Syarat :
C1: Semester Aktif Perkuliahan (Attribut Keuntungan)
C2: IPK (Attribut Keuntungan)
C3: Penghasilan Orang Tua (Attribut Biaya)
C4: Aktif Berorganisasi (Attribut Keuntungan)
Untuk bobot W=[3,4,5,4]
Adapun mahasiswa yang menjadi alternatif dalam pemberian beasiswa yaitu :
| No | Nama | C1 | C2 | C3 | C4 |
| 1 | Joko | VI | 3.7 | 1.850.000 | Aktif |
| 2 | Widodo | VI | 3.5 | 1.500.000 | Aktif |
| 3 | Simamora | VIII | 3.8 | 1.350.000 | Tidak Aktif |
| 4 | Susilawati | II | 3.9 | 1.650.000 | Tidak Aktif |
| 5 | Dian | IV | 3.6 | 2.300.000 | Aktif |
| 6 | Roma | IV | 3.3 | 2.250.000 | Aktif |
| 7 | Hendro | VI | 3.4 | 1.950.000 | Aktif |
Untuk pembobotan yang digunakan bisa mengacu pada bobot di bawah ini :
C1:Semester Aktif Perkuliahan
Semester II --> 1
Semester IV --> 2
Semester VI --> 3
Semester VIII --> 4
C2: IPK
IPK 3.00 - 3.249 --> 1
IPK 3.25 - 3.499 --> 2
IPK 3.50 - 3.749 --> 3
IPK 3.75 - 3.999 --> 4
IPK 4.00 --> 5
C3: Penghasilan Orang Tua
1.000.000 --> 1
1.400.000 --> 2
1.800.000 --> 3
2.200.000 --> 4
2.600.000 --> 5
C4: Aktif Berorganisasi
Aktif --> 2
Tidak Aktif --> 1
Penyelesaian :
1. Table pembobotan
No
|
Nama
|
C1
|
C2
|
C3
|
C4
|
1
|
Joko
|
3
|
3
|
3
|
2
|
2
|
Widodo
|
3
|
3
|
2
|
2
|
3
|
Simamora
|
2
|
4
|
1
|
1
|
4
|
Susilawati
|
1
|
4
|
2
|
1
|
5
|
Dian
|
1
|
3
|
4
|
2
|
6
|
Roma
|
2
|
2
|
4
|
2
|
7
|
Hendro
|
4
|
2
|
3
|
2
|
2. Mencari jumlah kriteria dengan matrik r
Maka Matrik R :
0,4522
|
0,3665
|
0,3905
|
0,4264
|
0,4522
|
0,3665
|
0,2603
|
0,4264
|
0,3015
|
0,4886
|
0,1301
|
0,2132
|
0,1507
|
0,4886
|
0,2603
|
0,2132
|
0,1507
|
0,3665
|
0,5207
|
0,4264
|
0,3015
|
0,2443
|
0,5207
|
0,4264
|
0,6030
|
0,2443
|
0,3905
|
0,4264
|
3. Mencari nilai y dengan cara mengalikan dengan nilai bobot. W = [4,4,5,3]
Y11 = 4(0,4522) = 1,8090
Y21 = 4(0,4522) = 1,8090
Y31 = 4(0,3015) = 1,2060
Y41 = 4(0,1507) = 0,6030
Y51 = 4(0,1507) = 0,6030
Y61 = 4(0,3015) = 1,2060
Y71 = 4(0,6030) = 2,4120
Y12 = 4(0,6030) = 1,4660
Y22 = 4(0,3665) = 1,4660
Y32 = 4(0,4886) = 1,9547
Y42 = 4(0,4886) = 1,9547
Y52 = 4(0,3665) = 1,4660
Y62 = 4(0,2443) = 0,9773
Y72 = 4(0,2443) = 0,9773
Y13 = 5(0,3905) = 1,9528
Y23 = 5(0,2603) = 1,3018
Y33 = 5(0,1301) = 1,6509
Y43 = 5(0,2603) = 1,3018
Y53 = 5(0,5207) = 2,6037
Y63 = 5(0,5207) = 2,6037
Y73 = 5(0,3905) = 1,9528
Y14 = 3(0,4264) = 1,2792
Y24 = 3(0,4264) = 1,2792
Y34 = 3(0,2132) = 0,6396
Y44 = 3(0,2132) = 0,6396
Y54 = 3(0,4264) = 1,2792
Y64 = 3(0,4264) = 1,2792
Y74 = 3(0,4264) = 1,2792
Maka Matrik Y :
1,8090
|
1,4660
|
1,9528
|
1,2792
|
1,8090
|
1,4660
|
1,3018
|
1,2792
|
1,2060
|
1,9547
|
0,6509
|
0,6396
|
0,6030
|
1,9547
|
1,3018
|
0,6396
|
0,6030
|
1,4660
|
2,6037
|
1,2792
|
1,2060
|
0,9773
|
2,6037
|
1,2792
|
2,4120
|
0,9773
|
1,9528
|
1,2792
|
4. Mencari nilai Y (Max)
Y1+ = Max {1,8090; 1,8090; 1,2060; 0,6030; 0,6030; 1,2060; 2,4120} = 2,4120
Y2+ = Max {1,4660; 1,4660; 1,9547; 1,9547; 1,4660; 0,9773; 0,9773} = 1,9547
Y3+ = Max {1,9528; 1,3018; 0,6509; 1,3018; 2,6037; 2,6037; 1,9528} = 2,6037
Y4+ = Max {1,2792; 1,2792; 0,6396; 0,6396; 1,2792; 1,2792; 1,2792} = 1,2792
5. Mencari nilai Y (Min)
Y1- = Min {1,8090; 1,8090; 1,2060; 0,6030; 0,6030; 1,2060; 2,4120} = 0,6030
Y2- = Min {1,4660; 1,4660; 1,9547; 1,9547; 1,4660; 0,9773; 0,9773} = 0,9773
Y3- = Min {1,9528; 1,3018; 0,6509; 1,3018; 2,6037; 2,6037; 1,9528} = 0,6509
Y4- = Min {1,2792; 1,2792; 0,6396; 0,6396; 1,2792; 1,2792; 1,2792} = 0,6396
6. Mencari nilai jarak antara alternative dengan solusi ideal
V1 = 1,5893/(1,5893+1,5157)=0,5118
V2 = 1,9486/(1,9486+1,0130)=0,6579
V3 = 2,2654/(2,2654+1,3651)=0,6239
V4 = 1,6279/(1,6279+2,0262)=0,4455
V5 = 0,8049/(0,8049+2,7064)=0,2292
V6 = 0,8790/(0,8790+2,4946)=0,2605
V7 = 2,0262/(2,0262+1,6279)=0,5544
Maka, yang berhak mendapatkan Beasiswa adalah V1, V2, V3, V4, dan V7
