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Operations Management and Strategic Management · Transportation

Initial Basic Feasible Solution Methods in Transportation

Updated 10 October 2026 · Fact-checked

An initial basic feasible solution (IBFS) is a first valid shipping plan that meets all supply and demand using exactly m + n − 1 allocations. You find it with the North-West Corner Rule, Least Cost Method or Vogel's Approximation Method (VAM). VAM usually gives the lowest starting cost.

Understand Initial Basic Feasible Solution Methods

A transportation problem asks how to ship goods from several sources (plants) to several destinations (warehouses) at the lowest total cost. Each source has a fixed supply. Each destination has a fixed demand. Each route has a cost per unit.

You cannot test every possible plan, so you start with one workable plan. This is the initial basic feasible solution. Feasible means all supply is shipped and all demand is met. Basic means the number of occupied cells is exactly m + n − 1, where m is the number of sources and n is the number of destinations.

The three methods differ in how they choose the cells. 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 most expensive start. The Least Cost Method allocates first to the cheapest cell, so it uses cost information. VAM looks at the penalty (the cost of not using the cheapest route) and allocates where a wrong choice would hurt most. It takes longer but is usually closest to the optimum.

The IBFS is not the final answer. A better solution may exist. The next step, the MODI test, checks whether the IBFS is optimal. In many exam questions you are asked only for the IBFS and its cost, so this step must be exact.

The problem must be balanced first: total supply must equal total demand. If it is not, add a dummy source or destination with zero costs.

Key rules to remember

Balanced condition
Σ supply = Σ demand
If not equal, add a dummy row or column with zero cost to balance it before starting.
Number of allocations in a basic solution
Occupied cells = m + n − 1
m = number of sources, n = number of destinations. Fewer cells means degeneracy.
Allocation rule
Allocation = min(remaining supply, remaining demand)
Used in all three methods. After allocating, cross out the exhausted row or column.
VAM penalty
Penalty = second-lowest cost − lowest cost (in a row or column)
Choose the row or column with the largest penalty, then allocate to its lowest-cost cell.
Total transportation cost
Total cost = Σ (units allocated × unit cost) over occupied cells
Report the cost in rupees, with the unit of cost per unit.

How to solve Initial Basic Feasible Solution Methods questions

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

  1. 1Write the cost matrix with supplies on the right and demands at the bottom. Check that total supply equals total demand. If not, add a dummy row or column with zero costs.
  2. 2Note the number of cells you must fill: m + n − 1.
  3. 3Pick the method named in the question. For NWCR, start at the top-left cell. For LCM, start at the lowest-cost cell. For VAM, work out the row and column penalties and pick the largest.
  4. 4Allocate the smaller of the remaining supply and the remaining demand to the chosen cell.
  5. 5Cross out the row or column that is exhausted. If both finish together, cross out only one and keep a zero allocation in the other so you do not lose a cell.
  6. 6Repeat steps 3 to 5 on the remaining table. For VAM, recalculate penalties every round.
  7. 7Count the occupied cells and check that there are m + n − 1.
  8. 8Multiply each allocation by its unit cost and add up to get the total cost. Show this as a line-by-line calculation.

Quickest way: Penalty-first shortcut for VAM under time pressure

When to use it: Use when the question asks for the best starting solution, or when you must write VAM within a few minutes.

  1. Write penalties in a small column on the right and a row below the table. Do not draw new tables each round.
  2. After each allocation, strike out the finished row or column, and only recompute penalties of lines that lost a cell.
  3. Tick the allocated cells as you go and write the quantity in the cell corner.
  4. At the end, check that the allocations add up to each supply and demand total before computing cost.
  5. For a quick cross-check, compare with the Least Cost answer. VAM should not be much worse.

Common mistakes in Initial Basic Feasible Solution Methods

  • Starting without checking that supply equals demand

    Students jump straight to the table.

    Fix: Total the supply and demand first. If unequal, add a dummy row or column with zero costs and state that you have done so.

  • Allocating the wrong quantity

    Students allocate the supply or the demand without comparing the two.

    Fix: Always write min(supply, demand) beside the cell and update both remaining figures.

  • Using the highest cost instead of the penalty in VAM

    Students confuse penalty with the difference between the highest and lowest cost.

    Fix: Penalty is second-lowest minus lowest cost in that row or column. Never use the highest cost.

  • Not recalculating VAM penalties after each allocation

    Students reuse the first set of penalties to save time.

    Fix: Recompute penalties on the remaining table after every allocation, because a crossed-out line changes the lowest and second-lowest costs.

  • Ending with fewer than m + n − 1 occupied cells

    A row and a column finish together and both get crossed out.

    Fix: Cross out only one of them and place a zero allocation in the next cell of the other. This keeps the solution basic for the MODI test.

  • Errors in the total cost

    Students multiply with costs from the wrong cells or skip a cell.

    Fix: List each occupied cell as units × cost on its own line, then add. Check that the units total equals total supply.

Worked examples

Example 1

A company has three plants P1, P2, P3 with supplies of 50, 70 and 30 units. Three warehouses W1, W2, W3 need 40, 60 and 50 units. Unit transport costs (₹) are: P1: 4, 6, 8; P2: 5, 3, 7; P3: 9, 2, 6. Find the initial basic feasible solution by the North-West Corner Rule and its cost.

Show the solution
  1. Total supply = 50 + 70 + 30 = 150. Total demand = 40 + 60 + 50 = 150. The problem is balanced. Cells needed = 3 + 3 − 1 = 5.
  2. Cell (P1, W1): min(50, 40) = 40. P1 has 10 left. W1 is finished.
  3. Cell (P1, W2): min(10, 60) = 10. P1 is finished. W2 needs 50 more.
  4. Cell (P2, W2): min(70, 50) = 50. W2 is finished. P2 has 20 left.
  5. Cell (P2, W3): min(20, 50) = 20. P2 is finished. W3 needs 30 more.
  6. Cell (P3, W3): min(30, 30) = 30. All supply and demand are met. Occupied cells = 5, as required.
  7. Cost = 40 × 4 + 10 × 6 + 50 × 3 + 20 × 7 + 30 × 6 = 160 + 60 + 150 + 140 + 180 = ₹690.

Answer: Allocations: P1→W1 40, P1→W2 10, P2→W2 50, P2→W3 20, P3→W3 30. Total cost = ₹690.

Example 2

For the same data as above, find the initial basic feasible solution using Vogel's Approximation Method and compare its cost with the North-West Corner solution.

Show the solution
  1. Round 1 penalties. Rows: P1 = 6 − 4 = 2, P2 = 5 − 3 = 2, P3 = 6 − 2 = 4. Columns: W1 = 5 − 4 = 1, W2 = 3 − 2 = 1, W3 = 7 − 6 = 1. The largest is 4 (row P3). Its lowest cost is 2 at W2. Allocate min(30, 60) = 30. P3 is finished. W2 has 30 left.
  2. Round 2 (rows P1, P2). Rows: P1 = 6 − 4 = 2, P2 = 5 − 3 = 2. Columns: W1 = 5 − 4 = 1, W2 = 6 − 3 = 3, W3 = 8 − 7 = 1. The largest is 3 (column W2). Its lowest cost is 3 at P2. Allocate min(70, 30) = 30. W2 is finished. P2 has 40 left.
  3. Round 3 (P1 and P2 against W1 and W3). Rows: P1 = 8 − 4 = 4, P2 = 7 − 5 = 2. Columns: W1 = 1, W3 = 1. The largest is 4 (row P1). Its lowest cost is 4 at W1. Allocate min(50, 40) = 40. W1 is finished. P1 has 10 left.
  4. Remaining: P1 has 10 and P2 has 40, W3 needs 50. Allocate P1→W3 = 10 and P2→W3 = 40. Occupied cells = 5.
  5. Cost = 30 × 2 + 30 × 3 + 40 × 4 + 10 × 8 + 40 × 7 = 60 + 90 + 160 + 80 + 280 = ₹670.
  6. Comparison: NWCR gives ₹690 and VAM gives ₹670. VAM is ₹20 cheaper because it uses the cost information. The Least Cost Method on this data also gives ₹670.

Answer: VAM allocations: P3→W2 30, P2→W2 30, P1→W1 40, P1→W3 10, P2→W3 40. Total cost = ₹670, which is ₹20 lower than the NWCR cost of ₹690.

Exam tips

  • Draw the table neatly and show each allocation round by round. Examiners give step marks for the method even if the final cost slips.
  • Always state the check: total supply = total demand, and occupied cells = m + n − 1.
  • For MCQs, know the one-line features: NWCR ignores cost, LCM picks the lowest cost cell, VAM uses penalties and is usually closest to optimal.
  • If the question gives an unbalanced table, add the dummy with zero cost first. Then solve as normal.
  • When the question asks to compare methods, compute all costs and state which is lowest. Note that the IBFS still needs a MODI test to confirm optimality.

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

Which method gives the best initial basic feasible solution?

VAM usually gives the lowest starting cost, then the Least Cost Method, then NWCR. This is a general tendency, not a rule. In some problems two methods give the same cost, as in the worked example.

What do I do if a row and column are exhausted together?

Cross out only one of them. Put a zero allocation in a cell of the remaining line so the number of occupied cells stays m + n − 1. This avoids a degenerate solution failing the MODI test.

How is VAM different from the Least Cost Method?

The Least Cost Method allocates to the single cheapest cell in the whole table. VAM first finds the row or column where a wrong choice would cost the most (the largest penalty), and then allocates to the cheapest cell in it.

Is the initial basic feasible solution the final answer?

Not necessarily. It is only a starting plan. You use the MODI method to test whether it is optimal and improve it if not. If the question asks only for the IBFS, stop after the cost.