SSTF Disk Scheduling Worked Example: Total Head Movement for the Given Request Queue

Verified Sources

May 27, 2026

Disk scheduling aims to reduce seek time by choosing a favorable service order among pending requests.2 In SSTF scheduling, the next request chosen is always the one with the minimum absolute distance from the current head position, which usually reduces total movement compared with FIFO service.2 However, SSTF may cause starvation for far-away requests under continuous load.

For this problem:

Disk cylinders: $0$ to $4999$
Current head position: $143$
Previous request: $125$
Pending queue in FIFO order: $86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130$

Under SSTF, FIFO order is not preserved after selection; instead, at each step we choose the pending request with the smallest distance from the current head.2

The total head movement is computed as the sum of absolute cylinder differences:

\text{Total movement} = \sum_{i=1}^{n} |H_i - H_{i-1}|

where $H_0 = 143$ and each subsequent $H_i$ is the next request selected by SSTF.

SSTF Algorithm | Disk Scheduling Algorithms - Explains SSTF as serving the request with the shortest seek time and contrasts it with FCFS. ↩ ↩² ↩³
Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩ ↩² ↩³ ↩⁴
Disk Scheduling Algorithms - GeeksforGeeks - Summarizes disk scheduling goals and highlights SSTF's starvation risk and fairness trade-offs. ↩

SSTF in Disk Scheduling with Example

Important Interpretation

The previous request at $125$ only indicates the recent direction of travel. In SSTF, the next request is still selected purely by minimum distance from the current head at $143$ .2

SSTF Algorithm | Disk Scheduling Algorithms - Explains SSTF as serving the request with the shortest seek time and contrasts it with FCFS. ↩
Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩

Applying SSTF to the Given Queue

1
Step 1
Current head position is $143$ . Pending requests are $86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130$ .
2
Step 2
Distances are: to $86$ is $57$ , to $130$ is $13$ , to $913$ is $770$ , to $948$ is $805$ , to $1022$ is $879$ , to $1470$ is $1327$ , to $1509$ is $1366$ , to $1750$ is $1607$ , and to $1774$ is $1631$ . The nearest is $130$ , so move $13$ cylinders.

Footnotes

Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩
3
Step 3
Remaining requests: $86, 1470, 913, 1774, 948, 1509, 1022, 1750$ . Distances from $130$ : to $86$ is $44$ , to $913$ is $783$ , to $948$ is $818$ , to $1022$ is $892$ , to $1470$ is $1340$ , to $1509$ is $1379$ , to $1750$ is $1620$ , to $1774$ is $1644$ . The nearest is $86$ , so move $44$ .
4
Step 4
Remaining requests: $1470, 913, 1774, 948, 1509, 1022, 1750$ . Distances from $86$ : to $913$ is $827$ , to $948$ is $862$ , to $1022$ is $936$ , to $1470$ is $1384$ , to $1509$ is $1423$ , to $1750$ is $1664$ , to $1774$ is $1688$ . The nearest is $913$ , so move $827$ .
5
Step 5
Remaining requests: $1470, 1774, 948, 1509, 1022, 1750$ . Distances from $913$ : to $948$ is $35$ , to $1022$ is $109$ , to $1470$ is $557$ , to $1509$ is $596$ , to $1750$ is $837$ , to $1774$ is $861$ . The nearest is $948$ , so move $35$ .
6
Step 6
Remaining requests: $1470, 1774, 1509, 1022, 1750$ . Distances from $948$ : to $1022$ is $74$ , to $1470$ is $522$ , to $1509$ is $561$ , to $1750$ is $802$ , to $1774$ is $826$ . The nearest is $1022$ , so move $74$ .
7
Step 7
Remaining requests: $1470, 1774, 1509, 1750$ . Distances from $1022$ : to $1470$ is $448$ , to $1509$ is $487$ , to $1750$ is $728$ , to $1774$ is $752$ . The nearest is $1470$ , so move $448$ .
8
Step 8
Remaining requests: $1774, 1509, 1750$ . Distances from $1470$ : to $1509$ is $39$ , to $1750$ is $280$ , to $1774$ is $304$ . The nearest is $1509$ , so move $39$ .
9
Step 9
Remaining requests: $1774, 1750$ . Distances from $1509$ : to $1750$ is $241$ , to $1774$ is $265$ . The nearest is $1750$ , so move $241$ .
10
Step 10
Only $1774$ remains. Move from $1750$ to $1774$ , which is $24$ cylinders.
11
Step 11
Total movement $= 13 + 44 + 827 + 35 + 74 + 448 + 39 + 241 + 24 = 1745$ cylinders.

The SSTF service sequence is therefore:

143 \rightarrow 130 \rightarrow 86 \rightarrow 913 \rightarrow 948 \rightarrow 1022 \rightarrow 1470 \rightarrow 1509 \rightarrow 1750 \rightarrow 1774

The corresponding head movements are:

From	To	Distance
$143$	$130$	$13$
$130$	$86$	$44$
$86$	$913$	$827$
$913$	$948$	$35$
$948$	$1022$	$74$
$1022$	$1470$	$448$
$1470$	$1509$	$39$
$1509$	$1750$	$241$
$1750$	$1774$	$24$

Thus,

13 + 44 + 827 + 35 + 74 + 448 + 39 + 241 + 24 = 1745

So, the disk arm moves a total of $1745$ cylinders to satisfy all pending requests using SSTF.2

SSTF Algorithm | Disk Scheduling Algorithms - Explains SSTF as serving the request with the shortest seek time and contrasts it with FCFS. ↩
Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩

Head Movement Per SSTF Step

Cylinder distance traveled at each selection step

SSTF Service Order for This Problem

Move to 130

Step 1

Nearest request to $143$ is $130$ , requiring $13$ cylinders of movement."

Move to 86

Step 2

Nearest request to $130$ is $86$ , requiring $44$ cylinders."

Move to 913

Step 3

After clearing nearby low cylinders, the nearest remaining request from $86$ is $913$ , requiring $827$ cylinders."

Move to 948

Step 4

Nearest request from $913$ is $948$ , requiring $35$ cylinders."

Move to 1022

Step 5

Nearest request from $948$ is $1022$ , requiring $74$ cylinders."

Move to 1470

Step 6

Nearest request from $1022$ is $1470$ , requiring $448$ cylinders."

Move to 1509

Step 7

Nearest request from $1470$ is $1509$ , requiring $39$ cylinders."

Move to 1750

Step 8

Nearest request from $1509$ is $1750$ , requiring $241$ cylinders."

Move to 1774

Step 9

Final remaining request is $1774$ , requiring $24$ cylinders."

Exam Strategy

In SSTF problems, do not follow the FIFO queue order. Recompute absolute distances from the current head after every move, because the nearest request changes dynamically.2

SSTF Algorithm | Disk Scheduling Algorithms - Explains SSTF as serving the request with the shortest seek time and contrasts it with FCFS. ↩
Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩

Common Mistake

Do not force the head to keep moving in the same direction just because the previous request was at $125$ . Direction matters in SCAN-like algorithms, but SSTF chooses only the minimum current seek distance.2

SSTF Algorithm | Disk Scheduling Algorithms - Explains SSTF as serving the request with the shortest seek time and contrasts it with FCFS. ↩
Disk Scheduling Algorithms - GeeksforGeeks - Summarizes disk scheduling goals and highlights SSTF's starvation risk and fairness trade-offs. ↩

Concept Checks and Edge Cases

Starting sequence: $143 \rightarrow 130 \rightarrow 86 \rightarrow 913 \rightarrow 948 \rightarrow 1022 \rightarrow 1470 \rightarrow 1509 \rightarrow 1750 \rightarrow 1774$

Total movement:

|143-130|+|130-86|+|86-913|+|913-948|+|948-1022|+|1022-1470|+|1470-1509|+|1509-1750|+|1750-1774|=1745

A useful interpretation is that SSTF greedily minimizes the next move, not necessarily the globally optimal total path.2 In this instance, it first clears the two nearby low-cylinder requests ( $130$ and $86$ ), then jumps to the closest remaining higher-cylinder cluster and continues locally from there. This local-choice behavior is characteristic of a greedy algorithm and explains both SSTF's efficiency and its fairness limitations.2

Therefore, the correct result for the given queue is:

\boxed{1745 \text{ cylinders}}

Program for SSTF Disk Scheduling Algorithm - GeeksforGeeks - Provides the SSTF procedure, formula style examples, and notes a straightforward $O(n^2)$ simulation approach. ↩ ↩²
Disk Scheduling Algorithms - GeeksforGeeks - Summarizes disk scheduling goals and highlights SSTF's starvation risk and fairness trade-offs. ↩ ↩²

Knowledge Check

Question 1 of 4

Q1Single choice

In SSTF disk scheduling, what criterion determines the next request to be served?

The request with the minimum absolute distance from the current head

The earliest request in FIFO order

The request in the current direction of motion

The largest cylinder number

Explore Related Topics

Graph Traversals: Breadth-First Search (BFS) vs. Depth-First Search (DFS)

This content contrasts Breadth‑First Search (BFS) and Depth‑First Search (DFS), outlining their traversal order, complexity, and typical use cases.

BFS uses a FIFO queue, visits nodes level by level (A→B→C→D→E→F); DFS uses a LIFO stack, dives deep (A→B→D→E→C→F).
Both run in $O(V+E)$ time; BFS may need $O(V)$ (or $O(b^d)$ ) space, while DFS typically uses $O(d)$ stack depth.
BFS guarantees the shortest path in unweighted graphs, suited for routing, web crawling, and level‑order serialization.
DFS excels in memory‑limited, wide graphs and in tasks like topological sort and cycle detection, but deep recursion can cause stack overflow.

The Banker's Algorithm: Deadlock Avoidance in Operating Systems

The Banker's Algorithm is a deadlock‑avoidance method that keeps a system in a safe state by checking each resource request against the maximum declared needs of processes.

Maintains Available, Max, Allocation, and Need matrices, where $\text{Need}[i][j]=\text{Max}[i][j]-\text{Allocation}[i][j]$ .
The Safety Algorithm uses vectors Work and Finish to find an execution order; if all processes finish, the state is safe.
The Resource‑Request Algorithm simulates allocation, runs the safety check, and commits only if the resulting state remains safe.
Time complexity of the safety check is $O(m \times n^{2})$ .
In practice the algorithm is rarely used because processes must predeclare maximum needs and the algorithm’s overhead is high.

Differentiating Rotating Storage Media: Constant Linear Velocity (CLV) vs. Constant Angular Velocity (CAV)

Rotating storage media use either Constant Angular Velocity (CAV) or Constant Linear Velocity (CLV) to control the relationship between angular speed  $\omega$ and linear speed  $v$ on the disk.

CAV: Fixed  $\omega$ (e.g., 7200 RPM), $v$ rises with radius, sectors per track stay constant → lower outer‑track density, constant transfer rate, minimal seek latency.
CLV: $\omega$ varies as $\omega(r)=v/r$ to keep $v$ constant, giving uniform sector size, higher outer‑track capacity, but slower seeks due to motor speed changes.
Zone Bit Recording (ZBR): Hybrid CAV that keeps $\omega$ constant while dividing the platter into zones with increasing sectors per track, boosting capacity and outer‑track throughput.
Mechanical limits: Very high‑speed CLV would require inner‑edge RPM > 10 000, causing vibration and disc failure, prompting a shift to CAV or hybrid modes.
Key formulas: $v=\omega r$ and $\omega(r)=\dfrac{v}{r}$ govern the trade‑offs between data density, transfer rate, and seek time.

Research more with Coursify

SSTF Disk Scheduling Worked Example: Total Head Movement for the Given Request Queue

AI Summary

Footnotes

SSTF in Disk Scheduling with Example

Important Interpretation

Footnotes

Applying SSTF to the Given Queue

Footnotes

Footnotes

Head Movement Per SSTF Step

SSTF Service Order for This Problem

Move to 130

Move to 86

Move to 913

Move to 948

Move to 1022

Move to 1470

Move to 1509

Move to 1750

Move to 1774

Exam Strategy

Footnotes

Common Mistake

Footnotes

Concept Checks and Edge Cases

Footnotes

Knowledge Check

Explore Related Topics