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Strategic Cost Management · Transportation

Initial Basic Feasible Solution Methods in Transportation

Updated 11 October 2026 · Fact-checked

An initial basic feasible solution (IBFS) is a first allocation of units from sources to destinations that meets all supply and demand and uses m + n − 1 occupied cells. You find it by the North-West Corner Rule, the Least Cost Method or Vogel's Approximation Method (VAM), then test it for optimality.

Understand Initial Basic Feasible Solution Methods

A transportation problem moves goods from several sources (plants, warehouses) to several destinations (markets) at the lowest total cost. Each route has a cost per unit. Each source has a fixed supply and each destination has a fixed demand.

Before you can improve a plan, you need a plan. That first plan is the initial basic feasible solution. Feasible means every supply is used up and every demand is met. Basic means the allocation uses exactly m + n − 1 occupied cells for m sources and n destinations (in a non-degenerate case). The solution is only a starting point. It may not be the cheapest.

There are three standard methods. The North-West Corner Rule ignores costs. It starts at the top-left cell and moves right or down. It is fast but usually gives the highest cost. The Least Cost Method allocates first to the cheapest cell, so it uses cost information and usually starts closer to the best answer.

Vogel's Approximation Method (VAM) looks at the penalty of not using the cheapest cell. The penalty of a row or column is the difference between its two smallest costs. The largest penalty is the most urgent, so you serve that row or column at its cheapest cell. VAM usually gives the best starting cost of the three, though this is a tendency and not a guarantee.

All three methods need a balanced problem, where total supply equals total demand. If not, add a dummy row or column with zero cost first. After the IBFS, you apply the MODI test to check optimality.

Key rules to remember

Balanced problem condition
Σ supply = Σ demand
If supply and demand differ, add a dummy destination or source with zero cost to balance.
Number of occupied cells in a basic feasible solution
m + n − 1
m = number of sources, n = number of destinations. Fewer occupied cells means degeneracy, so add a zero allocation.
Allocation in a chosen cell
Allocation = min(remaining supply of row, remaining demand of column)
Applies in all three methods. Cross out the row or column that is exhausted.
VAM penalty
Penalty = second-lowest cost − lowest cost (in a row or column)
Compute it only on rows and columns still open. Recompute after every allocation.
Total transportation cost
Total cost = Σ (units allocated × unit cost) over occupied cells
Use it to compare methods and to report the IBFS cost.

How to solve Initial Basic Feasible Solution Methods questions

Use this sequence for any IBFS question, whichever method the question names.

  1. 1Check that total supply equals total demand. If not, add a dummy row or column with zero costs and the balancing quantity.
  2. 2Note m + n − 1 so you know how many occupied cells to expect.
  3. 3Pick the method named. For NWC, start at the top-left cell. For Least Cost, start at the lowest-cost cell. For VAM, compute all row and column penalties and pick the largest.
  4. 4Allocate the smaller of the remaining supply and remaining demand to the chosen cell. Write the quantity in the cell.
  5. 5Cross out the row or column that is fully used. If both finish together, cross out one only and keep the other with zero to avoid losing a basic cell.
  6. 6Update the remaining supplies and demands. For VAM, recompute penalties on the open rows and columns. Repeat until all supply and demand are met.
  7. 7Count the occupied cells and compare with m + n − 1.
  8. 8Compute total cost as the sum of units × unit cost and state it clearly.

Quickest way: Penalty-first VAM with a running check

When to use it: When the question says 'use VAM' or asks for the best starting solution, and you have limited time.

  1. Write penalties for rows beside the table and for columns below it. Do this once per round.
  2. Circle the largest penalty. Break ties by choosing the row or column that contains the lowest cost cell.
  3. Allocate in its cheapest cell and strike out the finished line. If only one row or column is left, fill it by cost without more penalties.
  4. Tick off supplies and demands as you go, so the final totals check itself.
  5. Finish with a one-line cost calculation. Verify that the allocations in each row and column add to the given supply and demand.

Common mistakes in Initial Basic Feasible Solution Methods

  • Starting without balancing the problem.

    Students see the table and begin allocating straight away.

    Fix: Add the totals first. If unequal, add a dummy row or column with zero cost and the difference as its quantity.

  • Calculating VAM penalty as highest minus lowest cost.

    The idea of 'difference' is remembered, but not which costs.

    Fix: Use the two smallest costs in the row or column: second-lowest minus lowest.

  • Not recalculating penalties after each allocation.

    Students want to save time and reuse the first set.

    Fix: Cross out the finished line, then recompute penalties for what remains. Old penalties are no longer valid.

  • Crossing out both a row and a column when supply and demand finish together.

    Both are exhausted, so crossing out both seems logical.

    Fix: Cross out only one. Allocate a zero in a suitable cell later so that occupied cells equal m + n − 1.

  • Using the lowest cost in the whole table in the Least Cost Method after cells are crossed out.

    Students lose track of which rows and columns are closed.

    Fix: Only look at cells in rows and columns that are still open. Shade closed lines as you go.

  • Multiplying units by the wrong cost in the final total.

    Allocations are written in the cost table and the two numbers get confused.

    Fix: Write allocations in a separate table or in a clear corner of each cell. List each product in the total cost line.

Worked examples

Example 1

A company has three plants P1, P2, P3 with supplies of 40, 60 and 50 units. It has three markets M1, M2, M3 with demands of 45, 70 and 35 units. Unit transport costs (₹) are: P1: 8, 6, 10; P2: 9, 7, 4; P3: 3, 5, 11. Find the initial basic feasible solution and its cost by (a) the North-West Corner Rule and (b) the Least Cost Method.

Show the solution
  1. Total supply = 40 + 60 + 50 = 150. Total demand = 45 + 70 + 35 = 150. The problem is balanced. Expected occupied cells = 3 + 3 − 1 = 5.
  2. (a) NWC. Cell P1-M1: min(40, 45) = 40. P1 is exhausted. M1 needs 5 more.
  3. P2-M1: min(60, 5) = 5. M1 is met. P2 has 55 left.
  4. P2-M2: min(55, 70) = 55. P2 is exhausted. M2 needs 15 more.
  5. P3-M2: min(50, 15) = 15. M2 is met. P3 has 35 left.
  6. P3-M3: min(35, 35) = 35. All supply and demand are met. Five cells are occupied.
  7. NWC cost = 40 × 8 + 5 × 9 + 55 × 7 + 15 × 5 + 35 × 11 = 320 + 45 + 385 + 75 + 385 = ₹1,210.
  8. (b) Least Cost. The lowest cost is 3 at P3-M1. Allocate min(50, 45) = 45. M1 is met. P3 has 5 left.
  9. Among open cells (M2 and M3 columns), the lowest cost is 4 at P2-M3. Allocate min(60, 35) = 35. M3 is met. P2 has 25 left.
  10. Next lowest is 5 at P3-M2. Allocate min(5, 70) = 5. P3 is exhausted. M2 needs 65 more.
  11. Next lowest is 6 at P1-M2. Allocate min(40, 65) = 40. P1 is exhausted. M2 needs 25 more.
  12. Last cell P2-M2 gets 25. Five cells are occupied.
  13. Least Cost = 45 × 3 + 35 × 4 + 5 × 5 + 40 × 6 + 25 × 7 = 135 + 140 + 25 + 240 + 175 = ₹715.

Answer: NWC gives an initial cost of ₹1,210. The Least Cost Method gives ₹715. Both use 5 occupied cells.

Example 2

For the same data as above, find the initial basic feasible solution by Vogel's Approximation Method. Compare it with the NWC cost and state what you would do next.

Show the solution
  1. Round 1 row penalties (second-lowest minus lowest): P1: 8 − 6 = 2; P2: 7 − 4 = 3; P3: 5 − 3 = 2.
  2. Round 1 column penalties: M1: 8 − 3 = 5; M2: 6 − 5 = 1; M3: 10 − 4 = 6.
  3. The largest penalty is 6 in column M3. Its lowest cost is 4 at P2. Allocate min(60, 35) = 35. M3 is met. P2 has 25 left. Cross out column M3.
  4. Round 2 with columns M1 and M2. Row penalties: P1: 8 − 6 = 2; P2: 9 − 7 = 2; P3: 5 − 3 = 2. Column penalties: M1: 8 − 3 = 5; M2: 6 − 5 = 1.
  5. The largest penalty is 5 in column M1. Its lowest cost is 3 at P3. Allocate min(50, 45) = 45. M1 is met. P3 has 5 left. Cross out column M1.
  6. Only column M2 remains, with demand 70. Remaining supplies are P1 40, P2 25 and P3 5, which total 70. Allocate P1-M2 = 40, P2-M2 = 25 and P3-M2 = 5.
  7. Occupied cells = 5 = m + n − 1.
  8. VAM cost = 35 × 4 + 45 × 3 + 40 × 6 + 25 × 7 + 5 × 5 = 140 + 135 + 240 + 175 + 25 = ₹715.
  9. Compare: NWC ₹1,210 minus VAM ₹715 = ₹495 higher for NWC.

Answer: VAM gives an initial cost of ₹715, the same as the Least Cost Method here. This is ₹495 lower than NWC. The next step is to apply the MODI test to check whether ₹715 is optimal.

Exam tips

  • Show every allocation, the crossed-out lines and the final cost line. Marks are given for the method as well as the answer.
  • In VAM, write the penalty rows and columns for each round clearly. Examiners follow your logic from them.
  • Always check balance and the m + n − 1 count. Quote both in your answer.
  • In MCQs, you can often reject options by cost ordering. NWC is generally the costliest start and VAM generally the cheapest, but do not assume it for every problem.
  • When asked to compare methods, state costs and add one line on why: NWC ignores cost, Least Cost uses the lowest cell, VAM uses penalties.

Practice questions from Transportation

Initial Basic Feasible Solution Methods in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Initial Basic Feasible Solution Methods: frequently asked questions

What are the steps of Vogel's Approximation Method?

Find the penalty for each open row and column as the difference between its two smallest costs. Choose the largest penalty and allocate as much as possible in the cheapest cell of that line. Cross out the finished row or column, recompute penalties, and repeat until all supply and demand are met.

What is the difference between the Least Cost Method and VAM?

The Least Cost Method always picks the cheapest open cell in the whole table. VAM first measures which row or column would lose most if its cheapest cell were not used, and serves that one first. VAM usually gives a better starting cost but takes more work.

Why does the North-West Corner Rule give a poor starting solution?

It ignores costs completely and only follows the position of cells from the top-left. The allocation can therefore land on expensive routes. It is the quickest method and is useful when the question asks for it by name.

What if supply and demand finish together in a cell?

Cross out only the row or the column, not both. Keep the other line open with a zero allocation in a suitable cell. This keeps the number of occupied cells at m + n − 1, which you need for the MODI test.