公司含硫原料生产
作者：互联网
问题
某公司将4种不同含硫量的液体原料(分别记为甲、乙、丙、丁)混合生产两种产品
(分别记为A,B).按照生产工艺的要求,原料甲,乙、丁必须首先倒入混合池中混合,混
合后的液体再分别与原料丙混合生产A,B.已知原料甲、乙,丙、丁的含硫量分别是
3% , 1 % ,2% , 1 % ,进货价格(单位:千元/t)分别为6,16,10,15;产品A,B的含硫量分
别不能超过2.5% ,1.5% ,售价分别为9,15.根据市场信息,原料甲、乙,丙的供应没有
限制,原料丁的供应量最多为50t;产品A,B的市场需求量分别为100 t ,200 t.问应如何
安排生产？
假设
y
1
,
z
1
y_1,z_1
y1,z1分别为来自混合池和丙池的生成A的原料质量，
y
2
,
z
2
y_2,z_2
y2,z2分别为来自混合池和丙池的生成B的原料质量，混合池中
x
1
,
x
2
,
x
4
x_1,x_2,x_4
x1,x2,x4分别为甲乙丁所占的比例。
那么可构建模型如下
M a x q q u a d y = 9 ∗ ( y 1 + z 1 ) + 15 ∗ ( y 2 + z 2 ) − 10 ∗ ( z 1 + z 2 ) − 6 ∗ x 1 ∗ ( y 1 + y 2 ) − 16 ∗ x 2 ∗ ( y 1 + y 2 ) − 15 ∗ x 4 ∗ ( y 1 + y 2 ) S . T . = { x 4 ∗ ( y 1 + y 2 ) < = 50 y 1 + z 1 < = 100 y 2 + z 2 < = 200 ( 3 ∗ x 1 + 1 ∗ x 2 + 1 ∗ x 4 ) ∗ y 1 + 2 ∗ z 1 < = 2.5 ∗ ( y 1 + z 1 ) ( 3 ∗ x 1 + 1 ∗ x 2 + 1 ∗ x 4 ) ∗ y 2 + 2 ∗ z 2 < = 1.5 ∗ ( y 2 + z 2 ) x 1 + x 2 + x 4 = = 1 Max \\qquad y = 9*(y_1+z_1)+15*(y_2+z_2)-10*(z_1+z_2)-6*x_1*(y_1+y_2)-16*x_2*(y_1+y_2)-15*x_4*(y_1+y_2)\\ S.T. = \left\{ \begin{aligned} x_4*(y_1+y_2)<=50\\ y_1+z_1<=100\\ y_2+z_2<=200\\ (3*x_1+1*x_2+1*x_4)*y_1+2*z_1<=2.5*(y_1+z_1)\\ (3*x_1+1*x_2+1*x_4)*y_2+2*z_2<=1.5*(y_2+z_2)\\ x_1+x_2+x_4==1\\ \end{aligned} \right. Maxqquady=9∗(y1+z1)+15∗(y2+z2)−10∗(z1+z2)−6∗x1∗(y1+y2)−16∗x2∗(y1+y2)−15∗x4∗(y1+y2)S.T.=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧x4∗(y1+y2)<=50y1+z1<=100y2+z2<=200(3∗x1+1∗x2+1∗x4)∗y1+2∗z1<=2.5∗(y1+z1)(3∗x1+1∗x2+1∗x4)∗y2+2∗z2<=1.5∗(y2+z2)x1+x2+x4==1
import gurobipy as grb
from gurobipy import *
model = grb.Model("Make")
# 定义变量
y1 = model.addVar(name='y1')
y2 = model.addVar(name='y2')
z1 = model.addVar(name='z1')
z2 = model.addVar(name='z2')
x1 = model.addVar(name='x1')
x2 = model.addVar(name='x2')
x4 = model.addVar(name='x4')
# 添加约束
model.addConstr(x4*(y1+y2)<=50,"丁的供应量")
model.addConstr(y1+z1<=100,"A的需求量")
model.addConstr(y2+z2<=200,"B的需求量")
model.addConstr((3*x1+1*x2+1*x4)*y1+2*z1<=2.5*(y1+z1),"A含S占比")
model.addConstr((3*x1+1*x2+1*x4)*y2+2*z2<=1.5*(y2+z2),"B含S占比")
model.addConstr(x1+x2+x4==1,"比例约束")
model.Params.NonConvex = 2#将线性规划变为非线性规划 模型将按照求解MIP的逻辑来求解此类非凸问题。
model.setObjective( 9*(y1+z1)+15*(y2+z2)-10*(z1+z2)-6*x1*(y1+y2)-16*x2*(y1+y2)-15*x4*(y1+y2), grb.GRB.MAXIMIZE)
model.optimize()
if model.status == GRB.OPTIMAL:
model.printAttr('X')
model.printAttr('Slack')
Changed value of parameter NonConvex to 2
Prev: -1 Min: -1 Max: 2 Default: -1
Gurobi Optimizer version 9.1.2 build v9.1.2rc0 (win64)
Thread count: 4 physical cores, 8 logical processors, using up to 8 threads
Optimize a model with 3 rows, 7 columns and 7 nonzeros
Model fingerprint: 0x34e4e21e
Model has 6 quadratic objective terms
Model has 3 quadratic constraints
Coefficient statistics:
Matrix range [1e+00, 1e+00]
QMatrix range [1e+00, 3e+00]
QLMatrix range [5e-01, 3e+00]
Objective range [1e+00, 2e+01]
QObjective range [1e+01, 3e+01]
Bounds range [0e+00, 0e+00]
RHS range [1e+00, 2e+02]
QRHS range [5e+01, 5e+01]
Continuous model is non-convex -- solving as a MIP.
Presolve time: 0.00s
Presolved: 19 rows, 14 columns, 50 nonzeros
Presolved model has 6 bilinear constraint(s)
Variable types: 14 continuous, 0 integer (0 binary)
Root relaxation: objective 2.933333e+03, 7 iterations, 0.00 seconds
Nodes | Current Node | Objective Bounds | Work
Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time
0 0 2933.33333 0 6 - 2933.33333 - - 0s
0 0 550.00000 0 2 - 550.00000 - - 0s
0 0 550.00000 0 2 - 550.00000 - - 0s
0 0 525.00000 0 3 - 525.00000 - - 0s
0 0 450.00000 0 3 - 450.00000 - - 0s
H 0 0 450.0000000 450.00000 0.00% - 0s
Cutting planes:
RLT: 6
Explored 1 nodes (19 simplex iterations) in 0.09 seconds
Thread count was 8 (of 8 available processors)
Solution count 1: 450
Optimal solution found (tolerance 1.00e-04)
Best objective 4.500000000000e+02, best bound 4.500000000000e+02, gap 0.0000%
Variable X
-------------------------
y2 100
z2 100
x2 0.5
x4 0.5
Constraint Slack
-------------------------
A的需求量 100
标签：0s,00,name,原料,含硫,range,生产,model,addVar 来源： https://blog.csdn.net/Kilig___/article/details/120659736