Computer Science 9618/32 October/November 2022: question paper and mark scheme

Mark scheme

1(a) Mantissa Exponent 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1(b) Two marks for working 3 • correct calculation of exponent seen • correct application of exponent to mantissa seen One mark for correct answer Working: = 1.0110010011 × 29 //exponent = 9 = 1011001001.1 (moving bp 9 places to right) // evaluate two’s complement For example: –512 + 128 + 64 + 8 + 1 + 0.5 Answer: –310.5 // –3101/2 1(c) One mark per point 2 • Following an arithmetic/logical operation • … the result is too small to be precisely represented in the available system // When the number of bits is not enough / too small for the computer’s allocated word size / to represent the binary number

Mark scheme

2(a) One mark per point 2 • Code generation • Optimisation 2(b) One mark per point (Max 2) 2 • It checks that the code matches the grammar of the language // It checks that the tokens conform with the rules of the programming language • Syntax errors are reported • A parse tree is produced.

Mark scheme

3(a) One mark per point (Max 2) 2 • Protocols set a standard for communication • Protocols enable communication/compatibility between devices from different manufacturers/platforms • If two devices were sending messages to each other but using different protocols, they would not be able to communicate properly 3(b) One mark for each correct answer 2 Application (Layer) Transport Internet/Network (Layer) Link 3(c) One mark per point (Max 2) 2 • used by email clients to retrieve email messages // a pull protocol • from a mail server (over a TCP/IP connection) • keeps the server and client in sync (by not deleting the original email). // allows a copy of the email to be downloaded from the mail server.

Mark scheme

4(a) Marks as shown in the square brackets: 2 TYPE Aircraft = (C300, C350, D242, E757, X380) TYPE Aircraft = [1] (C300, C350, D242, E757, X380) [1] 4(b) One mark for each point (Max 4) 4 • TYPE Flight and ENDTYPE correct • DECLARE FlightNumber and DECLARE Destination as STRING • DECLARE DepartureDate as DATE • DECLARE AircraftType as Aircraft (correct data type from part 4(a)) Example answer: TYPE Flight DECLARE FlightNumber : STRING DECLARE Destination : STRING DECLARE DepartureDate : DATE DECLARE AircraftType : Aircraft ENDTYPE 4(c)(i) Example answer: 1 DECLARE Flight1 : Flight 4(c)(ii) One mark for each point (Max 3) 3 • Correct assignments of both string data values • Correct assignments of date data value • Correct assignments of enumerated data value Example answer: Flight1.FlightNumber ← "XA782" Flight1.Destination ← "Cambridge" Flight1.DepartureDate ← 12/12/2022 Flight1.AircraftType ← C350

Mark scheme

5 One mark for each description (Max 2) 6 • The emulation of a computer system / hardware and/or software • … using a host computer system. • Using guest operating system(s) for emulation. One mark for each correct benefit (Max 2) • Multiple guest operating systems / VMs can be used on the same computer. • Different instruction set architectures can be emulated on a single computer. • A virtual machine can crash without affecting the host machine. • There are security benefits // Trying a piece of suspicious software and if it is / has a virus, it will only infect the virtual machine. • Cost savings due to not needing to purchase extra hardware. • Can run legacy applications that are currently incompatible. One mark for each correct drawback (Max 2) • A virtual machine is less efficient / has poorer performance than real machines because of extra load on the host computer • Performance of the guest system cannot be adequately measured. • A virtual machine may be affected by any weaknesses of the host machine. • Costly and/or complex to maintain / implement / manage. • Cannot emulate some hardware.

Mark scheme

6(a) One mark for each point 2 • Symmetric encryption uses a single key and asymmetric encryption uses a pair of keys. • The symmetric single key is used by all, whereas only one of the keys for asymmetric encryption is available to everyone / one of the asymmetric encryption keys needs to be kept secret. 6(b) One mark for each point (Max 4) 4 • The organisation requests a certificate from a Certificate Authority (CA) • The organisation may send their public key to CA • The organisation gathers all the information required by the CA in order to obtain their certificate, which includes information to prove their identity • The CA verifies the organisation’s identity • The CA generates / issues the certificate including the organisation’s public key (and other information).

Mark scheme

7 One mark for each point 4 Supervised learning (Max 3 of 4) • Supervised learning allows data to be collected, or a data output produced, from the previous experience. • In supervised learning, known input and associated outputs are given // uses sample data with known outputs (in training) // uses labelled input data. • Able to predict future outcomes based on past data. Unsupervised learning (Max 3 of 4) • Unsupervised machine learning helps all kinds of unknown patterns in data to be found. • Unsupervised learning only requires input data to be given. • Uses any data // not trained on the right output // uses unlabelled input data.

Mark scheme

8(a) One mark for each point (Max 4) 4 • One correct NAND or NOR gate with two separate inputs and one output • Second correct logic gate of same type as first with two separate inputs and one output • Correct connections between logic gates • Correctly labelled inputs Example answers: S/Set R/Reset 8(a) Or R/Reset S/Set 8(b) To store a binary digit / (single) bit. 1 8(c) One mark for each point (Max 3) 3 • Correct application of De Morgan’s Law • Correct application of Double Negation Law or Distributive Law • Correct answer (A.B).(A.C).(B.D) (A.B) + (A.C) + (B.D) [1] _ _ _ (A.B) + (A.C) + (B.D) [1] _ _ A. (B + C) + B.D [1]

Mark scheme

9(a) One mark for each point (Max 3) 3 • Process scheduling allows more than one program/task to appear to be executed at the same time / enables multi- tasking / multiprogramming. • To allow high priority jobs to be completed first. • To keep the CPU busy all the time • … to ensure that all processes execute efficiently • … and to have reduced wait times for all processes / to ensure all processes have fair access to the CPU / prevent starvation of some processes. 9(b) One mark for each point (Max 2) for Shortest job first: 6 • Process are executed in ascending order of the amount of CPU time required // Short processes are executed first and followed by longer processes. • …which leads to an increased throughput (because more processes can be executed in a smaller amount of time). One mark for each point (Max 2) for Round robin: • Each process is served by the CPU for a fixed time/time slice (so all processes are given the same priority). • Starvation doesn’t occur (because for each round robin cycle, every process is given a fixed time/time slice to execute). One mark for each point (Max 2) for First come first served: • No complex logic, each process request is queued as it is received and executed one by one. • Starvation doesn’t occur (because every process will eventually get a chance to run) // less processor overhead.

Mark scheme

10(a) One mark for each correct OOP term definition 3 • Encapsulation – putting properties and methods inside a class // ensures sensitive data is hidden from users by hiding values of a structured data object inside a class. • Getter – method that is used to return the value of a property. • Setter – method that is used to update the value of a property. 10(b) One mark for each point 3 • properties correct • setters correct • getters correct SubstituteTeacher SubName : STRING Telephone : STRING InSchool : BOOLEAN SetSubName(StaffName : STRING) SetTelephone(Tel : STRING) SetInSchool(Present : BOOLEAN) GetSubName() GetTelephone() GetInSchool()

Mark scheme

11(a) One mark for each point 3 • Correct HeadPointer • Any three correct NextPointer • All six NextPointer correct HeadPointer 5 Index Flower NextPointer 1 Rose 0 2 Marigold 3 Foxglove 4 4 Iris 9 5 Daisy 3 6 Dahlia 7 Saxifrage 8 Lupin 1 9 Lily 8 10 Hydrangea 11(b) One mark for each correct line 5 Pointer ← HeadPointer Found ← 0 OUTPUT "Enter a flower name " INPUT FlowerName WHILE Pointer <> 0 IF Flower[Pointer] = FlowerName THEN Found ← Pointer Pointer ← 0 ELSE Pointer ← NextPointer[Pointer] ENDIF ENDWHILE IF Pointer = 0 THEN // IF Found <> 0 THEN OUTPUT Flower[Found], " is found" ELSE OUTPUT "The flower you wanted is not in the list" ENDIF 11(c) One mark for each point (Max 2) 2 • Include a free list pointer • …to reuse the unused space • …as a linked list of free space.