ojAlgo - 在优化中将变量表示为边界

Question

我一直在玩ojAlgo，到目前为止，我对它感到非常兴奋。我已经用它进行了一些研究，但我遇到了本文中描述的这个问题。

我使用的是 Kotlin 而不是 Java，但这应该不会导致任何问题。我被困在尝试将表达式输入到我的模型中，但以变量而不是文字数值为边界。我该如何输入？

这是我到目前为止的工作：

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

对于未来的读者，这是我提出的完整工作解决方案。

import org.ojalgo.optimisation.ExpressionsBasedModel
import org.ojalgo.optimisation.Variable
import java.math.RoundingMode

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(17.5).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(25).apply {
        set(variables["Pork-Premium"], 1)
        set(variables["Wheat-Premium"], 1)
        set(variables["Starch-Premium"], 1)
    }
    // Pe >= 0.4(Pe + We + Se)
    model.addExpression("EconomyPorkRatio").upper(0.0).apply {
        set(variables["Pork-Economy"], -0.6)
        set(variables["Wheat-Economy"], .4)
        set(variables["Starch-Economy"], .4)
    }
    // Pe >= 0.6(Pp + Wp + Sp)
    model.addExpression("PremiumPorkRatio").upper(0.0).apply {
        set(variables["Pork-Premium"], -0.4)
        set(variables["Wheat-Premium"], .6)
        set(variables["Starch-Premium"], .6)
    }
    // Se <= .25(Pe + We + Se)
    // Sp <= .25(Pp + Wp + Sp)
    sausageTypes.values.forEach {
        model.addExpression("${it}StarchRestriction").lower(0.0).apply {
            set(variables["Pork-$it"], .25)
            set(variables["Wheat-$it"], .25)
            set(variables["Starch-$it"], -0.75)
        }
    }
    // Pe + Pp <= 30
    // We + Wp <= 20
    // Se + Sp <= 17
    ingredients.values.forEach { ingredient ->
        model.addExpression("${ingredient}SupplyConstraint").upper(ingredient.availability).apply {
            sausageTypes.values.forEach { sausageType ->
                set(variables["$ingredient-$sausageType"], 1)
            }
        }
    }
    // Pe + Pp >= 23
    model.addExpression("ContractPorkRestriction").lower(23).apply {
        set(variables["Pork-Economy"], 1)
        set(variables["Pork-Premium"], 1)
    }

    // go!
    val result = model.minimise()
    println("OPTIMIZED COST: ${result.value}")

    model.variables.asSequence()
            .map { it.name }
            .zip(result.asSequence().map { it.setScale(3, RoundingMode.HALF_DOWN) })
            .forEach(::println)
}
data class Combo(val ingredient: Ingredient, val sausageType: SausageType) {
    override fun toString() = "$ingredient-$sausageType"
}
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
}

输出：

OPTIMIZED COST: 140.955
(Pork-Economy, 8.000)
(Pork-Premium, 15.000)
(Wheat-Economy, 5.125)
(Wheat-Premium, 3.750)
(Starch-Economy, 4.375)
(Starch-Premium, 6.250)

Answer 2

你不能

那样做。您不能直接对expr1 >= expr2建模。相反，您必须(expr1 - expr2) >= 0建模。在ojAlgo维基上有一个例子描述了如何对类似的问题进行建模： https://github.com/optimatika/ojAlgo/wiki/The-Diet-Problem