Cambridge International AS & A Level
COMPUTER SCIENCE 9608/21
Paper 2 Written Paper May/June 2020
MARK SCHEME
Maximum Mark: 75
Published
Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic.
This mark scheme is published to support teachers and students and should be read together with the
question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the
proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also
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provides the most likely acceptable alternative responses expected from students. Examiners usually review
the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the
June series, Examiners were unable to consider the acceptability of alternative responses, as there were no
student responses to consider.
Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However,
because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June
2020 series.
Cambridge International will not enter into discussions about these mark schemes.
Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge
IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level
components.
This document consists of 15 printed pages.
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9608/21 Cambridge International AS & A Level – Mark Scheme May/June 2020
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Generic Marking Principles
These general marking principles must be applied by all examiners when marking candidate answers.
They should be applied alongside the specific content of the mark scheme or generic level descriptors
for a question. Each question paper and mark scheme will also comply with these marking principles.
GENERIC MARKING PRINCIPLE 1:
Marks must be awarded in line with:
• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.
GENERIC MARKING PRINCIPLE 2:
Marks awarded are always whole marks (not half marks, or other fractions).
GENERIC MARKING PRINCIPLE 3:
Marks must be awarded positively:
• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit
is given for valid answers which go beyond the scope of the syllabus and mark scheme,
referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these
features are specifically assessed by the question as indicated by the mark scheme. The
meaning, however, should be unambiguous.
GENERIC MARKING PRINCIPLE 4:
Rules must be applied consistently e.g. in situations where candidates have not followed
instructions or in the application of generic level descriptors.
GENERIC MARKING PRINCIPLE 5:
Marks should be awarded using the full range of marks defined in the mark scheme for the question
(however; the use of the full mark range may be limited according to the quality of the candidate
responses seen).
GENERIC MARKING PRINCIPLE 6:
Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should
not be awarded with grade thresholds or grade descriptors in mind.
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Question Answer Marks
1(a) An algorithm is a solution to a problem expressed as: 2
a sequence of defined steps / stages / instructions / lines of code
1 mark for each underlined term (or equivalent)
1(b) • Allows the subroutine code to be called from many/multiple places 3
• Subroutine code may be (independently) tested and debugged
• If the subroutine task changes the change needs to be made only once
• Reduces unnecessary duplication / program lines
• Enables sharing of development between programmers
Or equivalent points that relate to a PROGRAM (not an algorithm)
Max 3
1(c) 3
One mark for each correct line to max 3
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Question Answer Marks
2(a) 5
Mark as follows:
1 One mark for all three boxes correctly labelled
2 One mark for selection diamond
3 One mark for passing value and return Boolean from PayByCard()
4 One mark for passing Value, AccountNUmber and AccountLimit to
PayByAccount()
5 One mark for passing CurrentBalance ByRef
2(b)(i) Trace table shows: 2
• 'A' is not treated as an upper case character (row 7)
• NumUpper not incremented as expected
• Incorrect final value for NumUpper (should be 1)
Max 2
2(b)(ii) One mark per point: 2
• Line number: 33
• Correction: IF NextChar >= 'A' AND NextChar <= 'Z'
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Question Answer Marks
2(b)(iii) CASE OF NextChar 4
>= 'a' AND <= 'z' : NumLower ← NumLower + 1
> 'A' AND <= 'Z' : NumUpper ← NumUpper + 1
OTHERWISE NumNonAlpha ← NumNonAlpha + 1
ENDCASE
One mark for CASE OF NextChar ... ENDCASE
One mark for each remaining line
Accept alternative range description. E.g. 'a' to 'z'
Accept corrected version for the second range.
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Question Answer Marks
3(a) PROCEDURE AddCredit(TopUp : REAL, PhoneNum : STRING) 7
DECLARE Multiple : REAL
DECLARE Balance : REAL
Multiple ← 1
Balance ← GetBalance(PhoneNum)
IF Balance > 10
THEN
Multiple ← 1.125
ELSE
IF Balance > 5
THEN
Multiple ← 1.1
ENDIF
ENDIF
TopUp ← TopUp * Multiple
SetBalance(PhoneNum, Balance + TopUp)
ENDPROCEDURE
1 mark for each of the following:
1 PROCEDURE heading and ending including parameters
2 Initialise Multiple
3 Assign value to Balance using GetBalance()
4 Check for Balance > 10 and assignment: Multiple ← 1.25
5 Check for Balance > 5 and assignment: Multiple ← 1.1
6 Assignment: TopUp ← TopUp * Multiple
7 Calling SetBalance()with correct parameters
Note:
MP6 could be included in MP7 statement
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Question Answer Marks
3(b) PROCEDURE Search(SearchString : STRING) 8
DECLARE Index, Msg : STRING
Msg ← "Found at:" //initial value
FOR Index ← 1 TO 100
IF NameList[Index, 1] = SearchString__
AND NameList[Index, 2] = "Active"
THEN
Msg ← Msg & " " & NUM_TO_STRING(Index)
ENDIF
ENDFOR
IF Msg = "Found at:" // no change to initial value
THEN
OUTPUT "Search String not found"
ELSE
OUTPUT Msg
ENDIF
ENDPROCEDURE
1 mark for each of the following:
1 PROCEDURE heading and ending including parameter
2 Declare local variables for Index and Msg and initialise Msg to
appropriate string
3 Loop structure
4 Compare SearchString to name (column 1)...
5 ... AND Compare status to "Active" (column 2) in a loop
6 Add Index to Msg when a match is encountered (using type conversion)
7 Condition to determine which string is output after loop
8 Correct output of single message
Note:
Credit alternative solutions for forming and checking a single output string
Question Answer Marks
4(a) A program fault is something that makes the program not do what it is 2
supposed to do under certain circumstances
One mark per underlined phrase or equivalent
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Question Answer Marks
4(b) Answers include the use of: 3
• Tried and tested (library) subroutines / code
• Modular programming techniques (to break the problem down and make
it easier to solve)
• Good programming practice (formatting, sensible variable names,
comments etc)
• IDE features (parameter type-checking, auto-complete)
Max 3
4(c) Syntax error: 3
A construct / statement in the source code that breaks the rules of the
language
Logic Error:
An error in the algorithm that causes the program not to behave as intended
Run-time:
A program performs an invalid operation / tries to divide by zero // enters an
infinite loop / stops unexpectedly
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Question Answer Marks
5(a)(i) PROCEDURE SortContacts() 8
DECLARE Temp : STRING
DECLARE FirstName, SecondName : STRING
DECLARE NoSwaps : BOOLEAN
DECLARE Boundary, J : INTEGER
Boundary ← 999
REPEAT
NoSwaps ← TRUE
FOR J ← 1 TO Boundary
FirstName ← RIGHT(Directory[J],__
LENGTH(Directory[J]) – 4)
SecondName ← RIGHT(Directory[J + 1],__
LENGTH(Directory[J + 1]) – 4)
IF FirstName > SecondName
THEN
Temp ← Directory[J]
Directory[J] ← Directory[J + 1]
Directory[J + 1] ← Temp
NoSwaps ← FALSE
ENDIF
ENDFOR
Boundary ← Boundary - 1
UNTIL NoSwaps = TRUE
ENDPROCEDURE
One mark per highlighted phrase
5(b) Description: 4
• uses a flag variable to stop the outer loop
• after no more swaps made during one pass of the inner loop
• the flag is reset before the inner loop starts, and set whenever a swap is
made
• decreases the loop size at end of inner loop (Boundary decremented)
Max 3 for description
Effective because:
• It prevents unnecessary iterations / passes through the array (i.e. when
the array is already sorted) // terminates the algorithm when all elements
are in order // reduces the number of unnecessary comparisons
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Question Answer Marks
6(a) PROCEDURE ListAvailable(StartTime : STRING) 8
DECLARE NumAvailable, Index : INTEGER
DECLARE TimeBack : STRING
DECLARE Available : BOOLEAN
NumAvailable ← 0
FOR Index ← 1 TO 10
Available ← FALSE // initialise
IF HireTime[Index] = "Available" // not on hire
THEN
Available ← TRUE // available now
ELSE
TimeBack ← AddTime(HireTime[Index],__
Duration[Index])
IF TimeBack < StartTime // < or <=
THEN
Available ← TRUE // will be available
ENDIF
ENDIF
IF Available = TRUE
THEN
OUTPUT "Boat " , Index , " is available"
NumAvailable ← NumAvailable + 1
ENDIF
ENDFOR
IF NumAvailable > 0
THEN
OUTPUT "There are " , NumAvailable ,__
" boats available."
ELSE
OUTPUT "Sorry, there are no boats available"
ENDIF
ENDPROCEDURE
1 mark for each of the following:
1 Procedure heading and ending including input parameter
2 Declare local variable for the count of available boats and initialise to 0
3 Loop through all 10 boats
4 Use of AddTime() to calculate TimeBack
5 Check for boats that are not on hire OR those due back in time in a loop
6 Increment count for number of available boats in a loop
7 Output a message for each available boat in a loop
8 Output both messages as appropriate outside a loop
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Question Answer Marks
6(b) 'Pseudocode' solution included here for development and clarification of mark 7
scheme.
Programming language example solutions appear in the Appendix.
PROCEDURE RecordHire(HBoatNumber, HDuration : INTEGER,__
HTime : STRING, HCost : REAL)
DECLARE FileLine : STRING
CONSTANT Comma = ','
HireTime[HBoatNumber] ← HTime
Duration[HBoatNumber] ← HDuration
Cost[HBoatNumber] ← HCost
DailyTakings ← DailyTakings + HCost
OPENFILE "HireLog.txt" FOR APPEND
FileLine ← NUM_TO_STRING(HBoatNumber) & Comma
FileLine ← FileLine & HTime & Comma
FileLine ← FileLine & NUM_TO_STRING(HDuration)
FileLine ← FileLine & Comma & NUM_TO_STRING(HCost)
WRITEFILE "HireLog.txt", FileLine
CLOSEFILE "HireLog.txt"
ENDFUNCTION
One mark for each of the following:
1 Procedure heading and ending (where appropriate), including input
parameters (order not essential)
2 Updating the three arrays from parameter values
3 Totalling DailyTakings
4 OPEN "HireLog.txt" in append mode
5 Creating file text line including separators
6 ....making use of type conversion as required
7 Writing the line to the file
8 Closing the file
Solutions may combine mark points 5 and 6 (and 7)
Max 7
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Question Answer Marks
6(c)(i) 'Pseudocode' solution included here for development and clarification of mark 2
scheme.
Programming language example solutions appear in the Appendix.
EndTime ← Addtime (BeginTime, 60)
One mark per underlined section (Space before bracket for mark scheme
clarification only)
6(c)(ii) One mark for each test: 2
For example:
Test 1
Start time value "10:00", Duration value 30
Expected new time value "10:30"
Test 2
Start time value "10:45", Duration value 30
Expected new time value "11:15"
String values (time) must be enclosed in quotation marks, integer values
(duration) must not. Penalise once then FT.
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Program Code Example Solutions
To be reviewed at STM
Q6(b)(i): Visual Basic
Sub RecordHire(HBoatNumber, HDuration As Integer, HTime As String, HCost As
Real)
Dim FileLine As String
Const Comma = ','
HireTime(HBoatNumber) = HTime
Duration(HBoatNumber) = HDuration
Cost(HBoatNumber) = HCost
DailyTakings = DailyTakings + HCost
FileOpen(1, "HireLog.txt", OpenMode.Append)
FileLine = CStr(HBoatNumber) & Comma
FileLine = FileLine & HTime & Comma
FileLine = FileLine & CStr(HDuration) & Comma
FileLine = FileLine & CStr(HCost)
Print(1, FileLine)
PrintLine(1)
Fileclose(1)
End Sub
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Q6(b)(i): Pascal
procedure RecordHire(HBoatNumber, HDuration : integer; HTime : string;
HCost : Real);
var
Fileline : string;
ThisFile: TextFile;
const Comma = ',';
begin
HireTime[HBoatNumber] := HTime;
Duration[HBoatNumber] := HDuration;
Cost[HBoatNumber] := HCost;
DailyTakings := DailyTakings + HCost;
AssignFile(Thisfile, "HireLog.txt");
FileLine := IntToStr(HBoatNumber) + Comma;
FileLine := FileLine + HTime + Comma;
FileLine := FileLine + IntToStr (HDuration) + Comma;
FileLine := FileLine + IntToStr (HCost);
writeln(ThisFile, FileLine);
CloseFile(ThisFile);
end;
Q6(b)(i): Python
def RecordHire(HBoatNumber, HDuration, HTime, HCost)
# FileLine : String
# File : File handle
Comma = ','
HireTime[HBoatNumber] = HTime
Duration[HBoatNumber] = HDuration
Cost[HBoatNumber] = HCost
DailyTakings = DailyTakings + HCost
File = Open("HireLog.txt", "a")
FileLine = Str(HBoatNumber) + Comma
FileLine = FileLine + HTime + Comma
FileLine = FileLine + Str(HDuration) + Comma
FileLine = FileLine + Str(HCost)
File.write(FileLine)
File.close
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Q6(c)(i): Visual Basic
EndTime = Addtime(BeginTime, 60)
Q6(c)(i): Pascal
EndTime := Addtime(BeginTime, 60)
Q6(c)(i): Python
EndTime = Addtime(BeginTime, 60)
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