使用分段线性逼近的单位承诺问题变为MIQP [英] unit commitment problem using piecewise-linear approximation become MIQP

问题描述

我尝试使用MILP(混合整数线性规划)来计算单位承诺问题. (单位承诺:试图找到最佳发电机调度的优化问题)

I try to use MILP (Mixed Integer Linear Programming) to calculate the unit commitment problem. (unit commitment: An optimization problem trying to find the best scheduling of generator)

有两个优化变量.

发电机功率: P (连续变量). 成本曲线上要使用的线段: BN (二进制变量). ，用于线性化生成器的二次成本函数.

Generator power :P(continuous variables). Which line segment on cost curve to use :BN(binary variable). ,Used to linearize the quadratic cost function of the generator.

一次只能打开一个线段.因此将有一个约束. Bn1 + Bn2 + Bn3 <= 1

Only one line segment can be opened at a time. So there will be a Constraint. Bn1 + Bn2 + Bn3 <=1

每个线段将具有自己的斜率和截距. 我想计算最低成本. 这个数学公式表示从1到H小时的总成本.

Each line segment will have its own slope and intercept. I want to calculate the minimum cost. This mathematical formula represents the sum of the cost of 1 to H hours.

这就是我的编码方式:sum(slope1 * p * Bn1 +拦截1 * Bn1 + slope2 * p * Bn2 +拦截2 * Bn2 +坡度3 * p * Bn3 +拦截3 * Bn3 )

This is how I code : sum(slope1* p * Bn1 +intercept1* Bn1 +slope2* p * Bn2 +intercept2* Bn2 +slope3* p * Bn3 +intercept3* Bn3 )

这样，两个优化变量将相乘.使问题从MILP变成MIQP. 我想问一下是否有任何方法可以保持我在MILP中的问题. 谢谢你. ps:我使用python API的ibm cplex解决优化问题

This way, the two optimization variables will be multiplied. Make the problem from MILP become to MIQP. I want to ask if there is any way can maintain my problem in MILP. thank you. ps : i use ibm cplex of python API to solve Optimization problem

推荐答案

您可以在docplex中使用分段线性.让我分享来自python中 zoo和bus的示例:

You could use piecewise linear in docplex. Let me share the example from the zoo and bus example in python:

from docplex.mp.model import Model

mdl = Model(name='buses')
nbbus40 = mdl.integer_var(name='nbBus40')
nbbus30 = mdl.integer_var(name='nbBus30')
mdl.add_constraint(nbbus40*40 + nbbus30*30 >= 300, 'kids')

#after 4 buses, additional buses of a given size are cheaper
f=mdl.piecewise(0, [(0, 0),(4,4)], 0.8)

mdl.minimize(f(nbbus40)*500 + f(nbbus30)*400)

mdl.solve()

mdl.export("c:\\buses.lp")

for v in mdl.iter_integer_vars():
    print(v," = ",v.solution_value)

给出

nbBus40  =  0
nbBus30  =  10.0

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