MIP-Tool

MIP-Tool is a package for Python-MIP.

Installation

pip install mip-tool

Example

Show/View model

from mip import Model, maximize
from mip_tool import show_model

m = Model()
x = m.add_var("x")
y = m.add_var("y")
m.objective = maximize(x + y)
m += x + 2 * y <= 16
m += 3 * x + y <= 18
show_model(m)

Output

\Problem name: 

Minimize
OBJROW: - x - y
Subject To
constr(0):  x + 2 y <= 16
constr(1):  3 x + y <= 18
Bounds
End

In Jupyter

from mip_tool.view import view_model

view_model(m)

Output

モデル

変数 x : 非負変数 y : 非負変数

目的関数 x + y → 最大化

制約条件 x + 2.0y ≦ 16.0 3.0x + y ≦ 18.0

Non-convex piecewise linear constraint

Maximize y which is on points of (-2, 6), (-1, 7), (2, -2), (4, 5).

import numpy as np
from mip import INF, Model, OptimizationStatus
from mip_tool import add_lines, show_model

m = Model(solver_name="CBC")
x = m.add_var("x", lb=-INF)
y = m.add_var("y", obj=-1)
curve = np.array([[-2, 6], [-1, 7], [2, -2], [4, 5]])
add_lines(m, curve, x, y)
m.verbose = 0
m.optimize()
assert m.status == OptimizationStatus.OPTIMAL
assert (x.x, y.x) == (-1, 7)
show_model(m)

Output

\Problem name: 

Minimize
OBJROW: - y
Subject To
constr(0):  x - w_0 - w_1 - w_2 = -2
constr(1):  y - w_0 + 3 w_1 -3.50000 w_2 = 6
constr(2):  - w_0 + z_0 <= -0
constr(3):  w_0 <= 1
constr(4):  - w_1 + 3 z_1 <= -0
constr(5):  w_1 -3 z_0 <= -0
constr(6):  w_2 -2 z_1 <= -0
Bounds
 x Free
 0 <= z_0 <= 1
 0 <= z_1 <= 1
Integers
z_0 z_1 
End

F example

Easy to understand using F.

attention: Change Model and Var when using mip_tool.func.

from mip_tool.func import F

m = Model(solver_name="CBC")
x = m.add_var("x")
y = m.add_var("y", obj=-1)
m += y <= F([[0, 2], [1, 3], [2, 2]], x)
m.verbose = 0
m.optimize()
print(x.x, y.x)  # 1.0 3.0

• y <= F(curve, x) and y >= F(curve, x) call add_lines_conv.
• y == F(curve, x) calls add_lines.

pandas.DataFrame example

attention: Change Series when using mip_tool.func.

import pandas as pd
from mip import Model, maximize, xsum
from mip_tool.func import addvars

A = pd.DataFrame([[1, 2], [3, 1]])
b = pd.Series([16, 18])
m = Model(solver_name="CBC")
x = addvars(m, A, "X", False)
m.objective = maximize(xsum(x))
m += A @ x <= b
m.verbose = 0
m.optimize()
print(x.astype(float))  # [4. 6.]

Expression m += A.T.apply(lambda row: xsum(row * x)) <= b may be faster than m += A @ x <= b.