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CMA Intermediate · Operations Management and Strategic Management

Transportation Problem for CMA Inter Operations Management

A transportation problem finds the cheapest way to ship goods from several sources to several destinations, given supply, demand and unit costs. You formulate a cost table, balance it, find a starting feasible solution (NWCR, LCM or VAM), then test and improve it with the MODI method until no negative opportunity cost remains.

What this chapter covers

Transportation is a special type of linear programming problem. A firm has factories or warehouses with fixed supply and customers or markets with fixed demand. Each route has a cost per unit. Your job is to decide how many units go on each route so that total cost is lowest and all supply and demand conditions are met.

The chapter has a clear method flow. First you formulate the problem as a table. Then you build an initial basic feasible solution using the North-West Corner Rule, the Least Cost Method or Vogel's Approximation Method. Then you check optimality with the MODI method (u-v method) and improve the solution by a loop of reallocation. Finally you handle special cases: degeneracy, unbalanced problems and prohibited routes.

In the paper, this chapter sits with the quantitative techniques of Operations Management, next to other optimisation topics. It is procedural, so it rewards practice. The same table skills help you in allocation-type problems and in understanding how cost minimisation decisions are made in operations and logistics.

Transportation is a numerical chapter with a fixed method, so a well-prepared student can score full step marks in a 14-mark question and can also pick up quick MCQs on rules such as the number of allocations, balancing and degeneracy. Marks are lost mainly through arithmetic slips and missed steps, not through difficult concepts, which makes this one of the more dependable chapters to invest in.

Transportation: topics in the order to study them

  1. 1Transportation Problem: Introduction and FormulationYou need the table layout, supply-demand logic and the balanced condition before any method makes sense.
  2. 2Initial Basic Feasible Solution MethodsEvery problem starts with a feasible allocation, and the quality of this start decides how many MODI iterations you need.
  3. 3Optimality Test using MODI MethodIt builds on a completed starting solution, so study it only after you can produce one quickly and correctly.
  4. 4Special Cases: Degeneracy, Unbalanced and Prohibited RoutesThese are variations of the main method, so they are easiest to learn once the standard procedure is solid.

How to prepare Transportation

Treat this chapter as a procedure you rehearse, not a theory you read. Speed and neatness come only from solving full problems by hand.

  1. Learn the table layout: sources in rows, destinations in columns, unit costs in cells, supply on the right and demand at the bottom. Check total supply equals total demand before you start.
  2. Solve three or four problems each with NWCR, Least Cost Method and VAM. For VAM, practise writing row and column penalties clearly, as the difference between the two lowest costs in each line.
  3. After each starting solution, count the allocated cells. For a non-degenerate solution there must be m + n − 1 of them, where m is the number of rows and n the number of columns.
  4. Practise MODI in a fixed routine: set one u or v to 0, compute the other values from allocated cells using u + v = cost, then find the opportunity cost for each empty cell as cost − (u + v).
  5. If any empty cell has a negative value, choose the most negative, draw the closed loop, shift the smallest quantity from the minus cells and recompute. Repeat until none is negative.
  6. Do separate drills for unbalanced problems (add a dummy row or column at zero cost), prohibited routes (assign a very high cost) and degeneracy (add a tiny quantity ε to an empty cell).
  7. Finish by solving past-style 14-mark problems in exam time, showing the table, total cost and a one-line conclusion on the optimal plan.

Common mistakes in Transportation

  • Starting without checking that supply equals demand.

    Fix: Total both sides first. If they differ, add a dummy row or column with zero costs and the difference as its quantity.

  • Wrong penalties in VAM, such as using the highest and lowest cost instead of the two lowest.

    Fix: Always take the two smallest costs in each row and column, subtract, and cross out exhausted rows or columns before recalculating.

  • Having fewer or more than m + n − 1 allocations and carrying on into MODI.

    Fix: Count allocations after the initial solution. If short, add ε to a suitable empty cell that does not form a closed loop with the allocated cells.

  • Errors in u and v values that spoil every opportunity cost.

    Fix: Write the u and v values beside the table, and verify each allocated cell satisfies u + v = cost before moving on.

  • Drawing the closed loop wrongly or shifting the wrong quantity.

    Fix: Use only horizontal and vertical turns at allocated cells, alternate + and − from the entering cell, and shift the smallest quantity among the minus cells.

  • Giving the final total cost without recomputing it or stating the plan.

    Fix: Recalculate total cost as Σ (units × unit cost) from the final table and write the optimal allocation and cost clearly to secure the final marks.

Last-day revision: Transportation

  • A transportation problem minimises total shipping cost subject to supply and demand constraints.
  • A problem is balanced when total supply equals total demand.
  • An unbalanced problem is balanced by adding a dummy source or destination with zero cost.
  • A non-degenerate basic feasible solution has exactly m + n − 1 allocations.
  • NWCR starts at the top-left cell and ignores costs, so it is fast but often far from optimal.
  • Least Cost Method allocates first to the cell with the lowest unit cost.
  • VAM uses penalties (difference of two lowest costs in each row and column) and usually gives the best starting solution.
  • In MODI, set one u or v to 0 and use u + v = cost for allocated cells.
  • Opportunity cost of an empty cell = cost − (u + v); for minimisation, the solution is optimal when all are zero or positive.
  • Improve a solution by drawing a closed loop from the most negative empty cell and shifting the smallest quantity on the minus corners.
  • Degeneracy arises when allocations are fewer than m + n − 1; add ε to an independent empty cell.
  • For a prohibited route, assign a very large cost so it receives no allocation.

Transportation practice questions

Transportation in other exams

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

Transportation: frequently asked questions

Which initial method should I use in the exam?

Use the method the question names. If it leaves the choice to you, VAM is usually best because it gives a solution close to optimal and saves MODI steps. It takes more time to write, so practise it well.

How do I know the solution is optimal in MODI?

Compute the opportunity cost of every empty cell as cost − (u + v). For a minimisation problem, if none is negative, the solution is optimal. A zero value indicates an alternative optimal solution.

What do I do if the problem is unbalanced?

Add a dummy destination if supply exceeds demand, or a dummy source if demand exceeds supply. Give the dummy the difference as its quantity and zero unit costs, then solve as usual.

How is a prohibited route handled?

Assign that cell a very large cost, often written as M, so that the method avoids allocating to it. Then solve the problem normally and confirm the final table has no allocation in that cell.