The initial basic feasible solution to the following transportation problem…

Concept — Vogel's approximation method builds an initial basic feasible solution from penalties. Every row and every column is given a penalty equal to the difference between the two smallest unit costs still available in that line, which measures what is given up when the cheapest cell of that line cannot be used. Allocation therefore starts in the line carrying the largest penalty, at that line's cheapest cell, for as much as the remaining supply or demand permits; the exhausted row or column is struck out and all penalties are recomputed on what is left.

Feasibility — whatever method produces it, an allocation is a basic feasible solution only when every row's shipments add up to that source's supply and every column's receipts add up to that destination's demand. In a balanced problem the two grand totals are equal, so the entire available quantity has to move, and a non-degenerate solution of an m × n table occupies m + n − 1 cells.

Application — the totals here balance: the supplies 30 + 50 + 20 = 100 match the demands 20 + 40 + 30 + 10 = 100, so no dummy row or column is needed. The rounds of the method run as follows.

1. Penalties on the full table — rows: S1 1 − 1 = 0, S2 2 − 1 = 1, S3 4 − 2 = 2; columns: D1 3 − 1 = 2, D2 2 − 2 = 0, D3 2 − 1 = 1, D4 4 − 1 = 3. The largest penalty is 3 at D4, whose cheapest cell costs 1 in row S2, so x24 = min(50, 10) = 10. D4 is satisfied and S2 keeps 40 units.

2. Recompute over D1 to D3 — rows: S1 0, S2 1, S3 2; columns: D1 2, D2 0, D3 1. The largest value 2 is tied between row S3 and column D1; the standard tie-break takes the line holding the cheaper cell, which is D1 at a cost of 1 in row S1. So x11 = min(30, 20) = 20, D1 is satisfied and S1 keeps 10 units.

3. Recompute over D2 and D3 — rows: S1 2 − 1 = 1, S2 3 − 2 = 1, S3 5 − 2 = 3; columns: D2 2 − 2 = 0, D3 2 − 1 = 1. The largest penalty is 3 at S3, whose cheapest remaining cell costs 2 in column D2, so x32 = min(20, 40) = 20. S3 is emptied and D2 still needs 20 units.

4. Two rows remain — S1 with 10 units and S2 with 40 units, against D2 20 and D3 30. Every penalty is now 1, so the cheapest available cell decides: it costs 1 at row S1, column D3, giving x13 = 10. S1 is emptied and D3 still needs 20 units.

5. One row remains — S2 covers everything left over, so x22 = 20 and x23 = 20.

Resulting allocation (a dash marks a cell that carries nothing):

Total cost = 20(1) + 10(1) + 20(3) + 20(2) + 10(1) + 20(2) = 180, spread over 6 occupied cells, which is exactly m + n − 1 = 3 + 4 − 1, so the solution is non-degenerate.

Cross-check — resolving the second-round tie the other way, in favour of row S3, allocates x32 = 20 first and then reproduces the same four remaining allocations — x11 = 20, x13 = 10, x22 = 20 and x23 = 20 — so the tie-break does not change the outcome. Reading the allocation table back also squares every row total with its supply and every column total with its demand.

Contrast with the printed choices — each of the three allocation sets on offer breaks the balance equations:

• the set carrying x21 = 20 next to x11 = 20 pushes 40 units into D1, which needs only 20, and moves just 10 of the 20 units waiting at S3;

• the set carrying x12 = 20 and x22 = 20 with no entry in row S3 sends 50 units out of S1, which holds only 30, and leaves all 20 units at S3 standing;

• the set that reproduces the right cells but writes x32 = 10 delivers only 30 units to D2, which needs 40, and moves 90 units in all at a cost of 160, not the 180 printed beside it.

None of the three printed sets is a feasible allocation at all, let alone the one Vogel's approximation method produces. That method's answer, x11 = 20, x13 = 10, x22 = 20, x23 = 20, x24 = 10, x32 = 20 at a cost of 180, is absent from the list, so the response that fits is “None of the above”. UGC's official updated answer key for this paper (December 2012, Paper III, Computer Science and Applications, question 28) records the same outcome.