Computer Science 9618/11 October/November 2022: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 1 Theory Fundamentals, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) (i) Convert the unsigned binary integer into denary.
00100111
Answer …… [1]
(ii) Convert the Binary Coded Decimal (BCD) into denary.
00100111
Answer …… [1]
(iii) Convert the 8-bit two’s complement binary integer into denary.
11100111
Answer …… [1]
(b) Perform the following binary subtraction. Show your working.
1 0 1 1 0 0 1 1
_ 0 1 1 1 0 1 0 1
[2]
(c) Give one similarity and two differences between the ASCII and Unicode character sets.
Similarity ……
Difference 1 ……
Difference 2 ……
[3]
(d) Sound samples are recorded and saved in a file.
(i) State what is meant by sampling rate.
…… [1]
(ii) Explain the effect of increasing the sampling resolution on the sound file.
…… [2]
Mark scheme
1(a)(i) 39 1
1(a)(ii) 27 1
1(a)(iii) −25 1
1(b) 1 mark for working, 1 mark for answer (0011 1110) 2
Working using borrowing:
1 10 10 1
0 10 0 0 10 10
1 0 1 1 0 0 1 1
0 1 1 1 0 1 0 1 -
0 0 1 1 1 1 1 0
Working using two’s complement:
1 0 1 1 0 0 1 1
Two’s complement 1 0 0 0 1 0 1 1 +
(1) 0 0 1 1 1 1 1 0
1 1
1(c) 1 mark for similarity, 2 marks for differences 3
Similarity (max 1):
• both can use 8 bits
• both represent each character using a unique code
• Unicode will contain all the characters that ASCII contains // ASCII is a
subset of Unicode
Differences (max 2):
• Unicode can go up to 32 bits per character whereas ASCII is 7 or 8 bits
Unicode can represent a wider range of characters than ASCII
• different languages are represented using Unicode, ASCII is only for one
language
1(d)(i) the number of samples taken per unit time / per second 1
1(d)(ii) 1 mark for each bullet point (max 2): 2
• increases the number of bits per sample // larger range of values
• which means that the file size increases
• makes the sound file more accurate //digital waveform closer to original
(analogue) waveform
• smaller quantisation errors
Draw one line from each security feature to its most appropriate description.
Security feature Description
converts data to an alternative form
firewall
redirects a user to a fake website
pharming
verifies the authenticity of data
anti-virus software
scans files on the hard drive for malicious software
encryption
accepts or rejects incoming and outgoing packets based on criteria
[4]
Mark scheme
2 1 mark for each correct line. 4
Security feature Description
converts data to an
alternative form
firewall
redirects a user to a false
website
pharming
verifies the authenticity of
data
anti-virus
software
scans files on the hard drive
for malicious software
encryption
accepts or rejects incoming
and outgoing packets based
on criteria
(a) Draw a logic circuit for the logic expression:
X = NOT ((NOT A AND NOT B) OR (NOT B AND NOT C))
A
B X
C
[2]
(b) Complete the truth table for the logic expression:
X = NOT ((NOT A AND NOT B) OR (NOT B AND NOT C))
Working space
A B C X
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
[2]
Mark scheme
3(a) 1 mark for each bullet point: 2
• NOT A AND NOT B and NOT B AND NOT C // A NOR B and B NOR C
• final OR and NOT gates (with correct inputs) // NOR gate (with correct
inputs)
3(b) 1 mark for each set of rows as highlighted: 2
A B C X
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
A photographer creates a relational database to store data about photographs taken at birthday parties.
The database, PHOTOGRAPHS, stores details of the customer, the party, the photographs taken and the cameras used.
The photographer has several cameras that are used for taking the photographs at the parties.
Each camera has a specific lens type (for example, XY32Z) and lighting type (for example, F1672).
Data about each photograph is stored in the database including the party at which it was taken, the time it was taken and the camera used.
The database has these four tables:
CUSTOMER(CustomerID, FirstName, LastName, Telephone)
PARTY(PartyID, CustomerID, PartyDate, StartTime)
PHOTO_DATA(PhotoID, PartyID, TimeTaken, CameraID)
CAMERA_DATA(CameraID, LensType, LightingType)
(a) Complete the entity-relationship (E-R) diagram for the database PHOTOGRAPHS.
PARTY CUSTOMER
PHOTO_DATA CAMERA_DATA
[3]
(b) The database is normalised and is in Third Normal Form (3NF).
Describe the characteristics of a database that is in Third Normal Form (3NF).
…… [3]
(c) The table shows some sample data for the table PHOTO_DATA.
PhotoID PartyID TimeTaken CameraID
ST23-56 BD987 08:34 NIK-02
ST23-57 BD987 08:55 NIK-02
ST23-60 BC08 09:01 CAN-01
ST23-61 BC08 10:23 CAN-12
ST23-62 BC08 10:56 NIK-01
(i) State what is meant by a tuple. Give an example of a tuple from PHOTO_DATA.
Tuple ……
Example ……
[2]
(ii) Complete the Structured Query Language (SQL) script to display the total number of photographs that have been taken using a camera with a camera ID starting with CAN.
SELECT …… (……)
FROM ……
WHERE CameraID LIKE …… ;
[4]
(d) Write an SQL script to include two new fields in CAMERA_DATA to store the number of photographs currently on the camera and the date the camera was last used.
…… [3]
Question 5 begins on page 10.
Mark scheme
4(a) 1 mark for each correct relationship: 3
• 1:M between CUSTOMER and PARTY
• 1:M between PARTY and PHOTO_DATA
• 1:M between CAMERA_DATA and PHOTO_DATA
PARTY CUSTOMER
PHOTO_DATA CAMERA_DATA
4(b) 1 mark for each bullet point (max 3): 3
• no repeating groups of attributes // data is atomic
• no partial key dependencies
• no non-key dependencies // no transitive dependencies
4(c)(i) 1 mark for the definition, 1 mark for the example: 2
• definition: a single row in a table
• example: from the PHOTO_DATA table
4(c)(ii) 1 mark for each correctly completed empty space: 4
• COUNT
• PhotoID
• PHOTO_DATA
• 'CAN*' // ‘CAN%’
SELECT COUNT(PhotoID)
FROM PHOTO_DATA
WHERE CameraID LIKE 'CAN*'; // WHERE CameraID LIKE
'CAN%';
4(d) 1 mark for each bullet point: 3
• ALTER TABLE CAMERA_DATA
• ADD NumberStored INTEGER
• , LastUsed DATE;
ALTER TABLE CAMERA_DATA
ADD NumberStored INTEGER, LastUsed DATE;
(a) State what is meant by the stored program concept in the Von Neumann model of a computer system.
…… [1]
(b) A Central Processing Unit (CPU) contains several special purpose registers and other components.
(i) State the role of the following registers.
Program Counter (PC) ……
Index Register (IX) ……
Status Register (SR) ……
[3]
(ii) Tick () one box in each row to identify the system bus used by each CPU component.
CPU component Data bus Address bus Control bus
System clock
Memory Address Register (MAR)
[1]
(iii) Describe the purpose of the Control Unit (CU) in a CPU.
…… [2]
(c) Describe the purpose of an interrupt in a computer system.
…… [2]
(d) Identify two causes of a software interrupt.
1 ……
2 ……
[2]
Mark scheme
5(a) Instructions and data are stored in the same memory space / in main 1
memory.
5(b)(i) 1 mark for each special purpose register: 3
Program Counter (PC):
• to store the address / location / memory location of the next instruction to
be fetched
Index Register (IX):
• to store a value that is added to an address to give another address
Status Register (SR):
• to store flags which are set by events // from the results of arithmetic and
logic operations and interrupt flags
5(b)(ii) 1 mark for both rows: 1
CPU component Data bus Address bus Control bus
System clock ✓
Memory Address Register (MAR) ✓
5(b)(iii) 1 mark for each bullet point (max 2): 2
• to coordinate / synchronise the actions of other components in the CPU
• to send / receive control signals along the control bus
• to manage the execution of instructions (in sequence)
• to control the communication between the components of the CPU
5(c) 1 mark for each bullet point: 2
• to send a signal from a device or process
• ... seeking the attention of the processor
5(d) 1 mark for each bullet point (max 2). 2
For example:
• division by zero // runtime error in a program
• attempt to access an invalid memory location
• array index out of bounds
• stack overflow
The following table shows part of the instruction set for a processor. The processor has one general purpose register, the Accumulator (ACC), and an Index Register (IX).
Instruction
Explanation
Opcode Operand
LDM #n Immediate addressing. Load the number n to ACC
LDD <address> Direct addressing. Load the contents of the location at the given address to ACC
LDX <address> Indexed addressing. Form the address from <address> + the contents of the index register. Copy the contents of this calculated address to ACC
LDR #n Immediate addressing. Load the number n to IX
STO <address> Store contents of ACC at the given address
ADD <address> Add the contents of the given address to the ACC
ADD #n Add the denary number n to the ACC
INC <register> Add 1 to the contents of the register (ACC or IX)
CMP #n Compare the contents of ACC with number n
JPE <address> Following a compare instruction, jump to <address> if the compare was True
JPN <address> Following a compare instruction, jump to <address> if the compare was False
OUT Output to the screen the character whose ASCII value is stored in
ACC
END Return control to the operating system
<address> can be an absolute or a symbolic address
# denotes a denary number, e.g. #123
B denotes a binary number, e.g. B01001101
(a) The current contents of main memory and selected values from the ASCII character set are given.
(i) Trace the program currently in memory using the trace table.
Instruction Memory address
ACC IX Output
Address Instruction address 100 101 110 111 112
77 LDR #0
0 0 66 65 35
78 LDX 110
79 CMP #35
80 JPE 92
81 ADD 100
82 STO 101
83 LDM #1
84 ADD 100
85 STO 100
86 INC IX
87 LDX 110
88 CMP #35
89 JPN 81
90 LDD 100
91 ADD #48
92 OUT
93 END
100 0
101 0
110 66
111 65
112 35
ASCII
Character value
49 1
50 2
51 3
52 4
65 A
66 B
67 C
68 D
[4]
(ii) The following instructions are repeated for your reference.
Instruction
Explanation
Opcode Operand
LDD <address> Direct addressing. Load the contents of the location at the given address to ACC
STO <address> Store contents of ACC at the given address
State the purpose of this part of an assembly language program.
LDD 100
STO 165
LDD 101
STO 100
LDD 165
STO 101
[1]
Question 6(b) begins on page 16.
(b) The following table shows another part of the instruction set for the processor.
Instruction
Explanation
Opcode Operand
AND #n Bitwise AND operation of the contents of ACC with the operand
AND Bn Bitwise AND operation of the contents of ACC with the binary number n
XOR #n Bitwise XOR operation of the contents of ACC with the operand
XOR Bn Bitwise XOR operation of the contents of ACC with the binary number n
OR #n Bitwise OR operation of the contents of ACC with the operand
OR Bn Bitwise OR operation of the contents of ACC with the binary number n
LSR #n Bits in ACC are shifted logically n places to the right. Zeros are introduced on the left-hand end
<address> can be an absolute or a symbolic address
# denotes a denary number, e.g. #123
B denotes a binary number, e.g. B01001101
(i) The current contents of the ACC are:
1 0 0 1 0 0 1 1
Show the result after the execution of the following instruction.
XOR B00011111
[1]
(ii) The current contents of the ACC are:
1 0 0 1 0 0 1 1
Show the result after the execution of the following instruction.
AND B11110000
[1]
(iii) The current contents of the ACC are:
1 0 0 1 0 0 1 1
Show the result after the execution of the following instruction.
OR B11001100
[1]
(iv) The current contents of the ACC are:
1 0 0 1 0 0 1 1
Show the result after the execution of the following instruction.
LSR #2
[1]
(c) Tick (3) one or more boxes in each row to indicate whether the task is performed in the first pass or the second pass of a two-pass assembler.
Task First pass Second pass
Remove comments.
Read the assembly language program one line at a time.
Generate the object code.
Check the opcode is in the instruction set.
[2]
Mark scheme
6(a)(i) 1 mark for each set of highlighted rows. 4
Instructio Memory address
ACC IX Output
n address 100 101 110 111 112
0 0 66 65 35
77 0
78 66
79
80
81
82 66
83 1
84
85 1
86 1
87 65
88
89
81 66
82
83 1
84 2
85 2
86 2
87 35
88
89
90 2
91 50
92 2
93
6(a)(ii) swaps the contents of memory address 100 and 101 1
6(b)(i) 1000 1100 1
6(b)(ii) 1001 0000 1
6(b)(iii) 1101 1111 1
6(b)(iv) 0010 0100 1
6(c) 1 mark for each pair of highlighted rows 2
Task First pass Second pass
Remove comments. ✓
Read the assembly language program one
✓ ✓
line at a time.
Generate the object code. ✓
Check the opcode is in the instruction set. ✓
(a) State two benefits to a programmer of using Dynamic Link Library (DLL) files.
1 ……
2 ……
[2]
(b) Memory management is one of the tasks performed by an Operating System (OS).
Describe the ways in which memory management organises and allocates Random Access
Memory (RAM).
…… [2]
(c) An Operating System may include a utility program to compress text files.
Describe one appropriate method of compressing a text file.
…… [3]
(d) Explain the reasons why increasing the amount of cache memory can improve the performance of a CPU.
…… [2]
(e) State the name of a peripheral device port that provides a physical connection in the computer for each of these peripherals.
3D printer ……
Monitor ……
[2]
Mark scheme
7(a) 1 mark for each benefit (max 2): 2
• (main) memory requirements for program are reduced as dynamic link
library is loaded only once / when required
• the executable file size is smaller because the executable does not
contain all the library routines
• maintenance not needed to be done by the programmer because the DLL
is separate from program
• no need to recompile the main program when changes are made to DLL
because changes / improvements/ error correction to the DLL file code
are done independently of the main program
7(b) 1 mark for each bullet point (max 2): 2
• RAM is assigned into blocks
• dynamic allocation of RAM to programs / processes
• reclaims unused blocks of RAM
• prevents two programs / processes occupying the same area of RAM at
the same time
• moves data from secondary storage when needed // manages paging,
segmentation and virtual memory
7(c) 1 mark for each bullet point (max 3): 3
• lossless compression
• Run Length Encoding
• repeated sequences of characters are replaced by
• ... a single copy of the character
• ... and a counter of the number of characters
7(d) 1 mark for each bullet point (max 2): 2
• cache is fast access memory (close to the CPU)
• cache stores frequently used instructions / data
• ... more cache means more instructions / data can be transferred faster
• ... less swapping between RAM and cache
• prevents the CPU idling while waiting for data
7(e) 1 mark for each device. 2
• 3D printer: USB port / COM port
• Monitor: HDMI / VGA / USB / DisplayPort
A Local Area Network (LAN) uses a bus topology.
Describe how Carrier Sense Multiple Access / Collision Detection (CSMA / CD) is used in a bus network.
…… [4]
Mark scheme
8 1 mark for bullet point (max 4): 4
• CSMA/CD is a protocol used to detect and prevent collisions in a bus
topology
• before transmitting, a device checks if the channel is busy
• If the channel is busy the device waits // if the channel is free the data is
sent
• because there is more than one computer connected to the same
transmission medium
• ... two workstations can start to transmit at the same time, causing a
collision
• If a collision is detected by the device, transmission is aborted / a
jamming signal is transmitted
• both devices wait a (different) random time and then try again
Many modern televisions are examples of embedded systems.
(a) Explain why these televisions are embedded systems.
…… [2]
(b) Embedded systems use Electrically Erasable Programmable ROM (EEPROM).
Describe one benefit of using EEPROMs in an embedded system.
…… [2]
Mark scheme
9(a) 1 mark for each bullet point (max 2): 2
• the embedded system is built into / integrated into the TV
• combination of hardware and software designed for a specific function
• The system is not easily changed/updated by the TV owner
9(b) 1 mark for each benefit and 1 mark for corresponding expansion (max 2). 2
For example:
• no additional equipment is needed to change
• ... enables firmware updates by non-technical users
• can be erased and reprogrammed several times
• ... so firmware can be updated
• ... can erase a particular byte or the whole EEPROM
• possible to reprogram / update
• ... without removing it from the device
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Where can I find the 9618/11 October/November 2022 mark scheme?+
On this page, free: the mark scheme for AS & A Level Computer Science 9618/11, October/November 2022 (9618_w22_ms_11), and the question paper (9618_w22_qp_11), to read or download, with each question's marking points written out underneath.
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Computer Science 9618/11 for October/November 2022 is out of 75 marks, with 1 hour 30 minutes to answer it.
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