The task involves solving a transportation problem using the potential method to maximize the objective function. Let's break the problem into key steps:

Problem Details:

1. Supplies: $S_1 = 23$, $S_2 = 45$, $S_3 = 32$
2. Demands: $D_1 = 30$, $D_2 = 50$, $D_3 = 42$
3. Cost Matrix: & 30 & 50 & 42 \\ \hline 23 & 6 & 3 & 5 \\ 45 & 5 & 6 & 9 \\ 32 & 1 & 2 & 5 \\ \end{array}$$

Steps to Solve:

1. Initial Feasibility Check:

   • Total Supply: $23 + 45 + 32 = 100$
   • Total Demand: $30 + 50 + 42 = 100$
   • Since supply equals demand, the problem is balanced.

2. Construct Initial Basic Feasible Solution (IBFS): Use the Northwest Corner Method, Least-Cost Method, or Vogel’s Approximation Method to set up an initial solution.

3. Optimize Using the Potential Method:

   • Calculate potentials ($u_i$ for rows and $v_j$ for columns).
   • Compute opportunity costs for non-basic cells: $c_{ij} - u_i - v_j$.
   • Identify if the solution is optimal or adjust the basic plan.

I’ll start calculating the IBFS using the Northwest Corner Method and then proceed with optimization.

Step 1: Initial Basic Feasible Solution

Using the Northwest Corner Method:

1. Allocate $\min(23, 30) = 23$ units from $S_1$ to $D_1$.

   • Remaining: $S_1 = 0$, $D_1 = 7$.

2. Allocate $\min(0, 7) = 7$ units from $S_2$ to $D_1$.

   • Remaining: $S_2 = 38$, $D_1 = 0$.

3. Allocate $\min(38, 50) = 38$ units from $S_2$ to $D_2$.

   • Remaining: $S_2 = 0$, $D_2 = 12$.

4. Allocate $\min(32, 12) = 12$ units from $S_3$ to $D_2$.

   • Remaining: $S_3 = 20$, $D_2 = 0$.

5. Allocate $\min(20, 42) = 20$ units from $S_3$ to $D_3$.

   • Remaining: $S_3 = 0$, $D_3 = 22$.

Would you like me to continue with the optimization step and solve for the optimal solution using the potential method?