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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2019
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2019 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Offsite storage to prevent data loss

System fundamentals

(a)

2 marks

Describe offsite data storage protection:
1) Backup copies of critical data files are physically transferred to a separate remote geographic facility or securely uploaded to remote cloud servers.
2) In the event of a local catastrophic disaster (such as a fire, flood, hardware theft, or localized ransomware incident), the primary facility's lost data can be restored from the undamaged offsite copy.

Mentor tip

Explain geographical isolation protecting against local disasters.

Question 2

3 marks

Purpose of cache memory

Computer organisation

(a)

3 marks

Explain the role and architecture of cache memory:
1) Cache is extremely fast, small SRAM located directly on (or very close to) the CPU core.
2) It buffers frequently accessed instructions and recently referenced data copied from slower main memory (RAM).
3) When the processor needs data, it checks cache first (cache hit); accessing cache takes vastly fewer clock cycles than querying RAM across the system bus, substantially minimizing processor idle wait states and boosting overall throughput.

Mentor tip

Mention high-speed SRAM, storing frequently used instructions/data, and reducing RAM access latency.

Question 3

3 marks

Colour representation and hexadecimal

Computer organisation

(a)

1 mark

State why hexadecimal is frequently used in computing:
Hexadecimal provides a much shorter, human-readable representation of binary values (each hex digit represents exactly 4 binary bits / 1 nibble), making memory addresses, color codes, and machine bytes easier to read, write, and debug without transcription errors.

Mentor tip

State that it is a compact, human-friendly representation of binary.

(b)

1 mark

Determine bits used for a non-primary colour (such as yellow, made from red + green + blue channels):
A 24-bit true color representation allocates 8 bits (1 byte) for Red, 8 bits for Green, and 8 bits for Blue: 8 + 8 + 8 = 24 bits.

8 bits (Red) + 8 bits (Green) + 8 bits (Blue) = 24 bits

Mentor tip

State 24 bits (or 3 bytes).

(c)

1 mark

Calculate the maximum number of unique colours in a 24-bit pixel:
Each channel has 256 levels (0 to 255): 256 x 256 x 256 = 2^24 = 16,777,216 distinct colours.

2^{24} = 16777216 unique colours (or 25 6^{3})

Mentor tip

Write 16,777,216 or 2^24.

Question 4

4 marks

Compound logic truth table

Computer organisation

(a)

4 marks

Formulate Boolean expressions for intermediate and output signals:
1) X = A XOR B (1 when A and B differ, 0 when equal).
2) Y = A NOR C = NOT (A OR C) (1 only when both A=0 and C=0).
3) Z = X OR (NOT Y). Note that NOT Y is 1 whenever Y=0.
Evaluate row by row:
Row 1: A=0, B=0, C=0 -> X=0, Y=1 -> NOT Y=0 -> Z = 0 OR 0 = 0
Row 2: A=0, B=0, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 3: A=0, B=1, C=0 -> X=1, Y=1 -> NOT Y=0 -> Z = 1 OR 0 = 1
Row 4: A=0, B=1, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 5: A=1, B=0, C=0 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 6: A=1, B=0, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 7: A=1, B=1, C=0 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 8: A=1, B=1, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1

X = A \oplus B, Y = \overline{A \lor C}, Z = X \lor \overline{Y} A 00001111 B 00110011 C 01010101 X 00111100 Y 10100000 Z 01111111

Mentor tip

Notice that Y=0 for 6 out of 8 rows; whenever Y=0, NOT Y=1, forcing Z=1.

Question 5

3 marks

Double linked list representation

Abstract data structures

(a)

3 marks

Identify components of a doubly linked list node:
Each node contains three fields: Previous Pointer (prev), Data field, and Next Pointer (next).
Structure of the list:
1) Head pointer references the first node 'Anne'.
2) Node 1: [prev: null | data: "Anne" | next: -> Node 2].
3) Node 2: [prev: -> Node 1 | data: "Lana" | next: -> Node 3].
4) Node 3: [prev: -> Node 2 | data: "Mary" | next: null].
5) Tail pointer references 'Mary' (or last next pointer terminates in null).

Mentor tip

Ensure bidirectional arrows connect adjacent nodes, with the first prev pointer and last next pointer set to null.

Question 6

3 marks

Abstraction in algorithm design

Computational thinking, problem-solving and programming

(a)

3 marks

Explain abstraction in algorithmic problem solving:
1) Filtering non-essential detail: Abstraction removes irrelevant real-world complexities and physical specifics, isolating only the core logical properties necessary to solve the problem.
2) Managing cognitive complexity: Decomposing large problems into generalized abstract components (procedures, objects, classes) enables programmers to focus on high-level logic without being overwhelmed by low-level implementation details.
3) Reusability and generalization: Abstract algorithms can be applied across different applications and hardware platforms without rewriting fundamental logic.

Mentor tip

Focus on hiding irrelevant details, managing complexity, and creating reusable generalized models.

Question 7

2 marks

Virtual memory definition

Computer organisation

(a)

2 marks

Define virtual memory operational architecture:
1) An operating system memory management technique that extends available primary memory by utilizing secondary storage (HDD/SSD) as temporary swap space.
2) Inactive memory pages or program segments are swapped between physical RAM and disk, allowing processes requiring more memory than physically installed RAM to execute seamlessly.

Mentor tip

State secondary storage acting as an extension of RAM via paging/swapping.

Question 8

5 marks

Trace table for iterative loop

Computational thinking, problem-solving and programming

(a)

5 marks

Trace the execution step by step:
Initial values: K = 1, N = 1, M = 2.
Iteration 1 (K = 1 < 5):
- output(1, 2)
- K = 1 + 1 = 2
- N = 1 + 2 = 3
- M = 2 * 2 = 4
Iteration 2 (K = 2 < 5):
- output(3, 4)
- K = 2 + 1 = 3
- N = 3 + 2 = 5
- M = 4 * 2 = 8
Iteration 3 (K = 3 < 5):
- output(5, 8)
- K = 3 + 1 = 4
- N = 5 + 2 = 7
- M = 8 * 2 = 16
Iteration 4 (K = 4 < 5):
- output(7, 16)
- K = 4 + 1 = 5
- N = 7 + 2 = 9
- M = 16 * 2 = 32
Loop check: K = 5 is not < 5. Loop terminates.
Final variables: K=5, N=9, M=32. Outputs produced: (1,2), (3,4), (5,8), (7,16).

K 12345 N 13579 M 2481632 OUTPUT 1, 2 3, 4 5, 8 7, 16

Mentor tip

Record values on the exact rows they are updated, and list outputs in sequence.

Questions 9–13 · 75 marks

Question 9

15 marks

System implementation and testing

System fundamentals

(a)

2 marks

Identify two organizational issues related to new system implementation:
1) User resistance to change and requirement for comprehensive employee retraining programs.
2) Temporary disruption to operational workflows, data migration overhead, or business restructuring.

Mentor tip

Mention user retraining and workflow disruption/resistance.

(b)

5 marks

Evaluate phased conversion vs direct changeover:
1) Phased conversion:
- Advantage: Implemented module by module or department by department; allows staff to adapt gradually, and if an issue occurs in a newly deployed module, the rest of the business continues operating normally.
- Disadvantage: Prolonged implementation period; complex temporary data bridging and communication interfaces must be maintained between old and new subsystems.
2) Direct changeover:
- Advantage: Immediate transition at a fixed cut-over date; low operational cost with zero duplication of effort and no bridging interfaces needed.
- Disadvantage: Extremely high risk; if the new system contains catastrophic flaws, operations halt immediately with no fallback system available.
Conclusion: Phased conversion is much safer for critical enterprise systems, whereas direct changeover is only suitable for non-critical systems with complete confidence in testing.

Mentor tip

Provide clear advantages and disadvantages for both strategies and give a concluding comparative judgment.

(c)

4 marks

(i) State one type of testing that involves users (1 mark):

Beta testing (or End-user operational testing / UAT).

(ii) Identify three consequences of inadequate testing (3 marks):

1) Undetected critical software bugs causing frequent system crashes, data corruption, or permanent data loss.

2) Severe financial losses due to operational downtime, customer churn, and emergency development bug-fixing costs.

3) Reputational damage, loss of client trust, and potential legal liabilities from failing service level agreements (SLAs).

Mentor tip

Name Beta or UAT for part (i), then list 3 distinct business/technical impacts for (ii).

(d)

4 marks

Discuss social and ethical issues associated with implementing a new computer system:
1) Social issues: Potential workforce redundancies or role displacements due to automation, leading to employee stress and job insecurity; digital divide challenges for staff struggling to adapt to digital tooling.
2) Ethical issues: Privacy concerns surrounding employee monitoring (e.g., tracking keystrokes, active hours, or location); ethical responsibility to protect sensitive corporate/customer data gathered by new platforms against data breaches.

Mentor tip

Address both social impacts (job loss, retraining stress) and ethical impacts (privacy, monitoring).

Question 10

15 marks

Network hardware, protocols, and security

Networks

(a)

4 marks

Outline two advantages of a school using a computer network:
1) Resource and hardware sharing: Multiple classrooms can share expensive peripheral equipment (centralized high-volume printers, scanners) and central network storage drives, substantially lowering hardware costs.
2) Centralized data management and collaboration: Central servers allow automated administrative backups, centralized software licensing, collaborative document editing, and controlled student access to shared educational resources from any school workstation.

Mentor tip

Detail how sharing hardware or centralized administration directly benefits school operations.

(b)

4 marks

Describe the purpose of network hardware components:
(i) Router (2 marks): An internetworking device that connects disparate networks (e.g., school LAN to the external Internet); it inspects destination IP addresses in incoming packets and forwards them along the optimal path using routing tables.
(ii) Network Interface Card / NIC (2 marks): A hardware controller inside a device that provides the physical and electrical interface to the network medium (Ethernet cable or Wi-Fi radio); it formats data into frames and contains the unique physical MAC address.

Mentor tip

Associate Router with IP routing between networks, and NIC with physical media connection and MAC addressing.

(c)

2 marks

Outline why protocols are necessary in network communications:
1) Protocols establish standardized syntactic, semantic, and synchronization rules for data transmission.
2) This guarantees that disparate computing hardware and operating systems can exchange data reliably, detect transmission errors, and manage flow control without misinterpreting transmitted signals.

Mentor tip

Mention standard rules, interoperability across different vendors, and error detection.

(d)

1 mark

Define data encryption:
The process of encoding plaintext data into ciphertext using a mathematical algorithm and an encryption key, such that unauthorized parties intercepting the transmission cannot read or decipher the data without the corresponding decryption key.

Mentor tip

Mention transforming plaintext into ciphertext using a key.

(e)

4 marks

Evaluate trusted MAC address filtering as a network security method:
1) How it works & Advantages (2 marks): The network access point maintains an authorized whitelist of physical MAC addresses; devices not on the approved whitelist are denied connection, providing a simple first-line perimeter defense against casual unauthorized users.
2) Limitations & Vulnerabilities (2 marks): MAC addresses are transmitted in unencrypted plaintext in every frame header; attackers running network sniffers can easily capture an authorized MAC address and spoof their own device's MAC address to bypass the filter. Additionally, administrative overhead is high when adding new legitimate devices.

Mentor tip

Contrast simple access restriction against MAC address spoofing vulnerabilities.

Question 11

15 marks

Display technologies, operating systems, and embedded control

System fundamentals
Resource management
Control systems

(a)

5 marks

(i) Define screen resolution (1 mark): The total number of individual physical pixels (width x height) that can be displayed on a screen (e.g., 1920x1080), or pixel density per unit area (PPI).

(ii) Describe two issues resulting from viewing the website on different devices (4 marks):

1) Layout and navigation distortion: On smaller smartphone screens with lower resolutions or vertical aspect ratios, content designed for wide desktop displays may cause awkward horizontal scrolling, illegible microscopic text, overlapping navigation buttons, or unclickable touch targets.

2) Bandwidth and asset optimization constraints: Serving high-resolution desktop images or heavy scripts to mobile devices on cellular connections increases page load latency, degrades mobile performance, and exhausts user mobile data quotas unless responsive image scaling is implemented.

Mentor tip

Define pixel dimensions for (i), then describe layout/usability issues and performance/data consumption for (ii).

(b)

4 marks

Explain the role of the operating system in managing hardware resources:
1) Processor / CPU scheduling: Allocates CPU time slices to concurrent processes using scheduling algorithms (e.g., Round Robin, Priority Scheduling), ensuring fair resource distribution and preventing process starvation.
2) Memory management: Allocates physical RAM blocks to running applications, protects memory spaces between processes, and manages paging/virtual memory when RAM is saturated.
3) Device and I/O management: Communicates with peripherals via device drivers, manages I/O buffers and interrupts to balance fast CPU cycles with slow peripheral transfers.

Mentor tip

Structure around CPU scheduling, memory management, and device/peripheral driver control.

(c)

6 marks

Describe the interaction between sensors, microprocessors, and output transducers in a washing machine:
1) Input sensing: Sensors continuously monitor physical variables (water level pressure sensor, water temperature thermistor, drum rotation speed tachometer, lid lock sensor).
2) Signal conversion: Analog sensor signals are converted to digital values via an Analogue-to-Digital Converter (ADC) and fed into the microprocessor.
3) Processing & Decision logic: The microprocessor compares current readings against the programmed wash cycle parameters stored in ROM/EEPROM (e.g., target water level or 40°C temperature).
4) Output transducer actuation: When setpoints are reached or changes are required, the microprocessor sends digital control signals (via DAC/relays) to output transducers: opening/closing water inlet solenoid valves, turning electric heater elements on/off, and driving electric motors for spinning and drain pumps.
5) Feedback loop: Continual sensor sampling forms a real-time feedback loop, stopping filling when water reaches full level and shutting off heaters when temperature reaches the setpoint.

Mentor tip

Explicitly identify specific sensors (temperature, water level), ADC, the microprocessor comparison, and specific transducers (valves, heating element, motor).

Question 12

15 marks

Stacks, binary trees, and data structures

Abstract data structures

(a)

2 marks

State two practical applications of stacks (LIFO structures):
1) Subroutine / function call management (call stack storing return addresses and local stack frames).
2) Undo/Redo mechanisms in text/graphics editors (or expression evaluation / syntax parsing / backtracking algorithms).

Mentor tip

Name function call stack and undo/redo operations.

(b)

6 marks

Explain the implementation of a static stack using a one-dimensional array:
1) Structure & Pointer: A fixed-size array STACK[0..MAX-1] is allocated with a pointer TOP initialized to -1 (or 0) indicating an empty stack.
2) Test for empty: isEmpty() returns TRUE if TOP == -1 (or TOP == 0).
3) Test for full: isFull() returns TRUE if TOP == MAX - 1 (or TOP == MAX).
4) Push operation: Before pushing, test isFull(). If full, signal stack overflow. Otherwise, increment TOP (TOP = TOP + 1) and insert new element at STACK[TOP].
5) Pop operation: Before popping, test isEmpty(). If empty, signal stack underflow. Otherwise, retrieve element at STACK[TOP], decrement TOP (TOP = TOP - 1), and return retrieved element.

Mentor tip

Explain the role of TOP, isEmpty check, isFull check, push algorithm, and pop algorithm.

(c)

4 marks

(i) State postorder traversal of the binary tree (1 mark):

Given binary search tree rooted at Hazel:

- Inorder is alphabetical: Elder, Elm, Hazel, Holly, Larch, Rowan.

- Hazel is root. Left child is Elder, right child Elm (Elder.right = Elm). Right child of Hazel is Larch, with left child Holly and right child Rowan.

- Postorder (Left -> Right -> Root): Elm, Elder, Holly, Rowan, Larch, Hazel.

(ii) Draw the binary tree after deleting root node 'Hazel' (3 marks):

- Deleting a node with two children requires replacing it with either its inorder predecessor (Elm) or inorder successor (Holly).

- If using inorder successor (Holly): Holly becomes the new root node.

- Left subtree remains rooted at Elder (with right child Elm).

- Right subtree of Holly has root Larch, which now has right child Rowan (Holly's former position is vacated).

Mentor tip

Specify postorder traversal sequence, and for deletion, promote Holly (or Elm) to new root.

(d)

3 marks

Compare static and dynamic data structures:
1) Size flexibility: Static structures have fixed bounds allocated at compile time; dynamic structures grow and shrink dynamically at runtime as elements are inserted or removed.
2) Memory efficiency & predictability: Static arrays occupy contiguous memory with predictable bounds and no pointer overhead, but risk wasted space or overflow; dynamic structures allocate non-contiguous heap nodes on demand, eliminating overflow limits but requiring extra memory for link pointers.
3) Access speed: Static arrays offer direct O(1) random index access; dynamic linked structures require sequential pointer traversal O(n) to reach elements.

Mentor tip

Compare across size flexibility, pointer memory overhead, and access speed.

Question 13

15 marks

Tri-diagonal matrices and recursive algorithms

Computational thinking, problem-solving and programming

(a)

1 mark

Locate element in matrix MAT at row index 3, column index 4:
Examine row 3 of MAT: [0, 0, -5, 6, 4, 0].
Index 4 is the super-diagonal element: 4.
Value is 4.

Mentor tip

Double check 0-based indices for row 3 and column 4.

(b)

8 marks

Construct an efficient algorithm for isValidMatrix(N, A):
Requirements: Dimensions N x N.
1) Diagonals: main diagonal (R==C), super-diagonal (C==R+1), sub-diagonal (C==R-1) must be non-zero.
2) All other positions (|R - C| > 1) must be exactly zero.
Efficient implementation traversing each row R from 0 to N-1 and column C from 0 to N-1:
method isValidMatrix(N, A):
loop R from 0 to N - 1
loop C from 0 to N - 1
DIFF = R - C
if DIFF >= -1 and DIFF <= 1 then
// On one of the three diagonals: must NOT be zero
if A[R][C] == 0 then
return False
end if
else
// Outside the three diagonals: must BE zero
if A[R][C] != 0 then
return False
end if
end if
end loop
end loop
return True
end isValidMatrix

Mentor tip

Use condition |R - C| <= 1 (or C == R, C == R-1, C == R+1) to classify entries cleanly.

(c)

4 marks

Trace recursive call X = mystery(MAT, 5):
Definition: mystery(A, R) returns A[R][R-1] + mystery(A, R-1) for R > 0, else 0.
Evaluate recursion expansion:
- R = 5: A[5][4] + mystery(MAT, 4)
- R = 4: A[4][3] + mystery(MAT, 3)
- R = 3: A[3][2] + mystery(MAT, 2)
- R = 2: A[2][1] + mystery(MAT, 1)
- R = 1: A[1][0] + mystery(MAT, 0)
- R = 0: returns 0
Look up sub-diagonal values A[R][R-1] from MAT:
- A[1][0] = 1
- A[2][1] = 9
- A[3][2] = -5
- A[4][3] = 7
- A[5][4] = 5
Unwind recursion summation:
X = 5 + 7 + (-5) + 9 + 1 + 0 = 17.

mystery (MAT, 5) = R = 1 \sum 5 A [R] [R - 1] = A [5] [4] + A [4] [3] + A [3] [2] + A [2] [1] + A [1] [0] = 5 + 7 + (- 5) + 9 + 1 = 17

Mentor tip

Write out each recursive term clearly: A[5][4] + A[4][3] + A[3][2] + A[2][1] + A[1][0].

(d)

2 marks

Deduce the general purpose of method mystery(A, R):
It computes and returns the sum of all elements along the sub-diagonal (the diagonal immediately below the main diagonal) of matrix A from row 1 up to row R.

Mentor tip

State that it calculates the sum of elements on the sub-diagonal.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Computational thinking, problem-solving and programming carries the most marks on this paper.

Computational thinking, problem-solving and programming23 marks · 23%
Q6, Q8, Q13
System fundamentals22 marks · 22%
Q1, Q9, Q11
Abstract data structures18 marks · 18%
Q5, Q12
Networks15 marks · 15%
Q10
Computer organisation12 marks · 12%
Q2, Q3, Q4, Q7
Control systems6 marks · 6%
Q11
Resource management4 marks · 4%
Q11

Mentor insights

Traps to watch for

01
NOR gate evaluation in compound truth table · Q4
A NOR C evaluates to 1 ONLY when both A = 0 and C = 0. When Y = 0, NOT Y evaluates to 1, which immediately makes output Z = 1.
02
Root node deletion in a binary search tree · Q12
When deleting a root node with two children, replace it with either the maximum in its left subtree (inorder predecessor) or minimum in its right subtree (inorder successor), then recursively delete that promoted node.
03
Sub-diagonal summation negative numbers handling · Q13
In mystery(MAT, 5), the recursive call sums elements where column is row - 1. When encountering negative elements like A[3][2] = -5, add the negative value directly (+ (-5) = -5) to avoid calculation errors.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2019
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 13

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2019 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Offsite storage to prevent data loss

System fundamentals

(a)

2 marks

Describe offsite data storage protection:
1) Backup copies of critical data files are physically transferred to a separate remote geographic facility or securely uploaded to remote cloud servers.
2) In the event of a local catastrophic disaster (such as a fire, flood, hardware theft, or localized ransomware incident), the primary facility's lost data can be restored from the undamaged offsite copy.

Mentor tip

Explain geographical isolation protecting against local disasters.

Question 2

3 marks

Purpose of cache memory

Computer organisation

(a)

3 marks

Explain the role and architecture of cache memory:
1) Cache is extremely fast, small SRAM located directly on (or very close to) the CPU core.
2) It buffers frequently accessed instructions and recently referenced data copied from slower main memory (RAM).
3) When the processor needs data, it checks cache first (cache hit); accessing cache takes vastly fewer clock cycles than querying RAM across the system bus, substantially minimizing processor idle wait states and boosting overall throughput.

Mentor tip

Mention high-speed SRAM, storing frequently used instructions/data, and reducing RAM access latency.

Question 3

3 marks

Colour representation and hexadecimal

Computer organisation

(a)

1 mark

State why hexadecimal is frequently used in computing:
Hexadecimal provides a much shorter, human-readable representation of binary values (each hex digit represents exactly 4 binary bits / 1 nibble), making memory addresses, color codes, and machine bytes easier to read, write, and debug without transcription errors.

Mentor tip

State that it is a compact, human-friendly representation of binary.

(b)

1 mark

Determine bits used for a non-primary colour (such as yellow, made from red + green + blue channels):
A 24-bit true color representation allocates 8 bits (1 byte) for Red, 8 bits for Green, and 8 bits for Blue: 8 + 8 + 8 = 24 bits.

8 bits (Red) + 8 bits (Green) + 8 bits (Blue) = 24 bits

Mentor tip

State 24 bits (or 3 bytes).

(c)

1 mark

Calculate the maximum number of unique colours in a 24-bit pixel:
Each channel has 256 levels (0 to 255): 256 x 256 x 256 = 2^24 = 16,777,216 distinct colours.

2^{24} = 16777216 unique colours (or 25 6^{3})

Mentor tip

Write 16,777,216 or 2^24.

Question 4

4 marks

Compound logic truth table

Computer organisation

(a)

4 marks

Formulate Boolean expressions for intermediate and output signals:
1) X = A XOR B (1 when A and B differ, 0 when equal).
2) Y = A NOR C = NOT (A OR C) (1 only when both A=0 and C=0).
3) Z = X OR (NOT Y). Note that NOT Y is 1 whenever Y=0.
Evaluate row by row:
Row 1: A=0, B=0, C=0 -> X=0, Y=1 -> NOT Y=0 -> Z = 0 OR 0 = 0
Row 2: A=0, B=0, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 3: A=0, B=1, C=0 -> X=1, Y=1 -> NOT Y=0 -> Z = 1 OR 0 = 1
Row 4: A=0, B=1, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 5: A=1, B=0, C=0 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 6: A=1, B=0, C=1 -> X=1, Y=0 -> NOT Y=1 -> Z = 1 OR 1 = 1
Row 7: A=1, B=1, C=0 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1
Row 8: A=1, B=1, C=1 -> X=0, Y=0 -> NOT Y=1 -> Z = 0 OR 1 = 1

X = A \oplus B, Y = \overline{A \lor C}, Z = X \lor \overline{Y} A 00001111 B 00110011 C 01010101 X 00111100 Y 10100000 Z 01111111

Mentor tip

Notice that Y=0 for 6 out of 8 rows; whenever Y=0, NOT Y=1, forcing Z=1.

Question 5

3 marks

Double linked list representation

Abstract data structures

(a)

3 marks

Identify components of a doubly linked list node:
Each node contains three fields: Previous Pointer (prev), Data field, and Next Pointer (next).
Structure of the list:
1) Head pointer references the first node 'Anne'.
2) Node 1: [prev: null | data: "Anne" | next: -> Node 2].
3) Node 2: [prev: -> Node 1 | data: "Lana" | next: -> Node 3].
4) Node 3: [prev: -> Node 2 | data: "Mary" | next: null].
5) Tail pointer references 'Mary' (or last next pointer terminates in null).

Mentor tip

Ensure bidirectional arrows connect adjacent nodes, with the first prev pointer and last next pointer set to null.

Question 6

3 marks

Abstraction in algorithm design

Computational thinking, problem-solving and programming

(a)

3 marks

Explain abstraction in algorithmic problem solving:
1) Filtering non-essential detail: Abstraction removes irrelevant real-world complexities and physical specifics, isolating only the core logical properties necessary to solve the problem.
2) Managing cognitive complexity: Decomposing large problems into generalized abstract components (procedures, objects, classes) enables programmers to focus on high-level logic without being overwhelmed by low-level implementation details.
3) Reusability and generalization: Abstract algorithms can be applied across different applications and hardware platforms without rewriting fundamental logic.

Mentor tip

Focus on hiding irrelevant details, managing complexity, and creating reusable generalized models.

Question 7

2 marks

Virtual memory definition

Computer organisation

(a)

2 marks

Define virtual memory operational architecture:
1) An operating system memory management technique that extends available primary memory by utilizing secondary storage (HDD/SSD) as temporary swap space.
2) Inactive memory pages or program segments are swapped between physical RAM and disk, allowing processes requiring more memory than physically installed RAM to execute seamlessly.

Mentor tip

State secondary storage acting as an extension of RAM via paging/swapping.

Question 8

5 marks

Trace table for iterative loop

Computational thinking, problem-solving and programming

(a)

5 marks

Trace the execution step by step:
Initial values: K = 1, N = 1, M = 2.
Iteration 1 (K = 1 < 5):
- output(1, 2)
- K = 1 + 1 = 2
- N = 1 + 2 = 3
- M = 2 * 2 = 4
Iteration 2 (K = 2 < 5):
- output(3, 4)
- K = 2 + 1 = 3
- N = 3 + 2 = 5
- M = 4 * 2 = 8
Iteration 3 (K = 3 < 5):
- output(5, 8)
- K = 3 + 1 = 4
- N = 5 + 2 = 7
- M = 8 * 2 = 16
Iteration 4 (K = 4 < 5):
- output(7, 16)
- K = 4 + 1 = 5
- N = 7 + 2 = 9
- M = 16 * 2 = 32
Loop check: K = 5 is not < 5. Loop terminates.
Final variables: K=5, N=9, M=32. Outputs produced: (1,2), (3,4), (5,8), (7,16).

K 12345 N 13579 M 2481632 OUTPUT 1, 2 3, 4 5, 8 7, 16

Mentor tip

Record values on the exact rows they are updated, and list outputs in sequence.

Questions 9–13 · 75 marks

Question 9

15 marks

System implementation and testing

System fundamentals

(a)

2 marks

Identify two organizational issues related to new system implementation:
1) User resistance to change and requirement for comprehensive employee retraining programs.
2) Temporary disruption to operational workflows, data migration overhead, or business restructuring.

Mentor tip

Mention user retraining and workflow disruption/resistance.

(b)

5 marks

Evaluate phased conversion vs direct changeover:
1) Phased conversion:
- Advantage: Implemented module by module or department by department; allows staff to adapt gradually, and if an issue occurs in a newly deployed module, the rest of the business continues operating normally.
- Disadvantage: Prolonged implementation period; complex temporary data bridging and communication interfaces must be maintained between old and new subsystems.
2) Direct changeover:
- Advantage: Immediate transition at a fixed cut-over date; low operational cost with zero duplication of effort and no bridging interfaces needed.
- Disadvantage: Extremely high risk; if the new system contains catastrophic flaws, operations halt immediately with no fallback system available.
Conclusion: Phased conversion is much safer for critical enterprise systems, whereas direct changeover is only suitable for non-critical systems with complete confidence in testing.

Mentor tip

Provide clear advantages and disadvantages for both strategies and give a concluding comparative judgment.

(c)

4 marks

(i) State one type of testing that involves users (1 mark):

Beta testing (or End-user operational testing / UAT).

(ii) Identify three consequences of inadequate testing (3 marks):

1) Undetected critical software bugs causing frequent system crashes, data corruption, or permanent data loss.

2) Severe financial losses due to operational downtime, customer churn, and emergency development bug-fixing costs.

3) Reputational damage, loss of client trust, and potential legal liabilities from failing service level agreements (SLAs).

Mentor tip

Name Beta or UAT for part (i), then list 3 distinct business/technical impacts for (ii).

(d)

4 marks

Discuss social and ethical issues associated with implementing a new computer system:
1) Social issues: Potential workforce redundancies or role displacements due to automation, leading to employee stress and job insecurity; digital divide challenges for staff struggling to adapt to digital tooling.
2) Ethical issues: Privacy concerns surrounding employee monitoring (e.g., tracking keystrokes, active hours, or location); ethical responsibility to protect sensitive corporate/customer data gathered by new platforms against data breaches.

Mentor tip

Address both social impacts (job loss, retraining stress) and ethical impacts (privacy, monitoring).

Question 10

15 marks

Network hardware, protocols, and security

Networks

(a)

4 marks

Outline two advantages of a school using a computer network:
1) Resource and hardware sharing: Multiple classrooms can share expensive peripheral equipment (centralized high-volume printers, scanners) and central network storage drives, substantially lowering hardware costs.
2) Centralized data management and collaboration: Central servers allow automated administrative backups, centralized software licensing, collaborative document editing, and controlled student access to shared educational resources from any school workstation.

Mentor tip

Detail how sharing hardware or centralized administration directly benefits school operations.

(b)

4 marks

Describe the purpose of network hardware components:
(i) Router (2 marks): An internetworking device that connects disparate networks (e.g., school LAN to the external Internet); it inspects destination IP addresses in incoming packets and forwards them along the optimal path using routing tables.
(ii) Network Interface Card / NIC (2 marks): A hardware controller inside a device that provides the physical and electrical interface to the network medium (Ethernet cable or Wi-Fi radio); it formats data into frames and contains the unique physical MAC address.

Mentor tip

Associate Router with IP routing between networks, and NIC with physical media connection and MAC addressing.

(c)

2 marks

Outline why protocols are necessary in network communications:
1) Protocols establish standardized syntactic, semantic, and synchronization rules for data transmission.
2) This guarantees that disparate computing hardware and operating systems can exchange data reliably, detect transmission errors, and manage flow control without misinterpreting transmitted signals.

Mentor tip

Mention standard rules, interoperability across different vendors, and error detection.

(d)

1 mark

Define data encryption:
The process of encoding plaintext data into ciphertext using a mathematical algorithm and an encryption key, such that unauthorized parties intercepting the transmission cannot read or decipher the data without the corresponding decryption key.

Mentor tip

Mention transforming plaintext into ciphertext using a key.

(e)

4 marks

Evaluate trusted MAC address filtering as a network security method:
1) How it works & Advantages (2 marks): The network access point maintains an authorized whitelist of physical MAC addresses; devices not on the approved whitelist are denied connection, providing a simple first-line perimeter defense against casual unauthorized users.
2) Limitations & Vulnerabilities (2 marks): MAC addresses are transmitted in unencrypted plaintext in every frame header; attackers running network sniffers can easily capture an authorized MAC address and spoof their own device's MAC address to bypass the filter. Additionally, administrative overhead is high when adding new legitimate devices.

Mentor tip

Contrast simple access restriction against MAC address spoofing vulnerabilities.

Question 11

15 marks

Display technologies, operating systems, and embedded control

System fundamentals
Resource management
Control systems

(a)

5 marks

(i) Define screen resolution (1 mark): The total number of individual physical pixels (width x height) that can be displayed on a screen (e.g., 1920x1080), or pixel density per unit area (PPI).

(ii) Describe two issues resulting from viewing the website on different devices (4 marks):

Mentor tip

Define pixel dimensions for (i), then describe layout/usability issues and performance/data consumption for (ii).

(b)

4 marks

Explain the role of the operating system in managing hardware resources:
1) Processor / CPU scheduling: Allocates CPU time slices to concurrent processes using scheduling algorithms (e.g., Round Robin, Priority Scheduling), ensuring fair resource distribution and preventing process starvation.
2) Memory management: Allocates physical RAM blocks to running applications, protects memory spaces between processes, and manages paging/virtual memory when RAM is saturated.
3) Device and I/O management: Communicates with peripherals via device drivers, manages I/O buffers and interrupts to balance fast CPU cycles with slow peripheral transfers.

Mentor tip

Structure around CPU scheduling, memory management, and device/peripheral driver control.

(c)

6 marks

Describe the interaction between sensors, microprocessors, and output transducers in a washing machine:
1) Input sensing: Sensors continuously monitor physical variables (water level pressure sensor, water temperature thermistor, drum rotation speed tachometer, lid lock sensor).
2) Signal conversion: Analog sensor signals are converted to digital values via an Analogue-to-Digital Converter (ADC) and fed into the microprocessor.
3) Processing & Decision logic: The microprocessor compares current readings against the programmed wash cycle parameters stored in ROM/EEPROM (e.g., target water level or 40°C temperature).
4) Output transducer actuation: When setpoints are reached or changes are required, the microprocessor sends digital control signals (via DAC/relays) to output transducers: opening/closing water inlet solenoid valves, turning electric heater elements on/off, and driving electric motors for spinning and drain pumps.
5) Feedback loop: Continual sensor sampling forms a real-time feedback loop, stopping filling when water reaches full level and shutting off heaters when temperature reaches the setpoint.

Mentor tip

Explicitly identify specific sensors (temperature, water level), ADC, the microprocessor comparison, and specific transducers (valves, heating element, motor).

Question 12

15 marks

Stacks, binary trees, and data structures

Abstract data structures

(a)

2 marks

State two practical applications of stacks (LIFO structures):
1) Subroutine / function call management (call stack storing return addresses and local stack frames).
2) Undo/Redo mechanisms in text/graphics editors (or expression evaluation / syntax parsing / backtracking algorithms).

Mentor tip

Name function call stack and undo/redo operations.

(b)

6 marks

Explain the implementation of a static stack using a one-dimensional array:
1) Structure & Pointer: A fixed-size array STACK[0..MAX-1] is allocated with a pointer TOP initialized to -1 (or 0) indicating an empty stack.
2) Test for empty: isEmpty() returns TRUE if TOP == -1 (or TOP == 0).
3) Test for full: isFull() returns TRUE if TOP == MAX - 1 (or TOP == MAX).
4) Push operation: Before pushing, test isFull(). If full, signal stack overflow. Otherwise, increment TOP (TOP = TOP + 1) and insert new element at STACK[TOP].
5) Pop operation: Before popping, test isEmpty(). If empty, signal stack underflow. Otherwise, retrieve element at STACK[TOP], decrement TOP (TOP = TOP - 1), and return retrieved element.