SUPSupply & productionStop 6 of 12
Multi-plant allocation
Deciding which plant should make how much of each product, given different costs, capacities and distances to customers.
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A company with several plants making similar products has to decide each month who makes what. The cheapest plant per unit isn’t automatically the answer, because freight, duties, capacity and changeovers all count.
Take shampoo made in two plants:
| Cost to make | Freight to North India | Freight to South India | Monthly capacity | |
|---|---|---|---|---|
| Plant A (Baddi) | ₹40 | ₹4 | ₹14 | 60,000 |
| Plant B (Chennai) | ₹43 | ₹15 | ₹3 | 50,000 |
North needs 50,000 units and South 40,000. Plant A is cheaper to run, but for the South it lands at ₹54 a unit against Plant B’s ₹46. The sensible split is that A serves the North (₹44 landed) and B serves the South (₹46 landed). Both fit within capacity, and the total comes to ₹40.4 lakh. Sending everything from A would cost more and would also exceed its capacity.
With dozens of products, plants and markets, this becomes a linear program: choose quantities that minimize total landed cost (production + freight + duties) while meeting demand and staying within each plant’s capacity. Real models add setup costs and minimum runs, which turn it into a mixed-integer problem, plus things like keeping a second source qualified.
Words to use
- Landed cost
- Production cost plus freight, duties and handling to the point of demand.
- Capacity constraint
- The limit on how much each plant can make in the period.
- Transportation problem
- The classic LP that assigns supply from sources to demand points at minimum cost.
Writing it as a linear program
- Decision: , how many units plant sends to market .
- Objective: minimize total landed cost, , where is the plant’s variable production cost and the freight and duties to market .
- Constraints: each market’s demand is met, ; each plant stays within capacity, ; and quantities are never negative.
Spreadsheet solvers handle small cases; planning systems and optimization solvers handle thousands of products, plants and markets.
Landed costs per unit: Plant A to North ₹44, A to South ₹54, Plant B to North ₹58, B to South ₹46. Capacities: A 60,000, B 50,000.
Now North needs 70,000 and South 40,000: 110,000 in total, exactly the combined capacity.
- A can’t cover the whole North, so the cheapest plan is: A sends 60,000 North, B sends 10,000 North and 40,000 South.
- Total cost: 60,000 × ₹44 + 10,000 × ₹58 + 40,000 × ₹46 = ₹50.6 lakh.
- Any other split costs more. Sending 10,000 from A to the South and 20,000 from B to the North, for example, costs ₹52.8 lakh.
Shadow prices: what more capacity is worth
The solver also reports a shadow price for each constraint: how much the total cost would fall if that constraint were relaxed by one unit. Here, one more unit of capacity at Plant A would replace a unit B sends North at ₹58 with one from A at ₹44, so A’s capacity is worth ₹14 a unit. If overtime at A costs less than ₹14 a unit extra, it pays. That turns a cost model into a capacity-investment argument.
When the model needs integers
Real allocation decisions include yes-or-no choices: whether to run a line this month (with a fixed cost), whether a product’s run meets its minimum batch, whether a market must be served by a single plant. These turn the linear program into a mixed-integer program. Solvers handle them well at planning scale, but solution times grow, so models are kept as simple as the decision allows.
Which costs belong in the model
Use the costs that actually change with the decision. For a monthly allocation, that is variable production cost (materials, energy, variable labour), freight, duties and taxes. A plant’s “standard cost” often includes fixed overheads such as depreciation, management salaries and rent, which don’t change with this month’s volume. Allocating on full standard cost can send volume away from a plant that is actually cheaper at the margin. Fixed costs matter in longer-term decisions, such as closing a line or a plant, which belong to network design.
Adding time
A multi-period model plans several months together: build stock at a plant with spare capacity in quiet months to cover a peak, weighed against holding cost. This links plant allocation with the level-versus-chase decision in S&OP, and the same solver can answer both.
Constraints that aren’t about cost
- Regulatory approvals: in pharmaceuticals and food, a plant may be approved only for certain markets or products.
- Service: some customers need short delivery times only a nearby plant can meet.
- Risk: keeping a second plant qualified and producing at a minimum level protects supply if the main plant fails.
- Taxes and duties: cross-border flows can change the landed cost sharply, and rules change.
- Carbon: some companies add emissions limits or an internal carbon price to the objective.
Making the answer usable
Planners need to trust the model. Show the landed cost per lane, the binding constraints and their shadow prices, and how the plan changes under different demand scenarios. A model that only outputs numbers, without the reasons, tends to be overridden.
Interview kit
Practise it out loud
Answer each question yourself before opening the model answer.
Multi-plant allocation decides how much of each product each plant makes for each market. The objective is minimum total landed cost, production plus freight plus duties, subject to meeting demand and each plant’s capacity. It’s a transportation-style linear program, extended with setup costs or minimum runs when needed, and the cheapest plant per unit often loses once freight is included.
Plant A makes at ₹40, Plant B at ₹43. Why might B still serve some markets?
Because freight can outweigh the production gap. If B is much closer to a market, its landed cost there can be lower. In the example, B lands at ₹46 in the South against ₹54 from A. Capacity limits at A may also force some volume to B.
How would you formulate this as an optimization problem?
Decision variables are the quantities from each plant to each market for each product. Minimize the sum of quantity times landed cost. Subject to: each market’s demand is met, and each plant’s total production stays within its capacity. Add binary setup variables if lines need minimum runs.
Connected stops
Where this idea feeds in, or gets its inputs from.
Sources and further reading
- Chopra, S. & Meindl, P. Supply Chain Management: Strategy, Planning, and Operation. Pearson. Chapters on network design and aggregate planning.
- Hillier, F. S. & Lieberman, G. J. Introduction to Operations Research. McGraw-Hill. Chapter on the transportation problem.
- Hillier, F. S., & Lieberman, G. J. (2021). Introduction to Operations Research (11th ed.). McGraw-Hill.