This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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To calculate the optimum cost element of transportation using the Vogel's Approximation Method (VAM), we first need to set up the transportation table. From the image, we can identify the following: Costs (C_ij): Row 1: 4, 1, 2, 6 Row 2: 8, 8, 3, 2 Row 3: 5, 9, 8, 7 Demand (D_j): D_1 = 1300 D_2 = 1000 D_3 = 1000 D_4 = 900 Total Demand = 1300 + 1000 + 1000 + 900 = 4200 Supply (S_i): S_2 = 1400 S_3 = 1400 The supply for S_1 is missing. Since the total demand is 4200, we assume the total supply must also be 4200 for a balanced problem. Therefore, S_1 = 4200 - (1400 + 1400) = 4200 - 2800 = 1400. The balanced transportation table is: |c|c|c|c|c|c|c| From/To & D_1 (1300) & D_2 (1000) & D_3 (1000) & D_4 (900) & Supply (S_i) \\ S_1 & 4 & 1 & 2 & 6 & 1400 \\ S_2 & 8 & 8 & 3 & 2 & 1400 \\ S_3 & 5 & 9 & 8 & 7 & 1400 \\ Demand (D_j) & 1300 & 1000 & 1000 & 900 & 4200 \\ We will now apply the VAM steps. Step 1: Calculate initial penalties. Calculate the difference between the two lowest costs for each row and column. |c|c|c|c|c|c|c| From/To & D_1 (1300) & D_2 (1000) & D_3 (1000) & D_4 (900) & Supply (S_i) & Row Penalty \\ S_1 & 4 & 1 & 2 & 6 & 1400 & 2-1=1 \\ S_2 & 8 & 8 & 3 & 2 & 1400 & 3-2=1 \\ S_3 & 5 & 9 & 8 & 7 & 1400 & 7-5=2 \\ Demand (D_j) & 1300 & 1000 & 1000 & 900 & & \\ Col Penalty & 5-4=1 & 8-1=7 & 3-2=1 & 6-2=4 & & \\ Step 2: Make the first allocation. The highest penalty is 7 (Column D_2). In D_2, the minimum cost is C_12=1. Allocate (S_1, D_2) = (1400, 1000) = 1000 units to cell (S_1, D_2). Update supply and demand: S_1 = 1400 - 1000 = 400, D_2 = 1000 - 1000 = 0. Column D_2 is satisfied. Step 3: Recalculate penalties and make the second allocation. |c|c|c|c|c|c| From/To & D_1 (1300) & D_3 (1000) & D_4 (900) & Supply (S_i) & Row Penalty \\ S_1 & 4 & 2 & 6 & 400 & 4-2=2 \\ S_2 & 8 & 3 & 2 & 1400 & 3-2=1 \\ S_3 & 5 & 8 & 7 & 1400 & 7-5=2 \\ Demand (D_j) & 1300 & 1000 & 900 & & \\ Col Penalty & 5-4=1 & 3-2=1 & 6-2=4 & & \\ The highest penalty is 4 (Column D_4). In D_4, the minimum cost is C_24=2. Allocate (S_2, D_4) = (1400, 900) = 900 units to cell (S_2, D_4). Update supply and demand: S_2 = 1400 - 900 = 500, D_4 = 900 - 900 = 0. Column D_4 is satisfied. Step 4: Recalculate penalties and make the third allocation. |c|c|c|c|c| From/To & D_1 (1300) & D_3 (1000) & Supply (S_i) & Row Penalty \\ S_1 & 4 & 2 & 400 & 4-2=2 \\ S_2 & 8 & 3 & 500 & 8-3=5 \\ S_3 & 5 & 8 & 1400 & 8-5=3 \\ Demand (D_j) & 1300 & 1000 & & \\ Col Penalty & 5-4=1 & 3-2=1 & & \\ The highest penalty is 5 (Row S_2). In S_2, the minimum cost is C_23=3. Allocate (S_2, D_3) = (500, 1000) = 500 units to cell (S_2, D_3). Update supply and demand: S_2 = 500 - 500 = 0, D_3 = 1000 - 500 = 500. Row S_2 is satisfied. Step 5: Recalculate penalties and make the fourth allocation. |c|c|c|c|c| From/To & D_1 (1300) & D_3 (500) & Supply (S_i) & Row Penalty \\ S_1 & 4 & 2 & 400 & 4-2=2 \\ S_3 & 5 & 8 & 1400 & 8-5=3 \\ Demand (D_j) & 1300 & 500 & & \\ Col Penalty & 5-4=1 & 8-2=6 & & \\ The highest penalty is 6 (Column D_3). In D_3, the minimum cost is C_13=2. Allocate (S_1, D_3) = (400, 500) = 400 units to cell (S_1, D_3). Update supply and demand: S_1 = 400 - 400 = 0, D_3 = 500 - 400 = 100. Row S_1 is satisfied. Step 6: Make the final allocations. Only row S_3 and columns D_1, D_3 remain. |c|c|c|c| From/To & D_1 (1300) & D_3 (100) & Supply (S_i) \\ S_3 & 5 & 8 & 1400 \\ Demand (D_j) & 1300 & 100 & \\ Allocate remaining demand to S_3. Allocate (S_3, D_1) = (1400, 1300) = 1300 units to cell (S_3, D_1). Update supply and demand: S_3 = 1400 - 1300 = 100, D_1 = 1300 - 1300 = 0. Allocate (S_3, D_3) = (100, 100) = 100 units to cell (S_3, D_3). Update supply and demand: S_3 = 100 - 100 = 0, D_3 = 100 - 100 = 0. All supply and demand are met. Step 7: Calculate the total transportation cost. The allocations are: X_12 = 1000 units at cost C_12=1 X_24 = 900 units at cost C_24=2 X_23 = 500 units at cost C_23=3 X_13 = 400 units at cost C_13=2 X_31 = 1300 units at cost C_31=5 X_33 = 100 units at cost C_33=8 Total Cost = (1000 × 1) + (900 × 2) + (500 × 3) + (400 × 2) + (1300 × 5) + (100 × 8) Total Cost = 1000 + 1800 + 1500 + 800 + 6500 + 800 Total Cost = 12400 The optimum cost element of transportation using VAM is 12400. Send me the next one 📸