couplr (version 1.0.10)

assignment_duals: Solve assignment problem and return dual variables

Description

Solves the linear assignment problem and returns dual potentials (u, v) in addition to the optimal matching. The dual variables provide an optimality certificate and enable sensitivity analysis.

Usage

assignment_duals(cost, maximize = FALSE)

Value

A list with class "assignment_duals_result" containing:

• match - integer vector of column assignments (1-based)

• total_cost - optimal objective value

• u - numeric vector of row dual variables (length n)

• v - numeric vector of column dual variables (length m)

• status - character, e.g. "optimal"

Arguments

cost

Numeric matrix; rows = tasks, columns = agents. NA or Inf entries are treated as forbidden assignments.

maximize

Logical; if TRUE, maximizes the total cost instead of minimizing.

Details

The dual variables satisfy the complementary slackness conditions:

• For minimization: u[i] + v[j] <= cost[i,j] for all (i,j)

• For any assigned pair (i,j): u[i] + v[j] = cost[i,j]

This implies that sum(u) + sum(v) = total_cost (strong duality).

Applications of dual variables:

• Optimality verification: Check that duals satisfy constraints

• Sensitivity analysis: Reduced cost c[i,j] - u[i] - v[j] shows how much an edge cost must decrease before it enters the solution

• Pricing in column generation: Use duals to price new columns

• Warm starting: Reuse duals when costs change slightly

See Also

assignment() for standard assignment without duals

Examples

cost <- matrix(c(4, 2, 5, 3, 3, 6, 7, 5, 4), nrow = 3, byrow = TRUE)
result <- assignment_duals(cost)

# Check optimality: u + v should equal cost for assigned pairs
for (i in 1:3) {
  j <- result$match[i]
  cat(sprintf("Row %d -> Col %d: u + v = %.2f, cost = %.2f\n",
              i, j, result$u[i] + result$v[j], cost[i, j]))
}

# Verify strong duality
cat("sum(u) + sum(v) =", sum(result$u) + sum(result$v), "\n")
cat("total_cost =", result$total_cost, "\n")

# Reduced costs (how much must cost decrease to enter solution)
reduced <- outer(result$u, result$v, "+")
reduced_cost <- cost - reduced
print(round(reduced_cost, 2))