Operating systems, scheduling, polling and interrupts – IB Computer Science exam questions with mark scheme

1Core4 marks

A games developer writes a program that saves files and plays sound. The program never refers to the specific hard drive model or sound card in the computer it runs on.

(a) Explain how the operating system makes this possible. [2]

(b) Identify the function of the operating system that is being performed in each of the following situations.

(i) A user is refused access to a folder owned by another user. [1]

(ii) A school network records how many pages each student prints so that they can be charged. [1]

(a) The OS provides an abstraction layer / hides the hardware details by offering standard system calls / APIs (e.g. 'open file', 'play sound') that programs use;(1)
the OS (through device drivers) translates these general requests into the specific instructions for that particular hardware, so the same program works on different hardware (accept: the OS manages the hardware on the program's behalf)(1)
(b)(i) Security (accept: access control / user permissions / authentication; reject 'file system' alone)(1)
(b)(ii) Accounting (accept: resource usage tracking / logging use of resources; reject 'device management' alone)(1)
Answer: (a) The OS hides hardware detail behind standard system calls; drivers convert them to device-specific commands. (b)(i) Security (ii) Accounting.
Examiner tip: 'The OS controls the hardware' scores nothing on its own; say that programs call a standard interface and the OS/drivers translate it for the specific device.

2StandardHL only4 marks

A single-core laptop is running a web browser, a music player and a word processor at the same time. The user also sends two documents to the same printer from different applications within one second of each other.

(a) Explain how the operating system allows the three applications to appear to run simultaneously on one core. [2]

(b) Explain how the operating system prevents resource contention when both applications try to use the printer. [2]

(a) The OS scheduler gives each process a short time slice/turn on the CPU in rotation (time-sharing), switching so quickly that they appear to run at the same time;(1)
at each switch the OS performs a context switch: saves the state (registers, program counter) of the current process (e.g. in its PCB) and restores the state of the next one, so each process resumes where it left off(1)
(b) The OS controls access to the printer so that only one process can use it at a time (accept: mutual exclusion / lock / the printer is allocated to one process);(1)
other requests are held in a queue / spooled to disk (print queue) and sent to the printer in turn when it becomes free, so the documents are not interleaved/mixed(1)
Answer: (a) Time slicing by the scheduler with context switches saving/restoring each process's state. (b) Printer allocated to one process at a time; other jobs spooled/queued and printed in turn.
Examiner tip: Many answers say 'the processes run at the same time'; on one core they do not, so you must mention rapid switching and saving/restoring state.

3Standard4 marks

Four processes arrive at a CPU that uses first-come first-served (FCFS) scheduling.

Process	Arrival time (ms)	Burst time (ms)
P1	0	8
P2	1	4
P3	2	2
P4	3	1

(a) Calculate the average waiting time for the four processes. Show your working. [2]

(b) Explain why FCFS gives a poor average waiting time for this set of processes. [2]

(a) Correct waiting times: P1 = 0, P2 = 7, P3 = 10, P4 = 11 ms (allow from correct start times 0, 8, 12, 14 minus arrival times);(1)
average = 28 / 4 = 7 ms (correct answer with no working scores 1 only; allow follow-through from one wrong waiting time)(1)
(b) A long process (P1, 8 ms) arrived first and must run to completion because FCFS is non-preemptive;(1)
so the short processes (P2, P3, P4) queue behind it / wait much longer than their own burst time (convoy effect), raising the average (accept: running shorter jobs first would reduce the average)(1)
Answer: (a) Waits 0, 7, 10, 11 ms; average 7 ms. (b) Long P1 runs first without interruption, so short jobs wait behind it (convoy effect).
Examiner tip: Waiting time is start time minus arrival time; students often forget to subtract the arrival time and get 0, 8, 12, 14.

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4Hard6 marks

Four processes, A, B, C and D, all arrive at time 0 ms in the order A, B, C, D. Their burst times are A = 7 ms, B = 4 ms, C = 2 ms and D = 5 ms. The operating system uses round robin scheduling with a time quantum of 3 ms. Ignore the time taken to switch between processes. The start of the Gantt chart is shown.

(a) Complete the Gantt chart from 6 ms to 18 ms, showing which process runs in each interval. [3]

(b) Calculate the average turnaround time of the four processes. [2]

(a) C 6–8 then D 8–11;(1)
A 11–14 then B 14–15;(1)
D 15–17 then A 17–18 (follow through within (a) only if one earlier slot is wrong but the round-robin rule is applied consistently)(1)
(b) Completion times A = 18, B = 15, C = 8, D = 17 ms (turnaround = completion time since all arrive at 0);(1)
average = 58 / 4 = 14.5 ms (allow follow-through from their Gantt chart)(1)
(c) Every process finishes within its first turn, so it behaves the same as FCFS / processes run to completion in order A, B, C, D (accept: no pre-emption takes place)(1)
Answer: (a) A 0–3, B 3–6, C 6–8, D 8–11, A 11–14, B 14–15, D 15–17, A 17–18. (b) 14.5 ms. (c) It becomes FCFS.
Examiner tip: When a process needs less than a full quantum (C needs 2 ms) it gives up the CPU early; students who give C a full 3 ms shift every later slot.

5Standard4 marks

A single-core CPU uses non-preemptive priority scheduling. A lower number means a higher priority.

Process	Arrival time (ms)	Burst time (ms)	Priority
W	0	5	3
X	1	3	1
Y	2	4	4
Z	3	2	2

(a) State the order in which the processes are executed. [1]

(b) Calculate the waiting time of process Y. [1]

(c) Explain one problem that could affect a low-priority process in a busy system using priority scheduling, and how the operating system could reduce it. [2]

(a) W, X, Z, Y (W runs first because it is the only process at time 0 and cannot be pre-empted)(1)
(b) 8 ms (Y starts at 10 ms, arrived at 2 ms)(1)
(c) Starvation / indefinite blocking: if higher-priority processes keep arriving, a low-priority process may never get the CPU;(1)
ageing: the OS gradually increases the priority of a process the longer it waits, so it eventually runs (accept: guarantee a minimum share/time slice to every priority level)(1)
Answer: (a) W, X, Z, Y. (b) 8 ms. (c) Starvation; solved by ageing.
Examiner tip: W starts first even though X has higher priority: in non-preemptive scheduling the running process is not interrupted when a higher-priority one arrives.

6Hard4 marks

A school server runs three types of process: interactive processes from students using the online learning platform, system processes such as the antivirus scanner, and overnight batch jobs that produce reports.

The technician is deciding between a single round robin queue and a multilevel queue.

(a) Describe how a multilevel queue scheduler would handle these processes. [2]

(b) Compare the suitability of the two approaches for this server. [2]

(a) Processes are permanently placed in separate queues according to their type (e.g. system, interactive, batch), each queue having its own priority;(1)
a queue is only serviced when higher-priority queues are empty / CPU time is shared between queues, and each queue can use its own algorithm (e.g. round robin for interactive, FCFS for batch)(1)
(b) Multilevel queue lets interactive/system processes be favoured so students get fast response, whereas a single round robin queue treats a long report job the same as a student's request, slowing response time;(1)
but multilevel queue is more complex to configure and batch jobs may starve during busy periods, whereas round robin is simple and fair / guarantees every process a turn (award for a reasoned point favouring round robin or a drawback of multilevel queue)(1)
Answer: (a) Separate fixed queues by process type with their own priority and algorithm. (b) MLQ gives better response for interactive work but is complex and risks starving batch jobs; RR is simple and fair but cannot prioritise.
Examiner tip: A compare question needs both approaches mentioned in each point; describing only the multilevel queue caps you at half marks in (b).

7Standard4 marks

A user presses a key while the CPU is running a spreadsheet program. The keyboard controller raises an interrupt.

Describe the sequence of steps taken from the moment the interrupt signal is received to the moment the spreadsheet program continues. [4]

The CPU checks for interrupts (at the end of the current fetch–decode–execute cycle) and, if the interrupt has a higher priority than the current task, accepts it;(1)
the current contents of the registers / program counter (the state of the spreadsheet process) are saved (e.g. pushed onto the stack);(1)
the address of the correct interrupt service routine (ISR / handler) is found (e.g. from an interrupt vector table) and the ISR runs, e.g. reading the key code into the keyboard buffer;(1)
the saved register values are restored / reloaded and the spreadsheet program resumes from where it stopped (ignore 'it carries on' without restoring state)(1)
Answer: Finish current instruction and check priority; save registers/PC; locate and run the ISR (store key in buffer); restore registers and resume.
Examiner tip: The marks for saving and restoring the process state are the ones most often missed; without them the program could not resume correctly.

8Hard7 marks

A network card on a server running at 2 GHz (2 × 10⁹ clock cycles per second) can be serviced by polling or by interrupts.

Polling: the CPU polls the card 50 000 times per second; each poll uses 400 clock cycles.
Interrupts: each arriving packet causes one interrupt; handling one interrupt (including saving and restoring state) uses 2 500 clock cycles.

(a) Calculate the percentage of CPU time used by polling. [2]

(b) Calculate the percentage of CPU time used by interrupts when 2 000 packets arrive per second. [1]

(d) Using your answers, explain why some network drivers switch from interrupts to polling during periods of very heavy traffic. [2]

(a) 50 000 × 400 = 2 × 10⁷ cycles per second;(1)
2 × 10⁷ ÷ 2 × 10⁹ × 100 = 1% (correct answer alone scores 2)(1)
(b) 2 000 × 2 500 = 5 × 10⁶ cycles per second, so 0.25%(1)
(c) Interrupt cost equals polling cost when rate × 2 500 = 2 × 10⁷;(1)
rate = 8 000 packets per second (accept 'more than 8 000 per second'; allow follow-through from (a))(1)
(d) Polling cost is fixed / does not depend on traffic, whereas interrupt cost grows with each packet, so at high event frequency (above about 8 000 packets/s) interrupts use more CPU time than polling;(1)
each interrupt has a large overhead (saving/restoring state, context switch), so handling many packets per poll is more efficient / avoids the CPU being swamped with interrupts (accept: at high load a poll almost always finds data waiting, so few polls are wasted)(1)
Answer: (a) 1% (b) 0.25% (c) 8 000 packets per second (d) Polling's cost is fixed while interrupt cost rises per event, so at very high event frequency polling is cheaper.
Examiner tip: Show the cycles-per-second step before dividing by the clock rate; a bare percentage with an arithmetic slip earns nothing.

9Stretch6 marks

Two embedded systems are being designed.

System 1: a battery-powered soil-moisture sensor in a field that reports a change in moisture level roughly twice a day and must run for two years without a battery change.

System 2: an airbag controller in a car that must respond to a crash sensor within a fixed, guaranteed time of 2 ms. The software is simple and the system runs from the car's power supply.

Evaluate the use of polling and interrupts for each system. [6]

System 1: events are very infrequent, so polling would waste CPU cycles checking a sensor that has almost always not changed;(1)
interrupts allow the processor to sleep/enter a low-power state between events and wake only when the sensor signals, which saves battery power;(1)
therefore interrupts are the better choice for System 1 (conclusion must be supported by power and/or event-frequency reasoning)(1)
System 2: polling at a fixed rate gives predictable, deterministic timing: the worst-case response is known (e.g. polling every 0.5 ms guarantees detection well inside 2 ms), which is needed in a safety-critical real-time system (accept: a highest-priority / non-maskable interrupt gives an immediate response with a known, bounded latency);(1)
power is not a constraint and the controller has little else to do, so the CPU overhead of polling is acceptable / interrupt timing can be less predictable if several interrupts arrive together or interrupts are disabled at the wrong moment (accept, for an interrupt-based answer: interrupts avoid wasted polling cycles and respond as soon as the sensor signals, without waiting for the next poll);(1)
therefore polling is justified for System 2 (accept a reasoned case for a high-priority interrupt with guaranteed latency, e.g. a non-maskable interrupt, if latency and predictability are both discussed)(1)
Answer: System 1: interrupts (infrequent events, CPU can sleep, saves battery). System 2: polling at a known rate gives predictable worst-case latency; overhead and power do not matter.
Examiner tip: Evaluate means reaching a judgement for each system; listing generic pros and cons of polling and interrupts without linking them to the scenario limits you to about half marks.

10HardHL only6 marks

The diagram shows three processes, P1, P2 and P3, and three resources, R1, R2 and R3, managed by an operating system. Each resource can be used by only one process at a time.

(a) Identify which processes are deadlocked. Justify your answer using the diagram. [2]

(b) Outline two of the conditions that must be present for this deadlock to occur. [2]

(a) P1 and P2 are deadlocked (P3 is not; reject any answer including P3);(1)
P1 holds R1 and is waiting for R2, while P2 holds R2 and is waiting for R1, so there is a circular wait and neither can ever continue (accept: there is a cycle P1 → R2 → P2 → R1 → P1); P3 holds R3 and is not waiting for anything, so it can finish(1)
(b) First condition correctly named and outlined, from: mutual exclusion – each resource can be used by only one process at a time; hold and wait – a process holds one resource while waiting for another; no pre-emption – a resource cannot be forcibly taken from the process holding it; circular wait – a closed chain of processes each waiting for a resource held by the next;(1)
a second, different condition from the list correctly named and outlined (names alone with no outline: max 1 for the pair)(1)
(c) Terminate/abort one of the deadlocked processes (e.g. P2) / pre-empt (forcibly take) one resource, e.g. R2 from P2;(1)
so the resource is released and allocated to the other process (P1), which can then complete; the aborted process is restarted / rolled back later (accept: roll back a process to a checkpoint and restart it)(1)
Answer: (a) P1 and P2: circular wait over R1 and R2; P3 is not involved. (b) e.g. hold and wait; circular wait. (c) Abort P2 (or pre-empt R2) so P1 gets R2, then restart P2.
Examiner tip: Read arrow directions carefully: an arrow from a resource to a process means 'held by', and only processes in a closed cycle are deadlocked.

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