优化 | 线性化：含绝对值的线性化
作者：刘兴禄， 清华大学
清华-伯克利深圳学院，博士在读
非线性整数规划模型
考虑下面的非线性整数规划
max 2 ∣ x 1 ∣ + ∣ x 2 ∣ s . t . ∣ x 1 ∣ + 2 ∣ x 2 ∣ ⩽ 8 − 5 ⩽ x 1 , x 2 ⩽ 5 \begin{aligned} \max \quad \,\,\,& 2|x_1| + |x_2| \\ s.t. \quad \,\,\,& |x_1| + 2|x_2| \leqslant 8 \\ & -5 \leqslant x_1, x_2 \leqslant 5 \end{aligned} maxs.t.2∣x1∣+∣x2∣∣x1∣+2∣x2∣⩽8−5⩽x1,x2⩽5
Gurobi求解代码
from gurobipy import *
model = Model('non-linear model')
x1 = model.addVar(lb=-5, ub=5, vtype=GRB.CONTINUOUS, name='x1')
x2 = model.addVar(lb=-5, ub=5, vtype=GRB.CONTINUOUS, name='x2')
x1_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x1_abs')
x2_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x2_abs')
model.setObjective(2 * x1_abs + x2_abs, GRB.MAXIMIZE)
model.addConstr(x1_abs + 2*x2_abs <= 8)
model.addGenConstrAbs(x1_abs, x1)
model.addGenConstrAbs(x2_abs, x2)
model.optimize()
print('x1:', x1.x)
print('x2:', x2.x)
print('x1_abs:', x1_abs.x)
print('x2_abs:', x2_abs.x)
求解结果
Solution count 1: 11.5
Optimal solution found (tolerance 1.00e-04)
Best objective 1.150000000000e+01, best bound 1.150000000000e+01, gap 0.0000%
x1: 5.0
x2: 1.5
x1_abs: 5.0
x2_abs: 1.5
绝对值的线性化技巧
这个比较容易理解，例如
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0.5
x_i = -0.5
xi=−0.5， 则
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x
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0.5
x_i^{+} = 0, x_i^{-} = 0.5
xi+=0,xi−=0.5， 因此有
x
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x
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−
x
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−
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0
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0.5
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0.5
x_i = x_i^{+} - x_i^{-} = 0 - 0.5 = -0.5
xi=xi+−xi−=0−0.5=−0.5；
∣
x
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∣
=
x
i
+
+
x
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−
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0
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0.5
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|x_i| = x_i^{+} + x_i^{-} = 0 + 0.5 = 0.5
∣xi∣=xi++xi−=0+0.5=0.5。
利用上面的技巧进行线性化
- 我们根据
得到，原模型可以线性化为
max 2 ( x 1 + + x 1 − ) + ( x 2 + + x 2 − ) s . t . ( x 1 + + x 1 − ) + 2 ( x 2 + + x 2 − ) ⩽ 8 x 1 = x 1 + − x 1 − x 2 = x 2 + − x 2 − x 1 abs = x 1 + + x 1 − x 2 abs = x 2 + + x 2 − x 1 abs ⩾ x 1 x 1 abs ⩾ − x 1 x 2 abs ⩾ x 2 x 2 abs ⩾ − x 2 − 5 ⩽ x 1 , x 2 ⩽ 5 0 ⩽ x i + , x i − , x i abs ⩽ 5 , ∀ i = 1 , 2 \begin{aligned} \max \quad \,\,\,& 2(x_1^{+} + x_1^{-}) + (x_2^{+} + x_2^{-}) \\ s.t. \quad \,\,\,& (x_1^{+} + x_1^{-}) + 2(x_2^{+} + x_2^{-}) \leqslant 8 \\ & x_1 = x_1^{+} - x_1^{-} \\ & x_2 = x_2^{+} - x_2^{-} \\ & x_1^{\text{abs}} = x_1^{+} + x_1^{-} \\ & x_2^{\text{abs}} = x_2^{+} + x_2^{-} \\ & x_1^{\text{abs}} \geqslant x_1 \\ & x_1^{\text{abs}} \geqslant -x_1 \\ & x_2^{\text{abs}} \geqslant x_2 \\ & x_2^{\text{abs}} \geqslant -x_2 \\ & -5 \leqslant x_1, x_2 \leqslant 5 \\ & 0 \leqslant x_i^{+}, x_i^{-}, x_i^{\text{abs}}\leqslant 5, \forall i = 1, 2 \end{aligned} maxs.t.2(x1++x1−)+(x2++x2−)(x1++x1−)+2(x2++x2−)⩽8x1=x1+−x1−x2=x2+−x2−x1abs=x1++x1−x2abs=x2++x2−x1abs⩾x1x1abs⩾−x1x2abs⩾x2x2abs⩾−x2−5⩽x1,x2⩽50⩽xi+,xi−,xiabs⩽5,∀i=1,2
下面我们用代码进行验证。
from gurobipy import *
model = Model('non-linear model')
x1 = model.addVar(lb=-5, ub=5, vtype=GRB.CONTINUOUS, name='x1')
x2 = model.addVar(lb=-5, ub=5, vtype=GRB.CONTINUOUS, name='x2')
x1_pos_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x1_pos_abs')
x1_neg_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x1_neg_abs')
x2_pos_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x2_pos_abs')
x2_neg_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x2_neg_abs')
x1_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x1_abs')
x2_abs = model.addVar(lb=0, ub=5, vtype=GRB.CONTINUOUS, name='x2_abs')
model.setObjective(2*(x1_pos_abs + x1_neg_abs) + (x2_pos_abs + x2_neg_abs), GRB.MAXIMIZE)
model.addConstr((x1_pos_abs + x1_neg_abs) + 2*(x2_pos_abs + x2_neg_abs) <= 8)
model.addConstr(x1 == x1_pos_abs - x1_neg_abs)
model.addConstr(x2 == x2_pos_abs - x2_neg_abs)
model.addConstr(x1_abs == x1_pos_abs + x1_neg_abs)
model.addConstr(x2_abs == x2_pos_abs + x2_neg_abs)
model.addConstr(x1_abs >= x1)
model.addConstr(x1_abs >= -x1)
model.addConstr(x2_abs >= x2)
model.addConstr(x2_abs >= -x2)
model.optimize()
print('x1:', x1.x)
print('x2:', x2.x)
print('x1_pos_abs:', x1_pos_abs.x)
print('x1_neg_abs:', x1_neg_abs.x)
print('x2_pos_abs:', x2_pos_abs.x)
print('x2_neg_abs:', x2_neg_abs.x)
求解结果为
Iteration Objective Primal Inf. Dual Inf. Time
0 1.6000000e+01 3.000000e+00 0.000000e+00 0s
2 1.1500000e+01 0.000000e+00 0.000000e+00 0s
Solved in 2 iterations and 0.01 seconds
Optimal objective 1.150000000e+01
x1: 5.0
x2: 1.5
x1_pos_abs: 5.0
x1_neg_abs: 0.0
x2_pos_abs: 1.5
x2_neg_abs: 0.0
可见结果是一致的。
总结
含有绝对值形式的线性化。
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