The following transportation problem : has a solution The above solution of a…

Check: Verify optimality using the MODI (u-v) method.

• Occupied (basic) cells and allocations: I→A = 1, II→A = 3, II→B = 0 (basic zero), III→B = 2, III→C = 2.

• Set potential u for row I = 0. Compute potentials from occupied cells: v_A = 50; u_II = 90 - v_A = 40; v_B = 45 - u_II = 5; u_III = 200 - v_B = 195; v_C = 50 - u_III = -145.

• Compute reduced costs (cost - (u + v)) for non-basic cells:

  • I→B: 30 - (0 + 5) = 25

  • I→C: 220 - (0 + -145) = 365

  • II→C: 170 - (40 + -145) = 275

  • III→A: 250 - (195 + 50) = 5

• All reduced costs are nonnegative, so no improvement is possible for this minimization problem.

Conclusion: The given allocation is an optimum solution.