A Level Computer Science 9618 Paper 31, Oct/Nov 2024

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Question paper

9618/31 · Oct/Nov 2024

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1

Numbers are stored in a computer using binary floating-point representation with: • 10 bits for the mantissa • 6 bits for the exponent • two’s complement form for both the mantissa and the exponent.

a

Calculate the normalised binary floating-point representation of +201.125 in this system. Show your working.

[3]
b

Calculate the denary value of the given normalised binary floating-point number. Show your working. Mantissa Exponent 1 0 1 0 1 1 0 0 1 1 0 0 0 1 0 1 Working Answer

[3]
2

Reduced Instruction Set Computers (RISC) is a type of processor. Identify four features of a RISC processor. 1 2 3 4

[4]
3
a

Describe circuit switching as a method of data transmission.

[3]
b

State one benefit and one drawback of circuit switching as a method of data transmission. Benefit Drawback

[2]
4

The TCP/IP protocol may be viewed as a stack that contains four layers: Application, Transport, Internet, Link. Describe how the layers of the TCP/IP protocol stack interact with each other.

[4]
5
a

Explain what is meant by a hashing algorithm in the context of file access.

[3]
b

The use of a hashing algorithm can result in the same storage location being identified for more than one record. Outline two methods of overcoming this issue. 1 2

[2]
6
a

Describe the user-defined data type set.

[3]
b

Write pseudocode statements to declare the set data type, SymbolSet, to hold the following set of mathematical operators, using the variable Operators. + – * / ^

[4]
7

The truth table for a logic circuit is shown.

a

Write the Boolean logic expression that corresponds to the given truth table as the sum-of-products. T =

[3]
b

Complete the Karnaugh map (K-map) for the given truth table.

[2]
c

Draw loop(s) around appropriate group(s) in the K-map to produce an optimal sum-of-products.

[2]
d(i)

Write the Boolean logic expression from your answer to part (c) as the simplified sum-of-products. T =

[2]
d(ii)

Use Boolean algebra to write your answer to part (d)(i) in its simplest form. T =

[1]
8
a

Describe the process of segmentation for memory management.

[4]
b

Explain what is meant by disk thrashing.

[3]
9

A veterinary surgery wants to create a class for individual pets. Some of the attributes required in the class are listed in the table. Attribute Data type Description PetID STRING unique ID assigned at registration PetType STRING type of pet assigned at registration OwnerTelephone STRING telephone number of owner assigned at registration DateRegistered DATE date of registration

Figure 1 of 2 for question 9
Figure 2 of 2 for question 9
a

State one reason why the attributes would be declared as PRIVATE.

[1]
b

Complete the class diagram for Pet, to include: • an attribute and data type for the name of the pet • an attribute and data type for the name of the owner • a method to create a Pet object and set attributes at the time of registration • a method to assign a pet ID • a method to assign the date of registration • a method to return the pet name • a method to return the owner’s telephone number.

[5]
10

Several syntax diagrams are shown.

Figure 1 of 3 for question 10
Figure 2 of 3 for question 10
Figure 3 of 3 for question 10
a

Complete the Backus-Naur Form (BNF) for the given syntax diagrams. <operator> ::= .................................................................................................................. ................................................................................................................................................... <label> ::= ......................................................................................................................... ................................................................................................................................................... <equation> ::= .................................................................................................................. ...................................................................................................................................................

[4]
b(i)

A new syntax rule, password, is required. It must begin with a letter or a symbol, followed by a digit and end with one or two symbols. Draw a syntax diagram for password.

[3]
b(ii)

Write the BNF for password.

[2]
11

The following diagram shows an ordered binary tree.

Figure 1 of 2 for question 11
Figure 2 of 2 for question 11
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.

[4]
b

A user-defined record structure is used to store the nodes of the linked list in part (a). Complete the diagram, using your answer for part (a).

[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 BinTree. 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 SearchTree(Item : STRING) ........................................................................ NowPtr ......................................................................................................................... WHILE NowPtr <> -1 IF ..................................................................................................................... THEN NowPtr BinTree[NowPtr].LeftPtr ELSE IF BinTree[NowPtr].Data < Item THEN ......................................................................................................................... ELSE RETURN NowPtr ENDIF ENDIF ENDWHILE RETURN NowPtr ENDFUNCTION

[4]

Mark scheme

9618/31 · Oct/Nov 2024

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1a · 3 marks

  • 1 mark

    correct answer // statement regarding number losing precision/rounding error

  • 1 mark

    number converted to binary 201.125 = 11001001.001 // 128 + 64 + 8 + 1 + 0.125 / 1/8 seen

  • 1 mark

    use of the exponent e.g. moving the binary point 8 places / × 2^8.

1b · 3 marks

  • 1 mark

    application of exponent to go from 1.010110011 to 101011.0011 // x 2^5 // movement of binary point 5 places seen

  • 1 mark

    –32 + 8 + 2 + 1 + .125 + .0625 // –32 + 8 + 2 + 1 + 1/8 + 1/16 seen // –1 + ¼ + 1/16 + 1/32 + 1/256 + 1/512 // –1 + 179/512 // –333/512

  • 1 mark

    –20.8125 // –20 13/16

2 · 4 marks

  • 1 mark

    low number of instruction formats //low number of instruction sets

  • 1 mark

    uses single-clock cycle instructions

  • 1 mark

    uses fixed length instructions

  • 1 mark

    uses many general-purpose registers

3a · 3 marks

  • 1 mark

    A dedicated circuit / channel is required

  • 1 mark

    The circuit is established before the transmission begins

  • 1 mark

    The circuit lasts for the whole of the transmission // The circuit is closed at the end of the transmission

3b · 2 marks

  • 1 mark

    Benefit: MP1 No need for data to be reassembled // data / frames arrive in the same order in which they were sent // MP2 Suitable for real time transmission // fast data transfer rate // MP3 The whole of the bandwidth is available

  • 1 mark

    Drawback: MP4 No other transmission can use the same circuit when it is in use // Bandwidth can be wasted as it cannot be used by other messages // MP5 Not secure // Can be intercepted as all data travelling along the same route // MP6 If there is a problem with the route the transmission ends // No other route is available without first doing the setup // MP7 The circuit is always there whether or not it’s being used // MP8 Can take time to set up before transmission starts.

4 · 4 marks

  • 1 mark

    Each layer can only accept input from the next higher layer or the next lower layer

  • 1 mark

    There is an interface between the adjacent layers which is the only interaction between layers

  • 1 mark

    Data is added to the headers as the frames/packets pass through the layers

  • 1 mark

    The interactions are carried out by installed software

5a · 3 marks

  • 1 mark

    A hashing algorithm is used in direct access methods on random and sequential files

  • 1 mark

    It is a mathematical formula

  • 1 mark

    … used to perform a calculation applied to the key field of the record being searched / stored

5b · 2 marks

  • 1 mark

    The record is stored in the next free memory space after the one identified by the hashing algorithm // Use linear progression

  • 1 mark

    An overflow area is set up and the record is stored in the next free memory space in the overflow area.

6a · 3 marks

  • 1 mark

    A set user-defined data type is a composite data type

  • 1 mark

    … which includes a list of unordered elements

  • 1 mark

    Set theory operations, such as intersection and union, can be applied to these elements

6b · 4 marks

  • 1 mark

    TYPE SymbolSet/Operators =

  • 1 mark

    SET OF CHAR

  • 1 mark

    DEFINE Operators/SymbolSet

  • 1 mark

    ('+', '–', '*', '/', '^') // : SymbolSet/Operators

7a · 3 marks

  • 1 mark

    First two correct products

  • 1 mark

    Next two correct products

  • 1 mark

    Final two correct products

7b · 2 marks

  • 2 marks

    Two marks if no errors present, One mark if one error present

7c · 2 marks

  • 1 mark

    Correct loop for group of 4 (columns 11, 10; rows 01, 11)

  • 1 mark

    Correct loop for wraparound group of 4 (columns 00, 10; rows 01, 11)

7d(i) · 2 marks

  • 1 mark

    Any correct Boolean term

  • 1 mark

    Boolean terms and operator correct and no other terms present

7d(ii) · 1 mark

  • 1 mark

    (T =) D. (A + B̅)

8a · 4 marks

  • 1 mark

    In segmented memory, the logical / virtual address space is broken into varying sized blocks called segments / sections.

  • 1 mark

    Each segment has a name and size.

  • 1 mark

    During execution segments from logical / virtual memory are loaded into physical memory.

  • 1 mark

    The address is specified by the user // … it contains the segment name and offset value.

8b · 3 marks

  • 1 mark

    Disk thrashing is a problem that may occur when virtual memory is being used.

  • 1 mark

    As the main memory fills up, more and more pages need to be swapped in and out of virtual memory.

  • 1 mark

    This swapping leads to a very high rate of hard disk access / excessive disk head movements.

9a · 1 mark

  • 1 mark

    To ensure that the attributes are only accessible using the class's own methods/within the class.

9b · 5 marks

  • 1 mark

    Two correct attributes with sensible names and correct data types.

  • 1 mark

    Constructor present.

  • 1 mark

    Two correct setters with exact names and appropriate parameters and data types.

  • 1 mark

    Two correct getters with appropriate names.

  • 1 mark

    Name assigned to pet name getter matches the attribute.

10a · 4 marks

  • 1 mark

    <operator> ::= + | – | * | /

  • 1 mark

    <label> ::= <letter><digit>|<letter><digit><digit>

  • 1 mark

    <equation> ::= <label> =

  • 1 mark

    <label><operator><label>

10b(i) · 3 marks

  • 1 mark

    begin with either a letter or a symbol

  • 1 mark

    end with either one or two symbols

  • 1 mark

    digit and all other connections and label correct.

10b(ii) · 2 marks

  • 1 mark

    <password> ::= <letter><digit><symbol>|

  • 1 mark

    <letter><digit><symbol><symbol>|<symbol><digit><symbol>|<symbol><digit><symbol><symbol>

11a · 4 marks

  • 1 mark

    Four additional nodes with correct data values

  • 1 mark

    Correct null pointers in all added nodes (6) with no extra null pointers where the arrow points to the next node

  • 1 mark

    Correct arrows to represent pointers joining parent nodes to child nodes

  • 1 mark

    All nodes in correct order and no extra data added to pointers.

11b · 4 marks

  • 1 mark

    Correct Red and Green rows

  • 1 mark

    Correct Yellow and Blue rows

  • 1 mark

    Correct Orange, Indigo and Violet rows

  • 1 mark

    Correct FreePtr with blank row 7

11c · 4 marks

  • 1 mark

    NowPtr  RootPtr and WHILE NowPtr <> -1

  • 1 mark

    IF BinTree[NowPtr].Data > Item THEN NowPtr  BinTree[NowPtr].LeftPtr

  • 1 mark

    ELSE IF BinTree[NowPtr].Data < Item THEN NowPtr  BinTree[NowPtr].RightPtr

  • 1 mark

    ELSE RETURN NowPtr and RETURN NowPtr after ENDWHILE

Questions

  1. 1
    Question 16 marks
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  2. 2
    Question 24 marks
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  3. 3
    Question 35 marks
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  4. 4
    Question 44 marks
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  5. 5
    Question 55 marks
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  6. 6
    Question 67 marks
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  7. 7
    Question 710 marks
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  8. 8
    Question 87 marks
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  9. 9
    Question 96 marks
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  10. 10
    Question 109 marks
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  11. 11
    Question 1112 marks
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