// (c) 2023-2025 Fair Isaac Corporation
import static com.dashoptimization.objects.Utils.sum;
//These imports are only for the parser.
import java.io.IOException;
import java.util.stream.IntStream;
import com.dashoptimization.ColumnType;
import com.dashoptimization.XPRSenumerations;
import com.dashoptimization.XPRSenumerations.ObjSense;
import com.dashoptimization.objects.LinExpression;
import com.dashoptimization.objects.Variable;
import com.dashoptimization.objects.XpressProblem;
/**
* Project planning: MIP example. A company has several projects that it must
* undertake in the next few months. Each project lasts for a given time (its
* duration) and uses up one resource as soon as it starts. The resource profile
* is the amount of the resource that is used in the months following the start
* of the project. For instance, project 1 uses up 3 units of resource in the
* month it starts, 4 units in its second month, and 2 units in its last month.
* The problem is to decide when to start each project, subject to not using
* more of any resource in a given month than is available. The benefit from the
* project only starts to accrue when the project has been completed, and then
* it accrues at BEN[p] per month for project p, up to the end of the time
* horizon.
*/
public class Pplan {
static final String[] PROJ = { "A", "B", "C" }; /* Set of projects */
static final int NM = 6; /* Time horizon (months) */
static final int[] MONTHS = IntStream.range(1, NM + 1).toArray(); /* Set of time periods (months) to plan for */
static final int[] DUR = { 3, 3, 4 }; /* Duration of project p */
static final int[] RESMAX = { 5, 6, 7, 7, 6, 6 }; /* Resource available in month m */
static final double[] BEN = { 10.2, 12.3, 11.2 }; /* Benefit per month once project finished */
static final int[][] RESUSE = /* Res. usage of proj. p in its t'th month */
{ { 3, 4, 2, 0, 0, 0 }, { 4, 1, 5, 0, 0, 0 }, { 3, 2, 1, 2, 0, 0 } };
public static void main(String[] args) throws IOException {
try (XpressProblem prob = new XpressProblem()) {
// Create the decision variables
// 1 if proj p starts in month m, else 0
Variable[][] start = prob.addVariables(PROJ.length, NM).withType(ColumnType.Binary)
.withName((p, m) -> String.format("start(%s,%d)", PROJ[p], MONTHS[m])).toArray();
// Each project starts once and only once
prob.addConstraints(PROJ.length, p -> sum(start[p]).eq(1));
// Resource availability. A project starting in month m is in its k-m month in
// month k:
// sum(p in PROJ, m in 0..k) RESUSE[p,k-m]*start[p,m] <= RESMAX[k]
prob.addConstraints(NM,
k -> sum(PROJ.length, p -> sum(k + 1, m -> start[p][m].mul(RESUSE[p][k - m]))).leq(RESMAX[k]));
/*
* Objective: Maximize Benefit If project p starts in month m, it finishes in
* month m+DUR[p]-1 and contributes a benefit of BEN[p] for the remaining
* NM-[m+DUR[p]-1] months:
*/
LinExpression obj = LinExpression.create();
for (int p = 0; p < PROJ.length; p++) {
for (int m = 0; m < NM - DUR[p]; m++) {
obj.addTerm(start[p][m], BEN[p] * (NM - m - DUR[p]));
}
}
prob.setObjective(obj, ObjSense.MAXIMIZE);
// Solve the problem
prob.mipOptimize("");
System.out.println("Problem status: " + prob.attributes().getMIPStatus());
if (prob.attributes().getMIPStatus() != XPRSenumerations.MIPStatus.SOLUTION
&& prob.attributes().getMIPStatus() != XPRSenumerations.MIPStatus.OPTIMAL)
throw new RuntimeException("optimization failed with status " + prob.attributes().getMIPStatus());
// Solution printing
System.out.println("Solution value is: " + prob.attributes().getObjVal());
double[] sol = prob.getSolution();
for (int p = 0; p < PROJ.length; p++)
for (int m = 0; m < NM; m++) {
if (start[p][m].getValue(sol) > 0.5)
System.out.println("Project " + PROJ[p] + " starts in month " + MONTHS[m]);
}
}
}
}
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// (c) 2023-2025 Fair Isaac Corporation
import static com.dashoptimization.objects.SOS.sos1;
import static com.dashoptimization.objects.Utils.sum;
//These imports are only for the parser.
import java.io.IOException;
import java.util.stream.IntStream;
import com.dashoptimization.XPRSenumerations;
import com.dashoptimization.XPRSenumerations.ObjSense;
import com.dashoptimization.objects.LinExpression;
import com.dashoptimization.objects.Variable;
import com.dashoptimization.objects.XpressProblem;
/**
* Project planning: Defining SOS1. A company has several projects that it must
* undertake in the next few months. Each project lasts for a given time (its
* duration) and uses up one resource as soon as it starts. The resource profile
* is the amount of the resource that is used in the months following the start
* of the project. For instance, project 1 uses up 3 units of resource in the
* month it starts, 4 units in its second month, and 2 units in its last month.
* The problem is to decide when to start each project, subject to not using
* more of any resource in a given month than is available. The benefit from the
* project only starts to accrue when the project has been completed, and then
* it accrues at BEN[p] per month for project p, up to the end of the time
* horizon.
*/
public class Pplan2 {
static final String[] PROJ = { "A", "B", "C" }; /* Set of projects */
static final int NM = 6; /* Time horizon (months) */
static final int[] MONTHS = IntStream.range(1, NM + 1).toArray(); /* Set of time periods (months) to plan for */
static final int[] DUR = { 3, 3, 4 }; /* Duration of project p */
static final int[] RESMAX = { 5, 6, 7, 7, 6, 6 }; /* Resource available in month m */
static final double[] BEN = { 10.2, 12.3, 11.2 }; /* Benefit per month once project finished */
static final int[][] RESUSE = /* Res. usage of proj. p in its t'th month */
{ { 3, 4, 2, 0, 0, 0 }, { 4, 1, 5, 0, 0, 0 }, { 3, 2, 1, 2, 0, 0 } };
public static void main(String[] args) throws IOException {
try (XpressProblem prob = new XpressProblem()) {
// Create the decision variables
// 1 if proj p starts in month m, else 0
Variable[][] start = prob.addVariables(PROJ.length, NM)
.withName((p, m) -> String.format("start(%s,%d)", PROJ[p], MONTHS[m])).toArray();
// Each project starts once and only once
prob.addConstraints(PROJ.length, p -> sum(start[p]).eq(1));
// Resource availability. A project starting in month m is in its k-m month in
// month k:
// sum(p in PROJ, m in 0..k) RESUSE(p,k-m)*start(p,m) <= RESMAX(k)
prob.addConstraints(NM,
k -> sum(PROJ.length, p -> sum(k + 1, m -> start[p][m].mul(RESUSE[p][k - m]))).leq(RESMAX[k]));
/*
* Define SOS-1 sets that ensure that at most one start(p,m) is non-zero for
* each project p. Use month index to order the variables
*/
prob.addConstraints(PROJ.length, p -> sos1(start[p], null, String.format("sos%d", p)));
/*
* Objective: Maximize Benefit If project p starts in month m, it finishes in
* month m+DUR[p]-1 and contributes a benefit of BEN[p] for the remaining
* NM-[m+DUR[p]-1] months:
*/
LinExpression obj = LinExpression.create();
for (int p = 0; p < PROJ.length; p++) {
for (int m = 0; m < NM - DUR[p]; m++) {
obj.addTerm(start[p][m], BEN[p] * (NM - m - DUR[p]));
}
}
prob.setObjective(obj, ObjSense.MAXIMIZE);
// Solve the problem
prob.mipOptimize("");
System.out.println("Problem status: " + prob.attributes().getMIPStatus());
if (prob.attributes().getMIPStatus() != XPRSenumerations.MIPStatus.SOLUTION
&& prob.attributes().getMIPStatus() != XPRSenumerations.MIPStatus.OPTIMAL)
throw new RuntimeException("optimization failed with status " + prob.attributes().getMIPStatus());
// Solution printing
System.out.println("Solution value is: " + prob.attributes().getObjVal());
double[] sol = prob.getSolution();
for (int p = 0; p < PROJ.length; p++)
for (int m = 0; m < NM; m++) {
if (start[p][m].getValue(sol) > 0.5)
System.out.println("Project " + PROJ[p] + " starts in month " + MONTHS[m]);
}
}
}
}
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