Variable and Constraint Management
- Variable duplication
Linear operations
- Absolute value
- 1-norm
Quadratic and Power operations
- Square
- 2-norm
- Powers
- p-norm
- Geometric mean
Exponentials and Logarithms
- log
- exp
- Entropy
- Relative entropy
- Log-sum-exp
Integer Modeling
- semicontinuous variable
- indicator variable
- Logical OR
- Logical NAND
Cardinality bound
Model assembly example
Problem Modification and Reoptimization
- Example: Production Planning
- - 改变线性约束矩阵
  - 增加变量
  - 加入新约束
  - 改变bounds
Reference

Variable and Constraint Management

Variable duplication

# x=y
def dup(M, x, y):M.constraint(Expr.sub(x, y), Domain.equalsTo(0.0))

Linear operations

Absolute value

t≥∣x∣t\geq |x|t≥∣x∣

def abs(M, t, x):M.constraint(Expr.add(t, x), Domain.greaterThan(0.0))M.constraint(Expr.sub(t, x), Domain.greaterThan(0.0))

1-norm

t≥∑i∣xi∣t\geq \sum_i|x_i|t≥∑i∣xi∣

def norm1(M, t, x):u=M.variable(x.getShape(), Domain.unbounded())abs(M, u, x) # u>=|x|M.constraint(Expr.sub(t, Expr.sum(u)), Domain.greaterThan(0.0))

Quadratic and Power operations

Square

t≥x2t\geq x^2t≥x2

def sq(M, t, x):M.constraint(Expr.hstack(0.5, t, x), Domain.inRotatedQCone())

2-norm

t≥∑ixi2t\geq \sqrt{\sum_i x_i^2}t≥∑ixi2

def norm2(M, t, x):M.constraint(Expr.vstack(t, x), Domain.inQCone())

Powers

t≥∣x∣p,p>1t\geq |x|^p, p\gt 1t≥∣x∣p,p>1

def pow(M, t, x, p):M.constraint(Expr.hstack(t, 1, x), Domain.inPPowerCone(1.0/p))

t≥1/xp,x>0,p>0t\geq 1/x^p, x>0, p>0t≥1/xp,x>0,p>0

def pow_inv(M, t, x, p):M.constraint(Expr.hstack(t, x, 1), Domain.inPPowerCone(1.0/(1.0+p)))

p-norm

t≥(∑i∣xi∣p)1/p,p>1t\geq (\sum_i |x_i|^p)^{1/p}, p>1t≥(∑i∣xi∣p)1/p,p>1

def pnorm(M, t, x, p):n=int(x.getSize())r=M.variable(n)M.constraint(Expr.sub(t, Expr.sum(r)), Domain.equalsTo(0.0))M.constraint(Expr.hstack(Var.repeat(t, n), r, x), Domain.inPPowerCone(1.0-1.0/p))

Geometric mean

t≥(x1×x2×⋯×xn)1/n,xi>0t\geq (x1\times x_2\times\dots\times x_n)^{1/n}, x_i>0t≥(x1×x2×⋯×xn)1/n,xi>0

def geo_mean(M, t, x):n=int(x.getSize())if n==1: abs(M, x, t)else:t2=M.variable()M.constraint(Expr.hstack(t2, x.index(n-1), t), Domain.inPPowerCone(1.0-1.0/n))geo_mean(M, t2, x.slice(0, n-1))

Exponentials and Logarithms

log

t≤log⁡x,x>0t\leq \log x, x>0t≤logx,x>0

def log(M, t, x):M.constraint(Expr.hstack(x, 1, t), Domain.inPExpCone())

exp

t≥ext\geq e^xt≥ex

def exp(M, t, x):M.constraint(Expr.hstack(t, 1, x), Domain.inPExpCone())

Entropy

t≥xlog⁡x,x>0t\geq x\log x, x>0t≥xlogx,x>0

def ent(M, t, x):M.constraint(Expr.hstack(1, x, Expr.neg(t)), Domain.inPExpCone())

Relative entropy

t≥xlog⁡x/y,x,y>0t\geq x\log x/y, x,y>0t≥xlogx/y,x,y>0

def relent(M, t, x, y):M.constraint(Expr.hstack(y, x, Expr.neg(t)), Domain.inPExpCone())

Log-sum-exp

log⁡∑iexi≤t\log\sum_i e^{x_i}\leq tlog∑iexi≤t

def logsumexp(M, t, x):n=int(x.getSize())u=M.variable(n)M.constraint(Expr.hstack(u, Expr.constTerm(n, 1.0), Expr.sub(x, Var.repeat(t, n))), Domain.inPExpCone())M.constraint(Expr.sum(u), Domain.lessThan(1.0))

Integer Modeling

semicontinuous variable

x∈{0}∪[a,b],b>a>0x\in\{0\}\cup [a, b], b>a>0x∈{0}∪[a,b],b>a>0

def semicontinuous(M, x, a, b):u=M.variable(x.getShape(), Domain.binary())M.constraint(Expr.sub(x, Expr.mul(a, u)), Domain.greaterThan(0.0))M.constraint(Expr.sub(x, Expr.mul(b, u)), Domain.lessThan(0.0))

indicator variable

x≠0⇒t=1x\neq 0\Rightarrow t=1x=0⇒t=1

def indicator(M, t, x):M.constraint(t, Domain.inRange(0, 1))t.makeInteger()abs(M, t, x)

Logical OR

逻辑或运算

# x or y
def logic_or(M, x, y):M.constraint(Expr.adds(x, y), Domain.greaterThan(1.0))
# x1 or .. or xn, x is a binary vector
def logic_or_vec(M, x):M.constraint(Expr.sum(x), Domain.greaterThan(1.0))

Logical NAND

not (x and y)，最多有一个指示标量为真

def logic_sos1(M, x):M.constraint(Expr.sum(x), Domain.lessThan(1.0))def logic_nand(M, x, y):M.constraint(Expr.add(x, y), Domain.lessThan(1.0))

Cardinality bound

最多有kkk个变量非0，且∣xi∣≤1|x_i|\leq 1∣xi∣≤1

def card(M, x, k):t=M.variable(x.getShape(), Domain.binary())abs(M, t, x)M.constraint(Expr.sum(t), Domain.lessThan(k))

Model assembly example

求解模型
max⁡−x2+y2+log⁡y−x1.5s.t.x≥y+3\begin{aligned} &\max -\sqrt{x^2+y^2}+\log y-x^{1.5} \\ &s.t.\quad x\geq y+3 \end{aligned} max−x2+y2+logy−x1.5s.t.x≥y+3
等价转换为
max⁡−t0+t1−t2s.t.{x≥y+3t0≥x2+y2t1≤log⁡yt2≥x1.5\begin{aligned} &\max -t_0+t_1-t_2\\ &s.t. \begin{cases} x\geq y+3\\ t_0\geq \sqrt{x^2+y^2}\\ t_1\leq \log y\\ t_2\geq x^{1.5} \end{cases} \end{aligned} max−t0+t1−t2s.t.⎩⎪⎪⎪⎨⎪⎪⎪⎧x≥y+3t0≥x2+y2t1≤logyt2≥x1.5
mosek建模代码

def example():M=Model()x=M.variable()y=M.variable()t=M.variable(3)M.constraint(Expr.sub(x, y), Domain.greaterThan(3.0))norm2(M, t.index(0), Var.stack(x, y))log(M, t.index(1), y)pow(M, t.index(2), x, 1.5)# 目标函数M.objective(ObjectiveSense.Maximize, Expr.dot(t, [-1, 1, -1]))

Problem Modification and Reoptimization

当需要求解一系列相似优化问题时，可以将求解问题之间的不同分为以下几类

加入新的约束和变量
改变现有约束条件

Example: Production Planning

建立模型
max⁡1.5x0+2.5x1+3.0x2s.t.{2x0+4x1+3x2≤1000003x0+2x1+3x2≤500002x0+3x1+2x2≤60000x0,x1,x2≥0\begin{aligned} &\max 1.5x_0+2.5x_1+3.0x_2\\ &s.t. \begin{cases} 2x_0+4x_1+3x_2\leq 100000\\ 3x_0+2x_1+3x_2\leq 50000\\ 2x_0+3x_1+2x_2\leq 60000\\ x_0, x_1, x_2\geq 0 \end{cases} \end{aligned} max1.5x0+2.5x1+3.0x2s.t.⎩⎪⎪⎪⎨⎪⎪⎪⎧2x0+4x1+3x2≤1000003x0+2x1+3x2≤500002x0+3x1+2x2≤60000x0,x1,x2≥0
mosek建立模型

c=[1.5, 2.5, 3.0]
A=[[2, 4, 3], [3, 2, 3], [2, 3, 2]]
b=[100000, 50000, 60000]
nvar, ncon=len(c), len(b)with Model() as M:x=M.variable('x', nvar, Domain.greaterThan(0.0))con=M.constraint(Expr.mul(A, x), Domain.lessThan(b))M.objective(ObjectiveSense.Maximize, Expr.dot(c, x))M.solve()

改变线性约束矩阵

设置矩阵约束系数A00=3A_{00}=3A00=3

x0=x.index(0)
con.index(0).update(Expr.mul(3.0, x0), x0)

增加变量

加入一条新的数据信息后对模型进行更新
新模型为
max⁡1.5x0+2.5x1+3.0x2+x3s.t.{3x0+4x1+3x2+4x3≤1000003x0+2x1+3x2≤500002x0+3x1+2x2+x3≤60000x0,x1,x2≥0\begin{aligned} &\max 1.5x_0+2.5x_1+3.0x_2+x_3\\ &s.t. \begin{cases} 3x_0+4x_1+3x_2+4x_3\leq 100000\\ 3x_0+2x_1+3x_2\leq 50000\\ 2x_0+3x_1+2x_2+x_3\leq 60000\\ x_0, x_1, x_2\geq 0 \end{cases} \end{aligned} max1.5x0+2.5x1+3.0x2+x3s.t.⎩⎪⎪⎪⎨⎪⎪⎪⎧3x0+4x1+3x2+4x3≤1000003x0+2x1+3x2≤500002x0+3x1+2x2+x3≤60000x0,x1,x2≥0

x3=M.variable('y', Domain.greaterThan(0.0))
x_new=Var.vstack(x, x3)
# 更新约束条件
con.update(Expr.mul(x3, [4, 0, 1]), x3)
# 更新目标函数
M.objective(ObjectiveSense.Maximize, Expr.dot(c+[1.0], x_new))

加入新约束

加入约束条件
x0+2x1+x2+x3≤30000x_0+2x_1+x_2+x_3\leq 30000 x0+2x1+x2+x3≤30000

con2=M.constraint(Expr.dot(x_new, [1, 2, 1, 1]), Domain.lessThan(30000))

改变bounds

根据下表更新upper bounds

Operation	Time available(old)	Time available(new)
Assembly	100000	80000
Polishing	50000	40000
Packing	60000	50000
Quality control	30000	22000

新模型为
max⁡1.5x0+2.5x1+3.0x2+x3s.t.{3x0+4x1+3x2+4x3+20000≤1000003x0+2x1+3x2+10000≤500002x0+3x1+2x2+x3+10000≤60000x0+2x1+x2+x3+8000≤30000x0,x1,x2≥0\begin{aligned} &\max 1.5x_0+2.5x_1+3.0x_2+x_3\\ &s.t. \begin{cases} 3x_0+4x_1+3x_2+4x_3+20000\leq 100000\\ 3x_0+2x_1+3x_2+10000\leq 50000\\ 2x_0+3x_1+2x_2+x_3+10000\leq 60000\\ x_0+2x_1+x_2+x_3+8000\leq 30000\\ x_0, x_1, x_2\geq 0 \end{cases} \end{aligned} max1.5x0+2.5x1+3.0x2+x3s.t.⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧3x0+4x1+3x2+4x3+20000≤1000003x0+2x1+3x2+10000≤500002x0+3x1+2x2+x3+10000≤60000x0+2x1+x2+x3+8000≤30000x0,x1,x2≥0

cAll=Constraint.vstack(con, con2)
cAll.update([20000, 10000, 10000, 8000])

Reference

mosek fusion python API

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