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Soal: Menggunakan Metode TOPSIS

Sebuah PTS di Kota Medan, akan memberikan beasiswa kepada 5 orang mahasiswanya. Adapun syarat pemberian beasiswa tersebut, yaitu harus memenuhi ketentuan berikut ini :

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=[4,4,5,3]

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	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	1.950.000	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:

- X1 = √(3^2+3^2+ 2^2+1^2+1^2+2^2+4^2 ) = √44 = 6,633
R11 = 3/6,633=0,4522                  R51 = 1/6,633=0,1507
R21 = 3/6,633=0,4522                R61 = 2/6,633=0,3015
R31 = 2/6,633=0,3015                R71 = 4/6,633=0,6030
R41 = 1/6,633=0,1507

- X2 = √(3^2+3^2+ 4^2+4^2+3^2+2^2+2^2 ) = √67 = 8,185
R12 = 3/8,815=0,3665                  R52 = 3/8,815=0,3665
R22 = 3/8,815=0,3665                R62 = 2/8,815=0,2443
R32 = 4/8,815=0,4886                R72 = 2/8,815=0,2443
R42 = 4/8,815=0,4886

- X3 = √(3^2+2^2+ 1^2+2^2+4^2+4^2+3^2 ) = √59 = 7,681
R13 = 3/7,681=0,3905                  R53 = 2/7,681=0,5207
R23 = 2/7,681=0,2603                R63 = 2/7,681=0,5207
R33 = 1/7,681=0,1301                R73 = 3/7,681=0,3905
R43 = 2/7,681=0,2603

- X4 = √(2^2+2^2+ 1^2+1^2+2^2+2^2+2) = √22 = 4,690
R14 = 2/4,690=0,4264                  R54 = 2/4,690=0,4264
R24 = 2/4,690=0,4264                R64 = 2/4,690=0,4264
R34 = 1/4,690=0,2132                R74 = 2/4,690=0,4264
R44 = 1/4,690=0,2132

0,4522 0,3665 0,3905 0,4264

0,4522 0,3665 0,2603 0,4264

0,3015 0,4886 0,1301 0,2132

R = 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

y_ij= w_ij r_ijw = [4,4,5,3]

y₁₁ = 4(0,4522) = 1,8090 y₁₃ = 5(0,3905) = 1,9528

y₂₁ = 4(0,4522) = 1,8090 y₂₃ = 5(0,2603) = 1,3018

y₃₁ = 4(0,3015) = 1,2060 y₃₃ = 5(0,1301) = 0,6509

y₄₁ = 4(0,1507) = 0,6030 y₄₃ = 5(0,2603) = 1,3018

y₅₁ = 4(0,1507) = 0,6030 y₅₃ = 5(0,5207) = 2,6037

y₆₁ = 4(0,3015) = 1,2060 y₆₃ = 5(0,5207) = 2,6037

y₇₁ = 4(0,6030) = 2,4120 y₇₃ = 5(0,3905) = 1,9528

y₁₂ = 4(0,3665) = 1,4660 y₁₄ = 3(0,4264) = 1,2792

y₂₂ = 4(0,3665) = 1,4660 y₂₄ = 3(0,4264) = 1,2792

y₃₂ = 4(0,4886) = 1,9547 y₃₄ = 3(0,2132) = 0,6396

y₄₂ = 4(0,4886) = 1,9547 y₄₄ = 3(0,2132) = 0,6396

y₅₂ = 4(0,3665) = 1,4660 y₅₄ = 3(0,4264) = 1,2792

y₆₂ = 4(0,2443) = 0,9773 y₆₄ = 3(0,4264) = 1,2792

y₇₂ = 4(0,2443) = 0,9773 y₇₄ = 3(0,4264) = 1,2792

1,8090 1,4660 1,9528 1,2792

1,8090 1,4660 1,3018 1,2792

1,2060 1,9547 0,6509 0,6396

R = 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

A⁺A^-

Y₁⁺= max 2,4120 Y₁^-= min 0,6030

Y₂⁺= max 1,9547 Y₂^-= min 0,9773

Y₃⁺= min 0,6509 Y₃^-= max 2,6037

Y₄⁺= max 1,2792 Y₄^-= min 0,6396

Di-=√(∑_(j=1)^n▒〖(y_ij- y_(i^+ ) 〗 )^2 )

D1-=√((1,8090-0,6030)^2+ (1,4660-0,9773)^2+(1,9528-2,6037)^2+ (1,2792-0,6396)^2 )
= √2,5261 = 1,5893

D2-=√((1,8090-0,6030)^2+ (1,4660-0,9773)^2+(1,3018-2,6037)^2+ (1,2792-0,6396)^2 )
= √3,7973 = 1,9486

D3-=√((1,2060-0,6030)^2+ (1,9547-0,9773)^2+(0,6509-2,6037)^2+ (0,6396-0,6396)^2 )
= √5,1324 = 2,2654

D4-=√((0,6030-0,6030+ (1,9547-0,9773)^2+(1,3018-2,6037)^2+ (0,6396-0,6396)^2 )
= √2,6501 = 1,6279

D5-=√((0,6030-0,6030)^2+ (1,4660-0,9773)^2+(2,6037-2,6037)^2+ (1,2792-0,6396)^2 )
= √0,6478 = 0,8049

D6-=√((1,2060-0,6030)^2+ (0,9773-0,9773)^2+(2,6037-2,6037)^2+ (1,2792-0,6396)^2 )
= √0,7727 = 0,8790

D7-=√((2,4120-0,6030)^2+ (0,9773-0,9773)^2+(1,9528-2,6037)^2+ (1,2792-0,6396)^2 )
= √4,1055= 2,0262

Di+ =√(∑_(j=1)^n▒〖(y_ij- y_(i^+ ) 〗 )^2 )

D1+=√((1,8090-0,6030)^2+ (1,4660-1,9547)^2+(1,9528-0,6509)^2+ (1,2792-1,2792)^2 )
= √2,2973 = 1,5157

D2+=√((1,8090-2,4120)^2+ (1,4660-1,9547)^2+(1,3018-0,6509)^2+ (1,2792-1,2792)^2 )
= √1,0261 = 1,0130

D3+=√((1,2060-2,4120)^2+ (1,9547-1,9547)^2+(0,6509-0,6509)^2+ (0,6396-1,2792)^2 )
= √1,8636 = 1,3651

D4+=√((0,6030-2,4120)^2+ (1,9547-1,9547)^2+(1,3018-0,6509)^2+ (0,6396-1,2792)^2 )
= √4,1055 = 2,0262

D5+=√((0,6030-2,4120)^2+ (1,4660-1,9547)^2+(2,6037-0,6509)^2+ (1,2792-1,2792)^2 )
= √7,3250 = 2,7064

D6+=√((1,2060-2,4120)^2+ (0,9773-1,9547)^2+(2,6037-0,6509)^2+ (1,2792-1,2792)^2 )
= √6,2233 = 2,4946

D7+=√((2,4120-2,4120)^2+ (0,9773-1,9547)^2+(1,9528-0,6509)^2+ (1,2792-1,2792)^2 )
= √2,6501 = 1,627

Vi = Di^-/Di^-+Di⁺

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

Berarti yang berhak mendapatkan Beasiswa yaitu V1, V2, V3, V4, dan V7

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