CSC264 · TU past paper

Operating Systems 2076 question paper

The complete TU 2076 exam paper for Operating Systems (CSC264), all 12 questions with solved model answers written to the mark scheme.

Past Papers2081 2080.1 2080 2079 2078 2076

Tap a question to open its answer.

110 marksNumericalProcess SchedulingHideAnswer
Interactive System Goals and Scheduling Algorithms
Interactive systems serve users who expect fast, predictable responses. Scheduling goals: - Response Time: Minimize time from request to first response. - Proportionality: Simple tasks should feel quick; user expectations must be met. - ...
210 marksPage Replacement AlgorithmsHideAnswer
How Second Chance page replacement algorithm differs from FIFO page replacement policy? Discuss the concept of Belady’s anomaly with suitable example.[10]
--- In FIFO (First-In-First-Out) page replacement, the page that has been in memory the longest is replaced first, regardless of how frequently or recently it has been used. Key characteristics of FIFO: - Pages are maintained in a queue;...
310 marksNumericalDisk SchedulingHideAnswer
What is the main objective of disk scheduling algorithms? Why SSTF is not practically feasible? Assume that we have disk with 100 tracks and currently head is at track number 35. What will be the seek time for the algorithms SCAN and LOOK for processing IO requests queue: 52, 67, 21, 11, 43, 85, 18, 75, 92, 8?[10]
Disk Scheduling: Objectives, SSTF Feasibility, SCAN and LOOK

1. Main Objective of Disk Scheduling Algorithms

The primary objective of disk scheduling is to minimize the total seek time (the time the disk arm spends moving between tracks), because seek time dominates disk access latency and the disk is one of the slowest components of a system.

Related objectives:
- Maximize throughput (requests served per unit time)
- Minimize disk arm movement by ordering requests intelligently
- Ensure fairness so no request starves
- Minimize average response and waiting time
2. Why SSTF Is Not Practically Feasible

SSTF (Shortest Seek Time First) always services the request nearest to the current head position.
- Starvation: Requests far from the head may wait indefinitely if closer requests keep arriving. This is the main reason it is not practically feasible.
- Unfairness: Requests to the innermost/outermost tracks are consistently delayed.
- Not globally optimal: Minimizing each local step does not minimize total head movement.
- Unpredictable response time: Hard to guarantee bounded waiting.
Because of starvation and unfairness, SSTF is attractive theoretically but unsuitable in practice.

3. Numerical Solution

Given data:
- Total tracks: 100 (tracks $0$ to $99$ )
- Current head position: 35
- Request queue: 52, 67, 21, 11, 43, 85, 18, 75, 92, 8
- Assumed initial direction: toward higher track numbers (standard assumption)
Sorted requests: 8, 11, 18, 21, 43, 52, 67, 75, 85, 92

SCAN (Elevator)

Head moves up, servicing all requests, reaches end track 99, then reverses to service lower requests.

Order: $35 \to 43 \to 52 \to 67 \to 75 \to 85 \to 92 \to 99 \to 21 \to 18 \to 11 \to 8$

From To Distance
35 43 8
43 52 9
52 67 15
67 75 8
75 85 10
85 92 7
92 99 7
99 21 78
21 18 3
18 11 7
11 8 3

Alternatively, total movement = (up: $99-35=64$ ) + (down: $99-8=91$ ) = $155$ .

$\text{Total Seek (SCAN)} = 8+9+15+8+10+7+7+78+3+7+3 = \mathbf{155 \text{ tracks}}$

LOOK

Same as SCAN but the head only goes as far as the last request (92) in the upward direction, not to track 99.

Order: $35 \to 43 \to 52 \to 67 \to 75 \to 85 \to 92 \to 21 \to 18 \to 11 \to 8$

From To Distance
35 43 8
43 52 9
52 67 15
67 75 8
75 85 10
85 92 7
92 21 71
21 18 3
18 11 7
11 8 3

Alternatively, total movement = (up: $92-35=57$ ) + (down: $92-8=84$ ) = $141$ .

$\text{Total Seek (LOOK)} = 8+9+15+8+10+7+71+3+7+3 = \mathbf{141 \text{ tracks}}$

Summary

Algorithm Total Seek Time
SCAN 155 tracks
LOOK 141 tracks

(If seek time per track is $t$ ms, multiply by $t$ : SCAN $=155t$ , LOOK $=141t$ .)
Disk Scheduling: Objectives, SSTF Feasibility, SCAN and LOOK

1. Main Objective of Disk Scheduling Algorithms

The primary objective of disk scheduling is to minimize the total seek time (the time the disk arm spends moving between tracks), because seek time dominates disk access latency and the disk is one of the slowest components of a system.

Related objectives:
- Maximize throughput (requests served per unit time)
- Minimize disk arm movement by ordering requests intelligently
- Ensure fairness so no request starves
- Minimize average response and waiting time
2. Why SSTF Is Not Practically Feasible

SSTF (Shortest Seek Time First) always services the request nearest to the current head position.
- Starvation: Requests far from the head may wait indefinitely if closer requests keep arriving. This is the main reason it is not practically feasible.
- Unfairness: Requests to the innermost/outermost tracks are consistently delayed.
- Not globally optimal: Minimizing each local step does not minimize total head movement.
- Unpredictable response time: Hard to guarantee bounded waiting.
Because of starvation and unfairness, SSTF is attractive theoretically but unsuitable in practice.

3. Numerical Solution

Given data:
- Total tracks: 100 (tracks $0$ to $99$)
- Current head position: 35
- Request queue: 52, 67, 21, 11, 43, 85, 18, 75, 92, 8
- Assumed initial direction: toward higher track numbers (standard assumption)
Sorted requests: 8, 11, 18, 21, 43, 52, 67, 75, 85, 92

SCAN (Elevator)

Head moves up, servicing all requests, reaches end track 99, then reverses to service lower requests.

Order: $35 \to 43 \to 52 \to 67 \to 75 \to 85 \to 92 \to 99 \to 21 \to 18 \to 11 \to 8$

From To Distance
35 43 8
43 52 9
52 67 15
67 75 8
75 85 10
85 92 7
92 99 7
99 21 78
21 18 3
18 11 7
11 8 3

Alternatively, total movement = (up: $99-35=64$) + (down: $99-8=91$) = $155$.

$$\text{Total Seek (SCAN)} = 8+9+15+8+10+7+7+78+3+7+3 = \mathbf{155 \text{ tracks}}$$

LOOK

Same as SCAN but the head only goes as far as the last request (92) in the upward direction, not to track 99.

Order: $35 \to 43 \to 52 \to 67 \to 75 \to 85 \to 92 \to 21 \to 18 \to 11 \to 8$

From To Distance
35 43 8
43 52 9
52 67 15
67 75 8
75 85 10
85 92 7
92 21 71
21 18 3
18 11 7
11 8 3

Alternatively, total movement = (up: $92-35=57$) + (down: $92-8=84$) = $141$.

$$\text{Total Seek (LOOK)} = 8+9+15+8+10+7+71+3+7+3 = \mathbf{141 \text{ tracks}}$$

Summary

Algorithm Total Seek Time
SCAN 155 tracks
LOOK 141 tracks

(If seek time per track is $t$ ms, multiply by $t$: SCAN $=155t$, LOOK $=141t$.)

From	To	Distance
35	43	8
43	52	9
52	67	15
67	75	8
75	85	10
85	92	7
92	99	7
99	21	78
21	18	3
18	11	7
11	8	3

From	To	Distance
35	43	8
43	52	9
52	67	15
67	75	8
75	85	10
85	92	7
92	21	71
21	18	3
18	11	7
11	8	3

Algorithm	Total Seek Time
SCAN	155 tracks
LOOK	141 tracks

From	To	Distance
35	43	8
43	52	9
52	67	15
67	75	8
75	85	10
85	92	7
92	99	7
99	21	78
21	18	3
18	11	7
11	8	3

From	To	Distance
35	43	8
43	52	9
52	67	15
67	75	8
75	85	10
85	92	7
92	21	71
21	18	3
18	11	7
11	8	3

Algorithm	Total Seek Time
SCAN	155 tracks
LOOK	141 tracks

45 marksOperating System StructuresAnswer

What are two modes of OS? Discuss different OS structures briefly. [5]

Two Modes of OS and OS Structures

Two Modes of OS (Kernel Modes)

Modern operating systems operate in two distinct processor modes to protect the system from faulty or malicious programs:

1. User Mode

In this mode, the executing program does not have direct access to hardware or memory.
Application programs (user programs) run in user mode.
If a user program needs to access hardware or perform privileged operations, it must request the OS through a system call.
Prevents user programs from accidentally or intentionally damaging the system.

2. Kernel Mode (Supervisor/Privileged Mode)

In this mode, the OS (kernel) has full access to all hardware and can execute any CPU instruction.
The kernel runs in this mode to perform critical tasks like memory management, process scheduling, and I/O operations.
Switching from user mode to kernel mode happens via system calls or interrupts.

The dual-mode operation protects the OS from user programs and protects users from each other.

OS Structures

Different approaches have been used to organize and structure an operating system:

1. Monolithic Structure (Simple / No Structure)

The entire OS runs as a single large program in kernel mode.
All OS services (file system, memory management, device drivers, etc.) are packed together.
Advantage: Fast and efficient due to no overhead of communication between components.
Disadvantage: Difficult to maintain, debug, and extend. A bug in one part can crash the entire system.
Example: Early UNIX, MS-DOS.

2. Layered Structure

The OS is divided into a number of layers (levels), each built on top of the lower one.
The bottom layer (Layer 0) is the hardware; the topmost layer is the user interface.
Each layer only uses functions and services of the layer below it.
Advantage: Easy to debug and maintain (one layer at a time).
Disadvantage: Defining the layers properly is difficult; performance overhead due to multiple layer traversals.
Example: THE operating system by Dijkstra.

3. Microkernel Structure

Only the most essential functions (like inter-process communication, basic scheduling, and memory management) are kept in the kernel.
All other services (file system, device drivers, etc.) run in user space as separate processes.
Advantage: More reliable and secure; a failure in one service does not crash the kernel.
Disadvantage: Performance overhead due to frequent message passing between user-space services and the kernel.
Example: MINIX, QNX.

4. Virtual Machine Structure

The OS creates an illusion of multiple independent machines on a single physical machine.
Each virtual machine gets its own (virtual) copy of the hardware.
Advantage: Multiple different OS environments can run simultaneously on the same hardware.
Disadvantage: Complex to implement; some overhead in virtualization.
Example: VMware, IBM VM/370.

5. Client-Server Structure (Modular Structure)

A variation of the microkernel approach where OS services are divided into server processes.
Client processes request services from server processes through message passing.
Advantage: Highly modular and flexible; suitable for distributed systems.
Disadvantage: Communication overhead between client and server processes.

Summary Table:

Structure	Key Idea	Example
Monolithic	Single large kernel	MS-DOS, UNIX
Layered	Hierarchical layers	THE OS
Microkernel	Minimal kernel + user-space services	MINIX, QNX
Virtual Machine	Multiple virtual OS instances	VMware
Client-Server	Services as separate server processes	Distributed OS

55 marksUser and Kernel Space ThreadsHideAnswer
When threads are better than processes? Explain the concept of user level threads in detail. [5]
--- Threads are preferred over processes in the following situations: Situation Reason ------ Resource sharing All threads share the same set of open files, child processes, and memory. Processes require separate resources. Speed and lig...
65 marksNumericalMonoprogramming vs. Multi-programmingHideAnswer
Differentiate between multi programming and Monoprogramming. What will be the CPU utilization with 6 processes with 60% IO waiting time are in memory? [5]
Number of processes: $n = 6$ - I/O wait fraction: $p = 60\% = 0.60$ --- Feature Monoprogramming Multiprogramming --------- Definition Only one program resides and executes in memory at a time Several programs reside in memory simultane...

Number of processes: $n = 6$ - I/O wait fraction: $p = 60% = 0.60$ --- Feature Monoprogramming Multiprogramming --------- Definition Only one program resides and executes in memory at a time Several programs reside in memory simultane...
75 marksFree Space ManagementHideAnswer
How can you manage free disk space? Explain the linked list approach of managing free disk space with example. [5]
The operating system maintains a free space list to keep track of disk blocks that are not allocated to any file or directory. This list is used to allocate space when new files are created and to reclaim space when files are deleted. Th...
85 marksHandling IOHideAnswer
When programmed IO is suitable than other IO handling techniques? Explain the process of IO handling using DMA. [5]
--- Programmed I/O is the simplest form of I/O where the CPU does all the work. It directly controls the I/O operation including sensing device status, sending read/write commands, and transferring data. Programmed I/O is suitable in the...
95 marksHandling DeadlocksHideAnswer
Differentiate between deadlock and starvation? Discuss the process of detecting deadlocks when there are multiple resources of each type. [5]
--- Aspect Deadlock Starvation --------- Definition A situation where a set of processes are permanently blocked, each holding a resource and waiting for a resource held by another process in the set A situation where a process waits ind...
105 marksMonitorsHideAnswer
What is problem associated with semaphores? Explain the concept of monitors in brief. [5]
Semaphores are a widely used synchronization mechanism, but they suffer from several serious problems: 1. Incorrect Use / Programming Errors: Semaphores require programmers to call wait() (down) and signal() (up) in the correct order. A ...
115 marksRelocation and ProtectionHideAnswer
Why program relocation and protection is important? Explain the technique of achieving program relocation and protection. [5]
Program Relocation is important because: - In a multiprogramming environment, multiple processes must reside in memory simultaneously. A program cannot always be loaded at the same fixed memory address, so it must be relocated to whateve...
125 marksResource - Allocation GraphHideAnswer
Write short notes on: a. Linux File System b. Resource Allocation Graph [5]
--- The Linux File System refers to how a Linux-based computer organizes, stores, and manages system files. It is basically a combination of directories (folders) that serve as placeholders for addresses of other files. - In Linux, there...

Keep going

Important questions + predictions

Full syllabus Flashcards and practice

Other papers

2081 paper 2080.1 paper 2080 paper 2079 paper 2078 paper

Operating Systems 2076 question paper

Disk Scheduling: Objectives, SSTF Feasibility, SCAN and LOOK

1. Main Objective of Disk Scheduling Algorithms

2. Why SSTF Is Not Practically Feasible

3. Numerical Solution

SCAN (Elevator)

LOOK

Summary

Disk Scheduling: Objectives, SSTF Feasibility, SCAN and LOOK

1. Main Objective of Disk Scheduling Algorithms

2. Why SSTF Is Not Practically Feasible

3. Numerical Solution

SCAN (Elevator)

LOOK

Summary

Two Modes of OS and OS Structures

Two Modes of OS (Kernel Modes)

1. User Mode

2. Kernel Mode (Supervisor/Privileged Mode)

OS Structures

1. Monolithic Structure (Simple / No Structure)

2. Layered Structure

3. Microkernel Structure

4. Virtual Machine Structure

5. Client-Server Structure (Modular Structure)

Disk Scheduling: Objectives, SSTF Feasibility, SCAN and LOOK

1. Main Objective of Disk Scheduling Algorithms

2. Why SSTF Is Not Practically Feasible

3. Numerical Solution

SCAN (Elevator)

LOOK

Summary