bottleneck_assignment: Solve the Bottleneck Assignment Problem

In couplr: Optimal Pairing and Matching via Linear Assignment

Solve the Bottleneck Assignment Problem

Description

Finds an assignment that minimizes (or maximizes) the maximum edge cost in a perfect matching. Unlike standard LAP which minimizes the sum of costs, BAP minimizes the maximum (bottleneck) cost.

Usage

bottleneck_assignment(cost, maximize = FALSE)

Arguments

cost	Numeric matrix; rows = tasks, columns = agents. NA or Inf entries are treated as forbidden assignments.
maximize	Logical; if TRUE, maximizes the minimum edge cost instead of minimizing the maximum (maximin objective). Default is FALSE (minimax).

Details

The Bottleneck Assignment Problem (BAP) is a variant of the Linear Assignment Problem where instead of minimizing the sum of assignment costs, we minimize the maximum cost among all assignments (minimax objective).

Algorithm: Uses binary search on the sorted unique costs combined with 'Hopcroft-Karp' bipartite matching to find the minimum threshold that allows a perfect matching.

Complexity: O(E * sqrt(V) * log(unique costs)) where E = edges, V = vertices.

Applications:

• Task scheduling with deadline constraints (minimize latest completion)

• Resource allocation (minimize maximum load/distance)

• Network routing (minimize maximum link utilization)

• Fair division problems (minimize maximum disparity)

Value

A list with class "bottleneck_result" containing:

• match - integer vector of length nrow(cost) giving the assigned column for each row (1-based indexing)

• bottleneck - numeric scalar, the bottleneck (max/min edge) value

• status - character scalar, e.g. "optimal"

See Also

assignment() for standard LAP (sum objective), lap_solve() for tidy LAP interface

Examples

# Simple example: minimize max cost
cost <- matrix(c(1, 5, 3,
                 2, 4, 6,
                 7, 1, 2), nrow = 3, byrow = TRUE)
result <- bottleneck_assignment(cost)
result$bottleneck  # Maximum edge cost in optimal assignment

# Maximize minimum (fair allocation)
profits <- matrix(c(10, 5, 8,
                    6, 12, 4,
                    3, 7, 11), nrow = 3, byrow = TRUE)
result <- bottleneck_assignment(profits, maximize = TRUE)
result$bottleneck  # Minimum profit among all assignments

# With forbidden assignments
cost <- matrix(c(1, NA, 3,
                 2, 4, Inf,
                 5, 1, 2), nrow = 3, byrow = TRUE)
result <- bottleneck_assignment(cost)

couplr documentation built on Jan. 20, 2026, 5:07 p.m.