Python 优化入门
Jasmin Ludolf
Content Developer
需求:
礼服生产:


$C$:面料成本 + S 先生工资 + T 女士机会成本
机会成本:选择缝纫而放弃其他职责的成本
$C=110g+240g+105g+75t+160t+35t$
$C=455g+270t$
| 成本 | 面料 | S 先生 | T 女士 |
|---|---|---|---|
| 礼服 | $\$110$ | $\$40/h \times 6h = \$240$ | $\$35/h \times 3h = \$105$ |
| 西装 | $\$75$ | $\$40/h \times 4h = \$160$ | $\$35/h \times 1h = \$35$ |
约束:
需求:$g\leq20$, $t\leq12$
供给:$6g+4t\leq40$, $3g+t\leq20$
from scipy.optimize import milp, Bounds, LinearConstraintresult = milp([-545, -330],integrality=[1, 1],bounds=Bounds([0, 0], [20, 12]),constraints=LinearConstraint([[6, 4], [3, 1]], ub=[40, 20]))
print(result.message)
print(f'The optimal number of gowns produced is: {result.x[0]:.2f}')
print(f'The optimal number of tuxedos produced is: {result.x[1]:.2f}')
Optimization terminated successfully. (HiGHS Status 7: Optimal)
The optimal number of gowns produced is: 6.00
The optimal number of tuxedos produced is: 1.00
result = milp([-545, -330],
bounds=Bounds([0, 0], [20, 12]),
constraints=LinearConstraint([[6, 4], [3, 1]], ub=[40, 20]))
...
The optimal number of gowns produced is: 6.67
The optimal number of tuxedos produced is: 0.00
建议解为 6.67 件礼服、0.00 套礼服西装 $\rightarrow$
四舍五入为 7 件礼服、0 套西装
截断为 6 件礼服、0 套西装
Python 优化入门