a2 Student Book Answers

8/15/2019 a2 Student Book Answers

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Student book answers

1.1 ........................................................................................................................................... 3

End of topic questions ..................................................................................................... 3

Exam practice questions ................................................................................................. 6

1.2 ........................................................................................................................................... 7

End of topic questions ..................................................................................................... 7Exam practice questions ................................................................................................. 9

1.3 ........................................................................................................................................... 9

End of topic questions ..................................................................................................... 9

1.4 ......................................................................................................................................... 14

End of topic questions ................................................................................................... 14

Exam practice questions ............................................................................................... 16

1.5 ......................................................................................................................................... 16



1.6 ......................................................................................................................................... 18



2.1 ......................................................................................................................................... 21


Exam practice questions ............................................................................................... 302.2 ......................................................................................................................................... 36


2.3 ......................................................................................................................................... 38



2.4 ......................................................................................................................................... 68


Exam practice questions ............................................................................................... 842.5 ......................................................................................................................................... 90



2.6 ......................................................................................................................................... 95


Exam practice questions ............................................................................................. 102

2.7 ....................................................................................................................................... 106End of topic questions ................................................................................................. 106


3.1 ....................................................................................................................................... 112

End of topic questions ................................................................................................. 112





5.1 ....................................................................................................................................... 121



5.2 ....................................................................................................................................... 122





6.1 ....................................................................................................................................... 137



6.2 ....................................................................................................................................... 142

End of topic questions ................................................................................................. 142Exam practice questions ............................................................................................. 146

6.3 ....................................................................................................................................... 148



6.4 ....................................................................................................................................... 150





7.2 ....................................................................................................................................... 159


7.3 ....................................................................................................................................... 160


7.4 ....................................................................................................................................... 166


7.6 ....................................................................................................................................... 168




1.1

End of topic quest ions

1 Information hiding means hiding design details behind a standard interface.

2 An interface is a device or combination of devices such as a keyboard/VDU situated between the

user and the internal operation of a machine and through which the user interacts with the machine.

3 Possible answers are:

a) By hiding the complexities of the machine behind a well-designed interface users are able to

learn how to use a machine with the minimum of training.

b) The operator of a machine does not need to know how the operation of the machine is achieved

only how to control it.

c) The internal components of the machine may be replaced by similarly functioning components

but designed in a different way, e.g. using a different technology, without the experience of the

user changing or the user needing to be re-trained.4 In order to make the manufacturing, and maintenance of a car manageable, a car is divided into

parts with particular interfaces hiding design decisions. Some car parts or components are listed

below:

Engine

Gearbox

Transmission (transfers power from engine to wheels)

Lights Pedals

This division into separate components or modules allows the car manufacturer to offer different

options. For example, a particular model may be offered with different capacities of engine, 1.8

litres, 2.5 litres, 3 litres. All three engines fit the same transmission, the same engine mounts, and

the same controls. This is possible because all three engines provide the same interface to the other

parts/modules that they connect to. The same applies for maintenance. An engine may be replaced

relatively easily because it is designed to fit the engine mounts, transmission system and so on.

5 A well-designed computer program has a solution in which the source code has been decomposedinto modules using the principle of information hiding. When a programmer is working on one

module he or she does not need to know the detail contained in other modules, only what other

modules can do for the programmer. Also changes are much easier to make with this approach

because the changes typically are local to a module rather than global changes across all modules.

6 a) Different objects can have identical interfaces

b) Sharing a common interface among many different objects means that users of the objects do not

need to be re-trained when changing from using one object to using another

c) To use a module or object a user needs to have no knowledge of the module’s or object’s

internal design

d) The internal design can be kept secret

e) Allows flexibility, one module may be replaced by another module with an identical interface

but a different internal design or manner in which the module’s function is carried out



and link these by lines representing the route. You have left out all the unnecessary detail on the

ordnance survey map in order to convey the essentials of the route in your sketch.

9

F ← GetF

C ← ConvertFtoC(F)

Display(C)

10 There are several ways of classifying plants. One way to classify plants is according to the way a

plant absorbs water. Plants are either vascular or non-vascular. Vascular plants have tube-like

structures that transport water from the roots to the stem to the leaves. Non-vascular plants absorb

water only through their surfaces.

Parallel veins Branching

network of veins

Non-

vascular

Angiosperms

Vascular

Monocots

Seeds on cones

Seeds in

flowers

Write('Input F: ')

Readln(FTemp)

GetF ← FTemp

ConvertFtoC ← (F - 32)* 5/9 Writeln('Celsius = ', C)

F ← GetF

ConvertTemperature

Display(C) C ← ConvertFtoC(F)

Parallel Branching



12 Assume that in a box there are 10 black socks and 12 blue socks and you need to get one pair of

socks of the same colour. Supposing you can take socks out of the box only once and only without

looking, how many socks do you have to pull out together? Use the Pigeonhole principle.

The correct answer is three. To have at least one pair of the same colour (m = 2 holes, one

per colour), using one pigeonhole per colour, you need only three socks (n = 3 objects). In

this example, if the first and second sock drawn are not of the same colour, the very next

sock drawn would complete at least one same-colour pair. (m =.

13

Columbia

Suriname

Argentina

Peru

French

Guyana

Bolivia

VenezuelaGuyana

Ecuador

Brazil

Paraguay

Chile

Uruguay



14

(a)

(b) 3

Exam pract ice quest ions

Question Number Answer Marks

1 Information hiding means hiding design details behind a

standard interface.

In software development, a programmer can use alibrary module without needing an intimate knowledge

of the module’s design. Information hiding usually

means that a software module is self-contained,

otherwise its design would have to be revealed. This

allows the internal design of a module to be changed

whilst isolating users from these changes, i.e. the

software that uses the module should still work and probably better than before because of the internal

changes to the module. Also, changes are much easier to

make with this approach because the changes typically are

local to a module rather than global changes across all

modules. If a whole software project is divided into

modules using the principle of information hiding then it is

possible for different programmers to work on different

modules at the same time so that the project can becompleted more quickly. Information hiding also creates

designs that can be kept secret more easily. From a user’s

point of view, information hiding means that when the

internal design of a module changes the user doesn’t have to

be retrained. Outwardly the module should still appear the

same

3

A

E

D

C

B



maintenance of a car manageable, a car is divided into parts

with particular interfaces hiding design decisions. This

division into separate components or modules allows the car

manufacturer to offer different options. For example, a

particular model may be offered with different capacities of

engine, 1.8 litres, 2.5 litres, 3 litres. All three engines fit the

same transmission, the same engine mounts, and the same

controls. This is possible because all three engines provide

the same interface to the other parts/modules that they

connect to.


2a Abstraction by generalisation proceeds from a detailed

picture of specific instances to a less detailed picture of

classes to which the instances belong or are examples of.

This is an example of classification-type generalisation.

Another type of generalisation is of the decomposition-type.

In this type, the abstraction proceeds in the direction of

more and more detail as in top-down program development.

1

2b Abstraction by representation is applied to problem solving.

Details are removed until it becomes possible to represent

the problem in a way that is possible to solve.

1

5 Question

Number

6 Answer 7 Marks

3a 5

3b 2 tables 2

1.2


1 Algorithms are considered to be procedural solutions to problems and problem solving is a human

B

C D

E

A



2 a) Twice as long

b) Eight times as long

c) 2n times as long

3 Using a standard unit of time measurement, such as a second, to estimate the running time of a

program implementing the algorithm, would not be sensible. The estimate would be dependent on the

speed of the computer, the quality of the program implementing the algorithm and the compiler used in

generating the machine code.

4 Compute the number of times the basic operation is executed on inputs of size n. The basicoperation is the operation in the algorithm contributing the most to the total running time.

5 a) The computational complexity of an algorithm measures how economical the algorithm is with

time and space.

b) Computational complexity of a problem is taken to be the worst-case complexity of the most

efficient algorithm, which solves the problem.

6 a) Time-complexity of an algorithm indicates how fast an algorithm runs.

b) Space-efficiency of an algorithm indicates how much memory an algorithm needs.

7 Input size is number of elements in the array, i.e. n. The innermost loop contains a single operation,

the comparison of two elements. Consider this as the algorithm’s basic operation. In the worst-case

scenario, the input is an array for which the number of element comparisons is the largest amongst

all arrays of size n. There are two cases for this worst case scenario:

a) arrays with no equal elements

b) arrays in which the last two elements are the only pairs of equal elements.

For such inputs, one comparison is made for each repetition of the innermost loop, i.e. for each

value of the loop’s variable j between limits i + 1 and n. This is repeated for each value of the outerloop, i.e. for each value of the loop’s variable i between its limits 1 and n - 1. Consider the case

when n = 6.

Element

no.

No. of

comparisons

Element in first column compared with

elements

1 5 2, 3, 4, 5, 6

2 4 3, 4, 5, 6

3 3 4, 5, 6

4 2 5, 6

5 1 6

In general, the number of comparisons becomes 1 + 2 + 3 + 4 + 5 + … + n

This has the sum

b) For large values of n this becomes2

2n

as (n - 1)n = n2 - n and for large n, n

2 dominates.

8 Asymptotic behaviour of a function is the behaviour of the function for very large values of n.

9 Given a function g (n) then big O of g , represents the class of functions that grow no faster than the

function g .

10 f = O(log 2 n)

n-1 n

2



f = O(2n)

f = O(n!)

11 (a)Exponential growth exhibits growth rates that vary according to the form k n, e.g. 2n where k = 2

and n = 1, 2, 3, etc.

(b) Polynomial growth exhibits growth rates that vary according to the form nk , e.g. n

3 where k = 3

and n = 1, 2, 3, 4, etc.

(c) An exponential time algorithm is one whose execution time grows exponentially with input size.

(d) A polynomial time algorithm is one whose execution time grows as a polynomial of input size.(e) A linear time algorithm is a polynomial time algorithm that executes in O(n) time.


Question 1 – Specimen Paper COMP 3 Q1

Question number Answer Marks

1a 1. A

2. C

3. B

1

1b Start at first item; and examine each succeeding item in turn;

Until item is found; or the end of the list reached;

A algorithm

max 3

O(n) as up to n items searched;

All items may be searched;

max 1

4

1c For each succeeding pair of items;

if they are out of sequence they are swapped;the process is repeated;

up to n-1 times; or until no more swaps are made;

max 3

All n items are compared up to n-1 times;

1

4

1.3


1 A finite state machine (FSM) consists of a set of input symbols (input symbol alphabet) and if it



3 00000000, S0

4

5 Mealy machine: an FSM that determines its outputs from the present state and from the inputs.

Moore machine: an FSM that determines its outputs from the present state only.

6

7 = 0

= 1

= 1

= 0c1

8 One-bit binary adder

9 An FSM without output is known as a finite state automaton (FSA) or finite automaton, plural finite

state automata. FSA are restricted to decision problems, i.e. problems that are solved by answeringYES or NO. Therefore, FSA do not write anything at all whilst processing input whereas an FSM

with output does.



10

No

Yes

No

Yes

No







11 An FSA is deterministic if every state has exactly one transition or edge leaving it for each of the

symbols in the alphabet. As there is exactly one edge for each symbol, an input word will cause the

exact same behaviour every time a deterministic finite automaton is run on the word. A finite

automaton can be non-deterministic if it does not have exactly one edge for each symbol of the

alphabet. A state can have no edges for a symbol or it can have multiple edges for each symbol of

the alphabet.

12 On reaching the end of the input if the FSA is in an accepting state then the FSA outputs a YES

response.

13 aa|b(a|b)*

1.4


1 11111



2 a)

b)

3 Adds 1 to the number on the tape

4 A Turing machine (TM) is a finite state machine that controls one or more tapes, where at least one

tape is of unbounded length (i.e. infinitely long).

5 Principle of universality: a universal machine is a machine capable of simulating any other

machine.6 A universal Turing machine, U, is an interpreter that reads the description <M> of any arbitrary

Turing machine M and faithfully executes operations on data D precisely as M does. For single-tape

Turing machines, imagine that <M> is written at the beginning of the tape, followed by D.

7 Coffee maker

8 A consequence of a universal Turing machine is that programs and data are really the same thing.

A program is just a sequence of symbols that looks like any other piece of input but when fed to a

universal machine this input wakes up and begins to compute. A computer downloads Java applets,

e-mail viruses as data but can then proceed to execute them as programs.9 A microprocessor is a universal machine. An embedded computer with a microprocessor at its core

can be programmed to perform as a microcontroller by choosing the right set of program

instructions. The program or programs are changed when a different functionality is required.

10 A task is computable if and only if it can be computed by a Turing machine.




Question 1 – Specimen Paper COMP3 Q6

Question

Number

Answer Marks

1a (i)

1 mark for correctly positioned arrow, ignore data on tape

(ii) 1

1

1

1b 2 marks for correct tape symbols, 1 mark for correct position of arrow 3

1c 3 + 1 or the successor to 3 1

1.5


1

a) The answer is not decidable

b) Undecidable or noncomputable

2

1024

10485761073741824

1099511627776

1125899906842624

1152921504606847000

1.329227995784916 × 10+36

3

a)

24 = 16

210

= 1024

18446744073709552000

1.33 × 1050

b) Intractable



c)

first option = £10.24 and second option = £11



5 15 days

6 NP-type problem

7 No



1a Let’s call the liars group A and the truth tellers group B. If

the native belongs to B, then since this native tells the truth

he/she contradicts themself by saying that they did not say

they were a truth teller when asked if they were a truth

teller. Therefore the native belongs to the liars group.

1

1b The problem posed in part (a) belongs to the family of logic

problems known as decision problems.

1

1c First is from the liars group. If the first is from the truth

teller’s group then the statement would be a contradiction.

If the first is from the liar’s group then he has lied and so both are not from the liar’s group. Therefore, the second

must be from the truth teller’s group.

1

The second from the truth teller’s group. 1

1d 1. No, there is only one haircutter and he cuts the hair of

everyone on the island who does not cut their own hair.

If the haircutter cut their hair then this would contradict

the statement in the previous sentence. Therefore, thehaircutter does not cut his own hair. So, the haircutter’s

hair must be cut by someone else. However, everyone

who cuts another’s hair is called a haircutter but the

island has only one haircutter. Therefore, the question

cannot be answered.

1

2. This type of problem is undecidable. 1

1e 16 1

210

1

264

1

Intractable 1

The planks would take up a large amount of space and it 2



2a Intractable means that no reasonable (polynomial) time

solution has yet been found. Processing will take an

unreasonable amount of time even for quite small data sets.

2

2b For a class of intractable problems it is possible to guess a

solution and check the solution in polynomial time, i.e. a

reasonable amount of time. School timetabling is one of

these problems. They are known as non-deterministic

polynomial type problems or NP-type problems.

2


3 The halting problem is the problem of determining without

executing a program whether it will halt on a given input. It

relates to answering the question of whether or not it is

possible to write a program that can, in general, test any

other program supplied with its input and determine, without

executing the program under test, if it will halt on this input.The answer to this question is no.

2

1.6


1

a) A valid string consists of one or more as or one or more cs followed by two bs[ll]

b) aaccbb, cacccbb

2 (a|c)a*c*bb

3

a) (0|1)*

b) 1+

c) 10*

d) (110|011)(0|1)*

e) (0|1)+0

f) 11(0|1)*11

4 x*y*z*

5 beean

6 bead7 10101

8 01296-433006 or 01793 234589

9 www.([_a-zA-Z\d\-]+(\.[_a-zA-Z\d\-]+)+)

10 li(c|s)en(c|s)e



11 Roman Numerals

12 Identifier

Letter

Digit

V

I

X

I

I

V

I



13 UserID

14

a) xy+

b) xy+2/

c) xy+x4-/

d) xy↑

e) xy+2↑

f) xy+xy-/

15

a) x * y

b) (x + y) * 6

c) (x + y) / (x - y)

d) (2 * x + 3 * y) /8

e) ((x +y) ↑ 2) * (6 - x)

Digit HDigit

L

D

R

Pa

Ph

A..Z A..Z




Question 1 – Specimen Paper COMP3 CPT4 Q7

Question

number

Answer Marks

1a 4

1b 5 6 2 ;*; +;

No Brackets;

Easy to Compute;

2

1

Question 2 - KB


2a Yes 1

2b No 1

2c Yes 1

2.1


Q1 This is to produce a table as Table 01 in the book.

Q2 & Q3 are just using the code in the book and running it.

Q4 & Q5 Library unit

Unit Unit2;

Interface

Type TClock = Class

Pri ate



Procedure SetTime(h, m : Integer);Procedure IncrementTime ; Virtual;

End ;

Type TAlarmClock = Class (TClock)

Private

AlarmHours: Integer;

AlarmMinutes: Integer;

Public

Function GetAlarmHours: Integer;

Function GetAlarmMinutes: Integer;Procedure SetAlarmTime(h, m : Integer);

End ;

Type TWatch = Class (TClock)

Private

DayOfWeek: Integer;

DayNumber: Integer;

Public

Function GetDayNumber: Integer;

Function GetDayName: String;Procedure SetDayOfWeek(d : Integer);

Procedure SetDayNumber(d : Integer);

Procedure IncrementTime; Override;

End ;

Implementation

Uses SysUtils;

Function TClock.GetHours: Integer;

Begin

GetHours := Hours;

End ;

Function TClock.GetMinutes: Integer;

Begin

GetMinutes := Minutes;

End ;

Procedure TClock.SetTime (h, m : Integer);

BeginHours := h;

Minutes := m;

End ;

Procedure TClock.IncrementTime;

Begin

Minutes := Minutes + 1;

Sleep(600); // wait one minute

If Minutes = 60Then

Begin

Hours := Hours + 1;

Minutes := 0;

If Hours = 24 Then Hours := 0;

End;



If Hours = 24Then

Begin

Hours := 0;

DayNumber := DayNumber + 1;

If DayNumber > 31 Then Daynumber := 1;

DayOfWeek := DayOfWeek + 1;

If DayOfWeek > 7 Then DayOfWeek := 1;

End ;

End ;

Function TWatch.GetDayNumber: Integer;

Begin

GetDayNumber := DayNumber;

End ;

Function TWatch.GetDayName: String;

Begin

Case DayOfWeek Of1: GetDayName := 'Sun';

2: GetDayName := 'Mon';

3: GetDayName := 'Tue';

4: GetDayName := 'Wed';

5: GetDayName := 'Thu';

6: GetDayName := 'Fri';

7: GetDayName := 'Sat';

End ;

End ;

Procedure TWatch.SetDayOfWeek (d : Integer);

Begin

DayOfWeek := d;

End ;

Procedure TWatch.SetDayNumber(d: Integer);

Begin

DayNumber := d;

End ;

Function TAlarmClock.GetAlarmHours: Integer;

Begin

GetAlarmHours := AlarmHours;

End ;

Function TAlarmClock.GetAlarmMinutes: Integer;

Begin

GetAlarmMinutes := AlarmMinutes;End ;

Procedure TAlarmClock.SetAlarmTime (h, m : Integer);

Begin

AlarmHours := h;

AlarmMinutes := m;



Q4)

Program AlarmClockExample;

{$APPTYPE CONSOLE}

Uses

SysUtils,

Unit2 in 'Unit2.pas';

Var AlarmClock: TAlarmClock;

Begin

AlarmClock := TAlarmClock.Create;

AlarmClock.SetTime(15,10);

AlarmClock.SetAlarmTime(15,15);

Repeat

AlarmClock.IncrementTime;

writeln(AlarmClock.GetHours,':',AlarmClock.GetMinutes);

Until (AlarmClock.GetAlarmHours = AlarmClock.GetHours)

And (AlarmClock.GetAlarmMinutes = AlarmClock.GetMinutes);Writeln('Wake up');;

AlarmClock.Free;

ReadLn;

End .

Q5)

Program WatchExample;

{$APPTYPE CONSOLE}

uses

SysUtils,

Unit2 in 'Unit2.pas';

Var Watch: TWatch;

Begin

Watch := TWatch.Create;

Watch.SetTime(23,50); // give a start time

Watch.SetDayNumber(15); // give a start date

Watch.SetDayOfWeek(7); // give a day of the week

Repeat

Watch.IncrementTime;// display the time and date

write(Watch.GetHours:2,':',Watch.GetMinutes:2);

WriteLn(' ',Watch.GetDayName,' ', Watch.GetDayNumber);

Until False;

Watch.Free;

ReadLn;



Q6 & Q7Class definition code:

Unit ClassDefs;

Interface

Uses Classes, StdCtrls;

Type TClock = Class (TLabel)

Private

Hours: Integer;Minutes: Integer;

Seconds: Integer; // only required for Exercise 2

Function IntToStrF (i: Integer): String;

Public

Constructor Create (AOwner:TComponent); Override;

Procedure IncrementTime;

Procedure DisplayTime;

End ;

Implementation

Uses SysUtils, StrUtils;

Constructor TClock.Create (AOwner: TComponent);

Begin

Inherited Create(AOwner);

Top := 20;

Left := 20;

Hours := 0;

Minutes := 0;

Seconds := 0; // only required for Exercise 2

Caption := '00:00:00';

End ;

Procedure TClock.IncrementTime;

Begin

Seconds := Seconds + 1; // only required for Exercise 2

If Seconds = 60 // only required for Exercise 2

ThenBegin


Seconds := 0; // only required for Exercise 2

If Minutes = 60

Then

Begin

Hours := Hours + 1;

Minutes := 0;

If Hours = 24Then Hours := 0;

End ;

End ;

End ;

Function TClock IntToStrF (i: Integer): String;



Else IntToStrF := StringI;End ;

Procedure TClock.DisplayTime;

Var StrMinutes, StrHours, StrSeconds : String[2];

Begin

StrMinutes := IntToStrF(Minutes);

StrHours := IntToStrF(Hours);

StrSeconds := IntToStrF(Seconds); // only required for Exercise 2

Caption := StrHours + ':' + StrMinutes + ':' + StrSeconds;End ;

End .

Code for the form:

Unit ClockForm;

Interface

Uses

Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls,

Forms,

Dialogs, ClassDefs, ExtCtrls;

Type

TForm1 = Class(TForm)

tmrClockTick: TTimer;

Procedure FormCreate(Sender: TObject);

Procedure tmrClockTickTimer(Sender: TObject);

Private

{ Private declarations }

Public

Clock : TClock;

End ;

Var

Form1: TForm1;

Implementation

{$R *.dfm}

Procedure TForm1.FormCreate(Sender: TObject);

Begin

Clock := TClock.Create(Self);

Clock.Parent := Self;tmrClockTick.Enabled := True;

tmrClockTick.Interval := 1000; // tick every second

End ;

Procedure TForm1.tmrClockTickTimer(Sender: TObject);

Begin



Q8)

Unit ClassDefs;

Interface

Uses Classes, StdCtrls;

Type TClock = Class (TLabel)

Private

Hours: Integer;

Minutes: Integer;Function IntToStrF (i: Integer): String;

Public

Constructor Create (AOwner:TComponent); Override;

Procedure IncrementTime;

Procedure DisplayTime;

Function GetHours: Integer;

Function GetMinutes: Integer;

Procedure SetTime(h, m : Integer);

End ;Type TAlarmClock = Class (TClock)

Private

AlarmHours: Integer;

AlarmMinutes: Integer;

Public

Function GetAlarmHours: Integer;

Function GetAlarmMinutes: Integer;

Procedure SetAlarmTime(h, m : Integer);

Procedure Ring;

End ;

Implementation

Uses SysUtils, StrUtils;

Constructor TClock.Create (AOwner: TComponent);

Begin

Inherited Create(AOwner);

Top := 20;Left := 20;

SetTime(0,0);

End ;

Procedure TClock.SetTime(h: Integer; m: Integer);

Begin

Hours := h;

Minutes := m;

End ;

Function TClock.GetHours: Integer;

Begin

GetHours := Hours;

End ;



Procedure TClock.IncrementTime;Begin


If Minutes = 60

Then

Begin

Hours := Hours + 1;

Minutes := 0;

If Hours = 24

Then Hours := 0;End ;

End ;

Function TClock.IntToStrF (i: Integer): String;

Var StringI: String[2];

Begin

StringI := IntToStr(i);

If Length(StringI)=1

Then IntToStrF := '0' + StringI

Else IntToStrF := StringI;

End ;

Procedure TClock.DisplayTime;

Var StrMinutes, StrHours : String[2];

Begin

StrMinutes := IntToStrF(Minutes);

StrHours := IntToStrF(Hours);

Caption := StrHours + ':' + StrMinutes;

End ;

Function TAlarmClock.GetAlarmHours: Integer;

Begin

GetAlarmHours := AlarmHours;

End ;

Function TAlarmClock.GetAlarmMinutes: Integer;

Begin

GetAlarmMinutes := AlarmMinutes;End ;

Procedure TAlarmClock.SetAlarmTime (h, m : Integer);

Begin

AlarmHours := h;

AlarmMinutes := m;

End ;

Procedure TAlarmClock.Ring;Begin

If (AlarmHours=Hours) And (AlarmMinutes=Minutes)

Then

Caption := 'Wake Up' // this can be elaborated on!!!

End ;



Uses

Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls,

Forms,

Dialogs, ClassDefs, ExtCtrls, StdCtrls;

Type

TForm1 = Class(TForm)

tmrClockTick: TTimer;

btnSetAlarm: TButton;procedure FormCreate(Sender: TObject);

procedure tmrClockTickTimer(Sender: TObject);

procedure btnSetAlarmClick(Sender: TObject);

Private

{ Private declarations }

Public

AlarmClock : TAlarmClock;

End ;

Var

Form1: TForm1;

Implementation

{$R *.dfm}

Procedure TForm1.btnSetAlarmClick(Sender: TObject);

Begin

AlarmClock.SetTime(15,10); // this can made interactive by copying

AlarmClock.SetAlarmTime(15,15); // values from textboxes

tmrClockTick.Enabled := True;

End ;

Procedure TForm1.FormCreate(Sender: TObject);

Begin

AlarmClock := TAlarmClock.Create(Self);

AlarmClock.Parent := Self;

tmrClockTick.Enabled := False;tmrClockTick.Interval := 600; // tick every minute

End ;

Procedure TForm1.tmrClockTickTimer(Sender: TObject);

Begin

AlarmClock.IncrementTime;

AlarmClock.DisplayTime;

AlarmClock.Ring;

End ;

End .



9 Inheritance diagram of the shapes classes

Exam pract ice quest ionsQuestion 1 – June 2008 CPT4 Q6

Question

number

Answer Marks

1a 2

1b Insert a SetColour procedure; A Function

into the Public section;

R make Colour Public 2

TShape

TRectangle TEllipse



1c Yacht = Class/sub-class (Boat)

(Public)

Procedure SetBoatDetails (Override)

Function GetMasts

Function GetEngine

Private

Masts: IntergerEngine: Boolean

End

A Procedure SetMasts and Procedure SetEngine/Procedure

AddnewYacht/SetYachtsDetails instead of Procedure SetBoatDetails

Mast and Engine must be private

P1 if extra functions/variables are included

R any diagramsI any parameters to methods

1

1

1

1

11

Question 2 – January 2008 CPT4 Q4

Question

number

Answer Marks

2a

1 mark for correct boxes

1 mark for correct lines

1 mark for correct line endings

3

2b

4



A any reasonable names for operations and data items

2c Add a new data item ShortLoan; of type Boolean; A loanlength;

integer;

A loantype; string;

Modify the code for the operations.

Max 2

Question

Number

Answer Marks

3a mouse click// mouse movement// keyboard operation// any interrupt; 1

3b event-driven programs service an event and wait for another;

non event-driven programs run to completion/ are sequential;

2

3c contains its own data/fields/variables/properties;

contains its own

operations/methods/functions/procedures/behaviours/code;

responds to messages;

A Based on a Class definition

Max 2

3d frame/form/window/button/check box/radio button/menu/text box;

A any sensible widget

R Plurals

1


Questionnumber Answer Marks

4a

1 mark for all three classes in appropriate single enclosures.

1 k f t i d d t i t di ti

2



R make Colour Public

4c Van = Class/ subclass (Vehicle), i.e. Clearly identify Van as a (sub)

class of vehicle (1 mark)

(Public)

Procedure SetVehicleDetails (Override) condone if not included

Function GetCapacity

Function GetTailLift

(penalise extra functions/procedures once.)

Private

Capacity : Integer/real/fixed/float

TailLift : Boolean

(penalise once if not private and once if extra variables listed.)

End

Max 6

1 mark

1 mark

1 mark

1 mark

1 mark



A Procedure SetCapacity and Procedure SetTailLift/

Procedure AddNewVan

instead of Procedure SetVehicleDetails

OR

Public class/subclass Van extends/inherits Vehicle (1 mark)

{

public void SetVehicleDetails

public int GetCapacity

public boolean/int GetTailLift

private int Capacity

private boolean/int TailLift

}

A public void SetCapacity and public void SetTailLift//

public void AddNewVan instead of public void

SetVehicleDetailsR any diagrams

I any parameters to methods

1

mark

1

mark

1

mark

1

mark

1

mark

1mark

6

marks

(10)


Question

number

Answer Marks

5a a class has properties/fields/attributes/characteristics and

methods/procedures/functions

f h l i i d i d f

1 mark



5b StockItem (=) Class // Class (=) StockItem; (A Object instead of Class)

Public (procedure DisplayDetails) (virtual)(virtual;abstract)

(procedure) SetLoan (virtual)(virtual;abstract)

Private; A protected

Title: String

OnLoan: Boolean

DateAcquired: String/DateTime/

End

Book = Class (StockItem) // Class Book extends/derives from

StockItem

// Book Sub-class: StockItem; A without keyword Class

1 mark for keywords Class and StockItem1 mark for keyword Public and correct methods

A text instead of string

1 mark for correct data fields Date & data types

don’t allow the other fields

Private

Author: StringISBN: String

Public

(Procedure) DisplayDetails (override)

End

CD = Class (StockItem) // Class CD extends/derives from StockItem //

CD Subclass: StockItem;

Private

Artist: String

PlayingTime: Integer/Real/Time/DateTime

Public

(Procedure) DisplayDetails (override)

End

No marks for a diagrammatic answer. I method parameters

Java version:

Public Class StockItem

1 mark

1 mark

1 mark1 mark

1 mark

1 mark

1 mark

If candidate declared

‘getters’ and

‘setters’ for the baseclass fields then

don’t have to have

DisplayDetails as a

base class method.



Public void displayDetails ();Public void setLoan ();

}

Public Class Book extends StockItem

{

Private string author;

Private string isbn;

Public void displayDetails ();

}

Public Class CD extends StockItem

{

Private string artist;

Private integer playingTime;

Public void displayDetails();}

Total (max 7)

2.2


2 Product := 1;

For i := 1 To n

Do Product := i * Product;

Factorial := Product;

3

Call number n Output Comment

Call number 1 5

4

Call number 2 4

3

Call number 3 3

2

Call number 4 2

1 1 Procedure call completed



Continue call 2 3 3 Procedure call completed

Return to previous call

Continue call 1 4 4 Procedure call completed

4

Call number n Output Comment

Call number 1 5

4 4

Call number 2 4

3 3

Call number 3 3

2 2

Call number 4 2

1 1 Procedure call completed


Continue call 3 2 Procedure call completed





5

Function Fibonacci(n: Integer): Integer;

Begin

If n=0

Then Fibonacci := 0Else

If n=1

Then Fibonacci := 1

Else Fibonacci := Fibonacci(n-1) + Fibonacci(n-2);

End ;

6

Function Power(a, n: Integer): Integer;

Begin

If n=0

Then Power := 1

Else Power := a * Power(a, n-1);

End ;

7 2n

1



2.3


Exercise 1

Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const MaxListSize=30;

Var

ExitChosen: Boolean = False;

MenuItem: Char;NameList: Array[1..MaxListSize] Of String;

NumberOfFriendsStored: Integer = 0;

Function ListEmpty: Boolean;

Begin

ListEmpty := NumberOfFriendsStored=0;

End;

Function ListFull: Boolean;

Begin

ListFull := NumberOfFriendsStored=MaxListSize;

End;

Procedure GetMenuChoice(Var Choice: Char);

Begin

WriteLn;WriteLn('Menu');

WriteLn;

WriteLn('1) Add a name to the list');

WriteLn('2) Show number of names in the list');

WriteLn('3) Show list of names');

WriteLn('4) Exit Program');

WriteLn;

Write('Enter your choice number: ');ReadLn(Choice);

End;

Procedure AddNameToList;

Var NewName: String; Finish: Boolean;



BeginRepeat

Write('Enter Name to be added to list (to finish typeXXX): ');

ReadLn(NewName);

If NewName = 'XXX'

Then

Finish := True

Else

Begin

NumberOfFriendsStored := NumberOfFriendsStored + 1;

NameList[NumberOfFriendsStored] := NewName;

End;

Until Finish Or ListFull;

If ListFull

Then WriteLn('Can''t add any more names');

End;

End;

Procedure PrintListOfNames;

Var i: Integer;

Begin

If Not ListEmpty

Then

For i := 1 To NumberOfFriendsStored

Do WriteLn(NameList[i]);

End;

Procedure OutputNumberOfNamesInList;

Begin

WriteLn('There are ', NumberOfFriendsStored, ' Names in thelist');

End;

Begin

AddNametoList;

Repeat

GetMenuChoice(MenuItem);

Case MenuItem Of

'1': AddNameToList;

'2': OutputNumberOfNamesInList;

'3': PrintListOfNames;

'4': ExitChosen := True;

Else WriteLn('Invalid menu choice. Try again');

End;

Until ExitChosen;



{$APPTYPE CONSOLE}

Uses

SysUtils;


Var


MenuItem: Char;NameList: Array[1..MaxListSize] Of String;



Begin

ListEmpty := NumberOfFriendsStored=0;

End ;


Begin


End ;

Procedure GetMenuChoice( Var Choice: Char);

Begin

WriteLn;WriteLn('Menu');

WriteLn;




WriteLn('4) Find a name');

WriteLn('5) Delete a name');


WriteLn;

Write('Enter your choice number: ');

ReadLn(Choice);

End ;


Var NewName: String; Finish: Boolean;

Begin

Finish := False;

If ListFull

Then WriteLn('Can''t add any more names')

Else

Begin



ThenFinish := True

Else

Begin


NameList[NumberOfFriendsStored] := NewName;

End ;


If ListFullThen WriteLn('Can''t add any more names');

End ;

End ;


Var i: Integer;

Begin

If Not ListEmptyThen


Do WriteLn(NameList[i]);

End ;


Begin

WriteLn('There are ', NumberOfFriendsStored, ' Names in thelist');

End ;

Procedure SearchFor(N: String; Var Pos: Integer);

Var Found, NotInList: Boolean;

Begin

Found := False;

NotInList := False;

Pos := 1;

Repeat

If NameList[Pos] = N

Then Found := True

Else Pos := Pos + 1;

If Pos > NumberOfFriendsStored

Then NotInList := True;

Until Found Or NotInList;

If NotInList

Then Pos := 0;

End ;

Procedure Delete(Pos: Integer);



Pos := Pos + 1;End ;

NameList[Pos] := ''; // clear space

NumberOfFriendsStored := NumberOfFriendsStored - 1;

End ;

Procedure FindName;

Var Name: String; Position: Integer;

BeginWrite('Please enter the name you wish to find: ');

ReadLn(Name);

SearchFor(Name, Position);

If Position=0

Then WriteLn('Sorry, the name is not in the list')

Else WriteLn(Name, ' is name number ', Position, ' in thelist');

End ;

Procedure DeleteName;

Var Name: String; Position: Integer;

Begin

Write('Please enter the name you wish to delete: ');

ReadLn(Name);


If Position=0


Else Delete(Position);

End ;

Begin

AddNametoList;

Repeat


Case MenuItem Of

'1': AddNameToList;



'4': FindName;

'5': DeleteName;



End ;

Until ExitChosen;

End .



Exercise 3Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;


Type TFriend = Record

Name: String;

Replied: Boolean;

ComingToParty: Boolean;

End ;

VarExitChosen: Boolean = False;

MenuItem: Char;

NameList: Array[1..MaxListSize] Of TFriend;



Begin

ListEmpty := NumberOfFriendsStored=0;End ;


Begin


End ;

Procedure GetMenuChoice( Var Choice: Char);Begin

WriteLn;

WriteLn('Menu');

WriteLn;



WriteLn('3) Print list of names');

WriteLn('4) Find a name');WriteLn('5) Delete a name');

WriteLn('6) Print list of all those definitely coming to theparty');

WriteLn('7) Print list of all those who have not replied yet');




Procedure AddNameToList; Var NewName: String; Finish: Boolean;

Begin

Finish := False;

If ListFull


Else

Begin

RepeatWrite('Enter Name to be added to list (to finish type

XXX): ');

ReadLn(NewName);

If NewName = 'XXX'

Then

Finish := True

Else

Begin


NameList[NumberOfFriendsStored].Name := NewName;

NameList[NumberOfFriendsStored].Replied := False;

NameList[NumberOfFriendsStored].ComingToParty :=False;

End ;


If ListFull


End ;

End ;


Var i: Integer;

Begin

If Not ListEmpty

Then


Do WriteLn(NameList[i].Name);

End ;


Begin

WriteLn('There are ', NumberOfFriendsStored, ' Names in the

list');End ;

Procedure SearchFor(N: String; Var Pos: Integer);

Var Found, NotInList: Boolean;

Begin



If NameList[Pos].Name = NThen Found := True

Else Pos := Pos + 1;

If Pos > NumberOfFriendsStored

Then NotInList := True;

Until Found Or NotInList;

If NotInList

Then Pos := 0;

End ;

Procedure Delete(Pos: Integer);

Begin

While Pos < NumberOfFriendsStored

Do

Begin

NameList[Pos] := NameList[Pos+1];

Pos := Pos + 1;

End ;

NameList[Pos].Name := ''; // clear space

NumberOfFriendsStored := NumberOfFriendsStored - 1;

End ;

Procedure FindName;

Var Name: String; Position: Integer; Answer: Char;

Begin

Write('Please enter the name you wish to find: ');

ReadLn(Name);


If Position=0


Else

Begin

Repeat

Write('Has ', Name, ' replied to the invite? (y/n) ');

ReadLn(Answer);

Until ((Answer = 'y') Or (Answer = 'n'));

If Answer='y'

Then

Begin

NameList[Position].Replied := True;

Repeat

Write('Is ', Name, ' coming to the party? (y/n) ');

ReadLn(Answer);

Until ((Answer = 'y') Or (Answer = 'n'));

If Answer='y'

Then NameList[Position].ComingToParty := True;



Var Name: String; Position: Integer;Begin


ReadLn(Name);


If Position=0


Else Delete(Position);

End ;

Procedure AttendanceList;

Var i: Integer;

Begin

WriteLn;

WriteLn('The following are definitely coming to the party:');

WriteLn('=================================================');

WriteLn;


Do

If NameList[i].ComingToParty

Then WriteLn(NameList[i].Name);

End ;

Procedure NoReplyList;

Var i: Integer;

Begin

WriteLn;

WriteLn('The following have not replied yet:');

WriteLn('===================================');

WriteLn;


Do

If Not NameList[i].Replied

Then WriteLn(NameList[i].Name);

End ;

Begin

AddNametoList;

Repeat


Case MenuItem Of

'1': AddNameToList;



'4': FindName;



End ;

Until ExitChosen;

End .



Exercise 4

Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const MaxListSize = 6;

Type TNode = Record

Data: String;

Pointer: Integer;

End ;

Var Node: Array[1..MaxListSize] Of TNode;

Start: Integer = 0; // list empty

NextFree: Integer;

MenuItem: Char;


Procedure InitialiseList;

Var i: Integer;Begin

NextFree := 1;

For i := 1 To MaxListSize - 1

Do

Node[i].Pointer := i+ 1;

Node[MaxListSize].Pointer := 0;

End ;


Begin

ListFull := NextFree=0;

End ;


Begin

ListEmpty := Start=0;End ;


Begin

WriteLn;




WriteLn;


ReadLn(Choice);

End ;

Procedure AddNodeToFrontOfList(Name: String);

Var Temp: Integer;

Begin

Node[NextFree].Data := Name;

Temp := NextFree;

NextFree := Node[NextFree].Pointer;

Node[Temp].Pointer := Start;

Start := Temp;

End ;

Procedure AddNodeIntoList(Name: String; ThisNode: Integer);

// the new node is to be inserted after ThisNode

Var Temp: Integer;

Begin

Node[NextFree].Data := Name;

Temp := NextFree;

NextFree := Node[NextFree].Pointer;

Node[Temp].Pointer := Node[ThisNode].Pointer;

Node[ThisNode].Pointer := Temp;

End ;

Procedure DeleteHeadOfList;

Var Temp: Integer;

Begin

Temp := Start;

Start := Node[Start].Pointer;

Node[Temp].Pointer := NextFree;

NextFree := Temp;

End ;

Procedure DeleteNodeWithinList(PreviousNode, ThisNode: Integer);

// ThisNode is the node to be deleted)

Var Temp: Integer;

Begin

Temp := NextFree;

NextFree := Node[PreviousNode].Pointer;

Node[PreviousNode].Pointer := Node[ThisNode].Pointer;

Node[ThisNode].Pointer := Temp;

End ;



Else

Begin

ThisNode := Start;

Repeat

WriteLn(Node[ThisNode].Data);

ThisNode := Node[ThisNode].Pointer;

Until ThisNode = 0;

End ;

End ;

Function InsertionPoint(N: String): Integer;

Var ThisNode, NextNode: Integer;

Begin

ThisNode := 0; // start at the beginning of the list

NextNode := Start;

While NextNode > 0 // not end of list

Do

Begin

ThisNode := NextNode;

NextNode := Node[NextNode].Pointer; // follow thepointer

End ;

InsertionPoint := ThisNode;

End ;

Procedure SearchFor(N: String; Var This, Previous: Integer);

Begin

This:= Start;

Previous:= 0;

While (This > 0) And (Node[This].Data <> N)

Do

Begin

Previous := This;

This := Node[This].Pointer;

End ;

If Node[This].Data <> N

Then This := 0; // name not found

End ;


Var Finish: Boolean; NewName: String; Position: Integer;

Begin

Finish := False;

If ListFull


Else



If NewName = 'XXX'

Then

Finish := True

Else

Begin

If ListEmpty

Then AddNodeToFrontOfList(NewName)

Else

Begin

Position := InsertionPoint(NewName);

If Position = 0

Then AddNodeToFrontOfList(NewName)

Else AddNodeIntoList(NewName, Position);

End ;

End ;


If ListFull


End ;

End ;

Procedure DeleteName;

Var Name: String; ThisNode, PreviousNode: Integer;

Begin

If ListEmpty

Then WriteLn('There are no names in the list')

Else

Begin


ReadLn(Name);

SearchFor(Name, ThisNode, PreviousNode);

If ThisNode=0


Else

If ThisNode = Start

Then DeleteHeadOFList

Else DeleteNodeWithinList(PreviousNode, ThisNode);

End ;

End ;

Procedure ShowArray;

// this is just to check the linked list is built up correctly

Var i: Integer;

Begin

WriteLn('Start: ', Start);

WriteLn('NextFree: ',NextFree);



InitialiseList;

Repeat


Case MenuItem Of

'1': AddNameToList;

'2': OutputListElements;

'3': DeleteName;


'6': ShowArray; // for testing purposes


End ;

Until ExitChosen;

End .




{$APPTYPE CONSOLE}

Uses

SysUtils;

Type

TNodePtr = ^TNode;

TNode = Record Data : String;

Ptr : TNodePtr;

End ;

Var

Start : TNodePtr;

MenuItem: Char;


Procedure InitList ( Var List: TNodePtr);

Begin

List := Nil;

End ;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;





WriteLn;


ReadLn(Choice);

End ;

Procedure AddItem ( Var List: TNodePtr; NewItem: String);

Var NewNode, ThisNode, NextNode: TNodePtr;

Begin

New (NewNode);

NewNode^.Data := NewItem;

NewNode^.Ptr := Nil;

If List = Nil

Then List := NewNode



Begin


NextNode := NextNode^.Ptr;

End ;

ThisNode^.Ptr := NewNode;

End ;

End ;

Procedure DeleteItem ( Var List: TNodePtr; WantedData: String);

Var ThisNode, PreviousNode: TNodePtr;

Begin

If List <> Nil

Then

Begin

ThisNode := List;

If List^.Data = WantedData // is it the first node ?

Then List := List^.Ptr

Else

Begin

While ThisNode^.Data <> WantedData

Do

Begin

PreviousNode := ThisNode;

ThisNode := ThisNode^.Ptr

End ;

PreviousNode^.Ptr := ThisNode^.Ptr;

End ;

Dispose (ThisNode);

End

Else WriteLn('List empty');

End ;

Procedure PrintList (List : TNodePtr);

Var ThisNode : TNodePtr;

Begin

If List <> Nil

Then

Begin

ThisNode := List;

Repeat

WriteLn (ThisNode^.Data);

ThisNode := ThisNode^.Ptr;

Until ThisNode = Nil;

End ;

End ;



GetName := Name;

End ;

Begin

InitList(Start);

Repeat


Case MenuItem Of

'1': AddItem(Start,GetName);

'2': PrintList(Start);

'3': DeleteItem(Start,GetName);



End ;

Until ExitChosen;

End .



Exercise 6

Program Project2;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Type

TNodePtr = ^TNode;

TNode = Record

Data : String;

Ptr : TNodePtr;

End ;

Var

Start : TNodePtr;

MenuItem: Char;



Begin

List := Nil;

End ;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;





WriteLn;


ReadLn(Choice);

End ;



Begin

New (NewNode);



If List = Nil



Do

Begin



End ;


End ;

End ;


Var ThisNode, PreviousNode: TNodePtr; Found: Boolean;

Begin

If List = Nil

Then

WriteLn('List empty')

Else

Begin

ThisNode := List;


Then

Begin

List := List^.Ptr; // Then adjust front pointer

Dispose (ThisNode);

End

Else

Begin

Found:= False;

While (ThisNode^.Ptr <> Nil) And Not Found

Do

Begin

PreviousNode := ThisNode;


Found := ThisNode^.Data = WantedData;

End ;

If Found

Then

Begin


Dispose (ThisNode);

End

Else WriteLn('Name not in list');

End ;

End ;

End ;




ThisNode := List;

Repeat




End ;

End ;

Function GetName: String;

Var Name: String;

Begin

Write('Enter a name: ');

ReadLn(Name);

GetName := Name;

End ;

Begin

InitList(Start);

Repeat


Case MenuItem Of






End ;

Until ExitChosen;

End .



Exercises 7, 8Program Project2;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Type

TNodePtr = ^TNode;

TNode = Record

Data : String;

Ptr : TNodePtr;

End ;

Var

Start : TNodePtr;

MenuItem: Char;



Begin

List := Nil;

End ;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;






WriteLn;


ReadLn(Choice);

End ;



Begin

New (NewNode);



If List = Nil



Do

Begin



End ;


End ;

End ;

Procedure SearchFor(N:String; List: TNodePtr; Var This, Previous:TNodePtr);

Var Found: Boolean;

Begin

This := List;

Previous := Nil;

Found:= False;

While (This^.Ptr <> Nil) And Not Found

Do Begin

Previous := This;

This := This^.Ptr;

Found := This^.Data = N;

End ;

If Not Found

Then This := Nil;

End ;



Begin

If List = Nil

Then


Else Begin

ThisNode := List;


Then

Begin


Dispose (ThisNode);

End Else

Begin

SearchFor(WantedData, List, ThisNode, PreviousNode);

If ThisNode = Nil



End ;

End ;

End ;

End ;



Begin

If List = Nil

Then


Else

Begin

ThisNode := List;

Repeat




End ;

End ;


Var Name: String;

Begin


ReadLn(Name);

GetName := Name;

End;

Procedure FindName(List: TNodePtr; Name: String);

Var ThisNode, PreviousNode: TNodePtr;

Begin

If List = Nil

Then WriteLn('No names in list')Else

Begin

SearchFor(Name, List, ThisNode, PreviousNode);

If ThisNode = Nil

Then WriteLn('Sorry, the name does not exist in thelist');

End;

End;

Begin

InitList(Start);

Repeat






End;

Until ExitChosen;

End .




{$APPTYPE CONSOLE}

Uses

SysUtils;

Type

TNodePtr = ^TNode;

TNode = Record

Data : String;

Ptr : TNodePtr;

End ;

Var

Start : TNodePtr;

MenuItem: Char;



Begin

List := Nil;

End ;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;






WriteLn;


ReadLn(Choice);

End ;

Function InsertionPoint(List: TNodePtr; NewItem: String): TNodePtr;

Var This, Next: TNodePtr;

Begin

This := Nil;

Next := List;

While (Next <> Nil) And (Next^.Data < NewItem)

Do

Begin





Begin

New (NewNode);



If List = Nil

Then List := NewNode

Else

Begin

ThisNode := InsertionPoint(List, NewItem);

If ThisNode = Nil

Then

Begin

NewNode^.Ptr := List;

List := NewNode;

End

Else

Begin

NewNode^.Ptr := ThisNode^.Ptr;


End ;

End ;

End ;

Procedure SearchFor(N:String; List: TNodePtr; Var This, Previous:TNodePtr);

Var Found: Boolean;

Begin

This := List;

Previous := Nil;

Found:= False;

While (This^.Ptr <> Nil) And Not FoundDo

Begin

Previous := This;

This := This^.Ptr;

Found := This^.Data = N;

End ;

If Not Found

Then This := Nil;End ;



Begin



Begin

ThisNode := List;


Then

Begin


Dispose (ThisNode);

End

Else

Begin

SearchFor(WantedData, List, ThisNode, PreviousNode);

If ThisNode = Nil

Then WriteLn('Name not in list')

Else

Begin


Dispose (ThisNode);

End ;

End ;

End ;

End ;



Begin

If List = Nil

Then


Else

Begin

ThisNode := List;

Repeat




End ;

End ;


Var Name: String;

Begin


ReadLn(Name);

GetName := Name;

End ;

Else



Else

Begin

SearchFor(Name, List, ThisNode, PreviousNode);

If ThisNode = Nil

Then WriteLn('Sorry, the name does not exist in thelist');

End ;

End ;

Begin InitList(Start);

Repeat


Case MenuItem Of



'3': FindName(Start, GetName);

'4': DeleteItem(Start,GetName);'5': ExitChosen := True;


End ;

Until ExitChosen;

End .





Question

number

Answer Marks

1 a) i) Data must be given in correct order

1 pheasant 2

2 teal 33 widgeon 5

START 4 partridge 1

5 woodpigeon 0

End Pointer can be blank

1 for correct START and END pointers;

1 for correctly numbered nodes and correct pointers (need all birds)

1

ii) 1 pheasant 7

2 teal 3

3 widgeon 54 partridge 1

5 woodpigeon 0

START 6 grouse 4

7 snipe 2

//correctly amended diagram

1 for grouse and snipe physically at end;

1 for correct pointers (if not as ms then clear and logical);

1

Max 2

b) The amount of memory taken up can vary;

//The size/length of the structure/linked list can vary;

At run time;

1

1

c) A heap/stack/a pool of available locations;

A pointer holds the address of the allocated block/next available location; 2



2.4


1a)

1b)



2Stack

Index Data

TopOfStackPointer 3 6

5

4

3 Cup

2 Saucer

1 Plate

3

Program StackExampleUsingLinearList;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const MaxStackSize=6;

Var

ExitChosen: Boolean;

MenuItem: Char;

Stack: Array[1..MaxStackSize] Of String;

TopOfStackPointer: Integer = 0;

Function StackEmpty: Boolean;Begin

StackEmpty := TopOfStackPointer=0;

End ;

Function StackFull: Boolean;

Begin

StackFull := TopOfStackPointer=MaxStackSize;

End ;

Procedure GetMenuChoice( Var Choice: Char);Begin

WriteLn;

WriteLn('Menu');

WriteLn;

WriteLn('1) Add an item to the stack');

WriteLn('2) Remove an item from the stack');

WriteLn('3) Show contents of stack');


WriteLn;


ReadLn(Choice);

End ;

Procedure AddItemToStack;



Begin

If StackFull

Then WriteLn('Can''t add to a full stack')

Else

Begin

TopOfStackPointer := TopOfStackPointer + 1;

Write('Enter item to be added to stack: ');

ReadLn(Stack[TopOfStackPointer]);

End ;End ;

Procedure RemoveItemFromStack;

Begin

If StackEmpty

Then WriteLn('Can''t take anything from an empty stack')

Else

Begin

WriteLn('Item popped off stack: ', Stack[TopOfStackPointer]);TopOfStackPointer := TopOfStackPointer - 1;

End ;

End ;

Procedure ShowContentsOfStack;

Var Pointer: Integer;

Begin

Pointer := MaxStackSize; // output stack with bottom of stacklast

While TopOfStackPointer < Pointer

Do

Begin

WriteLn(Pointer);

Pointer := Pointer − 1;

End ;

While Pointer > 0

Do

Begin

WriteLn(Pointer,' ',Stack[Pointer]);

Pointer := Pointer − 1;

End ;

End ;

Begin

ExitChosen := False;

RepeatGetMenuChoice(MenuItem);

Case MenuItem Of

'1': AddItemToStack;

'2': RemoveItemFromStack;

'3': ShowContentsOfStack;





End ;

Until ExitChosen;

End .

4

Program LinkedListStackExample;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const MaxStackSize=6;

Type TNode = Record Data: String;

Pointer: Integer;

End ;

Var


MenuItem: Char;

Stack: Array[1..MaxStackSize] Of TNode;

FreePtr: Integer;

TopOfStackPointer: Integer = 0;

Procedure InitialiseStack;

Var i: Integer;

Begin

FreePtr := 1;

For i := 1 To MaxStackSize − 1

Do

Stack[i].Pointer := i + 1;

Stack[MaxStackSize].Pointer := 0;

End ;

Function StackEmpty: Boolean;

Begin

StackEmpty := TopOfStackPointer=0;

End ;

Function StackFull: Boolean;Begin

StackFull := FreePtr=0;

End ;


Begin

W it L



WriteLn;

WriteLn('Menu');

WriteLn;





WriteLn;

Write('Enter your choice number: ');ReadLn(Choice);

End ;

Procedure AddItemToStack;

Var Temp: Integer;

Begin

If StackFull

Then WriteLn('Can''t add to a full stack')

ElseBegin

Temp := TopOfStackPointer;

TopOfStackPointer := FreePtr;

FreePtr := Stack[FreePtr].Pointer;


ReadLn(Stack[TopOfStackPointer].Data);

Stack[TopOfStackPointer].Pointer := Temp;

End ;

End ;

Procedure RemoveItemFromStack;

Var Temp: Integer;

Begin

If StackEmpty

Then WriteLn('Can''t take anything from an empty stack')

Else

Begin

Temp := TopOfStackPointer;

WriteLn('Item popped: ', Stack[TopOfStackPointer].Data);

TopOfStackPointer := Stack[TopOfStackPointer].Pointer;

Stack[Temp].Pointer := FreePtr;

FreePtr := Temp;

End ;

End ;

Procedure ShowContentsOfStack; Var ThisNode: Integer;

Begin

If StackEmpty

Then WriteLn('empty stack')

Else

Begin



Begin

ThisNode := TopOfStackPointer;

Repeat

WriteLn(Stack[ThisNode].Data);

ThisNode := Stack[ThisNode].Pointer;

Until ThisNode=0; // end of linked list reached

End ;

End ;

BeginExitChosen := False;

InitialiseStack;

Repeat


Case MenuItem Of

'1': AddItemToStack;

'2': RemoveItemFromStack;

'3': ShowContentsOfStack;'4': ExitChosen := True;


End ;

Until ExitChosen;

End .

5Program StackExampleUsingHeap;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Type

TNodePtr = ^TNode;

TNode = Record

Data : String;

Ptr : TNodePtr;

End ;

Var

TopOfStack : TNodePtr;


MenuItem: Char;

Procedure InitStack ( Var Stack: TNodePtr);

Begin

Stack := Nil;

End ;

Procedure AddItemToStack ( Var Stack: TNodePtr);

Var NewNode: TNodePtr; NewItem: String;

Begin




Write( Enter item to be added to stack: );

ReadLn(NewItem);

New (NewNode);



If Stack = Nil

Then

Stack := NewNode

ElseBegin

NewNode^.Ptr := Stack;

Stack := NewNode;

End ;

End ;

Procedure RemoveItemFromStack ( Var Stack: TNodePtr);

Var ThisNode: TNodePtr;

BeginIf Stack <> Nil

Then

Begin

WriteLn('Popped Data: ', Stack.Data);

ThisNode := Stack;

Stack := Stack^.Ptr;

Dispose (ThisNode);

End

Else WriteLn('Stack Empty');

End ;

Procedure ShowContentsOfStack (Stack : TNodePtr);


Begin

If Stack <> Nil

Then

Begin

ThisNode := Stack;

Repeat




End

Else WriteLn('Stack Empty');

End ;Procedure GetMenuChoice( Var Choice: Char);

Begin

WriteLn;

WriteLn('Menu');

WriteLn;




WriteLn( 1) Add an item to the stack );




WriteLn;


ReadLn(Choice);

End ;

BeginInitStack (TopOfStack);


Repeat


Case MenuItem Of

'1': AddItemToStack(TopOfStack);

'2': RemoveItemFromStack(TopOfStack);'3': ShowContentsOfStack(TopOfStack);



End ;

Until ExitChosen;

Readln;

End .

6

Program StackExampleOOP;

{$APPTYPE CONSOLE}

Uses

SysUtils, StackClass in 'StackClass.pas'; Var

MenuItem: Char;

Stack : TStack;


NewItem: String;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;





WriteLn;



;


ReadLn(Choice);

End ;

Begin

Stack := TStack.Create;

Repeat


Case MenuItem Of

'1': Begin


ReadLn(NewItem);

Stack.Push(NewItem);

End ;

'2': WriteLn('Item popped: ',Stack.Pop);'3': Stack.PrintStack;



End ;

Until ExitChosen;

End .

7a)

Queue

Index Data

FrontPointer 1 1 Jones

2 Smith

3 Peters

4 Franklin

BackPointer 5 5 Taylor

6

b)

Queue

Index Data

FrontPointer 3

1

2

3 Peters

4 Franklin




6

c)

Queue

Index Data

FrontPointer 3 1 Y

2 Z

3 Peters

4 Franklin


6 X

8

Program LinearQueue;

{$APPTYPE CONSOLE}

uses

SysUtils;

Const MaxQueueSize=6;

Var


MenuItem: Char;

Queue: Array[1..MaxQueueSize] Of String;BackPointer: Integer = 0;

Function QueueEmpty: Boolean;

Begin

QueueEmpty := BackPointer=0;

End ;

Function QueueFull: Boolean;

Begin

QueueFull := BackPointer=MaxQueueSize;End ;

Procedure GetMenuChoice(Var Choice: Char);

Begin

WriteLn;

WriteLn('Menu');

WriteLn;

WriteLn('1) Add an item to the queue');



WriteLn('2) Remove an item from the queue');

WriteLn('3) Show contents of queue');


WriteLn;


ReadLn(Choice);

End ;Procedure AddItemToQueue;

Begin

If QueueFull

Then WriteLn('Can''t add to a full queue')

Else

Begin

BackPointer := BackPointer + 1;

Write('Enter item to be added to queue: ');

ReadLn(Queue[BackPointer]);

End ;

End ;

Procedure ShuffleQueue(Ptr: Integer);

Var i: Integer;

Begin

For i := 2 To Ptr

Do Queue[i-1] := Queue[i];

Queue[Ptr]:=''; // clear the space

End ;

Procedure RemoveItemFromQueue;

Begin

If QueueEmpty

Then WriteLn('Can''t take anything from an empty queue')

Else

Begin

WriteLn('Item taken from queue: ', Queue[1]);

If BackPointer > 1

Then ShuffleQueue(BackPointer);

BackPointer := BackPointer - 1;

End ;

End ;

Procedure ShowContentsOfQueue;


BeginFor Pointer := 1 To MaxQueueSize

Do WriteLn(Pointer, Queue[Pointer]);

End ;

Begin


R t



Repeat


Case MenuItem Of

'1': AddItemToQueue;

'2': RemoveItemFromQueue;

'3': ShowContentsOfQueue;


Else WriteLn('Invalid menu choice. Try again');End ;

Until ExitChosen;

End .

9

Program CircularQueueExample;

{$APPTYPE CONSOLE}

usesSysUtils;

Const MaxQueueSize=6;

Var


MenuItem: Char;

Queue: Array[1..MaxQueueSize] Of String;

BackPointer: Integer = 0;

FrontPointer: Integer = 0;

NumberOfItemsInQueue: Integer = 0;

Function QueueEmpty: Boolean;

Begin

QueueEmpty := NumberOfItemsInQueue=0;

End ;

Function QueueFull: Boolean;

Begin

QueueFull := NumberOfItemsInQueue=MaxQueueSize;

End ;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;


WriteLn('2) Remove an item from the queue');

WriteLn('3) Show contents of queue');


WriteLn;


ReadLn(Choice);

End ;

Procedure IncrementPointer(Var Ptr: Integer);



Procedure IncrementPointer( Var Ptr: Integer);

Begin

Ptr := Ptr MOD MaxQueueSize + 1;

End ;

Procedure AddItemToQueue;

Begin

If QueueFull

Then WriteLn('Can''t add to a full queue')Else

Begin

If QueueEmpty

Then FrontPointer := 1;

IncrementPointer(BackPointer);


ReadLn(Queue[BackPointer]);

NumberOfItemsInQueue := NumberOfItemsInQueue + 1;

End ;

End ;


Begin

If QueueEmpty

Then WriteLn('Can''t take anything from an empty queue')

Else

Begin

WriteLn('Item taken from queue: ', Queue[FrontPointer]);

Queue[FrontPointer] := ''; // clear space

NumberOfItemsInQueue := NumberOfItemsInQueue - 1;

If NumberOfItemsInQueue = 0

Then

Begin

FrontPointer := 0;

BackPointer := 0;

End

Else

IncrementPointer(FrontPointer);

End ;

End ;



BeginFor Pointer := 1 To MaxQueueSize

Do

Begin

Write(Pointer, Queue[Pointer]);

If Pointer = FrontPointer

Then Write('<-- Front');

If Pointer = BackPointer



If Pointer = BackPointer

Then Write('<-- Back');

WriteLn;

End ;

End ;

Begin


RepeatGetMenuChoice(MenuItem);

Case MenuItem Of






End ;

Until ExitChosen;

End .

10

Program QueueExampleUsingHeap;

{$APPTYPE CONSOLE}

UsesSysUtils;

Type

TNodePtr = ^TNode;

TNode = Record

Data : String;

Ptr : TNodePtr;

End ; Var

FrontPointer : TNodePtr;

BackPointer : TNodePtr;


MenuItem: Char;

Procedure InitQueue;

Begin

FrontPointer := Nil;

BackPointer := Nil;

End ;

Procedure AddItemToQueue;

Var NewNode: TNodePtr; NewItem: String;

Begin

Write('Enter item to be added to Queue: ');

ReadLn(NewItem);New (NewNode);



New (NewNode);



If BackPointer = Nil

Then

Begin

FrontPointer := NewNode;

BackPointer := NewNode;End

Else

Begin

BackPointer^.Ptr := NewNode;

BackPointer := NewNode;

End ;

End ;


Var ThisNode: TNodePtr;

Begin

If FrontPointer <> Nil

Then

Begin

WriteLn('Popped Data: ', FrontPointer.Data);

ThisNode := FrontPointer;

Frontpointer := FrontPointer^.Ptr;

Dispose (ThisNode);

IF FrontPointer = Nil

Then BackPointer := Nil;

End

Else WriteLn('Queue Empty');

End ;


Var ThisNode : TNodePtr;Begin

If FrontPointer <> Nil

Then

Begin

ThisNode := FrontPointer;

Repeat


ThisNode := ThisNode^.Ptr;Until ThisNode = Nil;

End

Else WriteLn('Queue Empty');

End ;


BeginWriteLn;



WriteLn('Menu');

WriteLn;

WriteLn('1) Add an item to the Queue');

WriteLn('2) Remove an item from the Queue');

WriteLn('3) Show contents of Queue');


WriteLn;Write('Enter your choice number: ');

ReadLn(Choice);

End ;

Begin

InitQueue;


Repeat


Case MenuItem Of






End ;

Until ExitChosen;

Readln;

End .

11

Program OOPQueueExample;

{$APPTYPE CONSOLE}

Uses

SysUtils,

QueueClass In 'QueueClass.pas';

Var

Queue: TQueue;

MenuItem: Char;


NewItem: String;


Begin

WriteLn;

WriteLn('Menu');

WriteLn;


WriteLn('2) Remove an item from the queue');WriteLn('3) Show contents of queue');




WriteLn;


ReadLn(Choice);

End ;

Begin

Queue := TQueue.Create;Repeat


Case MenuItem Of

'1': Begin


ReadLn(NewItem);

Queue.Add(NewItem);

End ;

'2': WriteLn('Item popped: ',Queue.Remove);

'3': Queue.PrintQueue;



End ;

Until ExitChosen;

End .


Question 1 – May 2007 CPT1 Q5

Question

number

Answer Marks

1a Last (item) in, is the first (item) out/first (item) in is the last (item) out;R. LIFO / FILO

1

1b i) 1



All items in the correct locations

ii)

Correct three items // ft from an incorrect (i) including 605 as the first

location used;

A. .R. and .Y. entries indicated in some way as . deleted.

iii)

Correct list of five items // ft from an incorrect (i) + a correct ft (ii) including605 as

the first location used;

1

1

1c i) Queue ; A. First In . First Out FIFO / LILO

ii) Items are removed/popped from the stack (one at a time) (and items are

then added to the queue);

iii) Items leave the queue on a .first in-first out. basis; A. from the front of

the queueiv) .Y., .R., .E., .V., .A. on the queue;

.Y., .R., .E., .V., .A. on the final stack;

A. using 701 for the first queue location

1

1

1

2


Question

number

Answer Marks

2a Next item to be added is at position/location/address (Tail + 1); Max 1



2a Next item to be added is at position/location/address (Tail + 1);

Position/location/address Tail is the last item in the queue; R. .points to the

end of the queue.

Max 1

2b Cat // item at position Head; 1

2c

Snake + Eel + Shark at positions 3,4,5;Tail points to 5;

Head points to 2;

I. Dog and Cat crossed through

1

11

2d Tail will eventually reach position 99 (A. 100);

Head will eventually reach 99 (A. 100);

Memory/queue will become full;

Space is not re-useable

Max 2

Question 3 – June 2005 CPT1 Q7

Question

Number

Answer Marks

3a First In First Out; or by description

Last In First Out; or by description 2

3b 2

3c Reverse the contents of a queue/list;

Push all contents of queue/list onto stack then pop them off into a new

queue/list;

Procedure/function calls;

2

Local variables;

Parameters;Return address;



Volatile environment; A register contents

State 1 Describe 1

3d List of elements inserted into tree;

to allow rapid/fast searching of the data;

to output sorted/ordered data;

2


Question

number

Answer Marks

4 Last In First Out; 1

4b) i) 1

4b) ii) 1

4b) iii) 1

4b) iv) 1

4c) To reverse elements/pass parameters/store volatile environment;

A store return address

1


Questionnumber

Answer Marks



5a) i) Empty entries waste memory // Maximum size // fixed size; 1

5a) ii) Memory used by pointers//takes more time to add/delete notes//indirect

access takes more time; R programming difficulties1

5b) Place next item in first location/location 0/location 1//Implement a

circular array/queue//allow wraparound;1

5c) IsFull/IsQueueFull; 1


Question

number

Answer Marks

6a 3

6b AddItem//Add; 1

6c i) Full/FullQueue;

ii) No memory used for pointers;I Faster

R Easier to program

iii) Size is limited by array size;

memory wasted when not full;

1

1

2


Question

number

Answer Marks

7a) i) Empty entries waste space // Maximum/fixed/static size

A stack may overflow

ii) Space used by pointers // more complex to program

1

1

7b) i) The size of the stack/amount of data is known/limited/predictable

Memory saved since no pointers (if not given in a (ii))

R easier to program

ii) The size of the stack is unknown//The stack is volatile/number of items fluctuates widely;

1

1


Question

number

Answer Marks

8a)4



8a)

8b)2

8c)In a linear queue data is static, so queue ‘moves’ through storage/In a FIFO structure storage locations are only used once;

In a circular queue, the locations will be re-used;

Thus a circular queue has a more efficient use of memory;

1 mark for each of 2 points

1

1

1

2.5




1 The number of actors that the actor at the vertex has starred with in movies.

2

3 A graph is a diagram consisting of circles called vertices, joined by lines called edges or arcs.

Each edge joins exactly two vertices.

4

5a)



b) Three cages: one for A, one for D and one for B, C and E

6a)

Vertex 1 2 3 4 5

1 0 1 1 0 1

2 1 0 0 1 1

3 1 0 0 1 1

4 0 1 1 0 1

5 1 1 1 1 0

b)

Vertex Adjacent vertices

1 2, 3, 5

2 1, 4, 5

3 1, 4, 5

4 2, 3, 5

5 1, 2, 3, 4

7a)

Vertex 1 2 3 4 5

1 0 0 1 0 0

2 1 0 0 0 0



2 1 0 0 0 0

3 0 0 0 0 1

4 0 1 1 0 0

5 1 1 0 1 0

b)

Vertex Adjacent vertices

1 3

2 1

3 5

4 2, 3

5 1, 2, 4

8a)

Vertex 1 2 3 4 5

1 ∞ ∞ 1 ∞ ∞

2 20 ∞ ∞ ∞ ∞

3 ∞ ∞ ∞ ∞ 81

4 ∞ 30 -3 ∞ ∞

5 5 8 ∞ 54 ∞

b)

Vertex AdjacentVertices

1 3 3 0

2 1 1 20

3 5 5 81

4 2, 3 2 30 3 −3

5 1, 2, 4 1 5 2 8 4 54

9a) When many vertex pairs are connected by edges, then the adjacency matrix is best because

it doesn't waste much space, and it indicates whether an edge exists with one access (rather than

following a list).b) When the graph is sparse, that is, not many of its vertex pairs have edges between them, the

adjacency list is best.



10

11 Imagine a graph of four vertices connected by four edges to create a cycle as follows:

If one edge is removed then the graph becomes a tree – a tree is a connected graph with no cycles.

Generalising, a tree of n vertices must have n−1 edges.

12 A tree is a connected undirected graph with no cycles. A rooted tree is a tree in which one

vertex has been designated as the root and every edge is directed away from the root.

13 Vertices visited in the following order EGFDABC


Question 1 – Specimen Paper COMP 3 Q3



Question

number

Answer Marks

1a

or

Max 3

1b i) Max size only limited by memory;Only uses memory it requires;

ii) No pointers required;

max 1

1

1c i) Adjacency matrix;

ii) Faster to insert/delete; fixed size;

1

2

2 6



2.6


1 Dry-run of iterative algorithm searching for Davidson

ItemSought First Last Midpoint List[Midpoint] ItemFound SearchFailed

Davidson 1 15 8 King False False

1 7 4 Farmer

1 3 2 Bond

3 3 3 Clark True

2 Dry-run of recursive algorithm searching for Davidson

Cal

l

ItemSough

t

Firs

t

Las

t

Midpoin

t

List[Midpoint

]

ItemFoun

d

SearchFaile

d

1 Davidson 1 15 8 King False False

2 Davidson 1 7 4 Farmer False False

3 Davidson 1 3 2 Bond False False

4 Davidson 3 3 3 Clark False True

Four calls are made before the search fails.

3a)

Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const Max = 13; Var NameToFind: String;

Var NumberOfIterations: Integer = 0;

Var NumberOfProcedureCalls: Integer = 0;

Var SortedList: Array[0..Max] Of String = ('Alan', 'Ben','Chris', 'David', 'Eve', 'Fred', 'Greg', 'Heidi','Ida', 'Jude', 'Karl', 'Les', 'Mike', 'Noel');

// could read a text file of many names into an array to testinstead

Procedure BinarySearch1 ( Var List: Array Of String; First, Last:Integer; ItemSought: String);

Var ItemFound, SearchFailed: Boolean; MidPoint: Integer;

Begin

ItemFound := False;

SearchFailed := False;

RepeatNumberOfIterations := NumberOfIterations + 1; // for stats

only

Midpoint := (First + Last) DIV 2; // integer part of division



If List[Midpoint] = ItemSought

Then ItemFound := True

Else

If First >= Last

Then SearchFailed := True

Else

If List[Midpoint] > ItemSought

Then Last := Midpoint − 1

Else First := Midpoint + 1

Until ItemFound Or SearchFailed;

If SearchFailed

Then WriteLn('No such name')

Else WriteLn('Name found at position ',MidPoint);

End ;

c) The iterative program uses less memory as it does not have the procedure call overheads.

Each procedure call saves return address, parameters and local variables on the stack (in

main memory).

4

First Last CurrentPointer CurrentValue Pointer List

[1] [2] [3] [4] [5] [6]

1 8 2 72 1 87 72 28 45 59 36

1

0

87 87 28 45 59 36

72 87 28 45 59 36

3 28 2 72 87 28 45 59 36

2 72 87 87 45 59 36

1

0

72 72 87 45 59 36

28 72 87 45 59 36

4 45 3 28 72 87 45 59 36

3 28 72 87 87 59 36

21

28 72 72 87 59 36

28 45 72 87 59 36

5 59 4 28 45 72 87 59 36

4 28 45 72 87 87 363

2

28 45 72 72 87 36



28 45 59 72 87 36

6 36 5 28 45 59 72 87 36

5 28 45 59 72 87 87

4 28 45 59 72 72 87

3 28 45 59 59 72 87

2

1

28 45 45 59 72 87

28 36 45 59 72 87

5a) & b) Answer to insertion sort:Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const Max = 7;

Type TList = Array[1..Max] Of Integer;

Var UnsortedList: TList ;

Procedure OutputList(List:TList; First, Last: Integer); Var i: Integer;

Begin

Write('(');

For i := First To Last−1

Do Write(List[i], ' ,');

WriteLn(List[Last],')');

End ;

Procedure Sort (List: TList; First, Last: Integer);

Var CurrentPointer, Pointer: Integer; CurrentValue: Integer;

Begin

For CurrentPointer := First + 1 To Last

Do

Begin

CurrentValue := List[CurrentPointer];

Pointer := CurrentPointer − 1;

Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,

Pointer:6);

OutputList(List, First, Last);

While (List[Pointer] > CurrentValue) AND (Pointer > 0)

Do

Begin

List[Pointer+1] := List[Pointer];

Pointer := Pointer−

1;Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,

Pointer:6);




End ;

List[Pointer+1] := CurrentValue;

Write(First:6, Last:6, CurrentPointer:6, CurrentValue:6,Pointer:6);


End ;

End ;

Begin

UnsortedList[1]:=56;





UnsortedList[6]:=99;UnsortedList[7]:=17;

OutputList(UnsortedList,1,Max);

Sort(UnsortedList,1,Max);

ReadLn;

End .

6 Bubble sort programProgram Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;

Const Max = 7;


Var UnsortedList: TList;

Procedure OutputList(List:TList; NumberOfItems: Integer); Var i: Integer;

Begin

Write('(');

For i := 1 To NumberOfItems-1


WriteLn(List[NumberOfItems],')');

End ;

Procedure BubbleSort(List: TList; NumberOfItems: Integer); Var NoMoreSwaps: Boolean; Temp, Element: Integer;

Begin

Repeat

NoMoreSwaps := True;

For Element := 1 To NumberOfItems − 1

DoIf List[Element] > List[Element+1]

Then

Begin



NoMoreSwaps := False;

Temp := List[Element];

List[Element] := List[Element + 1];

List[Element + 1] := Temp;

OutputList(List,NumberOfItems);

End ;

Until NoMoreSwaps;

End ;

Begin








OutputList(UnsortedList,Max);

BubbleSort(UnsortedList,Max);

ReadLn;

End .

The bubble sort program does 11 swaps of elements using the given test data. Each

swap consists of 3 assignments. This means that 33 assignments are required.

The insertion sort program uses 22 assignments for the same test data.

i) using the list: 12, 23, 45, 56, 67, 99, 17

The insertion sort uses 17 assignments.

The bubble sort uses 15 assignments.

ii) using the list: 99, 67, 56, 45, 23, 17, 12

The insertion sort uses 66 assignments.

The bubble sort uses 66 assignments.

8

Program Project1;

{$APPTYPE CONSOLE}

Uses

SysUtils;Const Max = 7;


Var UnsortedList: TList ;

Procedure OutputList(List:TList; First, Last: Integer);

Var i: Integer;

Begin Write('(');

For i := First To Last-1




WriteLn(List[Last],')');

End ;

Procedure Swap(Var A, B : Integer);

Var Temp: Integer;

Begin

Temp := A;

A := B;

B := Temp;

End ;

Procedure QuickSort ( Var List: TList; First, Last:Integer);

Var PivotValue, LeftPointer, RightPointer, Pivot: Integer;

Begin

If First < Last {if there are elements in the list}

Then

Begin

PivotValue := List[First];

LeftPointer := First + 1;

RightPointer := Last;

Pivot := 0;

While LeftPointer <= RightPointer

Do

Begin While (List[LeftPointer] < PivotValue) And (LeftPointer

<= RightPointer)

Do

Begin

LeftPointer := LeftPointer + 1;

End ;

While (List[RightPointer] > PivotValue) And (LeftPointer<= RightPointer)

Do

Begin

RightPointer := RightPointer − 1;

End ;

If LeftPointer < RightPointer

Then

Swap (List[LeftPointer], List[RightPointer]);

End ;

Pivot := RightPointer;Swap (List[First], List[Pivot]);

QuickSort (List, First, Pivot − 1);

QuickSort (List, Pivot + 1, Last);

End ;

End ;

BeginUnsortedList[1]:=56;



U t dLi t[4] 12







OutputList(UnsortedList,1,7);

QuickSort(UnsortedList,1,7);

OutputList(UnsortedList,1,7);

ReadLn;

End .

Quick sor t qu estions

1

List Comment

First Last Pivot

value

Left

pointer

Right

pointer

Pivot [1] [2] [3] [4] [5] [6] [7]

1 7 56 2 7 56 23 67 12 45 99 17 Swap left &

right

3 7 56 23 17 12 45 99 67 Swap right &

pivot

6 5 5 45 23 17 12 56 99 671 4 45 2 4 45 23 17 12 No value

larger than

pivot value

5 4 4 12 23 17 45

1 3 12 2 3 12 23 17 No value

smaller than

pivot value2 1 1 12 23 17

1 0 Last < First

return from

call

2 3 23 3 3 23 17 No value

larger than

pivot value

4 3 3 17 23

2 2 Last = First

Return from

call

4 3 Last < First

return from

call

5 4 Last < First



5 4 Last < First

return from

call

6 7 99 7 7 99 67 No value

larger thanpivot value

8 7 7 67 99

6 6 Last = First

return from

call

8 7 Last < First

return fromcall


Question 1 – Jan 2007 CPT4 Q7

Question

number

Answer Marks

1a)

1 mark for first row (12, 1, 9)

2 marks for second row (1 mark for each 5)2 marks for third row (3 and 3)

2 marks for fourth row (1 mark for Lower = 4, 1 mark for upper = 4)

8

1 mark for correct return value

1b) Find the position of 12/ a number in the array// search for 12/ a number in the

array;

1




Question

number

Answer Marks

2a) 271;

The required item might be the 271st one/last one/ not be present//

Every item accessed;

1

1



2b) 9;

Each comparison halves the number of items to be accessed//

271 lies between 28 and 29

1

1

2c)

1 mark for Count1

1 mark for Count2

1 mark for Temp

(ii) (Bubble) sort the items into ascending order;

(ii) Reduce the number of tests each pass// stop when no swaps occur during

a pass//

Add a flag NoSwaps to indicate when no swaps occur// change loop control

to Repeat until no swaps// sort variable sized array;

5

1

1


Question

number

Answer Marks

3a) A procedure/routine which calls itself//is defined in terms of itself; R re-

entrant;

A function instead of procedure R program

1



iteration Talked Out (no mark)

3b) i)

Accept True in row 3

Marks in each row all three/two parts correct

Accept empty cell to mean: same as in previous row.

Stop marking when logic goes wrong

7

3b) ii) Binary search;;

Search;

R any other type of search 2


Question

number

Answer Marks

4a) i) 8 1

4a) ii) Each time a comparison is made in a binary search the number of items

to be searched/list is halved;

//137 lies between 27 and 28

Could have (ii) even if (i) incorrect

1

1

4b) i)137 1

4b) ii) In a linear search of 137 items, the required item might be the 137th

one;

need a termination - must explain why 137 is the maximum

4


Question

number

Answer Marks

5a) 1, 17, 9, 21, 15, 23;(2 if all right, 1 if 4 of 6)

If > misinterpreted, follow through for 1 mark

2

5b) A bubble sort 1

5c) To detect when all the numbers have been sorted

Efficiency (to stop procedure repeating unnecessarily);

R to detect when numbers have switched

1



2.7


1a entities: lorry, queue, loading bay

Attributes of lorry: waiting, being unloadedAttributes of queue: lorries in queue, queue empty

Attributes of loading bay: loading bay free, loading bay in use

b states: no lorry waiting, no lorry being unloadedno lorry waiting, lorry being unloaded

lorries waiting in queue, lorry being unloaded

possible events: a lorry arrives, a lorry leaves the loading bay possible activities: unloading a lorry, inter-arrival time of lorries

c

Lorries Arriving

LorryBeing Unloaded

LorriesQueuing

Loading BayFree

d possible variables: arrival times of lorries, loading times of lorries, number of loading

bays

2

Master

clock

(hours)

Lorries

arriving

Lorries being

unloaded

Lorries in

queue

Lorry hours

in queue

Loading bay

status

05:00 - - 0 0 Free

06:00 Lorry1 Lorry1 0 0 In use

07:00 Lorry1 0 0 In use


09:00 Lorry2 0 1 In use10:00 Lorry3 Lorry2 1 2 In use














3

Master

clock(hours

)

Lorries

arriving

Lorries

beingunloaded

in Bay 1

Lorries

beingunloaded

in Bay 2

Lorries

inqueue

Loading

bayhours

free

Loading

Bay 1status

Loading

bay 2status

05:00 - - 0 2 Free Free

06:00 Lorry1 Lorry1 0 3 In use Free



06:00 Lorry1 Lorry1 0 3 In use Free

07:00 Lorry1 0 4 In use Free

08:00 Lorry2 Lorry1 Lorry2 0 4 In use In use

09:00 Lorry2 0 5 Free In use

10:00 Lorry3 Lorry3 Lorry2 0 5 In use In use11:00 Lorry3 0 6 In use Free


13:00 Lorry4 0 7 Free In use




4a entities: passenger, queue, check-in desk

Attributes for passenger: waiting, checking in

Attributes for queue: passengers in queue, queue emptyAttributes for check-in desk: in use, free

b states: no passenger waiting, check-in desk free

no passenger waiting, passenger checking in passengers waiting, passenger checking in

events: passenger arrives, passenger finishes checking in

activities: checking in a passenger, inter-arrival time of passengers

c

Master

clock

(minutes)

Passenger

arriving

Passenger

checking in

Passengers

in queue (25

at start)

Passenger

minutes in queue

(from check-indesk opening)

Check-in

desk

status

1 Passenger1 24 24 In use




5 Passenger26 Passenger3 23 117 In use

6 Passenger3 23 140 In use7 Passenger4 22 162 In use




































42 Passenger35 Passenger21 14 763 In use43 Passenger22 13 776 In use


















g






























6 passengers are still waiting to check in half an hour before take off.

If the check-in process takes longer, there would be even more passengers notchecked in.Queues would be even longer if more passengers arrive during the peak half-hour.

A second check-in desk would then become necessary.

Exam pract ice quest ionsQuestion

number

Answer Marks

1a A computer model is used to manipulate variables and observe results. 1

1b We may consider simulation because experimenting with the real 1



1b We may consider simulation because experimenting with the real

system might cause unknown irremovable effects, or because we canexperiment with a model more cheaply and at a time of our choosing.

1

1c Possible examples include queuing for service, traffic flow, population growth. 1

Question

number

Answer Marks

2a The entities are: ship, queue, berth.

The attributes for ship are: waiting, being unloaded.The attributes for berth are: berth free, berth in use.

The attributes for queue are: ships in queue, queue empty.2b Possible states are:

no ship waiting in queue, berth free

no ship waiting in queue, ship being unloaded in berthships waiting in queue, ship being unloaded in berth.

Possible events are:

a ship arrives

a ship leaves the berth.

Possible activities are:the unloading of a ship

the inter-arrival time of ships.

2cShips

ArrivingShip

Being Unloaded

ShipsQueuing

BerthFree

2d

Masterclock

(hours)

Shiparriving

Ship beingunloaded

Ships inqueue

Ship hoursin queue

Berthstatus

1 - - 0 - Free

2 Ship1 Ship1 0 - In use



p p

3 Ship1 0 0 In use

4 Ship2 Ship1 1 1 In use

5 Ship2 0 1 In use


7 Ship2 1 3 In use


9 Ship3 1 5 In use


2e Possible variables: ship arrival times, unloading times, number of berths, maximum space for ships queuing.

3.1


Qu. Mantissa Exponent m x 2e Denary

1 a 0.1011 10010 0001102=610 0.101110012 x 26 = 101110.012 46.25

b 1.0101 01000 0001002=410 -0.1010112 x 24 = -01010.112 -10.75

c 0.1100 00000 1111112=-110 0.112 x 2-1 = 0.0112 0.375

d 1.0100 00000 1111012=-3

10 -0.11

2 x 2-3 = -0.00011

2 -0.09375

2 a 1.10100 01000 101002=2010 1.10100012 x 25 = 110100.012 52.25

b 1.01100 00000 011112=1510 1.0112 x 20 = 1.0112 1.375

c 1.01010 00000 011102=1410 1.01012 x 2-1 = 0.101012 0.65625

d -1.10001 11010 101102=2210 -1.1000111012 x 27 = -11000111.012 -199.25

e -1.01100 00000 011012=1310 -1.0112 x 2-2 = -0.010112 -0.34375

f 0.00000 00000 111112 special case infinity +∞

g 0.00000 00000 000002 special case zero 0

h -0.00000 00000 111112 special case infinity -∞

3 a Mantissa Exponent

5.5 0.10112 x 23 0.1011 00000 000011

0.375 0.112 x 2-1 0.1100 00000 111111

9 75 1 0110012 x 24 1 0110 01000 000100



-9.75 1.0110012 x 24 1.0110 01000 000100

-0.1875 1.012 x 2-2 1.0100 00000 111110

3 b Sign Exponent Mantissa

5.5 1.0112 x 22 0 10001 01 1000 0000

0.375 1.12 x 2-2 0 01101 10 0000 0000

-9.75 -1.001112 x 23 1 10010 00 1110 0000

-0.1875 -1.12 x 2-3 1 01100 10 0000 0000

4 Precision means the maximum number of significant digits we can represent. Precision is

important with floating point numbers because there are an infinite number of values between

any two given real numbers. These have to be mapped to a finite number of representations

(bit patterns). Therefore some values will be approximated.

5 36.1 is between 1.0010000011 x 25 and 1.0010000100 x 25

1.0010000011 x 25 = 36.09375

1.0010000100 x 25 = 36.125

Rounding to the nearest value, means 36.1 is represented by 1.0010000011 x 25 This introduces an absolute error or 0.00625 or a relative error of 0.000173

6 1.0110000000 x 212 x 27 = 1.0110000000 x 219

The largest possible value in the exponent field in minifloat format is 111102 = 30. As the

exponent is stored in excess-15 mode, the largest exponent possible is 15. Therefore the

result of this calculation will cause the exponent field to contain 11111 and set the mantissa to

0 (this means +∞).

7 1.0110000000 x 2-9 x 27 = 1.0110000000 x 2-16

The smallest possible value in the exponent field in minifloat format is 000002 = 0. As the

exponent is stored in excess-15 mode, the smallest exponent possible is -15. Therefore the

result of this calculation will cause the exponent field to contain 0 and set the mantissa to 0

(this means 0).

8 1.10100 000002 x 212 = 1101000000000.02

+ 1.01000 000002 x 2-7 = + 000000000000.0000001012

1101000000000.0000001012 = 1.1010000000000000001012 x 212

However, there are only 11 bits for the mantissa, so the result is 1.10100000002 x 212

Note that the effect of adding a very small number to a large number does not change theresult.


Question 1 – January 2008 CPT4 Q1Question

NumberAnswer Marks

1a BF2 1



1b -;1038 2

(1 mark

for sign, 1mark for

value)1c +;191;.125; A 1/8 3

1d -;2;.03125; A 1/32

(If incorrect part marks as follows

1 mark for complemented mantissa 01000001

1 mark for moving binary point 2 places)

3

1e To maximise precision in a given number of bits;

A to maximise accuracy in a given number of bitsTo minimise rounding errors;

To allow a wider range of values to be stored;

Max 2


Question

NumberAnswer Marks

2a 140 1/4 ;; one mark for correct integer part,

140.25;; one mark for correct fractional part

2

2b i) -14.5;;; give 2 marks for 14.5

partial marks for workings if result incorrect:

1 mark for negative number;1 mark for x24 (accept 16 instead of 24);

A showing that binary point moves 4 places right;

ii)leftmost 2 digits/bits are different;

a significant bit is stored after the binary point;

bit after point different from bit before point;

3



p p

(negative number) starts with 10. (positive number starts with 01)..;

A the first bit after the sign bit is a .0.;

A The second bit is a .0.;

A an answer that clearly implies a .0. follows the .1.

iii) to maximise accuracy/precision for a given number of bits

// to minimise rounding errors;

A more accurate/precise for a given number of bits;

a given number can only be expressed in one way in a given number

of bits

// a given number can only be expressed in one way in a given format;

to simplify arithmetic/logical operations;

I range

max 1

1


Question

numberAnswer Marks

3a) BE4; must be capital letters 1

b) 190.25 / 190 1/4 ;; one mark for correct integer part,

one mark for correct fractional partone mark for correct working(e.g. correct place values)

3

c) -1052;; 1 mark for workings if result incorrect

1 mark for sign, 1 mark for 1052

2

d) i) -8.25 / -81/4;;; partial marks for workings if result incorrect

1 mark for sign, 1 mark for moving binary point 4 places or showing

24

3

ii) starts with 1 0the first 2 binary digits are different;

a significant bit is stored after the (implied) binary point;

bit after (implied) binary point different from bit before binary point;

A all leading 1.s have been removed // there are no leading 1.s;

R there are no leading zeros

1

Question 4 – June 2004 CPT4 Q4(b)

Question Answer Marks

number

4a) i) 4



(if the answer is wrong give:moving e places to right / exponent processing 2

e or equivalent: 1

markcorrectly identifying negative number: 1 mark follow through if binary representation wrong

ii) To maximise precision in a given number of bits // to minimiserounding errors

// to have just one representation of the decimal number // to simplify

arithmetic operations: 1 mark

A to maximise accuracy in a given number of bits;

9


Question

number

Answer Marks

5a) i) positive 1

ii) < 2-2

1

5b) Correct answer 194.5 or 194 ½

working

If wrong answer, method marks as follows:

exponent 2

8

[clearly defined] 1 markapplication of shift / *28 from correct start point 1 mark

correct interpretation of bits 1 mark

(Basically here, if it is a little inaccurate, give 2 marks, if quite

inaccurate but slightly correct, give 1 mark.)

2

1

c) i) Processing fixed point numbers is quicker than floating point / less

processing is required;More accurate / greater precision;

1

ii) Where the possible range of numbers to be stored is limited / small;

Where number is of a set format / processing integers /

Working with currency;Where maximum precision is required

1



4.1


1 System programs and application programs.

2 Hide the complexities of the hardware from the user.

Manage the hardware resources in order to provide for an orderly and controlled allocation of

the processors, memories and I/O devices among the various programs competing for them,

manage the storage of data.

3 Processors

a) Storage

b) Input/output devices

c) Data

4 Processor scheduling

a) Memory management

b) I/O management

c) File management5 Virtual machine: this is the apparent machine that the operating system presents to the user,

achieved by hiding the complexities of the hardware behind layers of operating system

software.

6 Application program interface: this is a layer of software that allows application programs to

call on the services of the operating system.

7 A standard application program interface (API) allows a software developer to write an

application on one computer and have a high degree of confidence that it will run on another

computer of the same type, even if the specifications of the latter are different. The programshould continue to run when hardware upgrades and updates occur because the operating

system, not the application, manages the hardware and the distribution of its resources.

8 Command Line and Graphical user interface.

Some embedded systems require no operating system because they function as very simple

controllers. Others such as mobile phones do require an operating system. However, a mobile

phone operating system has to manage a different set of resources. The resources include the

keypad, the screen, the address book, the phone dialler, the battery and the network connection.

Operating systems for general purpose computers have to manage resources such as printers, harddrives, DVD R/W drives, high resolution flat screen visual display units, mouse and standard

QWERTY keyboard.


Question 1 – January 2003 CPT2 Q5Question number Answer Marks

1a) OS hides complexities of hardware from the user; 1

1b) Any three @ 1 each

Processor(s)/cpu(s);

Max 3



( ) p ( );

Memory/IAS/Main memory;

Disk (space)/backing store; A Hard disk/drive //Floppy

disk (drive)//Secondary storage

I/O devices//peripherals; R examples

File space; A files R data

R programs

4.2


1a) Interactive operating system: An operating system in which the user and the computer are in

direct two-way communication.

1b) Real time operating system: inputs are processed in a timely manner so that the output can

affect the source of the inputs.

1c) Network operating system: in this system, a layer of software, added to the operating system

of a computer connected to the network, intercepts commands that reference resources

elsewhere on network, e.g. File server and then redirects the request to the remote resource in

a manner completely transparent to the user.

2 Both the airline reservation system and the system to land an aircraft:

have to support application programs which are non-sequential in nature, i.e. programs which

do not have a START- PROCESS - END structure;

have to deal with a number of events which happen in parallel and at unpredictable momentsin time;

have to carry out processing and produce a response within a specified interval of time.

In the case of the airline reservation system Airline reservation system - up to a 1000

messages per second can arrive from any one of 11000-12000 terminals, situated all over the

world. The response time must be less than 3 seconds.

In the case of the aircraft landing system - up to 1000 signals per second can arrive from

sensors attached to the system being controlled. The response time must be less than one

thousandth of a second.a) C: - local hard drive; D: - DVD R/W drive; E: - flash memory drive; N: - network drive

b) Network operating system

c) Washing machine; microwave oven; central heating controller

d) Cable TV set-top box; broadband router; computer network firewall

3 The mobile phone operating system will be different from that of a desktop computer becausethe kind of resources available to a mobile phone is different. In particular, the energy, the

physical context and the mobility of the user, and the limited amount of some resources such

as memory or CPU. An operating system for a mobile phone needs to manage a limited

amount of energy from the battery, a limited amount of memory and a processor of limited

capability. It needs to manage a network connection in a fairly autonomous way that

continually sends a message packet to base stations so that mobile phone location can be



continually sends a message packet to base stations so that mobile phone location can be

identified. It needs to manage a keyboard that supports predictive text, a physically small

display, an address book, a phone dialler and a battery of limited power. It needs to meet real

time constraints because, for example, multimedia applications need to run. It needs to takeinto account the instability of mobile phone networks (bandwidth, connection, etc).

4 The O.S. for a smartphone needs to be more powerful than that of an ordinary mobile phone

because the smartphone offers advanced capabilities with PC-functionality beyond a typical

mobile phone. The operating system must support advanced features like e-mail and Internet

capabilities and/or a full keyboard. The OS must have the capability to allow a user to install

new applications on the smartphone.

5 The operating system must take on the tasks of the BIOS in a PC and has to be designed to

run on processors with low clock frequency and a main memory of limited capacity. Theoperating system must use various techniques to save energy and must cater for short reaction

times. The operating system must support character input via a stylus and touch sensitive

screen and should support high-resolution colour rendition with resolutions of approximately

640x480 pixels.

6 An embedded computer system is a dedicated computer system with a limited or non-existent

user interface and designed to operate completely or largely autonomously from within othermachinery, e.g. a motorcar. An embedded computer system in a motorcar could be set up to

manage the engine of the car carrying out a range of tasks one of which is controlling theignition of the gases in the cylinders.

7 The embedded computer O.S. must manage multiple tasks that need to meet specific time

constraints. In the simplest system, a low-level piece of code switches between tasks or

threads based on a timer (connected to an interrupt). Operating systems for embedded

systems are designed to work with the constraints of limited memory size and limited

processor performance. In portable embedded system the operating system must also take

account of limited battery life.

The O.S. on which embedded systems rely has had to become more sophisticated asthe hardware of embedded systems has become more complex with every

generation, and more features have been added. The role of an O.S. is to manage the

resources and to hide the complexities of the hardware. In the case of embedded

computer systems, it required to hide the complexities from the application

programs so making it possible to create new applications without consuming an

excessive amount of development time. With more tasks to manage and subject to

strict timing constraints a greater reliance on getting this support from the O.S. has

become necessary. This in turn requires greater sophistication in the O.S.


Quest ion 1 - KB


1 An embedded computer system’s operating system has a

limited number of resources to manage unlike a desktop

operating system, most notably a limited or non-existent

user-interface and does not have to support the execution of

a large number of different applications On the contrary the

2



a large number of different applications. On the contrary the

embedded system’s operating system has to support theexecution of a limited number of applications.

Quest ion 2 - KB


2 A server operating system is an operating system optimised

to provide one or more specialised services to networked

clients such as file storage, domain control, running

applications. They are optimised for the service that they

carry out. This leads to a near optimum performance thatwould not be achieved if a server were carrying out a large

amount of general-purpose processing.

2

Quest ion 3 - KB


3 1 - It has one set of tasks to perform; 2 - Very

straightforward input to expect (a numbered keypad and a

few pre-set buttons); 3 - Simple, never-changing hardware to

control.

3

Quest ion 4 - KB


4 Any three from:

1) They have to support application programs which are non-

sequential in nature, i.e. programs which do not have aSTART- PROCESS - END structure;

2) They have to deal with a number of events which happen

in parallel and at unpredictable moments in time;

3) They have to carry out processing and produce a response

within a specified interval of time;

4) Some systems are safety-critical meaning they must be

fail-safe and guarantee a response within a specified time

interval.

3

5.1End of topic quest ions

1



Blind Person Guide Dogprovided with

Guide Dog Blind Personallocated to

2

Person Carkeeps

Car Personregistered to

3

Borrower BookCopyborrows

BookTitle has

Loanmakes of


Please see the Teacher Notes for Section 5.3: SQL for all of the answer to the

Examination-style questions.

5.2





1 Course (CourseCode, CourseName)

Student (StudentID, StudentName, StudentAddress, StudentDateOfBirth)

CourseSet (CourseCode, SetNo, TeacherInitials)

Enrolment (StudentID, CourseCode, SetNo)

2 Borrower (BorrowerID, BorrowerName, BorrowerAddress)

BookTitle (ISBN, Title, Author)

BookCopy (AccessionNumber, ISBN)

Loan (BorrowerID, AccessionNumber, LoanDate)


Please see the Teacher Notes for Section 5.3: SQL for all of the answer to the

Examination-style questions.

5.3


1a)

CREATE DATABASE CourseEnrolment

CREATE TABLE Course

(

CourseCode VARCHAR(5),

CourseName VARCHAR(20),

PRIMARY KEY (CourseCode)

);

CREATE TABLE Student

(

StudentID VARCHAR(10),

StudentName VARCHAR(25),

StudentAddress VARCHAR(50),

StudentDateOfBirth Date,

PRIMARY KEY (StudentID)

);CREATE TABLE CourseSet

(


SetNo INT,

TeacherInitials VARCHAR(3),

PRIMARY KEY (CourseCode, SetNo)

FOREIGN KEY (CourseCode) REFERENCES Course (CourseCode));

CREATE TABLE Enrolment

(

StudentID VARCHAR(10),


SetNo INT,

PRIMARY KEY (StudentID, CourseCode),



PRIMARY KEY (StudentID, CourseCode),

FOREIGN KEY (StudentID) REFERENCES Student (StudentID),

FOREIGN KEY (CourseCode, SetNo) REFERENCES CourseSet (CourseCode,

SetNo)

);

1 b) SELECT * FROM Course

1 c) SELECT CourseCode, CourseName, StudentID, StudentName

FROM Enrolment, Student, Course

WHERE Enrolment.CourseCode = Course.CourseCode

AND Enrolment.StudentID = Student.StudentID

ORDER BY CourseCode

2a)

CREATE DATABASE BookLoanSystem

CREATE TABLE Borrower

(

BorrowerID VARCHAR(5),

BorrowerName VARCHAR(20),

BorrowerAddress VARCHAR(50),

PRIMARY KEY (BorrowerID)

);CREATE TABLE BookCopy

(

AccessionNumber INT,

ISBN VARCHAR(13),

PRIMARY KEY (AccessionNumber),

Foreign Key (ISBN) REFERENCES BookTitle(ISBN)

);

CREATE TABLE BookTitle

( ISBN VARCHAR(13),

Title VARCHAR(30),

Author VARCHAR(25),

PRIMARY KEY (ISBN)

);

CREATE TABLE Loan

(

BorrowerID VARCHAR(5),

AccessionNumber INT,

LoanDate Date,PRIMARY KEY (AccessionNumber, LoanDate),

Foreign Key (BorrowerID) REFERENCES Borrower(BorrowerID),

Foreign Key (AccessionNumber) REFERENCES BookCopy(AccessionNumber)

);

2 b)

SELECT BorrowerID, BorrowerNameFROM Borrower, Loan

WHERE Borrower.BorrowerID = Loan.BorrowerID

2 c)SELECT BorrowerID, BorrowerName

COUNT(Loan.AccessionNumber) AS BookCount

FROM Borrower Loan



FROM Borrower, Loan

WHERE Borrower.BorrowerID = Loan.BorrowerID

Group By BorrowerID



Question

number

Answer Marks

1a i) Recipe table; A Figure 2;

ii) Why : contains multiple values in Ingredients field/attribute/column;// data in Ingredients column not atomic // repeating groups.

1

1

1b i) fully normalised :

every attribute is dependent on the key, the whole key and nothing but the

key;

OR (tables contain no repeating groups of attributes,) no partial

dependencies;no non-key dependencies; A rely on instead of depend on;

OR if (and only if) every determinant in the relation is a candidate key.

ii) Why : to aid consistency of data // to avoid potential data inconsistency

problems;

// to eliminate data inconsistency // to minimise data duplication;

// to eliminate data redundancy; A reduce instead of eliminate;

R saving space.

2

1

1c i) Recipe (RecipeID, Dish, PrepTime, CookTime, NoOfServings,

CookInstructions);

ii) FoodItem (FoodItemID, FoodItemName, PackSize, Price);

iii) RecipeIngredient(FoodItemID, RecipeID, Quantity).

(1 mark for each correct field, 1 mark for correct primary key.

1

1

4

Take off 1 mark for every extra field included.)

1d SELECT FoodItemName, Quantity, PackSize, Price

FROM FoodItem, RecipeId = RecipeIngredient, RecipeWHERE (Recipe.RecipeId = RecipeIngreident.RecipeId)

AND (RecipedIngredient.FoodItemId = FoodItem.FoodItemId)

AND (Recipe.Dish = “Feat Salad”)

ORDER BY FoodItemName

1

11

1

1

1

Max 5



field names F/T P1 for fieldname.tablename P1 tbl prefix

A ORDER BY FoodItemName

A Dish instead of Recipe.Dish

A “feta salad” instead of “Feta Salad” A #feta salad# instead of “Feta

Salad”


Question

number

Answer Marks

2a Copyright, Designs and Patents Act (1998); if other laws included T.O. 1

2b Boxes for correct entities: SoftwareLicence SoftwareInstallation (1 mark)

Correct degree of relationship: 1 to many (1 mark)

Suitable name for relationship: (1 mark)

3



2c Any sensible field length accepted except for SoftwareID, ComputerID,

StaffID

i) SoftwareID VARCHAR(10) PRIMARY KEY (NOT NULL)

// SoftwareID VARCHAR(10) PRIMARY KEY(SoftwareID);

SoftwareName VARCHAR(30)

Version VARCHAR(10)

Supplier VARCHAR(20)

DatePurchased DATE

ExpiryDate/DateValidTo DATE

NoOfLicences INT

ii )

SoftwareID VARCHAR(10)

ComputerID VARCHAR(6)

DateInstalled DATE

StaffID VARCHAR(3)

PRIMARY KEY (SoftwareID, ComputerID)

FOREIGN KEY (SoftwareID)

REFERENCES Software Licence(SoftwareID)

I NOT NULL

(1 mark for any 3 attributes correctP1 if extra symbols used.

Ignore spaces and case in attribute

names.)

(A char/string/text/alphanumeric

Instead of VARCHARA Date/Time instead of DateA Integer instead of INT

BOD any attributes which are

clearly more than 1 word.)

(1 mark for any 2 attributes correct.)

(If not DDL, give 1mark if composite

key identified.)

2d

SELECT ComputerID, SoftwareName, Version;

FROM SoftwareLicence, SoftwareInstallation;WHERE SoftwareLicence.SoftwareID=SoftwareInstallation.SoftwareID;

ORDER BY ComputerID; A ASC or DESC

4

Extra attributes: T.O.

Accept (instead of FROM WHERE):

FROM SoftwareLicence INNER JOIN



A LEFT JOIN

Question 3 – June 2006 CPT 5 Q5

Question

number

Answer Marks

3a i)

R one to many relationship

ii)

Entity-Relationship Diagram; A E-R diagram; A E-R D R E-A-R diagram

2

1

FROM SoftwareLicence INNER JOIN

SoftwareInstallation ON

SoftwareLicence.SoftwareID =

SoftwareInstallation.SoftwareID

P1 for other spurious

punctuation inc.semicolons

Table names prefixed with tbl, P1

If table name and attribute transposed, P1

3b 6



3c 6



F/T with attribute names

P1 for tbl prefixP1 if table name after attribute name

I extra punctuation

Quest ion 4 – Janu ary 2005 CPT5 Q9

Questionnumber

Answer Marks

4a I Minor spelling

i)

1



ii)

iii)

1

1

4b Penalise table name. field name in reverse order once

R Quotes and additional constructs

I Table names unless in wrong order or wrongly expressed

i)

Select FirstName, Surname

From Student

ii)

1

I table name

unless incorrect



Select Student.FirstName, Student.Surname

MarkAwarded.Mark;

From Student, MarkAwareded;

Where MarkAwarded.LifecyclePhaseID = 1;

And Student.StudentID =

MarkAwarded.StudentID ;

Order By Student Surname;

5


Question

number

Answer Marks

5a (NB Take note of labelling inside boxes because candidate’s positioning of

labels may be opposite to that shown below.)

Alternative symbols:

Order By Student.Surname Asc

A Ascending

Asc/Ascending must be in correct position

A OrderBy

1

Accept minor mis-

spelling or spaces between parts of

entity name.

A. plural names

I. Box outlines



i)

ii)

iii)

iv)

1

1

1

5b ( R. Tbl in front of table name – penalise once.)

i)

Select Surname

From SwimmerWhere SwimmerNo = 6;

1Select Swimmer.Surname isOK

I. Brackets surroundingattributes

ii)

Select SwimmerNoFrom GalaRaceSwimmer

Where (RaceNo = 5); And (GalaNo = 2);

Order By TimeRecordedForRace;

3

I. Brackets surrounding

attributes, table names in

front of attributes unlessincorrect

A. Criteria without

brackets

R. Extra attributes,

A. Asc or Ascending in correct

place i.e. after

Ti R d dF R



iii)

Select Swimmer.SurnameFrom Swimmer, GalaRace;

Where (GalaNo = 4);

And (GalaRace.SwimmerNoOfWinner

= Swimmer.SwimmerNo);

3

Total 11

tables, criteria in thissolution but be careful because candidates may

give an alternative

involving extra

TimeRecordedForRace

R. Asc/Ascending in any otherosition and/or with other words

A. Select

Swimmer.Surname,GalaRace.RaceNo From …

OrA. Select Surname, RaceNo

From…

Brackets may be omitted.A. GalaRace.GalaNo = 4

And SwimmerNoOfWinner

Select Surname

From SwimmerWhere SwimmerNo In;

(Select SwimmerNoOfWinner

From GalaRace;Where GalaNo = 4);

R. = in lace of In


Question

number

Answer Marks

6a i) 1

1



ii)

L other entities

1

6b i)

Select Book.Title

From Book ;

Where BookISBN = “1-57820-082-2”;

ii)Don’t need Book in Select

Select Book.AuthorName, Book.ISBN ;

From Book, BookCopy ;

Where (Book.ISBN = BookCopy.ISBN) ;

And )BookCopy.AccessionNumber = 1234) ;

2

4

A. Title

A. ISBN = “1-57820-

Any extra attributes lose mark where extra attributes

used

R. 1-57820-082-2 Need quotesA. “1-57820-082-2”

R. TblBook – enalise once

Penalise TblBook/TblBookCopy

once

R. quotes on 1234

Any extra attributes lose markwhere extra attributes used

Brackets non-essential. Ma see

A. BoodCopy.ISBN in placeof BOOK.ISBN

A. AccessionNumber in

place of

A. In for =



C Mail-merge//Mail-merging 1

Total 9


Question

number

Answer Marks

7a

(NB: don’t allow relationship between Ward and PatientMedicalCondition.)

3

7b (For each extra attr ibute lose one mark.)

i)Ward(WardName, NurseInCharge, NoOfBeds)

ii)

1

2

A. NumberOBeds, NameOfNurseInCharge, NurseInChargeName

R. WardId, Name, NurseName, NameOfNurse, BedNo,



Patient(PatientNo, Surname, Forename, Address, DOB, Gender, WardName)

iii)

MedicalConditionID(MedicalConditionNumber, Name,

RecommendedStandardTreatment)

iv)

PatientMedicalCondition( PatientNo. MedicalConditionNo)

1

7c (Accept tbl in fr ont of table name.)

Select Patient.Forename, Patient.Surname, PatientMedicalCondition,

MedicalConditionNo

From Patient, PatientMedicalCondition

Where Patient.WardName = ‘Victoria’

And Patient.PatientNo = PatientMedicalCondition.PatientNo

A. Forename, Surname, MedicalConditionNo, WardName

3

Total 12

A. PatientId, PatientNumber, PatientSurname, PatientForename,PatientAddress, DateOfBirth, PatientDateOfBirth,

PatientGender, Sex, PatientSex

A. MedicalConditionId, MedicalConditionNumber,

MedicalconditionName, ConditionName, StandardTreatment,Treatment, RecommendedTreatment

R. ConditionNumber ConditionID

A. Attributes rejected in (ii) and (iii) for PatientNo and

MedicalConditionNo

R. If attributes used are not consistent with (ii) and (iii)

6.1

End of topic quest ions1 In serial data transmission, single bits are sent one after another along a single wire

2 In parallel data transmission, bits are sent down several wires simultaneously.

a) The transmission of data between computer and printer is controlled by the printer via the

Ready/Busy wire. If the printer is ready to receive, the Ready/Busy wire is set by the

printer at 5 volts, otherwise it is set to 0 volts. The computer reads the state of Ready/Busy

wire. If set to Busy, the computer does not send data. If the signal on this wire is set to

Ready, the computer places signals representing the data onto the data wires.



Ready, the computer places signals representing the data onto the data wires.

b) A short while afterwards, the computer sets the voltage on the Strobe wire to 5 volts. The

printer detects this strobe signal voltage and starts to read the data on the data wires at the

same time it sets the Ready/Busy wire status to busy. When the printer is finished reading

data, it sets the Ready/Busy wire back to Ready.

c) Serial data transmission is used for long-distance communication. The need for only one

signal pathway each way is the main reason for this use. This makes it easy to regenerate

the signal, which can reduce significantly in strength over long distances. It makes it easier

to route through telecommunication switches when routing the signal. It also saves on the

cost of cabling.

d) Parallel data transmission is used over short distances because it is difficult to keep the

voltages on the eight wires ’in-line’ with each other (the problem is known as skew)

beyond a certain distance. This can lead to the voltage on each wire being read incorrectly.

It is also expensive to run eight or more wires over long distances especially if the signals

on the wires need to be switched onto a different path, for example, in digital switching

centres such as telephone exchanges. Therefore, parallel communication has been restricted

to computer to printer connections and computer buses.

3 The Baud rate is the rate at which signals on a wire /line may change. One baud is onesignal change per second. The bit rate is the number of bits transmitted per second. The bit

rate can be greater than the baud rate if each signal can code more than one bit. For

example, if each signal contains 2 bits and the baud rate is 100 baud (100 signals sent per

second) then the bit rate is 2 × 100 bits/s.

4 The greater the bandwidth of the transmission system, the higher is the bit rate that can be

transmitted over that system

5 Latency refers to the time delay that can occur between the moment something is initiated

and the moment its first effect begins. In a wide area network involving satellites,significant time delay occurs because of the physical distance between the ground stations

involved and the geostationary satellite. Requesting and receiving a web page can involve a

considerable time delay even though the bit rate of the uplink and downlink to the satellite

is high, that is, the bandwidth is large. With a round trip distance in excess of 143,200

kilometres, speed of microwaves 3 × 108 metres/second, the propagation time delay is

approximately 0.4 seconds.

6 In asynchronous serial data transmission, the arrival of data cannot be predicted by the

receiver, instead a start bit is used to signal the arrival of data and to synchronise thetransmitter and receiver temporarily.

a) The arrival of data at the receiver is signalled by a special bit called a start bit.

b) The data bits are sent immediately following the start bit, then a parity bit if the data bits

are protected by a parity bit and finally, the transmitter attaches a stop bit. Thus, the start

bit and the stop bit frame a transmission of a few data bits encoding a single character, for

example, the character ‘A’. The time interval for the stop bit allows the receiver to dealwith the received bits, that is, transfer these into the RAM of the computer.

7 A communication protocol is a set of pre-agreed signals, codes and rules to be used for data

and information exchange between computers or a computer and a peripheral device such

as a printer that ensure that the communication is successful.

8 The handshaking protocol in serial data transmission involves exchange of the following

signals between sender and receiver, for example, between computer and a printer.

Assuming that the printer is present and switched on, the handshaking protocol for



computer serial port to serial port of a printer is as follows:

The sending device, the computer, enquires whether the receiving device is ready to

receive. The sending device waits for a response that indicates that the receiving device, the

printer, is ready to receive. On receipt of this signal, the sending device coordinates the

sending of the data and informs that receiver that it is sending the data. The sender then

waits for the receiver to become ready to receive more data.9 The parity bit is an extra bit added to each 7 or 8 data bits that are transmitted. This parity

bit is calculated from the 7 or 8 data bits and is either a 1 or a 0 depending on whether even

parity or odd parity is chosen.

Even parity:

The parity bit is set to either 1 or 0 so that the number of 1s across data bits and parity bit is

an even number.

Com uter

Computer

Computer

Printer

Printer

Printer

Clear to send pin 8 on printer

Start bit

Clear to send pin 8 on printer

Computer Printer Request to send pin 7 on computerYes, I am

Computer Printer Stop bitThat’s it

Are you ready?

Here it is

I am ready again

Computer Printer Clear to send pin 8 on printerBusy

e.g.

Odd parity:

The parity bit is set to either 1 or 0 so that the number of 1s across data bits and parity bit is

an odd number.

e.g.

10 The sending computer generates the correct parity bit and attaches it to the end of the data bits

as they are transmitted. The receiving computer regenerates a parity bit from the received data

0 0 1 1 0 0 1 1 0

Parity Bit Data Bits

0 0 1 1 0 0 1 1 0


1 0 1 1 0 0 1 1 0




bits, which it checks against the received parity bit. If the two are different, an error duringtransmission has occurred.

11 Baseband mode of operation dedicates the whole bandwidth of the transmission medium to

one data channel at a time (a single channel system).

12 Broadband mode of operation employs the bandwidth of the transmission medium to carry

several data streams at the same time (a multi-channel system).

a) Baseband tends to be used over short distances such as found in local area networks

where high performance can be achieved with this approach at low cost.

b) Broadband is used for long distance communication because long-distance

communications media are expensive to install and maintain. It would be wasteful if each

media path could support only a single data stream



Question

number

Answer Marks

1a Name: Start bit;

Purpose: Synchronise receiver;

Name: Parity bit;

Purpose: Perform parity check// check for errors in transmission;

A Prevent errors

Name: Stop bit;

Purpose: Allow start bit to be recognised// Allow receiver to process

received bits;A Indicates end of data

2

2

2

1b the number of signal/voltage changes per second;

A rate at which signals are sent; A rate at which voltage changes;

number of bits per second/unit of time;

R the rate at which bits are sent (question paraphrased)

Range of frequencies a channel can handle;

A maximum line speed; A maximum transmission speed

1

1

1

1c A signal can contain one or more bits;

Bit rate can be higher than baud rate;

bit rate = baud rate * number of bits per signal change

2


Question

number

Answer Marks

2a Bits are sent along a single wire/line; bits are sent one after

another;Max 1




Question

number

Answer Marks

3a Several bits are transmitted simultaneously/at the same time; downseveral wires; 1

A several bits of data; R dataR down two wires A diagram A a byte/word at a time;

1

R Bits of (the) data

2b i Data bit;

Parity (bit);

Signal to start data transfer/strobe;

Signal ‘ready to receive data’/busy;

Signal to acknowledge data transfer/Complete;

Out of paper/ink/error ;

On-line/off-line;

Handshaking//control signal/status signal (BUT only if not byexampleabove);

Ground;ii Transmission over long distances;

When a high data transfer rate is required;

A No driver is available

Max 2

Max 1

2c Data is transmitted intermittently (rather than as a steady stream);

Sender and receiver are only synchronised when data is beingsent //start bitsynchronises the receiver;

R Description only of start and stop bits

1

3b Exchange of signals between devices/devices communicating with each

other;

To establish their readiness to send or receive data – a method of ensuring

that both the sender and receiver are ready before transmission begins;

2


Question

number

Answer Marks

4a) i) Baseband: Single/data signal sent at a time Or single

message/packet/frame sent at a time Or uses single channel Or one4



transmission at a time

A Single stream of data;

Over full bandwidth (of the cable) Or occupying full bandwidth (of the

cable)

Or signal uses all available frequencies;

(ignore any additional references which are bits, 0/1 in any part of the

above)

R Single bits sent at a time

R Only works over short distances

4a) ii) Broadband: Several/data signals sent simultaneously Or several

messages/packets/frames simultaneously Or more than one signal

occupies bandwidth;

Each at a different frequency or in a different channel Or in a different

time slot;

OrMultiple channels used;

Each at a different frequency Or in a different time slot;

Or signal (or equivalent) uses only one frequency Or signal (or

equivalent) uses only part of the bandwidth;;

R Fast connection

R Video, sound and text….

R ADSL, cable examples, etc4b) i) Two reasons:

Wide area networks expensive to install;

Wide area networks expensive to maintain;

Wide area networks involve long distances;

Can allow multiple data stream to keep costs down;

Many channels needed to cope with high volume of traffic

Or enables more users to use network without experiencing congestion;

R Faster

Max 2

R Can work over longer distances

R More than one user will want to use it simultaneously

R Cheaper, more efficient


Question

number

Answer Marks

5a bits transferred simultaneously I concurrently; R data R bytes

bits sent down many wires at the same time; A bits of data

A a clear diagram;



5b) i) data get skewed; timing of bits becomes different / out of line;

A over longer distances the data may not be correct;

A too expensive because of amount of wires/cables/lines;

R signal decays R corrupted data

Max 1

5b) ii) use serial transmission

6.2


1 a) Local Area Network or LAN consists of linked computers in close proximity.

b) A standalone computer is a computer that is not networked. It therefore requires its own

peripherals, such as printer and application software, to be installed locally.

c) A wide area network is a set of links that connect together geographically remote

computers.

2 The term ‘topology’, in the context of networking, refers to the shape, configuration or

structure of the inter-connections connecting devices to the network.

3 An internet is a collection of local area networks and computers that are inter-connected by a

wide area network



4 A computer communicates on the network through a network interface card or network

adapter. A network adapter plugs into the motherboard of a computer and into a network

cable. Network adapters perform all the functions required to communicate on a network.They convert data from the form stored in the computer to the form transmitted or received

on the cable.

5 In the Ethernet protocol, each network card is assigned a unique address called its MAC

address. MAC is an acronym for Media Access Control. A MAC address is a 48-bit address,

for example 00-02-22-C9-54-13, expressed in hexadecimal and separated into six bytes. Part

of the MAC address identifies the manufacturer. Each network card manufacturer has been

allocated a block of MAC addresses to assign to their cards.

6 The collision problem occurs in bus networks when two computers transmit onto the bus at

the same time. The pulses of voltages from each will eventually collide resulting in higher

voltage swings. A computer attempting to read these pulses will fail to do so correctly.

When this happens, a collision is said to have occurred and the bus becomes unusable for

the duration of the transmissions from both computers.

7 Coaxial cable bus networks are segmented so as to reduce collisions and therefore improve

the performance of the network. A network segment in Ethernet is a run of Ethernet cable,

for example coaxial cable, to which is attached a number of workstations. Non-switched

Ethernet bus networks, for example coaxial cable, are often split into smaller sections, called

segments, in order to improve their performance. Performance in a non-switched Ethernetnetwork can drop significantly as more stations are added to it. This is because in Ethernet,

the cable along which data travels is shared by all stations connected to it. If lots of stations

have data to transmit, the network gets congested, and many collisions occur. Segmentation

is one solution to congestion on an Ethernet network.

8 The central switch in a switched Ethernet local area network ensures that collisions cannot

occur by forming a temporary unshared connection between sender and receiver. Frames

sent by other computers arrive on dedicated cables and are queued until an unshared

temporary connection can be made for these. The switch is thus making and breakingtemporary connections between each sender and receiver at high speed. As there is never

more than one frame travelling along each connection, collisions cannot occur.

9 In a thin-client network, all processing takes place in a central server, the clients are dumb

terminals with little or no processing power or local hard disk storage.

10 A thick- or rich-client network is one in which the applications are run in the client

workstations. Therefore, client workstations must have local processing power.11 Bus and star topologies appear very similar in the way that they are physically wired using

the current switch-based hardware. Even thin client systems, which can be considered to

resemble a traditional star network, use an Ethernet bus switch to connect a central server to

nodes. In a traditional star topology network, each link from node to central computer is an

independent link. Each link is therefore secure from eavesdropping by other nodes. If a link

to a node goes down, the other links and nodes are unaffected. However, if the central

computer goes down, then the whole network will fail. In a true star-based network, the

speed of each link to the central computer should remain high because the links are not



speed of each link to the central computer should remain high because the links are notshared. Traffic between nodes in a switched-based bus network will not be adversely

affected if a node goes down unless, of course, the traffic involves the broken node or the

node is a domain server that validates users when they attempt to log in. Neither will

unplugging a network cable in a switched-based bus network affect the rest of the network.

In a coaxial cable bus network, a break in the cable stops the whole network from working.

All connected nodes are able to read the frames travelling on the coaxial cable bus network.

Therefore, coaxial cable bus networks are not secure against eavesdropping. The frames in a

coaxial cable Ethernet bus networks can collide when multiple nodes send at the same time.

This results in the network showing a noticeable slowing of performance. Althoughcollisions between frames in switch-based Ethernet bus networks cannot occur, performance

can be affected when traffic volumes are high because the buffers in the switches suffer

overflow. A wireless network is a broadcast network and so is less secure than a cabled

switch-based Ethernet network unless wireless encryption is enabled. With a wireless-based

network without encryption, it is possible to eavesdrop on traffic intended for other

computers. A wireless network can also suffer congestion because the channels are shared.

a) A peer-to-peer network (P2P) is one that has no dedicated servers. All computers are

equal and therefore known as peers.b) A server is a computer that provides shared resources to network users.

c) A client is a computer that uses the services provided by a server.

d) In server-based networking resource security, administration and other functions are

provided by dedicated servers.

e) In Web 2.0, software becomes a service that is accessed over the internet.

f) Web services are self-contained, modular applications that can be described, published,

located, and invoked over a network, generally, the web.g) Software as a Service (SaaS, typically pronounced 'sass') is a model of software

deployment where an application is hosted as a service provided to customers across the

internet.

h) In web services architecture, all components in the system are services.

i) Ajax is a web technology that allows only the part of a web page that needs updating to

be fetched from the web server.

j) A wireless network refers to any type of local area network in which the nodes

(computers or computing devices, often portable devices) are not connected by wires butinstead use radio waves to transmit data back and forth between the computers.

k) Wi-Fi is the trademark for various IEEE 802.11 technologies that support wireless

networking of home and business networks.

l) Bluetooth is a wireless protocol for exchanging data over short distances from fixed and

mobile devices.12 Google’s search engine is an example of Web 2.0. Google began its life as a native web

application that has never been sold or packaged for its customers. Instead, the web

application, Google’s search engine, is delivered as a service, with customers paying,

directly or indirectly, for the use of that service. None of the trappings of the old software

industry are present. There are no scheduled software releases, just continuous improvement.

There is no licensing or sale, just usage. There is no porting to different platforms so that

customers can run the software on their own equipment, just a massively scalable collection

of commodity PCs running open-source operating systems plus Google-written applicationsd tiliti th t t id th t t G l ’ d t t h it



of commodity PCs running open source operating systems plus Google written applicationsand utilities that no one outside the company ever gets to see. Google’s data centres where it

all happens are closely guarded and visitors are not welcome. The software never needs to

be distributed but only performed. Google's service is not a server – though it is delivered by

a massive collection of internet servers. Neither is it a browser, though it is experienced by

the user within the browser. Nor does its flagship search service even host the content that it

enables users to find. This content is located on the web in the web pages. Google’s software

finds the links in these web pages and using a very clever algorithm arrives at a page rank

index for pages on the web.

13 A router is a device that receives packets or datagrams from one host (computer) or router

and uses the destination IP address that they contain to pass on the packets, correctly

formatted, to another host (computer) or router.

14 A gateway is a device used to connect networks using different protocols so that information

can be passed from one system to another successfully

15 When a computer, X, in the English local area network (LAN) wishes to send to a computer,Y, in the Australian LAN, it knows immediately that Y is on a different network. It therefore

sends to the inward facing port of the gateway that it is directly connected to. This gateway

then reforms the frame so that it is compatible with the internet and sends this frame on port

B to the internet. The internet is a collection of routers. The public IP address of Y’s LAN is

used by the routers to route the wide area network frame to the other side of the world.When the frame eventually arrives at the outward facing port of the gateway for the

Australian LAN, this gateway reforms the frame before sending it into the LAN via the

inward facing port. Both gateways have two network cards, one for the outward facing port

and one for the inward facing port.

First Hop

Second Hop

Third Hop




Question 1 – June 2002 CPT5 Q4 b) ii) and iii)

Question

number

Answer Marks

1a First mark: (More packet) collisions take place; Stations attempting to send at

the same time; Each station broadcasts to every other one; Some stations may

be attempting to broadcast at the same time;

Second mark: Packets need to be sent again;

Station has to retransmit (after a random delay);

Max 2

1b First mark: Exclusive bus connection (made temporarily between sender and

receiver)

Or data transfers take place in turns

Or switch connects just sender and receiver

Or collision domain limited to two stations

Or switch splits bus LAN into several smaller segments

Or switch allocates a time slot to each transmission

Or each host/computer/station/workstation/node has its own link on which

one packet at most can travel

Or switch allows a dedicated connection/pathway to be set up when a

computer wishes to send information to another

Or switch separates work stations (in different segments) and only passes

packets between segments when necessary

Or switch (ports) act as bridges which segment network;

Second mark: therefore collisions cannot occur (between two stations) Or

collisions reduced

2


Question

number

Answer Marks

2a A set of rules/procedures 1

2b Bus; R Ethernet on its own 1

2c Twisted pair//coaxial (cable)//optical fibre//fibre optic; 1

2d Need first octet or first and second octet or first, second and third octet to be

identical. Also must have four octets.

For example:

3



192.168.0.1

192.168.0.2

One mark for four octets;

One mark for same LAN;

2e (Use candidates example from D)

i 192.168.0;

ii 1 or 2;

1

2f a (unique) address/identifier assigned to network card // (unique) hardware

address/identifier;

1

2g Any two tasks @ one each

Allocation of port numbers;

Routing a packet/frame/segment to correct application/service;Splitting messages/data into packets // Disassembling messages //

Assembling packets;

Adding TCP headers // Adding sequence nos;

Error handling // sets parity bits;

Checking that transmission successful;

Resending transmission if necessary;

A Sets packet size

2

2h Any one of the following applications for one mark;

Telnet;

Internet browser;

http (client) // web server;

email;

FTP;

TFTP;SMTP;

1

R Non-networked applications such as word processor

2i Internet registry // Internet registrar;

A I.P. registry/registrar

1

Question 3 – 2010 Specimen question paper COMP3 Q5

Question

number

Answer Marks

3a A 192.71.0.1; A any valid number instead of 1

B 192.71.0.2; A any valid number instead of 2

C 192.71.1.1; A any valid number instead of 1

3



6.3


1 The common gateway interface or CGI: this is a gateway between the web server and a

web-server extension that tells the server how to send information to a web-serverextension, and what the server should do after receiving information from a web-server

extension.

2 A web-server extension: this is a program written in native code, that is, an executable or

a script that is interpreted by an interpreter running on the web server that extends the

functionality of the web server and allows it to generate content at the time of the HTTP

request.

3 The common gateway interface or CGI is a gateway between the web server and a web-

server extension. The CGI specification tells the server how to send information to aweb-server extension, and what the server should do after receiving information from a

web server extension. In its simplest form, the gateway consists of two objects, the

request object and the response object. An HTTP request sent by a web browser such as

Get / is placed in the request object in the gateway. The web-server extension accesses

the request object and processes its contents before formulating a response, which is

placed in the response object in the gateway. The web server then accesses the response

object to pass its contents back to the web browser.

4 Get /webpage.asp?myname=fred&age=65 When the post method is used by the browser or an HTTP application then the data, for

example myname=fred&age=6, is passed in the message body while when the Get

method is used the data is passed with the command in the address bar of the browser as

shown in the answer to question 4.

6 Dynamic web page content means content that is generated at the time of receipt of the

web browser request.

7 A script is executed on the server to determine the parts of the web page that are

dynamically created at the time of receipt of the request.8 www.educational-computing.co.uk?surname=bond&age=6

9 Perl, php, VBScript

10 Internet service providers (ISPs) are able to read an interpreted server-side script because

it is high level and therefore can monitor scripts and prevent scripts that could be

harmful from executing. This is not the case with server-side executables which arecompiled code presented in low-level form, making it difficult for ISPs to understand

whether the executable could be harmful.

11

http://www.educational-computing.co.uk/?surname=bond&age=6





12

Create a connection

Select a database

Perform a database query

Use returned data, if any

Close connection

13 Delphi, php, VBScript


Question 1 – January 2003 CPT5 Q7b(i) and (ii)


1b) i) A Fred;

B James;

1

1

1b) ii) A Fred;

B James;

1

1

6 4



6.4


1

Virus

Name

When Type of

attack

Type of damage Number of

computers affected

Melissa

virus

1999 Macro

virus

A document was created with the virus

in it and anyone who opened it would

‘catch’ the virus. The virus would then

send itself by email to the first 50

people in the person’s address book.

This made the virus replicate at a fast

rate

Thousands of

computers

worldwide. Melissa

shut down internet

mail systems that got

clogged with infected

emails

The creator of this

virus was jailed in2002

2005 Cross-

scripting

virus

a malicious website could load another

website into another frame or window,

then use Javascript to read/write data on

the other website

MySpace and Yahoo

sites affected

Love Bug 2000 worm flooded the internet with emails withthe subject, ILOVEYOU. The body of

the deceptive email read, "Kindly check

the attached love letter coming from

me." When opened, the e-mail wreaked

havoc on computers, replicating it

automatically, sending copies to

everyone in the user’s address book,

and damaging computer files, such as

MP3s.

affected 80 percentof businesses in

Australia and the

United States

Code Red 2001 worm operated in three stages – scanning,

flooding and sleeping. During the

scanning phase, the worm searched for

vulnerable computers and ran damagingcomputer code on them. Next, in the

flooding phase, the worm sent bogusdata packets to the White House

website. The White House, however,

changed their website’s IP address and

was therefore able to avoid the attack.

Blaster 2003 worm wreaked havoc on Microsoft XP,

Windows NT 4.0 and Windows Server2003 users. The worm spread quickly,

http://en.wikipedia.org/wiki/Internet

http://en.wikipedia.org/wiki/Internet



p q y,

checking for vulnerable computers and

then sending itself to those machines.

The worm was intended to attack

Microsoft’s update website. Some users

found that their computers were

sluggish, but otherwise may have been

unaware that they had been infected.

2 Here is what one bank says:

- We’ll NEVER send you time sensitive emails to confirm or change your security

details, account numbers, card numbers, PINs or expiry dates.

- We’ll NEVER ask you to input ALL of your security when logging onto internet

banking.

- We’ll NEVER send you an email containing a link to a log in page.

You probably get mountains of email every day. The chances are a fair bit of it’s

unsolicited but comes with halfway plausible return addresses.

More and more of these emails contain attachments carrying viruses, worms, or

Trojan horses which will harm your system if they go undetected.

We’re also seeing a lot more ‘ phishing’ emails. This is where a mail is sent from a

third party, saying they are from your bank, asking you to click on a link to

confirm/update your security information. The page you get directed to usually looks

exactly like the bank ’s page and when you enter your account and securityinformation, you may get forwarded to the real login page, or have to click another

link to login to your account.

What really happens is you actually send your account and security details to

someone who wants to get away with your money.

Some junk email uses social engineering to tell you about a contest you’ve won, or

details of a product you might fancy. Pretty obviously, the sender wants you to

interact with the email in a way that’s financially beneficial to them.Our emails never ask for personal details. The only time we’ll need those is when

you’re securely logged on to our banking or enrolment.

But if you do get mail asking for personal information it’s best to delete it right

away. There could be a dodgy dealer on the other end.

Please DON’T respond to any such email. Remember - we would never ask a

customer to re-register their security details. And we’ll never ask for your securitydetails until you’re safely logged into your account on our website.

The best advice we can give for emails you’re unsure of is, don’t open them. If you

use an email program like Outlook, Eudora or Notes, it’s a good idea to turn off your

‘Preview Pane’ too. Finally, if you receive an attachment you weren’t expecting, or

you don’t know whom it’s from, don’t open them. If you haven’t received an email

virus or worm yet, you will. Sorry, but it’s only a matter of time.

Every time you receive a message with an attachment, test it with these criteria. The Know test: Is the email from someone that you know?



y

The Received test: Have you received email from this sender before?

The Expect test: Were you expecting email with an attachment from this sender?

The Sense test: Does email from the sender make sense? For example, would the sender - let’s

say your mother - send you an email message with the Subject line, "Here you go, ;o)"

containing a message with the attachment AnnaKournikova.jpg.vbs? A message like that

probably doesn’t make sense. In fact, it’s an instance of the Anna Kournikova worm and

reading it can damage your system.

The Virus test: Does this email contain a virus? To determine this, you need to install and use

an anti-virus program. Alternatively, use an email service that checks viruses for you.

Some email systems provide additional benefits at no extra cost. Hotmail uses the McAfee scan to look for

viruses in your email attachments before they send you it. If the email looks like it contains a virus you will

be given a warning before you open it.

3 Strong passwords use combinations of uppercase and lowercase letters, numbers, and punctuation, they aren’t usually found in any dictionary. For example using ‘river’

would be a weak password, whereas ‘r!V3r_78’ would be much stronger .



Question

Number

Answer Marks

1a If you send the key with the message, anyone can decrypt the message

key would need to be sent by means other than email, otherwise anyone could

intercept the key and use it do decrypt the message;

1

1b Jill.s public key;

Jill.s private key;

1

1

1c the message data is hashed into a message digest; the message digest is

encrypted; with the sender.s private key;

Jill’s software decrypts the signature; using Jack ’s public key; contained in3

digital certificate sent with message; to verify Jack.s public key; decrypt digital

certificate using Certificate Authority.s (trusted third party.s) public key; Jill’s

software then hashes the document data into a message digest; If recalculated

message digest is the same as the original message digest (decryptedsignature); then Jill knows that the signed data has not been changed;Max 4

I decryption of message


Question

Number

Answer Marks

2a converting/transforming from plain text into ciphertext/secret code;

A bl d A i i / i / di

max 1



A scrambled; A transposition / conversion / coding

the sender processes the message prior to transmission so that if it is

accidentally or

deliberately intercepted while it is being transferred it will be incomprehensible

to

the intercepting party;

Data coded so that unauthorised users can.t read or access the data;

2b b) i) B’s public key;

ii) B’s private key;

1

1

2c i) a hashing function is applied to the text of the message;

the result/message digest is encrypted;

using B’s private key;

A the data generated is added to the end of the message;

A message/date stamp is used to produce digital signature;ii) A uses Certificate Authority’s public key;

to verify B’s public key;

digital signature is decrypted;

using B’s public key;

the hashing function is applied to the text of the message;

the result of the hashing function is compared with the digital signature;

if they are the same the message is authentic;

max 3

max 4


Question

number

Answer Marks

3a E-mail may pass through many computers/servers if it travels over a

network, each computer can make a copy/can be accessed;

When a message arrives at its destination, it waits until the intended

recipient picks it up. During this time the message is vulnerable to being

read or copied by the computer ’s operator;

Electronic eavesdropping of telephone wires and local area networks is

possible;

With e-mail alterations lave no trace (no physical damage) wheras with

paper alterations leave a physical mark;

Max 1

3b) i) E-mail encrypted using public key;

Recipient’s private key used to decrypt e-mail;

2

ii) E-mail encrypted by sender using private key;

Recipient decrypts e-mail using sender’s public key;

2

7.1




1 The Humber Bridge solved the problem for commuters of getting from Hull to Grimsby across

the Humber River, thus reducing the distance that they had to travel considerably.

2 Any three from:

a) The solution must cope with a volume of traffic crossing the bridge per hour up to …

b) The solution must cover a distance across water of …

c) The solution must be capable of supporting a total weight of …

d) The solution must connect to the existing road network in the area.

3 a) a ferry between the south and north shores and vice versa

b) a hovercraft between the south and north shores and vice versa

c) a bridge between the south and north shores and vice versa

4 because the requirements were not precise enough for the bridge designers to work with.5 problem definition

feasibility study

analysis

high-level design

low-level design

implementation and testing

maintenance6a) What is the proposed system to do?

What are the problems with the current way of doing things?

What data/information is recorded in the current system?

How much data is recorded at present?

What data/information is to be recorded in the proposed system?

How much data will the proposed system record?

How frequently will the data need to be updated?Will new records need to be added or old ones deleted?

How often? Will the changes come in batches, or in ones and twos?

How important is the data/information that is recorded?

What processes or functions are performed by the current system?

What processes or functions are to be performed by the new system?

When should these be done and where?

What special algorithms do these processes use, e.g. calculation of compound

interest? Which processes should be executed manually?

What are the inputs to the current system and what inputs will be required for the

proposed system? Ask to see any input documents that are used in the current

system.

What are the outputs from the current system and what outputs will be requiredfrom the proposed system? Establish if hard copy output is required.

H ft ill t t b i d? A k t t t f th t



How often will outputs be required? Ask to see some output from the current

system.

What computing resources does the end user possess?

Is the end user prepared to purchase software/hardware resources?

Is security an issue?

Should there be limited access to some or all parts of the proposed system?

How are exceptions and errors handled in the current system?

What errors and exceptions should be reported in the proposed system?

How should they be reported? Should anything else be done?

Are there any constraints on hardware, software, data, methods of working, cost,

time, and so on?

Does the user have a particular solution in mind?

Do you have some suggestions to make to the user?

b) renewal subscription letters

reminder letters

membership lists

member’s record

membership categories list and fees

payment records

c) Interviewing staff – facts can be gathered directly from the people who have

direct experience of the present system. Full and detailed answers can be obtained

by pursuing particular lines of questioning.

Examining existing paperwork, documentation, records and procedure manuals

for the current system – can be used to identify the data that is used in the current

system, the information that is produced by the current system and the procedures

that are carried out.

Using a questionnaire to survey opinions of staff enables the same set of questions

to be asked to many people. A carefully designed questionnaire can be a very quick

and cheap way to obtain specific answers to specific questions from a large number

of people.

Observation of operation of the current system enables current methods of

working to be examined and necessary exceptions to the normal pattern of

working to be noted.

7 A prototype is an early or trial working version of the proposed system developed to test possible solutions.

Prototyping is used to (any three):

clarify requirements

perform risk analysis

find a solution to a particular problem

check whether a solution can handle the workload

test a solution within the proposed environment.



Through a process of refinement to progress, in conjunction with users, to a final working

system. This process is called evolutionary, iterative development:

to discover errors in design

to discover problems.

8 Once one phase is complete, it is not revisited. The customer doesn’t get to see thedeveloping product. Therefore, it is more difficult to discover, and expensive to fix, errors

in a previous stage. For example, analysis can be incomplete or wrong in places because

there is little opportunity to evaluate the consequences of the analysis and also as the

project progresses, requirements may change. The customer doesn’t get to see the product

until it is ready for delivery. The communication path between the customer and the

developers is very long, that is, indirect, because developers talk to analysts who talk to

customers. This means that there is greater potential for the developer to misinterpret the

customer’s requirements. There is no opportunity for the customer to correct the

developer’s error. The waterfall model communicates by written documents forwardsthrough the cycle but no analyst can express the requirements that they know about 100 per

cent completely and correctly through writing. Customers express themselves in everyday

English whereas developers use technical language. The lack of interaction between

customer and developer in the waterfall model may result in a gulf appearing between the

developer and the customer which may result in imperfect communication and a product

that is far removed from what the customer wanted.

9 In this model, ‘water is allowed to flow up hill’, that is, the analysis, design and

implementation phases can be revisited, which is not the case with the waterfall model.10 The spiral model is more appropriate because you are not yet a professional software

developer and you will be working on your own.

11 Volumetrics refers to measuring/assessing the volume of data that a system will be required

to process and store. Analysis needs to provide developers with volumetric information in

order for the developers to make appropriate design decisions.

12 It is important that the set of objectives is agreed with the end user(s)/customer. However, it

is very easy for the analyst and the end user(s) to overlook something important that should

be included. A list is not always the best way of visualising what it is that needs to be done.Analysis must also try to represent what needs to be done pictorially. One way that this can

be done is using a data flow diagram.

Context diagram

Level 1 DFD

Level 2 DFD

13 Business algorithms are algorithms supplied by the end user. These are algorithms used bythe existing system or would be used if a system existed. An example is an algorithm for

calculating interest on an investment, for example, if a customer invests £100 how much

interest would be earned in the first year of the investment and in subsequent years?

14 At least 10:

Bubble sort

Insertion sort

Shell sortMerge sort

H t



Heap sort

Quick sort

Bucket sort

Radix sort

Distribution sort

Shuffle sort15 a) specification

b) design

c) specification

d) design

e) specification

16 association, aggregation (composition/containment), inheritance.

Inheritance class diagram shows the relationship between two object types inwhich one object type is a kind of the other object type, for example, a car is a

kind of vehicle.

Aggregation or composition or containment class diagram shows the relationship

between two object types in which one object type has a component which is of

the other object type, for example, a car has an engine or contains an engine.

Association class diagram shows the relationship when it is not inheritance and itis not aggregation, for example, a snake uses the ground to get from A to B. The

relationship is just an association because a snake is not a kind of ground and asnake does not contain ground. It is, in fact, a uses relationship.

17 Analyst doesn’t fully understand the problem to be solved and therefore does not provide a

clear statement of the problem to be solved. Consequently, a different problem is solved.

Analyst fails to obtain a complete picture of the customer’s needs/requirements.

Analyst fails to specify in the clearest terms the customer’s requirements to the

customer so customer agrees to the development of a solution that is not exactly

what they want.

Analyst fails to communicate to the developer precisely what the system must do

and therefore the developer develops a solution which doesn’t match the

customer’s expectations.


Question

number

Answer Marks

1 Problem definition, feasibility study, analysis, design,

implementation, testing, maintenance.

3

2a Any three from: Interview, survey, observation, examination

of the paperwork.

3



p p

2b 1. Interviewing

Interviewing is useful because facts can be

gathered directly from the people who have

direct experience of the present system. Full

and detailed answers can be obtained by

pursuing particular lines of questioning

2. Examination of existing paperwork

Examination of the existing paper work,

documentation, records and procedure

manuals can be used to identify the data that

is used in the current system, the information

that is produced by the current system and

the procedures that are carried out.3. Survey

Questionnaires enable the same set of questions to

be asked to many people. A carefully designed

questionnaire can be a very quick and cheap way to

obtain specific answers to specific questions from a

large number of people.

4. Observation

Observation of the current practice enables current

methods of working to be examined and necessary

exceptions to the normal pattern of working to be noted.

3

3a Specification 1

3b Design 1

3c Design 1

3d Specification 1

7.2


1



2

7.3

Exam pract ice quest ionsQuestion 1 – June 2008 CPT5 Q1eStartDateTime EndDateTime Mileage OverdueHours Normal Erroneous Boundary

01/12/07 06:00 01/12/07

15:30

15 2 1

16/12/07 18:00 12/12/07

09:00

237 3

04/12/07 23:00 04/12/07

08:30

5 2 1



03/12/07 08:00 03/12/07

09:00

0 0 1

01/12/07 06:00 01/12/07

15:30

0 1.5 1 I

01/12/07 06:00 01/12/0715:30

0 -2 1 I

04/12/07 08:30 05/12/07

23:00

57 0 1 A

01/12/07 06:00 01/12/07

15:30

15 3

Question 2 –

January 2008 CPT5 Q3Question

number

Answer Marks

2a normal data: accept valid dates within 120 years before present (i.e.

23/01/2008)

birthday sometime before today’s date;

birthday sometime after today’s date;

reason: to check the routine takes into account whether birthday has already

been or not;A 29/2 as a special day to test for

A day/month instead of birthday

The data values should be any dates within 120 years before the present. If the

reason is just given as ‘they are normal dates’ when both dates are before or

after 23/1 of any year then 2 marks max.

3

2b boundary data:

a birthday just before boundary;a birthday just after boundary;

4

a birthday exactly on boundary;

A ‘yesterday’, ‘today’ ‘tomorrow’

reason: to check that age is calculated exactly, taking into account whether

birthday is past, now or future;

// this is the oldest you can be // this is the youngest you can be;

22/1 – 24/1 Any year within 1887 – 2008

2c R anything that is not valid date format;

erroneous data: a date after today’s date; reason: can’t have an age for someone not born yet;

or

erroneous data: a date which makes a person over 120 years old;(i.e. Before 24/1/1887)

A ‘tomorrow’;

2



reason: no person expected to be over this age;

A outside any expected values // outside range;

Remember that wrong formats or invalid dates are not acceptable answers.

Give mark for correct reason even if no acceptable date value given.


Question

number

Answer Marks

3a beta testing : a test prior to commercial release

// testing that involves sending the product to beta test sites outside the

company for real-world exposure;

must convey external testing

A the last stage of testing;

1

8



R error messages3b factors to help maintenance: 1 mark per factor (with some relevant

explanation) to max 3

structured program code // no GOTO statements // using iteration/selection

constructs appropriately;

local variables; procedures/functions; with interfaces/parameters;

modules/units;

layout/indentation/white space;

meaningful identifier names; self-documenting code //

comments/annotation;

object oriented programming // use of classes;

use of pre-tested routines/library routines; error logs;

3

3c types of testing. Any 2 of:

system/integration testing;

functional/black box testing;

structural/white box testing;

acceptance testing;

unit/module/subsystem testing;

Alpha testing;

A Dry running/walk-through/inspection testing; R bottom-up/top-down

t ti

2



testing.


Question

number

Answer Marks

4 any normal/typical/valid data;: 1<=Hours ,=35; A 0,= Hours ,= 48

boundary/extreme values;: eg Hours = 35, 36 & Hours = 0, 48, 49; A -1

test that it correctly calculates premium rate;: Hours >35;

any erroneous/invalid data; any negative value; any non-integer data; any

value over 48;

1 mark for test data, 1 mark for justification

If justification does not match test data, then 1 mark max.

6


Question

number

Answer Marks

5 1 mark for test data, 1 mark for justification for 3 sets of test data

£1, 50p, 20p, 10p, 5p; simple change of a single coin;

£3.85, 15p, 25p, 30p, 35p, 60p, 65p, 70p, 75p, 80p, 85p; change made up of

one of several coins;

40p, 45p, 90p, 95p; change made up of more than one of the same coins:

2x20p;

0p; boundary data: no change / zero coins;5p; minimum change that can be given;

6

£1.95; maximum change that can be given // extreme/boundary value;

a negative amount; although the routine that calculates how much change is

due should not allow erroneous change, this routine should still test for

erroneous input;3p; an amount that is not a multiple of 5p // erroneous data;

not all values need to be listed

R justifications not referring to scenario

R answers which seem to test the coins inserted f the calculation of change


QuestionNumber

Answer Marks

6 Any two points at 1 mark each: 2



Any two points at 1 mark each:

Bugs/Errors/Mistakes in software/system/code/program/it;

Problem NE R data errors (T.O.)

Requirements change // adding new tasks;

Parameters change e.g. VAR rate, No of users adjusted, No of licences change;Performance needs tuning // buffer size needs adjusting // indexing needs to beswitched off

Or on // indexes need to be rebuilt;

Hardware is changed;

Software / system is updated // upgrades;

Adaptive/Corrective/Perfective maintenance not enough without explanation

“Keeping up to date” NE

2


Question

Number

Answer Marks

7 Top-down testing;

Bottom-up testing;

Black Box testing;White Box testing;

Dry run/walk-through;

Unit/Module testing; A Prototyping;

R Integration/Acceptance/Alpha/Beta/System/Performance/Compatibility

testing;

R anything clearly late in the development cycle;

Mark first 3 responses only

3

BUT

beware of expansion on same line.


Question

number

Answer Marks

8 Any two at 1 mark each:

Top-down (testing);

Bottom-up (testing);

White Box (testing);

Integration/Integrated (testing);Interface/Stub (testing);

Acceptance (testing);

Alpha (testing);

Max 2



Alpha (testing);

Beta (testing);

Walkthrough/Dry run/Tracing/Desk checking:

Compatibility (testing):

Performance (testing):

System (testing);

Unit/Module (testing);

R. Functional (testing)

R. Structural (testing)

R. Test cases and examples of test cases


Question

number

Answer Marks

9a N.B. Must be an automated system (=control) or a system that provides real

time data (=monitoring)

Flight control software;

Software controlling life support systems;

Software controlling hazardous materials with potential for exposure to

humans;

Software controlling mechanical equipment which could cause death

through impact/crushing/cutting;Any software which provides information to operators where an inaccuracy

or misinterpretation of the data could result in death/injury through an

incorrect decision;

A Air traffic control, railway signalling system, traffic lights, heart rate

monitor, drip feed controller for administering drugs.

Any two at

one mark

each

2

9b) i) Acceptance testing is specified/performed by customer (against originalspecifications);

1

9b) ii) Poorly/incorrectly specified system//inadequate/inaccurate systems analysis;Poor training of staff using system//staff use system incorrectly;

1

Situation outside specification occurs or example which relates explicitly to

specification or similar, e.g. more users attempt to log on than should;

A Virus has entered system//malicious misuse

R Design flaws, hardware failure, inadequate testing

R Data corrupted

9b) iii) N.B. Name required

System testing;

Alpha testing

Beta testing;

Performance testing;

A Black Box testing, White Box testing, integration testing, unit/module

testing, top-down testing, interface testingR Phased testing, bottom-up testing

Any one for

one mark

1




Question 10 January 2004 CPT5 Q8

Question

number

Answer Marks

10a) i) Strategy: White Box testing;

Justification: Need to follow a path through the code (to discover the

error);

2

10a) ii) ii) While r >= y 1 mark for identifying(in some way) While statement,

1 mark for correct condition r >= y

R. q := 0 changed to q := 1

2

10b r never changes or an alternative that describes/implies the same thing;

A. gets stuck in an infinite loop1

10c If y > 0 Then x = y * (result q) + (remainder r);

R. y cannot equal zeroA. y must always be greater than zero

1

10d) Acceptance is specified/performed by “customer” ;

against original specifications

1

7.4


Question 1 – January 2006 CPT5 Q5 c) and d)

Question

number

Answer Marks

1ae) Method

1 mark

Justification

1 mark

Parallel To check against the pupil’s food diary to ensure the data

2

collected is correct;

A if new system fails, they have the old one to fall back

on;

Direct It would take too long to check against pu pil’s food diaries

// not a critical system // cheaper than parallel // cheaper

than running two systems;

Pilot The new system may not work very well and should be

piloted in one school first (before being rolled out to all

schools in the LEA);

Phased Use the data collection module before the diet checking

one, in case the system refuses to allow acceptable

choices, causing chaos at the checkouts;

R explanation of pilot1b data files will have to be entered into the system; R

conversion/transfer of data3



users / cafeteria staff / pupils will have to be trained; R printing

labels

new hardware has to be installed/purchased; new software has to be

installed;

I licences

smart cards will have to be updated; pupils will have to be issuedwith smart cards;

define/create new operating procedures

A produce documentation/user guide/system guide


Question

number

Answer Marks

2a i. Old system and new system operate alongside each other/in parallel

until new system proved; (require time limited trial)

ii. Parts of old system gradually replaced in stages by new system;

Do not give mark for a pilot

1

1

2b Information/Data/Files may have to beconverted/copied/reformatted/modified so that compatible with new

system;

Users/operators will have to be trained so that they can use the new

system//Staff needed/hired to maintain new system;

Old data archived;

Make full backup before changing to new system;

2

Hardware replaced/upgraded;

System software replaced/upgraded;

2c NB Emphasis is on changes not performance

Is it possible/How easy is it/How long will it take to correct an error in

software;

Is it possible/How easy is it/How long will it take to change parameters in

system, e.g. VAT rate;

Is it possible/How easy is it/How long will it take to change system to cope

with more users;Is it possible/How easy it is/How long will it take to change system to cope

with more terminals/workstations;

Is it possible/How easy is it/How long will it take to change system to cater

3



Is it possible/How easy is it/How long will it take to change system to cater

for more software licences;

Is it possible/How easy is it/How long will it take to change system to work

with different hardware;

Is it possible/How easy is it/How long will it take to update/upgrade

system;

Ho extensive is support documentation;

What is skill level of support staff;

What is availability of support staff;

Can operators/users configure system/change settings;

How long will support be available for;

Is source code available;

A One reference only to documentation;

R How easy is it to add new features/expand system;

R What is standing of developers;

7.6


1 Evaluation is systematic assessment of whether something meets its objectives or

specifications and how well it meets the latter in terms of effectiveness, useability,

maintainability.

2 Effectiveness in meeting requirements: learnability, useability, maintainability.

3 In addition to requiring criteria for judging effectiveness evaluation of effectiveness requires:

a) planning

b) planning time

c) execution time.

4 Monitor number of registered users; monitor number of stored files; monitor frequency of file

uploads/downloads; monitor frequency of use by registered users.5 Observe users – class of 7-year-olds – using the system: measure improvement in word

spelling as software is used; note how quickly 7-year-olds learn to use the system; note how

easily the 7-year-olds navigate the software; note how long 7-year-olds take to step through

the instructions and to answer.

6 Useability refers to the ease with which a User Interface can be used by its intended audience

to achieve defined goals.

7 Target acquisition time

LatencyReadability

Use of metaphors

Navigability



Target acquisition time: Latency, e.g. moving the mouse pointer to a position over a button.

a) Latency: How long does a user who has initiated an action have to wait before the action is

completed, e.g. a button is clicked to execute an SQL statement that returns a dataset

matching the query?

b) Readability: How easy is it for a user to read commands/guidance text?

c) Use of metaphors: Visual pictures/icons to enable users to intuitively grasp what can be done.

d) Navigability: How easy is it to navigate the user interface?

8 Maintainability of software refers to how easy or difficult it is to fix bugs, change parameters

and respond to changing requirements.

9 Any four from:

a) Is the software modular?

o Procedures/Functions are used that perform a single task and are self-contained.

o Collections of related procedures are placed in independent units that can be compiled

separately.

o Information hiding is supported by the use of separate units.

b) Is the code self-documenting?

o Are meaningful identifier names used for variable names, procedure/function names, etc?

o Is indentation used to indicate the structure of the code.c) Blocks of code have one entry point and one exit point.

d) Is the use of global variables across units minimised?

e) Is good use made of local variables?

f) Are parameter values passed in and out of procedures through procedure interfaces?

g) Is an object-oriented programming approach used?

o Is inheritance used to allow functionality to be extended?

o Is polymorphism used to offer different functionality depending on object type?

h) Is the database created from a DDL script (which itself has been produced using a CASE

tool) so that re-engineering the database can be done using the CASE tool?

i) Does querying of the database use SQL?

j) Is the code commented where clarification is needed?k) The documentation that accompanies the software is also essential to the maintenance of this

software. Evaluating maintainability of software requires consideration of the following:

l) Is the project supported by analysis and design data dictionaries?

m) Are entities and entity-relationships documented?

n) Is the software accompanied by:

o dataflow diagrams

o system flow charts

o other relevant diagrams?

o) Are algorithms documented?



10 Evaluation should take place early in the project when prototyping for the purpose of

clarifying the customer’s requirements. Evaluation should take place when designing so that

the effectiveness of different approaches can be made before committing to one particular

design. This should include types of interface. When a particular user interface type has been

settled on, evaluation of different mock-ups of this user interface should be made. This is still part of the design stage. Evaluation should take place to assess useability and navigability of

the user interface for a range of users. This takes place during the later stages of

implementation when the design is fairly stable. When the final solution is produced further

effectiveness evaluation is necessary to confirm that the design has achieved what it set out to

achieve.

Nelson Thornes is responsible for the solution(s) given and they may not constitute the

only possible solution(s).

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a2 Student Book Answers