Let's address your transportation problem step by step. ### **Given Data** **Sources (Supply):** 3 **Destinations (Demand):** 4 **Unit transportation costs:** | | D1 | D2 | D3 | D4 | Supply | |---|----|----|----|----|--------| | S1| 6 | 7 | 5 | 12 | 5 | | S2| 9 |10 | 9 | 6 | 14 | | S3| 3 | 4 | 4 | 2 | 10 | **Demands:** | D1 | D2 | D3 | D4 | |----|----|----|----| | 8 | 8 |11 | 8 | --- ### **(a) Linear Programming Formulation** Let \( x_{ij} \) = units to be shipped from source \( i \) to destination \( j \). - \( i = 1,2,3 \) (sources) - \( j = 1,2,3,4 \) (destinations) #### **Decision Variables:** \( x_{ij} \ge 0 \), and (if integer) \( x_{ij} \) integer. #### **Objective Function:** Minimize total transportation cost: \[ \text{Minimize } Z = 6x_{11} + 7x_{12} + 5x_{13} + 12x_{14} + 9x_{21} + 10x_{22} + 9x_{23} + 6x_{24} + 3x_{31} + 4x_{32} + 4x_{33} + 2x_{34} \] #### **Subject to:** **Supply constraints:** \[ x_{11} + x_{12} + x_{13} + x_{14} = 5 \qquad (\text{S1}) \] \[ x_{21} + x_{22} + x_{23} + x_{24} = 14 \qquad (\text{S2}) \] \[ x_{31} + x_{32} + x_{33} + x_{34} = 10 \qquad (\text{S3}) \] **Demand constraints:** \[ x_{11} + x_{21} + x_{31} = 8 \qquad (\text{D1}) \] \[ x_{12} + x_{22} + x_{32} = 8 \qquad (\text{D2}) \] \[ x_{13} + x_{23} + x_{33} = 11 \qquad (\text{D3}) \] \[ x_{14} + x_{24} + x_{34} = 8 \qquad (\text{D4}) \] **Non-negativity:** \[ x_{ij} \ge 0, \text{ and integer (if required)} \] --- ### **(b) Excel Solver Setup** #### **Step-by-Step:** 1. **Input the cost matrix in a 3x4 table.** 2. **Create variable cells for \( x_{ij} \) (12 variables).** 3. **Set up row sums for supply and column sums for demand.** 4. **Objective cell:** SUMPRODUCT(costs, variables). 5. **Constraints:** - Row sums = supply. - Column sums = demand. - All variables ≥ 0 and integer (if integer solution required). 6. **Set Solver to minimize the objective cell.** 7. **Solve.** --- ### **(c) Transportation Algorithm (u-v Method / MODI)** #### **Step 1: Check for Balance** \[ \text{Total Supply} = 5+14+10 = 29 \\ \text{Total Demand} = 8+8+11+8 = 35 \] **This problem is unbalanced.** - Total supply = 29, total demand = 35. - **Add a dummy supply** with supply 6 (35-29) and zero costs. **Update cost table:** | | D1 | D2 | D3 | D4 | Supply | |---|----|----|----|----|--------| | S1| 6 | 7 | 5 | 12 | 5 | | S2| 9 |10 | 9 | 6 | 14 | | S3| 3 | 4 | 4 | 2 | 10 | | D | 0 | 0 | 0 | 0 | 6 | --- #### **Step 2: Initial Feasible Solution (e.g., Northwest Corner)** **Northwest Corner Method:** - S1-D1: min(5,8) = 5. S1 exhausted, D1 left: 3. - S2-D1: min(14,3) = 3. S2 left: 11, D1 fulfilled. - S2-D2: min(11,8) = 8. S2 left: 3, D2 fulfilled. - S2-D3: min(3,11) = 3. S2 exhausted, D3 left: 8. - S3-D3: min(10,8) = 8. S3 left: 2, D3 fulfilled. - S3-D4: min(2,8) = 2. S3 exhausted, D4 left: 6. - D-D4: min(6,6) = 6. Dummy supply exhausted. **Table:** | | D1 | D2 | D3 | D4 | Supply | |---|----|----|----|----|--------| | S1| 5 | | | | 0 | | S2| 3 | 8 | 3 | | 0 | | S3| | | 8 | 2 | 0 | | D | | | | 6 | 0 | |Demand| 0 | 0 | 0 | 0 | | **Allocated cells: (S1-D1), (S2-D1), (S2-D2), (S2-D3), (S3-D3), (S3-D4), (D-D4).** --- #### **Step 3: MODI (u-v) Table** Let’s assign \( u_1 = 0 \) (arbitrary). For each allocated cell: \( u_i + v_j = c_{ij} \) - (S1, D1): \( u_1 + v_1 = 6 \) ⇒ \( v_1 = 6 \) - (S2, D1): \( u_2 + v_1 = 9 \) ⇒ \( u_2 + 6 = 9 \) ⇒ \( u_2 = 3 \) - (S2, D2): \( u_2 + v_2 = 10 \) ⇒ \( 3 + v_2 = 10 \) ⇒ \( v_2 = 7 \) - (S2, D3): \( u_2 + v_3 = 9 \) ⇒ \( 3 + v_3 = 9 \) ⇒ \( v_3 = 6 \) - (S3, D3): \( u_3 + v_3 = 4 \) ⇒ \( u_3 + 6 = 4 \) ⇒ \( u_3 = -2 \) - (S3, D4): \( u_3 + v_4 = 2 \) ⇒ \( -2 + v_4 = 2 \) ⇒ \( v_4 = 4 \) - (D, D4): \( u_4 + v_4 = 0 \) ⇒ \( u_4 + 4 = 0 \) ⇒ \( u_4 = -4 \) So: - \( u_1 = 0 \) - \( u_2 = 3 \) - \( u_3 = -2 \) - \( u_4 = -4 \) (dummy supply) - \( v_1 = 6 \) - \( v_2 = 7 \) - \( v_3 = 6 \) - \( v_4 = 4 \) --- #### **Step 4: Check for Optimality** Calculate reduced costs for unallocated cells: \( \Delta_{ij} = c_{ij} - (u_i + v_j) \) | | D1 | D2 | D3 | D4 | |---|----|----|----|----| |S1 |6-(0+6)=0|7-(0+7)=0|5-(0+6) = -1|12-(0+4)=8| |S2 |*allocated*|*allocated*|*allocated*|6-(3+4)= -1| |S3 |3-(-2+6)= -1|4-(-2+7)= -1|*allocated*|*allocated*| |D |0-(-4+6)= -2|0-(-4+7)= -3|0-(-4+6)= -2|*allocated*| If any \( \Delta_{ij} < 0 \), the solution is not optimal. Several cells have negative \( \Delta_{ij} \) (e.g., S1-D3, S2-D4, S3-D1, S3-D2, D-D1, D-D2, D-D3). --- **You would now select the most negative \( \Delta_{ij} \) (here, D-D2 = -3) and perform the stepping stone method to improve the solution.** Keep repeating the u-v/method of potentials and adjust allocations until all \( \Delta_{ij} \geq 0 \). --- ### **Summary Table** #### **(a) LP Model:** See formulation above. #### **(b) Excel:** - Set up cost matrix, variable cells, sum constraints, and use Solver. #### **(c) MODI (u-v):** - **Balance the problem.** - **Initial solution (e.g., Northwest Corner)** - **Compute u and v.** - **Check for optimality.** - **Iterate with stepping stone method.** --- If you want to see the full iterations for part (c) or get the optimal solution with values, let me know! (The process can be lengthy but can be provided.)