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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9691 COMPUTING 9691/11 Paper 1…
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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9691 COMPUTING 9691/11 Paper 1 (Written Paper), maximum raw mark 75 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 1 (a) buffer – any one from: temporary storage area used to hold data before being transferred allows for difference in working speeds (of processors and peripheral devices) interrupt – any one from: signal sent to the processor/CPU (which causes break in the execution of current routine) [2] (b) (i) Any three points from: data is transferred from (primary) memory to printer buffer when the buffer is full, the processor can carry on with other tasks printer buffer is emptied to printer when printer buffer is empty, printer sends an interrupt to the processor requesting more data to be sent according to priorities [3] (ii) Any two points from: first (block) of data sent to the first buffer whilst this data is being printed by the printer next block of data is sent to the second buffer when the first buffer is empty data from the second buffer is then printed meanwhile more data is then sent to the first buffer this continues until all data has been processed by the printer [2] © Cambridge International Examinations 2013 Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 2 (a) 1 mark for naming type of media + 1 mark for description/examples magnetic media surface coated with magnetic material magnetic properties altered to represent 1s and 0s used by hard disks, magnetic tapes, floppy disks optical media surface coated with light sensitive material read/written by lasers CDs use one spiral track used by DVD-RAM, CD-R, CDROM, CDRW, blu-ray disc solid state media uses millions of tiny transistors where movement of electrons controlled within a microchip has no moving parts used by memory sticks, MP3 players, cameras/mobile phones [6] (b) (i) Any two DIFFERENT points from: RAM contents can be altered/written to holds data/program currently in use volatile memory/temporary memory/contents lost when switched off usually has a greater memory capacity than ROM ROM contents can be read only/can’t be altered holds bootstrap/BIOS/system data non-volatile memory/permanent memory/retains contents when switched off [2] (ii) Any two points from: needs RAM to store instructions given by the user needs RAM to temporarily store program controlling car needs RAM to store current radio frequencies to control car needs ROM to store the factory settings/basic instructions needs ROM to store radio frequencies (etc.) understood by remote controller needs ROM to store start up routines when car switched on [2] © Cambridge International Examinations 2013 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 3 (a) (i) Any one point from: transmission is sent in one direction only along a single data line Reject single CABLE [1] (ii) Any one point from: transmission can be in both directions at same time along several data lines/one data line per bit [1] (b) (i) letter bytes adjusted for even parity C 1 1 0 0 0 0 1 1 O 1 1 0 0 1 1 1 1 M 0 1 0 0 1 1 0 1 P 0 1 0 1 0 0 0 0 U 0 1 0 1 0 1 0 1 T 1 1 0 1 1 0 0 0 I 1 1 0 0 1 0 0 1 N 0 1 0 0 1 1 1 0 G 0 1 0 0 0 1 1 1 (–1 mark for each error in the first column) [2] (ii) 0 1 0 1 1 1 0 0 [1] (iii) Any three points from: character “P” flagged as having odd parity (row 4 in diagram) parity byte sent with data i.e. 0 1 0 1 1 1 0 0 column 5 also has odd parity (or equivalent) faulty bit must be in row 4 and column 5 idea of auto correction of fault (in row 4, column 5) (Check if diagram has been annotated to show faulty bit) [3] © Cambridge International Examinations 2013 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 4 (a) 1 mark for name of device + 1 mark for reason scanner to produce an electronic/digital map version of the passport photograph (scans) into computer readable format digital camera/video camera to produce an electronic image of the passenger’s face produces a similar format to the scanned image [4] (b) (i) 1 mark for each point use of a pressure sensor sends data back to computer system need for conversion to digital form (ADC) computer calculates weight based on sensor data this calculated value is compared / (or equivalent) to stored values sends weight back to a small screen at check-in if weight exceeds airline limit, operator warned at check-in by, e.g., a flashing screen or beeping sound / if weight below limit luggage accepted [3] (ii) 1 mark per point (labels printed in form of) a barcode barcode is unique use of barcode reader/scanner barcode is used as a key field in passenger record barcode read at each stage ….. ... and this data is stored in passenger record thus allowing tracking/whereabouts of luggage at any stage [2] © Cambridge International Examinations 2013 Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 5 (a) 1 mark for naming type of diagram + 2 marks for description/examples data flow diagram (DFD) graphical representation of flow of data through the system shows what data is input and output from the system shows where data comes from and goes to and where it is stored contains process, flow, store and terminators systems flow chart/diagram way of displaying how decisions are made to control events shows devices (e.g. disk drives) shows media use for input/output and storage (e.g. paper output) shows what files are used in the system Program flowchart/Jackson structured diagram shows method of solution shows data inputs and outputs shows modularisation If NO valid diagram stated then cannot get features for that part of the answer REJECT flowchart alone (accept examples/diagrams as part of explanation reflecting description above) [3] © Cambridge International Examinations 2013 Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 (b) 1 mark for each item + 1 mark for reason purpose of the system this allows another programmer to know what the aim of the software is program listing/coding if an error occurs in the original program this allows another programmer to alter the coding to cure the fault flowchart/algorithm these will show the logical steps in the program or system so another programmer can follow the logic input/output formats allows another programmer to test the system to see if it produces expected results hardware and software requirements useful if a system needs to be upgraded so another analyst can decide exactly what hardware or software would need to be changed/ ensure new hardware is compatible list of variables/data dictionary this allows another programmer to understand what the variables represent so they can understand the coding file structures this is important so that another programmer knows if they are fixed length etc. and how the data is arranged in the file test plans/sample runs/test data this is important so that another programmer can see how the original software performed and to see how it was tested validation rules knowledge of these rules will let a programmer know what error trapping exists within the software so that new data follows the same rules meaning of error messages this helps another programmer understand what is going wrong with the software [4] © Cambridge International Examinations 2013 Page 8 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 6 (a) (i) 1 mark for each correct output device + 1 mark for each reason production dot matrix printer factory environment could be dirty/dusty/damp need to produce stickers from continuous stationery design office large computer screen/monitor need to be able to see new designs very clearly if screens are CRT it is possible to use light pens (graph) plotters to produce “blue-prints” which can be used to build prototypes often necessary to produce full size drawings of new product 3D printers these allow production of solid prototypes/models that work reduces cost of tooling up to make a real working example marketing laser printers producing large numbers of brochures/catalogues/flyers quiet operation in office environment high speed operation when producing large print runs screens to show dialogue/script when answering customer queries headphones to hear customer queries finance dot matrix/impact printer pay slips produced on continuous stationery “hidden information” in ready sealed envelopes laser/inkjet printer produces payslip for later sealing none secure NE to produce hard copy of a wage slip [8] (b) Any three points from: large coloured graphics showing process green colour to show ON and red colour to show OFF flashing/highlighted colours used to indicate error/attention required layout needs to allow links to other screens showing other part of process layout should be clear and easy to use ability to “click on”/choose graphic to show status of the item use of input devices such as touch screens, trackerballs and keyboards/keypads to allow selection etc. use of output devices such as large screens, printers and speakers/ beepers to produce plant data and warn of problems [3] © Cambridge International Examinations 2013 Page 9 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 7 (a) (i) Any two points from: (use of a 2-dimensional array) Grid [1:3, 1:3] use 1 value for “X” and 0 value for “O” or characters “X” and “O” each square in game corresponds to position in array ….. ... e.g. 1 value in Grid [1, 1], 0 value in Grid [2,1] and so on each time a square is “filled in” the position is identified and appropriate value stored in correct position in array Grid [2] (ii) integer/char [1] (iii) 1/null/space [1] (b) Any four points from: each array position is scanned each row is checked e.g. first row: Grid [1, 1], Grid [2, 1] then Grid [3, 1] and so on computer checks to see if all three values in the row are the same … … if they are all 1s then “X” has won; if all 0s then “O” has won * each column is checked e.g. first column: Grid [1 ,1], Grid [1, 2] then Grid [1, 3] and so on computer checks to see if all three values in the column are the same … if they are all 1s then “X” has won; if all 0s then “O” has won * each diagonal is checked e.g. Grid [1, 1], Grid [2, 2] then Grid [3,3] OR Grid [3, 1], Grid [2, 2] then Grid [1, 3] computer checks to see if all three values in diagonal are the same … … if they are all 1s then “X” has won; if all 0s then “O” has won * this scan is carried out each time a square has been filled in * this answer can only be accepted once If an algorithmic approach has been applied mark along the following lines: initialise flag before checking array concept of looping for a row concept of looping for a column recognising position in an array check symbol in adjacent cells is the same flag or recognition of change in content [4] (c) 1 mark for device + 1 mark for reason touch screen user only has to touch a square on screen to make choice computer knows player is “X” and puts an “X” where screen touched mouse/trackerball/trackpad user moves cursor/arrow and points to chosen square “click” to confirm and “X” is placed in chosen square keyboard move cursor/arrow by using ↑ ↓ ← → keys press ENTER key when chosen square found [2] © Cambridge International Examinations 2013 Page 10 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9691 11 8 (a) record key is used the key value is subjected to arithmetic algorithm to give/calculate the location/address of the record [3] (b) 1 mark for chosen method + 1 mark for description use of overflow/bucket area used any record that is subject to a collision is placed, serially, in overflow area set flag when overflow / bucket is in use use of linked lists original location acts as head of list and points to a list of any records that have been subject to a collision use of next location after occupied one is used this continues until empty location is found [2] © Cambridge International Examinations 2013 age Pa e 11 1 Mar M rk Sc S he eme e Syl S labbus s Pa ape er GC GCE AS S/A A LEV L VELL–MMay y/Jun ne 20 2 13 96991 11 9 ( (a) 1m marrk for f eaach co orre ectt logicc gate e (a acc cep pt oth o er log gic ga ate no omen ncla ature)) A X B C Iff a ca and dida ate e ha nly on as on ne iinp put to AN NDD ga ate e orr an nOOR R ga ate heyy lose the ma e th mark forr th hat ga ate 4] [4 ( ) (b) A B C X } 0 0 0 1 0 0 1 0 1 ma ark k } 0 1 0 1 1 ma ark k 0 1 1 1 1 1 0 0 0 1 1 0 } 1 ma ark k 1 1 1 1 0 1 1 1 } 1 ma ark k [4 4] © Ca amb brid dge e In nterrnattion nal Ex xam mina atio ons s 20 013 3
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