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 (AttributKeuntungan)
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 alternative dalam pemberian beasiswa yaitu :
Syarat :
C1: Semester Aktif Perkuliahan (AttributKeuntungan)
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 alternative 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
|
TidakAktif
|
|
4
|
Susilawati
|
II
|
3.9
|
1.650.000
|
TidakAktif
|
|
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
|
Penyelesain
:
|
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
|
Matrik
:
X= [█(■(3& 3& 3@3& 3& 2@ 2& 4& 1) ■( 2@ 2 @ 1 )@■( 1& 4& 2 @ 1& 3&4 @ 2& 2&4 ) ■( 1 @2@2)@■(4& 2& 3 ) 2)]
Untuk : C1
X_1= √(3^2+3^2+2^2+1^2+1^2+2^2+4^2 ) =√44 =6.633
r_11= 3/6.633 = 0.4522
r_12= 3/6.633 = 0.4522
r_13= 2/6.633 = 0.3015
r_14= 1/6.633 = 0.1507
r_15= 1/6.633 = 0.1507
r_16= 2/6.633 = 0.3015
r_17= 4/6.633 = 0.6030
Untuk : C2
X_2= √(3^2+3^2+4^2+4^2+3^2+2^2+2^2 ) =√67 = 8.185
r_21= 3/8.185 = 0.3665
r_22= 3/8.185 = 0.3655
r_23= 4/8.185 = 0.4886
r_24= 4/8.185 = 0.4886
r_25= 3/8.185 = 0.3665
r_26= 2/8.185 = 0.2443
r_27= 2/8.185 = 0.2443
X_1= √(3^2+3^2+2^2+1^2+1^2+2^2+4^2 ) =√44 =6.633
r_11= 3/6.633 = 0.4522
r_12= 3/6.633 = 0.4522
r_13= 2/6.633 = 0.3015
r_14= 1/6.633 = 0.1507
r_15= 1/6.633 = 0.1507
r_16= 2/6.633 = 0.3015
r_17= 4/6.633 = 0.6030
Untuk : C2
X_2= √(3^2+3^2+4^2+4^2+3^2+2^2+2^2 ) =√67 = 8.185
r_21= 3/8.185 = 0.3665
r_22= 3/8.185 = 0.3655
r_23= 4/8.185 = 0.4886
r_24= 4/8.185 = 0.4886
r_25= 3/8.185 = 0.3665
r_26= 2/8.185 = 0.2443
r_27= 2/8.185 = 0.2443
Untuk : C3
X_3= √(3^2+2+1^2+2^2+4^2+4^2+3^2 ) =√59 =7.681
r_31= 3/7.681 = 0.3905
r_32= 2/7.681 = 0.2603
r_33= 1/7.681 = 0.1301
r_34= 2/7.681 = 0.2603
r_35= 4/7.681 = 0.5207
r_36= 4/7.681 = 0.5207
r_37= 3/7.681 = 0.3905
Untuk : C4
X_4= √(2^2+2^2+1^2+1^2+2+2^2+2^2 ) =√22 =4.690
r_41= 2/4.690 = 0.4264
r_42= 2/4.690 = 0.4264
r_43= 1/4.690 = 0.2132
r_44= 1/4.690 = 0.2132
r_45= 2/4.690 = 0.4264
r_46= 2/4.690 = 0.4264
r_47= 2/4.690 = 0.4264
X_3= √(3^2+2+1^2+2^2+4^2+4^2+3^2 ) =√59 =7.681
r_31= 3/7.681 = 0.3905
r_32= 2/7.681 = 0.2603
r_33= 1/7.681 = 0.1301
r_34= 2/7.681 = 0.2603
r_35= 4/7.681 = 0.5207
r_36= 4/7.681 = 0.5207
r_37= 3/7.681 = 0.3905
Untuk : C4
X_4= √(2^2+2^2+1^2+1^2+2+2^2+2^2 ) =√22 =4.690
r_41= 2/4.690 = 0.4264
r_42= 2/4.690 = 0.4264
r_43= 1/4.690 = 0.2132
r_44= 1/4.690 = 0.2132
r_45= 2/4.690 = 0.4264
r_46= 2/4.690 = 0.4264
r_47= 2/4.690 = 0.4264
Matrik R ternomalisasi :
█(■(0.4522&0.3665@0.4522&0.3655 ) ■(0.3905 & 0.4264 @0.2603 &0.4264)@■(0.3015&0.4886@0.1507&0.4886 ) ■(0.1301 & 0.2132 @0.2603 &0.2132)@■(0.1507&0.3665@0.3015&0.2443 ) ■(0.5207 & 0.4264 @0.5207 &0.4264)@■( 0.6030 &0.2443 ) ■(0.3905& )0.4264)
█(■(0.4522&0.3665@0.4522&0.3655 ) ■(0.3905 & 0.4264 @0.2603 &0.4264)@■(0.3015&0.4886@0.1507&0.4886 ) ■(0.1301 & 0.2132 @0.2603 &0.2132)@■(0.1507&0.3665@0.3015&0.2443 ) ■(0.5207 & 0.4264 @0.5207 &0.4264)@■( 0.6030 &0.2443 ) ■(0.3905& )0.4264)
W=[4,4,5,3]
y_11=(4).0,4522 = 1.8088
y_12=(4).0,4522 = 1.8088
y_13=(4).0,3015 = 1.206
y_14=(4).0,1507 = 0.6028
y_15=(4).0,1507 = 0.6028
y_16=(4).0,3015 = 1.206
y_17=(4).0,6030 = 2.412
y_21=(4).0,3665 = 1.466
y_22=(4).0,3665 = 1.466
y_23=(4).0,4886 = 1.9544
y_24=(4).0,4886 = 1.9544
y_25=(4).0,3665 = 1.466
y_26=(4).0,2443 = 0.9772
y_27=(4).0,2443 = 0.9772
y_31=(5).0,3905 = 1.9525
y_32=(5).0,2603 = 1.3015
y_33=(5).0,1301 = 0.6505
y_34=(5).0,2603 = 1.3015
y_35=(5).0,5207 = 2.6035
y_36=(5).0,5207 = 2.6035
y_37=(5).0,3905 = 1.9525
y_41=(3).0,4264 = 1.2792
y_42=(3).0,4264 = 1.2792
y_43=(3).0,2132 = 0.6396
y_44=(3).0,2132 = 0.6396
y_45=(3).0,4264 = 1.2792
y_46=(3).0,4264 = 1.2792
y_47=(3).0,4264 = 1.2792
Matrik y ternomalisasi terbobot :
█(■(1.8088&1.466@1.8088&1.466 ) ■(1.9525 & 1.2792 @1.3015 &1.2792)@■(1.206&1.9544@0.6028&1.9544) ■(0.6505 & 0.6396 @1.3015 &0.6396)@■(0.6028&1.466@1.206&0.9772 ) ■(2.6035 & 1.2792 @2.6035 &1.2792)@■(2.412 & 0.9772) ■(1.9525& ) 1.2792)
Solusi ideal positif (A^+) :
Y_1^+= Max {1.8088, 1.8088, 1.206, 0.6028, 0.6028, 1.206, 2.412}= 2.412
Y_2^+= Max {1.466, 1.466, 1.9544, 1.9544, 1.466, 0.9772, 0.9772} = 1.9544
Y_3^-= Min {1.9525,1.3015, 0.6505, 1.3015, 2.6035, 2.6035, 1.9525} = 0.6505
Y_4^+= Max {1.2792, 1.2792,0.6396,0.6396 ,1.2792, 1.2792, 1.2792,} = 1.2792,
A^+={2.412,1.9544,0.6505,1.2792}
Solusi ideal negative (A^_) :
Y_1^-= Min {1.8088, 1.8088, 1.206, 0.6028, 0.6028, 1.206, 2.412}= 0.6028
Y_2^-= Min {1.466, 1.466, 1.9544, 1.9544, 1.466, 0.9772, 0.9772} = 0.9772
Y_3^-= Max {1.9525,1.3015, 0.6505, 1.3015, 2.6035, 2.6035, 1.9525} = 2.6035
Y_4^-= Min {1.2792, 1.2792,0.6396,0.6396 ,1.2792, 1.2792, 1.2792,} = 0.6396
A^-={0.6028,0.9772,2.6035,0.6396}
Jarak Altenatif terbobot dengan solusi ideal positif :
D_1^+=√(〖(1.8088-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(1.9525-0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√2.2975 = 1.5157
D_2^+=√(〖(1.8088-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(1.3015 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√1.0261 = 1.0130
D_3^+=√(〖(1.206-2.412)〗^2+〖(1.9544-1.9544)〗^2+〖(0.6505-0.6505)〗^2+〖(0.6396-1.2792)〗^2 )
=√1.8635 = 1.3651
D_4^+ √(〖(0.6028-2.412)〗^2+〖(1.9544-1.9544)〗^2+〖(1.3015-0.6505)〗^2+〖(0.6396-1.2792)〗^2 )
=√4.1060 = 2.0263
D_5^+=√(〖(0.6028-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(2.6035 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√7.3259 = 2.7066
D_6^+=√(〖(1.206-2.412)〗^2+〖(0.9772-1.9544)〗^2+〖(2.6035 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√6.2235 = 2.4947
D_7^+=√(〖(2.412 -2.412)〗^2+〖(0.9772-1.9544)〗^2+〖(1.9525-0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√2.6501 = 1.6279
Jarak Altenatif terbobot dengan solusi ideal Negatif :
D_1^-= √(〖(1.8088-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(1.9525-2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√2.5262 = 1.5894
D_2^-=√(〖(1.8088-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(1.3015 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√3.7976 = 1.9487
D_3^-=√(〖(1.206-0.6028)〗^2+〖(1.9544-0.9772)〗^2+〖(0.6505-2.6035)〗^2+〖(0.6396-0.6396)〗^2 )
=√5.1329 = 2.2650
D_4^-=√(〖(0.6028-0.6028)〗^2+〖(1.9544-0.9772)〗^2+〖(1.3015-2.6035)〗^2+〖(0.6396-0.6396)〗^2 )
=√2.6501 = 1.6279
D_5^-=√(〖(0.6028-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(2.6035 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√0.6480 = 0.8049
D_6^-=√(〖(1.206-0.6028)〗^2+〖(0.9772-0.9772)〗^2+〖(2.6035 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√0.7729 = 0.8791
D_7^-=√(〖(2.412 -0.6028)〗^2+〖(0.9772-0.9772)〗^2+〖(1.9525-2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√4.1060 = 2.0263
V_i= (D_i^-)/(D_i^-+D_i^+ )
V_1= 1.5894/(1.5894+1.5157) =1.5894/3.1051 = 0.5118
V_2= 1.9487/(1.9487+1.0130) =1.9487/2.9617 = 0.6579
V_3= 2.2650/(2.2650+1.3651) =2.2650/3.6301 = 0.6239
V_4= 1.6279/(1.6279+2.0263) =1.6279/3.6542 = 0.4454
V_5= 0.8049/(0.8049+2.7066) =0.8049/3.5115 = 0.2292
V_6= 0.8791/(0.8791+2.4947) =0.8791/3.3738 = 0.2605
V_7= 2.0263/(2.0263+1.6279) =2.0263/3.6542 = 0.5545
y_11=(4).0,4522 = 1.8088
y_12=(4).0,4522 = 1.8088
y_13=(4).0,3015 = 1.206
y_14=(4).0,1507 = 0.6028
y_15=(4).0,1507 = 0.6028
y_16=(4).0,3015 = 1.206
y_17=(4).0,6030 = 2.412
y_21=(4).0,3665 = 1.466
y_22=(4).0,3665 = 1.466
y_23=(4).0,4886 = 1.9544
y_24=(4).0,4886 = 1.9544
y_25=(4).0,3665 = 1.466
y_26=(4).0,2443 = 0.9772
y_27=(4).0,2443 = 0.9772
y_31=(5).0,3905 = 1.9525
y_32=(5).0,2603 = 1.3015
y_33=(5).0,1301 = 0.6505
y_34=(5).0,2603 = 1.3015
y_35=(5).0,5207 = 2.6035
y_36=(5).0,5207 = 2.6035
y_37=(5).0,3905 = 1.9525
y_41=(3).0,4264 = 1.2792
y_42=(3).0,4264 = 1.2792
y_43=(3).0,2132 = 0.6396
y_44=(3).0,2132 = 0.6396
y_45=(3).0,4264 = 1.2792
y_46=(3).0,4264 = 1.2792
y_47=(3).0,4264 = 1.2792
Matrik y ternomalisasi terbobot :
█(■(1.8088&1.466@1.8088&1.466 ) ■(1.9525 & 1.2792 @1.3015 &1.2792)@■(1.206&1.9544@0.6028&1.9544) ■(0.6505 & 0.6396 @1.3015 &0.6396)@■(0.6028&1.466@1.206&0.9772 ) ■(2.6035 & 1.2792 @2.6035 &1.2792)@■(2.412 & 0.9772) ■(1.9525& ) 1.2792)
Solusi ideal positif (A^+) :
Y_1^+= Max {1.8088, 1.8088, 1.206, 0.6028, 0.6028, 1.206, 2.412}= 2.412
Y_2^+= Max {1.466, 1.466, 1.9544, 1.9544, 1.466, 0.9772, 0.9772} = 1.9544
Y_3^-= Min {1.9525,1.3015, 0.6505, 1.3015, 2.6035, 2.6035, 1.9525} = 0.6505
Y_4^+= Max {1.2792, 1.2792,0.6396,0.6396 ,1.2792, 1.2792, 1.2792,} = 1.2792,
A^+={2.412,1.9544,0.6505,1.2792}
Solusi ideal negative (A^_) :
Y_1^-= Min {1.8088, 1.8088, 1.206, 0.6028, 0.6028, 1.206, 2.412}= 0.6028
Y_2^-= Min {1.466, 1.466, 1.9544, 1.9544, 1.466, 0.9772, 0.9772} = 0.9772
Y_3^-= Max {1.9525,1.3015, 0.6505, 1.3015, 2.6035, 2.6035, 1.9525} = 2.6035
Y_4^-= Min {1.2792, 1.2792,0.6396,0.6396 ,1.2792, 1.2792, 1.2792,} = 0.6396
A^-={0.6028,0.9772,2.6035,0.6396}
Jarak Altenatif terbobot dengan solusi ideal positif :
D_1^+=√(〖(1.8088-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(1.9525-0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√2.2975 = 1.5157
D_2^+=√(〖(1.8088-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(1.3015 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√1.0261 = 1.0130
D_3^+=√(〖(1.206-2.412)〗^2+〖(1.9544-1.9544)〗^2+〖(0.6505-0.6505)〗^2+〖(0.6396-1.2792)〗^2 )
=√1.8635 = 1.3651
D_4^+ √(〖(0.6028-2.412)〗^2+〖(1.9544-1.9544)〗^2+〖(1.3015-0.6505)〗^2+〖(0.6396-1.2792)〗^2 )
=√4.1060 = 2.0263
D_5^+=√(〖(0.6028-2.412)〗^2+〖(1.466-1.9544)〗^2+〖(2.6035 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√7.3259 = 2.7066
D_6^+=√(〖(1.206-2.412)〗^2+〖(0.9772-1.9544)〗^2+〖(2.6035 -0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√6.2235 = 2.4947
D_7^+=√(〖(2.412 -2.412)〗^2+〖(0.9772-1.9544)〗^2+〖(1.9525-0.6505)〗^2+〖(1.2792-1.2792)〗^2 )
=√2.6501 = 1.6279
Jarak Altenatif terbobot dengan solusi ideal Negatif :
D_1^-= √(〖(1.8088-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(1.9525-2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√2.5262 = 1.5894
D_2^-=√(〖(1.8088-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(1.3015 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√3.7976 = 1.9487
D_3^-=√(〖(1.206-0.6028)〗^2+〖(1.9544-0.9772)〗^2+〖(0.6505-2.6035)〗^2+〖(0.6396-0.6396)〗^2 )
=√5.1329 = 2.2650
D_4^-=√(〖(0.6028-0.6028)〗^2+〖(1.9544-0.9772)〗^2+〖(1.3015-2.6035)〗^2+〖(0.6396-0.6396)〗^2 )
=√2.6501 = 1.6279
D_5^-=√(〖(0.6028-0.6028)〗^2+〖(1.466-0.9772)〗^2+〖(2.6035 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√0.6480 = 0.8049
D_6^-=√(〖(1.206-0.6028)〗^2+〖(0.9772-0.9772)〗^2+〖(2.6035 -2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√0.7729 = 0.8791
D_7^-=√(〖(2.412 -0.6028)〗^2+〖(0.9772-0.9772)〗^2+〖(1.9525-2.6035)〗^2+〖(1.2792-0.6396)〗^2 )
=√4.1060 = 2.0263
V_i= (D_i^-)/(D_i^-+D_i^+ )
V_1= 1.5894/(1.5894+1.5157) =1.5894/3.1051 = 0.5118
V_2= 1.9487/(1.9487+1.0130) =1.9487/2.9617 = 0.6579
V_3= 2.2650/(2.2650+1.3651) =2.2650/3.6301 = 0.6239
V_4= 1.6279/(1.6279+2.0263) =1.6279/3.6542 = 0.4454
V_5= 0.8049/(0.8049+2.7066) =0.8049/3.5115 = 0.2292
V_6= 0.8791/(0.8791+2.4947) =0.8791/3.3738 = 0.2605
V_7= 2.0263/(2.0263+1.6279) =2.0263/3.6542 = 0.5545
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