Operating System
AKTU 2023-24 scheme (B.Tech 2nd Year, AICTE model curriculum) · 100 marks: 70 end-sem theory (3 hrs; Section A: 7 short questions x 2 = 14; Section B: any 3 of 5 x 7 = 21; Section C: 5 questions, one per unit, attempt 1 of 2 parts each, x 7 = 35) + 30 internal (CT 20 + TA 10). 4 credits (3-1-0). Lab BCS451 Operating System Lab (0-0-2, 1 credit) is marked separately out of 100.
One-Night Pass Pack
Monolithic: the whole OS (file system, scheduler, drivers) runs as one large program in kernel mode; fast but hard to maintain (early UNIX, MS-DOS, Linux). Layered: OS split into layers 0 (hardware) to N (user interface); each layer uses only the layer below, so it is easy to debug but slower. Microkernel: only basic services (IPC, scheduling, low level memory) stay in the kernel; the rest run as user-space servers using message passing (Mach, QNX). Microkernel is more reliable and secure; monolithic is faster because there is less message passing.
Full-marks tip: Draw three block diagrams (one big kernel box, stacked layers, small kernel with servers around it) and finish with a comparison table on speed, reliability, size and example. Do not forget to give one real OS example for each.
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Unit-wise syllabus
U1Introduction
Weightage: Not officially published (Section C has one 7-mark question per unit; Sections A and B are not split equally)
U2Concurrent Processes
Weightage: Not officially published (Section C has one 7-mark question per unit; Sections A and B are not split equally)
U3CPU Scheduling and Deadlock
Weightage: Not officially published (Section C has one 7-mark question per unit; Sections A and B are not split equally)
U4Memory Management
Weightage: Not officially published (Section C has one 7-mark question per unit; Sections A and B are not split equally)
U5I/O Management, Disk Scheduling and File System
Weightage: Not officially published (Section C has one 7-mark question per unit; Sections A and B are not split equally)
Curated video lectures
Free lectures from well-known educators, matched to units. Videos load only when you tap “Watch here”, to save data.
Operating System complete playlist (all units)
KnowledgeGATE by Sanchit Sir · Playlist · 82 videos · Hindi
Exam Guide Radar
Standard textbook
Silberschatz, Galvin and Gagne, Operating System Concepts, Wiley (first text in the AKTU BCS401 list; Dhamdhere, Stallings and Deitel are also listed)
Exam guide students buy
Quantum Series: Operating System (BCS401) by Quantum Publications
Unit-wise short and long answers with solved AKTU papers. Buy the edition that matches your scheme code (BCS/BAS, not the old KCS/KAS).
Local market tip
Buy the Quantum printed for code BCS401 (2023-24 scheme onward), not the older KCS401 edition, from bookshops near your college around April before the even-semester exams; stock sells out, so check early.
Lab Viva Survival
BCS451: CPU scheduling programs (FCFS, SJF, priority, multilevel queue)
- Q. What is the difference between turnaround time and waiting time?
- → Turnaround time = completion - arrival. Waiting time = turnaround - burst, the time spent in the ready queue.
- Q. Which algorithm gives the minimum average waiting time?
- → SJF (or SRTF when preemptive) is optimal for average waiting time, but burst times must be known.
- Q. What is the convoy effect?
- → In FCFS, short processes wait behind one long CPU-bound process, which lowers CPU and device use.
- Q. What is starvation and how is it fixed in priority scheduling?
- → A low priority process may never run. Aging slowly raises the priority of processes that wait long.
BCS451: Banker's algorithm and resource allocation graph
- Q. What is a safe state?
- → A state where some order (safe sequence) exists in which every process can get its maximum need and finish.
- Q. Which matrices does Banker's algorithm use?
- → Available, Max, Allocation and Need, where Need = Max - Allocation.
- Q. How is a wait-for graph made from a RAG?
- → Remove the resource nodes and draw Pi to Pj if Pi waits for a resource held by Pj; a cycle means deadlock.
BCS451: Bounded buffer (producer-consumer) and readers-writers using semaphores
- Q. What are the initial values of the semaphores in bounded buffer?
- → mutex = 1, empty = buffer size n, full = 0.
- Q. What is the difference between a binary and a counting semaphore?
- → A binary semaphore takes only 0 or 1 (like a mutex lock); a counting semaphore can take any non-negative value to count resources.
- Q. What problem does the first readers-writers solution have?
- → Writers may starve because readers keep entering while at least one reader is inside.