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It is an easier and faster method to be able to manage allocations from several sources to several marketing areas. With the following steps:
- Arrange the needs, capacity of each source, and transportation costs into a matrix.
- Find the difference of the two smallest costs, namely the first smallest cost and the second smallest cost for each row and column in the matrix (Cij).
- Fill in one of the rectangles included in the selected column or row, namely the rectangle with the lowest cost among the other rectangles in that column/row. Fill in as many as possible.
- Remove rows/columns that are already fully filled (full capacity) so that they cannot be filled again. Then pay attention to columns and rows that have not been filled/allocated.
- Re-determine the cost differences in step 2 for the unallocated columns and rows. Repeat steps 3 through 5, until all rows and columns are fully allocated.
- Once everything is filled in, calculate the transportation costs.
- If there are 2 cost difference values that are the same, look at the rectangle that is included in the column or row that has the largest value. If this rectangle has the lowest cost among the rectangles in the row or column, then fill in the maximum allocation on this rectangle. If the cost is not the lowest, then choose the rectangle to be filled based on one of the selected rows or selected columns.
Getting to Know the Vogel Approximation Method
Example
Using case studies that have been worked on using the Stepping Stone and MODI methods previously.
Previous matrices (Stepping Stone and/or MODI)
1. Arrange the needs, capacity of each source, and transportation costs into a matrix.
Arrange the allocations into a table
Arrange the alloxans into a table (vogel)
According to Vogel's method, the solution is optimal, but this is not necessarily the case according to other methods, so it must be tested first using the following method:
1. Stepping Stone
a. Calculating Zij of Non-Basic Variables
Cell (1,2) Loop (1,1) (3,1) (3,2) (1,2)
Z12 = C11-C31+C32-C12
Z12 = 3-9+2-10
Z12 = -14
Cell (1,3) Loop (1,1) (2,1) (2,3) (1,3)
Z13 = C11-C21+C23-C13
Z13 = 3-7+1-8
Z13 = -11
Cell (2,2) Loop (2,1) (3,1) (3,2) (2,2)
Z22 = C21-C31+C32-C22
Z22 = 7-9+2-5
Z22 = -5
Cell (3,3) Loop (3,1) (2,1) (2,3) (3,3)
Z31 = C31-C21+C23-C33
Z31 = 9-7+1-7
Z31 = -4
Since the non-basic variable Zij is already ≤ 0, it is optimal.
2. MODI
Determine the row and column values using the formula: Ri+Kj = Cij
MODI matrix for Vogel's
2.3. 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, "Vogel's Approximation", compiled by Minawarti, ST.
UAS Vogel's Approximation Method
It is known that the transportation problem of ASM Bakery Company in production has 3 factories ASM1, ASM2, ASM3, each day can produce 30, 40, 50 boxes of bread containing various flavors. To sell it, a counter is opened at MM, Amplaz, GM which can sell 10, 20, 90 boxes of bread each day. The shipping cost (in IDR 10,000) from ASM1 to MM, Amplaz, GM is 8, 10, 11 respectively, from ASM2 to MM, Amplaz, GM is 9, 7, 10 respectively, ASM3 is 11, 12, 9 respectively. Determine the optimal allocation of bread boxes from factories ASM1, ASM2, ASM3 to MM, Amplaz, GM to obtain the lowest shipping cost. Solve the transportation problem above using the Stepping Stone or MODI or VAM method (if using VAM, test the final result using the Stepping Stone or MODI method). (Point 40%).
UAS Vogel Method, ES EF LS LF, and Matrix
Completion
The solution to the 2nd case study is HERE .
1. Using Vogel's Method
In this case I use variables A1, A2, A3 for the factory names (ASM1, ASM2, ASM3), and G1, G2, G3 for the destinations (MM, Amplaz, GM).
Arranging Allocations into a Table / Matrix
According to Vogel's method, the solution is optimal, but this is not necessarily the case according to other methods, so it must be tested first using the following method:
2. Testing to Stepping Stone
a. Calculating Zij of Non-Basic Variables
Cell (1,2) Loop (2,2) (2,3) (1,3) (1,2)
Z12 = C22-C23+C13-C12
Z12 = 7-10+11-10
Z12 = -2
Cell (2,1) Loop (2,3) (1,3) (1,1) (2,1)
Z21 = C23-C13+C11-C21
Z21 = 10-11+8-9
Z21 = -2
Cell (3,1) Loop (3,3) (1,3) (1,1) (3,1)
Z31 = C33-C13+C11-C31
Z31 = 9-11+8-11
Z31 = -5
Cell (3,2) Loop (3,3) (2,3) (2,2) (3,2)
Z32 = C33-C23+C22-C32
Z32 = 9-10+7-12
Z32 = -6
Since the non-basic variable Zij is already ≤ 0, it is optimal.
3. Testing to MODI
Determine the row and column values using the formula: Ri+Kj = Cij
MODI matrix for Vogel's
Calculating the Improvement Index with the Formula Cij-Ri-Kj = improvement index.
| Segi Empat | Cij – Ri – Kj | Indeks Perbaikan |
|------------|---------------|------------------|
| A1G2 | 10 – 0 – 8 | 2 |
| A2G1 | 9 – (-1) – 8 | 2 |
| A3G1 | 11 – (-2) – 8 | 5 |
| A3G2 | 12 – (-2) – 8 | 6 |Because the improvement index has reached ≥ 0, it is optimal.
- Cost = 10 8 + 20 11 + 20 7 + 20 10 + 50*9
- Cost = 80 + 220 + 140 + 200 + 450
- Cost = 1,090 (scale 1:10,000,-)
- Cost = 1,090*Rp10,000,-
- Cost = Rp. 10,900,000,-
Conclusion
So the allocation of units of goods from factories ASM1, ASM2, ASM3 to warehouses MM, AMPLAZ, GM is optimal with a minimum cost of Rp. 10,900,000.