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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2021
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 2021 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

Primary storage vs secondary storage distinctions

Computer organisation

2 marks

Volatility: Primary storage (RAM) is volatile and loses all stored data when power is disconnected, whereas secondary storage (SSD, HDD) is non-volatile and permanently retains data without power.
Direct CPU access and speed: Primary memory is directly addressable by the CPU over the system bus with nanosecond access latencies, whereas secondary storage cannot be directly addressed by the CPU and operates orders of magnitude slower.

Mentor tip

Contrast volatility and direct CPU addressability/access speed.

Question 2

1 mark

Binary to hexadecimal number conversion

Computer organisation

1 mark

Split binary number 10001010 into two 4-bit nibbles:
- Upper nibble 1000 = 8.
- Lower nibble 1010 = 10, which corresponds to hexadecimal digit A.
Combining nibbles yields 8A.

1000101 0_{2} = 8A_{16}

Mentor tip

Group into four-bit nibbles: 1000 is 8, 1010 is A.

Question 3

4 marks

Sorting algorithm definition and Bubble vs Selection sort distinction

Computational thinking, problem-solving and programming

(a)

2 marks

An algorithm that systematically rearranges elements of an unordered collection or array into a specified linear order (ascending or descending) according to a predefined comparison key.

Mentor tip

Define systematic rearrangement of elements according to a key.

(b)

2 marks

Bubble sort repeatedly compares adjacent pairs and swaps them immediately if they are out of order (gradually bubbling items to the end), whereas selection sort scans the entire unsorted partition to identify the absolute minimum element and performs at most one swap per pass.

Mentor tip

Contrast adjacent pairwise swapping against finding the global minimum per pass.

Question 4

2 marks

Compound Boolean logic evaluation

Computer organisation

2 marks

Given values: X = 5, Y = 3, A = TRUE.
Expression: ((X > 5) XOR A) AND (Y + 2 > 4)
1) Evaluate (X > 5): 5 > 5 is FALSE.
2) Evaluate ((X > 5) XOR A): FALSE XOR TRUE evaluates to TRUE.
3) Evaluate (Y + 2 > 4): 3 + 2 = 5, and 5 > 4 is TRUE.
4) Evaluate final conjunction: TRUE AND TRUE evaluates to TRUE.

(FALSE \oplus TRUE) \land (5 > 4) = TRUE \land TRUE = TRUE

Mentor tip

Show evaluation of each sub-expression step-by-step.

Question 5

3 marks

Binary tree characteristics and root deletion in binary search tree

Abstract data structures

(a)

1 mark

In a binary tree, every node has a maximum degree of two (at most two children: left and right), whereas in a general non-binary tree, a node can have an arbitrary number of children.

Mentor tip

State the strict limit of at most two children per node.

(b)

2 marks

Given BST with root 3, left child 1, right child 5, and left child of 5 is 4.
Deleting the root (node 3):
- Replace root 3 with its inorder successor (the smallest node in the right subtree), which is 4 (or its inorder predecessor 1).
- Using inorder successor 4: Node 4 becomes the new root.
- Left child of 4 remains 1.
- Right child of 4 remains 5 (with its original left child 4 removed).
Resulting BST: Root is 4, left child is 1, right child is 5.

4 ↙ ↘ 15

Mentor tip

Replace the deleted root with its inorder successor (4) to maintain valid binary search tree ordering.

Question 6

3 marks

Network transmission advantages of data compression

Networks

3 marks

Reduced bit volume: Compression algorithms eliminate data redundancy and encode files into significantly fewer bits.
Faster transmission duration: Smaller byte payloads transmit across the network in less time, reducing latency for end users.
Bandwidth conservation: Lower aggregate traffic minimizes network congestion and reduces data transfer costs over metered connections.

Mentor tip

Connect file size reduction to shorter transfer times and conserved network capacity.

Question 7

7 marks

Subprograms as abstraction, concurrent development, and update notifications

Computational thinking, problem-solving and programming
System fundamentals

(a)

2 marks

Hiding implementation complexity: A subprogram encapsulates complex multi-step algorithms behind a simple, well-defined identifier and parameter interface.
Callers only need to understand what arguments to pass and what result is returned, without needing to know internal logic details.

Mentor tip

Explain that subprograms conceal complex inner mechanics behind clean procedural interfaces.

(b)

3 marks

Advantage: Accelerates time-to-market by allowing specialized developer teams to build distinct modules (e.g. database, UI, network layer) simultaneously in parallel.
Disadvantage / Risk: Requires strict interface contract adherence; if inter-module API standards are ambiguous or change mid-development, complex integration defects and costly rework occur during system assembly.

Mentor tip

Balance accelerated development velocity against the risk of integration mismatches.

(c)

2 marks

In-application pop-up notification / splash modal: The application queries a central version repository on launch and prompts the user with release notes and an instant update button.

Mentor tip

Mention in-app update banners or automated email notification to registered accounts.

Question 8

3 marks

Operating system peripheral device management

Resource management

3 marks

Device driver abstraction: Translates high-level operating system I/O instructions into specific hardware control signals understood by peripheral controllers.
Interrupt handling: Listens for hardware interrupt signals raised by peripherals, executing dedicated Interrupt Service Routines (ISRs) to transfer data promptly.
Buffering and spooling: Manages temporary memory buffers (e.g. print spoolers, keyboard buffers) to smooth out speed mismatches between fast CPUs and slow peripheral devices.

Mentor tip

Explain the role of device drivers, interrupt service routines, and I/O buffer management.

Questions 9–13 · 75 marks

Question 9

15 marks

Requirement gathering methodologies, prototyping, and rigorous testing

System fundamentals

(a)(i)

2 marks

Scalability and quantitative reach: Surveys can be distributed digitally to hundreds of dispersed stakeholders simultaneously, gathering broad quantitative feedback at minimal cost.

Mentor tip

Highlight wide geographic reach and low distribution cost.

(a)(ii)

2 marks

Inflexible questions and lack of depth: Fixed survey questions prevent dynamic follow-up probing, often yielding superficial responses and low completion rates.

Mentor tip

Point out the inability to ask clarifying follow-up questions.

(b)

1 mark

Direct stakeholder interviews (or direct workplace observation).

Mentor tip

State interviews or workplace observation.

(c)

4 marks

1) Early requirement clarification: Provides users with an interactive, tangible model to test workflows, exposing misunderstandings and missing specifications early.
2) Higher stakeholder engagement and confidence: Users actively participate in design iterations, fostering buy-in and reducing post-launch user resistance.

Mentor tip

Detail two distinct benefits like early requirement validation and reduced design risk.

(d)

3 marks

Iterative refinement: Initial prototypes frequently reveal unexpected user needs or technical constraints that were not anticipated in the first analysis pass.
Subsequent analysis and design cycles allow developers to adapt specifications based on real user interaction data, converging on a robust final design.

Mentor tip

Explain that real user interaction exposes hidden requirements that necessitate iterative revision.

(e)

3 marks

Preventing catastrophic failure and data corruption: Rigorous testing under varied operational loads detects edge-case defects and vulnerabilities before deployment.
Unchecked errors in production can cause business disruption, severe financial losses, and irreversible data corruption.

Mentor tip

Connect rigorous pre-deployment testing to operational reliability and risk mitigation.

Question 10

15 marks

Enterprise data governance, network security, and disaster recovery

System fundamentals
Networks

(a)

4 marks

Role-Based Access Control (RBAC): Assigning permissions strictly according to job roles, ensuring staff can only view or edit files essential to their daily tasks.
Multi-Factor Authentication (MFA) and biometric verification: Requiring two or more independent credentials before granting access to sensitive databases.

Mentor tip

Describe two concrete access control measures such as RBAC and multi-factor authentication.

(b)

4 marks

Next-generation firewall and intrusion prevention systems (IPS): Inspecting inbound and outbound packet headers and payloads to filter malicious traffic.
End-to-end data encryption across network segments: Encrypting internal data transmissions (via TLS/IPsec) to prevent packet eavesdropping.

Mentor tip

Cover firewall/IPS implementation and cryptographic traffic encryption.

(c)

4 marks

Automated periodic backups: Restoring intact records from scheduled differential or immutable off-site backups created before corruption occurred.
Database transaction logging and rollback: Using write-ahead transaction logs to roll back incomplete or damaged states to the last known consistent checkpoint.

Mentor tip

Explain restoring from off-site backups and replaying/rolling back transaction logs.

(d)

3 marks

Secure remote connectivity: Employees traveling or working from home can access confidential corporate servers across public internet connections through an encrypted tunnel, safeguarding corporate secrets.

Mentor tip

Highlight encrypted remote access to internal resources over public networks.

Question 11

15 marks

GPS satellite telemetry, real-time tracking, and urban transit applications

Networks
System fundamentals

(a)

4 marks

Satellite constellation radio broadcast: A network of orbiting GPS satellites continuously transmit synchronized time-stamped radio signals containing their exact orbital ephemeris.
Receiver signal acquisition: A ground GPS receiver calculates the time delay for signals to arrive from multiple visible satellites.
Trilateration calculation: By knowing signal propagation speed (speed of light) and time differences from at least 4 satellites, the receiver solves geometric equations to calculate its exact 3D latitude, longitude, and elevation.

Mentor tip

Detail satellite time-stamped signals, signal propagation delay calculation, and 4-satellite trilateration.

(b)(i)

1 mark

Estimated arrival time / minutes until arrival (or delay duration and platform number).

Mentor tip

State estimated time of arrival or platform number.

(b)(ii)

2 marks

Wheel tachometers / trackside RFID sensors: Complement GPS signals in tunnels and urban canyons where satellite line-of-sight is blocked, providing dead-reckoning position updates.

Mentor tip

Mention trackside sensors or wheel odometer sensors for coverage when satellite signals are obstructed.

(c)

4 marks

Live walking directions and walking time to the optimal station entrance based on the user's current GPS coordinates.
Real-time countdown and platform departure alert tailored to whether the user can reach the platform before the train departs.

Mentor tip

Describe two user-centric features combining phone GPS with train GPS telemetry.

(d)

4 marks

Advantages: Enhances passenger satisfaction with predictable real-time schedules; enables transit dispatchers to optimize fleet routing and respond rapidly to delays.
Disadvantages: Vulnerable to signal attenuation in tunnels, underground stations, and deep urban canyons; satellite telemetry hardware and cellular backhaul incur continuous maintenance costs.

Mentor tip

Evaluate operational efficiency against signal obstruction and maintenance dependencies.

Question 12

15 marks

Circular linked list manipulation, sum calculation, and queue implementation

Abstract data structures

(a)

3 marks

Original list: [20] -> [5] -> [10] -> [20] (circular, head points to 20).
Operation 1: Insert 30 at beginning. New list: [30] -> [20] -> [5] -> [10] -> [30]. Head points to 30.
Operation 2: Delete last node (node 10). Node 5 now points back to head (node 30). List: [30] -> [20] -> [5] -> [30].
Operation 3: Delete first node (node 30). Head updates to 20. Last node (5) now points back to 20.
Resulting circular linked list: [20] -> [5] -> [20], with head pointing to node 20.

head \to [20] \to [5] \to [20]

Mentor tip

Track pointer updates through each step: inserting 30 at head, deleting 10 from tail, then deleting 30 from head.

(b)

2 marks

In a circular linked list, the last node is identified by the condition that its next pointer references the head node (i.e. CURRENT.next == head).

Mentor tip

State that the last node's next pointer equals the head pointer.

(c)

4 marks

Algorithm to sum circular linked list:
1) Check if head is null; if so, return 0.
2) Initialize SUM = head.value, and pointer CURRENT = head.next.
3) loop while CURRENT != head:
- SUM = SUM + CURRENT.value
- CURRENT = CURRENT.next
4) end loop.
5) Return SUM.

Mentor tip

Process the head value and traverse until CURRENT wraps back to head.

(d)

4 marks

Arrays:
- Contiguous memory allocation with constant O(1) random access via index subscripts.
- Fixed size defined at declaration; resizing requires allocating a new memory block and copying all elements.
Linked Lists:
- Non-contiguous memory allocation; elements can be inserted and deleted dynamically in O(1) time without shifting elements.
- Sequential traversal requires O(N) access time to reach an arbitrary index; extra memory overhead for pointer references.

Array: O (1) access, Linked list: O (1) insert/delete with known pointer

Mentor tip

Compare memory layout (contiguous vs pointer-linked) and complexity of indexing vs dynamic insertion.

(e)

2 marks

Print spooler queue managing document print jobs.
Operating system CPU process ready queue managing active threads.

Mentor tip

Give two standard queue applications such as print spooling or OS process scheduling.

Question 13

15 marks

Bus transit triangular distance matrix validation and lookup algorithm

Computational thinking, problem-solving and programming

(a)

1 mark

Kiko is station index 3; Longlines is station index 8.
In the lower triangular array, lookup row 8, column 3: ROUTE_X_DISTANCES[8][3] = 5.9 km.

ROUTE_X_DISTANCES [8] [3] = 5.9 km

Mentor tip

Row index must be greater than column index in a lower triangular matrix: row 8, column 3.

(b)

5 marks

Algorithm to validate ROUTE_X_DISTANCES (10 x 10 matrix):
1) Initialize VALID = TRUE.
2) Nested loop: R from 0 to 9, C from 0 to 9.
3) If R <= C (on or above main diagonal):
- If ROUTE_X_DISTANCES[R][C] != 0 then VALID = FALSE.
4) Else (R > C, below main diagonal):
- If ROUTE_X_DISTANCES[R][C] <= 0 then VALID = FALSE.
5) Output 'Valid' if VALID is true, else 'Invalid'.

VALID = TRUE loop R from 0 to 9 loop C from 0 to 9 if R \leq C then if ROUTE_X_DISTANCES [R] [C] \neq = 0 then VALID = FALSE else if ROUTE_X_DISTANCES [R] [C] \leq 0 then VALID = FALSE end if end loop end loop output VALID

Mentor tip

Partition checks by diagonal: R <= C must be zero, R > C must be strictly positive.

(c)

6 marks

Algorithm to find distance between two station names:
1) Input NAME1, NAME2.
2) Initialize POS1 = -1, POS2 = -1.
3) Search ROUTE_X_NAMES (size 10):
- loop I from 0 to 9: if ROUTE_X_NAMES[I] == NAME1 then POS1 = I; if ROUTE_X_NAMES[I] == NAME2 then POS2 = I.
4) If POS1 == -1 or POS2 == -1 then output 'Station not found'.
5) Else:
- If POS1 == POS2 then output 0.
- Else if POS1 > POS2 then output ROUTE_X_DISTANCES[POS1][POS2].
- Else output ROUTE_X_DISTANCES[POS2][POS1].

if POS1 = - 1 or POS2 = - 1 then output "Station not found" else if POS1 = POS2 then output 0 else if POS1 > POS2 then output ROUTE_X_DISTANCES [POS1] [POS2] else output ROUTE_X_DISTANCES [POS2] [POS1] end if

Mentor tip

Ensure the larger index is used as the row subscript when querying the lower-triangular matrix.

(d)

3 marks

Identify the station indices P1 and P2 in the route array.
Ensure P1 < P2 (swapping if necessary).
Sum the transit minutes in ROUTE_X_TIMES from index P1 + 1 up to P2.
The resulting sum represents the cumulative travel time between the two stations.

Total Time = i = P_{1} + 1 \sum P_{2} ROUTE_X_TIMES [i]

Mentor tip

Explain that summing consecutive interval travel times from min(P1,P2) + 1 to max(P1,P2) gives total elapsed travel time.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals32 marks · 32%
Q7, Q9, Q10, Q11
Networks21 marks · 21%
Q6, Q10, Q11
Computational thinking, problem-solving and programming21 marks · 21%
Q3, Q7, Q13
Abstract data structures18 marks · 18%
Q5, Q12
Computer organisation5 marks · 5%
Q1, Q2, Q4
Resource management3 marks · 3%
Q8

Mentor insights

Traps to watch for

01
Inorder successor selection upon BST root deletion · Q5
When deleting a root with two children, replace it with the smallest node in its right subtree (inorder successor, 4). Replacing it with an arbitrary child breaks the binary search property.
02
Circular linked list loop termination condition · Q12
Because a circular linked list has no null termination pointer, loop termination must check when the moving pointer equals head again. Using a while pointer != null check creates an infinite loop.
03
Lower triangular matrix index ordering · Q13
In a lower triangular distance matrix, values exist only where row index > column index. Querying with row < column accesses the zero-filled upper triangle. Always use max(P1, P2) as row and min(P1, P2) as column.

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 2021
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 2021 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

Primary storage vs secondary storage distinctions

Computer organisation

2 marks

Volatility: Primary storage (RAM) is volatile and loses all stored data when power is disconnected, whereas secondary storage (SSD, HDD) is non-volatile and permanently retains data without power.
Direct CPU access and speed: Primary memory is directly addressable by the CPU over the system bus with nanosecond access latencies, whereas secondary storage cannot be directly addressed by the CPU and operates orders of magnitude slower.

Mentor tip

Contrast volatility and direct CPU addressability/access speed.

Question 2

1 mark

Binary to hexadecimal number conversion

Computer organisation

1 mark

Split binary number 10001010 into two 4-bit nibbles:
- Upper nibble 1000 = 8.
- Lower nibble 1010 = 10, which corresponds to hexadecimal digit A.
Combining nibbles yields 8A.

1000101 0_{2} = 8A_{16}

Mentor tip

Group into four-bit nibbles: 1000 is 8, 1010 is A.

Question 3

4 marks

Sorting algorithm definition and Bubble vs Selection sort distinction

Computational thinking, problem-solving and programming

(a)

2 marks

An algorithm that systematically rearranges elements of an unordered collection or array into a specified linear order (ascending or descending) according to a predefined comparison key.

Mentor tip

Define systematic rearrangement of elements according to a key.

(b)

2 marks

Mentor tip

Contrast adjacent pairwise swapping against finding the global minimum per pass.

Question 4

2 marks

Compound Boolean logic evaluation

Computer organisation

2 marks

Given values: X = 5, Y = 3, A = TRUE.
Expression: ((X > 5) XOR A) AND (Y + 2 > 4)
1) Evaluate (X > 5): 5 > 5 is FALSE.
2) Evaluate ((X > 5) XOR A): FALSE XOR TRUE evaluates to TRUE.
3) Evaluate (Y + 2 > 4): 3 + 2 = 5, and 5 > 4 is TRUE.
4) Evaluate final conjunction: TRUE AND TRUE evaluates to TRUE.

(FALSE \oplus TRUE) \land (5 > 4) = TRUE \land TRUE = TRUE

Mentor tip

Show evaluation of each sub-expression step-by-step.

Question 5

3 marks

Binary tree characteristics and root deletion in binary search tree

Abstract data structures

(a)

1 mark

In a binary tree, every node has a maximum degree of two (at most two children: left and right), whereas in a general non-binary tree, a node can have an arbitrary number of children.

Mentor tip

State the strict limit of at most two children per node.

(b)

2 marks

Given BST with root 3, left child 1, right child 5, and left child of 5 is 4.
Deleting the root (node 3):
- Replace root 3 with its inorder successor (the smallest node in the right subtree), which is 4 (or its inorder predecessor 1).
- Using inorder successor 4: Node 4 becomes the new root.
- Left child of 4 remains 1.
- Right child of 4 remains 5 (with its original left child 4 removed).
Resulting BST: Root is 4, left child is 1, right child is 5.

4 ↙ ↘ 15

Mentor tip

Replace the deleted root with its inorder successor (4) to maintain valid binary search tree ordering.

Question 6

3 marks

Network transmission advantages of data compression

Networks

3 marks

Reduced bit volume: Compression algorithms eliminate data redundancy and encode files into significantly fewer bits.
Faster transmission duration: Smaller byte payloads transmit across the network in less time, reducing latency for end users.
Bandwidth conservation: Lower aggregate traffic minimizes network congestion and reduces data transfer costs over metered connections.

Mentor tip

Connect file size reduction to shorter transfer times and conserved network capacity.

Question 7

7 marks

Subprograms as abstraction, concurrent development, and update notifications

Computational thinking, problem-solving and programming
System fundamentals

(a)

2 marks

Hiding implementation complexity: A subprogram encapsulates complex multi-step algorithms behind a simple, well-defined identifier and parameter interface.
Callers only need to understand what arguments to pass and what result is returned, without needing to know internal logic details.

Mentor tip

Explain that subprograms conceal complex inner mechanics behind clean procedural interfaces.

(b)

3 marks

Advantage: Accelerates time-to-market by allowing specialized developer teams to build distinct modules (e.g. database, UI, network layer) simultaneously in parallel.
Disadvantage / Risk: Requires strict interface contract adherence; if inter-module API standards are ambiguous or change mid-development, complex integration defects and costly rework occur during system assembly.

Mentor tip

Balance accelerated development velocity against the risk of integration mismatches.

(c)

2 marks

In-application pop-up notification / splash modal: The application queries a central version repository on launch and prompts the user with release notes and an instant update button.

Mentor tip

Mention in-app update banners or automated email notification to registered accounts.

Question 8

3 marks

Operating system peripheral device management

Resource management

3 marks

Device driver abstraction: Translates high-level operating system I/O instructions into specific hardware control signals understood by peripheral controllers.
Interrupt handling: Listens for hardware interrupt signals raised by peripherals, executing dedicated Interrupt Service Routines (ISRs) to transfer data promptly.
Buffering and spooling: Manages temporary memory buffers (e.g. print spoolers, keyboard buffers) to smooth out speed mismatches between fast CPUs and slow peripheral devices.

Mentor tip

Explain the role of device drivers, interrupt service routines, and I/O buffer management.

Questions 9–13 · 75 marks

Question 9

15 marks

Requirement gathering methodologies, prototyping, and rigorous testing

System fundamentals

(a)(i)

2 marks

Scalability and quantitative reach: Surveys can be distributed digitally to hundreds of dispersed stakeholders simultaneously, gathering broad quantitative feedback at minimal cost.

Mentor tip

Highlight wide geographic reach and low distribution cost.

(a)(ii)

2 marks

Inflexible questions and lack of depth: Fixed survey questions prevent dynamic follow-up probing, often yielding superficial responses and low completion rates.

Mentor tip

Point out the inability to ask clarifying follow-up questions.

(b)

1 mark

Direct stakeholder interviews (or direct workplace observation).

Mentor tip

State interviews or workplace observation.

(c)

4 marks

1) Early requirement clarification: Provides users with an interactive, tangible model to test workflows, exposing misunderstandings and missing specifications early.
2) Higher stakeholder engagement and confidence: Users actively participate in design iterations, fostering buy-in and reducing post-launch user resistance.

Mentor tip

Detail two distinct benefits like early requirement validation and reduced design risk.

(d)

3 marks

Iterative refinement: Initial prototypes frequently reveal unexpected user needs or technical constraints that were not anticipated in the first analysis pass.
Subsequent analysis and design cycles allow developers to adapt specifications based on real user interaction data, converging on a robust final design.

Mentor tip

Explain that real user interaction exposes hidden requirements that necessitate iterative revision.

(e)

3 marks

Preventing catastrophic failure and data corruption: Rigorous testing under varied operational loads detects edge-case defects and vulnerabilities before deployment.
Unchecked errors in production can cause business disruption, severe financial losses, and irreversible data corruption.

Mentor tip

Connect rigorous pre-deployment testing to operational reliability and risk mitigation.

Question 10

15 marks

Enterprise data governance, network security, and disaster recovery

System fundamentals
Networks

(a)

4 marks

Role-Based Access Control (RBAC): Assigning permissions strictly according to job roles, ensuring staff can only view or edit files essential to their daily tasks.
Multi-Factor Authentication (MFA) and biometric verification: Requiring two or more independent credentials before granting access to sensitive databases.

Mentor tip

Describe two concrete access control measures such as RBAC and multi-factor authentication.

(b)

4 marks

Next-generation firewall and intrusion prevention systems (IPS): Inspecting inbound and outbound packet headers and payloads to filter malicious traffic.
End-to-end data encryption across network segments: Encrypting internal data transmissions (via TLS/IPsec) to prevent packet eavesdropping.

Mentor tip

Cover firewall/IPS implementation and cryptographic traffic encryption.

(c)

4 marks

Automated periodic backups: Restoring intact records from scheduled differential or immutable off-site backups created before corruption occurred.
Database transaction logging and rollback: Using write-ahead transaction logs to roll back incomplete or damaged states to the last known consistent checkpoint.

Mentor tip

Explain restoring from off-site backups and replaying/rolling back transaction logs.

(d)

3 marks

Mentor tip

Highlight encrypted remote access to internal resources over public networks.

Question 11

15 marks

GPS satellite telemetry, real-time tracking, and urban transit applications

Networks
System fundamentals

(a)

4 marks

Satellite constellation radio broadcast: A network of orbiting GPS satellites continuously transmit synchronized time-stamped radio signals containing their exact orbital ephemeris.
Receiver signal acquisition: A ground GPS receiver calculates the time delay for signals to arrive from multiple visible satellites.
Trilateration calculation: By knowing signal propagation speed (speed of light) and time differences from at least 4 satellites, the receiver solves geometric equations to calculate its exact 3D latitude, longitude, and elevation.

Mentor tip

Detail satellite time-stamped signals, signal propagation delay calculation, and 4-satellite trilateration.

(b)(i)

1 mark

Estimated arrival time / minutes until arrival (or delay duration and platform number).

Mentor tip

State estimated time of arrival or platform number.

(b)(ii)

2 marks

Wheel tachometers / trackside RFID sensors: Complement GPS signals in tunnels and urban canyons where satellite line-of-sight is blocked, providing dead-reckoning position updates.

Mentor tip

Mention trackside sensors or wheel odometer sensors for coverage when satellite signals are obstructed.

(c)

4 marks

Live walking directions and walking time to the optimal station entrance based on the user's current GPS coordinates.
Real-time countdown and platform departure alert tailored to whether the user can reach the platform before the train departs.

Mentor tip

Describe two user-centric features combining phone GPS with train GPS telemetry.

(d)

4 marks

Advantages: Enhances passenger satisfaction with predictable real-time schedules; enables transit dispatchers to optimize fleet routing and respond rapidly to delays.
Disadvantages: Vulnerable to signal attenuation in tunnels, underground stations, and deep urban canyons; satellite telemetry hardware and cellular backhaul incur continuous maintenance costs.

Mentor tip

Evaluate operational efficiency against signal obstruction and maintenance dependencies.

Question 12

15 marks

Circular linked list manipulation, sum calculation, and queue implementation

Abstract data structures

(a)

3 marks

Original list: [20] -> [5] -> [10] -> [20] (circular, head points to 20).
Operation 1: Insert 30 at beginning. New list: [30] -> [20] -> [5] -> [10] -> [30]. Head points to 30.
Operation 2: Delete last node (node 10). Node 5 now points back to head (node 30). List: [30] -> [20] -> [5] -> [30].
Operation 3: Delete first node (node 30). Head updates to 20. Last node (5) now points back to 20.
Resulting circular linked list: [20] -> [5] -> [20], with head pointing to node 20.

head \to [20] \to [5] \to [20]

Mentor tip

Track pointer updates through each step: inserting 30 at head, deleting 10 from tail, then deleting 30 from head.

(b)

2 marks

In a circular linked list, the last node is identified by the condition that its next pointer references the head node (i.e. CURRENT.next == head).

Mentor tip

State that the last node's next pointer equals the head pointer.

(c)

4 marks

Algorithm to sum circular linked list:
1) Check if head is null; if so, return 0.
2) Initialize SUM = head.value, and pointer CURRENT = head.next.
3) loop while CURRENT != head:
- SUM = SUM + CURRENT.value
- CURRENT = CURRENT.next
4) end loop.
5) Return SUM.

Mentor tip

Process the head value and traverse until CURRENT wraps back to head.

(d)

4 marks

Arrays:
- Contiguous memory allocation with constant O(1) random access via index subscripts.
- Fixed size defined at declaration; resizing requires allocating a new memory block and copying all elements.
Linked Lists:
- Non-contiguous memory allocation; elements can be inserted and deleted dynamically in O(1) time without shifting elements.
- Sequential traversal requires O(N) access time to reach an arbitrary index; extra memory overhead for pointer references.

Array: O (1) access, Linked list: O (1) insert/delete with known pointer

Mentor tip

Compare memory layout (contiguous vs pointer-linked) and complexity of indexing vs dynamic insertion.

(e)

2 marks

Print spooler queue managing document print jobs.
Operating system CPU process ready queue managing active threads.

Mentor tip

Give two standard queue applications such as print spooling or OS process scheduling.

Question 13

15 marks

Bus transit triangular distance matrix validation and lookup algorithm

Computational thinking, problem-solving and programming

(a)

1 mark

Kiko is station index 3; Longlines is station index 8.
In the lower triangular array, lookup row 8, column 3: ROUTE_X_DISTANCES[8][3] = 5.9 km.

ROUTE_X_DISTANCES [8] [3] = 5.9 km

Mentor tip

Row index must be greater than column index in a lower triangular matrix: row 8, column 3.

(b)

5 marks

Algorithm to validate ROUTE_X_DISTANCES (10 x 10 matrix):
1) Initialize VALID = TRUE.
2) Nested loop: R from 0 to 9, C from 0 to 9.
3) If R <= C (on or above main diagonal):
- If ROUTE_X_DISTANCES[R][C] != 0 then VALID = FALSE.
4) Else (R > C, below main diagonal):
- If ROUTE_X_DISTANCES[R][C] <= 0 then VALID = FALSE.
5) Output 'Valid' if VALID is true, else 'Invalid'.

VALID = TRUE loop R from 0 to 9 loop C from 0 to 9 if R \leq C then if ROUTE_X_DISTANCES [R] [C] \neq = 0 then VALID = FALSE else if ROUTE_X_DISTANCES [R] [C] \leq 0 then VALID = FALSE end if end loop end loop output VALID

Mentor tip

Partition checks by diagonal: R <= C must be zero, R > C must be strictly positive.

(c)

6 marks

Algorithm to find distance between two station names:
1) Input NAME1, NAME2.
2) Initialize POS1 = -1, POS2 = -1.
3) Search ROUTE_X_NAMES (size 10):
- loop I from 0 to 9: if ROUTE_X_NAMES[I] == NAME1 then POS1 = I; if ROUTE_X_NAMES[I] == NAME2 then POS2 = I.
4) If POS1 == -1 or POS2 == -1 then output 'Station not found'.
5) Else:
- If POS1 == POS2 then output 0.
- Else if POS1 > POS2 then output ROUTE_X_DISTANCES[POS1][POS2].
- Else output ROUTE_X_DISTANCES[POS2][POS1].

if POS1 = - 1 or POS2 = - 1 then output "Station not found" else if POS1 = POS2 then output 0 else if POS1 > POS2 then output ROUTE_X_DISTANCES [POS1] [POS2] else output ROUTE_X_DISTANCES [POS2] [POS1] end if

Mentor tip

Ensure the larger index is used as the row subscript when querying the lower-triangular matrix.

(d)

3 marks

Identify the station indices P1 and P2 in the route array.
Ensure P1 < P2 (swapping if necessary).
Sum the transit minutes in ROUTE_X_TIMES from index P1 + 1 up to P2.
The resulting sum represents the cumulative travel time between the two stations.

Total Time = i = P_{1} + 1 \sum P_{2} ROUTE_X_TIMES [i]

Mentor tip

Explain that summing consecutive interval travel times from min(P1,P2) + 1 to max(P1,P2) gives total elapsed travel time.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. System fundamentals carries the most marks on this paper.

System fundamentals32 marks · 32%
Q7, Q9, Q10, Q11
Networks21 marks · 21%
Q6, Q10, Q11
Computational thinking, problem-solving and programming21 marks · 21%
Q3, Q7, Q13
Abstract data structures18 marks · 18%
Q5, Q12
Computer organisation5 marks · 5%
Q1, Q2, Q4
Resource management3 marks · 3%
Q8

Mentor insights

Traps to watch for

01
Inorder successor selection upon BST root deletion · Q5
When deleting a root with two children, replace it with the smallest node in its right subtree (inorder successor, 4). Replacing it with an arbitrary child breaks the binary search property.
02
Circular linked list loop termination condition · Q12
Because a circular linked list has no null termination pointer, loop termination must check when the moving pointer equals head again. Using a while pointer != null check creates an infinite loop.
03
Lower triangular matrix index ordering · Q13
In a lower triangular distance matrix, values exist only where row index > column index. Querying with row < column accesses the zero-filled upper triangle. Always use max(P1, P2) as row and min(P1, P2) as column.

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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