Level 1 — RecognitionOperating Systems

Operating Systems

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Difficulty Level: 1 — Recognition (MCQ, Matching, True/False with justification) Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) — 12 marks

Choose the single best answer.

Q1. When a user program executes a system call, the CPU transitions from:

(a) kernel mode to user mode
(b) user mode to kernel mode via a trap
(c) user mode to kernel mode via an interrupt from a device
(d) kernel mode to kernel mode

Q2. Which of the following is NOT saved during a context switch?

(a) Program counter
(b) CPU registers
(c) The contents of the hard disk
(d) Stack pointer

Q3. A fork() call in a parent process returns:

(a) 0 to both parent and child
(b) the child's PID to the parent and 0 to the child
(c) 0 to the parent and the child's PID to the child
(d) −1 to both on success

Q4. In the Coffman conditions for deadlock, which is NOT one of the four?

(a) Mutual exclusion
(b) Hold and wait
(c) Preemption
(d) Circular wait

Q5. For processes with burst times, which scheduling algorithm gives the provably minimum average waiting time (non-preemptive)?

(a) FCFS
(b) Round Robin
(c) SJF
(d) Priority (non-preemptive)

Q6. Internal fragmentation occurs primarily in:

(a) contiguous allocation with variable partitions
(b) fixed-size partitioning / paging
(c) linked file allocation
(d) segmentation only

Q7. RAID level that provides striping with no redundancy:

(a) RAID 0
(b) RAID 1
(c) RAID 5
(d) RAID 6

Q8. A test-and-set instruction is useful for building a mutex because it:

(a) disables all interrupts permanently
(b) atomically reads and sets a memory word in one indivisible step
(c) is faster than any load instruction
(d) never causes busy waiting

Q9. The main difference between a container and a virtual machine is that a container:

(a) runs its own full guest kernel
(b) shares the host OS kernel using namespaces and cgroups
(c) requires a type-1 hypervisor
(d) cannot isolate processes

Q10. A page fault occurs when:

(a) the page table is full
(b) a referenced page is valid and present in a frame
(c) a referenced page is not currently in physical memory
(d) the CPU is in user mode

Q11. In C-SCAN disk scheduling, after reaching the end of the disk the head:

(a) reverses direction servicing requests
(b) jumps back to the beginning and continues in the same direction
(c) stops
(d) services the nearest request regardless of direction

Q12. Journaling in a file system primarily improves:

(a) read throughput
(b) crash consistency / recovery
(c) internal fragmentation
(d) CPU utilization

Section B — Matching (1 mark each) — 8 marks

Q13. Match each item in Column X to the correct description in Column Y.

Column X	Column Y
1. Microkernel	A. Table mapping virtual pages to physical frames
2. PCB	B. Minimal kernel; services run in user space
3. Page table	C. Multiple hard links point to one file's data
4. Inode	D. Stores per-process state, registers, PID
5. Copy-on-write	E. Structure holding file metadata + block pointers
6. Hard link	F. Parent and child share pages until one writes
7. Semaphore `P()`	G. Distributes threads across CPUs
8. Load balancing	H. Decrement/wait operation

Write answers as 1→_, 2→_, …

Section C — True/False WITH Justification (2 marks each) — 10 marks

1 mark for correct T/F, 1 mark for a correct justification. No justification = max 1 mark.

Q14. "Round Robin scheduling with a very large time quantum behaves like FCFS."

Q15. "The Banker's algorithm is a deadlock detection technique used after deadlock has occurred."

Q16. "A user-level threading library that maps many threads to one kernel thread can run those threads truly in parallel on a multicore CPU."

Q17. "Best-fit memory allocation always eliminates external fragmentation."

Q18. "In demand paging, a valid–invalid bit in the page table is used to detect that a page is not in memory."

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (b). A system call issues a software trap that switches the CPU into kernel mode so privileged code can run. (Not (c): a device interrupt is asynchronous, unrelated to a program's syscall.)

Q2 — (c). Context switch saves the volatile CPU state (PC, registers, stack pointer) into the PCB. Disk contents are persistent storage, not part of CPU context.

Q3 — (b). fork() returns the child's PID to the parent and 0 to the child, letting each branch identify itself.

Q4 — (c). The four Coffman conditions are mutual exclusion, hold-and-wait, no preemption, and circular wait. "Preemption" is the opposite of the actual condition.

Q5 — (c). Shortest Job First minimizes average waiting time among non-preemptive schedules (classic optimality result).

Q6 — (b). Fixed-size blocks/pages waste space inside the last allocated block ⇒ internal fragmentation. Variable partitions cause external fragmentation.

Q7 — (a). RAID 0 stripes for performance/capacity with zero redundancy.

Q8 — (b). Test-and-set is atomic: it reads the old value and writes 1 indivisibly, so only one thread sees the "was 0" result and enters the critical section.

Q9 — (b). Containers share the host kernel, isolating via namespaces (visibility) and cgroups (resource limits); VMs run full guest kernels on a hypervisor.

Q10 — (c). A page fault = referenced page absent from physical memory; the OS then fetches it (demand paging).

Q11 — (b). C-SCAN sweeps one direction, jumps to the start, and repeats — giving more uniform wait times than SCAN.

Q12 — (b). The journal records intended changes first, so after a crash the FS replays/rolls back to a consistent state.

Section B (1 mark each)

Q13: 1→B, 2→D, 3→A, 4→E, 5→F, 6→C, 7→H, 8→G

Microkernel = minimal kernel, services in user space (B)
PCB = per-process state/registers/PID (D)
Page table = virtual page → frame map (A)
Inode = file metadata + block pointers (E)
Copy-on-write = shared pages until a write (F)
Hard link = multiple names → same data (C)
P() = wait/decrement (H)
Load balancing = spread threads across CPUs (G)

Section C (2 marks each: 1 T/F + 1 justification)

Q14 — TRUE. With a quantum larger than every burst, no process is ever preempted, so each runs to completion in arrival order — identical to FCFS.

Q15 — FALSE. Banker's algorithm is a deadlock avoidance method: it checks whether granting a request leaves the system in a safe state before deadlock occurs; detection is a separate technique.

Q16 — FALSE. Many-to-one mapping means the kernel schedules only one kernel thread, so at most one user thread runs at a time — no true parallelism on multicore.

Q17 — FALSE. Best-fit chooses the smallest adequate hole to reduce waste, but leftover holes still accumulate; external fragmentation remains (and best-fit tends to leave many tiny unusable holes). Compaction is needed to eliminate it.

Q18 — TRUE. Each page-table entry has a valid–invalid (present) bit; a reference to an invalid-marked page triggers a trap so the OS handles the page fault.

[
  {"claim": "SJF minimizes avg waiting time: for bursts 6,8,7,3 SJF order 3,6,7,8 gives lower avg wait than FCFS order 6,8,7,3",
   "code": "bursts_fcfs=[6,8,7,3]\n\ndef avg_wait(order):\n    t=0; total=0\n    for b in order:\n        total+=t\n        t+=b\n    return total/len(order)\nfcfs=avg_wait([6,8,7,3])\nsjf=avg_wait(sorted([6,8,7,3]))\nresult = (sjf < fcfs) and (abs(sjf-7)<1e-9)"},
  {"claim": "RAID 0 usable capacity fraction with n disks equals 1 (no redundancy), RAID 1 mirror equals 1/2",
   "code": "raid0=1\nraid1=Rational(1,2)\nresult = (raid0==1) and (raid1==Rational(1,2))"},
  {"claim": "Coffman conditions count is exactly 4",
   "code": "conds={'mutual_exclusion','hold_and_wait','no_preemption','circular_wait'}\nresult = len(conds)==4"},
  {"claim": "test-and-set returns old value; sequence of two calls on lock starting 0 yields (0 then 1)",
   "code": "lock=[0]\n\ndef tas():\n    old=lock[0]; lock[0]=1; return old\nr1=tas(); r2=tas()\nresult = (r1==0) and (r2==1)"}
]