Strategic Cost Management · Transportation
Transportation Problem Formulation and Basics for CMA Final
Updated 11 October 2026 · Fact-checked
A transportation problem finds the cheapest way to ship a single product from several sources with fixed supply to several destinations with fixed demand. You list supplies, demands and unit costs in a matrix, check whether total supply equals total demand, and add a dummy row or column if it does not.
Understand Transportation Problem Formulation and Basics
A transportation problem is a special linear programming problem. A firm has a product at several origins (factories or warehouses). Each origin has a fixed supply. The product must go to several destinations (markets or depots). Each destination has a fixed demand. Shipping one unit from an origin to a destination has a known cost. You must decide how many units to send on each route so that total transportation cost is lowest.
The data goes into a cost matrix. Rows are origins, columns are destinations. Each cell holds the unit cost for that route. The last column shows supply and the last row shows demand. The unknowns are the quantities shipped, usually written xij for units sent from origin i to destination j.
The model rests on some assumptions: one homogeneous product; supply and demand are known and fixed; the unit cost on each route is constant, so total cost on a route is unit cost × quantity (cost is linear); and goods move directly from origin to destination, with no limit on route capacity unless the question states one.
A problem is balanced when total supply equals total demand. It is unbalanced when they differ. If supply exceeds demand, add a dummy destination with demand equal to the excess. If demand exceeds supply, add a dummy origin with supply equal to the shortage. Dummy cells normally carry zero cost, because nothing is actually shipped. Unless the question says otherwise, treat the dummy as a way to record unused stock or unmet demand.
Formulation comes first. Solving by an initial basic feasible solution method and the MODI test comes after, and those are separate topics. In the exam, a clean formulation earns marks even if later arithmetic slips.
Key rules to remember
- Objective function
- Minimise Z = ΣΣ cij × xij (i = 1 to m, j = 1 to n)
- cij is the unit cost from origin i to destination j. xij is the quantity shipped. Z is total transportation cost.
- Supply constraints
- Σj xij = ai for each origin i
- Shipments from an origin equal its supply. In an unbalanced case with a dummy, this holds after balancing.
- Demand constraints
- Σi xij = bj for each destination j
- Receipts at a destination equal its demand.
- Non-negativity
- xij ≥ 0 for all i, j
- You cannot ship a negative quantity.
- Balance condition
- Σ ai = Σ bj
- Balanced if true. If Σ ai > Σ bj add a dummy destination; if Σ ai < Σ bj add a dummy origin.
- Number of variables and constraints
- Variables = m × n; constraints = m + n
- m origins and n destinations, after balancing.
- Basic feasible solution size
- Allocated cells = m + n − 1
- A non-degenerate basic feasible solution has exactly this many occupied cells. Fewer means degeneracy.
How to solve Transportation Problem Formulation and Basics questions
Use this order for any question that asks you to formulate or set up a transportation problem.
- 1Identify the origins and destinations and note the supply at each origin and the demand at each destination.
- 2Write the unit costs into an m × n matrix, rows for origins, columns for destinations. Mark any prohibited route with a very large cost M, or as not allowed.
- 3Add total supply and total demand.
- 4If they are equal, the problem is balanced. If not, add a dummy destination (supply greater) or a dummy origin (demand greater) for the difference, with zero cost in its cells unless the question gives penalty or storage costs.
- 5Define the variables xij and write the objective function Z = ΣΣ cij xij to be minimised.
- 6Write the supply constraints, demand constraints and non-negativity condition.
- 7Check that total supply equals total demand in the final table and that the number of cells and constraints is right.
- 8State the next step if asked: find an initial basic feasible solution, then test for optimality.
Quickest way: Total-first balance check
When to use it: Use when time is short in a case-based MCQ or a short formulation question.
- Add the supply column and the demand row first, before touching costs.
- Compute the difference. This gives the size of the dummy at once.
- Decide the dummy's side: extra supply gives a dummy column; extra demand gives a dummy row.
- Draw the table with a zero-cost dummy line and write the totals on the margins.
- For MCQs, remember the counts: m × n variables, m + n constraints, m + n − 1 basic cells.
Common mistakes in Transportation Problem Formulation and Basics
Solving an unbalanced problem without adding a dummy
Students rush to allocate and skip the total check.
Fix: Always total supply and demand first. Never begin allocation until the table is balanced.
Adding the dummy on the wrong side
Confusion between surplus supply and shortage of supply.
Fix: Surplus supply means unsold stock, so add a dummy destination (column). Shortage means unmet demand, so add a dummy origin (row).
Putting rows and columns the wrong way round
Students copy the data order from the question rather than origins as rows.
Fix: Rows are origins with supply on the right. Columns are destinations with demand at the bottom. Label them clearly.
Giving a dummy cell a non-zero cost without being told
Students think every cell must have a cost.
Fix: Use zero for dummy cells unless the question gives storage, penalty or shortage costs. If it does, use those figures.
Ignoring a prohibited route
A statement such as 'no supply from plant B to depot 2' is easy to miss in the case text.
Fix: Put a very large cost M in that cell so it is never chosen, and underline such conditions as you read.
Miscounting the basic cells needed
Students recall m + n instead of m + n − 1.
Fix: Count after balancing, including dummies, and use m + n − 1.
Worked examples
Example 1
A company has two plants, P1 with 40 units and P2 with 60 units, and three depots, D1, D2 and D3, needing 30, 45 and 25 units. Unit costs in ₹ are: P1 to D1, D2, D3: 8, 6, 10; P2 to D1, D2, D3: 9, 12, 7. Formulate the problem.
Show the solution
- Total supply = 40 + 60 = 100 units. Total demand = 30 + 45 + 25 = 100 units.
- Supply equals demand, so the problem is balanced and no dummy is needed.
- Cost matrix: rows P1 and P2, columns D1, D2, D3, costs as given, supply 40 and 60, demand 30, 45, 25.
- Let xij be units sent from Pi to Dj. Objective: Minimise Z = 8x11 + 6x12 + 10x13 + 9x21 + 12x22 + 7x23.
- Supply constraints: x11 + x12 + x13 = 40 and x21 + x22 + x23 = 60.
- Demand constraints: x11 + x21 = 30, x12 + x22 = 45, x13 + x23 = 25.
- Non-negativity: all xij ≥ 0. Variables = 2 × 3 = 6; constraints = 5; a basic feasible solution needs 2 + 3 − 1 = 4 occupied cells.
Answer: Balanced problem (100 = 100). Minimise Z = 8x11 + 6x12 + 10x13 + 9x21 + 12x22 + 7x23 subject to the supply and demand equations above and xij ≥ 0.
Example 2
Three warehouses W1, W2, W3 hold 50, 70 and 30 units. Four stores S1, S2, S3, S4 need 40, 30, 50 and 20 units. Unit costs in ₹: W1: 5, 8, 6, 9; W2: 7, 4, 5, 6; W3: 6, 3, 8, 4. Make the problem balanced and state the table.
Show the solution
- Total supply = 50 + 70 + 30 = 150 units.
- Total demand = 40 + 30 + 50 + 20 = 140 units.
- Supply exceeds demand by 10 units, so add a dummy destination S5 with demand 10.
- Costs in the dummy column are zero for W1, W2 and W3, since nothing is shipped and no storage cost is given.
- The balanced table is 3 × 5: columns S1 to S5, demands 40, 30, 50, 20, 10, total 150; supplies 50, 70, 30, total 150.
- Variables = 3 × 5 = 15; constraints = 3 + 5 = 8; basic cells required = 3 + 5 − 1 = 7.
- Any units allocated to S5 represent stock left unsent at that warehouse.
Answer: Unbalanced (150 > 140). Add dummy destination S5 with demand 10 and zero costs. The balanced 3 × 5 table has total 150 on both sides and needs 7 occupied cells in a basic feasible solution.
Exam tips
- In MCQs, the first thing examiners test is the balance check. Add totals before reading anything else.
- In a case scenario, look for prohibited routes, storage costs or shortage penalties. These change the dummy costs and the M entries.
- When asked to formulate, write the objective function and all constraints, not just the table. Marks are given for each part.
- State the type of problem (balanced or unbalanced) and the dummy you added in one clear line. This is easy credit.
- Present the table neatly with labelled rows, columns, supply and demand, because later steps depend on it.
Practice questions from Transportation
- A transportation problem has 3 sources and 4 destinations. An initial feasible solution by the North-West Corner method has only 5 occupied …
- A cement firm ships from plants P1, P2, P3 (supply 40, 30, 30 tonnes) to markets D1, D2, D3 (demand 30, 40, 30 tonnes). Unit costs in rupees…
- A transportation problem has two sources and three destinations. Unit costs in rupees are S1: 8, 6, 10 and S2: 9, 7, 5 for D1, D2, D3. The c…
- A transport planner at a Pune distribution firm has a balanced transportation problem with 3 godowns (origins) and 4 retail hubs (destinatio…
- A firm has three plants (A, B, C) and three markets (D1, D2, D3). Unit costs (Rs) are A: 8, 5, 9; B: 6, 10, 7; C: 11, 12, 13. Under Vogel's …
Transportation Problem Formulation and Basics in other exams
The same ground in other exams, if you are preparing for more than one or want another angle on it.
Transportation Problem Formulation and Basics: frequently asked questions
What is a transportation problem in operations research?
It is a linear programming problem of moving one product from several origins with fixed supplies to several destinations with fixed demands at minimum total cost. The decision is how many units to send on each route.
What is the difference between balanced and unbalanced transportation problems?
In a balanced problem, total supply equals total demand. In an unbalanced one they differ. You balance it by adding a dummy destination if supply is greater, or a dummy origin if demand is greater.
What cost do I give the dummy row or column?
Zero, unless the question gives storage, penalty or shortage costs. Then use those figures. A very large cost M is used only for routes that are not allowed.
Why does a basic feasible solution have m + n − 1 occupied cells?
A balanced problem has m + n constraints, but one is redundant because total supply equals total demand. So only m + n − 1 are independent, which gives that many basic variables.