SCIP求解整数规划实例

SCIP求解整数规划实例

• 使用开源的求解器scip
• 以python接口求解整数规划问题demo

1. 整数规划问题：
   $$\begin{gathered} minZ=5x_1+6x_2+23x_3+5x_4+24x_5+6x_6+23x_7+5x_8 \\ \{\begin{array}{l}2\ast x_1+x_2+x_3+x_4>=100\\ 2\ast x_2+x_3+3\ast x_5+2\ast x_6+x_7>=150\\ x_1+x_3+3\ast x_4+2\ast x_6+3\ast x_7+5\ast x_8>=100\\ x_j>=0且为整数;j=1,2,...,8\end{array} \end{gathered}$$

2. demo.py

import pyscipopt as opt

def test_opt():
    model = opt.Model()

    # 定义变量
    x1 = model.addVar(vtype='I', name='x1')
    x2 = model.addVar(vtype='I', name='x2')
    x3 = model.addVar(vtype='I', name='x3')
    x4 = model.addVar(vtype='I', name='x4')
    x5 = model.addVar(vtype='I', name='x5')
    x6 = model.addVar(vtype='I', name='x6')
    x7 = model.addVar(vtype='I', name='x7')
    x8 = model.addVar(vtype='I', name='x8')

    # 定义约束
    model.addCons(2*x1 + x2 + x3 + x4 >= 100)
    model.addCons(2*x2+ x3 + 3*x5 + 2*x6 + x7 >= 150)
    model.addCons(x1 + x3 + 3*x4 + 2*x6 + 3*x7 + 5*x8 >= 100)

    # 设置目标函数
    model.setObjective(5*x1 + 6*x2 + 23*x3 + 5*x4 + 24*x5 + 6*x6 + 23*x7 + 5*x8)
    model.setMinimize()
    model.hideOutput()

    # 优化求解
    model.optimize()

    # 求解状态
    m_gap = model.getGap()
    m_status = model.getStatus()
    print('m_status = ', m_status)
    print('m_gap =', m_gap)

    # 求解值
    print('目标函数值: ', model.getObjVal())
    print('x1 = ', model.getVal(x1))
    print('x2 = ', model.getVal(x2))
    print('x3 = ', model.getVal(x3))
    print('x4 = ', model.getVal(x4))
    print('x5 = ', model.getVal(x5))
    print('x6 = ', model.getVal(x6))
    print('x7 = ', model.getVal(x7))
    print('x8 = ', model.getVal(x8))

if __name__ == '__main__':
    # 求解简单整数规划问题
    test_opt()

3. 输出结果