I'm new to operation research. I'm confusing about the sign of reduced cost and I have a few questions that I'd like to ask. Suppose that I solve a maximization problem, considering a variable x with positive coefficient, then

  1. What is the sign of the reduced cost of this variable if in the solution x > 0 ?
  2. If x > 0 in the solution, then is x a basic variable ?
  3. How do you interpret the sign of reduced cost in a maximization problem ?

Thank you for your time


1 Answer 1


The reduced cost of a nonbasic variable $x$ is the net rate of change to the objective if you increase the value of $x$ by a small amount and optimally adjust the other variables to preserve feasibility. Regardless of whether you are maximizing or minimizing, a positive reduced cost means increasing $x$ will increase the objective value and a negative reduced cost means increasing $x$ will decrease the objective value. The reduced cost of a basic variable is always 0.

Assuming that the problem has been solved to optimality, and assuming that the domain of $x$ is $[0, \infty)$ (typical for most LP formulations), then if $x>0$ in the optimal solution it will be a basic variable, and if $x$ is basic the reduced cost will be 0.

If $x$ has a nonzero lower bound $\ell > 0$ that is expressed as a bound on the variable rather than as a constraint (meaning that the constraint matrix does not include $x\ge \ell$), and if $x=\ell$ in the solution, then it may be nonbasic even though it is positive. The same is true if $x$ has a finite upper bound $u$ expressed as a bound rather than a constraint and $x=u$ in the optimal solution. This is because there are versions of the simplex algorithm that handle a lower bound $\ell$ by substituting $\hat{x}=x - \ell$ for $x$ in the model, and handle an upper bound $u$ by substituting $\hat{x} = u - x$ for $x$ when $x$ hits its upper bound.

  • $\begingroup$ Thank you for your answer. Let say if the solution of the LP is optimal, then is it safe to say that the reduced cost of variable $x$ will be negative ? Since increasing $x$ cannot obtain a better solution, while decreasing $x$ will make the objective function smaller $\endgroup$ Commented Apr 27, 2023 at 3:26
  • $\begingroup$ No, that is incorrect on two counts. First, the reduced cost of a basic variable is always 0, so if $x$ is basic its reduced cost cannot be negative. Second, if LP has multiple optima, the reduced cost of a nonbasic variable that is positive in a different optimal solution will be 0. $\endgroup$
    – prubin
    Commented Apr 27, 2023 at 16:28
  • $\begingroup$ Yes I agree with if $x$ is basic then its reduced cost is 0. I was talking about a case in which $x$ is non basic but its value is >0 in the optimal solution (because there's an lower bound and upper bound on $x$), then the sign of the reduced cost will always be negative ? $\endgroup$ Commented Apr 28, 2023 at 14:16
  • $\begingroup$ Not necessarily if there are multiple optima. $\endgroup$
    – prubin
    Commented Apr 28, 2023 at 19:18
  • $\begingroup$ Could you explain more if it's not the case if there are multiple optima ? My thought is that if the $rcost(x) > 0$, its mean that increase $x$ will increase the objective value, but the objective value is already optimal, so it cant be increase anymore. $\endgroup$ Commented Apr 29, 2023 at 3:10

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