Penentuan Warung Makan

Perhitungan lengkap metode TOPSIS
 Belajar TOPSIS
Matriks Keputusan (X)
ALTERNATIFC1C2C3C4C5
A1 — Warung Mang Ujang71000069150
A2 — Warung Cinta Rasa91100088250
A3 — Warung Padang6900057120
A4 — Warung Bugis9600078100
KriteriaRasa MakananHarga MakananPelayananSuasanaJarak dalam meter
Atributbenefitcostbenefitbenefitcost
Bobot (W)45233
1
Normalisasi Bobot Kriteria
Menjadikan total bobot ΣW = 1
\[ W_j = \dfrac{W_j}{\sum W_j} \]
W1 =
44 + 5 + 2 + 3 + 3
=
417,00
= 0,2353
W2 =
54 + 5 + 2 + 3 + 3
=
517,00
= 0,2941
W3 =
24 + 5 + 2 + 3 + 3
=
217,00
= 0,1176
W4 =
34 + 5 + 2 + 3 + 3
=
317,00
= 0,1765
W5 =
34 + 5 + 2 + 3 + 3
=
317,00
= 0,1765

Bobot ternormalisasi ini (Wj) akan digunakan pada Tahap 3 untuk membentuk matriks ternormalisasi terbobot.

2
Normalisasi Matriks Keputusan (R)
rij = xij / √Σxij²
\[ r_{ij} = \dfrac{x_{ij}}{\sqrt{\sum_{i=1}^{m} x_{ij}^2}} \]

Langkah 2a — Menghitung penyebut (akar jumlah kuadrat) tiap kolom kriteria:

√Σxi1² = √(7² + 9² + 6² + 9²) = √247,00 = 15,7162
√Σxi2² = √(10000² + 11000² + 9000² + 6000²) = √338.000.000,00 = 18.384,7763
√Σxi3² = √(6² + 8² + 5² + 7²) = √174,00 = 13,1909
√Σxi4² = √(9² + 8² + 7² + 8²) = √258,00 = 16,0624
√Σxi5² = √(150² + 250² + 120² + 100²) = √109.400,00 = 330,7567

Langkah 2b — Menghitung setiap elemen matriks ternormalisasi R:

r11 = 7 / 15,7162 = 0,4454
r12 = 10000 / 18.384,7763 = 0,5439
r13 = 6 / 13,1909 = 0,4549
r14 = 9 / 16,0624 = 0,5603
r15 = 150 / 330,7567 = 0,4535
r21 = 9 / 15,7162 = 0,5727
r22 = 11000 / 18.384,7763 = 0,5983
r23 = 8 / 13,1909 = 0,6065
r24 = 8 / 16,0624 = 0,4981
r25 = 250 / 330,7567 = 0,7558
r31 = 6 / 15,7162 = 0,3818
r32 = 9000 / 18.384,7763 = 0,4895
r33 = 5 / 13,1909 = 0,3790
r34 = 7 / 16,0624 = 0,4358
r35 = 120 / 330,7567 = 0,3628
r41 = 9 / 15,7162 = 0,5727
r42 = 6000 / 18.384,7763 = 0,3264
r43 = 7 / 13,1909 = 0,5307
r44 = 8 / 16,0624 = 0,4981
r45 = 100 / 330,7567 = 0,3023

Matriks Ternormalisasi (R):

ALTERNATIFC1C2C3C4C5
A10,44540,54390,45490,56030,4535
A20,57270,59830,60650,49810,7558
A30,38180,48950,37900,43580,3628
A40,57270,32640,53070,49810,3023
3
Matriks Ternormalisasi Terbobot (Y)
yij = Wj × rij
\[ y_{ij} = w_j \cdot r_{ij} \]
y11 = 0,2353 × 0,4454 = 0,1048
y12 = 0,2941 × 0,5439 = 0,1600
y13 = 0,1176 × 0,4549 = 0,0535
y14 = 0,1765 × 0,5603 = 0,0989
y15 = 0,1765 × 0,4535 = 0,0800
y21 = 0,2353 × 0,5727 = 0,1347
y22 = 0,2941 × 0,5983 = 0,1760
y23 = 0,1176 × 0,6065 = 0,0714
y24 = 0,1765 × 0,4981 = 0,0879
y25 = 0,1765 × 0,7558 = 0,1334
y31 = 0,2353 × 0,3818 = 0,0898
y32 = 0,2941 × 0,4895 = 0,1440
y33 = 0,1176 × 0,3790 = 0,0446
y34 = 0,1765 × 0,4358 = 0,0769
y35 = 0,1765 × 0,3628 = 0,0640
y41 = 0,2353 × 0,5727 = 0,1347
y42 = 0,2941 × 0,3264 = 0,0960
y43 = 0,1176 × 0,5307 = 0,0624
y44 = 0,1765 × 0,4981 = 0,0879
y45 = 0,1765 × 0,3023 = 0,0534

Matriks Ternormalisasi Terbobot (Y):

ALTERNATIFC1C2C3C4C5
A10,10480,16000,05350,09890,0800
A20,13470,17600,07140,08790,1334
A30,08980,14400,04460,07690,0640
A40,13470,09600,06240,08790,0534
4
Solusi Ideal Positif (A♠) & Negatif (A♣)
Nilai terbaik & terburuk tiap kolom kriteria
\[ A^+_j=\begin{cases}\max_i y_{ij} & \text{benefit}\\ \min_i y_{ij} & \text{cost}\end{cases} \qquad A^-_j=\begin{cases}\min_i y_{ij} & \text{benefit}\\ \max_i y_{ij} & \text{cost}\end{cases} \]
Kolom C1 (Rasa Makanan, benefit): nilai = [0,1048, 0,1347, 0,0898, 0,1347]
A♠1 = MAX (benefit) = 0,1347  |  A♣1 = MIN (benefit) = 0,0898
Kolom C2 (Harga Makanan, cost): nilai = [0,1600, 0,1760, 0,1440, 0,0960]
A♠2 = MIN (cost) = 0,0960  |  A♣2 = MAX (cost) = 0,1760
Kolom C3 (Pelayanan, benefit): nilai = [0,0535, 0,0714, 0,0446, 0,0624]
A♠3 = MAX (benefit) = 0,0714  |  A♣3 = MIN (benefit) = 0,0446
Kolom C4 (Suasana, benefit): nilai = [0,0989, 0,0879, 0,0769, 0,0879]
A♠4 = MAX (benefit) = 0,0989  |  A♣4 = MIN (benefit) = 0,0769
Kolom C5 (Jarak dalam meter, cost): nilai = [0,0800, 0,1334, 0,0640, 0,0534]
A♠5 = MIN (cost) = 0,0534  |  A♣5 = MAX (cost) = 0,1334

Ringkasan Solusi Ideal:

 C1C2C3C4C5
A♠ (ideal positif)0,13470,09600,07140,09890,0534
A♣ (ideal negatif)0,08980,17600,04460,07690,1334
5
Jarak ke Solusi Ideal (D♠ & D♣)
Jarak Euclidean tiap alternatif
\[ D^+_i=\sqrt{\sum_{j=1}^{n}(y_{ij}-A^+_j)^2} \qquad D^-_i=\sqrt{\sum_{j=1}^{n}(y_{ij}-A^-_j)^2} \]

A1 — Warung Mang Ujang

D♠1 = √((0,1048-0,1347)² + (0,1600-0,0960)² + (0,0535-0,0714)² + (0,0989-0,0989)² + (0,0800-0,0534)²) = √0,0060 = 0,0776
D♣1 = √((0,1048-0,0898)² + (0,1600-0,1760)² + (0,0535-0,0446)² + (0,0989-0,0769)² + (0,0800-0,1334)²) = √0,0039 = 0,0624

A2 — Warung Cinta Rasa

D♠2 = √((0,1347-0,1347)² + (0,1760-0,0960)² + (0,0714-0,0714)² + (0,0879-0,0989)² + (0,1334-0,0534)²) = √0,0129 = 0,1137
D♣2 = √((0,1347-0,0898)² + (0,1760-0,1760)² + (0,0714-0,0446)² + (0,0879-0,0769)² + (0,1334-0,1334)²) = √0,0029 = 0,0534

A3 — Warung Padang

D♠3 = √((0,0898-0,1347)² + (0,1440-0,0960)² + (0,0446-0,0714)² + (0,0769-0,0989)² + (0,0640-0,0534)²) = √0,0056 = 0,0751
D♣3 = √((0,0898-0,0898)² + (0,1440-0,1760)² + (0,0446-0,0446)² + (0,0769-0,0769)² + (0,0640-0,1334)²) = √0,0058 = 0,0764

A4 — Warung Bugis

D♠4 = √((0,1347-0,1347)² + (0,0960-0,0960)² + (0,0624-0,0714)² + (0,0879-0,0989)² + (0,0534-0,0534)²) = √0,0002 = 0,0142
D♣4 = √((0,1347-0,0898)² + (0,0960-0,1760)² + (0,0624-0,0446)² + (0,0879-0,0769)² + (0,0534-0,1334)²) = √0,0153 = 0,1235
6
Nilai Preferensi (V) & Perangkingan
Vi = D♣ / (D♠ + D♣)
\[ V_i = \dfrac{D^-_i}{D^+_i + D^-_i} \]
V1 = 0,0624 / (0,0776 + 0,0624) = 0,4456
V2 = 0,0534 / (0,1137 + 0,0534) = 0,3197
V3 = 0,0764 / (0,0751 + 0,0764) = 0,5044
V4 = 0,1235 / (0,0142 + 0,1235) = 0,8972
Hasil Perangkingan
🥈
#2
A3 Warung Padang
V = 0,5044
🥇
#1
A4 Warung Bugis
V = 0,8972
🥉
#3
A1 Warung Mang Ujang
V = 0,4456
RankAlternatifD♠D♣Nilai Preferensi (V)
1 TERBAIKA4 — Warung Bugis0,01420,12350,8972
2A3 — Warung Padang0,07510,07640,5044
3A1 — Warung Mang Ujang0,07760,06240,4456
4A2 — Warung Cinta Rasa0,11370,05340,3197

Berdasarkan perhitungan metode TOPSIS, alternatif terpilih adalah A4 (Warung Bugis) dengan nilai preferensi V = 0,8972, karena memiliki jarak terdekat terhadap solusi ideal positif sekaligus jarak terjauh terhadap solusi ideal negatif.

Aplikasi Pembelajaran SPK — Metode TOPSIS © 2026