Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
