Penyelesaian Soal Metode Topsis - UTS

Penyelesaian Soal Metode Topsis - UTS
Penyelesaian Soal Metode Topsis - UTS

Studi Kasus

Untuk membantu masyarakat agar memiliki rumah layak huni, pemerintah secara rutin menyalurkan dana bantuan untuk melakukan program bedah rumah. Dana bantuan ini diprioritaskan diberikan pada masyarakat dengan golongan tidak mampu. Setiap tahunnya, Desa Sariharjo mendapatkan Dana Bantuan Bedah Rumah sebesar Rp. 10.000.000 untuk memperbaiki rumah salah satu warga di Desa Sariharjo.

Pendataan yang dilakukan oleh Kepala Desa Sariharjo tahun ini didapatkan 5 rumah yang kondisinya sudah tidak layak huni yaitu Rumah Bapak Poniran, Rumah Ibu Ani, Rumah Bapak Wasito, Rumah Bapak Kus, Rumah Ibu Yatemi. Kriteria yang digunakan untuk melakukan pemilihan rumah mana yang berhak mendapatkan bantuan ada 5 yaitu:

1. Jumlah penghasilan per bulan (Cost)
2. Jumlah tanggungan keluarga (Benefit)
3. Kondisi atap rumah (Benefit)
  • Beri nilai 3 jika kerusakan atap rumah rusak berat dan membahayakan penghuni rumah,
  • Beri nilai 2 jika kerusakan atap rumah sedang, dimana dalam jangka waktu yang cukup lama dapat roboh,
  • Beri nilai 1 jika kerusakan atap rumah ringan.
4. Kondisi dinding rumah (Benefit)
  • Beri nilai 5 jika dinding rumah masih belum permanen / anyaman bambu,
  • Beri nilai 4 jika dinding rumah sudah permanen namun kondisinya rapuh,
  • Beri nilai 2 jika dinding rumah sudah permanen dengan kondisi rangka baik namun belum plester,
  • Beri nilai 0 jika dinding sudah permanen dengan kondisi baik dan sudah di cat.
5. Komunikasi di Lingkungan Masyarakat (Benefit)
  • Beri nilai 3 jika warga yang bersangkutan aktif dalam kegiatan desa,
  • Beri nilai 2 jika warga yang bersangkutan cukup aktif dalam kegiatan desa, dan
  • Beri nilai 0 jika warga yang bersangkutan tidak aktif dalam kegiatan desa.
Bobot dari masing-masing kriteria dinilai dari skala 1 sampai dengan 5
  • 1 = Sangat rendah,
  • 2 = Rendah,
  • 3 = Cukup,
  • 4 = Tinggi,
  • 5 = Sangat Tinggi.
Pengambil keputusan memberikan bobot untuk tiap kriteria adalah W = (5, 4, 4, 5, 3).

Berdasarkan hasil penilaian Kepala Desa, kondisi masing-masing rumah digambarkan dalam tabel di bawah ini:
Alternatif
Kriteria
penghasilan per-bulan
(…*100.000)
Tanggungan keluarga
Kerusakan Rumah
Status Rumah
Komunikasi/ Peran Aktif
Bapak Poniran
4
4
3
5
3
Ibu Ani
3
2
3
5
2
Bapak Wasito
4.5
5
2
2
3
Bapak Kus
3.5
3
3
5
3
Ibu Yatemi
3
2
3
5
2

Dengan metode TOPSIS tentukan siapakah dari ke-5 orang warga yang berhak mendapat kesempatan untuk dibedah rumahnya?

Penyelesaian


1. Menentukan Kriteria dan Sifat
Nama Kriteria
Sifat
Bobot
C1
Cost
5
C2
Benefit
4
C3
Benefit
4
C4
Benefit
5
C5
Benefit
3

2. Menentukan Rating Kecocokan
Alternatif
Kriteria
C1
C2
C3
C4
C5
A1
4
4
3
5
3
A2
3
2
3
5
2
A3
4.5
5
2
2
3
A4
3.5
3
3
5
3
A5
3
2
3
5
2

3. Menentukan Matriks Keputusan Ternormalisasi
X1
X2
X3
X4
X5
4
4
3
5
3
3
2
3
5
2
4.5
5
2
2
3
3.5
3
3
5
3
3
2
3
5
2

\[\text{ }\!\!|\!\!\text{ x1 }\!\!|\!\!\text{ }\!\!|\!\!\text{ }=\sqrt{\text{(}{{\text{4}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}\text{ + 4}\text{.}{{\text{5}}^{\text{2}}}\text{+ 3}\text{.}{{\text{5}}^{\text{2}}}\text{+ }{{\text{3}}^{\text{2}}}\text{) }}\text{= 8}\text{.154753215}\]
r11 = x11 / |x1| = 4 / 8.154753215 = 0,490511472

r21 = x21 / |x1| = 3 / 8.154753215 = 0,367883604

r31 = x31 / |x1| = 4.5 / 8.154753215 = 0,551825406

r41 = x41 / |x1| = 3.5 / 8.154753215 = 0,429197538

r51 = x51 / |x1| = 3 / 8.154753215 = 0,367883604

\[\text{ }\!\!|\!\!\text{  x2  }\!\!|\!\!\text{  }\!\!|\!\!\text{ }=\sqrt{\text{(}{{\text{4}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{2}}}\text{ + }{{\text{5}}^{\text{2}}}\text{+ }{{\text{3}}^{\text{2}}}\text{+ }{{\text{2}}^{\text{2}}}\text{) }}\text{= 7}\text{.615773106}\]
r12 = x12 / |x2| = 4 / 7.615773106 = 0,525225731

r22 = x22 / |x2| = 2 / 7.615773106 = 0,262612866

r32 = x32 / |x2| = 5 / 7.615773106 = 0,656532164

r42 = x42 / |x2| = 3 / 7.615773106 = 0,393919299

r52 = x52 / |x2| = 2 / 7.615773106 = 0,262612866

\[\text{ }\!\!|\!\!\text{  }\!\!|\!\!\text{  x3  }\!\!|\!\!\text{  }\!\!|\!\!\text{ }=\sqrt{\text{(}{{\text{3}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{2}}}\text{+ }{{\text{3}}^{\text{2}}}\text{+ }{{\text{3}}^{\text{2}}}\text{) }}\text{= 6}\text{.324555320}\]
r13 = x13 / |x3| = 3 / 6.324555320 = 0,474341649

r23 = x23 / |x3| = 3 / 6.324555320 = 0,474341649

r33 = x33 / |x3| = 2 / 6.324555320 = 0,316227766

r43 = x43 / |x3| = 3 / 6.324555320 = 0,474341649

r53 = x53 / |x3| = 3 / 6.324555320 = 0,474341649

\[\text{ }\!\!|\!\!\text{  }\!\!|\!\!\text{  x4  }\!\!|\!\!\text{  }\!\!|\!\!\text{ }=\sqrt{\text{(}{{\text{5}}^{\text{2}}}\text{ + }{{\text{5}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{2}}}\text{+ }{{\text{5}}^{\text{2}}}\text{+ }{{\text{5}}^{\text{2}}}\text{) }}\text{= 10}\text{.198039027}\]
r14 = x14 / |x4| = 5 / 10.198039027 = 0,490290338

r24 = x24 / |x4| = 5 / 10.198039027 = 0,490290338

r34 = x34 / |x4| = 2 / 10.198039027 = 0,196116135

r44 = x44 / |x4| = 5 / 10.198039027 = 0,490290338

r54 = x54 / |x4| = 5 / 10.198039027 = 0,490290338

\[\text{ }\!\!|\!\!\text{  }\!\!|\!\!\text{  x5  }\!\!|\!\!\text{  }\!\!|\!\!\text{ }=\sqrt{\text{(}{{\text{3}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}\text{+ }{{\text{3}}^{\text{2}}}\text{+ }{{\text{2}}^{\text{2}}}\text{) }}\text{= 5}\text{.916079783}\]
r15 = x15 / |x5| = 3 / 5.916079783 = 0,507092553

r25 = x25 / |x5| = 2 / 5.916079783 = 0,338061702

r35 = x35 / |x5| = 3 / 5.916079783 = 0,507092553

r45 = x45 / |x5| = 3 / 5.916079783 = 0,507092553

r55 = x55 / |x5| = 2 / 5.916079783 = 0,338061702

Matriks Ternormalisasi R
0,490511472
0,525225731
0,474341649
0,490290338
0,507092553
0,367883604
0,262612866
0,474341649
0,490290338
0,338061702
0,551825406
0,656532164
0,316227766
0,196116135
0,507092553
0,429197538
0,393919299
0,474341649
0,490290338
0,507092553
0,367883604
0,262612866
0,474341649
0,490290338
0,338061702

4. Perkalian Antara Bobot Dengan Nilai Setiap Atribut (Matriks R)
Diperoleh matriks Y sebagai berikut.
0,490511472*5
0,525225731*4
0,474341649*4
0,490290338*5
0,507092553*3
0,367883604*5
0,262612866*4
0,474341649*4
0,490290338*5
0,338061702*3
0,551825406*5
0,656532164*4
0,316227766*4
0,196116135*5
0,507092553*3
0,429197538*5
0,393919299*4
0,474341649*4
0,490290338*5
0,507092553*3
0,367883604*5
0,262612866*4
0,474341649*4
0,490290338*5
0,338061702*3

menjadi,
2.452557358
2.100902926
1.897366596
2.451451689
1.521277659
1.839418018
1.050451463
1.897366596
2.451451689
1.014185106
2.759127028
2.626128657
1.264911064
0.980580676
1.521277659
2.145987688
1.575677194
1.897366596
2.451451689
1.521277659
1.839418018
1.050451463
1.897366596
2.451451689
1.014185106

5. Menentukan Matriks Solusi Ideal Positif & Negatif
 Nama kriteria
Sifat kriteria
Y+
Y-
C1 = penghasilan per-bulan
Cost
Min = 1.839418018
Max = 2.759127028
C2 = tanggungan keluarga
Benefit
Max = 2.626128657
Min = 1.050451463
C3 = kerusakan rumah
Benefit
Max = 1.897366596
Min = 1.264911064
C4 = status rumah
Benefit
Max = 2.451451689
Min = 0.980580676
C5 = peran aktif
Biaya/Cost.
Max = 1.521277659
Min = 1.014185106
Dapat disimpulkan :
A+ =
{1.839418018; 2.626128657; 1.897366596; 2.451451689; 1.521277659}
A- =
{2.759127028; 1.050451463; 1.264911064; 0.980580676; 1.014185106}

6.a. Jarak Antara Alternatif Ai Dengan Solusi Ideal Positif
\[D{{1}^{+}}=\sqrt{\begin{smallmatrix} {{\text{(}2.452557358\text{-}1.839418018\text{)}}^{\text{2}}}\text{+(}2.100902926\text{-}2.626128657{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.897366596{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}2.451451689\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.521277659{{\text{)}}^{\text{2}}} \end{smallmatrix}}=0.807342504\]
\[D{{2}^{+}}=\sqrt{\begin{smallmatrix} {{\text{(}1.839418018\text{-}1.839418018\text{)}}^{\text{2}}}\text{+(}1.050451463\text{-}2.626128657{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.897366596{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}2.451451689\text{)}}^{\text{2}}}\text{+(}1.014185106\text{-}1.521277659{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.655264776\]
\[D{{3}^{+}}=\sqrt{\begin{smallmatrix} {{\text{(}2.759127028\text{-}1.839418018\text{)}}^{\text{2}}}\text{+(}2.626128657\text{-}2.626128657{{\text{)}}^{\text{2}}}\text{+(}1.264911064\text{-}1.897366596{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}0.980580676\text{-}2.451451689\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.521277659{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.846436081\]
\[D{{4}^{+}}=\sqrt{\begin{smallmatrix} {{\text{(}2.145987688\text{-}1.839418018\text{)}}^{\text{2}}}\text{+(}1.575677194\text{-}2.626128657{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.897366596{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}2.451451689\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.521277659{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.094272927\]
\[D{{5}^{+}}=\sqrt{\begin{smallmatrix} {{\text{(}1.839418018\text{-}1.839418018\text{)}}^{\text{2}}}\text{+(}1.050451463\text{-}2.626128657{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.897366596{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}2.451451689\text{)}}^{\text{2}}}\text{+(}1.014185106\text{-}1.521277659{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.655264776\]

6.b. Jarak Antara Alternatif Ai Dengan Solusi Ideal Negatif
\[D{{1}^{-}}=\sqrt{\begin{smallmatrix} {{\text{(}2.452557358\text{-}2.759127028\text{)}}^{\text{2}}}\text{+(}2.100902926\text{-}1.050451463{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.264911064{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}0.980580676\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.014185106{{\text{)}}^{\text{2}}} \end{smallmatrix}}=2.004504336\]
\[D{{2}^{-}}=\sqrt{\begin{smallmatrix} {{\text{(}1.839418018\text{-}2.759127028\text{)}}^{\text{2}}}\text{+(}1.050451463\text{-}1.050451463{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.264911064{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}0.980580676\text{)}}^{\text{2}}}\text{+(}1.014185106\text{-}1.014185106{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.846436081\]
\[D{{3}^{-}}=\sqrt{\begin{smallmatrix} {{\text{(}2.759127028\text{-}2.759127028\text{)}}^{\text{2}}}\text{+(}2.626128657\text{-}1.050451463{{\text{)}}^{\text{2}}}\text{+(}1.264911064\text{-}1.264911064{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}0.980580676\text{-}0.980580676\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.014185106{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.655264776\]
\[D{{4}^{-}}=\sqrt{\begin{smallmatrix} {{\text{(}2.145987688\text{-}2.759127028\text{)}}^{\text{2}}}\text{+(}1.575677194\text{-}1.050451463{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.264911064{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}0.980580676\text{)}}^{\text{2}}}\text{+(}1.521277659\text{-}1.014185106{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.863439378\]
\[D{{5}^{-}}=\sqrt{\begin{smallmatrix} {{\text{(}1.839418018\text{-}2.759127028\text{)}}^{\text{2}}}\text{+(}1.050451463\text{-}1.050451463{{\text{)}}^{\text{2}}}\text{+(}1.897366596\text{-}1.264911064{{\text{)}}^{\text{2}}}\text{+} \\ {{\text{(}2.451451689\text{-}0.980580676\text{)}}^{\text{2}}}\text{+(}1.014185106\text{-}1.014185106{{\text{)}}^{\text{2}}} \end{smallmatrix}}=1.846436081\]

7. Menentukan Nilai Preferensi Untuk Setiap Alternatif
\[V1=\frac{D_{1}^{-}}{D_{1}^{-}+D_{1}^{+}}=\frac{2.004504336}{2.004504336+0.807342504}=0.712878208\]
\[V2=\frac{D_{2}^{-}}{D_{2}^{-}+D_{2}^{+}}=\frac{1.846436081}{1.846436081+1.655264776}=0.527296921\]
\[V3=\frac{D_{3}^{-}}{D_{3}^{-}+D_{3}^{+}}=\frac{1.655264776}{1.655264776+1.846436081}=0.472703079\]
\[V4=\frac{D_{4}^{-}}{D_{4}^{-}+D_{4}^{+}}=\frac{1.863439378}{1.863439378+1.094272927}=0.630027260\]
\[V5=\frac{D_{5}^{-}}{D_{5}^{-}+D_{5}^{+}}=\frac{1.846436081}{1.846436081+1.655264776}=0.527296921\]

8. Perangkingan
  1. Bpk. Poniran = 0.712878208
  2. Bpk. Kus = 0.630027260
  3. Ibu Ani = 0.527296921
  4. Ibu Yatemi = 0.527296921
  5. Bpk. Wasito = 0.472703079

Posting Komentar

Lebih baru Lebih lama

نموذج الاتصال