Computer Science 9618/12 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Computer Science (9618), Paper 1 Theory Fundamentals, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
When designing computer systems, it is important to consider the security, integrity and privacy of the data.
Draw one line from each measure to indicate whether it keeps data secure or protects the integrity of data.
Measure
Firewall
Double entry
Data Security
Presence check
Data Integrity
Access rights
Password
[2]
Mark scheme
1 1 mark for 3 correct lines only from Data Security 2
1 mark for 2 correct lines only from Data Integrity
Firewall
Double entry Data Security
Presence check
Access rights Data Integrity
Password
(a) Complete the truth table for each of the following two logic gates.
A B Output
A 0 0
B 0 1
1 0
1 1
A B Output
A 0 0
B 0 1
1 0
1 1
[2]
(b) Draw a logic circuit for the following logic expression.
X = NOT(NOT(A AND B)AND C)
A
X
B
C
[2]
Mark scheme
2(a) 1 mark for each completely correct truth table 2
NOR
A B Output
0 0 1
0 1 0
1 0 0
1 1 0
NAND
A B Output
0 0 1
0 1 1
1 0 1
1 1 0
2(b) 1 mark for each correct bullet point 2
• NOT (A AND B) // A NAND B
• NOT the result AND C // the result NAND C
OR
Andy likes to play computer games.
(a) Andy uses several input devices to play the games. These include a keyboard and a microphone.
Describe the principal operation of a microphone.
…… [3]
(b) Andy plays some of the computer games over the internet. He has several devices that connect wirelessly to the router in his house.
(i) Identify the topology of Andy’s home network. Justify your choice.
Topology ……
Justification ……
[2]
(ii) The router has a wireless access point (WAP) to allow the devices to connect wirelessly.
Identify three functions of the router in Andy’s network.
1 ……
2 ……
3 ……
[3]
Mark scheme
3(a) 1 mark for each bullet point to max 3 3
• The microphone has a diaphragm / ribbon
• The incoming sound waves cause vibrations of the diaphragm
• … causing a coil to move past a magnet // causing a magnet to move
past a coil (dynamic microphone) // changing the capacitance (condenser
microphone) // deforms the crystal (crystal microphone)
• An electrical signal is produced
3(b)(i) 1 mark for identification of star topology 2
1 mark for justification
Devices are connected directly to the router independently
// all devices are only connected to the router
3(b)(ii) 1 mark for each correct function to max 3 3
• To receive packets from devices or the Internet
• To forward / route packets to the destination
• To find the destination of the packet
• To assign / allocate private IP addresses to devices on LAN
• To store / update / maintain a routing table
• To find the most efficient path to the destination
• To maintain a table of MAC and IP addresses
A register stores the following binary number:
1 1 0 0 1 1 0 1
(a) The binary value in the register represents an unsigned binary integer.
Convert the unsigned binary integer into denary.
…… [1]
(b) The binary value in the register represents a two’s complement binary integer.
Convert the two’s complement binary integer into denary.
…… [1]
(c) The binary value in the register represents a hexadecimal number.
Convert the binary number into hexadecimal.
…… [1]
(d) State why the value in the register cannot be interpreted as a Binary Coded Decimal (BCD).
…… [1]
(e) The binary contents of two registers are:
Register 1 0 0 1 1 1 1 0 1
Register 2 0 0 1 0 1 1 0 1
(i) Add the contents of Register 1 and Register 2. Show your working.
Answer ……
[2]
(ii) Subtract the contents of Register 2 from the contents of Register 1. Show your working.
Answer ……
[2]
Mark scheme
4(a) 205 1
4(b) – 51 1
4(c) CD 1
4(d) 1 mark for: 1
The denary value in each group of 4 bits is greater than 9
// the denary value in each nibble is greater than 9
4(e)(i) 1 mark for working, 1 mark for answer 2
0011 1101
+0010 1101
0110 1010
111 1 1
4(e)(ii) 1 mark for working, 1 mark for answer 2
0011 1101
+1101 0011 (two’s complement)
0001 0000
1111 111
Riya has created the following logo as a vector graphic.
R
(a) Complete the table by writing a description of each vector graphic term and give an example for this logo.
Term Description Example from logo
Property
Drawing list
[4]
(b) Riya takes a photograph using a digital camera. The photograph is stored as a bitmap image.
(i) Describe two differences between a vector graphic and a bitmap image.
1 ……
2 ……
[4]
(ii) Riya needs to email the photograph. She compresses the photograph before sending it using an email.
Describe two lossy methods that Riya can use to compress the image.
Method 1 ……
Method 2 ……
[4]
Mark scheme
5(a) 1 mark for each description,1 mark for each valid example 4
Term Description Example from logo
data about the shapes e.g.
// defines one aspect of the black line // white fill // black fill
Property
appearance of the drawing //solid (line) // font of letter //
object colour of triangle
the list of shapes involved in e.g.
an image triangle // capital letter R //
Drawing
// a list that stores the rectangle // line
list
command/description required
to draw each object
5(b)(i) 1 mark for each bullet point to max 2 for each difference 4
• Bitmap made up of pixels // bitmap is made of colours stored for
individual pixels
• Vector graphic store a set of instructions about how to draw the shape
• When bitmap is enlarged the pixels get bigger and it pixelates
• When vector is enlarged it is recalculated and does not pixelate
• Bitmap files are usually bigger than vector graphics files because of the
need to store data about each pixel
• Vector graphics have smaller file size because they contain just the
instructions to create the shapes
• Bitmap images can be compressed with significant reduction in file size
• Vector graphic images do not compress well because of little redundant
data
5(b)(ii) 1 mark for each bullet point to max 2 for each method 4
• Reduce bit depth
• ... reduces the number of bits per colour / pixel which means each pixel
has fewer bits
• Reduce colour palette // reduce number of colours
• ... fewer colours mean fewer bits needed to store each colour
• Reduce image resolution
• ... fewer pixels per unit measurement means less binary to store
A shop sells plants to customers. The shop manager has a relational database to keep track of the sales.
The database, PLANTSALES, has the following structure:
PLANT(PlantName, QuantityInStock, Cost)
CUSTOMER(CustomerID, FirstName, LastName, Address, Email)
PURCHASE(PurchaseID, CustomerID)
PURCHASE_ITEM(PurchaseID, PlantName, Quantity)
(a) The database is normalised.
(i) The table lists the following three stages of normalisation:
• The first stage is from a database that is not normalised (0NF) to First Normal Form
(1NF).
• The second stage is from 1NF to Second Normal Form (2NF).
• The third stage is from 2NF to Third Normal Form (3NF).
Tick () one box in each row to identify the appropriate stage for each task.
Normalisation stage
Task
0NF to 1NF 1NF to 2NF 2NF to 3NF
Remove any partial key dependencies
Remove any repeating groups of attributes
Remove any non-key dependencies
[2]
(ii) Draw an entity-relationship (E-R) diagram for the database PLANTSALES.
PLANT CUSTOMER
PURCHASE_ITEM PURCHASE
[3]
(b) The shop manager uses a Database Management System (DBMS).
Describe the purpose and contents of the data dictionary in the DBMS.
…… [3]
(c) The shop manager uses both Data Definition Language (DDL) and Data Manipulation
Language (DML) statements to create and search the database.
(i) Complete the DML statements to return the total number of items purchased with the purchase ID of 3011A.
SELECT SUM(……)
FROM ……
WHERE …… = ……;
[4]
(ii) Write DDL statements to include a field in the table PURCHASE to store the date of the order.
…… [3]
Mark scheme
6(a)(i) 1 mark for 1 tick in the correct place 2
2 marks for all 3 ticks correct
Normalisation stage
Task
0NF to 1NF to 2NF to
1NF 2NF 3NF
Remove any partial key dependencies
Remove any repeating groups of attributes
Remove any non-key dependencies
6(a)(ii) 1 mark for each correct relationship 3
PLANT CUSTOMER
PURCHASE_ITEM PURCHASE
6(b) 1 mark for description of purpose 3
• Stores metadata about the database
1 mark for each example of contents to max 2
e.g.
• field / attribute names
• table name
• validation rules
• data types
• primary keys // foreign keys
• relationships
6(c)(i) 1 mark for each correctly completed space 4
SELECT SUM(Quantity)
FROM PURCHASE_ITEM
WHERE PurchaseID = "3011A";
6(c)(ii) 1 mark per bullet point 3
• ALTER TABLE PURCHASE
• ADD OrderDate
• Suitable data type, e.g. DATE
ALTER TABLE PURCHASE
ADD OrderDate DATE;
A computer has system software.
(a) The Operating System handles interrupts.
Tick () one box in each row to identify whether each event is an example of a hardware interrupt or a software interrupt.
Event Hardware interrupt Software interrupt
Buffer full
Printer is out of paper
User has pressed a key on the keyboard
Division by zero
Power failure
Stack overflow
[3]
(b) Describe the file management tasks that an Operating System performs.
…… [4]
(c) Identify two utility programs that can be used to improve the performance of a computer and state how they improve the performance.
1 ……
2 ……
[4]
Mark scheme
7(a) 1 mark per pair of rows (shaded & unshaded) 3
Event Hardware Software
Interrupt Interrupt
Buffer full
1
Printer is out of paper
User has pressed a key on the keyboard
1
Division by zero
Power failure
1
Stack overflow
7(b) 1 mark for each bullet point to max 4 4
• Storage space is divided into file allocation units
• Space is allocated to particular files
• Maintains / creates directory structures
• Specifies the logical method of file storage (e.g. FAT or NTFS)
• Provides file naming conventions
• Controls access // implements access rights // implements password
protection // Makes file sharing possible
• Specifies tasks that can be performed on a file (e.g. open, close, delete,
copy, create, move etc.)
7(c) 1 mark for identifying program 1 mark for description, max 2 per program 4
e.g.
• Defragmentation
• Less time is taken to access files because each one is contiguous so
there is less head movement
• Virus checker
• makes more RAM available for programs to run
• ... because it removes software that might be taking up memory /
replicating
• Disk repair / Disk contents analysis
• preventing bad sectors being used because it identifies / marks them
• reduces access times by optimising storage
• Disk/system clean up
• releases storage by removing unwanted / temporary files
The Von Neumann model for a computer system uses registers.
(a) Describe the role of the following special purpose registers in the fetch-execute (F-E) cycle.
(i) Memory Address Register (MAR) ……
Memory Data Register (MDR) ……
[4]
(ii) Another special purpose register is the Index Register.
Identify one other special purpose register used in the Von Neumann model for a computer system.
…… [1]
(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
STO <address> Store the contents of ACC at the given address
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>
JPN <address> Following a compare instruction, jump to <address> if the compare was False
JMP <address> Jump to the given address
IN Key in a character and store its ASCII value in ACC
OUT Output to the screen the character whose ASCII value is stored in ACC
END Return control to the operating system
XOR #n Bitwise XOR operation of the contents of ACC with the operand
XOR <address> Bitwise XOR operation of the contents of ACC with the contents of <address>
OR #n Bitwise OR operation of the contents of ACC with the operand
OR <address> Bitwise OR operation of the contents of ACC with the contents of <address>
AND #n Bitwise AND operation of the contents of ACC with the operand
AND <address> Bitwise AND operation of the contents of ACC with the contents of <address>
LSL #n Bits in ACC are shifted logically n places to the left. Zeros are introduced on the right hand end
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 symbolic address
# denotes a denary number, e.g. #123
The current contents of main memory are shown:
Address Data
100 01010101
101 11110000
102 00001111
103 00000000
104 11111111
(i) In the following table, each row shows the current contents of the ACC in binary and the instruction that will be performed on those contents.
Complete the table by writing the new contents of the ACC after the execution of each instruction.
Current contents of the ACC Instruction New contents of the ACC
01010101 XOR 101
11110000 AND 104
00001111 LSL #4
11111111 OR 102
[4]
(ii) The following table contains five assembly language instruction groups.
Write an appropriate assembly language instruction for each instruction group, using the given instruction set. The first one has been completed for you.
Instruction Group Instruction
Data movement LDM #2
Input and output of data
Arithmetic operations
Unconditional and conditional instructions
Compare instructions
[4]
(iii) The opcode LDM uses immediate addressing. The opcode LDD uses direct addressing.
Identify and describe one additional mode of addressing.
Mode of addressing ……
Description ……
[2]
Mark scheme
8(a)(i) 1 mark for each bullet point to max 2 for each register 4
MAR
• Stores the next address to be fetched
• ... held in the Program Counter (PC)
• The data at this address is then fetched
MDR
• Stores the data from the address pointed to by the MAR
• The data in it is copied to the Current Instruction Register (CIR)
8(a)(ii) 1 mark for a correct register 1
e.g.
Program Counter (PC)
Current Instruction Register (CIR)
Status register
Interrupt register
8(b)(i) 1 mark for each correct answer 4
Current contents of New contents of
Instruction
the ACC the ACC
01010101 XOR 101 1010 0101
11110000 AND 104 1111 0000
00001111 LSL #4 1111 0000
11111111 OR 102 1111 1111
8(b)(ii) 1 mark for each correct instruction 4
Instruction Group Instruction
Data movement LDM #2
Input and output of data IN / OUT
Arithmetic Operations INC ACC / INC IX
Unconditional and conditional instructions JPN 100 / JMP 100
Compare instructions CMP 100
8(b)(iii) 1 mark for name, 1 mark for description 2
• Indirect addressing
• the address to be used is at the given address
• Relative addressing
• the address to be used is an offset number of locations away, relative to
the address of the current instruction
• Indexed addressing
• form the address from the given address plus the contents of the index
register
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On this page, free: the mark scheme for AS & A Level Computer Science 9618/12, October/November 2021 (9618_w21_ms_12), and the question paper (9618_w21_qp_12), to read or download, with each question's marking points written out underneath.
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Computer Science 9618/12 for October/November 2021 is out of 75 marks, with 1 hour 30 minutes to answer it.
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