Jawaban Quizz Online Materi Sistem Pendukung Keputusan Dengan Metode Topsis.
Suatu kelurahan mendapatkan Bantuan
Langsung Tunai dari pemerintah untuk masing masing kepala keluarga dengan
syarat ketentuan sebagai berikut :
C1 : Jumlah Tanggungan
C2 : Pendapatan Kepala Keluarga
C3 : Luas Bangunan Rumah
C4 : Memiliki KK
Pilihlah 5 alternatif KK yang akan mendapatkan bantuan dari beberapa KK berikut ini menggunakan Metode TOPSIS:
C1 : Jumlah Tanggungan
C2 : Pendapatan Kepala Keluarga
C3 : Luas Bangunan Rumah
C4 : Memiliki KK
Pilihlah 5 alternatif KK yang akan mendapatkan bantuan dari beberapa KK berikut ini menggunakan Metode TOPSIS:
|
Nama KK
|
C1
|
C2
|
C3
|
C4
|
|
Aldyan
|
4
|
2.350.000
|
100M2
|
Tidak Ada
|
|
Hendro
|
5
|
3.050.000
|
50M2
|
Ada
|
|
Joko
|
3
|
3.350.000
|
70M2
|
Ada
|
|
Doni
|
4
|
2.550.000
|
90M2
|
Ada
|
|
Dono
|
6
|
2.850.000
|
120M2
|
Ada
|
|
Kasino
|
3
|
2.650.000
|
80M2
|
Ada
|
|
Susanto
|
2
|
3.350.000
|
150M2
|
Tidak Ada
|
Pembobotan dari kriteria diatas dapat dilihat dibawah ini :
C1 : Jumlah Tanggungan (Attribut Keuntungan)
1-2 : 1
3-4 : 2
5-6 : 3
C2 : Pendapatan Kepala Keluarga (Attribut Biaya)
2.000.000 : 1
2.400.000 : 2
2.800.000 : 3
3.200.000 : 4
3.600.000 : 5
C3 : Luas Bangunan Rumah (Attribut Biaya)
50-70 : 1
71-90 : 2
91-110 : 3
111-130 : 4
131-150 : 5
C4 : Memiliki KK (Attribut Keuntungan)
Ada : 2
Tidak Ada : 1
Bobot
W : 4,5,4,3
Penyelesaian :
1.
Tabel Kecocokan :
|
Nama KK
|
C1
|
C2
|
C3
|
C4
|
|
benefit
|
cost
|
cost
|
benefit
|
|
|
Aldyan
|
2
|
1
|
3
|
1
|
|
Hendro
|
3
|
3
|
1
|
2
|
|
Joko
|
2
|
4
|
1
|
2
|
|
Doni
|
2
|
2
|
2
|
2
|
|
Dono
|
3
|
3
|
4
|
2
|
|
Kasino
|
2
|
2
|
2
|
2
|
|
Susanto
|
1
|
4
|
5
|
1
|
2.
Membuat Matriks Keputusan :
r11=2/5.916079783= 0.338061702
r21=3/5.916079783=0.507092553
r31=2/5.916079783= 0.338061702 r41=2/5.916079783= 0.338061702
r51=3/5.916079783=0.507092553 r61=2/5.916079783= 0.338061702
r71=1/5.916079783= 0.169030851
r12=1/7.681145748=0.130188911 r22=3/7.681145748=0.390566733
r32=4/7.681145748=0.520755644 r42=2/7.681145748=0.260377822
r52=3/7.681145748=0.390566733 r62=2/7.681145748=0.260377822
r72=4/7.681145748=0.520755644
r13=3/7.745966692=0.387298335 r23=1/7.745966692=0.129099445
r33=1/7.745966692=0.129099445 r43=2/7.745966692=0.25819889
r53=4/7.745966692=0.516397779 r63=2/7.745966692=0.25819889
r73=5/7.745966692=0.645497224
r14=1/4.69041576=0.213200716 r24=2/4.69041576=0.426401433
r34=2/4.69041576=0.426401433 r44=2/4.69041576=0.426401433
r54=2/4.69041576=0.426401433 r64=2/4.69041576=0.426401433
r74=1/4.69041576=0.213200716
3. Matriks r Ternormalisasi :
|
0.338061702
|
0.130188911
|
0.387298335
|
0.213200716
|
|
0.507092553
|
0.390566733
|
0.129099445
|
0.426401433
|
|
0.338061702
|
0.520755644
|
0.129099445
|
0.426401433
|
|
0.338061702
|
0.260377822
|
0.25819889
|
0.426401433
|
|
0.507092553
|
0.390566733
|
0.516397779
|
0.426401433
|
|
0.338061702
|
0.260377822
|
0.25819889
|
0.426401433
|
|
0.169030851
|
0.520755644
|
0.645497224
|
0.213200716
|
4.
Menentukan
Matrik Solusi Ideal Positip Dan Negatif :
|
y11= 4 *
0.338061702=1.352246808
|
y12= 5 *
0.130188911=0.650944555
|
|
y21= 4 *
0.507092553=2.028370211
|
y22= 5 *
0.390566733=1.952833665
|
|
y31= 4 *
0.338061702=1.352246808
|
y32= 5 *
0.520755644=2.60377822
|
|
y41= 4 *
0.338061702=1.352246808
|
y42= 5 *
0.260377822=1.30188911
|
|
y51= 4 *
0.507092553=2.028370211
|
y52= 5 *
0.390566733=1.952833665
|
|
y61= 4 *
0.338061702=1.352246808
|
y62= 5 *
0.260377822=1.30188911
|
|
y71= 4 *
0.169030851=0.676123404
|
y72= 5 *
0.520755644=2.60377822
|
|
y13= 4 * 0.387298335=1.549193338
|
y14= 3 *
0.213200716=0.639602149
|
|
y23= 4 * 0.129099445=0.516397779
|
y24= 3 *
0.426401433=1.279204298
|
|
y33= 4 * 0.129099445=0.516397779
|
y34= 3 *
0.426401433=1.279204298
|
|
y43= 4 * 0.25819889=1.032795559
|
y44= 3 *
0.426401433=1.279204298
|
|
y53= 4 * 0.516397779=2.065591118
|
y54= 3 *
0.426401433=1.279204298
|
|
y63= 4 * 0.25819889=1.032795559
|
y64= 3 *
0.426401433=1.279204298
|
|
y73= 4 * 0.645497224=2.581988897
|
y74= 3 *
0.213200716=0.639602149
|
a.Matriks
y Ternormalisasi Terbobot :
|
1.352246808
|
0.650944555
|
1.549193338
|
0.639602149
|
|
2.028370211
|
1.952833665
|
0.516397779
|
1.279204298
|
|
1.352246808
|
2.60377822
|
0.516397779
|
1.279204298
|
|
1.352246808
|
1.30188911
|
1.032795559
|
1.279204298
|
|
2.028370211
|
1.952833665
|
2.065591118
|
1.279204298
|
|
1.352246808
|
1.30188911
|
1.032795559
|
1.279204298
|
|
0.676123404
|
2.60377822
|
2.581988897
|
0.639602149
|
b. Mencari nilai Max dan Min
y1+ = Max{1.352246808; 2.028370211; 1.352246808; 1.352246808; 2.028370211; 1.352246808; 0.676123404}=2.028370211
y2+ = Min{0.650944555; 1.952833665; 2.60377822; 1.30188911; 1.952833665;
1.30188911; 2.60377822}=0.650944555
y3+ = Min{1.549193338; 0.516397779; 0.516397779; 1.032795559; 2.065591118;
1.032795559; 2.581988897}=0.516397779
y4+ = Max{0.639602149; 1.279204298; 1.279204298; 1.279204298; 1.279204298;
1.279204298; 0.639602149}=1.279204298
y1- = Min{1.352246808; 2.028370211; 1.352246808; 1.352246808; 2.028370211;
1.352246808; 0.676123404}=0.676123404
y2- = Max{0.650944555; 1.952833665; 2.60377822; 1.30188911; 1.952833665;
1.30188911; 2.60377822}=2.60377822
y3- = Max{1.549193338; 0.516397779; 0.516397779; 1.032795559; 2.065591118;
1.032795559; 2.581988897}=2.581988897
y4- = Min{0.639602149; 1.279204298; 1.279204298; 1.279204298; 1.279204298;
1.279204298; 0.639602149}=0.639602149
|
A+
|
2.028370211
|
0.650944555
|
0.516397779
|
1.279204298
|
|
A-
|
0.676123404
|
2.60377822
|
2.581988897
|
0.639602149
|
5.
Menentukan jarak antara nilai setiap
alternative dengan maktrik solusi ideal
positip dan negatif :
6. Mencari Refrensi :
5 alternatif KK yang akan mendapatkan bantuan adalah:
|
Nama KK
|
V
|
|
Aldyan
|
0.624303633
|
|
Hendro
|
0.669068134
|
|
Joko
|
0.522970398
|
|
Doni
|
0.675246012
|
|
Kasino
|
0.675246012
|
