方法作為Microsoft Solver Foundation中的目標函數？

Question

以下是MSF的典型優化配方：

using Microsoft.SolverFoundation.Services;

SolverContext context = SolverContext.GetContext();
Model model = context.CreateModel();

//decisions
Decision xs = new Decision(Domain.Real, "Number_of_small_chess_boards");
Decision xl = new Decision(Domain.Real, "Number_of_large_chess_boards");

model.AddDecisions(xs, xl);

//constraints
model.AddConstraints("limits", 0 <= xs, 0 <= xl);
model.AddConstraint("BoxWood", 1 * xs + 3 * xl <= 200);
model.AddConstraint("Lathe", 3 * xs + 2 * xl <= 160);

//Goals
model.AddGoal("Profit", GoalKind.Maximize, 5 * xs + 20 * xl);

// This doesn't work!
// model.AddGoal("Profit", GoalKind.Maximize, objfunc(xs, xl));


Solution sol = context.Solve(new SimplexDirective());
Report report = sol.GetReport();
Console.WriteLine(report);

是否可以使用單獨的方法而不是像“5 * xs + 20 * xl”這樣的語句作為目標函數？ 例如，如下所示的方法？ 怎么樣？

// this method doesn't work!
static double objfunc(Decision x, Decision y)
{
    return 5 * x.ToDouble() + 20 * y.ToDouble();
}

Answer 1

它是如此簡單：

 static Term objfunc(Decision x, Decision y)
        {
            return 5 * x + 20 * y;
        }

該函數不必返回double ，而是返回Term 。

Answer 2

注意該函數不返回數字答案，而是一種評估答案的方法。 如果你要重新編碼為Term Test = 5 * x + 20 * y; String strTest = Test.ToString（）;

然后strTest會像（Add（mult（5，x），mult（20，y））;