We are asked to check the optimality of the given basic feasible solution (BFS) for the transportation problem and, if it is not optimal, to find the optimal solution and its cost. Let’s solve this step by step manually. Step 1: Identify the given BFS and supply/demand From the table: | | D1 | D2 | D3 | D4 | Supply | | ------ | ------ | ----- | ----- | ----- | ------ | | S1 | 6 | 1(25) | 9 | 3(45) | 70 | | S2 | 11 | 5(5) | 2(50) | 8 | 55 | | S3 | 10(85) | 12(5) | 4 | 7 | 90 | | Demand | 85 | 35 | 50 | 45 | | * The numbers in parentheses are the allocated quantities in the BFS. * Allocated cells: (S1,D2)=25, (S1,D4)=45, (S2,D2)=5, (S2,D3)=50, (S3,D1)=85, (S3,D2)=5 Step 2: Check degeneracy * Number of allocations = 6 * Required for non-degenerate BFS = m + n − 1 = 3 + 4 − 1 = 6 ✅ * Non-degenerate BFS Step 3: Calculate uᵢ and vⱼ (MODI method) Let u₁,u₂,u₃ be row potentials and v₁,v₂,v₃,v₄ column potentials. Start with u₁ = 0. 1. For (S1,D2): c₁₂ = 1 → u₁ + v₂ = 1 → 0 + v₂ = 1 → v₂ = 1 2. For (S1,D4): c₁₄ = 3 → u₁ + v₄ = 3 → 0 + v₄ = 3 → v₄ = 3 3. For (S2,D2): c₂₂ = 5 → u₂ + v₂ = 5 → u₂ + 1 = 5 → u₂ = 4 4. For (S2,D3): c₂₃ = 2 → u₂ + v₃ = 2 → 4 + v₃ = 2 → v₃ = −2 5. For (S3,D1): c₃₁ = 10 → u₃ + v₁ = 10 → u₃ + v₁ = 10 6. For (S3,D2): c₃₂ = 12 → u₃ + v₂ = 12 → u₃ + 1 = 12 → u₃ = 11 → then v₁ = 10 − 11 = −1 So potentials: u = \[0, 4, 11], v = \[−1, 1, −2, 3] Step 4: Calculate the reduced costs Δᵢⱼ = cᵢⱼ − (uᵢ + vⱼ) * For non-basic cells: 1. (S1,D1): Δ₁₁ = 6 − (0 + (−1)) = 6 − (−1) = 7 ✅ 2. (S1,D3): Δ₁₃ = 9 − (0 + (−2)) = 9 − (−2) = 11 ✅ 3. (S2,D1): Δ₂₁ = 11 −