Tabel Rating
Kecocokan Alternaif
Alternatif
|
Kriteria
|
|||
C1
|
C2
|
C3
|
C4
|
|
ALDYAN
|
2
|
2
|
3
|
1
|
HENDRO
|
3
|
4
|
1
|
2
|
JOKO
|
2
|
5
|
1
|
2
|
DONI
|
2
|
3
|
2
|
2
|
DONO
|
3
|
4
|
4
|
2
|
KASINO
|
2
|
3
|
2
|
2
|
SUSANTO
|
1
|
5
|
5
|
1
|
Bobot W=[ 3,5,4,4
]
Dimana : C1=Jumlah Tanggungan(Attribut
Keuntungan)
C2=Pendapatan Kepala Keluarga(Attribut Biaya)
C3=Luas Bangunan Rumah(Attribut Biaya)
C4=Memiliki KK(Attribut Keuntungan)
Penyelesaian:
Membuat
matrik ternormalisasi
Untuk C1
X1 = √22+32+22+22+32+22+12
=
√35
= 5.9160
R11 = 2 / 5.9160
= 0.3380
R21 = 3 / 5.9160
= 0.5070
R31 = 2 / 5.9160
= 0.3380
R41 = 2 / 5.9160
= 0.3380
R51 = 3 / 5.9160
= 0,5070
R61 = 2 / 5.9160
= 0.3380
R71 = 1 / 5.9160
= 0.1690
Untuk C2
X2 = √22+42+52+32+42+32+52
=
√104
= 10.1980
R12 = 2 / 10.1980
= 0.1961
R22 = 4 / 10.1980
= 0.3922
R32 = 5 / 10.1980
= 0.4902
R42 = 3 / 10.1980
= 0.2941
R52 = 4 / 10.1980
= 0.3922
R62 = 3 / 10.1980
=
0.2941
R72 = 5 / 10.1980
=
0.4902
Untuk C3
X3 = √32+12+12+22+42+22+52
=
√60
= 7.7459
R13 = 3 / 7.7459
=
0.3873
R23 = 1 / 7.7459
=
0.1291
R33 = 1 / 7.7459
=
0.1291
R43 = 2 / 7.7459
=
0.2582
R53 = 4 / 7.7459
=
0.5164
R63 = 2 / 7.7459
=
0.2582
R73 = 5 / 7.7459
=
0.6455
Untuk C4
X4 = √12+22+22+22+22+22+12
=
√22
= 4.6904
R14 = 1 / 4.6904
=
0.2132
R24 = 2 / 4.6904
=
0.4264
R34 = 2 / 4.6904
=
0.4264
R44 = 2 / 4.6904
=
0.4264
R54 = 2 / 4.6904
=
0.4264
R64 = 2 / 4.6904
=
0.4264
R74 = 1 / 4.6904
=
0.2132
Matriks R
ternormalisasi
0.3380 0.1961 0.3873 0.2132
0.5070 0.3922 0.1291 0.4264
0.3380 0.4902 0.1291 0.4264
0.3380 0.2941 0.2582 0.4264
0.5070 0.3922 0.5164 0.4264
0.3380 0.2941 0.2582 0.4264
0.1690 0.4902 0.6455 0.2132
Membuat matrik keputusan ternormalisasi terbobot
Yij =
Wi Rij dimana Bobot W=[
3, 5, 4, 4 ]
Y11
= 3(0.3380) = 1.014
Y21
= 3(0.5070) = 1.521
Y31
= 3(0.3380) = 1.014
Y41
= 3(0.3380) = 1.014
Y51
= 3(0.5070) = 1.521
Y61
= 3(0.3380) = 1.014
Y71
=3(0.1690) = 0.507
Y12=5(0.1961) = 0.9805
Y22=5(0.3922) =1.961
Y32=5(0.4902) = 2.451
Y42=5(0.2941) = 1.4705
Y52=5(0.3922) =1.961
Y62=5(0.2941) =1.4705
Y72=5(0.4902) =2.451
Y13=4(0.3873) =1.5492
Y23=4(0.1291) =0.5164
Y33=4(0.1291) =0.5164
Y43=4(0.2582) = 1.0328
Y53=4(0.5164) = 2.0656
Y63=4(0.2582) =1.0328
Y73=4(0.6455) =2.582
Y14=4(0.2132) = 0.8528
Y24=4(0.4264) = 1.7056
Y34=4(0.4264) = 1.7056
Y44=4(0.4264) = 1.7056
Y54=4(0.4264) = 1.7056
Y64=4(0.4264) = 1.7056
Y74=4(0.2132) = 0.8528
Matriks Y
ternormalisasi
1.014 0.9805 1.5492 0.8528
1.521 1.961 0.5164 1.7056
1.014
2.451 0.5164 1.7056
1.014
1.4705 1.0328 1.7056
1.521 1.961 2.0656 1.7056
1.014 1.4705 1.0328 1.7056
0.507 2.451 2.582 0.8528
Solusi
Ideal Positif (A+)
Y1 + = MAX {1.014 ; 1.521 ; 1.014 ; 1.014 ; 1.521 ; 1.014
; 0.507 }
= 1.521
Y2 + = MIN {0.9805
; 1.961 ; 2.451 ; 1.4705 ; 1.961 ; 1.4705 ; 2.451} = 0.9805
Y3 + = MIN {1.5492 ; 0.5164 ; 0.5164 ; 1.0328
; 2.0656 ; 1.0328 ; 2.582} =
0.5164
Y4 + = MAX {0.8528 ; 1.7056 ; 1.7056 ; 1.7056 ; 1.7056 ;
1.7056 ; 0.8528} = 1.7056
A+ = (1.521 ; 0.9805 ; 0.5164 ;
1.7056)
Solusi
Ideal Negatif (A- )
Y1 - = MIN {1.014
; 1.521 ; 1.014 ; 1.014 ; 1.521 ; 1.014 ; 0.507 } = 0.507
Y2 - = MAX {0.9805
; 1.961 ; 2.451 ; 1.4705 ; 1.961 ; 1.4705 ; 2.451} = 2.451
Y3 - = MAX {1.5492 ; 0.5164 ; 0.5164 ; 1.0328
; 2.0656 ; 1.0328 ; 2.582} =
2.582
Y4 - = MIN {0.8528
; 1.7056 ; 1.7056 ; 1.7056 ; 1.7056 ; 1.7056 ; 0.8528} = 0.8528
A- = (0.507 ; 2.451 ; 2.582 ; 0.8528)
Jarak
Alternatif Terbobot dengan Solusi Ideal Positif
D1+ = √(1.014-1.521)2 + (0.9805-0.9805)2 + (1.5492-0.5164)2 + (0.8528-1.7056)2
=
√2.0509
=
1.4321
D2+ = √(1.521-1.521)2 + (1.961-0.9805)2 + (0.5164-0.5164)2 + (1.7056-1.7056)2
=
√0.9613
=
0.9805
D3+ = √(1.014-1.521)2 + (2.451-0.9805)2 + (0.5164-0.5164)2 + (1.7056-1.7056)2
=
√2.4192
=
1.5554
D4+ = √(1.014-1.521)2 + (1.4705-0.9805)2 + (1.0328-0.5164)2 + (1.7056-1.7056)2
=
√0.7637
=
0.8739
D5+ = √(1.521-1.521)2 + (1.961-0.9805)2 + (2.0656-0.5164)2 + (1.7056-1.7056)2
=
√3.3613
=
1.8334
D6+ = √(1.014-1.521)2 + (1.4705-0.9805)2 + (1.0328-0.5164)2 + (1.7056-1.7056)2
=
√0.7637
=
0.8739
D7+ = √(0.507-1.521)2 + (2.451-0.9805)2 + (2.582-0.5164)2 + (0.8528-1.7056)2)
=
√8.1841
=
2.8608
Jarak
Alternatif Terbobot dengan Solusi Ideal Negatif
D1- = √(1.014-0.507)2 + (0.9805-2.451)2 + (1.5492-2.582)2 + (0.8528-0.8528)2
=
√3.4856
=
1.8671
D2- = √(1.521-0.507)2 + (1.961-2.451)2 + (0.5164-2.582)2 + (1.7056-0.8528)2
=
√6.2620
=
2.5024
D3-= √(1.014-0.507)2 + (2.451-2.451)2 + (0.5164-2.582)2 + (1.7056-0.8528)2
=
√5.2509
=
2.2915
D4- = √(1.014-0.507)2 + (1.4705-2.451)2 + (1.0328-2.582)2 + (1.7056-0.8528)2
=
√4.3455
=
2.0846
D5- = √(1.521-0.507)2 + (1.961-2.451)2 + (2.0656-2.582)2 + (1.7056-0.8528)2
=
√2.2620
=
1.5040
D6- = √(1.014-0.507)2 + (1.4705-2.451)2 + (1.0328-2.582)2 + (1.7056-0.8528)2
=
√4.3455
=
2.0846
D7- = √(0.507-0.507)2 + (2.451-2.451)2 + (2.582-2.582)2 + (0.8528-0.8528)2)
=
√0
=
0
Nilai
Preferensi untuk Setiap Alternatif (Vi)
Vi = Di- / Di- + D1+
V1 = 1.8671 / (1.8671 + 1.4321) = 0.5659
V2 = 2.5024 / (2.5024 + 0.9805) = 0.7184
V3 = 2.2915/ (2.2915 + 1.5554) = 0.5956
V4 = 2.0846/ (2.0846 + 0.8739) =
0.7046
V5 = 1.5040 / (1.5040 + 1.8334) = 0.4506
V6 = 2.0846 / (2.0846 + 0.8739) = 0.7046
V7 = 0/ (0 + 2.8608) =
0
Jadi, 5
alternatif KK yang akan mendapatkan Bantuan Langsung Tunai dari pemerintah adalah Aldyan,Hendro,Joko,Doni dan Kasino
Nama :Andika Brahmana
NPM :12110348
Kelas :TI-M1210
Nama :Andika Brahmana
NPM :12110348
Kelas :TI-M1210
