kotlin - ojAlgo-在优化中将变量表示为边界？

Question

我一直在与ojAlgo玩耍，到目前为止，我对此感到非常激动。我已经对其进行了一些研究，但是使用problem described in this article遇到了麻烦。

我使用的是Kotlin而不是Java，但这不会引起任何问题。我一直试图在模型中输入表达式，但是要限制在变量而不是文字数值上。我该如何输入？

到目前为止，这是我的工作:

import org.ojalgo.optimisation.ExpressionsBasedModel
import org.ojalgo.optimisation.Variable


fun main(args: Array<String>) {

    val model = ExpressionsBasedModel()

    val ingredients = sequenceOf(
            Ingredient("Pork", 4.32, 30),
            Ingredient("Wheat", 2.46, 20),
            Ingredient("Starch", 1.86, 17)
    ).map { it.name to it }
     .toMap()

    val sausageTypes = sequenceOf(
            SausageType("Economy", .40),
            SausageType("Premium", .60)
    ).map { it.description to it }
     .toMap()

    // Map concatenated string keys to variables
    val variables = ingredients.values.asSequence().flatMap { ingredient ->
        sausageTypes.values.asSequence()
                .map { type -> Combo(ingredient,type)}
    }.map { it.toString() to Variable.make(it.toString()).lower(0).weight(it.ingredient.cost) }
     .toMap()

    // add variables to model
    model.addVariables(variables.values)

    // Pe + We + Se = 350 * 0.05
    model.addExpression("EconomyDemand").level(350.0 * 0.05).apply {
        set(variables["Pork-Economy"], 1)
        set(variables["Wheat-Economy"], 1)
        set(variables["Starch-Economy"], 1)
    }

    // Pp + Wp + Sp = 500 * 0.05
    model.addExpression("PremiumDemand").level(500.0 * 0.05).apply {
        set(variables["Pork-Premium"], 1)
        set(variables["Wheat-Premium"], 1)
        set(variables["Starch-Premium"], 1)
    }

    // Pe >= 0.4(Pe + We + Se) 
    // compile error?
    model.addExpression("EconomyGovRestriction").upper(variables["Pork-Economy"]).apply {
        set(variables["Pork-Economy"], .4)
        set(variables["Wheat-Economy"], .4)
        set(variables["Starch-Economy"], .4)
    }
}

data class Combo(val ingredient: Ingredient, val sausageType: SausageType) {
    override fun toString() = "$sausageType-$ingredient"
}

data class SausageType(val description: String, val porkRequirement: Double) {
    override fun toString() = description
}

data class Ingredient(val name: String, val cost: Double, val availability: Int) {
    override fun toString() = name
}

Answer 1

你不能那样做。您不能直接建模expr1 >= expr2。相反，您必须建模(expr1 - expr2) >= 0。 ojAlgo Wiki上有一个示例，描述了如何为类似问题建模:https://github.com/optimatika/ojAlgo/wiki/The-Diet-Problem