When you solve a linear program, you usually focus on the optimal objective value and the values of the decision variables. But the reduced costs in the sensitivity report quietly answer a crucial question: “Why are some variables zero, and what would it take for them to become attractive?”

Reduced costs tell you how far a variable’s objective coefficient (profit, cost, contribution margin) must change before that variable enters the optimal solution. This is essential in product mix, transportation, and portfolio models where many potential options are currently excluded.

On this page, we’ll define reduced costs, interpret them in both maximization and minimization problems, and walk through a concrete numerical example.

• Linear Programming & Optimization
• Business Analytics Overview
• Inventory & Supply Chain Analytics
• Decision Analysis & Simulation

1. What Is a Reduced Cost?

In a linear program, each decision variable has an objective coefficient (profit per unit, cost per unit, etc.). At the optimal solution:

• Variables that are positive (basic variables) have a reduced cost of 0.
• Variables that are zero (nonbasic variables) typically have a nonzero reduced cost.

Intuitively, the reduced cost of a zero variable measures how much its objective coefficient would need to improve before it becomes profitable (in a maximization problem) or cheap enough (in a minimization problem) to enter the basis.

1.1 Maximization Problems

For a maximization problem:

• If a variable is at zero and has a positive reduced cost, its profit per unit must increase by at least that amount before it becomes attractive to produce.
• If the reduced cost is zero, the variable is already in the basis or is an alternative optimal variable.

1.2 Minimization Problems

For a minimization problem:

• If a variable is at zero and has a negative reduced cost, its cost per unit must decrease (become more negative) by at least that amount before it becomes attractive to include.
• Again, a reduced cost of zero means the variable is in the basis or part of an alternative optimum.

In both cases, reduced costs quantify the “distance” from the current solution to a solution where that variable becomes active.

2. Why Reduced Costs Matter in Practice

• Product mix decisions: Identify which products are currently unprofitable and how much their margins must improve to be worth producing.
• Transportation and logistics: See which routes or shipping options are not used and how much their costs must drop to become competitive.
• Portfolio and project selection: Understand which projects or investments are “almost” attractive and how much their returns must increase.
• Model refinement: Spot variables that will never be used under realistic parameter ranges, helping you simplify or refocus the model.

Reduced costs are especially powerful when you have many candidate variables (products, routes, projects) but only a few are selected in the optimal solution.

3. Numerical Example: Product Mix with an Unused Product

Consider a firm that can produce three products: \(x_1\), \(x_2\), and \(x_3\). They all use the same two resources: machine time and labor.

3.1 Model Formulation

Decision variables

• \(x_1\): units of Product 1
• \(x_2\): units of Product 2
• \(x_3\): units of Product 3

Objective (maximize profit)

Profit per unit:

• Product 1: 40
• Product 2: 30
• Product 3: 25

Maximize: \[ Z = 40x_1 + 30x_2 + 25x_3 \]

Constraints

• Machine time: each unit of Product 1 uses 2 hours, Product 2 uses 1 hour, Product 3 uses 1.5 hours, and there are 100 hours available. \[ 2x_1 + x_2 + 1.5x_3 \le 100 \]
• Labor: each unit of Product 1 uses 1 hour, Product 2 uses 1 hour, Product 3 uses 0.5 hours, and there are 80 hours available. \[ x_1 + x_2 + 0.5x_3 \le 80 \]
• Nonnegativity: \(x_1, x_2, x_3 \ge 0\)

3.2 Solving the LP (Ignoring \(x_3\) Initially)

First, suppose we only consider \(x_1\) and \(x_2\) (set \(x_3 = 0\)) and solve:

Maximize: \[ Z = 40x_1 + 30x_2 \] subject to: \[ 2x_1 + x_2 \le 100 \] \[ x_1 + x_2 \le 80 \] \[ x_1, x_2 \ge 0 \]

As in a standard two-variable LP, we find the intersection of the two constraints:

2x₁ + x₂ = 100
x₁ + x₂ = 80
  

Subtract the second equation from the first:

(2x₁ + x₂) - (x₁ + x₂) = 100 - 80
x₁ = 20
  

Substitute into \(x_1 + x_2 = 80\):

20 + x₂ = 80  ⇒  x₂ = 60
  

So the optimal solution (ignoring \(x_3\)) is:

• \(x_1 = 20\)
• \(x_2 = 60\)
• \(x_3 = 0\)
• Objective value: \(Z = 40(20) + 30(60) = 800 + 1800 = 2600\)

At this solution:

• Machine time used: \(2(20) + 60 = 100\) (binding)
• Labor used: \(20 + 60 = 80\) (binding)

4. Interpreting the Reduced Cost of Product 3

Now we bring \(x_3\) back into the model. Suppose a Solver sensitivity report for this full three-product model shows:

• Optimal solution still: \(x_1 = 20\), \(x_2 = 60\), \(x_3 = 0\)
• Reduced cost of \(x_3\): 5

Because this is a maximization problem and \(x_3 = 0\), a positive reduced cost of 5 means:

Product 3’s profit per unit must increase by at least 5 (from 25 to 30) before it becomes attractive to produce in the optimal solution.

In other words, at the current profit of 25, including any positive amount of \(x_3\) would reduce the objective value relative to the current optimal basis. Only if its profit rises to 30 or more will the Simplex algorithm consider bringing \(x_3\) into the basis.

4.1 Managerial Interpretation

• If marketing or R&D can increase Product 3’s contribution margin by at least 5 per unit, it may become part of the optimal product mix.
• If that improvement is unrealistic, managers know that Product 3 is not worth producing under the current cost and price structure.
• Reduced costs thus provide a clear threshold for when a currently unused option becomes viable.

4.2 Alternative Optima and Zero Reduced Costs

If a variable is at zero but has a reduced cost of 0, this indicates a potential alternative optimal solution. You could increase that variable from zero and adjust others while keeping the same optimal objective value.

In practice, this can reveal flexibility in your plan: multiple different mixes of products or routes that all achieve the same best objective value.

5. Reduced Costs vs. Shadow Prices

It’s easy to confuse reduced costs with shadow prices, but they answer different questions:

• Shadow prices (dual values): How much does the objective improve if I relax a constraint’s RHS by one unit?
• Reduced costs: How much must a variable’s objective coefficient improve before that variable enters the optimal solution?

Together, they give a full picture of the LP’s economics:

• Shadow prices value resources (constraints).
• Reduced costs value options (variables) that are currently unused.

6. Why Reduced Costs Are Essential for Analysts

For students and practitioners of linear programming, reduced costs are not just a technical detail: they are a diagnostic tool for understanding and explaining the solution.

• Explain zero variables: Instead of saying “the model chose not to produce Product 3,” you can say “Product 3 would need 5 more units of profit per unit to be worth producing.”
• Guide strategic changes: Reduced costs tell you exactly how much improvement is needed to make a new product, route, or project viable.
• Screen candidate options: In large models, you can quickly identify which variables are hopelessly uncompetitive under any realistic scenario.

If your course or project also covers duality, shadow prices, or sensitivity analysis more broadly, reduced costs are the natural complement: they show how the current optimal solution “pushes back” against adding new options.

To connect this with other topics in your syllabus, you can explore:

• Linear Programming & Optimization
• Parent Business Analytics
• Inventory & Supply Chain Analytics
• Decision Analysis & Decision Trees