Suatu PTS dikota medan, akan memberikan
beasiswa kepada 5 orang mahasiswanya adapun syarat pemberian beasiswa tersebut.
Yaitu harus memenuhi ketentuan sebagai berikut:
Syarat:
C1
= semester aktif perkuliahan
C2
= IPK
C3
= penghasilan orang tua
C4
= aktif organisasi
Bobot
W = [4, 4, 5, 3]
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
|
IV
|
3,8
|
1.350.000
|
T. AKTIF
|
4
|
SUSILAWATI
|
II
|
3,9
|
1.650.000
|
T. AKTIF
|
5
|
DIAN
|
II
|
3,6
|
2.300.000
|
AKTIF
|
6
|
ROMA
|
IV
|
3,3
|
2.250.000
|
AKTIF
|
7
|
HENDRO
|
VII
|
3,4
|
1.950.000
|
AKTIF
|
PEMBOBOTAN:
C1 :
II : 1
IV : 2
VI : 3
VIII : 4
C2 :
3,0
: 1
3,25
: 2
3,5
: 3
3,75
: 4
4,0
: 5
C3 :
1.000.000
: 1
1.400.000
: 2
1.800.000
: 3
2.200.000
: 4
2.600.000
: 5
C4 :
AKTIF
: 2
T.
AKTIF : 1
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 :
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 r45
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
r11 = 3 = 3 =
0,75
max{3,3,1,1,2,4} 4
r21 =
3 = 3 =
0,75
max{3,3,2,1,1,2,4} 4
r31 = 2 = 2 =
0,5
max{3,3,2,1,1,2,4} 4
r41 = 1 = 1 = 0,25
max{3,3,2,1,1,2,4} 4
r51 = 1 = 1 = 0,25
max{3,3,2,1,1,2,4} 4
r61 = 2 = 2 = 0,5
max{3,3,2,1,1,2,4} 4
r71 = 4 = 4 = 1
max{3,3,2,1,1,2,4} 4
Untuk
c2
r12 = 3 = 3 = 0,75
max{3,3,4,4,3,2,2} 4
r22 = 3 = 3 = 0,75
max{3,3,4,4,3,2,2} 4
r32 = 4 = 4 = 1
max{3,3,4,4,3,2,2} 4
r42 = 4 = 4 = 1
max{3,3,4,4,3,2,2} 4
r52 = 3 = 3 = 0,75
max{3,3,4,4,3,2,2} 4
r62 = 2 = 3 = 0,5
max{3,3,4,4,3,2,2} 4
r72 = 2 = 3 = 0,5
max{3,3,4,4,3,2,2} 4
Untuk
c3
r13 = max{3,2,1,2,4,4,3}= 4 =
1,333
3 3
r23 = max{3,2,1,2,4,4,3}
= 4 = 2
2 2
r33 = max{3,2,1,2,4,4,3}
= 4 = 4
1 1
r43 = max{3,2,1,2,4,4,3}
= 4 = 2
2 2
r53 = max{3,2,1,2,4,4,3}= 4 = 1
4 4
r63 = max{3,2,1,2,4,4,3}= 4 = 1
4 4
r73 = max{3,2,1,2,4,4,3}= 4 =
1,333
3 3
Untuk
c4
r14 = 2
= 2 = 1
max{2,2,1,1,2,2,2} 2
r24 =
2 = 2 = 1
max{2,2,1,1,2,2,2} 2
r34 = 1 = 1 = 0,5
max{2,2,1,1,2,2,2} 2
r44 = 1 = 1 = 0,5
max{2,2,1,1,2,2,2} 2
r54 = 2 = 2 = 1
max{2,2,1,1,2,2,2} 2
r64 = 2 = 2 = 1
max{2,2,1,1,2,2,2 2
r74 = 2 = 2 = 1
max{2,2,1,1,2,2,2 2
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
perangkaian
Bobot
W = [4, 4, 5, 3]
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+4+3 = 12
v7
= 4(1)+4(0,5)+5(1,333)+3(1)
4+2+6,665+3 = 15,665
