Optimal Storage on Tapes

Given n programs stored on a computer tape and length of each program i is L_i where 1<=i<=n, find the order in which the programs should be stored in the tape for which the Mean Retrieval Time (MRT given as   \frac{1}{n} \sum_{i=1}^{n} \sum_{j=1}^{i} L_j) is minimized.

Example:

Input : n = 3
        L[] = { 5, 3, 10 }
Output : Order should be { 3, 5, 10 } with MRT = 29/3

Prerequisites: Magnetic Tapes Data Storage



Let us first break down the problem and understand what needs to be done.

A magnetic tape provides only sequential access of data. In an audio tape/cassette, unlike a CD, a fifth song from the tape can’t be just directly played. The length of the first four songs must be traversed to play the fifth song. So in order to access certain data, head of the tape should be positioned accordingly.

Now suppose there are 4 songs in a tape of audio lengths 5, 7, 3 and 2 mins respectively. In order to play the fourth song, we need to traverse an audio length of 5 + 7 + 3 = 15 mins and then position the tape head.
Retrieval time of the data is the time taken to retrieve/access that data in its entirety. Hence retrieval time of the fourth song is 15 + 2 = 17 mins.

Now, considering that all programs in a magnetic tape are retrieved equally often and the tape head points to the front of the tape every time, a new term can be defined called the Mean Retrieval Time (MRT).

Let’s suppose that the retrieval time of program i is T_i. Therefore, T_i = \sum_{j=1}^{i} L_j
MRT is the average of all such T_i. Therefore MRT = \frac{1}{n} \sum_{i=1}^{n} T_i , or MRT = \frac{1}{n} \sum_{i=1}^{n} \sum_{j=1}^{i} L_j

The sequential access of data in a tape has some limitations. Order must be defined in which the data/programs in a tape are stored so that least MRT can be obtained. Hence the order of storing becomes very important to reduce the data retrieval/access time.
Thus, the task gets reduced – to define the correct order and hence minimize the MRT, i.e. to minimize the term \sum_{i=1}^{n} \sum_{j=1}^{i} L_i

For e.g. Suppose there are 3 programs of lengths 2, 5 and 4 respectively. So there are total 3! = 6 possible orders of storage.

Order Total Retrieval Time Mean Retrieval Time
1 1 2 3 2 + (2 + 5) + (2 + 5 + 4) = 20 20/3
2 1 3 2 2 + (2 + 4) + (2 + 4 + 5) = 19 19/3
3 2 1 3 5 + (5 + 2) + (5 + 2 + 4) = 23 23/3
4 2 3 1 5 + (5 + 4) + (5 + 4 + 2) = 25 25/3
5 3 1 2 4 + (4 + 2) + (4 + 2 + 5) = 21 21/3
6 3 2 1 4 + (4 + 5) + (4 + 5 + 2) = 24 24/3

It’s clear that by following the second order in storing the programs, the mean retrieval time is least.

In above example, the first program’s length is added ‘n’ times, the second ‘n-1’ times…and so on till the last program is added only once. So, careful analysis suggests that in order to minimize the MRT, programs having greater lengths should be put towards the end so that the summation is reduced. Or, the lengths of the programs should be sorted in increasing order. That’s the Greedy Algorithm in use – at each step we make the immediate choice of putting the program having the least time first, in order to build up the ultimate optimized solution to the problem piece by piece.

Below is the implementation:

C++

// CPP Program to find the order
// of programs for which MRT is
// minimized
#include <bits/stdc++.h>

using namespace std;

// This functions outputs the required
// order and Minimum Retrieval Time
void findOrderMRT(int L[], int n)
{
    // Here length of i'th program is L[i]
    sort(L, L + n);

    // Lengths of programs sorted according to increasing
    // lengths. This is the order in which the programs
    // have to be stored on tape for minimum MRT
    cout << "Optimal order in which programs are to be"
            "stored is: ";
    for (int i = 0; i < n; i++)
        cout << L[i] << " ";
    cout << endl;

    // MRT - Minimum Retrieval Time
    double MRT = 0;
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = 0; j <= i; j++)
            sum += L[j];
        MRT += sum;
    }
    MRT /= n;
    cout << "Minimum Retrieval Time of this"
           " order is " << MRT;
}

// Driver Code to test above function
int main()
{
    int L[] = { 2, 5, 4 };
    int n = sizeof(L) / sizeof(L[0]);
    findOrderMRT(L, n);
    return 0;
}

Java

// Java Program to find the order
// of programs for which MRT is
// minimized
import java.io.*;
import java .util.*;

class GFG 
{

// This functions outputs 
// the required order and 
// Minimum Retrieval Time
static void findOrderMRT(int []L, 
                         int n)
{
    // Here length of 
    // i'th program is L[i]
    Arrays.sort(L);

    // Lengths of programs sorted 
    // according to increasing lengths.
    // This is the order in which 
    // the programs have to be stored
    // on tape for minimum MRT
    System.out.print("Optimal order in which " + 
              "programs are to be stored is: ");
    for (int i = 0; i < n; i++)
        System.out.print(L[i] + " ");
        System.out.println();

    // MRT - Minimum Retrieval Time
    double MRT = 0;
    for (int i = 0; i < n; i++) 
    {
        int sum = 0;
        for (int j = 0; j <= i; j++)
            sum += L[j];
        MRT += sum;
    }
    MRT /= n;
    System.out.print( "Minimum Retrieval Time" + 
                    " of this order is " + MRT);
}

// Driver Code
public static void main (String[] args) 
{
    int []L = { 2, 5, 4 };
    int n = L.length;
    findOrderMRT(L, n);
}
}

// This code is contributed
// by anuj_67.

C#

// C# Program to find the 
// order of programs for 
// which MRT is minimized
using System;

class GFG 
{

// This functions outputs 
// the required order and 
// Minimum Retrieval Time
static void findOrderMRT(int []L, 
                         int n)
{
    // Here length of 
    // i'th program is L[i]
    Array.Sort(L);

    // Lengths of programs sorted 
    // according to increasing lengths.
    // This is the order in which 
    // the programs have to be stored
    // on tape for minimum MRT
    Console.Write("Optimal order in " +   
                  "which programs are" + 
                  " to be stored is: ");
    for (int i = 0; i < n; i++)
        Console.Write(L[i] + " ");
        Console.WriteLine();

    // MRT - Minimum Retrieval Time
    double MRT = 0;
    for (int i = 0; i < n; i++) 
    {
        int sum = 0;
        for (int j = 0; j <= i; j++)
            sum += L[j];
        MRT += sum;
    }
    MRT /= n;
    Console.WriteLine("Minimum Retrieval " + 
                  "Time of this order is " + 
                                       MRT);
}

// Driver Code
public static void Main () 
{
    int []L = { 2, 5, 4 };
    int n = L.Length;
    findOrderMRT(L, n);
}
}

// This code is contributed
// by anuj_67.

Output:

Optimal order in which programs are to be stored are: 2 4 5 
Minimum Retrieval Time of this order is 6.33333

Time complexity of the above program is the time complexity for sorting, that is O(n^2) (Since std::sort() operates in O(n^2))
But the time complexity can be reduced to O(n * logn) by avoiding two loops and can be done in this manner.

for (int i = 0; i < n; i++)
    MRT += (n - i) * L[i];


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