Question

Barr the Bear has started a business to sell fish to Poe and his fellow penguins. The penguin customers submit many fish orders, but Barr can only process one order at a time. Suppose that Barr currently has orders from n penguin customers (label them as 1, 2, . . . , n). Customer i’s order takes ti time to complete. Barr is going to process each penguin’s order one by one, and the scheduling of orders can be described as a permutation of the customers. Let Ci denote the completion time of order i. For instance, if customer j’s order is the first to be completed, then we would have Cj = tj (assume Barr begins processing orders at time 0); if customer k’s order is the second to be completed after that, then Ck = Cj + tk = tj + tk, and so on. Each customer is of different importance to Barr’s business, and we denote this relative weight by wi (for customer i). Barr wishes to minimize the weighted sum of the completion times of n orders, Pn i=1 wi · Ci . Intuitively, the more important a customer is, the sooner Barr wishes to complete the customer’s order. Design an O(n log n) algorithm to solve this problem to help Barr. You are given a set of n orders with a processing time ti and a weight wi for each customer i (assume ti , wi are positive integers). You want to decide an ordering of the customer orders so as to minimize the weighted sum of the completion times. Prove correctness of your algorithm (i.e., optimality) and analyze its runtime.

Answer 1

Answer:

Algorithm below designed using Java

Explanation:

import java.util.SortedMap;

import java.util.TreeMap;

import java.util.Vector;

public class SchedulingProblem {

static Vector<Integer> processTime = new Vector<>();

static Vector<Integer> processWeight = new Vector<>();

static SortedMap<Integer, Double> timeWeightRatio = new TreeMap<Integer, Double>() {

};

 

static int number = 2;

public static void main(String[] args) {

processTime.add(1);

processWeight.add(10);

processTime.add(3);

processWeight.add(2);

for (int i = 0; i < number; i++) { // O(N)

timeWeightRatio.put(i, processWeight.get(i) / (processTime.get(i) * 1.0)); // O(log(n))

}

// ------------ O(nlog(n))

int sum = 0, tmpTime = 0;

while (timeWeightRatio.size() > 0) { // O(n)

tmpTime += processTime.get(timeWeightRatio.firstKey()); // O(n)

sum += processWeight.get(timeWeightRatio.firstKey()) * tmpTime; // O(n)

System.out.print(processWeight.get(timeWeightRatio.firstKey()) + " * " + tmpTime + ((timeWeightRatio.size() > 1) ? " + " : " = " + sum));

timeWeightRatio.remove(timeWeightRatio.firstKey()); // O(log(n))

}

 

// >>>> O(nlog(n))

}

}