Computer Science 9618/32 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 3 Advanced Theory, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Describe how packet switching is used to transmit messages across a network.
…… [3]
(b) State two benefits and two drawbacks of packet switching as a method of transmitting messages across a network.
Benefit 1 ……
Benefit 2 ……
Drawback 1 ……
Drawback 2 ……
[4]
Mark scheme
1(a) One mark per mark point (Max 3) 3
MP1 The data to be transmitted is divided into equal sized packets
MP2 A packet header is attached to each packet containing key information
MP3 … such as source/destination IP addresses, packet number, etc
MP4 Packets are transmitted independently
MP5 … and may travel though different routes/paths to the destination
MP6 Routes are determined using a routing table//Packets take the optimum route depending on congestion
MP7 The packets usually arrive out of order
MP8 The packets are reassembled in the correct order at the destination // The packets are re-ordered using the
sequence number/the header
MP9 If packets are missing/corrupted a re-transmission request is sent / packets are re-sent.
1(b) One mark for each benefit (Max 2) 4
MP1 Packets are more likely to arrive because they can be re-routed if a problem occurs with one of the routes//Packets
are more likely to arrive because if a packet is lost, it can be re-transmitted
MP2 Bandwidth can be shared allowing packets from different messages to share the same path
MP3 Considered secure as the packets generally travel via different routes
MP4 High data transmission rate is possible
One mark for each drawback (Max 2)
MP5 Time delay because packets need to be re-ordered/reassembled at the destination//Time delay caused by missing
packets needing to be re-sent//Time delay because it has to share the bandwidth of the circuit / channel with
other packets
MP6 Requires a complex algorithm to function
MP7 Needs lots of RAM to handle large amounts of data.
(a) Describe serial file organisation as a method of storing data records in a file.
…… [2]
(b) State one example of a use for serial file organisation.
…… [1]
Mark scheme
2(a) One mark per mark point (Max 2) 2
MP1 Records are stored one after the other as they are collected // records are stored in chronological order
MP2 New records are appended to the end of the file.
2(b) One mark for a use (Max 1) 1
MP1 Creating unsorted / temporary transaction files
MP2 Creating data logging files
(a) Describe the user‑defined data type record.
…… [3]
(b) A programmer defines a record, Order, to store the following data:
• account number
• order number
• order price
• order date.
Write pseudocode statements to define this record.
…… [4]
Mark scheme
3(a) One mark per mark point (Max 3) 3
MP1 A user-defined record data type is a composite data type
MP2 It uses other data types in its definition to form a single new data type
MP3 The data types referenced may be primitive data types from a programming language or they may be other user-
defined data types.
MP4 Includes related items
MP5 Includes a fixed number of items.
3(b) One mark for TYPE Order and ENDTYPE correct 4
One mark for correct use of DECLARE in all declarations seen
One mark for the two shaded declarations
One mark for the two unshaded declarations
Example answer
TYPE Order
DECLARE AccountNumber : INTEGER
DECLARE OrderNumber : INTEGER
DECLARE OrderPrice : REAL
DECLARE OrderDate : DATE
ENDTYPE
Numbers are stored in a computer using binary floating‑point representation with:
• 12 bits for the mantissa
• 4 bits for the exponent
• two’s complement form for both the mantissa and the exponent.
(a) Calculate the denary value of the given normalised binary floating‑point number.
Show your working.
Mantissa Exponent
0 1 0 0 0 1 1 1 0 1 1 1 0 1 1 1
Working ……
Answer ……
[2]
(b) Calculate the normalised binary floating‑point representation of – 49.1875 in this system.
Show your working.
Mantissa Exponent
Working ……
[4]
Mark scheme
4(a) One mark for working 2
• application of exponent to mantissa to go from 0.10001110111 to 01000111.0111 //
moving the binary point 7 places //
multiplying by 27/128 in the fractions method //
64 + 4 + 2 +1 + .25+ .125 + .0625 seen
One mark for correct answer
• 71.4375
// 717/16
4(b) One mark per mark point (Max 2) 4
• correct mantissa – exact answer only
• correct exponent – exact answer only
Mantissa Exponent
1 0 0 1 1 1 0 1 1 0 1 0 0 1 1 0
One mark per mark point for working (Max 2)
• number converted to binary e.g., positive binary version of 49.1875 = 0110001.0011 //
two’s complement version bits flipped and 1 added = 1001110.1101 //
–64 + 8 + 4 + 2 + .5 + .25 + .0625 //
–64 + 14.8125
• use of the exponent e.g. moving the binary point 6 places / × 26.
(a) Name and describe two protocols used by the Application Layer of the TCP/IP protocol suite.
Protocol 1 ……
Description ……
Protocol 2 ……
Description ……
[4]
(b) Explain the purpose and function of the Application Layer in the TCP/IP protocol suite.
…… [3]
Mark scheme
5(a) One mark for each correct name (Max 2) 4
One mark for each correct corresponding expansion (Max 2)
MP1 HTTP/HTTPS
MP2 For sending and receiving / transferring web pages / hypertext
MP3 FTP
MP4 For sending and receiving files over a network / between devices
MP5 POP3
MP6 Pull protocol / for receiving / downloading emails
MP7 IMAP
MP8 Pull protocol / for receiving / downloading emails
MP9 SMTP
MP10 Push protocol / for sending / uploading emails
MP11 BitTorrent
MP12 Peer-to-peer file sharing over a network
5(b) One mark per mark point (Max 3) 3
MP1 The application layer provides access to all the programs that exchange data // Interacts directly with user.
MP2 … used by, for example, web browsers, server software.
MP3 Communicates/enables data transfer to/from Transport layer // It allows applications to access the services used in
other TCP/IP layers.
MP4 It defines the protocols that any application uses to allow the exchange of data.
The truth table for a logic circuit is shown.
INPUT OUTPUT
A B C D X
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
(a) Write the Boolean logic expression that corresponds to the given truth table as the
sum‑of‑products.
X = ……
…… [3]
(b) Complete the Karnaugh map (K‑map) for the given truth table.
AB
CD
00 01 11 10
00
01
11
10
[2]
(c) Draw loop(s) around appropriate group(s) in the K‑map to produce an optimal sum‑of‑products.
[2]
(d) Write the Boolean logic expression from your answer to part (c) as the simplified sum‑of‑products.
X = ……
…… [2]
Mark scheme
6(a) Three marks for all eight correct products and no additional products 3
Two marks for five, six or seven correct products
One mark for three or four correct products
(X =) A. B. C. D + A. B. C. D + A. B. C. D. + A. B. C. D + A. B. C. D + A. B. C. D + A. B. C. D + A. B. C. D
6(b) Two marks if no errors present 2
One mark if one error present
AB
CD 00 01 11 10
00 0 0 0 0
01 1 1 1 1
11 0 0 0 0
10 1 1 1 1
6(c) One mark for each correct loop (Max 2) 2
AB
CD 00 01 11 10
00 0 0 0 0
01 1 1 1 1
11 0 0 0 0
10 1 1 1 1
6(d) One mark for each mark point (Max 2) 2
MP1 Any correct relevant Boolean term
MP2 Boolean terms with correct operator + and no other terms present
(X =) C. D + C. D // C. D. + C. D
Several syntax diagrams are shown.
odd even
1 0
3 2
5 4
7 6
9 8
symbol letter
% A
£ B
# C
@ D
$ E
number odd odd
even
(a) State why each number is invalid for the given syntax diagrams.
21
Reason ……
123
Reason ……
[2]
(b) Complete the Backus‑Naur Form (BNF) for the given syntax diagrams.
<symbol> ::= ……
<number> ::= ……
[2]
(c) A new syntax rule, code, is required. It must begin with a letter, followed by one or two numbers, and end with a symbol.
(i) Draw a syntax diagram for code.
[3]
(ii) Write the BNF for code.
…… [2]
Mark scheme
7(a) One mark per mark point (Max 2) 2
MP1 21 - a number must begin with an odd digit, 2 is even
MP2 123 - a number can only be one or two digits in length not three
7(b) One mark per mark point (Max 2) 2
MP1 <symbol> ::= % | £ | # | @ | $
MP2 <number> ::= <odd>|<odd><even>|<odd><odd>
7(c)(i) One mark per mark point (Max 3) 3
MP1 letter, number and symbol all included in correct order: letter first, followed by number, finishing with symbol
MP2 provision for one or two numbers including relevant connectors
MP3 all other connections and label correct and no additional data
Example answer
code
letter number number symbol
7(c)(ii) One mark per correct line (Max 2) 2
<code> ::= <letter><number><symbol>|
<letter><number><number><symbol>
OR
<code> ::= <letter><number><number><symbol>|
<letter><number><symbol>
Alternative Answer
One mark per correct line (Max 2)
<digits> ::= <number>|<number><number>
<code> ::= <letter><digits><symbol>
Complex Instruction Set Computer (CISC) is a type of processor.
Identify four features of a CISC processor.
1 ……
2 ……
3 ……
4 ……
[4]
Mark scheme
8 One mark per mark point (Max 4) 4
MP1 many instruction formats possible
MP2 large instruction set
MP3 many addressing modes available
MP4 uses variable length/multi-operation instructions
MP5 multi-clock cycle instructions
MP6 complex decoding of instructions
MP7 uses complex circuits
MP8 makes frequent use of cache memory
MP9 uses programmable control unit // uses micro-programmed control unit // uses hardwired control unit
MP10 hardware needs to be able to handle more complex instructions convert into sub-instructions // Design emphasis is
on the hardware.
(a) The kernel is the central component of an Operating System (OS).
Outline how the kernel of an OS acts as an interrupt handler.
…… [2]
(b) (i) State what is meant by the term multi‑tasking in an Operating System.
…… [1]
(ii) Describe how multi‑tasking is implemented in an Operating System.
…… [2]
Mark scheme
9(a) One mark per mark point (Max 2) 2
MP1 the kernel receives a signal when an interrupt is generated
MP2 the kernel checks the priority and reviews the status/priority of the current interrupts
MP3 system enters kernel mode if the type of interrupt is of higher priority than the current process
MP4 the kernel consults the interrupt dispatch table / IDT
MP5 … and saves the state of the interrupted process / contents of the registers on the kernel stack
MP6 the kernel restores the process state e.g. contents of registers once the interrupt is serviced
9(b)(i) One mark per mark point (Max 1) 1
MP1 multi-tasking allows computers to carry out / seem to carry out more than one process at a time
9(b)(ii) One mark per mark point (Max 2) 2
MP1 processor time/common hardware and resources is/are shared between tasks
MP2 scheduling is used to decide on the processes to be carried out to ensure multi-tasking operates correctly / efficiently
/ without clashes
MP3 one task of a higher priority can interrupt another task that is currently running
Objects and classes form the basic structure of Object‑Oriented Programming (OOP).
(a) Outline the structure of a class.
…… [3]
(b) Give three differences between an object and a class.
1 ……
2 ……
3 ……
[3]
Mark scheme
10(a) One mark per mark point (Max 3) 3
MP1 Attributes/properties …
MP2 … with their data types // … are variables bound to the class
MP3 Methods …
MP4 … that are subroutines / functions / procedures that act upon the attributes
MP5 Getters/setters …
MP6 … are methods that can fetch / update the contents of attributes
MP7 A constructor …
MP8 … that is used to create instances / objects of this class.
10(b) One mark per mark point (Max 3) 3
MP1 A class is only defined once but many objects can be created from that class // A class is a template / blueprint from
which objects are created // An object is an instance of a class
MP2 No memory is allocated when a class is defined, but objects are allocated memory space whenever they are created
MP3 A class cannot be manipulated as it is not available in the memory, but objects can be manipulated
MP4 A class is defined but an object is declared / created / instantiated.
MP5 A class can use inheritance. An object cannot.
This binary tree shows an ordered list of integers.
25
4 36
1 16 64
9 49
(a) A linked list of nodes is used to store the data. Each node consists of a left pointer, the data and a right pointer.
−1 is used to represent a null pointer.
Complete this linked list to represent the given binary tree organisation.
RootPtr
LeftPtr Data RightPtr
25
4
–1 1 –1
[4]
(b) A 2D array is used to store the nodes of the linked list in part (a).
Complete the diagram using your answer for part (a).
RootPtr Index LeftPtr Data RightPtr
0 0 25
1 4
2 36
3 1
4 16
5 64
FreePtr 6 9
7 49
8
[4]
(c) The linked list in part (a) is implemented using a 1D array of records. Each record contains a left pointer, data and a right pointer.
The following pseudocode represents a function that searches for an element in the array of records LinkList. It returns the index of the record if the element is found, or it returns a null pointer if the element is not found.
Complete the pseudocode for the function.
FUNCTION SearchList(Item : INTEGER)……
NullPtr −1
…… RootPtr
WHILE NowPtr <> NullPtr
IF LinkList[NowPtr].Data < Item THEN
NowPtr LinkList[NowPtr].RightPtr
ELSE
IF …… THEN
NowPtr ……
ELSE
RETURN NowPtr
ENDIF
ENDIF
ENDWHILE
RETURN NullPtr
ENDFUNCTION
[4]
Mark scheme
11(a) One mark per mark point (Max 4) 4
MP1 Five additional nodes with correct data values
MP2 Correct null pointers in all nodes (7) – no extra null pointers where the arrow points to the next node
MP3 Correct arrows to represent pointers joining parent nodes to child nodes – must come from the correct left or right
pointer not the middle
MP4 All nodes in correct order and no additional data in left/right pointer boxes.
RootPtr
LeftPtr Data RightPtr
25
4 -1 36
-1 1 -1 16 -1 64 -1
-1 9 -1 -1 49 -1
11(b) First eight rows of table (Max 3) 4
Three marks for all eight correct rows
Two marks for six or seven correct rows
One mark for three, four or five correct rows
Last row of table and FreePtr (Max 1)
One mark for correct FreePtr with last row of table completely blank
RootPt RightP
Index LeftPtr Data
r tr
0 0 1 25 2
1 3 4 4
2 -1 36 5
3 -1 1 -1
4 6 16 -1
5 7 64 -1
FreePt
6 -1 9 -1
r
8 7 -1 49 -1
8
11(c) One mark for any correct row (Max 4) 4
FUNCTION SearchList(Item : INTEGER) RETURNS INTEGER
NullPtr ← -1
NowPtr ← RootPtr
WHILE NowPtr <> NullPtr
IF LinkList[NowPtr].Data < Item THEN
NowPtr ← LinkList[NowPtr].RightPtr
ELSE
IF LinkList[NowPtr].Data > Item THEN
NowPtr ← LinkList[NowPtr].LeftPtr
ELSE
RETURN NowPtr
ENDIF
ENDIF
ENDWHILE
RETURN NullPtr
ENDFUNCTION
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On this page, free: the mark scheme for AS & A Level Computer Science 9618/32, October/November 2024 (9618_w24_ms_32), and the question paper (9618_w24_qp_32), to read or download, with each question's marking points written out underneath.
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Computer Science 9618/32 for October/November 2024 is out of 75 marks, with 1 hour 30 minutes to answer it.
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