Soal Quiz Online Klik Disini
Penyelesaian:
No
|
Nama
|
C1
|
C2
|
C3
|
C4
|
1
|
Joko
|
3
|
3
|
3
|
2
|
2
|
Widodo
|
3
|
3
|
2
|
2
|
3
|
Simamora
|
4
|
4
|
1
|
1
|
4
|
Susilawati
|
1
|
4
|
2
|
1
|
5
|
Dian
|
2
|
3
|
4
|
2
|
6
|
Roma
|
2
|
2
|
4
|
2
|
7
|
Hendro
|
3
|
2
|
3
|
2
|
Bobot W=[ 3, 4, 5, 4 ]
Untuk C1
X1 = √32+32+42+12+22+22+32
= √9+9+16+1+4+4+9
= √52
= 7.2111
R11 = 3 / 7.2111
= 0.4160
R21 = 3 / 7.2111
= 0.4160
R31 = 4 / 7.2111
= 0.5547
R41 = 1 / 7.2111
= 0.1386
R51 = 2 / 7.2111
= 0.2773
R61 = 2 / 7.2111
= 0.2773
R71 = 3 / 7.2111
= 0.4160
Untuk C2
X2 = √32+32+42+42+32+22+22
= √9+9+16+16+9+4+4
= √67
= 8.1853
R12 = 3 / 8.1853
= 0.3665
R22 = 3 / 8.1853
= 0.3665
R32 = 4 / 8.1853
= 0.4886
R42 = 4 / 8.1853
= 0.4886
R52 = 3 / 8.1853
= 0.3665
R62 = 2 / 8.1853
= 0.2443
R72 = 2 / 8.1853
= 0.2443
Untuk C3
X3 = √32+22+12+22+42+42+32
= √9+4+1+4+16+16+9
= √59
= 7.6811
R13 = 3 / 7.6811
= 0.3905
R23 = 2 / 7.6811
= 0.2603
R33 = 1 / 7.6811
= 0.1301
R43 = 2 / 7.6811
= 0.2603
R53 = 4 / 7.6811
= 0.5207
R63 = 4 / 7.6811
= 0.5207
R73 = 3 / 7.6811
= 0.3905
Untuk C4
X4 = √22+22+12+12+22+22+22
= √4+4+1+1+4+4+4
= √22
= 4.6904
R14 = 2 / 4.6904
= 0.4264
R24 = 2 / 4.6904
= 0.4264
R34 = 1 / 4.6904
= 0.2132
R44 = 1 / 4.6904
= 0.2132
R54 = 2 / 4.6904
= 0.4264
R64 = 2 / 4.6904
= 0.4264
R74 = 2 / 4.6904
= 0.4264
Matriks R ternormalisasi
0.4160 0.3665 0.3905 0.4264
0.4160 0.3665 0.2603 0.4264
0.5547 0.4886 0.1301 0.2132
0.1386 0.4886 0.2603 0.2132
0.2773 0.3665 0.5207 0.4264
0.2773 0.2443 0.5207 0.4264
0.4160 0.2443 0.3905 0.4264
Yij = wiRij dimana Bobot W=[ 3, 4, 5, 4 ]
Y11 = 3(0.4160) = 1.248
Y21 = 3(0.4160) = 1.248
Y31 = 3(0.5547) = 1.6641
Y41 = 3(0.1386) = 0.4158
Y51 = 3(0.2773) = 0.8319
Y61 = 3(0.2773) = 0.8319
Y71 =3(0.4160) = 1.248
Y12=4(0.3665) = 1.466
Y22=4(0.3665) =1.466
Y32=4(0.4886) = 1.9544
Y42=4(0.4886) = 1.9544
Y52=4(0.3665) =1.466
Y62=4(0.2443) =0.9772
Y72=4(0.2443) =0.9772
Y13=5(0.3905) =1.9525
Y23=5(0.2603) =1.3015
Y33=5(0.1301) =0.6505
Y43=5(0.2603) = 1.3015
Y53=5(0.5207) = 2.6035
Y63=5(0.5207) =2.6035
Y73=5(0.3905) =1.9525
Y14=4(0.4264) = 1.7056
Y24=4(0.4264) = 1.7056
Y34=4(0.2132) = 0.8528
Y44=4(0.2132) = 0.8528
Y54=4(0.4264) = 1.7056
Y64=4(0.4264) = 1.7056
Y74=4(0.4264) = 1.7056
Matriks Y ternormalisasi
1.248 1.466 1.9525 1.7056
1.248 1.466 1.3015 1.7056
1.6641 1.9544 0.6505 0.8528
0.4158 1.9544 1.3015 0.8528
0.8319 1.466 2.6035 1.7056
0.8319 0.9772 2.6035 1.7056
1.248 0.9772 1.9525 1.7056
Solusi Ideal Positif (A+)
Y1 + = MAX {1.248 ; 1.248 ; 1.6641; 0.4158; 0.8319; 0.8319; 1.248 } = 1.6641
Y2 + = MAX {1.466 ; 1.466 ; 1.9544; 1.9544; 1.466 ; 0.9772; 0.9772} = 1.9544
Y3 + = MIN {1.9525; 1.3015; 0.6505; 1.3015; 2.6035; 2.6035; 1.9525} = 0.6505
Y4 + = MAX {1.7056; 1.7056; 0.8528; 0.8528; 1.7056; 1.7056; 1.7056} = 1.7056
A+ = (1.6641 ; 1.9544 ; 0.6505; 1.7056)
Solusi Ideal Positif (A- )
Y1 - = MIN {1.248 ; 1.248 ; 1.6641; 0.4158; 0.8319; 0.8319; 1.248 } = 0.4158
Y2 - = MIN {1.466 ; 1.466 ; 1.9544; 1.9544; 1.466 ; 0.9772; 0.9772} = 0.9772
Y3 - = MAX {1.9525; 1.3015; 0.6505; 1.3015; 2.6035; 2.6035; 1.9525} = 2.6035
Y4 - = MIN {1.7056; 1.7056; 0.8528; 0.8528; 1.7056; 1.7056; 1.7056} = 0.8528
A- = (0.4158 ; 0.9772 ; 2.6035 ; 0.8528)
Jarak Alternatif Terbobot dengan Solusi Ideal Positif
D1+ = √(1.248-1.6641)2 + (1.466-1.9544)2 + (1.9525-0.6505)2 + (1.7056-1.7056)2
= √2.1068
= 1.4514
D2+ = √(1.248-1.6641)2 + (1.466-1.9544)2 + (1.3015-0.6505)2 + (1.7056-1.7056)2
= √0.8354
= 0.9140
D3+ = √(1.6641-1.6641)2 + (1.9544-1.9544)2 + (0.6505-0.6505)2 + (0.8528-1.7056)2
= √0.7272
= 0.8527
D4+ = √(0.4158-1.6641)2 + (1.9544-1.9544)2 + (1.3015-0.6505)2 + (0.8528-1.7056)2
= √2.7093
= 1.6459
D5+ = √(0.8319-1.6641)2 + (1.466-1.9544)2 + (2.6035-0.6505)2 + (1.7056-1.7056)2
= √4.7453
= 2.1783
D6+ = √(0.8319-1.6641)2 + (0.9772-1.9544)2 + (2.6035-0.6505)2 + (1.7056-1.7056)2
= √5.4616
=2.3370
D7+ = √(1.248-1.6641)2 + (0.9772-1.9544)2 + (1.9525-0.6505)2 + (1.7056-1.7056)2)
= √2.8232
= 1.6802
Jarak Alternatif Terbobot dengan Solusi Ideal Negatif
D1- = √(1.248-0.4158)2 + (1.466-0.9772)2 + (1.9525-2.6035)2 + (1.7056-0.8528)2)
= √2.0825
= 1.4430
D2- = √(1.248-0.4158)2 + (1.466-0.9772)2 + (1.3015-2.6035)2 + (1.7056-0.8528)2)
= √3.3539
= 1.8313
D3- = √(1.6641-0.4158)2 + (1.9544-0.9772)2 + (0.6505-2.6035)2 + (0.8528-0.8528)2)
= √6.3273
= 2.5154
D4- = √(0.4158-0.4158)2 + (1.9544-0.9772)2 + (1.3015-2.6035)2 + (0.8528-0.8528)2
= √2.6501
= 1.6279
D5- = √(0.8319-0.4158)2 + (1.466-0.9772)2 + (2.6035-2.6035)2 + (1.7056-0.8528)2)
= √1.1393
= 1.0673
D6- = √(0.8319-0.4158)2 + (0.9772-0.9772)2 + (2.6035-2.6035)2 + (1.7056-0.8528)2
= √0.9004
=0.9488
D7- = √(1.248-0.4158)2 + (0.9772-0.9772)2 + (1.9525-2.6035)2 + (1.7056-0.8528)2
= √1.8436
= 1.3577
Nilai Preferensi untuk Setiap Alternatif (Vi)
V1 = 1.4430 / (1.4430 + 1.4514) = 0.4985
V2 = 1.8313 / (1.8313 + 0.9140) = 0.6670
V3 = 2.5154/ (2.5154 + 0.8527) = 0.7468
V4 = 1.6279/ (1.6279 + 1.6459) = 0.4972
V5 = 1.0673 / (1.0673 + 2.1783) = 0.3295
V6 = 0.9488 / (0.9488 + 2.3370) = 0.2887
V7 = 1.3577/ (1.3577 + 1.6802) = 0.4469
Jadi, 5 mahasiswa yang berhak untuk mendapatkan beasiswa adalah Joko, Widodo, Simamora, Susilawati, dan Hendro.
Nama : Iswahyuni
NPM : 1111706
Kelas : TI-P1101
By : Cut Iswahyuni
