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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Session: May 2016
Marks: 70
Duration: 1 hour 30 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 May 2016 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

2 marks

Maintenance challenges of legacy systems

System fundamentals

(a)

2 marks

Define a legacy system as an outdated computer hardware setup or obsolete software application that remains in active operational use.
Explain one maintenance problem: modifying legacy systems is difficult and costly due to missing or inaccurate documentation, obsolete programming languages, and a shortage of developers with specialized legacy expertise.

Mentor tip

Clarify that legacy systems use outdated technology and explicitly link to missing documentation or scarce programming expertise.

Question 2

2 marks

Concept and purpose of user operational sign-off testing

System fundamentals

(a)

2 marks

Define User Operational Sign-off Testing (UAT): providing end users with a functional release or prototype of the system in a real or simulated operational environment.
Explain its purpose: verifying that the software satisfies actual business requirements, meets user expectations, and gathers feedback on interface usability before final deployment.

Mentor tip

Highlight that actual end users test the software against business requirements before final sign-off.

Question 3

4 marks

Printed documentation versus online user support

System fundamentals

(a)

4 marks

State one advantage of printed documentation: Portability and physical availability — printed manuals do not require an active internet connection, power supply, or functioning display, making them accessible during complete hardware or power failures.
Elaborate on printed advantage: Physical reference cards or setup guides can be kept right next to new peripheral hardware (e.g., printers or scanners) during physical assembly.
State one disadvantage of printed documentation: Lack of interactivity and difficult searchability — hardcopy manuals cannot be hyperlinked, cannot easily be searched via keywords, and cannot be updated dynamically without reprinting.
Elaborate on printed disadvantage: Printed manuals can rapidly become obsolete as software receives patches and interface changes, leading to outdated instructions and environmental paper waste.

Mentor tip

Provide one advantage (e.g., constant offline availability) and one disadvantage (e.g., difficulty updating, poor searchability), expanding each fully.

Question 4

2 marks

Role and mechanism of a failover system

System fundamentals

(a)

2 marks

Identify that a failover system maintains a redundant, standby secondary server or system constantly synchronized with the primary platform.
Describe how it operates: when a hardware fault, power disruption, or software crash disables the primary system, the failover mechanism seamlessly switches operations to the secondary system automatically, avoiding disruption to end users.

Mentor tip

Emphasize automatic takeover by a redundant standby system during primary failure to preserve availability.

Question 5

2 marks

Role of the Control Unit (CU) in the CPU

Computer organisation

(a)

2 marks

State the primary function of the Control Unit: fetching instructions from main memory (RAM) and decoding them into micro-operations and control signals.
Outline its coordination role: directing the timing and movement of data between CPU registers, the Arithmetic Logic Unit (ALU), and input/output interfaces across the control bus.

Mentor tip

Focus on instruction fetching/decoding and issuing synchronization control signals to CPU components.

Question 6

2 marks

Cache memory mechanism for processor speedup

Computer organisation

(a)

2 marks

Describe cache location and speed: Cache memory is ultra-fast static RAM (SRAM) integrated directly on or adjacent to the CPU core, offering significantly faster access times than system RAM.
Explain how it accelerates execution: By staging frequently executed loop instructions and recently accessed data in cache, the CPU avoids high-latency bus trips to main memory, reducing idle wait cycles.

Mentor tip

Explain that holding frequently used data/instructions in fast on-chip SRAM minimizes latency compared to accessing RAM.

Question 7

2 marks

Operating system features for video game execution

Resource management

(a)

2 marks

Identify an operating system feature: Memory management and dedicated graphics driver interfacing.
Explain its importance for gaming: Video games render high-resolution textures and complex 3D assets in real time, requiring the OS to efficiently allocate large RAM buffers, manage GPU memory transfers, and maintain steady frame rates without stuttering.

Mentor tip

Mention memory management or device/GPU driver coordination required to handle continuous high-bandwidth frame rendering.

Question 8

3 marks

Truth table construction for A XOR (B OR C)

Computer organisation

(a)

3 marks

Evaluate the sub-expression (B OR C) for all 8 input combinations of binary inputs A, B, and C.
Apply the exclusive-OR (XOR) operator between input A and the intermediate result (B OR C), outputting 1 if and only if exactly one operand is true:
Row 1: A=0, B=0, C=0 -> B OR C = 0 -> 0 XOR 0 = 0
Row 2: A=0, B=0, C=1 -> B OR C = 1 -> 0 XOR 1 = 1
Row 3: A=0, B=1, C=0 -> B OR C = 1 -> 0 XOR 1 = 1
Row 4: A=0, B=1, C=1 -> B OR C = 1 -> 0 XOR 1 = 1
Row 5: A=1, B=0, C=0 -> B OR C = 0 -> 1 XOR 0 = 1
Row 6: A=1, B=0, C=1 -> B OR C = 1 -> 1 XOR 1 = 0
Row 7: A=1, B=1, C=0 -> B OR C = 1 -> 1 XOR 1 = 0
Row 8: A=1, B=1, C=1 -> B OR C = 1 -> 1 XOR 1 = 0

A 00001111 B 00110011 C 01010101 B \lor C 01110111 A \oplus (B \lor C) 01111000

Mentor tip

Evaluate (B OR C) first as an intermediate column before applying the XOR operation with A.

Question 9

2 marks

Hexadecimal 3B representation in an 8-bit register

Computer organisation

(a)

2 marks

Convert hexadecimal digit 3 into 4-bit binary: 3_{16} = 0011_2.
Convert hexadecimal digit B (decimal 11) into 4-bit binary: B_{16} = 1011_2.
Concatenate the two 4-bit nibbles into an 8-bit byte: 00111011.

3 B_{16} = (00111011)_{2}

Mentor tip

Convert each hex character into its 4-bit nibble and ensure all 8 bits including leading zeros are present.

Question 10

4 marks

Trace table for iterative division algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Initialize variables: Given N = 139, L = 3.
Initial values before loop: D = 139 div 3 = 46, Z = 1, B = false.
Iteration 1: Condition Z < L (1 < 3) is true.
Execute loop body: D = 46 div 3 = 15; Z = 1 + 1 = 2; B = NOT false = true.
Iteration 2: Condition Z < L (2 < 3) is true.
Execute loop body: D = 15 div 3 = 5; Z = 2 + 1 = 3; B = NOT true = false.
Iteration 3: Condition Z < L (3 < 3) is false; loop terminates.
Evaluate post-loop condition: if (D != 0 AND B) then output(D, B) else output(Z, NOT B).
Check condition: D = 5 != 0 is true, but B = false. Thus (D != 0 AND B) is false.
Execute else branch: output(Z, NOT B) -> output(3, NOT false) -> output(3, true).

D 46155 Z 123 B false true false Z < L ? true true false output 3, true

Mentor tip

Carefully trace each update: D divides by L, Z increments, and B toggles boolean values on every iteration.

Questions 11–13 · 45 marks

Question 11

15 marks

University examination record system upgrade

System fundamentals

(a)

2 marks

Identify aspect 1: Access permissions and security control — establishing strict authorization hierarchies (e.g., student read-only access for personal transcripts vs exam officer administrative write access).
Identify aspect 2: Data volume, archiving schedules, and storage scalability — accommodating rapid record expansion for current students while archiving two-year-old student papers and preserving historical numerical grades.

Mentor tip

Highlight access privileges and data volume/frequency of access between current and past student records.

(b)

3 marks

Explain direct observation: Analysts directly shadow office staff as they interact with the existing system during everyday administrative routines.
Detail benefits of first-hand insight: It reveals practical workflows, data bottlenecks, user operational errors, and actual usage patterns that staff might overlook or underreport in questionnaires.
Explain how this shapes requirements: Direct qualitative evidence exposes unstated pain points, ensuring the proposed system retains effective current features while rectifying inefficiency and validation flaws.

Mentor tip

Direct observation provides unbiased, first-hand evidence of real user workflows and systemic bottlenecks.

(c)

3 marks

State the purpose of creating a prototype: Presenting a functional early model to the examination office to validate core workflows, forms, and user interfaces before full-scale coding begins.
Explain feedback and iterative refinement: Allowing administrative staff to test navigation, identify missing requirements, and suggest adjustments early, significantly lowering redesign costs.
Accelerate consensus and development: Ensuring technical developers and end-user stakeholders share an identical understanding of system specifications before committing resources.

Mentor tip

Explain that prototypes elicit early user feedback, clarify functional requirements, and prevent costly late-stage revisions.

(d)

4 marks

Problem 1 - Data loss and corruption during transfer: Moving extensive exam records and student grades across network protocols or storage drives introduces risks of incomplete transmissions, corrupted fields, or packet drops, leaving academic records truncated or invalid.
Discussion 1: Resolving data corruption requires robust validation algorithms (such as hash checksum verification), pre-migration database backups, and data integrity audits before decommissioning legacy databases.
Problem 2 - Incompatibility of database formats and data structures: Legacy databases may use obsolete field encodings, deprecated data types, or proprietary schemas that do not map directly into the modern database schema.
Discussion 2: Resolving format mismatches requires writing and thoroughly testing Extraction, Transformation, and Loading (ETL) scripts, which can lead to project delays and elevated implementation costs if exceptions occur.

Mentor tip

Name data loss/corruption and schema/format incompatibility, discussing consequences and validation requirements for each.

(e)

3 marks

Outline the economic cost of parallel running: The organization must finance duplicate infrastructure, running both old servers and new hardware concurrently.
Explain staffing and operational expenses: Staff must enter all grades and examination records twice into both systems to verify parity, demanding substantial overtime wages or temporary staffing.
Discuss risk mitigation value: Although operationally expensive, this investment eliminates the catastrophic economic liability of complete university data loss or operational shutdown should the new system fail unexpectedly.

Mentor tip

Focus on duplicate operational costs (staffing, power, dual entry) balanced against catastrophic failure insurance.

Question 12

15 marks

College campus network and access policies

Networks

(a)

4 marks

Explain reliability of wireless transmission: Wi-Fi signals fluctuate based on physical distance from the access point, attenuation through building walls, electromagnetic interference from surrounding electronic appliances, and radio channel saturation when many students connect simultaneously.
Explain reliability of Ethernet transmission: Wired connections provide consistent, dedicated bandwidth with physical shielding inside twisted-pair cabling, making data transmission stable and immune to external radio interference.
Compare physical vulnerability: Ethernet links remain reliably connected provided cables and physical RJ-45 ports remain undamaged, whereas Wi-Fi experiences intermittent signal drops due to dynamic environmental interference.

Mentor tip

Contrast Wi-Fi susceptibility to radio interference, wall attenuation, and user density with the shielded stability of wired Ethernet.

(b)

4 marks

Feature 1 - Strong End-to-End Encryption: All payload packets traveling through the public internet are encrypted using symmetric ciphers (e.g., AES) so that any packet intercepted by a third party remains indecipherable.
Feature 2 - Tunneling and Encapsulation: VPN protocols (e.g., IPsec, OpenVPN) encapsulate entire original packets inside new IP headers with secure outer routing, concealing internal college IP addresses and packet metadata from eavesdroppers.
Feature 3 - Mutual Authentication: Both the client and the campus VPN gateway must authenticate using credentials, multi-factor tokens, or digital certificates before an authorized communication tunnel is established.

Mentor tip

Select encryption and tunneling/authentication as the two security features, explaining how each prevents unauthorized inspection.

(c)

1 mark

Identify one essential security protocol technology for VPN implementation: IPsec (Internet Protocol Security), TLS (Transport Layer Security), or SSL (Secure Sockets Layer).

Mentor tip

State a recognized tunneling/security protocol such as IPsec, TLS, or SSL.

(d)

6 marks

Advantage 1: Universal device synchronization and collaboration — cloud platforms allow students to co-edit shared project documents simultaneously across personal smartphones, home PCs, and college laptops without manual file copying.
Advantage 2: Modern feature availability and zero maintenance — commercial web services update continuously with advanced toolsets and abundant cloud storage exceeding what the college internal IT department can provision.
Disadvantage 1: Data sovereignty, security, and confidentiality risks — uploading sensitive research, student projects, or personal identification to third-party cloud servers exposes student data to external platform data breaches or unclear intellectual property terms.
Disadvantage 2: Academic misconduct and compliance issues — public blogs and unmonitored external platforms can facilitate plagiarism, defamatory postings, or cyberbullying outside the moderation and safeguarding purview of college IT policy.

Mentor tip

Balance collaborative flexibility and ubiquitous access against institutional compliance, data ownership, and moderation challenges.

Question 13

15 marks

Half-marathon team registration and results aggregation

Computational thinking, problem-solving and programming

(a)

1 mark

Analyze team constraints: Total participants = 450; each team must contain at least 3 participants.
Calculate minimum capacity needed for collection TNAMES:
Maximum possible number of distinct teams occurs when every team has the minimum allowed size of 3 members: 450 / 3 = 150 teams.

Minimum size of TNAMES = \frac{450}{3} = 150

Mentor tip

Divide total participants by minimum team size (450 / 3 = 150) to find maximum possible distinct teams.

(b)

1 mark

From the problem diagram, the first participant for team QA registered at index 11.
In the array TEAM at index 11, the pointer points to the second team member: TEAM[11] = 73.
Therefore, the second participant for team QA is located at position 73 in the PARTICIPANTS array.

Mentor tip

Look up TEAM[11] which holds the index pointer 73 into PARTICIPANTS for the next teammate.

(c)

7 marks

Sub-part (i) [3 marks] - Find first participant index MINP and initialize TEAM head:
Pseudocode:
FOUND = false
loop while FOUND == false AND P < 450
if PARTICIPANTS[P] == CURRENT then
FOUND = true
MINP = P
T = MINP
else
P = P + 1
end if
end loop
Sub-part (ii) [4 marks] - Find remaining team members and close the pointer path:
Pseudocode:
P = MINP + 1
loop while P < 450
if PARTICIPANTS[P] == CURRENT then
TEAM[T] = P
T = P
end if
P = P + 1
end loop
TEAM[T] = MINP // Close cycle back to first member

Mentor tip

In (c)(i) search for the first match to set MINP and T; in (c)(ii) scan remaining indices, linking successive matches and pointing the final member back to MINP.

(d)

6 marks

Algorithm to determine the winning team:
1) Initialize tracking variables: BEST_TEAM = "", LOWEST_TIME = 999999 (or suitable high sentinel value).
2) Reset TNAMES collection iterator to beginning: TNAMES.resetNext().
3) Loop while TNAMES.hasNext():
CURRENT_TEAM = TNAMES.getNext()
Find first team member index in PARTICIPANTS; let START = that index.
Initialize dynamic list/array MEMBER_TIMES.
Traverse the team cycle in TEAM array starting at START:
Read participant finishing time: TIME = TIMING[CURRENT_INDEX].
If TIME > 120 minutes (or did not finish), assign penalty default: TIME = 1000.
Append TIME to MEMBER_TIMES.
Advance CURRENT_INDEX = TEAM[CURRENT_INDEX] until CURRENT_INDEX == START.
Pass MEMBER_TIMES to subroutine: TEAM_SCORE = sum3best(MEMBER_TIMES).
If TEAM_SCORE < LOWEST_TIME then:
LOWEST_TIME = TEAM_SCORE
BEST_TEAM = CURRENT_TEAM
4) Output BEST_TEAM and LOWEST_TIME.

Mentor tip

Traverse each team via the circular TEAM linked path, gather finishing times from TIMING[], call sum3best(), and track the minimum score.

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 fundamentals25 marks · 36%
Q1, Q2, Q3, Q4, Q11
Computational thinking, problem-solving and programming19 marks · 27%
Q10, Q13
Networks15 marks · 21%
Q12
Computer organisation9 marks · 13%
Q5, Q6, Q8, Q9
Resource management2 marks · 3%
Q7

Mentor insights

Traps to watch for

01
Trace table loop termination and condition branch · Q10
When Z increments to 3, the loop condition Z < 3 is false and terminates immediately. The post-loop check evaluates (D != 0 AND B); because B is false, execution falls into the else branch, outputting (Z, NOT B) = (3, true).
02
TEAM closed path cyclic termination · Q13
When constructing the TEAM array, the final participant of each team must point back to MINP to close the cycle. Forgetting to link TEAM[T] = MINP leaves an open path and breaks traversal loops.
03
XOR truth table parenthesized sub-expression · Q8
Ensure (B OR C) is evaluated first before XORing with A. When A = 1 and (B OR C) = 1, 1 XOR 1 yields 0.

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 SL · Paper 1

Computer Science, Standard Level

Session: May 2016
Marks: 70
Duration: 1 hour 30 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 May 2016 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

2 marks

Maintenance challenges of legacy systems

System fundamentals

(a)

2 marks

Define a legacy system as an outdated computer hardware setup or obsolete software application that remains in active operational use.
Explain one maintenance problem: modifying legacy systems is difficult and costly due to missing or inaccurate documentation, obsolete programming languages, and a shortage of developers with specialized legacy expertise.

Mentor tip

Clarify that legacy systems use outdated technology and explicitly link to missing documentation or scarce programming expertise.

Question 2

2 marks

Concept and purpose of user operational sign-off testing

System fundamentals

(a)

2 marks

Define User Operational Sign-off Testing (UAT): providing end users with a functional release or prototype of the system in a real or simulated operational environment.
Explain its purpose: verifying that the software satisfies actual business requirements, meets user expectations, and gathers feedback on interface usability before final deployment.

Mentor tip

Highlight that actual end users test the software against business requirements before final sign-off.

Question 3

4 marks

Printed documentation versus online user support

System fundamentals

(a)

4 marks

State one advantage of printed documentation: Portability and physical availability — printed manuals do not require an active internet connection, power supply, or functioning display, making them accessible during complete hardware or power failures.
Elaborate on printed advantage: Physical reference cards or setup guides can be kept right next to new peripheral hardware (e.g., printers or scanners) during physical assembly.
State one disadvantage of printed documentation: Lack of interactivity and difficult searchability — hardcopy manuals cannot be hyperlinked, cannot easily be searched via keywords, and cannot be updated dynamically without reprinting.
Elaborate on printed disadvantage: Printed manuals can rapidly become obsolete as software receives patches and interface changes, leading to outdated instructions and environmental paper waste.

Mentor tip

Provide one advantage (e.g., constant offline availability) and one disadvantage (e.g., difficulty updating, poor searchability), expanding each fully.

Question 4

2 marks

Role and mechanism of a failover system

System fundamentals

(a)

2 marks

Identify that a failover system maintains a redundant, standby secondary server or system constantly synchronized with the primary platform.
Describe how it operates: when a hardware fault, power disruption, or software crash disables the primary system, the failover mechanism seamlessly switches operations to the secondary system automatically, avoiding disruption to end users.

Mentor tip

Emphasize automatic takeover by a redundant standby system during primary failure to preserve availability.

Question 5

2 marks

Role of the Control Unit (CU) in the CPU

Computer organisation

(a)

2 marks

State the primary function of the Control Unit: fetching instructions from main memory (RAM) and decoding them into micro-operations and control signals.
Outline its coordination role: directing the timing and movement of data between CPU registers, the Arithmetic Logic Unit (ALU), and input/output interfaces across the control bus.

Mentor tip

Focus on instruction fetching/decoding and issuing synchronization control signals to CPU components.

Question 6

2 marks

Cache memory mechanism for processor speedup

Computer organisation

(a)

2 marks

Describe cache location and speed: Cache memory is ultra-fast static RAM (SRAM) integrated directly on or adjacent to the CPU core, offering significantly faster access times than system RAM.
Explain how it accelerates execution: By staging frequently executed loop instructions and recently accessed data in cache, the CPU avoids high-latency bus trips to main memory, reducing idle wait cycles.

Mentor tip

Explain that holding frequently used data/instructions in fast on-chip SRAM minimizes latency compared to accessing RAM.

Question 7

2 marks

Operating system features for video game execution

Resource management

(a)

2 marks

Identify an operating system feature: Memory management and dedicated graphics driver interfacing.
Explain its importance for gaming: Video games render high-resolution textures and complex 3D assets in real time, requiring the OS to efficiently allocate large RAM buffers, manage GPU memory transfers, and maintain steady frame rates without stuttering.

Mentor tip

Mention memory management or device/GPU driver coordination required to handle continuous high-bandwidth frame rendering.

Question 8

3 marks

Truth table construction for A XOR (B OR C)

Computer organisation

(a)

3 marks

Evaluate the sub-expression (B OR C) for all 8 input combinations of binary inputs A, B, and C.
Apply the exclusive-OR (XOR) operator between input A and the intermediate result (B OR C), outputting 1 if and only if exactly one operand is true:
Row 1: A=0, B=0, C=0 -> B OR C = 0 -> 0 XOR 0 = 0
Row 2: A=0, B=0, C=1 -> B OR C = 1 -> 0 XOR 1 = 1
Row 3: A=0, B=1, C=0 -> B OR C = 1 -> 0 XOR 1 = 1
Row 4: A=0, B=1, C=1 -> B OR C = 1 -> 0 XOR 1 = 1
Row 5: A=1, B=0, C=0 -> B OR C = 0 -> 1 XOR 0 = 1
Row 6: A=1, B=0, C=1 -> B OR C = 1 -> 1 XOR 1 = 0
Row 7: A=1, B=1, C=0 -> B OR C = 1 -> 1 XOR 1 = 0
Row 8: A=1, B=1, C=1 -> B OR C = 1 -> 1 XOR 1 = 0

A 00001111 B 00110011 C 01010101 B \lor C 01110111 A \oplus (B \lor C) 01111000

Mentor tip

Evaluate (B OR C) first as an intermediate column before applying the XOR operation with A.

Question 9

2 marks

Hexadecimal 3B representation in an 8-bit register

Computer organisation

(a)

2 marks

Convert hexadecimal digit 3 into 4-bit binary: 3_{16} = 0011_2.
Convert hexadecimal digit B (decimal 11) into 4-bit binary: B_{16} = 1011_2.
Concatenate the two 4-bit nibbles into an 8-bit byte: 00111011.

3 B_{16} = (00111011)_{2}

Mentor tip

Convert each hex character into its 4-bit nibble and ensure all 8 bits including leading zeros are present.

Question 10

4 marks

Trace table for iterative division algorithm

Computational thinking, problem-solving and programming

(a)

4 marks

Initialize variables: Given N = 139, L = 3.
Initial values before loop: D = 139 div 3 = 46, Z = 1, B = false.
Iteration 1: Condition Z < L (1 < 3) is true.
Execute loop body: D = 46 div 3 = 15; Z = 1 + 1 = 2; B = NOT false = true.
Iteration 2: Condition Z < L (2 < 3) is true.
Execute loop body: D = 15 div 3 = 5; Z = 2 + 1 = 3; B = NOT true = false.
Iteration 3: Condition Z < L (3 < 3) is false; loop terminates.
Evaluate post-loop condition: if (D != 0 AND B) then output(D, B) else output(Z, NOT B).
Check condition: D = 5 != 0 is true, but B = false. Thus (D != 0 AND B) is false.
Execute else branch: output(Z, NOT B) -> output(3, NOT false) -> output(3, true).

D 46155 Z 123 B false true false Z < L ? true true false output 3, true

Mentor tip

Carefully trace each update: D divides by L, Z increments, and B toggles boolean values on every iteration.

Questions 11–13 · 45 marks

Question 11

15 marks

University examination record system upgrade

System fundamentals

(a)

2 marks

Identify aspect 1: Access permissions and security control — establishing strict authorization hierarchies (e.g., student read-only access for personal transcripts vs exam officer administrative write access).
Identify aspect 2: Data volume, archiving schedules, and storage scalability — accommodating rapid record expansion for current students while archiving two-year-old student papers and preserving historical numerical grades.

Mentor tip

Highlight access privileges and data volume/frequency of access between current and past student records.

(b)

3 marks

Explain direct observation: Analysts directly shadow office staff as they interact with the existing system during everyday administrative routines.
Detail benefits of first-hand insight: It reveals practical workflows, data bottlenecks, user operational errors, and actual usage patterns that staff might overlook or underreport in questionnaires.
Explain how this shapes requirements: Direct qualitative evidence exposes unstated pain points, ensuring the proposed system retains effective current features while rectifying inefficiency and validation flaws.

Mentor tip

Direct observation provides unbiased, first-hand evidence of real user workflows and systemic bottlenecks.

(c)

3 marks

State the purpose of creating a prototype: Presenting a functional early model to the examination office to validate core workflows, forms, and user interfaces before full-scale coding begins.
Explain feedback and iterative refinement: Allowing administrative staff to test navigation, identify missing requirements, and suggest adjustments early, significantly lowering redesign costs.
Accelerate consensus and development: Ensuring technical developers and end-user stakeholders share an identical understanding of system specifications before committing resources.

Mentor tip

Explain that prototypes elicit early user feedback, clarify functional requirements, and prevent costly late-stage revisions.

(d)

4 marks

Problem 1 - Data loss and corruption during transfer: Moving extensive exam records and student grades across network protocols or storage drives introduces risks of incomplete transmissions, corrupted fields, or packet drops, leaving academic records truncated or invalid.
Discussion 1: Resolving data corruption requires robust validation algorithms (such as hash checksum verification), pre-migration database backups, and data integrity audits before decommissioning legacy databases.
Problem 2 - Incompatibility of database formats and data structures: Legacy databases may use obsolete field encodings, deprecated data types, or proprietary schemas that do not map directly into the modern database schema.
Discussion 2: Resolving format mismatches requires writing and thoroughly testing Extraction, Transformation, and Loading (ETL) scripts, which can lead to project delays and elevated implementation costs if exceptions occur.

Mentor tip

Name data loss/corruption and schema/format incompatibility, discussing consequences and validation requirements for each.

(e)

3 marks

Outline the economic cost of parallel running: The organization must finance duplicate infrastructure, running both old servers and new hardware concurrently.
Explain staffing and operational expenses: Staff must enter all grades and examination records twice into both systems to verify parity, demanding substantial overtime wages or temporary staffing.
Discuss risk mitigation value: Although operationally expensive, this investment eliminates the catastrophic economic liability of complete university data loss or operational shutdown should the new system fail unexpectedly.

Mentor tip

Focus on duplicate operational costs (staffing, power, dual entry) balanced against catastrophic failure insurance.

Question 12

15 marks

College campus network and access policies

Networks

(a)

4 marks

Explain reliability of wireless transmission: Wi-Fi signals fluctuate based on physical distance from the access point, attenuation through building walls, electromagnetic interference from surrounding electronic appliances, and radio channel saturation when many students connect simultaneously.
Explain reliability of Ethernet transmission: Wired connections provide consistent, dedicated bandwidth with physical shielding inside twisted-pair cabling, making data transmission stable and immune to external radio interference.
Compare physical vulnerability: Ethernet links remain reliably connected provided cables and physical RJ-45 ports remain undamaged, whereas Wi-Fi experiences intermittent signal drops due to dynamic environmental interference.

Mentor tip

Contrast Wi-Fi susceptibility to radio interference, wall attenuation, and user density with the shielded stability of wired Ethernet.

(b)

4 marks

Feature 1 - Strong End-to-End Encryption: All payload packets traveling through the public internet are encrypted using symmetric ciphers (e.g., AES) so that any packet intercepted by a third party remains indecipherable.
Feature 2 - Tunneling and Encapsulation: VPN protocols (e.g., IPsec, OpenVPN) encapsulate entire original packets inside new IP headers with secure outer routing, concealing internal college IP addresses and packet metadata from eavesdroppers.
Feature 3 - Mutual Authentication: Both the client and the campus VPN gateway must authenticate using credentials, multi-factor tokens, or digital certificates before an authorized communication tunnel is established.

Mentor tip

Select encryption and tunneling/authentication as the two security features, explaining how each prevents unauthorized inspection.

(c)

1 mark

Identify one essential security protocol technology for VPN implementation: IPsec (Internet Protocol Security), TLS (Transport Layer Security), or SSL (Secure Sockets Layer).

Mentor tip

State a recognized tunneling/security protocol such as IPsec, TLS, or SSL.

(d)

6 marks

Advantage 1: Universal device synchronization and collaboration — cloud platforms allow students to co-edit shared project documents simultaneously across personal smartphones, home PCs, and college laptops without manual file copying.
Advantage 2: Modern feature availability and zero maintenance — commercial web services update continuously with advanced toolsets and abundant cloud storage exceeding what the college internal IT department can provision.
Disadvantage 1: Data sovereignty, security, and confidentiality risks — uploading sensitive research, student projects, or personal identification to third-party cloud servers exposes student data to external platform data breaches or unclear intellectual property terms.
Disadvantage 2: Academic misconduct and compliance issues — public blogs and unmonitored external platforms can facilitate plagiarism, defamatory postings, or cyberbullying outside the moderation and safeguarding purview of college IT policy.

Mentor tip

Balance collaborative flexibility and ubiquitous access against institutional compliance, data ownership, and moderation challenges.

Question 13

15 marks

Half-marathon team registration and results aggregation

Computational thinking, problem-solving and programming

(a)

1 mark

Analyze team constraints: Total participants = 450; each team must contain at least 3 participants.
Calculate minimum capacity needed for collection TNAMES:
Maximum possible number of distinct teams occurs when every team has the minimum allowed size of 3 members: 450 / 3 = 150 teams.

Minimum size of TNAMES = \frac{450}{3} = 150

Mentor tip

Divide total participants by minimum team size (450 / 3 = 150) to find maximum possible distinct teams.

(b)

1 mark

From the problem diagram, the first participant for team QA registered at index 11.
In the array TEAM at index 11, the pointer points to the second team member: TEAM[11] = 73.
Therefore, the second participant for team QA is located at position 73 in the PARTICIPANTS array.

Mentor tip

Look up TEAM[11] which holds the index pointer 73 into PARTICIPANTS for the next teammate.

(c)

7 marks

Sub-part (i) [3 marks] - Find first participant index MINP and initialize TEAM head:
Pseudocode:
FOUND = false
loop while FOUND == false AND P < 450
if PARTICIPANTS[P] == CURRENT then
FOUND = true
MINP = P
T = MINP
else
P = P + 1
end if
end loop
Sub-part (ii) [4 marks] - Find remaining team members and close the pointer path:
Pseudocode:
P = MINP + 1
loop while P < 450
if PARTICIPANTS[P] == CURRENT then
TEAM[T] = P
T = P
end if
P = P + 1
end loop
TEAM[T] = MINP // Close cycle back to first member

Mentor tip

In (c)(i) search for the first match to set MINP and T; in (c)(ii) scan remaining indices, linking successive matches and pointing the final member back to MINP.

(d)

6 marks

Algorithm to determine the winning team:
1) Initialize tracking variables: BEST_TEAM = "", LOWEST_TIME = 999999 (or suitable high sentinel value).
2) Reset TNAMES collection iterator to beginning: TNAMES.resetNext().
3) Loop while TNAMES.hasNext():
CURRENT_TEAM = TNAMES.getNext()
Find first team member index in PARTICIPANTS; let START = that index.
Initialize dynamic list/array MEMBER_TIMES.
Traverse the team cycle in TEAM array starting at START:
Read participant finishing time: TIME = TIMING[CURRENT_INDEX].
If TIME > 120 minutes (or did not finish), assign penalty default: TIME = 1000.
Append TIME to MEMBER_TIMES.
Advance CURRENT_INDEX = TEAM[CURRENT_INDEX] until CURRENT_INDEX == START.
Pass MEMBER_TIMES to subroutine: TEAM_SCORE = sum3best(MEMBER_TIMES).
If TEAM_SCORE < LOWEST_TIME then:
LOWEST_TIME = TEAM_SCORE
BEST_TEAM = CURRENT_TEAM
4) Output BEST_TEAM and LOWEST_TIME.

Mentor tip

Traverse each team via the circular TEAM linked path, gather finishing times from TIMING[], call sum3best(), and track the minimum score.

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 fundamentals25 marks · 36%
Q1, Q2, Q3, Q4, Q11
Computational thinking, problem-solving and programming19 marks · 27%
Q10, Q13
Networks15 marks · 21%
Q12
Computer organisation9 marks · 13%
Q5, Q6, Q8, Q9
Resource management2 marks · 3%
Q7

Mentor insights

Traps to watch for

01
Trace table loop termination and condition branch · Q10
When Z increments to 3, the loop condition Z < 3 is false and terminates immediately. The post-loop check evaluates (D != 0 AND B); because B is false, execution falls into the else branch, outputting (Z, NOT B) = (3, true).
02
TEAM closed path cyclic termination · Q13
When constructing the TEAM array, the final participant of each team must point back to MINP to close the cycle. Forgetting to link TEAM[T] = MINP leaves an open path and breaks traversal loops.
03
XOR truth table parenthesized sub-expression · Q8
Ensure (B OR C) is evaluated first before XORing with A. When A = 1 and (B OR C) = 1, 1 XOR 1 yields 0.

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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