目标函数:min y 约束条件: y = 0, 0 <= x < 20 y = 60, 20 <= x < 30 当x=15时 将这个模型用VB.NET调用Cplex 建模求解
时间: 2024-03-24 21:38:59 浏览: 124
转化为标准形式步骤-线形规划课件
以下是用VB.NET调用Cplex进行建模和求解的示例代码:
```
Imports ILOG.CPLEX
Module Module1
Sub Main()
Try
Dim cplex As New Cplex()
' 定义变量
Dim x As INumVar = cplex.NumVar(0, 30, NumVarType.Float, "x")
Dim y As INumVar = cplex.NumVar(0, 60, NumVarType.Float, "y")
' 定义目标函数
Dim obj As ILinearNumExpr = cplex.LinearNumExpr()
obj.AddTerm(1.0, y)
cplex.AddMinimize(obj)
' 定义约束条件
Dim c1 As IRange = cplex.AddGe(y, 0)
Dim c2 As IRange = cplex.AddLe(y, 60)
' 定义分段函数
Dim f1 As IRange = cplex.AddLe(y, 0)
Dim f2 As IRange = cplex.AddLe(x, 20)
Dim f3 As IRange = cplex.AddLe(x, 30)
' 将约束条件和分段函数组合起来
cplex.Add(cplex.IfThen(f1, c1))
cplex.Add(cplex.IfThen(f2, c2))
cplex.Add(cplex.IfThen(f3, c2))
' 求解模型
If cplex.Solve() Then
Console.WriteLine("Solution status = " + cplex.GetStatus().ToString())
Console.WriteLine("Solution value = " + cplex.ObjValue.ToString())
Console.WriteLine("x = " + cplex.GetValue(x).ToString())
Console.WriteLine("y = " + cplex.GetValue(y).ToString())
End If
cplex.End()
Catch e As ILOG.Concert.Exception
Console.WriteLine("Concert Error: " + e.ToString())
End Try
End Sub
End Module
```
在上述代码中,我们首先定义了两个变量 x 和 y,然后定义了目标函数和约束条件。接着,我们定义了分段函数,并将其与约束条件组合起来。最后,我们使用 cplex.Solve() 方法求解模型,并输出最优解。
当 x=15 时,模型的最优解为 y=0,x=15。
阅读全文