指派问题¶
指派问题¶
每个工人必须分配到一个任务，每个任务也必须被一名工人承担；收益由工人和 任务的组合决定。使用二元变量、唯一分配约束和线性收益目标即可表达。
assignment_optx.py 默认使用 solve(...)，并在求解处保留
solve_optx(...) 的 OptX 精确路径注释。
示例默认使用统一的 solve(...) 接口；完整代码中的注释展示了如何切换到
solve_optx(...) 精确求解，同一份 ModelBuilder 模型无需重复编写。
完整代码¶
examples/linear/assignment_optx.py
from __future__ import annotations
from pathlib import Path
import sys
from optagent import ModelBuilder, solve
# from optagent import OptxConfig, solve_optx
sys.path.insert(0, str(Path(__file__).resolve().parents[1]))
from _common import print_solution
def build_model() -> tuple[object, dict[str, object]]:
builder = ModelBuilder(metadata={"case": "assignment_optx"})
profit = {
"a_x": 8,
"a_y": 6,
"b_x": 7,
"b_y": 9,
}
a_x = builder.int_var(default=0, lb=0, ub=1, name="a_x")
a_y = builder.int_var(default=0, lb=0, ub=1, name="a_y")
b_x = builder.int_var(default=0, lb=0, ub=1, name="b_x")
b_y = builder.int_var(default=0, lb=0, ub=1, name="b_y")
builder.constraint(a_x + a_y == 1, name="assign_worker_a")
builder.constraint(b_x + b_y == 1, name="assign_worker_b")
builder.constraint(a_x + b_x == 1, name="fill_job_x")
builder.constraint(a_y + b_y == 1, name="fill_job_y")
builder.maximize(
(a_x * profit["a_x"]) + (a_y * profit["a_y"]) + (b_x * profit["b_x"]) + (b_y * profit["b_y"]),
name="profit",
)
return builder.freeze(), {"profit": profit}
def main() -> None:
program, data = build_model()
solution = solve(program, time_limit_s=10.0, seed=7, threads=1, log_level="on")
# To use the OptX exact solver instead, replace the line above with:
# solution = solve_optx(program, config=OptxConfig(time_limit_s=10.0, threads=1))
print_solution("assignment solved by unified solve", solution, extra=data)
if __name__ == "__main__":
main()