question 1. · page 2 mark scheme syllabus paper cambridge international a level october/november...

Mark Scheme Syllabus Paper

Cambridge International A Level – October/November 2015 9608 41

© Cambridge International Examinations 2015

1 (a) (i)

A F

5

B C E G

5 15 5 53

[max. 7] (ii) 1 – 2 – 3 – 5 – 6 – 7 – 9 – 8 – 10 1–5 scores 1 6–10 scores 1 [2] (iii) 43 weeks [1] (b) (i) week number 25 [1] (ii) week number 32 [1] (c) To see what activities can be done in parallel // show dependencies To record changes to project timings [max. 1]

QUESTION 1.




1 (a) (i) Activity

A

B

C

D

E

F

G

H

J

K

L

M

N

Week Number 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

1 mark for each square [2] (ii) week number 18 Allow follow through [1]

QUESTION 2.




(b) (i)

Activity Description Weeks to complete

A Write requirement specification 1

B Produce program design 1

C Write module code 7

D Module testing 2

E Integration testing 2

F Alpha testing 2

G Install software and carry out acceptance testing 2

H Research and order hardware 1

J Install delivered hardware 3

K Write technical documentation 4

L Write user training guide 2

M Train users on installed hardware and software 1

N Sign off final system 1

Activity

A

B

C

D

E

F

G

H

J

K

L

M

N

Week Number

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

1 mark per activity (but 1 mark for activity M and N) Notes: C must be after E (1 or 2 later is ok) D, E, F correct relative to C J must start in week 20 (allow 21, 22) G must come after the end of J (f.t.) K finishes after or at same time as F L finishes at the same time as G and after the end of J (or 1-2 weeks later) M starts when everything else has finished. N after or at same time as M [9]

(ii) week number: 26 Allow f.t. [1]




2 (a) parent(ali, ahmed). parent(meena, ahmed).

Accept statements in either order Wrong capitalisation minus 1 mark [2] (b) P =

ahmed

aisha

Ignore capitalisation Deduct 1 mark for every extra result [2]

(c) mother(M, gina).

Accept parent(M, gina) AND female(M). Accept a comma instead of AND

Reject mother(M, gina) IF female(M) AND parent(M, gina).

Ignore capitalisation [1] (d) father(F, C) IF

male(F) AND parent(F, C).

(1) (1) [2] (e) brother(X, Y) IF

male(X)AND [1]

parent(A, X) AND [1]

parent(A, Y) [1]

AND NOT X=Y. [1]

Accept any variable for A, but it must be the same in both places Accept father/mother instead of parent Ignore capitalisation




3 (a)

Student

StudentName : STRING DateOfBirth : DATETIME ……………………………………………………………………………………… ShowStudentName() ShowDateOfBirth() ………………………………………………………………………………………

FullTimeStudent PartTimeStudent

Address: STRING TelephoneNumber : STRING ………………………………………………………………………………………

NumberOfCourses : INTEGER Fee : Currency FeePaid : BOOLEAN

Constructor() ShowAddress() ShowTelephoneNumber() ………………………………………………………………………………………

Constructor()ShowNumberOFCourses() ShowFee() ShowFeePaid()

Mark as follows: Base class: – dateOfBirth declaration and associated method in Student – constructor Subclasses: – telephoneNumber declaration and associated method in FullTimeStudent – NumberOFCourses declaration and associated method in PartTimeStudent – fee declaration and associated method in PartTimeStudent – feepaid declaration and associated method in PartTimeStudent – constructor method in PartTimeStudent – inheritance arrows Ignore data types, ignore other methods/attributes Ignore brackets after methods [Max 7]




(b) (i) Mark as follows (parts to be ignored in grey): If no programming language stated, map to 1 of the 3 below (or check in Q1ai) Class header & ending (watch out these may be squashed into the next clip) Ignore methods 2 attributes with correct data types No mark if subclass properties shown here Attributes required: StudentName DateOfBirth (accept variations e.g. DoB) Pascal TYPE Student = CLASS PUBLIC Procedure ShowStudentName(); Procedure ShowDateOfBirth(); PRIVATE StudentName : STRING; DateOfBirth : TDateTime; // accept string reject Date END;

Python class Student : def __int__(self) : self.__StudentName = "" self.__DateOfBirth = "" # date(1,1,2015) def ShowStudentName() : pass def ShowDateOfBirth() : pass

Ignore __ before attributes

VB.NET Class Student Public Sub ShowStudentName() End Sub Public Sub ShowDateOfBirth() End Sub Private StudentName As String Private DateOfBirth As Date ‘ accept string End Class

(Ignore: must inherit) Ignore Private/protected/public Don’t give a mark if using DIM [2]




(ii) Mark as follows: – Class header and showing superclass

– Properties (Do not award this mark if properties from base class included here) Data types must be correct – Methods (Do not award this mark if methods from base class included here) must show heading and ending of procedure/function declaration

Ignore PUBLIC, PRIVATE Pascal TYPE FullTimeStudent = CLASS (Student) PUBLIC Procedure ShowAddress(); Procedure ShowTelephoneNumber(); PRIVATE Address : STRING; TelephoneNumber : STRING; // reject integer END;

Python class FullTimeStudent(Student) : def __init__(self) : self.__Address = "" self.__TelephoneNumber = "" def ShowAddress() : pass def ShowTelephoneNumber() : pass

VB.NET Class FullTimeStudent : Inherits Student Public Sub ShowAddress() End Sub Public Sub ShowTelephoneNumber() End Sub Private Address As String Private TelephoneNumber As String ‘ reject integer End Class

No mark if using DIM [3]




(iii) 1 mark per statement to max 3 Missing string delimiters: penalise once Accept use of constructor

Pascal NewStudent := FullTimeStudent.Create; NewStudent.StudentName := 'A.Nyone'; NewStudent.DateOfBirth := EncodeDate(1990, 11,12);//:= ‘11/12/1990’ NewStudent.TelephoneNumber := '099111';

Alternative NewStudent := FullTimeStudent.Create(‘A.Nyone’, ‘12/11/1990’, ‘099111’);

Python NewStudent = FullTimeStudent() NewStudent.StudentName = "A.Nyone" NewStudent.DateOfBirth = "12/11/1990" NewStudent.TelephoneNumber = "099111"

Alternative NewStudent = FullTimeStudent(‘A.Nyone’, ‘12/11/1990’, ‘099111’)

VB.NET Dim NewStudent As FullTimeStudent = New FullTimeStudent() NewStudent.StudentName = "A.Nyone" NewStudent.DateOfBirth = #11/12/1990# NewStudent.TelephoneNumber = "099111"

Alternative Dim NewStudent As FullTimeStudent = New FullTimeStudent(“A.Nyone”, “12/11/1990”, “099111”)

[Max 3]




4 (a) FUNCTION Hash(Key : STRING) RETURNS INTEGER DECLARE Number : INTEGER

Number � ASCII(LEFTSTRING(Key,1))

// Number � ASCII(Key[1])

Number � Number – 64

RETURN Number

// Result � Number // Hash � Number ENDFUNCTION

Accept ASC instead of ASCII

Accept LEFT instead of LEFTSTRING

Key can be a different identifier but must be the same in both places [5] (b) (i)

Dictionary

Index Key Value

1

2

3 Computer Rechner

4 Disk Platte

5 Error Fehler

6 File Datei

7

8

: :

: :

1999

2000

Ignore spelling mistakes 1 mark for 2 correct pairs entered in correct slots [2]

(ii) Collision / synonym / space already occupied / same index in array

Overwrites previous key-value pair reject error [Max 2] (iii) Create an overflow area

The 'home' record has a pointer to others with the same key // linked list

OR Store the overflow record at the next available address … in sequence (= next available) OR Re-design the hash function …. // write a different/another algorithm to generate a wider range of indexes // enlarging storage space // to create fewer collisions [2]




(iv) Mark as follows:

Check whether slot is empty: IF Dictionary[Index,1] ”” // != ‘’ // > NULL // >

NONE

If not: update index: THEN Index �

…to find an empty slot (loop / follow pointer / go to overflow area) reject FOR loop Insert code between lines 20 and 30

21 WHILE Dictionary[Index,1] > ""

22 Index � Index + 1 23 IF Index > 2000 24 THEN

25 Index � 1 26 ENDIF

27 ENDWHILE [4]

5 (a) (i)

Memory Address

Accumulator 509 510 511 512

0 7 3 0 0

7 7

0

1 1

7

14 14

1

2 2

14

21 21

2

3 3

3 marks 1 mark 1 mark

If values changed in column 509 or 510 don’t give marks for 511/512 [5] (ii) stores the counter value for ….// acts as a control variable/counter How many times the loop has been performed // control the loop Ignore re-stating the steps [2] (b) LDM #12 (must be instruction before storage) STO 509 (must be final instruction) 1 mark for each instruction [2]




6 (a) 1 mark for structure header/ending 1 mark for each field correct, take away 1 mark for additional fields Python answers will use a class

Pascal TYPE StockItem = RECORD ProductCode : String; // accept integer Price : Currency; // accept real NumberInStock : Integer; END;

Python class StockItem : def __init__(self) : self.ProductCode = "" # = 0 self.Price = 0.0 # = 0 self.NumberInStock = 0

VB.NET STRUCTURE StockItem Dim ProductCode As String ‘ accept integer Dim Price As Decimal ‘ Double/single Dim NumberInStock As Integer END STRUCTURE

VB6 Type StockItem ProductCode As String ‘ accept integer Price As Currency ‘ Double/single NumberInStock As Integer

END Type [4]

9608/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED

October/November 2017

© UCLES 2017 Page 7 of 15

Question Answer Marks

5(a)(i) PERT / GANTT 1

5(a)(ii) 1 mark per bullet to max 3 For example:

• Calculate total minimum time required for project • Identify milestones • Task dependencies • Provides the critical path analysis • Identify which tasks need to be prioritised • Determine when to begin specific tasks/stages • Identify slack time • Identify when resources need allocating • Identify tasks that can be completed in parallel

3

5(b)(i) Integration 1

5(b)(ii) Beta / acceptance 1


6(a) 1 mark per bullet to max 6 • Declaring a class with the name animal • Declaring variables for across, down and score (all Integers) • as private/protected • Correct constructor header and ending • Randomly generating an across between 0–39 inc. in constructor • Randomly generating a down between 0–39 inc. in constructor • Initialising Score to zero in constructor • Correct get for Across • Correct set for Across

6

QUESTION 3.





5(c)(ii) 1 mark per bullet point to max 2: Example: • If there is any delay to a task which is part of the critical path • then the overall project will be delayed • Gives the earliest possible completion time • this allows you to organise/allocate resources (efficiently) • Frequently recalculate the critical path • to see if there are any delays/new critical path has arisen • Can identify where there is slack in activities • so they can start later without affecting the critical path

2


6(a)(i) 1 mark per bullet point: • Declaring a record type // class etc. • Declaring Country as a string • Declaring Pointer as an integer Example: TYPE ListElement DECLARE Country : STRING DECLARE Pointer : INTEGER ENDTYPE

3

QUESTION 4.





1(a)(i) 1 mark for each correct Activity and time • B 3 • C 10 (following B) • D 7 (following B) • E 3 (following C and D) • G 2 (following F) • H 2 (following F)

6

1(a)(ii) The shortest time to complete the project // the sequence of activities that must be completed to avoid delaying the project 1

1(a)(iii) 1 mark for identify, max 1 for description • GANTT • A table that has time across the top and activities on the left, boxes are coloured to show dependencies and find

critical path // colour in the boxes to show the length of time for each task

2

1 2 3 5

8

6

9

4

7

A

1

F 1

B

3

C

10

A

1

D 7

H 2

G2

QUESTION 5.





1(b)(i) A, B, C and D in correct places with no alteration to start and end pointer

A B D C

1

1(b)(ii) 1 mark per bullet point

• correct jobs in correct order • correct location of start pointer • correction location of new end pointer

F G H D C E

3

1(b)(iii) 1 mark from: • An error message would be generated

1

Start Pointer End Pointer

Start Pointer End Pointer





1(b)(iv) 1 mark for each correct line FUNCTION Remove RETURNS STRING DECLARE PrintJob : STRING IF StartPointer = EndPointer THEN RETURN "Empty" ELSE PrintJob ← Queue[StartPointer] IF StartPointer = 5 THEN StartPointer ← 0 ELSE StartPointer ← StartPointer + 1 ENDIF RETURN PrintJob ENDIF ENDFUNCTION

4

1(b)(v) 1 mark per bullet point • A stack is Last In First Out (LIFO) while a queue is First In First Out (FIFO) • The queue removes and returns the element at start pointer // item is removed from the start/head // • A stack would remove and return the element at end pointer // item is removed from the end

2





1(c) 1 mark for each correct transition with arrow

5

Printer idle

Printer active

Job unsuccessful

Print job received

Print job successful

Print job sent

Print job in progress

Timeout

Error message

Print job in progress

Check print queue

Print job added to queue

Start





1(d)(i) 1 mark per bullet • Way of modelling logic • Show all possible outputs // shows every possible outcome // all outcomes • based on the inputs • Determine which action to take in specific conditions // how different conditions affect the actions/outcomes

2

1(d)(ii) 1 mark for each row Accept Y/X/ticks as long as clear which are Y. Accept N/X/– for empty spaces

Rules

Conditions

Document printed, but quality is poor Y Y Y Y N N N N

Error light is flashing on printer Y Y N N Y Y N N

Document printed, but paper size is incorrect Y N Y N Y N Y N

Actions

Check connection from computer to printer X

Check ink status X X X X

Check if there is a paper jam X

Check paper size selected X X X X

4





1(d)(iii) 1 mark each for each correct column Accept –/X for empty spaces. Accept Y/X/Ticks as long as clear which are used

Rules

Conditions

Document printed, but quality is poor Y Y N N N

Error light is flashing on printer Y N

Document printed, but paper size is incorrect Y N Y N N

Actions

Check connection from computer to printer X

Check ink status X X

Check if there is a paper jam X

Check paper size selected X X

5





1(e)(i) 1 mark per bullet point to max 4 • Method header and close (where necessary) with three parameters • Initialised PrintID, FirstName, LastName and Credits • to the parameters • Initialised Credits to 50 PYTHON def__init__(self, NewFN, NewLN, NewPrintID): self.__PrintID = NewPrintID self.__FirstName = NewFN self.__LastName = NewLN self.__Credits = 50 PASCAL Constructor NewPrintAccount.Create(NewFN, NewLN, NewPrintID); begin PrintID := NewPrintID; FirstName = NewFN; LastName = NewLN; Credits := 50; end; VB Public Sub New(NewFN, NewLN, NewPrintID As String) PrintID = NewPrintID FirstName = NewFN LastName = NewLN Credits = 50 End Sub

4





1(e)(ii) 1 mark per bullet point • method/procedure header (and close where appropriate) taking a parameter • FirstName is set to parameter PYTHON def __SetFirstName(self, NewFirstName): self.__FirstName = NewFirstName PASCAL procedure SetFirstName(newFirstName : String); begin FirstName := newFirstName; end; VB public sub SetFirstName(NewFirstName As String) FirstName = NewFirstName End Sub

2





1(e)(iii) 1 mark per bullet point • concatenates FirstName, space and LastName • function/method header without parameter and returns (generated) value PYTHON def __GetName(self): return(self.__FirstName + " " + self.__LastName) PASCAL function GetName(); begin result := FirstName + " " + LastName end; VB public function GetName() As String return(FirstName & " " & LastName) End Function

2





1(e)(iv) 1 mark per each correct bullet point from: • Procedure/method header and close (where necessary) passing MoneyInput • At least 3 constants (e.g. freecredit10, freecredit20, twenty, 10, creditperdollar) • If MoneyInput9 and MoneyInput 19 then calculate MoneyInput * 25 + 50 • All three correct calculations add to Credits, not overwrite • Efficient IF (i.e. elseif) PYTHON def __AddCredits(self, MoneyInput): CreditPerDollar = 25 FreeCredit10 = 25 FreeCredit20 = 50 Twenty = 20 Ten = 10 if MoneyInput >= Twenty: Credits = Credits + (MoneyInput * CreditPerDollar) + FreeCredit20 elif MoneyInput >= Ten: Credits = Credits + (MoneyInput * CreditPerDollar) + FreeCredit10 else: Credits = Credits + (MoneyInput * CreditPerDollar)

6





1(e)(iv) PASCAL procedure AddCredits(MoneyInput : Real); const CreditPerDollar = 25 const FreeCredit10 = 25 const FreeCredit20 = 50 const Twenty = 20 const Ten = 10 begin If MoneyInput > = Twenty Then Credits := Credits + (MoneyInput * CreditPerDollar) + FreeCredit20; Else If MoneyInput > = Ten Then Credits := Credits + (MoneyInput * CreditPerDollar) + FreeCredit10; Else Credits := Credits + (MoneyInput * CreditPerDollar); end; VB.NET Public Sub AddCredits(MoneyInput As Integer) Const CreditPerDollar As Integer = 25 Const FreeCredit10 As Integer = 25 Const FreeCredit20 AS integer = 50 Const Twenty As Integer = 20 Const Ten AS Integer = 10 If MoneyInput > = Twenty Then Credits = Credits + (MoneyInput * CreditPerDollar) + FreeCredit20 Else If MoneyInput > = Ten Then Credits = Credits + (MoneyInput * CreditPerDollar) + FreeCredit10 Else Credits = Credits + (MoneyInput * CreditPerDollar) End If End Sub





1(e)(v) 1 mark per bullet • Declaring StudentAccounts as array of 1000 elements • of type PrintAccount DECLARE StudentAccounts ARRAY[0:999] OF PrintAccount

2





1(e)(vi) 1 mark per bullet point to max 8 • Generating ID with ‘1’ at the end all in lowercase from parameters • Loop through array to last occupied element • Check if the PrintID already exists • using GetPrintID() • increment number at end of PrintID • Create a new instance of PrintAccount • sending FirstName, LastName, PrintID as parameters • adding new account to StudentAccounts at position NumberStudents • Increment NumberStudents VB.NET Sub CreateId(firstName, lastName) Dim count As Integer Dim PrintID = Left(firstname, 3).ToLower & Left(lastname, 3).ToLower & “1” Dim studentAdd As Integer = 0 If numberStudents 0 Then For x = 0 To numberStudents - 1 If studentAccounts(x).getPrintID() = username Then PrintID = PrintID + 1 username = Left(firstname, 3).ToLower & Left(lastname, 3).ToLower & PrintID.ToString End If Next studentAdd = numberStudents End If studentAccounts(studentAdd) = New printAccount(firstname, lastname, username) numberStudents = numberStudents + 1

8





1(e)(vi) Python def CreateID(firstname, lastname): count = 0 PrintID = firstname[0:3].lower() + lastname[-3].lower() + "1" StudentAdd = 0 if numberStudents != 0: for x in range(0, numberStudents): if studentAccounts[x].getPrintID() == username: PrintID = PrintID + 1 username = firstname[0:3].lower() + lastname[0:3].lower + str(PrintID) studentAdd = numberStudents studentAccounts[studentAdd] = printAccount(firstname, lastname, username) numberStudents = numberStudents + 1 Pascal procedure CreateID(firstname : String, lastname: String); var count : Integer; studentAdd : Integer; PrintID : String; begin studentAdd := 0; PrintID := LowerCase(substr(firstname,0,3)) + LowerCase(substr(lastname,0,3))+ "1"; if numberStudents 0: ror x := 0 To numberStudents - 1; if studentAccounts[x].getPrintID() = username: PrintID := PrintID + 1; username := LowerCase(substr(firstname, 3) + LowerCase(substr(lastname,0,3))+str(PrintID); studentAdd := numberStudents; studentAccounts[studentAdd] := printAccount.Create(firstname, lastname, username); numberStudents := numberStudents + 1





1(a)(i) 1 mark each bullet point: • B 1 • C 3 (following B) • D 10 (following C) • G 3 (following D) and nothing on dummy • E 7 (following C) and nothing on dummy • H1 in position • J 2 (following H) and nothing on dummy

7

1(a)(ii) 1 mark: • The next activity is dependent on the previous but there is no activity

1

1 2A

13 4

5

8

7

6

9

11 10

F

2

2

B

1

C I

3

D

1

G3

E

I

7

H1

J2

I

6

QUESTION 6.





1(b) 1 mark per row

Rules

Conditions

Public Holiday Y Y Y Y N N N N

Hours >= 160 Y Y N N Y Y N N

Pension Y N Y N Y N Y N

Actions

3% bonus payment X X X X

5% bonus payment X X X X

4% Pension payment X X X X

3





1(c)(i)

Employee

EmployeeID : STRING Name : STRING Address : STRING DateOfBirth : DATE

Constructor () GetEmployeeID() GetName() GetAddress() GetDateOfBirth() SetEmployeeID() SetName() SetAddress() SetDateOfBirth()

SalaryEmployee ApprenticeshipEmployee

MonthlyPayment : CURRENCYHoursThisMonth : REAL PublicHoliday : BOOLEAN Pension : BOOLEAN

HourlyRate : CURRENCY/REAL HoursThisWeek : REAL/INTEGER

Constructor() GetMonthlyPayment() GetPension() GetPublicHoliday() GetHoursThisMonth() SetMonthlyPayment() SetPension() SetPublicHoliday() SetHoursThisMonth()

Constructor() SetHoursThisWeek() GetHourlyRate() GetHoursThisWeek() SetHourlyRate()

3 1 mark per bullet point • inheritance (arrow filled or unfilled) • constructor and SetHoursThisWeek method for

ApprenticeshipEmployee • HourlyRate and HoursThisWeek attributes for

ApprenticeshipEmployee





1(c)(ii) 1 mark per bullet point • Constructor header and close (where necessary) • All 4 values sent as parameters (ID, Name, Address, DateOfBirth) with any • Attributes/properties set to a value • that are the parameters Example code: Pascal Constructor Employee.init(ID, NewName, NewAddress, NewDateOfBirth : String); begin EmployeeID := ID; Name := NewName; Address := NewAddress; DateOfBirth := NewDateOfBirth; end; Python def __init__(self, ID, NewName, NewAddress, NewDateOfBirth): self.__EmployeeID = ID self.__Name = NewName self.__Address = NewAddress self.__DateOfBirth = NewDateOfBirth VB.NET Public Sub New(ID, NewName, NewAddress, NewDateOfBirth) EmployeeID = ID Name = NewName Address = NewAddress DateOfBirth = NewDateOfBirth End Sub

4





1(c)(iii) 1 mark per bullet point • Get method header and close (where needed) with no parameters • Returns the attribute/property EmployeeID Example code: Pascal Function Employee.GetEmployeeID() : String; result := EmployeeID; end; Python def GetEmployeeID(self): return self.__EmployeeID VB.NET Public Function GetEmployeeID() As String Return EmployeeID End Function

2





1(c)(iv) 1 mark per bullet point • Set method/procedure header and close (where needed) with parameter • Sets EmployeeID to value of parameter Example code: Pascal procedure Employee.SetEmployeeID(NewID: String); EmployeeID := NewID end; Python def SetEmployeeID(self, NewID): self.__EmployeeID = NewID VB.NET Public Sub SetEmployeeID(NewID) EmployeeID = NewID End Sub

2





1(c)(v) 1 mark per bullet point • Set method/function header and close (where needed) with parameter • Checking value of parameter for both true and false • If valid – setting Pension to parameter and returning True • If not valid – not setting Pension and returning False Example code: Pascal Function salaryEmployee.SetPension(NewPension) : boolean; IF NewPension = true or NewPension = false THEN Pension := NewPension; Result := true; ELSE Result := false; end; Python def SetPension(self, NewPension): if NewPension == True Or NewPension == False: self.__Pension = NewPension return True else: return False VB.NET Public Function SetPension(NewPension) AS Boolean If NewPension = True Or NewPension = False Then Pension = NewPension Return True Else Return False End If End Function

4





1(c)(vi) 1 mark per bullet point to max 8 • Function header and close (where needed) with at least one parameter • Constants used for hours bonus, month bonus, holiday bonus, pension cost(at least 3) • Checking if Hours is > = 160, calculating bonus payment (monthlypay * 0.05) • Checking if pension is true, calculation pension to pay (monthlypay * 0.04) • Checking if public holiday is true, calculation of bonus payment (monthlypay * 0.03) • all 3 Hours, Pension and PublicHoliday accessed from the parameter using Get methods

• Adding both bonus payments to basic salary and deducting pension from salary • basic salary accessed by using GetMonthlyPayment with the parameter • Outputting the total final bonus and pension with appropriate message • Returning the new salary Example code: Pascal Function CalculateMonthlySalary(TheEmployee : SalaryEmployee) : real; var BonusPayment : real; PensionPayment : real; BasicSalary : real; const HoursBonus : real = 0.05; HoursMonthBonus : integer = 160; PensionCost : real = 0.04; PublicHolidayBonus : real = 0.03; begin BonusPayment :=0; PensionPayment :=0; BasicSalary :=0;

8





1(c)(vi) BasicSalary := TheEmployee.GetMonthlyPayment(); IF TheEmployee.GetHoursThisMonth() >= HoursMonthBonus THEN BonusPayment := BasicSalary * HoursBonus; IF TheEmployee.GetPension() = True THEN PensionPayment := BasicSalary * PensionCost; IF TheEmployee.GetPublicHoliday() = True THEN BonusPayment := BonusPayment + BasicSalary * PublicHolidayBonus; writeln("The pension payment is " & PensionPayment); writeln ("The bonus payment is " & BonusPayment); MonthlySalary := BasicSalary + BonusPayment - PensionPayment; result := MonthlySalary; end; Python def CalculateMonthlySalary(TheEmployee): BonusPayment = 0 PensionPayment = 0 HoursBonus = 0.05 HoursMonthBonus = 160 PensionCost = 0.04 PublicHolidayBonus = 0.03 BasicSalary = TheEmployee.GetMonthlyPayment() if TheEmployee.GetHoursThisMonth() >= HoursMonthBonus: BonusPayment = BasicSalary * HoursBonus





1(c)(vi) if TheEmployee.GetPension() == true: PensionPayment = BasicSalary * PensionCost if TheEmployee.GetPublicHoliday() == true: BonusPayment = BonusPayment + BasicSalary * PublicHolidayBonus print("The pension payment is ", str(PensionPayment)) print("The bonus payment is ", str(BonusPayment)) MonthlySalary = BasicSalary + BonusPayment - PensionPayment return MonthlySalary VB.NET Public Function CalculateMonthlySalary(TheEmployee As SalaryEmployee) As Double Dim BonusPayment As Single = 0 Dim PensionPayment As Single = 0 Dim MonthlySalary As Single = 0 Const HoursBonus As Single = 0.05 Const HoursMonthBonus As Integer = 160 Const PensionCost As Single = 0.04 Const PublicHolidayBonus As Single = 0.03 Dim BasicSalary As Double = TheEmployee.GetMonthlyPayment() If TheEmployee.GetHoursThisMonth() >= HoursMonthBonus Then BonusPayment = BasicSalary * HoursBonus End If If TheEmployee.GetPension() = True Then PensionPayment = BasicSalary * PensionCost End If If TheEmployee.GetPublicHoliday() = True Then BonusPayment = BonusPayment + BasicSalary * PublicHolidayBonus End If





1(c)(vi) Console.WriteLine("The pension payment is " & PensionPayment) Console.WriteLine("The bonus payment is " & BonusPayment) MonthlySalary = BasicSalary + BonusPayment - PensionPayment Return MonthlySalary End Function

1(d) Polymorphism 1

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