Computer Science 9618/13 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 1 Theory Fundamentals, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Anya scans an image into her computer for a school project.
(a) The scanned image is a bitmapped image.
(i) Complete the following table to describe the two terms about graphics.
Term Description
Pixel ……
File header ……
[2]
(ii) The image is scanned with an image resolution of 1024 × 512 pixels, and a colour depth of 8 bits per pixel.
Calculate an estimate for the file size, giving your answer in mebibytes. Show your working.
Working ……
Answer …… mebibytes
[3]
(b) The image is compressed using lossless compression.
Identify one method of lossless compression that can be used to compress the image and describe how the method will reduce the file size.
Lossless compression method ……
Description ……
[3]
(c) One of the colours used in the image has the hexadecimal colour code:
#FC238A
FC is the amount of red, 23 is the amount of green and 8A is the amount of blue in the colour.
(i) Convert the hexadecimal code FC into denary.
…… [1]
(ii) The amount of green in binary is 00100011. This has the denary number 15 added to it to create a second colour.
Add the denary number 15 to the binary number 00100011 and give your answer in binary.
Perform the addition in binary. Show your working.
Working ……
Answer (in binary) ……
[3]
(iii) Hexadecimal 23 in two’s complement representation is 00100011. The denary number
10 needs to be subtracted from this value.
Subtract the denary number 10 from the two’s complement representation 00100011.
Give your answer in binary. Show your working.
Working ……
Answer (in binary) ……
[3]
(d) Anya made sure that the image was not subject to any copyright before scanning it.
Describe what is meant by copyright.
…… [2]
Mark scheme
1(a)(i) 1 mark for each description 2
Pixel:
• A single square of one colour
• The smallest addressable element in an image
File header:
• Data about the bitmap image (e.g. number of colours)
1(a)(ii) 1 mark per bullet point for working, 1 mark for answer 3
Working:
• 1024 × 512 = 524 288 pixels/bytes
• 524288 / 1024 / 1024
Answer:
0.50 mebibytes
1(b) 1 mark for naming method, 1 mark per description to max 2 3
• Run-length encoding
• Replace sequences of the same colour pixel
• … with colour code and number of identical pixels
1(c)(i) 252 1
1(c)(ii) 1 mark per bullet point 3
• Converting 15 to binary 0000 1111
• Method for addition
• Final answer
0010 0011
+ 0000 1111
0011 0010
1 111
1(c)(iii) 1 mark per bullet point 3
• Converting −10 to two's complement binary 1111 0110
• Adding values
• Final answer 0001 1001
10 = 0000 1010
–10 = 1111 0110
0010 0011
+ 1111 0110
0001 1001
11 11
1(d) 1 mark per bullet point to max 2 2
• The formal and legal rights to ownership // intellectual property rights
• Protects against unauthorised reproduction of work
• Provides for legal right of redress
Bingwen’s computer comes with an Operating System and utility software.
(a) Draw one line from each utility software to its correct description.
Utility software Description
Scans software for errors and repairs the problems
Moves parts of files so that each file is
Disk formatter contiguous in memory
Defragmentation Creates a copy of data that is no longer required
Back-up Sets up a disk so it is ready to store files
Disk repair Scans for errors in a disk and corrects them
Creates a copy of data in case the original is lost
[4]
(b) Identify four key management tasks that the Operating System will perform.
1 ……
2 ……
3 ……
4 ……
[4]
Mark scheme
2(a) 1 mark for each correct line 4
2(b) 1 mark per bullet point to max 4 4
• memory management
• file management
• security management
• hardware / device / peripheral / resources management
• input/output management
• process management
• error checking and recovery
• provision of a platform for software
• provision of a user interface
A processor has one general purpose register, the Accumulator (ACC), and several special purpose registers.
(a) Complete the following description of the role of the registers in the fetch-execute cycle by writing the missing registers.
The …… holds the address of the next instruction
to be loaded. This address is sent to the …… .
The …… holds the data fetched from this address.
This data is sent to the …… and the Control Unit
decodes the instruction’s opcode.
The …… is incremented.
[5]
(b) 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
Indirect addressing: The address to be used is at the given address. Load the
LDI <address> contents of this second address to ACC
Indexed addressing. Form the address from <address> + the contents of the
LDX <address>
Index Register. Copy the contents of this calculated address to ACC
LDR #n Immediate addressing. Load the number n to IX
MOV <register> Move the contents of the accumulator to the given register (IX)
STO <address> Store contents of ACC at the given address
ADD <address> Add the contents of the given address to the ACC
INC <register> Add 1 to the contents of the register (ACC or IX)
CMP <address> Compare the contents of ACC with the contents of <address>
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
JMP <address> Jump to the given address
OUT Output to the screen the character whose ASCII value is stored in ACC
END Return control to the operating system
Bits in ACC are shifted logically n places to the left. Zeros are introduced on
LSL #n the right hand end
Bits in ACC are shifted logically n places to the right. Zeros are introduced on
LSR #n the left hand end
<address> can be an absolute address or a symbolic address
# denotes a denary number, e.g. #123
The current contents of the main memory and selected values from the ASCII character set are shown.
Address Instruction ASCII code table (selected codes only)
200 LDD 365 ASCII code Character
201 CMP 366 65 A
202 JPE 209 66 B
203 INC ACC 67 C
204 STO 365 68 D
205 MOV IX
206 LDX 365
207 OUT
208 JMP 200
209 END
365 1
366 3
367 65
368 66
IX 0
Complete the trace table for the program currently in main memory.
Instruction Memory address
ACC IX Output address 365 366 367 368
1 3 65 66 0
[6]
(c) (i) The Accumulator currently contains the binary number:
0 0 1 1 0 1 0 1
Write the contents of the Accumulator after the processor has executed the following instruction:
LSL #2
[1]
(ii) The Accumulator currently contains the binary number:
0 0 1 1 0 1 0 1
Identify the mathematical operation that the following instruction will perform on the contents of the accumulator.
LSR #3
…… [1]
Mark scheme
3(a) 1 mark for each completed statement 5
The Program Counter holds the address of the next instruction to be loaded.
This address is sent to the Memory Address Register.
The Memory Data Register holds the data fetched from this address. This
data is sent to the Current Instruction Register and the Control Unit
decodes the instruction’s opcode.
The Program Counter is incremented.
3(b) 1 mark for each shaded set of values 6
Instruction Memory address
ACC IX Output
address 365 366 367 368
1 3 65 66 0
200 1
201
202
203 2
204 2
205 2
206 65
207 A
208
200 2
201
202
203 3
204 3
205 3
206 66
207 B
208
200 3
201
202
209
3(c)(i) 1
1 1 0 1 0 1 0 0
3(c)(ii) 1 mark for correct answer 1
The number is divided by 8 (and only whole number retained)
Melinda and her friends set up a peer-to-peer network between their computers to share data.
(a) Describe the key features of a peer-to-peer network.
…… [2]
(b) Describe two drawbacks to Melinda and her friends of using a peer-to-peer network.
1 ……
2 ……
[4]
(c) Melinda connects her laptop to the internet through her router.
(i) Tick (3) one box in each row to identify whether the task is performed by the router or not.
Performed by Not performed
Task router by router
Receives packets from devices
Finds the IP address of a Uniform Resource Locator (URL)
Directs each packet to all devices attached to it
Stores the IP and/or MAC address of all devices attached to it
[2]
(ii) Melinda mainly uses the internet to watch films and play computer games.
Tick (3) one box to identify whether Melinda should connect to the router using a wired or wireless network and justify your choice.
Wired
Wireless
Justification ……
…… [3]
(d) Melinda sends emails from her webmail account (email account accessed through a website).
Explain whether Melinda is using the internet, or the World Wide Web (WWW), or both.
…… [3]
Mark scheme
4(a) 1 mark per bullet point to max 2 2
• All computers are of equal status
• Each computer provides access to resources and data // data is
distributed
• Computers can communicate and share resources
• Each computer is responsible for its own security
4(b) 1 mark per bullet point to max 2 per drawback 4
• Reduced security // no central management of security
• … only as secure as the weakest computer on the network
• … each computer is at risk from viruses from other computers
• No central management of backup
• ... if the data from one computer is not backed up it is lost to all of them
• No central management of files/software
• … consistency may be difficult to maintain
• … each computer may have different software from the others
• Individual computers may respond slower
• … because they are being accessed by other computers
• In order to share files etc. all the computers involved need to be switched
on
• ... so the files etc. may not be always available
4(c)(i) 1 mark for first 2 ticks, 1 mark for last 2 (shaded) 2
Performed Not performed
Task
by router by router
Receives packets from devices
Finds the IP address of a Uniform
Resource Locator (URL)
Directs each packet to all devices
attached to it
Stores the IP and/or MAC address of all
devices attached to it
4(c)(ii) 1 mark per bullet point for justification up to max 3 3
No mark for identification of wired/wireless
Wired
• Faster connection // higher bandwidth
• …. needed as she is downloading/streaming large files
• … less time waiting / less latency / fewer delays
• More reliable / stable connection
• … is less susceptible to issues with distance/walls/interference
• More secure
Wireless
• Freedom of movement
• … can move between different rooms with a mobile device and still
receive/transmit data
• … no need of a physical connection
• Easily expanded if friends want to access the same network
• Less cabling / expertise is needed
• ... making the initial setup less expensive
4(d) 1 mark for identifying that she is using both. 3
1 mark per bullet point for justification
• using internet because sending data on the infrastructure
• using WWW because accessing a website (that is stored on a web
server operated by the webmail) that is part of the WWW
Kiara has a washing machine and a refrigerator.
(a) She has an embedded system in her washing machine.
Describe what is meant by an embedded system, using the washing machine as an example.
…… [2]
(b) The washing machine’s embedded system makes use of both Random Access Memory
(RAM) and Read Only Memory (ROM).
State the purpose of RAM and ROM within the washing machine’s embedded system.
RAM ……
ROM ……
[2]
(c) The temperature in her refrigerator must be kept between 4 and 6 degrees Celsius.
The microprocessor in the refrigerator turns on the cooling if the temperature is too high, and turns off the cooling if the temperature is too low.
Explain why the system in the refrigerator is a control and not a monitoring system.
…… [2]
Mark scheme
5(a) 1 mark per bullet point to max 2 2
• Definition: Microprocessor/microcontroller within a larger system //
microprocessor/microcontroller that performs one specific task
• Example: e.g. Embedded system in washing machine only controls the
programs for the washing cycle // it is part of the washing machine but
does not perform any other function within it
5(b) 1 mark for RAM, 1 mark for ROM 2
RAM:
• Store the choices/wash program the user has entered // stores the data
read from the sensors // stores the time left in the program // by example
ROM:
• Store the start-up instructions (for the washing cycles)
5(c) 1 mark per bullet point 2
• The system uses feedback
• The system causes the temperature to change // produces an action
Each of the following algorithms performs data validation.
State the type of validation check that each of the algorithms performs.
(a)
INPUT x
IF x < 0 OR x > 10 THEN
OUTPUT "Invalid"
ENDIF
…… [1]
(b)
INPUT x
IF x = "" THEN
OUTPUT "Invalid"
ENDIF
…… [1]
(c)
INPUT x
IF NOT(x = "Red" OR x = "Yellow" OR x = "Blue") THEN
OUTPUT "Invalid"
ENDIF
…… [1]
Mark scheme
6(a) Range (check) 1
6(b) Presence (check) 1
6(c) Existence (check) 1
Bobby and Kim are discussing databases.
(a) Bobby tells Kim that a file-based approach is usually better than a relational database.
Explain why Bobby is incorrect.
…… [3]
(b) Bobby has a shop that sells products to customers. His database will store data about his customers, their payment details, orders and the products he sells. Customers will have login details to access their accounts. The database will update customers’ payment and login details without keeping any historical records.
(i) Give one example of each of the following relationships from Bobby’s database.
one-to-one
one-to-many
many-to-many
[3]
(ii) Tick (3) one box to identify the relationship that cannot be directly implemented in a normalised relational database.
Relationship Tick (3)
one-to-one
one-to-many
many-to-many
[1]
(iii) Bobby wants to name his database SHOPORDERS.
Write a Data Definition Language (DDL) statement to define a new database with the name SHOPORDERS.
…… [1]
(c) A database has a data dictionary.
Give three items that are stored in a data dictionary.
1 ……
2 ……
3 ……
[3]
Mark scheme
7(a) 1 mark per bullet point to max 3 3
• Flat-file has more data redundancy
• … because the same data is stored many times // data is stored in
different tables which are linked
• There is program-data dependence with flat-files
• ... because any changes to the structure of the data means the programs
that access that data have to be re-written
• Flat-file has more data inconsistency // worse data integrity
• ... because duplicated data might be stored differently //…because when
data is updated in one place, it is not updated everywhere
• It is not easy to perform complex searches /queries
• … because a new program has to be written each time
• Flat files could have a lack of privacy
• … as user views cannot easily be implemented
7(b)(i) 1 mark for each correct example 3
one-to-one
• e.g. customer to payment details // customer to login details
one-to-many
• e.g. customer to order
many-to-many
• e.g. order to product // customer to product
7(b)(ii) 1 mark 1
Relationship Tick ()
one-to-one
one-to-many
many-to-many
7(b)(iii) 1 mark 1
CREATE DATABASE SHOPORDERS;
7(c) 1 mark per item to max 3 3
• table name
• field name // attribute
• data type
• type of validation
• Primary Key
• Foreign Key
• relationships
Tick (3) one box in each row to identify the logic gate that each statement describes.
Statement AND NAND NOR XOR OR
The output is 1 only when both inputs are 1
The output is 1 only when both inputs are different
The output is 1 only when both inputs are 0
[3]
Mark scheme
8 1 mark per correct row 3
Statement AND NAND NOR XOR OR
The output is 1 only when
both inputs are 1
The output is 1 only when
both inputs are different
The output is 1 only when
both inputs are 0
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On this page, free: the mark scheme for AS & A Level Computer Science 9618/13, May/June 2021 (9618_s21_ms_13), and the question paper (9618_s21_qp_13), to read or download, with each question's marking points written out underneath.
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Computer Science 9618/13 for May/June 2021 is out of 75 marks, with 1 hour 30 minutes to answer it.
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Yes. Upload your answers to 9618/13/M/J/21 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.