Computer Science 9618/11 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 1 Theory Fundamentals, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Images are being created to advertise holidays.
Some of the images are bitmap images and some are vector graphics.
(a) Complete the table by defining the image terms.
Term Definition
Drawing list ……
Pixel ……
Colour depth ……
[3]
(b) The bitmap images are photographs of the holiday locations.
(i) Colour depth and image resolution are both included in the file header of a bitmap image.
Identify two other items that could be included in the file header of each photograph.
1 ……
2 ……
[2]
(ii) One of the photographs has a bit depth of 8 bytes and an image resolution of 1500 pixels wide and 3000 pixels high.
Calculate the file size of the photograph in megabytes. Show your working.
Working ……
File size …… MB
[2]
(c) The photographs are compressed before they are uploaded to a web server.
Customers download the photographs from this web server.
(i) Explain the reasons why compressing the photographs will benefit the customers.
…… [3]
(ii) An image can be compressed using run‑length encoding (RLE).
Explain the reasons why RLE may not reduce the file size of a bitmap image.
Give one example in your answer.
…… [3]
Mark scheme
1(a) 1 mark for each correct definition 3
Term Definition
Drawing list All the drawing objects in an image // a list
that stores the commands required to draw
each object
Pixel The smallest part of the image // one square
/ dot of one colour
Colour depth The number of bits per pixel // determines
the number of colours that can be
represented in the image
1(b)(i) 1 mark each 2
Examples:
Confirmation that it is a bitmap // file type
Compression type
Location/offset of data within the file
Dimensions e.g.100 × 100 pixels
1(b)(ii) 1 mark for working; 1 mark for answer 2
Working: (1500 * 3000 * 8) / 1000 / 1000
Answer: 36 MB
1(c)(i) 1 mark each 3
Examples:
The customers will be able to download the photographs in less time
…and they will take less of the customer's bandwidth
The photographs will take up less space on the customer's storage
medium
…therefore the customers can store more images
…and will have more space for other files
1(c)(ii) 1 mark each to max 2 for explanation; 1 mark for an image related example 3
RLE stores a colour and the number of times it occurs consecutively
An image may not have many sequences of the same colour
It would need to store each colour and then the count/number 1 which
adds data
Example:
Red-Green-Blue would become Red 1 Green 1 Blue 1
An organisation uses a database to store data about the types of bird that people have seen.
(a) The database is managed using a Database Management System (DBMS).
(i) State what is meant by a data dictionary and give one example of an item typically found in a data dictionary.
Definition ……
Example ……
[2]
(ii) State what is meant by data integrity and give one example of how this is implemented in a database.
Definition ……
Example ……
[2]
(b) The database, Birds, stores information about the types of bird and the people who have seen them.
Data about each bird seen is stored with its location and data about the person who saw the bird.
Database Birds has the following tables:
BIRD_TYPE(BirdID, Name, Size)
BIRD_SEEN(SeenID, BirdID, Date, Location, PersonID)
PERSON(PersonID, FirstName, LastName, EmailAddress)
(i) Complete the table by identifying two foreign keys and the database table where each is found.
Foreign key Database table
[2]
(ii) The database Birds has been normalised.
Draw one line from each Normal Form to the most appropriate definition.
Normal Form Definition
First Normal Form (1NF) All fields are fully dependent on the primary key.
Second Normal Form (2NF) There are no repeating groups of attributes.
Third Normal Form (3NF) There are no partial dependencies.
[1]
(iii) Part of the database table BIRD_TYPE is shown:
BirdID Name Size
0123 Blackbird Medium
0035 Jay Large
0004 Raven Large
0085 Robin Small
The database only supports these data types:
• character
• varchar
• Boolean
• integer
• real
• date
• time
Write a Structured Query Language (SQL) script to define the table Bird_Type.
…… [4]
(iv) The database tables are repeated here for reference:
BIRD_TYPE(BirdID, Name, Size)
BIRD_SEEN(SeenID, BirdID, Date, Location, PersonID)
PERSON(PersonID, FirstName, LastName, EmailAddress)
Complete the SQL script to return the number of birds of each size seen by the person with the ID of J_123.
SELECT BIRD_TYPE.Size, …… (BIRD_TYPE.BirdID)
AS NumberOfBirds
FROM BIRD_TYPE, ……
WHERE …… = "J_123"
AND BIRD_TYPE.BirdID = ……
…… BIRD_TYPE.Size;
[5]
Mark scheme
2(a)(i) 1 mark for definition: 2
Data about the data in the database // data about the structure of the
database // metadata for a database
1 mark for a suitable example
Examples:
table names
data types
field names
2(a)(ii) 1 mark for definition 2
Methods of making sure the data is consistent
1 mark for example
Examples:
Enforcing referential integrity
If data in one table is deleted/edited all tables are updated // cascading
update/delete
Validation/verification rules
2(b)(i) 1 mark for each field name and table 2
Foreign key Database table
BirdID BIRD_SEEN
PersonID BIRD_SEEN
2(b)(ii) 1 mark for all 3 correct lines 1
Normal Form Definition
First Normal Form All fields are fully
(1NF) dependent on the primary
key.
Second Normal There are no repeating
Form (2NF) groups of attributes.
Third Normal Form There are no partial
(3NF) dependencies.
2(b)(iii) 1 mark each 4
CREATE TABLE start and end bracket
Bird ID as CHAR/VARCHAR
Name and size as VARCHAR/CHAR
Bird ID as primary key
Example answer:
CREATE TABLE BIRD_TYPE(
BirdID CHAR(4) NOT NULL,
Name VARCHAR(9),
Size VARCHAR(6),
PRIMARY KEY (BirdID)
);
2(b)(iv) 1 mark for each correctly completed space 5
SELECT BIRD_TYPE.Size, COUNT(BIRD_TYPE.BirdID) AS
NumberOfBirds
FROM BIRD_TYPE, BIRD_SEEN
WHERE BIRD_SEEN.PersonID = "J_123"
AND BIRD_TYPE.BirdID = BIRD_SEEN.BirdID
GROUP BY BIRD_TYPE.Size;
A computer has an Operating System (OS).
(a) Describe how the Operating System manages the peripheral hardware devices of the computer.
…… [4]
(b) Hardware management is one key management task carried out by the Operating System.
Identify two other key management tasks carried out by the Operating System.
1 ……
2 ……
[2]
(c) The Operating System has utility software including defragmentation software.
Explain how defragmentation can improve the performance of the computer.
…… [3]
(d) The computer stores data in binary form.
(i) State the difference between a kibibyte and a kilobyte.
…… [1]
(ii) Convert the denary number 964 into Binary Coded Decimal (BCD).
…… [1]
(iii) Convert the positive binary integer 11110010 into hexadecimal.
…… [1]
(iv) Give the smallest and largest two’s complement binary number that can be represented using 8 bits.
Smallest ……
Largest ……
[2]
(v) Add the following two binary integers using binary addition. Show your working.
10110000
+00011011
[2]
(vi) Show the result of a 3‑place right logical shift on the binary number:
11001100
…… [1]
Mark scheme
3(a) 1 mark each to max 4 4
Examples:
Installs device drivers
… to allow communication between peripherals and computer
Sends data and receives data to and from peripherals
… such as to an output device and from an input device/by example
Handles buffers for transfer of data
… to ensure smooth transfer between devices that transmit and receive
at different speeds
Manages interrupts / signals from the device
3(b) 1 mark each to max 2 2
Memory management
File management
Security management
Process management
Error checking and recovery
3(c) 1 mark each to max 3 3
Rearranges blocks of individual files (on the HDD) so they are
contiguous // moves the free space together
Accessing each file is faster
…because there is no need to search for the next fragment / block of
the file
…so less head movement is needed
3(d)(i) 1 mark from 1
Kibibyte is 1024 bytes and kilobyte is 1000 bytes
Kibibyte is binary prefix and kilobyte is denary prefix
3(d)(ii) 1001 0110 0100 1
3(d)(iii) F2 1
3(d)(iv) Smallest: 10000000 2
Largest: 01111111
3(d)(v) 1 mark for working 2
1 mark for answer
10110000
+00011011
11001011
11
3(d)(vi) 00011001 1
A networked closed‑circuit television (CCTV) system in a house uses sensors and cameras to detect the presence of a person. It then tracks the person and records a video of their movements.
Data from the CCTV cameras is transmitted to a central computer.
(a) This computer has both Read Only Memory (ROM) and Random Access Memory (RAM).
(i) Describe the contents of the ROM in the central computer.
…… [2]
(ii) The central computer has Dynamic RAM (DRAM).
Identify two advantages of using DRAM instead of Static RAM (SRAM).
1 ……
2 ……
[2]
(b) The central computer stores the video files on secondary storage.
Describe two reasons why magnetic storage is more appropriate than solid state storage for this computer.
1 ……
2 ……
[4]
(c) The CCTV system uses Artificial Intelligence (AI) to identify the presence of a person in the house and to track their movements.
Describe how AI is used in this system.
…… [3]
(d) The CCTV cameras are connected to a network and transfer their data wirelessly to the central computer.
(i) Each device on the network has an IP address.
Complete the description of IP addresses.
An IPv4 address contains …… groups of digits. Each group is
represented in …… bits and the groups are separated by full stops.
An IPv6 address contains …… groups of digits. Each group is
represented in …… bits. Multiple groups that only contain zeros
can be replaced with a …… .
[5]
(ii) The network makes use of subnetting.
Describe two benefits of subnetting a network.
1 ……
2 ……
[4]
Mark scheme
4(a)(i) 1 mark each to max 2 2
Stores the bootstrap program // start-up instructions for the central
computer // BIOS
Stores the start-up instructions for the CCTV system/cameras //
firmware for CCTV
Stores the kernel of the Operating System // stores parts of the
Operating System
4(a)(ii) 1 mark each to max 2 2
Costs less per unit
Higher storage density
Simple design – uses fewer transistors
4(b) 1 mark for reason, 1mark for application/justification 4
The computer will have a large number of read/write operations
because it is working all the time
.. magnetic storage has more longevity
Magnetic storage costs less per storage unit
… videos are large files and therefore very large storage capacity is
required
4(c) 1 mark each to max 3 3
Examples:
Uses image recognition
Monitors every image taken to identify matching
images/shapes/features to a 'person' …
… starts recording to secondary storage/permanently when a person is
identified
System identifies direction of movement of person and uses this to
decide where/how to move the camera/record
System identifies other cameras to start recording based on direction of
movement
4(d)(i) 1 mark for each term 5
An IPv4 address contains 4 groups of digits. Each group is represented in 8
bits and the groups are separated by full stops.
An IPv6 address contains 8 groups of digits. Each group is represented in
16 bits. Multiple groups that only contain zeros can be replaced with a :: //
double colon.
4(d)(ii) 1 mark for identification, 1mark for expansion 4
e.g.
Reduce amount of traffic in a network // improve network speed
Data stays in its subnet so it does not travel as far
Improves network security
.. so that not all devices can access all areas of the network
Allows for easier maintenance
… because only one subnetwork may need taking down/changing while
the rest of the network can continue
(a) Draw the logic circuit for this logic expression:
T = (NOT A OR B) XOR (C NAND D)
A
B
T
C
D
[2]
(b) Describe the function of the NAND and NOR logic gates.
NAND ……
NOR ……
[2]
Mark scheme
5(a) 1 mark for 2 gates 2
2 marks for all 4 gates
5(b) 1 mark each 2
NAND
0 is only output when both inputs are 1 // 1 is only output when none, or
(either) one of the inputs is 1
NOR
1 is only output when both inputs are 0 // 0 is only output when (either) one
or both inputs are 1
An interrupt is generated when a key is pressed on a computer keyboard.
Explain how the computer handles this interrupt.
…… [5]
Mark scheme
6 1 mark each to max 5 5
An interrupt flag is raised in the (interrupt) register
At the end of the current FE cycle // at the start of the next FE cycle
The system checks the interrupt register for higher priority interrupts
than current process
If true, it stores the current contents of the registers on the stack
The appropriate interrupt service routine (ISR) for the key press is
called
The input data from the keyboard is processed
The contents of the registers are restored from the stack
... and control is passed back to previous process
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9618/11 May/June 2023 mark scheme?+
On this page, free: the mark scheme for AS & A Level Computer Science 9618/11, May/June 2023 (9618_s23_ms_11), and the question paper (9618_s23_qp_11), to read or download, with each question's marking points written out underneath.
How many marks is 9618/11 out of?+
Computer Science 9618/11 for May/June 2023 is out of 75 marks, with 1 hour 30 minutes to answer it.
Can I get my 9618/11 answers marked?+
Yes. Upload your answers to 9618/11/M/J/23 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.