Example 4: Blending Problem

Problem Description

Given a set of products and raw materials, each raw material has a specified available quantity, each product has a profit, each product has a maximum production quantity, and each product requires multiple types of raw materials.

	Raw Material A	Raw Material B
Available Quantity	$24$	$8$

	Product A	Product B
Profit	$5$	$4$

	Product A	Product B
Max Yield	$3$	$2$

	Raw Material A	Raw Material B
Product A	$6$	$1$
Product B	$4$	$2$

Determine the production quantity of each product to maximize total profit, while satisfying the following conditions:

1. The difference in production quantity between any two products must not exceed one unit.

Mathematical Model

Variables

$x_p$: Production quantity of product $p$.

Intermediate Values

1. Total Profit

$$\text{Profit}=\sum _{p\in P}{\text{Profit}}_p\cdot x_p$$

2. Raw Material Usage

$${\text{Use}}_m=\sum _{p\in P}{\text{Use}}_{pm}\cdot x_p,\mathrm{\forall }m\in M$$

Objective Function

1. Maximize Total Profit

$$max\text{Profit}$$

Constraints

1. Production Quantity Limit

$$\text{s.t.}{x}_p\le {\text{Yield}}_p^{\text{Max}},\mathrm{\forall }p\in P$$

2. Usage Quantity Limit

$$\text{s.t.}{\text{Use}}_m\le {\text{Available}}_m,\mathrm{\forall }m\in M$$

3. Production Difference Limit

$$\text{s.t.}{x}_p-x_{p^{\prime }}\le {\text{Diff}}^{\text{Max}},\mathrm{\forall }(p,p^{\prime })\in (P^2-\mathrm{\Delta }P)$$

Expected Results

Product A production quantity is $\frac{8}{3}$ units, Product B production quantity is $\frac{5}{3}$ units.

Code Implementation

kotlin

import fuookami.ospf.kotlin.utils.math.*
import fuookami.ospf.kotlin.utils.concept.*
import fuookami.ospf.kotlin.utils.functional.*
import fuookami.ospf.kotlin.utils.multi_array.*
import fuookami.ospf.kotlin.core.frontend.variable.*
import fuookami.ospf.kotlin.core.frontend.expression.monomial.*
import fuookami.ospf.kotlin.core.frontend.expression.polynomial.*
import fuookami.ospf.kotlin.core.frontend.expression.symbol.*
import fuookami.ospf.kotlin.core.frontend.inequality.*
import fuookami.ospf.kotlin.core.frontend.model.mechanism.*
import fuookami.ospf.kotlin.core.backend.plugins.scip.*

data class Material(
    val available: Flt64
) : AutoIndexed(Material::class)

data class Product(
     val profit: Flt64,
    val maxYield: Flt64,
    val use: Map<Material, Flt64>
) : AutoIndexed(Product::class)

val materials: List<Material> = ... // Raw material data
val products: List<Product> = ...  // Product data
val maxDiff = Int64(1)

// Create model instance
val metaModel = LinearMetaModel("demo4")

// Define variables
val x = RealVariable1("x", Shape1(products.size))
for (p in products) {
    x[p].name = "${x.name}_${p.index}"
}
metaModel.add(x)

// Define intermediate values
val profit = LinearExpressionSymbol(sum(products) { 
    p -> p.profit * x[p] 
}, "profit")
metaModel.add(profit)

val use = LinearIntermediateSymbols1("use", Shape1(materials.size)) { m, _ ->
    val material = materials[m]
    val ps = products.filter { it.use.contains(material) }
    LinearExpressionSymbol(
        sum(ps) { p -> p.use[material]!! * x[p] },
        "use_${m}"
    )
}
metaModel.add(use)

// Define objective function
metaModel.maximize(profit, "profit")

// Define constraints
for (p in products) {
    x[p].range.ls(p.maxYield)
}

for (m in materials) {
    metaModel.addConstraint(use[m] leq m.available)
}

for (p1 in products) {
    for (p2 in products) {
        if (p1.index == p2.index) {
            continue
        }
        metaModel.addConstraint((x[p1] - x[p2]) leq maxDiff)
    }
}

// Call solver to solve
val solver = ScipLinearSolver()
when (val ret = solver(metaModel)) {
    is Ok -> {
        metaModel.tokens.setSolution(ret.value.solution)
    }

    is Failed -> {}
}

// Parse results
val solution = HashMap<Material, Flt64>()
for (token in metaModel.tokens.tokens) {
    if (token.result!! eq Flt64.one && token.variable.belongsTo(x)) {
        solution[materials[token.variable.vectorView[0]]] = token.result!!
    }
}

Complete Implementation Reference:

• Kotlin