaan 27 yaitu Peluang Query untuk “he AND
hs” dengan Diberikan Objek u
i
ˆ Q
ℜ
he,hs |u
1
=
1 1
min ,
, 1
1, min
1 u he hs
u = =
∑ ∑
ˆ Q
ℜ
he,hs |u
2
=
2 2
min ,
, 0,
min 1
u he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
3
=
3 3
min ,
, 1
1, min
1 u he hs
u = =
∑ ∑
ˆ Q
ℜ
he,hs |u
4
=
4 4
min ,
, 0,
min 1
u he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs| u
5
=
5 5
min ,
, 0,
min 1
u he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
6
=
6 6
min ,
, 0,
min 1
u he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
7
=
7 7
min ,
, 1
1, min
1 u he hs
u = =
∑ ∑
ˆ Q
ℜ
he,hs |u
8
=
8 8
min ,
, 0,
min 1
u he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
9
=
9 9
min ,
, 1
1, min
1 u he hs
u = =
∑ ∑
ˆ Q
ℜ
he,hs |u
10
=
10 10
min ,
, 0.1
0.1, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
11
=
10 10
min ,
, 0,
min 1
u he hs
u = =
∑ ∑
ˆ Q
ℜ
he,hs |u
12
=
12 12
min ,
, 0.1
0.1, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
13
=
13 13
min ,
, 0.8
0.8, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
14
=
14 14
min ,
, 0.1
0.1, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
15
=
15 15
min ,
, 0.8
0.8, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
16
=
16 16
min ,
, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
17
=
17 17
min ,
, 0.8
0.8, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
18
=
18 18
min ,
, 1
1, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
19
=
19 19
min ,
, 0,
min 1
u he hs
u =
=
∑ ∑
ˆ Q
ℜ
he,hs |u
20
=
20 20
min ,
, 0,
min 1
u he hs
u =
=
∑ ∑
ˆ Q
ℜ
he,hs |u
21
=
21 21
min ,
, 0.8
0.8, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
22
=
22 22
min ,
, 0.1
0.1, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
23
=
23 23
min ,
, 0.8
0.8, min
1 u
he hs u
= =
∑ ∑
ˆ Q
ℜ
he,hs |u
24
=
24 24
min ,
, 0.
min 1
u he hs
u =
=
∑ ∑
LAMPIRAN 10 Perhitungan dengan Menggunakan Persamaan 30 yaitu Peluang Query Objek u
i
dengan Diberikan “he OR hs”
ˆ Q
ℜ
u
1
|he,hs =
1
maxmin ,
, 1
0.076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
2
|he,hs =
2
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
3
|he,hs =
3
maxmin ,
, 1
0.076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
4
|he,hs =
4
maxmin ,
, 0,
max ,
13.2 u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
5
|he,hs =
5
maxmin ,
, 0,
max ,
13.2 u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
6
|he,hs =
6
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
7
|he,hs =
7
maxmin ,
, 1
0.076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
8
|he,hs =
8
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
9
|he,hs =
9
maxmin ,
, 1
0.076, max
, 13.2
u he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
10
|he,hs =
10
maxmin ,
, 0.5
0.0379, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
11
|he,hs =
11
maxmin ,
, 0,
max ,
13.2 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
12
|he,hs =
12
maxmin ,
, 1
, max
, 13.2
u he hs
he hs =
∑ ∑
ˆ Q
ℜ
u
13
|he,hs =
13
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
14
|he,hs =
14
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
15
|he,hs =
15
min ,
, 1
0.076, min
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
16
|he,hs =
16
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
17
|he,hs =
17
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
18
|he,hs =
18
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
19
|he,hs =
19
maxmin ,
, 0,
max ,
13.2 u
he hs he hs
= =
∑ ∑
ˆ Q
ℜ
u
20
|he,hs =
20
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
21
|he,hs =
21
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
22
|he,hs =
22
maxmin ,
, 0.1
0.0076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
23
|he,hs =
23
maxmin ,
, 1
0.076, max
, 13.2
u he hs
he hs =
=
∑ ∑
ˆ Q
ℜ
u
24
|he,hs =
24
maxmin ,
, 0.1
0.0076. max
, 13.2
u he hs
he hs =
=
∑ ∑
LAMPIRAN 11 Perhitungan dengan Menggunakan Persamaan 30 yaitu Peluang Query untuk “he OR hs”
dengan Diberikan Objek u
i
ˆ Q
ℜ
he,hs |u
1
=
1 1
maxmin ,
, max
u he hs u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
2
=
2 2
maxmin ,
, max
u he hs u
∑ ∑
= 0.1,
ˆ Q
ℜ
he,hs |u
3
=
3 3
maxmin ,
, max
u he hs u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
4
=
4 4
maxmin ,
, max
u he hs u
∑ ∑
= 0,
ˆ Q
ℜ
he,hs |u
5
=
5 5
maxmin ,
, max
u he hs u
∑ ∑
= 0,
ˆ Q
ℜ
he,hs |u
6
=
6 6
maxmin ,
, max
u he hs u
∑ ∑
= 0.1,
ˆ Q
ℜ
he,hs |u
7
=
7 7
maxmin ,
, max
u he hs u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
8
=
8 8
maxmin ,
, max
u he hs u
∑ ∑
= 0.1,
ˆ Q
ℜ
he,hs |u
9
=
9 9
maxmin ,
, max
u he hs u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
10
=
10 10
maxmin ,
, max
u he hs
u
∑ ∑
= 0.5,
ˆ Q
ℜ
he,hs |u
11
=
11 11
maxmin ,
, max
u he hs
u
∑ ∑
= 0,
ˆ Q
ℜ
he,hs |u
12
=
12 12
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
13
=
13 13
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
14
=
14 14
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
15
=
15 15
min ,
, min
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
16
=
16 16
maxmin ,
, max
u he hs
u
∑ ∑
= 0.1,
ˆ Q
ℜ
he,hs |u
17
=
17 17
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
18
=
18 18
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
he,hs |u
19
=
19 19
maxmin ,
, max
u he hs
u
∑ ∑
= 0,
ˆ Q
ℜ
he,hs |u
20
=
20 20
maxmin ,
, max
u he hs
u
∑ ∑
= 0.1,
ˆ Q
ℜ
u
21
|he,hs =
21 21
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
u
22
|he,hs =
22 22
maxmin ,
, max
u he hs
u
∑ ∑
= 0.1,
ˆ Q
ℜ
u
23
|he,hs =
23 23
maxmin ,
, max
u he hs
u
∑ ∑
= 1,
ˆ Q
ℜ
u
24
|he,hs =
24 24
maxmin ,
, max
u he hs
u
∑ ∑
= 0.1.
APLIKASI RELASI PELUANG BERSYARAT FUZZY PADA SISTEM INFORMASI FUZZY
NIKEN WIDIASTUTI
DEPARTEMEN MATEMATIKA FAKULTAS MATEMATIKA DAN ILMU PENGETAHUN ALAM
INSTITUT PERTANIAN BOGOR BOGOR
2008
ABSTRAK
NIKEN WIDIASTUTI . Aplikasi Relasi Peluang Bersyarat Fuzzy pada Sistem Informasi Fuzzy.
Dibimbing oleh SRI NURDIATI dan N. K. KUTHA ARDANA.
Dalam kehidupan sehari-hari seringkali ditemui suatu fenomena yaitu data mengandung sesuatu yang tidak akurat. Data yang tidak akurat tersebut dapat berupa kata-kata manusia yang
bersifat relatif. Pada kasus ini, himpunan fuzzy dapat digunakan untuk merepresentasikan data yang tidak akurat tersebut dengan derajat keakuratan data yang berbeda. Pada tulisan ini
diperkenalkan relasi peluang bersyarat fuzzy atau fuzzy conditional probability relations FCPR yang digunakan untuk merepresentasikan relasi kemiripan antara dua himpunan fuzzy yang tidak
perlu simetris atau transitif. Konsep FCPR yang dibahas difokuskan pada relasi kemiripan yang lemah yang merupakan tipe khusus pada relasi fuzzy biner dengan perumuman relasi kemiripan.
Sistem informasi fuzzy yang digunakan adalah tabel data fuzzy sederhana yang merupakan aplikasi dari knowledge discovery and data mining KDD. Dengan memanfaatkan derajat dari dasar
kemiripan FCPR, aplikasi FCPR yaitu konsep Į-objek redundan, ketergantungan atribut, pendekatan data reduksi dan proyeksi, dan pendekatan data query. Perhitungan berdasarkan FCPR
berguna untuk menentukan derajat kemiripan dari dua kata yang tidak perlu simetris atau transitif. Konsep Į-objek redundan sangat penting untuk mereduksi angka dari aturan keputusan dengan
adanya tabel keputusan. Konsep ketergantungan atribut berdasarkan FCPR sangat penting untuk manganalisa ketergantungan dari atribut. Aplikasi pendekatan data reduksi dan proyeksi
digunakan untuk menemukan relasi di antara anak himpunan fuzzy dari partisi fuzzy dan menghasilkan fuzzy integrity constraints. Aplikasi pendekatan data query digunakan untuk
menghasilkan relasi fuzzy query dengan adanya tabel keputusan.
Kata kunci : Relasi peluang bersyarat fuzzy, fungsi ketergantungan fuzzy FFD, fuzzy integrity constraints FIC, knowledge discovery and data mining KDD, data query.
ABSTRACT
NIKEN WIDIASTUTI . The Applications of Fuzzy Conditional Probability Relations in Fuzzy
Information Systems. Supervised by SRI NURDIATI and N. K. KUTHA ARDANA.
In our daily life we often find a phenomenon related to an imprecise data. The imprecise data can be in the form of relative human words. In this case, fuzzy sets can be used to represent the
imprecise data in which preciseness degrees of data are intuitively different. This paper introduced a
fuzzy conditional probability relations FCPR which is used to represent a similarity relation between two fuzzy sets, The two fuzzy sets may not necessarily be symmetric or transitive. The
concept of FCPR which is explained in this paper was focused on a weak relation similarity which turned out to be a special type of fuzzy binary relation generalizing similarity relation. Fuzzy
information system which is used in this paper was a simple fuzzy data table that was an application of
knowledge discovery and data mining KDD. By using degrees of similarity FCPR, the application of FCPR consists of Į-redundant object concept, a concept of dependency
of attribute, approximate data reduction and projection, and approximate data query.
Calculation based on FCPR is used to determine degrees of similarity which
may not necessarily be symmetric or transitive. The concept of
Į -redundant object is very important for the purpose of reducing the
number of decision rules concerning a decision table. The concept of dependency attribute based on the FCPR is very important for purpose of analyzing dependency of attribute. Application of
approximate data reduction and projection is used to find relations among fuzzy subsets as results of fuzzy partition and provide fuzzy integrity constraints. Application of approximate data query is
used to design a fuzzy query relation in which the present the decision table.
Keywords : F uzzy conditional probability relations
, fuzzy functional dependency FFD, fuzzy integrity constraints FIC, knowledge discovery and data mining KDD, data
query.
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