Given an array of disk track numbers and initial head position, our task is to find the total number of seek operations done to access all the requested tracks if Shortest Seek Time First (SSTF) is a disk scheduling algorithm is used.
The basic idea is the tracks that are closer to the current disk head position should be serviced first in order to minimize the seek operations is basically known as Shortest Seek Time First (SSTF).
Advantages of Shortest Seek Time First (SSTF)
- Better performance than the FCFS scheduling algorithm.
- It provides better throughput.
- This algorithm is used in Batch Processing systems where throughput is more important.
- It has a less average response and waiting time.
Disadvantages of Shortest Seek Time First (SSTF)
- Starvation is possible for some requests as it favours easy-to-reach requests and ignores the far-away processes.
- There is a lack of predictability because of the high variance of response time.
- Switching direction slows things down.
Algorithm
Step 1: Let the Request array represents an array storing indexes of tracks that have been requested. ‘head’ is the position of the disk head.
Step 2: Find the positive distance of all tracks in the request array from the head.
Step 3: Find a track from the requested array which has not been accessed/serviced yet and has a minimum distance from the head.
Step 4: Increment the total seek count with this distance.
Step 5: Currently serviced track position now becomes the new head position.
Step 6: Go to step 2 until all tracks in the request array have not been serviced.
Example:
Request sequence = {176, 79, 34, 60, 92, 11, 41, 114}
Initial head position = 50
The following chart shows the sequence in which requested tracks are serviced using SSTF.
Therefore, the total seek count is calculated as:

SSTF (Shortest Seek Time First)
= (50-41)+(41-34)+(34-11)+(60-11)+(79-60)+(92-79)+(114-92)+(176-114)
= 204
which can also be directly calculated as: (50-11) + (176-11)
Implementation
The implementation of SSTF is given below. Note that we have made a node class having 2 members. ‘distance’ is used to store the distance between the head and the track position. ‘accessed’ is a boolean variable that tells whether the track has been accessed/serviced before by the disk head or not.
C++
#include <bits/stdc++.h>
using namespace std;
void calculatedifference( int request[], int head,
int diff[][2], int n)
{
for ( int i = 0; i < n; i++)
{
diff[i][0] = abs (head - request[i]);
}
}
int findMIN( int diff[][2], int n)
{
int index = -1;
int minimum = 1e9;
for ( int i = 0; i < n; i++)
{
if (!diff[i][1] && minimum > diff[i][0])
{
minimum = diff[i][0];
index = i;
}
}
return index;
}
void shortestSeekTimeFirst( int request[],
int head, int n)
{
if (n == 0)
{
return ;
}
int diff[n][2] = { { 0, 0 } };
int seekcount = 0;
int seeksequence[n + 1] = {0};
for ( int i = 0; i < n; i++)
{
seeksequence[i] = head;
calculatedifference(request, head, diff, n);
int index = findMIN(diff, n);
diff[index][1] = 1;
seekcount += diff[index][0];
head = request[index];
}
seeksequence[n] = head;
cout << "Total number of seek operations = "
<< seekcount << endl;
cout << "Seek sequence is : " << "\n" ;
for ( int i = 0; i <= n; i++)
{
cout << seeksequence[i] << "\n" ;
}
}
int main()
{
int n = 8;
int proc[n] = { 176, 79, 34, 60, 92, 11, 41, 114 };
shortestSeekTimeFirst(proc, 50, n);
return 0;
}
|
Java
class node {
int distance = 0 ;
boolean accessed = false ;
}
public class SSTF {
public static void calculateDifference( int queue[],
int head, node diff[])
{
for ( int i = 0 ; i < diff.length; i++)
diff[i].distance = Math.abs(queue[i] - head);
}
public static int findMin(node diff[])
{
int index = - 1 , minimum = Integer.MAX_VALUE;
for ( int i = 0 ; i < diff.length; i++) {
if (!diff[i].accessed && minimum > diff[i].distance) {
minimum = diff[i].distance;
index = i;
}
}
return index;
}
public static void shortestSeekTimeFirst( int request[],
int head)
{
if (request.length == 0 )
return ;
node diff[] = new node[request.length];
for ( int i = 0 ; i < diff.length; i++)
diff[i] = new node();
int seek_count = 0 ;
int [] seek_sequence = new int [request.length + 1 ];
for ( int i = 0 ; i < request.length; i++) {
seek_sequence[i] = head;
calculateDifference(request, head, diff);
int index = findMin(diff);
diff[index].accessed = true ;
seek_count += diff[index].distance;
head = request[index];
}
seek_sequence[seek_sequence.length - 1 ] = head;
System.out.println( "Total number of seek operations = "
+ seek_count);
System.out.println( "Seek Sequence is" );
for ( int i = 0 ; i < seek_sequence.length; i++)
System.out.println(seek_sequence[i]);
}
public static void main(String[] args)
{
int arr[] = { 176 , 79 , 34 , 60 , 92 , 11 , 41 , 114 };
shortestSeekTimeFirst(arr, 50 );
}
}
|
Python3
def calculateDifference(queue, head, diff):
for i in range ( len (diff)):
diff[i][ 0 ] = abs (queue[i] - head)
def findMin(diff):
index = - 1
minimum = 999999999
for i in range ( len (diff)):
if ( not diff[i][ 1 ] and
minimum > diff[i][ 0 ]):
minimum = diff[i][ 0 ]
index = i
return index
def shortestSeekTimeFirst(request, head):
if ( len (request) = = 0 ):
return
l = len (request)
diff = [ 0 ] * l
for i in range (l):
diff[i] = [ 0 , 0 ]
seek_count = 0
seek_sequence = [ 0 ] * (l + 1 )
for i in range (l):
seek_sequence[i] = head
calculateDifference(request, head, diff)
index = findMin(diff)
diff[index][ 1 ] = True
seek_count + = diff[index][ 0 ]
head = request[index]
seek_sequence[ len (seek_sequence) - 1 ] = head
print ( "Total number of seek operations =" ,
seek_count)
print ( "Seek Sequence is" )
for i in range (l + 1 ):
print (seek_sequence[i])
if __name__ = = "__main__" :
proc = [ 176 , 79 , 34 , 60 ,
92 , 11 , 41 , 114 ]
shortestSeekTimeFirst(proc, 50 )
|
C#
using System;
public class node
{
public int distance = 0;
public Boolean accessed = false ;
}
public class SSTF
{
public static void calculateDifference( int []queue,
int head, node []diff)
{
for ( int i = 0; i < diff.Length; i++)
diff[i].distance = Math.Abs(queue[i] - head);
}
public static int findMin(node []diff)
{
int index = -1, minimum = int .MaxValue;
for ( int i = 0; i < diff.Length; i++)
{
if (!diff[i].accessed && minimum > diff[i].distance)
{
minimum = diff[i].distance;
index = i;
}
}
return index;
}
public static void shortestSeekTimeFirst( int []request,
int head)
{
if (request.Length == 0)
return ;
node []diff = new node[request.Length];
for ( int i = 0; i < diff.Length; i++)
diff[i] = new node();
int seek_count = 0;
int [] seek_sequence = new int [request.Length + 1];
for ( int i = 0; i < request.Length; i++)
{
seek_sequence[i] = head;
calculateDifference(request, head, diff);
int index = findMin(diff);
diff[index].accessed = true ;
seek_count += diff[index].distance;
head = request[index];
}
seek_sequence[seek_sequence.Length - 1] = head;
Console.WriteLine( "Total number of seek operations = "
+ seek_count);
Console.WriteLine( "Seek Sequence is" );
for ( int i = 0; i < seek_sequence.Length; i++)
Console.WriteLine(seek_sequence[i]);
}
public static void Main(String[] args)
{
int []arr = { 176, 79, 34, 60, 92, 11, 41, 114 };
shortestSeekTimeFirst(arr, 50);
}
}
|
Javascript
function calculatedifference(request, head, diff, n) {
for (let i = 0; i < n; i++) {
diff[i][0] = Math.abs(head - request[i]);
}
}
function findMIN(diff, n) {
let index = -1;
let minimum = 1e9;
for (let i = 0; i < n; i++) {
if (!diff[i][1] && minimum > diff[i][0]) {
minimum = diff[i][0];
index = i;
}
}
return index;
}
function shortestSeekTimeFirst(request, head, n) {
if (n == 0) {
return ;
}
let diff = new Array(n);
for (let i = 0; i < n; i++) {
diff[i] = new Array(2);
}
let seekcount = 0;
let seeksequence = new Array(n + 1);
seeksequence[0] = head;
for (let i = 0; i < n; i++) {
calculatedifference(request, head, diff, n);
let index = findMIN(diff, n);
diff[index][1] = 1;
seekcount += diff[index][0];
head = request[index];
seeksequence[i + 1] = head;
}
console.log( "Total number of seek operations = " + seekcount);
console.log( "Seek sequence is : " );
for (let i = 0; i <= n; i++) {
console.log(seeksequence[i]);
}
}
let n = 8;
let proc = [176, 79, 34, 60, 92, 11, 41, 114];
shortestSeekTimeFirst(proc, 50, n);
|
Output
Total number of seek operations = 204
Seek Sequence: 50, 41, 34, 11, 60, 79, 92, 114, 176
Time Complexity: O(N^2)
Auxiliary Space: O(N)
FAQs on SSTF Disk Scheduling Algorithm
1. Will Starvation occur for some requests in SSTF?
Yes, when a continuos stream of requests are coming then it will lead to starvation.
2. Whether SSTF is suitable for minimizing disk arm movement?
Yes, SSTF is helpful in minimizing disk arm movement and it also helps in reducing seek time.
3. In case of two requests having the same seek time, which request SSTF will follow?
If multiple requests have the same seek time, SSTF can choose any of them, depending on the implementation.