this is business analytics question. (using gurobi) The days-off scheduling problem must be solved routinely by businesses that operate 6 or 7 days a week. Examples include hospitals, airlines, municipal transportation companies, and the postal service. The most common example is the (5,7)-cyclic staffing problem. The objective of it is to minimize the cost of assigning workers to a 7-day cyclic schedule so that 1) Sufficient workers are available every day. 2) Each person works 5 consecutive days and is idle to the remaining 2 days. Here is the table showing the cost of having an employee for each day and the number of employees required for each day. For example, the pattern that one works from Sunday to Thursday costs 90+604=330. Formulate and solve an integer programming problem to represent this problem. How many employees work Monday-Friday?.

1. This is to test a balanced tree. I need help testing an unbalanced tree the same way
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.*;
public class Test {
public static final String INPUT = System.getProperty("user.dir") + "/input/";
public static final String OUTPUT = System.getProperty("user.dir") + "/output/";
public static void main(String[] args) throws FileNotFoundException {
TreeMap map = new TreeMap();
long startTime;
long totalTime = 0;
for(int i = 1; i <= 77;i++) {
Scanner scnr = new Scanner(new File("c:/Users/tvazq/IdeaProjects/hw2/resources/" + i +
".okpuncs"));
while(scnr.hasNext()) {
String word = scnr.next();
startTime = System.nanoTime();
if(!map.containsKey(word))map.put(word, 0);
map.put(word, 1 + map.get(word));
totalTime += (System.nanoTime() - startTime);
}
}
PrintWriter out = new PrintWriter("c:/Users/tvazq/IdeaProjects/hw2/part2" +
"frequencies.txt");
startTime = System.nanoTime();
Set allWords = map.keySet();
totalTime += (System.nanoTime() - startTime);
for(String word: allWords) {
startTime = System.nanoTime();
int frequency = map.get(word);
totalTime += (System.nanoTime() - startTime);
out.printf("%st%dn", word, frequency);

2. out.flush();
}
out.close();
System.out.printf("%.3f ms", totalTime/1e6);
}
}
UnbalancedTree.java
package part2;
import java.util.ArrayList;
import java.util.List;
class BinaryNode {
E element;
BinaryNode left;
BinaryNode right;
BinaryNode(E element) {
this.element = element;
}
}
class OrderedKeyValue implements Comparable {
String key;
int value;
OrderedKeyValue(String key, int value) {
this.key = key;
this.value = value;
}
@Override
public int compareTo(OrderedKeyValue other) {
return this.key.compareToIgnoreCase(other.key);
}
}

3. class UnbalancedTreeMap {
BinaryNode root;
public UnbalancedTreeMap() {
root = null;
}
public int get(String key) {
BinaryNode current = root;
while (current != null) {
int comparison = key.compareToIgnoreCase(current.element.key);
if (comparison < 0) {
current = current.left;
} else if (comparison > 0) {
current = current.right;
} else {
return current.element.value;
}
}
return 0;
}
public int put(String key, int value) {
BinaryNode parent = null;
BinaryNode current = root;
int comparison = 0;
while (current != null) {
comparison = key.compareToIgnoreCase(current.element.key);
if (comparison < 0) {
parent = current;
current = current.left;
} else if (comparison > 0) {
parent = current;
current = current.right;
} else {

4. int oldValue = current.element.value;
current.element.value = value;
return oldValue;
}
}
OrderedKeyValue newElement = new OrderedKeyValue(key, value);
if (parent == null) {
root = new BinaryNode<>(newElement);
} else {
if (comparison < 0) {
parent.left = new BinaryNode<>(newElement);
} else {
parent.right = new BinaryNode<>(newElement);
}
}
return 0;
}
public String[] keySet() {
List keys = new ArrayList<>();
inOrderTraversal(root, keys);
return keys.toArray(new String[0]);
}
private void inOrderTraversal(BinaryNode node, List keys) {
if (node == null) {
return;
}
inOrderTraversal(node.left, keys);
keys.add(node.element.key);
inOrderTraversal(node.right, keys);
}
}