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Sagot :
### Step 1: Formulate the Initial Problem
We need to transport products from 3 farms to 4 destinations with the cost per unit provided. The goal is to minimize the total transportation cost. Here are the given data:
- Supply:
- FARM 1: 7 units
- FARM 2: 9 units
- FARM 3: 18 units
- Demand:
- D1: 5 units
- D2: 8 units
- D3: 7 units
- D4: 14 units
- Cost Matrix:
[tex]\[ \begin{array}{cccc} 19 & 30 & 50 & 10 \\ 70 & 30 & 40 & 60 \\ 40 & 8 & 70 & 20 \\ \end{array} \][/tex]
### Step 2: Start with the Least Cost Method (LCM)
Step 2.1: Initialize the Matrix
We initialize an allocation matrix to keep track of the number of units transported between farms and destinations.
Step 2.2: Allocation Process
- Find the least cost cell (10 in FARM1 to D4):
- Allocate min(supply, demand) = min(7, 14) = 7 units.
- Adjust Supply and Demand:
- Supply FARM1: 7 - 7 = 0
- Demand D4: 14 - 7 = 7
- Find Next Least Cost Cell (8 in FARM3 to D2):
- Allocate min(supply, demand) = min(18, 8) = 8 units.
- Adjust Supply and Demand:
- Supply FARM3: 18 - 8 = 10
- Demand D2: 8 - 8 = 0
- Find Next Least Cost Cell (19 in FARM1 to D1), no units allocated to any cell in FARM1 as its supply is 0.
- Find (30 in FARM2 to D1):
- Allocate min(supply, demand) = min(9, 5) = 5 units.
- Adjust Supply and Demand:
- Supply FARM2: 9 - 5 = 4
- Demand D1: 5 - 5 = 0
- Find (30 in FARM2 to D3):
- Allocate min(supply, demand) = min(4, 7) = 4 units.
- Adjust Supply and Demand:
- Supply FARM2: 4 - 4 = 0
- Demand D3: 7 - 4 = 3
- Find (20 in FARM3 to D4):
- Allocate min(supply, demand) = min(10, 7) = 7 units.
- Adjust Supply and Demand:
- Supply FARM3: 10 - 7 = 3
- Demand D4: 7 - 7 = 0
- Find (40 in FARM3 to D1):
- Allocate min(supply, demand) = min(3, 3) = 3 units.
- Adjust Supply and Demand at last as all other demands are satisfied.
### Step 3: Compile the Final Allocation Table
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline FARM1 & 0 & 0 & 0 & 7 \\ \hline FARM2 & 5 & 0 & 4 & 0 \\ \hline FARM3 & 3 & 8 & 3 & 7 \\ \hline \end{array} \][/tex]
### Step 4: Calculate the Total Transportation Cost
Using the allocation and cost matrix:
[tex]\[ \text{Total Cost} = (0 \times 19) + (0 \times 30) + (0 \times 50) + (7 \times 10) + (5 \times 70) + (0 \times 30) + (4 \times 40) + (0 \times 60) + (3 \times 40) + (8 \times 8) + (3 \times 70) + (7 \times 20) \][/tex]
Simplifying the calculation:
[tex]\[ \text{Total Cost} = 0 + 0 + 0 + 70 + 350 + 0 + 160 + 0 + 120 + 64 + 210 + 140 = 814 \][/tex]
Thus, the optimal allocation using the Least Cost Method results in a total transportation cost of 814.
We need to transport products from 3 farms to 4 destinations with the cost per unit provided. The goal is to minimize the total transportation cost. Here are the given data:
- Supply:
- FARM 1: 7 units
- FARM 2: 9 units
- FARM 3: 18 units
- Demand:
- D1: 5 units
- D2: 8 units
- D3: 7 units
- D4: 14 units
- Cost Matrix:
[tex]\[ \begin{array}{cccc} 19 & 30 & 50 & 10 \\ 70 & 30 & 40 & 60 \\ 40 & 8 & 70 & 20 \\ \end{array} \][/tex]
### Step 2: Start with the Least Cost Method (LCM)
Step 2.1: Initialize the Matrix
We initialize an allocation matrix to keep track of the number of units transported between farms and destinations.
Step 2.2: Allocation Process
- Find the least cost cell (10 in FARM1 to D4):
- Allocate min(supply, demand) = min(7, 14) = 7 units.
- Adjust Supply and Demand:
- Supply FARM1: 7 - 7 = 0
- Demand D4: 14 - 7 = 7
- Find Next Least Cost Cell (8 in FARM3 to D2):
- Allocate min(supply, demand) = min(18, 8) = 8 units.
- Adjust Supply and Demand:
- Supply FARM3: 18 - 8 = 10
- Demand D2: 8 - 8 = 0
- Find Next Least Cost Cell (19 in FARM1 to D1), no units allocated to any cell in FARM1 as its supply is 0.
- Find (30 in FARM2 to D1):
- Allocate min(supply, demand) = min(9, 5) = 5 units.
- Adjust Supply and Demand:
- Supply FARM2: 9 - 5 = 4
- Demand D1: 5 - 5 = 0
- Find (30 in FARM2 to D3):
- Allocate min(supply, demand) = min(4, 7) = 4 units.
- Adjust Supply and Demand:
- Supply FARM2: 4 - 4 = 0
- Demand D3: 7 - 4 = 3
- Find (20 in FARM3 to D4):
- Allocate min(supply, demand) = min(10, 7) = 7 units.
- Adjust Supply and Demand:
- Supply FARM3: 10 - 7 = 3
- Demand D4: 7 - 7 = 0
- Find (40 in FARM3 to D1):
- Allocate min(supply, demand) = min(3, 3) = 3 units.
- Adjust Supply and Demand at last as all other demands are satisfied.
### Step 3: Compile the Final Allocation Table
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & D1 & D2 & D3 & D4 \\ \hline FARM1 & 0 & 0 & 0 & 7 \\ \hline FARM2 & 5 & 0 & 4 & 0 \\ \hline FARM3 & 3 & 8 & 3 & 7 \\ \hline \end{array} \][/tex]
### Step 4: Calculate the Total Transportation Cost
Using the allocation and cost matrix:
[tex]\[ \text{Total Cost} = (0 \times 19) + (0 \times 30) + (0 \times 50) + (7 \times 10) + (5 \times 70) + (0 \times 30) + (4 \times 40) + (0 \times 60) + (3 \times 40) + (8 \times 8) + (3 \times 70) + (7 \times 20) \][/tex]
Simplifying the calculation:
[tex]\[ \text{Total Cost} = 0 + 0 + 0 + 70 + 350 + 0 + 160 + 0 + 120 + 64 + 210 + 140 = 814 \][/tex]
Thus, the optimal allocation using the Least Cost Method results in a total transportation cost of 814.
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