1. **Problem Statement:** Solve the transportation problem using Vogel's Approximation Method (VAM) for the given cost matrix, supply, and demand. 2. **Given Data:** Cost matrix: $$\begin{matrix} & 1 & 2 & 3 & 4 \\ 1 & 10 & 3 & 9 & 15 \\ 2 & 8 & 5 & 4 & 6 \\ 3 & 12 & 11 & 5 & 3 \end{matrix}$$ Supply: $100, 200, 100$ Demand: $50, 120, 150, 80$ 3. **Vogel's Approximation Method (VAM) Overview:** - Calculate penalties for each row and column as the difference between the two lowest costs. - Select the row or column with the highest penalty. - Allocate as much as possible to the lowest cost cell in that row or column. - Adjust supply and demand and cross out fulfilled rows or columns. - Repeat until all supply and demand are allocated. 4. **Step-by-step Solution:** **Iteration 1:** - Row penalties: - Row 1: costs = 3, 9, 10, 15; penalty = 9 - 3 = 6 - Row 2: costs = 4, 5, 6, 8; penalty = 5 - 4 = 1 - Row 3: costs = 3, 5, 11, 12; penalty = 5 - 3 = 2 - Column penalties: - Col 1: costs = 8, 10, 12; penalty = 10 - 8 = 2 - Col 2: costs = 3, 5, 11; penalty = 5 - 3 = 2 - Col 3: costs = 4, 5, 9; penalty = 5 - 4 = 1 - Col 4: costs = 3, 6, 15; penalty = 6 - 3 = 3 - Highest penalty is 6 (Row 1). - Allocate to lowest cost in Row 1: cost 3 at (1,2). - Allocate min(supply 100, demand 120) = 100. - Update supply Row 1 = 0, demand Col 2 = 20. **Iteration 2:** - Remove Row 1 (supply 0). - New penalties: - Row 2: costs = 4, 5, 6; penalty = 5 - 4 = 1 - Row 3: costs = 3, 5, 11; penalty = 5 - 3 = 2 - Col 1: costs = 8, 12; penalty = 12 - 8 = 4 - Col 2: costs = 20 demand left, costs = 5, 11; penalty = 11 - 5 = 6 - Col 3: costs = 4, 5; penalty = 5 - 4 = 1 - Col 4: costs = 6, 3; penalty = 6 - 3 = 3 - Highest penalty is 6 (Col 2). - Allocate to lowest cost in Col 2: cost 5 at (2,2). - Allocate min(supply 200, demand 20) = 20. - Update supply Row 2 = 180, demand Col 2 = 0. **Iteration 3:** - Remove Col 2 (demand 0). - New penalties: - Row 2: costs = 4, 6; penalty = 6 - 4 = 2 - Row 3: costs = 3, 11, 3; penalty = 11 - 3 = 8 - Col 1: costs = 8, 12; penalty = 12 - 8 = 4 - Col 3: costs = 4, 5; penalty = 5 - 4 = 1 - Col 4: costs = 6, 3; penalty = 6 - 3 = 3 - Highest penalty is 8 (Row 3). - Allocate to lowest cost in Row 3: cost 3 at (3,4). - Allocate min(supply 100, demand 80) = 80. - Update supply Row 3 = 20, demand Col 4 = 0. **Iteration 4:** - Remove Col 4 (demand 0). - New penalties: - Row 2: costs = 4, 6; penalty = 6 - 4 = 2 - Row 3: costs = 3, 11; penalty = 11 - 3 = 8 - Col 1: costs = 8, 12; penalty = 12 - 8 = 4 - Col 3: costs = 4, 5; penalty = 5 - 4 = 1 - Highest penalty is 8 (Row 3). - Allocate to lowest cost in Row 3: cost 3 at (3,1). - Allocate min(supply 20, demand 50) = 20. - Update supply Row 3 = 0, demand Col 1 = 30. **Iteration 5:** - Remove Row 3 (supply 0). - New penalties: - Row 2: costs = 4, 6; penalty = 6 - 4 = 2 - Col 1: cost = 8; penalty = 0 (only one cost) - Col 3: cost = 4; penalty = 0 - Highest penalty is 2 (Row 2). - Allocate to lowest cost in Row 2: cost 4 at (2,3). - Allocate min(supply 180, demand 150) = 150. - Update supply Row 2 = 30, demand Col 3 = 0. **Iteration 6:** - Remove Col 3 (demand 0). - Remaining supply Row 2 = 30, demand Col 1 = 30. - Allocate 30 at (2,1) cost 8. - Update supply Row 2 = 0, demand Col 1 = 0. 5. **Final Allocation:** $$\begin{matrix} & 1 & 2 & 3 & 4 & Supply \\ 1 & 0 & 100 & 0 & 0 & 100 \\ 2 & 30 & 20 & 150 & 0 & 200 \\ 3 & 20 & 0 & 0 & 80 & 100 \\ Demand & 50 & 120 & 150 & 80 & \\ \end{matrix}$$ 6. **Calculate Total Cost:** $$\text{Total Cost} = 0\times10 + 100\times3 + 0\times9 + 0\times15 + 30\times8 + 20\times5 + 150\times4 + 0\times6 + 20\times12 + 0\times11 + 0\times5 + 80\times3$$ $$= 0 + 300 + 0 + 0 + 240 + 100 + 600 + 0 + 240 + 0 + 0 + 240 = 1720$$ **Answer:** The minimum transportation cost using Vogel's Approximation Method is $1720$.