The TOPSIS method is based on the concept that the best selected alternative not only has the shortest distance from the positive ideal solution but also has the longest distance from the negative ideal solution.
TOPSIS method
A. Stages in the TOPSIS Method:
1. Determine criteria and characteristics
The criteria that will be used as a reference in decision making, namely Ci and the nature of each criterion.
2. Determine the suitability rating
Rating the suitability of each alternative on each criterion.
3. Create a normalized decision matrix
TOPSIS requires a performance rating of each alternative Ai on each normalized criterion Cj, namely:
4. Multiplication of the weight by the value of each attribute
This multiplication to form the Y matrix can be determined based on the normalized weight ranking (yij) as follows:
with i=1,2,...,m and j=1,2,...,n
5. Determine the positive ideal solution matrix and the negative ideal solution matrix.
6. Determine the distance between the value of each alternative with the positive and negative ideal solution matrix.
The distance between alternative Ai and the positive ideal solution is formulated as:
The distance between alternative Ai and the negative ideal solution is formulated as:
7. Determine the preference value for each alternative.
The preference value for each alternative (Vi) is given as:
A larger Vi value indicates that alternative Ai is preferred.
Case study
Example questions (taken from the book by Sri Kusumadewi, et al. 2006):
A company in the Special Region of Yogyakarta (DIY) wants to build a warehouse that will be used as a place to temporarily store its production results. There are 3 locations that will be alternatives, namely:
- A1 = Climbing,
- A2 = Kalasan,
- A3 = Big City.
There are 5 criteria that are used as a reference in decision making, namely: decision making, namely:
- C1 = distance to the nearest market (km),
- C2 = population density around the location (people/km2);
- C3 = distance from factory (km);
- C4 = distance to existing warehouse (km);
- C5 = land price for the location (1,000,000 Rp/m2).
The level of importance of each criterion is also assessed from 1 to 5, namely:
- 1 = Very low,
- 2 = Low,
- 3 = Enough,
- 4 = Height,
- 5 = Very High.
The decision maker assigns preference weights to each criterion as follows:
W = (5, 3, 4, 4, 2)
Completion
1. Determine criteria and characteristics
The criteria that will be used as a reference in decision making, namely Ci and the nature of each criterion.
Provide a value for the level of importance of each criterion with the following provisions:
- 1 : not important
- 2 : not too important
- 3 : quite important
- 4 : important
- 5 : very important
The initial weight value (w) is used to indicate the relative importance of each subcriterion.
| Nama Kriteria | Sifat | Alasan | Bobot |
|-------------------------------------------------------|----------------------|------------------------------------------------------------------------------------------------------------------------|-------|
| C1 = jarak dengan pasar terdekat | Biaya / Cost | Karena posisi pabrik yang diharapkan adalah dekat dari pasar agar proses distribusi barang tidak memakan biaya mahal | 5 |
| C2 = kepadatan penduduk di sekitar lokasi (orang/km2) | Biaya / Cost | Karena posisi pabrik yang diharapkan adalah jauh dari perumahan penduduk agar proses distribusi barang tidak terganggu | 3 |
| C3 = jarak dari pabrik (km) | Biaya / Cost | Semakin dekat jarak gudang dengan pabrik, maka akan semakin menguntungkan bagi perusahaan | 4 |
| C4 = jarak dengan gudang yang sudah ada (km); | Keuntungan / Benefit | Semakin jauh jarak gudang dengan posisi gudang sebelumnya, maka akan semakin menguntungkan bagi perusahaan | 4 |
| C5 = harga tanah untuk lokasi (1.000.000 Rp/m2) | Biaya / Cost | Semakin murah harga tanah maka akan semakin menguntungkan bagi perusahaan | 2 |2. Determine the suitability rating
Rating the suitability of each alternative to each criterion, by assigning a criterion value to all alternatives.
| Alternatif | Kriteria | | | | |
|------------|----------|------|----|----|-----|
| | C1 | C2 | C3 | C4 | C5 |
| A1 | 0.75 | 2000 | 18 | 50 | 500 |
| A2 | 0.50 | 1500 | 20 | 40 | 450 |
| A3 | 0.90 | 2050 | 35 | 35 | 800 |3. Create a normalized decision matrix
TOPSIS requires a performance rating of each alternative Ai on each normalized criterion Cj, namely:
| X1 | X2 | X3 | X4 | X5 |
|------|------|----|----|-----|
| 0.75 | 2000 | 18 | 50 | 500 |
| 0.50 | 1500 | 20 | 40 | 450 |
| 0.90 | 2050 | 35 | 35 | 800 |
Normalized matrix R:
| | | | | |
|--------|--------|--------|--------|--------|
| 0,5888 | 0,6186 | 0,4077 | 0,6852 | 0,4784 |
| 0,3925 | 0,4640 | 0,4530 | 0,5482 | 0,4305 |
| 0,7066 | 0,6341 | 0,7928 | 0,4796 | 0,7654 |4. Multiplication of the weight by the value of each attribute
This multiplication to form the Y matrix can be determined based on the normalized weight ranking (yij) as follows:
| | | | | |
|----------|----------|----------|----------|----------|
| 0,5888*5 | 0,6186*3 | 0,4077*4 | 0,6852*4 | 0,4784*2 |
| 0,3925*5 | 0,4640*3 | 0,4530*4 | 0,5482*4 | 0,4305*2 |
| 0,7066*5 | 0,6341*3 | 0,7928*4 | 0,4796*4 | 0,7654*2 |
| | | | | |
|--------|--------|--------|--------|--------|
| 2,9440 | 1,8558 | 1,6309 | 2,7408 | 0,9567 |
| 1,9627 | 1,3919 | 1,8121 | 2,1926 | 0,8611 |
| 3,5328 | 1,9022 | 3,1712 | 1,9185 | 1,5308 |5. Determine the positive ideal solution matrix and the negative ideal solution matrix.
| Nama kriteria | Sifat kriteria | Y+ | Y- |
|--------------------------------------------------------|----------------------|-----------------------------------------------|-----------------------------------------------|
| C1 = jarak dengan pasar terdekat (km), | Biaya/Cost. | Min{2,9440;1,9627;3,5328} = 1,9627 | Max{2,9440;1,9627;3,5328} = 3,5328 |
| C2 = kepadatan penduduk di sekitar lokasi (orang/km2); | Biaya/Cost. | Min{1,8558;1,3919;1,9022} = 1,3919 | Max{1,8558;1,3919;1,9022} = 1,9022 |
| C3 = jarak dari pabrik (km); | Biaya/Cost. | Min{1,6309;1,8121;3,1712} = 1,6309 | Max{1,6309;1,8121;3,1712} = 3,1712 |
| C4 = jarak dengan gudang yang sudah ada (km); | Keuntungan /Benefit. | Max{2,7408;2,1926;1,9185} = 2,7408 | Min{2,7408;2,1926;1,9185} = 1,9185 |
| C5 = harga tanah untuk lokasi (1.000.000 Rp/m2). | Biaya/Cost. | Min{0,9567;0,8611;1,5308} = 0,8611 | Max{0,9567;0,8611;1,5308} = 1,5308 |
| Dapat disimpulkan : | | A+ = {1,9627; 1,3919; 1,6309; 2,7408; 0,8611} | A- = {3,5328; 1,9022; 3,1712; 1,9185; 1,5308} |6.a. The distance between the alternative Ai and the positive ideal solution
6.b. The distance between the alternative Ai and the negative ideal solution
7. Determine the preference value for each alternative.
The preference value for each alternative (Vi) is given as:
Ranking
- A2 Kalasan 0.6689
- A1 Hitting 0.6392
- A3 Big City 0
Reference:
TOPSIS Method Module, compiled by Andri Syafrianto, S.Kom., M.Cs
Netizens
Q1: ERIXON SARUKSUK Apr 28, 2017, 22:44:00 complicated 
haha,, but ELECTREE seems like it too
A1: Maybe it's just because I'm not used to it, bro.
Topsis Method Question Solving - UTS
Case study
To help people have decent homes, the government routinely distributes aid funds to carry out home renovation programs. This aid fund is prioritized for people with low incomes. Every year, Sariharjo Village receives a Home Renovation Fund of Rp. 10,000,000 to repair the house of one of the residents in Sariharjo Village.
Topsis Method Question Solving - UTS
The data collection conducted by the Head of Sariharjo Village this year found 5 houses that were no longer habitable, namely Mr. Poniran's House, Mrs. Ani's House, Mr. Wasito's House, Mr. Kus's House, Mrs. Yatemi's House. There are 5 criteria used to select which houses are eligible for assistance, namely:
1. Amount of income per month (Cost)
2. Number of family dependents (Benefits)
3. Condition of the roof of the house (Benefit)
- Give a score of 3 if the roof of the house is badly damaged and endangers the occupants of the house,
- Give a score of 2 if the damage to the roof of the house is moderate, where it could collapse over a long period of time.
- Give a score of 1 if the damage to the roof of the house is light.
4. Condition of the house walls (Benefit)
- Give a score of 5 if the walls of the house are still not permanent / woven bamboo,
- Give a score of 4 if the walls of the house are permanent but in fragile condition,
- Give a score of 2 if the walls of the house are permanent with a good frame condition but have not been plastered.
- Give a score of 0 if the wall is permanent, in good condition and has been painted.
5. Communication in the Community Environment (Benefit)
- Give a score of 3 if the resident concerned is active in village activities,
- Give a score of 2 if the resident concerned is quite active in village activities, and
- Give a score of 0 if the resident concerned is not active in village activities.
The weight of each criterion is assessed on a scale of 1 to 5.
- 1 = Very low,
- 2 = Low,
- 3 = Enough,
- 4 = Height,
- 5 = Very High.
The decision maker assigns a weight to each criterion is W = (5, 4, 4, 5, 3) .
Based on the results of the Village Head's assessment, the condition of each house is described in the table below:
| 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 |Using the TOPSIS method, determine which of the 5 residents has the right to have their house renovated?
Completion
1. Determine Criteria and Characteristics
| Nama Kriteria | Sifat | Bobot |
|---------------|---------|-------|
| C1 | Cost | 5 |
| C2 | Benefit | 4 |
| C3 | Benefit | 4 |
| C4 | Benefit | 5 |
| C5 | Benefit | 3 |2. Determine the Match Rating
| 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. Determining the Normalized Decision Matrix
| 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 |
Normalized Matrix 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. Multiplication of Weights by the Values of Each Attribute (R Matrix)
The Y matrix is obtained as follows.
| 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 |become,
| 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. Determining Positive & Negative Ideal Solution Matrix
| 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. Distance Between Alternative Ai and Positive Ideal Solution
6.b. Distance Between Alternative Ai and Negative Ideal Solution
7. Determine the Preference Value for Each Alternative
8. Ranking
- Mr. Poniran = 0.712878208
- Mr. Kus = 0.630027260
- Mrs. Ani = 0.527296921
- Mrs. Yatemi = 0.527296921
- Mr. Wasito = 0.472703079