Sebuah PTS, di Medan akan memberikan
beasiswa kepada 5 mahasiswa, adapun syarat pemberian beasiswa tersebut,yaitu
harus memenuhi ketentuan dibawah ini :
Syarat :
C1 = Semester Aktif (Alternatif
Keuntungan)
C2 = IPK ((Alternatif Keuntungan)
C3 = Penghasilan Orang Tua(Alternatif
Biaya)
C4 = Aktif Organisasi(Alternatif
Keuntungan)
Untuk bobot W= [4,4,5,3]
Adapun mahasiswa yang menjadi
alternative dalam pemberian beasiswa adalah :
|
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.
|
Siamamora
|
IV
|
3,8
|
1.350.000
|
Tidak Aktif
|
|
4.
|
Susilawati
|
II
|
3,9
|
1.650.000
|
Tidak Aktif
|
|
5.
|
Dian
|
II
|
3,6
|
2.300.000
|
Aktif
|
|
6.
|
Roma
|
IV
|
3,3
|
2.250.000
|
Aktif
|
|
7.
|
Hendro
|
VIII
|
3,4
|
1.900.000
|
Aktif
|
C1 C3
Semester II = 1 1.000.000
= 1
Semester IV = 2 1.400.000
= 2
Semester VI = 3 1.800.000
= 3
Semester VIII = 4 2.200.000 = 4
2.600.000
= 5
C2
IPK
= 3.00-3.249 = 1
IPK
= 3.25-3.4999=2 C4
IPK
= 3.50-3.749 = 3 Aktif
= 2
IPK
= 3.75-3.999 = 4 Tidak
Aktif = 1
IPK = 4 = 5
|
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
|
Bobot W= [4,4,5,3]
Penyelesaian : Identik
dengan
3 3
3 2 r11 r12 r13 r14
3 3 2
2 r21 r22 r23 r24
2 4 1
1 r31 r32 r33 r34
1 4 2
1 r41 r42 r43 r44
1 3 4
2 r51 r52 r53 r54
2 2 4
2 r61 r62 r63 r64
4 2 3
2 r71 r72 r73 r74
Untuk
C1 Untuk
C2
r11
= 3 =
3/4 = 0,75 r12 = 3 =
3/4 = 0.75
max {3,3,2,1,1,2,4} max {3,3,4,4,3,2,2}
r21
= 3 =3/4
= 0.75
r22
= 3 =
3/4 = 0.75
max {3,3,2,1,1,2,4} max {3,3,4,4,3,2,2}
r31
= 2 =2/4
= 0.5
r32 = 4 = 4/4 = 1
max {3,3,2,1,1,2,4} max {3,3,4,4,3,2,2}
r41
= 1
=1/4 = 0.25 r42 = 4 =
4/4 = 1
max {3,3,2,1,1,2,4} max {3,3,4,4,3,2,2}
r51
= 1 =1/4
= 0.25 r52 = 3 = 3/4 = 0,75
max {3,3,2,1,1,2,4} max {3,3,4,4,3,2,2}
r61
= 2 =
2/4 = 0.5
r62 = 2 = 2/4 = 0.5
max {3,3,2,1,1,2,4,}
max {3,3,4,4,3,2,2}
r71
= 4
=
4/4 = 1 r72 = 2 =
2/4 = 0.5
max {3,3,2,1,1,2,4,} max {3,3,4,4,3,2,2}
Untuk
C3 Untuk
C4
r13
= {3,2,1,2,4,4,3} = 4/3 = 1,333 r14
= 2 =
2/2 = 1
min 3 max {2,2,1,1,2,2,2}
r23
= {3,2,1,2,4,4,3} = 4/2 = 2
r24
= 2 =
2/2 = 1
min 2 max {2,2,1,1,2,2,2}
r33
= {3,2,1,2,4,4,3} = 4/1 = 4
r34 = 1 = 1/2 = 0,5
min 1 max {2,2,1,1,2,2,2}
r43
={3,2,1,2,4,4,3}
=
4/2 = 2 r44 = 1 =
1/2 = 0,5
min 2 max {2,2,1,1,2,2,2}
r53
= {3,2,1,2,4,4,3} = 4/4 =1 r54 = 2 =
2/2 = 1
min
4
max
{2,2,1,1,2,2,2}
r63
= {3,2,1,2,4,4,3} = 4/4 = 1 r64 = 2 =
2/2 = 1
min 4 max
{2,2,1,1,2,2,2}
r73
={3,2,1,2,4,4,3} = 4/3 = 1,333 r74 = 2 = 2/2 = 1
min 3
max {2,2,1,1,2,2,2}
Sehingga Matriks X
0,75 0,75 1,333 1
0,75 0,75 2 1
0,5 1 4 0,5
0, 25 1 2 0,5
0,25 0,75 1 1
0,5 0,5 1 1
1 0,5 1,333 1
Proses Perangkingan
V1 = 4 (0,75) + 4 ( 0,75) + 5 (1.333) +
3 (1)
= 3+3+6.665+3 = 15,665
V2 = 4 (0,75) + 4 ( 0,75) + 5 (2) + 3
(1)
= 3+3+10+3 = 19
V3 = 4 (0,5) + 4 (1) + 5 (4) + 3 (0,5)
= 2+4+20+1,5 = 27,5
V4 = 4 (0,25) + 4 (1) + 5 (2) + 3 (0,5)
= 1+4+10+1,5 = 16,5
V5 = 4 (0,25) + 4 ( 0,75) + 5 (1) + 3
(1)
= 1+3+5+3 = 12
V6 = 4 (0,5) + 4 ( 0,5) + 5 (1) + 3 (1)
= 2+2+5+3 = 12
= 4+2+6,665+3 = 15,665
Jadi, Mahasiswa yang berhak mendapat
beasiswa adalah
1.
V3 = 27,5 = Simamora
2.
V2 = 19 =
Widodo
3.
V4 = 16,5 = Susilawati
4.
V1 = 15,665 = Joko
5.
V7 = 15,665 = Hendro
