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A Level Computer Science 9618 Paper 33, Oct/Nov 2024

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9618/33 · 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. Mantissa Exponent Working

[3]
b

Calculate the denary value of the given normalised binary floating-point number. Show your working. Mantissa: 1 0 1 0 1 1 0 0 1 1 Exponent: 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. INPUT OUTPUT A B C D T 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1

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. Pet PetID : STRING PetType : STRING OwnerTelephone : STRING DateRegistered : DATE .............................................................. : .............................................................. .............................................................. : .............................................................. ................................................................................................................................. ................................................................................................................................. ................................................................................................................................. ................................................................................................................................. .................................................................................................................................

[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

9618/33 · 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

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

  • 1 mark

    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 // Not secure // Can be intercepted as all data travelling along the same route // If there is a problem with the route the transmission ends // No other route is available without first doing the setup // The circuit is always there whether or not it’s being used // 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

    Two correct products

  • 1 mark

    Four correct products

  • 1 mark

    Six correct products: (T =) Ā.B̄.C̄.D + Ā.B̄.C.D + Ā.B.C̄.D + Ā.B.C.D + A.B̄.C̄.D + A.B.C.D

7b · 2 marks

  • 2 marks

    Correctly completed Karnaugh map (Two marks if no errors present, one mark if one error present)

7c · 2 marks

  • 1 mark

    Correct loop for column 00 (rows 01 and 11) overlapping/grouped with column 10 (rows 01 and 11) / wrap-around loop

  • 1 mark

    Correct loop for columns 11 and 10 (rows 01 and 11)

7d(i) · 2 marks

  • 1 mark

    Any correct Boolean term

  • 1 mark

    Boolean terms and operator correct and no other terms present: (T =) A.D + B̄.D // B̄.D + A.D

7d(ii) · 1 mark

  • 1 mark

    (T =) D.(A.B̄)

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 // MP5 … it contains the segment name and offset value. // MP6 Segments are numbered // MP7 … and this number is used as an index in the segment map table. // MP8 The offset value determines the size of the segment. // MP9 A segment map table maps logical / virtual addresses to physical addresses / contains the segment number and offset.

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

  • 1 mark

    NowPtr ← BinTree[NowPtr].LeftPtr

  • 1 mark

    NowPtr ← BinTree[NowPtr].RightPtr

  • 1 mark

    RETURN NowPtr (inside or after the loop)

Questions

  1. 1
    Question 16 marks
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  2. 2
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  3. 3
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  4. 4
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  5. 5
    Question 55 marks
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  6. 6
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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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