@Joseinnewworld struck again — 5 more #NFTs added to his vault around midnight 🌙🔥 At this point, he’s basically the midnight guardian of my collection 😂
— NFToa (@nftoa_) July 18, 2025
Huge thanks, as always 🙌 #NFTcollector #LateNightLegend #eCash $XEC #CryptoWeek pic.twitter.com/8KMs5PsPZH
It is a development of the stepping stone method, because the determination of empty rectangles that can save costs is done with a more certain and precise procedure and this method can achieve optimal results faster. Here are the steps:
Getting to Know the Modified Distribution (MODI) Method
1. Fill in the first table
Fill in the first table from the top left corner to the bottom right.
2. Determine row and column values
The row and column values are determined based on the equation (Ri+Kj = Cij). The first row is always given a value of 0, while the other row and column values are determined based on the calculation results obtained, then the column value associated with the stone quadrilateral can be found using the formula Ri+Kj = Cij.
3. Calculate the improvement index
The improvement index is the value of the ari rectangle (empty rectangle), find it using the formula Cij-Ri-Kj = improvement index.
4. Choosing a starting point for change
A rectangle that has a negative improvement index means "if allocated (filled in) it will be able to reduce the amount of transportation costs. The rectangle that is the starting point for change is the rectangle whose index is marked NEGATIVE with a LARGE NUMBER".
5. Improve allocation
Give a positive (+) sign to the selected rectangle. Choose one nearest rectangle that is filled and in the same row, one nearest rectangle that is filled and in the same column; give a negative (-) sign to these two rectangles. Then choose one rectangle that is in the same row or column with the 2 rectangles that are marked negative (-), and give this rectangle a positive (+) sign. Next, move the allocation from the rectangle that is marked negative (-) to the one marked positive (+) as much as the smallest content of the rectangle that is marked negative (-).
6. Repeat steps starting from step 2.
Repeat the steps above starting from step 2 until the lowest cost is obtained. If there is still a negative improvement index, it means that the allocation can still be changed to reduce transportation costs. If there is no more negative index, it means that it is optimal.
Example
Using sample questions that have been reviewed using the STEPPING STONE method .
1. Fill in the first table
Fill the first table from the top left corner to the bottom right corner
2. Determine row and column values
Formula Ri+Kj = Cij
Determine row and column values
3. Calculate the improvement index
Formula Cij-Ri-Kj = improvement index.
| Segi Empat | Cij – Ri – Kj | Indeks Perbaikan |
|------------|---------------|------------------|
| P1G2 | 10 – 0 – 1 | 9 |
| P1G3 | 8 – 0 – (-3) | 11 |
| P3G1 | 9 – 10 – 3 | -4 |
| P3G2 | 2 – 10 – 1 | -9 |4. Choosing a starting point for change
Selecting the improvement index with the largest negative value = -9 = P3G2.
5. Improve allocation
Fix allocation
Moving allocations
6.a Repeat the steps starting from step 2.
6.a.2. Determining row and column values
Formula Ri+Kj = Cij
Determine row and column values (2nd)
6.a.3. Calculating the improvement index
Formula Cij-Ri-Kj = improvement index.
| Segi Empat | Cij – Ri – Kj | Indeks Perbaikan |
|------------|---------------|------------------|
| P1G2 | 10 – 0 – (-8) | 18 |
| P1G3 | 8 – 0 – (-3) | 11 |
| P2G2 | 5 – 4 – (-8) | 9 |
| P3G1 | 9 – 10 – 3 | -4 |6.a.4. Selecting a starting point for change
Selecting the improvement index with the largest negative value = -4 = P3G1.
6.a.5. Improve allocation
Fix 2nd allocation
Moving the 2nd allocation
6.b Repeat steps starting from step 2
6.b.2. Determining row and column values
Formula Ri+Kj = Cij
Determine the value of the 3rd row and column
6.b.3. Calculating the improvement index
Formula Cij-Ri-Kj = improvement index.
| Segi Empat | Cij – Ri – Kj | Indeks Perbaikan |
|------------|---------------|------------------|
| P1G2 | 10 – 0 – (-4) | 14 |
| P1G3 | 8 – 0 – (-3) | 11 |
| P2G2 | 5 – 4 – (-4) | 5 |
| P3G3 | 7 – 6 – (-3) | 4 |Because the improvement index has reached ≥ 0, it is optimal.
Biaya = 50*3 + 0*7 + 40*1 + 10*9 + 20*2
Biaya = 150 + 0 + 40 + 90 + 40
Biaya = 320
Biaya = 320*Rp100.000,-
Biaya = Rp32.000.000,-Conclusion
So the allocation of units of goods from factories P1, P2, P3 to warehouses G1, G2, G3 in one month is optimal with a minimum cost of IDR 32,000,000.
Reference
Operations Research Module, CHAPTER 3 Transportation Methods, "MODI Method", compiled by Minawarti, ST.