Simple Additive Weighting Method (SAWM)

One of the problem solving methods in SPK is by using the Simple Additive Weighting (SAW) method. The SAW method is also often known as the weighted summation method.

Simple Additive Weighting (SAW) Method

The basic concept of the SAW method is to find the weighted sum of the performance ratings on each alternative from all attributes (Fishburn, 1967) [3]. This SAW method requires decision makers to determine the weight for each attribute. The total score for an alternative is obtained by adding up all the results of the multiplication between the ratings (which can be compared across attributes) and the weight of each attribute. The rating of each attribute must be dimensionless in the sense that it has gone through the previous matrix normalization process. The SAW method requires a normalization process of the decision matrix (X) to a scale that can be compared with all existing alternative ratings (Kusumadewi, 2006) [4].

SAW Completion Steps are as follows:

1. Determine the criteria

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

That is the rating of the suitability of each alternative for each criterion.

3. Create a decision matrix

The decision matrix is ​​created based on the criteria (Ci), then the matrix is ​​normalized based on the equation that is adjusted to the type of attribute (profit attribute or cost attribute) so that the normalized matrix R is obtained.

The formula for performing this normalization is:

SAW Method Normalization Formula

Information:

• Max xij = maximum value of each column
• Min xij = minimum value of each column
• xij = rows and columns of the matrix
• rij = normalized performance rating of alternative Ai on attribute Cj; i =1,2,...m and
• j = 1,2,...,n.

4. Final result

The final result is the conclusion obtained from the ranking process, namely the sum of the multiplication of the normalized matrix R with the weight vector so that the largest value is obtained which is selected as the best alternative (Ai) as a solution.

The preference value for each alternative (Vi) is given as:

Formula-Preference-Value-Each-Alternative

Information:

• Vi = Final value of the alternative
• wj = Predetermined weight
• rij = Matrix normalization

A larger Vi value indicates that the alternative Ai is more preferred.

Case Study Example

A higher education institution will select one of its employees to be promoted as the head of the Innovation Center unit. There are 4 criteria used to conduct the assessment, namely:

• C1 = knowledge test (insight) about building a system
• C2 = system design test
• C3 = personality test
• C4 = leadership test

The university leadership gives weight to each criterion as follows:

• C1 = 35%
• C2 = 25%
• C3 = 25%
• C4 = 15%

There are 6 employees who are candidates (alternatives) to be promoted as unit heads, namely:

• A1 = Indra Buana, ST, M.Kom
• A2 = Roni Pranowo, S.Kom., M.Kom
• A3 = Putri Herlina, S.Si., M.Kom
• A4 = Dani Ariatmanto, S.Kom., M.Kom
• A5 = Ratna Pradhana, ST, M.Cs
• A6 = Mira Santika, S.Si., M.Sc.

Completion

Using the SAW method.

1. Determine the criteria

The criteria that will be used as a reference in decision making, namely Ci and the nature of each criterion.

| Kriteria | Nama Kriteria                                              | Bobot               | Sifat   |
|----------|------------------------------------------------------------|---------------------|---------|
| C1       | Tes pengetahuan (wawasan) tentang pembangunan suatu sistem | 35% = 35/100 = 0.35 | Benefit |
| C2       | Tes perancangan suatu sistem                               | 25% = 25/100 = 0.25 | Benefit |
| C3       | Tes kepribadian                                            | 25% = 25/100 = 0.25 | Benefit |
| C4       | Tes kepemimpinan                                           | 15% = 15/100 = 0.15 | Benefit |

2. Determine the suitability rating

Rating the suitability of each alternative on each criterion. At this stage, the manager will provide an assessment of each candidate on each of the criteria that have been set. The range of values ​​given is 0 -- 100.

| Alternatif | Kriteria |    |    |    |
|------------|----------|----|----|----|
|            | C1       | C2 | C3 | C4 |
| Indra      | 70       | 50 | 80 | 60 |
| Roni       | 50       | 60 | 82 | 70 |
| Putri      | 85       | 55 | 80 | 75 |
| Dani       | 82       | 70 | 65 | 85 |
| Ratna      | 75       | 75 | 85 | 74 |
| Mira       | 62       | 50 | 75 | 80 |

3. Create a decision matrix

Decision matrix based on criteria (Ci).

Decision matrix explanation

The result,

|    |    |    |    |
|----|----|----|----|
| 70 | 50 | 80 | 60 |
| 50 | 60 | 82 | 70 |
| 85 | 55 | 80 | 75 |
| 82 | 70 | 65 | 85 |
| 75 | 75 | 85 | 74 |
| 62 | 50 | 75 | 80 |

Then perform matrix normalization based on the equation adjusted to the type of attribute (profit attribute or cost attribute) to obtain the normalized matrix R.

COLUMN 1

Line 1

r11 = x11 / max {x11;x21;x31;x41;x51;x61}
r11 = 70 / max {70;50;85;82;75;62}
r11 = 70 / 85
r11 = 0.82

Line 2

r21 = x21 / max {x11;x21;x31;x41;x51;x61}
r21 = 50 / max {70;50;85;82;75;62}
r21 = 50 / 85
r21 = 0.59

Line 3

r31 = x31 / max {x11;x21;x31;x41;x51;x61}
r31 = 85 / max {70;50;85;82;75;62}
r31 = 85 / 85
r31 = 1

Line 4

r41 = x41 / max {x11;x21;x31;x41;x51;x61}
r41 = 82 / max {70;50;85;82;75;62}
r41 = 82 / 85
r41 = 0.96

Line 5

r51 = x51 / max {x11;x21;x31;x41;x51;x61}
r51 = 75 / max {70;50;85;82;75;62}
r51 = 75 / 85
r51 = 0.88

Line 6

r61 = x61 / max {x11;x21;x31;x41;x51;x61}
r61 = 62 / max {70;50;85;82;75;62}
r61 = 62 / 85
r61 = 0.73

COLUMN 2

Line 1

r12 = x12 / max {x12;x22;x32;x42;x52;x62}
r12 = 50 / max {50;60;55;70;75;50}
r12 = 50 / 75
r12 = 0.67

Line 2

r22 = x22 / max {x12;x22;x32;x42;x52;x62}
r22 = 60 / max {50;60;55;70;75;50}
r22 = 60 / 75
r22 = 0.8

Line 3

r32 = x32 / max {x12;x22;x32;x42;x52;x62}
r32 = 55 / max {50;60;55;70;75;50}
r32 = 55 / 75
r32 = 0.73

Line 4

r42 = x42 / max {x12;x22;x32;x42;x52;x62}
r42 = 70 / max {50;60;55;70;75;50}
r42 = 70 / 75
r42 = 0.93

Line 5

r52 = x52 / max {x12;x22;x32;x42;x52;x62}
r52 = 75 / max {50;60;55;70;75;50}
r52 = 75 / 75
r52 = 1

Line 6

r62 = x62 / max {x13;x23;x33;x43;x53;x63}
r62 = 50 / max {50;60;55;70;75;50}
r62 = 50 / 75
r62 = 0.67

COLUMN 3

Line 1

r13 = x13 / max {x13;x23;x33;x43;x53;x63}
r13 = 80 / max {80;82;80;65;85;75}
r13 = 80 / 85
r13 = 0.94

Line 2

r23 = x23 / max {x13;x23;x33;x43;x53;x63}
r23 = 82 / max {80;82;80;65;85;75}
r23 = 82 / 85
r23 = 0.96

Line 3

r33 = x33 / max {x13;x23;x33;x43;x53;x63}
r33 = 80 / max {80;82;80;65;85;75}
r33 = 80 / 85
r33 = 0.94

Line 4

r43 = x43 / max {x13;x23;x33;x43;x53;x63}
r43 = 65 / max {80;82;80;65;85;75}
r43 = 65 / 85
r43 = 0.76

Line 5

r53 = x53 / max {x13;x23;x33;x43;x53;x63}
r53 = 85 / max {80;82;80;65;85;75}
r53 = 85 / 85
r53 = 1

Line 6

r63 = x63 / max {x13;x23;x33;x43;x53;x63}
r63 = 75 / max {80;82;80;65;85;75}
r63 = 75 / 85
r63 = 0.88

COLUMN 4

Line 1

r14 = x14 / max {x14;x24;x34;x44;x54;x64}
r14 = 60 / max {60;70;75;85;74;80}
r14 = 60 / 85
r14 = 0.71

Line 2

r24 = x24 / max {x14;x24;x34;x44;x54;x64}
r24 = 70 / max {60;70;75;85;74;80}
r24 = 70 / 85
r24 = 0.82

Line 3

r34 = x34 / max {x14;x24;x34;x44;x54;x64}
r34 = 75 / max {60;70;75;85;74;80}
r34 = 75 / 85
r34 = 0.88

Line 4

r44 = x44 / max {x14;x24;x34;x44;x54;x64}
r44 = 85 / max {60;70;75;85;74;80}
r44 = 85 / 85
r44 = 1

Line 5

r54 = x54 / max {x14;x24;x34;x44;x54;x64}
r54 = 74 / max {60;70;75;85;74;80}
r54 = 74 / 85
r54 = 0.87

Line 6

r64 = x64 / max {x14;x24;x34;x44;x54;x64}
r64 = 80 / max {60;70;75;85;74;80}
r64 = 80 / 85
r64 = 0.94

The normalization result matrix obtained:

Normalization Matrix

The result,

|      |      |      |      |
|------|------|------|------|
| 0.82 | 0.67 | 0.94 | 0.71 |
| 0.59 | 0.8  | 0.96 | 0.82 |
| 1    | 0.73 | 0.94 | 0.88 |
| 0.96 | 0.93 | 0.76 | 1    |
| 0.88 | 1    | 1    | 0.87 |
| 0.73 | 0.67 | 0.88 | 0.94 |

4. Final result 

The final result is obtained from the ranking process, namely the sum of the multiplication of the normalized matrix R with the weight vector so that the largest value is obtained which is selected as the best alternative (Ai) as a solution.

The ranking process uses the weights given by the decision maker:

w = [0,35 ; 0,25 ; 0,25 ; 0,15]

The ranking results obtained using the 2nd formula are:

v1 = (0,35* 0,82) + (0,25* 0,67) + (0,25* 0,94) + (0,15* 0,71) = 0,80
v2 = (0,35*0,59) + (0,25*0,8) + (0,25*0,96) + (0,15*0,82) = 0,77
v3 = (0,35*1) + (0,25*0,73) + (0,25*0,94) + (0,15*0,88) = 0,90
v4 = (0,35*0,96) + (0,25*0,93) + (0,25*0,76) + (0,15*1) = 0,91
v5 = (0,35*0,91) + (0,25*1) + (0,25*1) + (0,15*0,87) = 0,94 <-----jawaban
v6 = (0,35*0,73) + (0,25*0,67) + (0,25*0,88) + (0,15*0,94) = 0,78

The highest value is in v5 so that alternative A5 is the alternative chosen as the best alternative. in other words, Mrs. Ratna Pradhana, ST, M.Cs will be chosen as the head of the innovation center unit.

Exercises

Simple Additive Weighting (SAW) Method Problem Solving

Reference:

3rd Meeting Module MCDM & SAW, compiled by Andri Syafrianto, S.Kom., M.Cs

Netizens

• Bundle
• Oct 19, 2020

Q1: YANIX NOPHITHA 26 Apr 2017, 17:33:00 cool, easy to understand

A1: Hii YANIX NOPHITHA, welcome to the bundet, happy to be able to provide benefits, greetings admin.

• anjibari
• Sep 28, 2021

thanks for the easy to understand explanation

Case study:

A consumer wants to buy a house in the Special Region of Yogyakarta (DIY). There are 5 locations that will be alternatives, namely:

Simple Additive Weighting (SAW) Method Problem Solving

Alternative 1

• It is the Mutiara Palagan Housing Complex located on Jalan Tentara Pelajar, Palagan km. 10.
• Condition: Residential area on privately owned land, Freehold status, price 350 million, good water and air conditions, Public facilities have security posts, type of building is an improved quality type.

Alternative 2 

• It is the Kirana Mulia 1 Housing Complex located on Jalan Damai, Palagan km. 9.
• Condition: Residential area on privately owned land, Freehold status, price 250 million, good water and air conditions, Public facilities have security posts, type of building is an improved quality type.

Alternative 3 

• It is the Puri Rajawali Kamdanen which is located on Jalan Damai Palagan km. 9.
• Condition: Residential area on privately owned land, Freehold status, price 300 million, good water and air conditions, Public facilities have security posts, type of building is an improved quality type.

Alternative 4 

• It is the Taman Indah Housing Complex located on Jalan Tentara Pelajar, Palagan km. 10.
• Condition: Residential area in a non-housing area, village land status, price 200 million, water conditions are not good but air conditions are good, does not have public facilities, type of building is an improved quality type.

Final

• It is the Taman Bunga Housing Complex located on Jalan Damai, Palagan km. 9.5.
• Condition: Residential area in a non-housing area, village land status, price 180 million, water conditions are not good but air conditions are good, does not have public facilities, type of building is an improved quality type.

There are 6 criteria that are used as a reference in decision making, namely:

1. Suitability of Land Use

The suitability of land use referred to in this point is related to the function of residential land before it is used as a settlement. The suitability of this land use can be divided into:

• a. Residential area on privately owned land
• b. Residential area on state-owned land
• c. Residential areas in disaster-prone areas
• d. Residential areas in non-residential designated areas

Classification at this point can be divided into:

• Good: if point a is the initial condition of the housing land. If the alternative is in this condition, the value that will be given is 1
• Bad: if the location of the housing land is in conditions b, c, and d. If the alternative is in this condition, the value that will be given is 0.1.

2. Land and Building Ownership Status

Classification at this point can be divided into areas with land status and ownership:

• a. Ownership Rights
• b. Rent
• c. Have no rights (Wild)

Classification at this point can be divided into:

• Good: if point a is the initial condition of the housing land. If the alternative is in this condition, the value that will be given is 1
• Bad: if the location of the housing land is in conditions b, and c. If the alternative is in this condition, the value that will be given is 0.5.

3. Housing Prices

4. Environmental Conditions

The environmental component variables that are the main considerations for residents in choosing a housing location are water quality and air quality. Classification at this point can be divided into:

• Good: if the water and air quality is good. Give a score of 1 for housing that is in good environmental conditions.
• Sufficient: if one of the variables (air or water) is not of good quality, give a score of 0.5 for housing that is in a fairly good environmental condition.
• Bad: if both variables are not in good condition. Give a value of 0.1 for housing in poor environmental conditions.

5. Condition of Public Facilities

Adequate

A housing is said to have adequate facilities and infrastructure if there are at least 3 important public facilities, namely: a place of worship, a playground, and a security post. Give a score of 1 for housing that is in a condition of adequate public facilities.

Currently

A housing is said to have moderate facilities and infrastructure if there are at least 2 or 1 important public facilities, namely: a place of worship or a security post. Give a value of 0.5 for housing that is in a moderate public facility condition.

Not enough

A housing is said to have inadequate facilities and infrastructure if there are no public facilities in the housing. Give a score of 0.1 for housing that is in a condition of inadequate public facilities.

6. House Condition and Building Type

a. Standard type

• 1-storey building has the number of rooms and building facilities according to the type
• has a floor with cement pavement
• brick/crystal block walls have not been finished/painted
• tile roof with standard color
• have electricity connection and clean water source
• has a sewage drain and septic tank.

b. Type of quality improvement

• The floor has been upgraded to ceramic flooring
• The walls have been enhanced with a wall paint finish
• The bathroom/toilet already uses ceramic floors and walls
• The roof tiles already use a certain color.

Classification at this point can be divided into:

• Good: if point a is the condition of a housing building. If the alternative is in this condition, the value that will be given is 0.25
• Bad: if the condition of the housing building is in condition b. If the alternative is in this condition, the value that will be given is 1.

Completion

1. Determine the criteria that will be used as a reference in decision making, namely Ci and the nature of each criterion.

The level of importance of each criteria weight is assessed from a range of 1 to 5, namely:

• 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.

| Kriteria | Nama Kriteria                         | Bobot | Sifat   |
|----------|---------------------------------------|-------|---------|
| C1       | Kesesuaian peruntukan lahan           | 5     | Benefit |
| C2       | Status kepemilikan tanah dan bangunan | 5     | Benefit |
| C3       | Harga perumahan                       | 3     | Cost    |
| C4       | Kondisi lingkungan                    | 4     | Benefit |
| C5       | Kondisi fasilitas umum                | 3     | Benefit |
| C6       | Kondisi rumah dan jenis bangunan      | 5     | Benefit |

2. Determine the suitability rating of each alternative for each criterion.

| Alternatif | Kriteria |     |     |     |     |    |
|------------|----------|-----|-----|-----|-----|----|
|            | C1       | C2  | C3  | C4  | C5  | C6 |
| 1          | 1        | 1   | 3.5 | 1   | 0.5 | 1  |
| 2          | 1        | 1   | 2.5 | 1   | 0.5 | 1  |
| 3          | 1        | 1   | 3   | 1   | 0.5 | 1  |
| 4          | 0.1      | 0.5 | 2   | 0.5 | 0.1 | 1  |
| 5          | 0.1      | 0.5 | 1.8 | 0.5 | 0.1 | 1  |

3. Create a decision matrix based on criteria (Ci)

|     |     |     |     |     |   |
|-----|-----|-----|-----|-----|---|
| 1   | 1   | 3.5 | 1   | 0.5 | 1 |
| 1   | 1   | 2.5 | 1   | 0.5 | 1 |
| 1   | 1   | 3   | 1   | 0.5 | 1 |
| 0.1 | 0.5 | 2   | 0.5 | 0.1 | 1 |
| 0.1 | 0.5 | 1.8 | 0.5 | 0.1 | 1 |

Then perform matrix normalization based on the equation adjusted to the type of attribute (profit attribute or cost attribute) to obtain the normalized matrix R.

4. Create a normalization matrix

Normalization Process:

Column 1 ( Benefit → rij = xij / max (x11;x21;x31;x41;x51) )

r11 = 1 / max (1;1;1;0.1;0.1)
r11 = 1 / 1
r11 = 1 r21
= 1 / max (1;1;1;0.1;0.1)
r21 = 1 / 1
r21 = 1
r31 = 1 / max (1;1;1;0.1;0.1)
r31 = 1 / 1
r31 = 1
r41 = 0.1 / max (1;1;1;0.1;0.1)
r41 = 0.1 / 1
r41 = 0.1
r51 = 0.1 / max (1;1;1;0.1;0.1)
r51 = 0.1 / 1
r51 = 0.1

Column 2 ( Benefit → rij = xij / max (x12;x22;x32;x42;x52) )

r12 = 1 / max (1;1;1;0.5;0.5)
r12 = 1 / 1
r12 = 1
r22 = 1 / max (1;1;1;0.5;0.5)
r22 = 1 / 1
r22 = 1
r32 = 1 / max (1;1;1;0.5;0.5)
r32 = 1 / 1
r32 = 1
r42 = 0.5 / max (1;1;1;0.5;0.5)
r42 = 0.5 / 1
r42 = 0.5
r52 = 0.5 / max (1;1;1;0.5;0.5)
r52 = 0.5 / 1
r52 = 0.5

Column 3 ( Cost → rij = min (x13;x23;x33;x43;x53) / xij)

r13 = min (3.5;2.5;3;2;1.8) / 3.5
r13 = 1.8 / 3.5
r13 = 0.5
r23 = min (3.5;2.5;3;2;1.8) / 2.5
r23 = 1.8 / 2.5
r23 = 0.7
r33 = min (3.5;2.5;3;2;1.8) / 3
r33 = 1.8 / 3
r33 = 0.6
r43 = min (3.5;2.5;3;2;1.8) / 2 
r43 = 1.8 / 2
r43 = 0.9
r53 = min (3.5;2.5;3;2;1.8) / 1.8 
r53 = 1.8 / 1.8
r53 = 1

Column 4 ( Benefit → rij = xij / max (x14;x24;x34;x44;x54) )

r14 = 1 / max (1;1;1;0.5;0.5)
r14 = 1 / 1
r14 = 1 r24
= 1 / max (1;1;1;0.5;0.5)
r24 = 1 / 1
r24 = 1
r34 = 1 / max (1;1;1;0.5;0.5)
r34 = 1 / 1
r34 = 1
r44 = 0.5 / max (1;1;1;0.5;0.5)
r44 = 0.5 / 1
r44 = 0.5
r54 = 0.5 / max (1;1;1;0.5;0.5)
r54 = 0.5 / 1
r54 = 0.5

Column 5 ( Benefit → rij = xij / max (x15;x25;x35;x45;x55) )

r15 = 0.5 / max (0.5; 0.5;0.5;0.1;0.1 )
r15 = 0.5 / 0.5
r15 = 1 r25 = 0.5 / max (0.5;0.5;0.5;0.1;0.1) r25 = 0.5 / 0.5 r25 = 1 r35 = 0.5 / max (0.5;0.5;0.5;0.1;0.1) r35 = 0.5 / 0.5 r35 = 1 r45 = 0.1 / max (0.5;0.5;0.5;0.1;0.1) r45 = 0.1 / 0.5 r45 = 0.2 r55 = 0.1 / max (0.5;0.5;0.5;0.1;0.1) r55 = 0.1 / 0.5 r55 = 0.2

Column 6 ( Benefit → rij = xij / max (x16;x26;x36;x46;x56) )

r16 = 1 / max (1;1;1;1;1)
r16 = 1 / 1
r16 = 1
r26 = 1 / max (1;1;1;1;1)
r26 = 1 / 1
r26 = 1
r36 = 1 / max (1;1;1;1;1)
r36 = 1 / 1
r36 = 1
r46 = 1 / max (1;1;1;1;1)
r46 = 1 / 1
r46 = 1
r56 = 1 / max (1;1;1;1;1)
r56 = 1 / 1
r56 = 1

The normalization result matrix obtained:

|     |     |     |     |     |   |
|-----|-----|-----|-----|-----|---|
| 1   | 1   | 0.5 | 1   | 1   | 1 |
| 1   | 1   | 0.7 | 1   | 1   | 1 |
| 1   | 1   | 0.6 | 1   | 1   | 1 |
| 0.1 | 0.5 | 0.9 | 0.5 | 0.2 | 1 |
| 0.1 | 0.5 | 1   | 0.5 | 0.2 | 1 |

5. Ranking

The final result is obtained from the ranking process, namely the sum of the multiplication of the normalized matrix R with the weight vector so that the largest value is obtained which is selected as the best alternative (Ai) as a solution.

Predetermined weight:

w = [5; 5; 3; 4; 3; 5]

Ranking Process:

Step 1

|       |       |   |       |   |       |   |       |   |       |   |     |
|-------|-------|---|-------|---|-------|---|-------|---|-------|---|-----|
| V1 =  | 5*1   | + | 5*1   | + | 3*0.5 | + | 4*1   | + | 3*1   | + | 5*1 |
| V2 =  | 5*1   | + | 5*1   | + | 3*0.7 | + | 4*1   | + | 3*1   | + | 5*1 |
| V3 =  | 5*1   | + | 5*1   | + | 3*0.6 | + | 4*1   | + | 3*1   | + | 5*1 |
| V4 =  | 5*0.1 | + | 5*0.5 | + | 3*0.9 | + | 4*0.5 | + | 3*0.2 | + | 5*1 |
| V5 =  | 5*0.1 | + | 5*0.5 | + | 3*1   | + | 4*0.5 | + | 3*0.2 | + | 5*1 |

Step 2

|      |     |   |     |   |     |   |   |   |     |   |   |
|------|-----|---|-----|---|-----|---|---|---|-----|---|---|
| V1 = | 5   | + | 5   | + | 2.5 | + | 4 | + | 3   | + | 5 |
| V2 = | 5   | + | 5   | + | 2.1 | + | 4 | + | 3   | + | 5 |
| V3 = | 5   | + | 5   | + | 1.8 | + | 4 | + | 3   | + | 5 |
| V4 = | 0.5 | + | 2.5 | + | 2.7 | + | 2 | + | 0.6 | + | 5 |
| V5 = | 0.5 | + | 2.5 | + | 3   | + | 2 | + | 0.6 | + | 5 |

Step 3

|      |      |
|------|------|
| V1 = | 24.5 |
| V2 = | 24.1 |
| V3 = | 23.8 |
| V4 = | 13.3 |
| V5 = | 13.6 |

Ranking

1. = V1
2. = V2
3. = V3
4. = V5
5. = V4

The highest value is in V1, so ALTERNATIVE 1 located in Mutiara Palagan Housing on Jalan Tentara Pelajar, Palagan km. 10, is a house worth buying.

Reference:

3rd Meeting Module MCDM & SAW, compiled by Andri Syafrianto, S.Kom., M.Cs

Netizens

• Mike
• Jun 2, 2021

What software is used to complete this method, bro? Thx

• Bundle
• Jun 6, 2021
• Edited

Mike  , ideally the software is made by IT graduates, but for just prototyping and testing algorithms... you can use MS Excel or spreadsheet... both are powerful enough to implement the algorithm into a formula.