A-level COMPUTER SCIENCE 7517/2 Paper 2 Mark scheme June 2022 Version: 1.0 Final *226A7517/2/MS* MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 2

A-level COMPUTER SCIENCE 7517/2 Paper 2 Mark scheme June 2022 Version: 1.0 Final *226A7517/2/MS* MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 2 Mark schemes are prepared by the Lead Assessment Wr ...

A-level COMPUTER SCIENCE 7517/2 Paper 2 Mark scheme June 2022 Version: 1.0 Final *226A7517/2/MS* MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk Copyright information AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. Copyright © 2022 AQA and its licensors. All rights reserved. MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 3 Level of response marking instructions Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme. Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content. Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks. MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 4 A-level Computer Science Paper 2 June 2022 To Examiners: • When to award '0' (zero) when inputting marks on CMI+ A mark of 0 should be awarded where a candidate has attempted a question but failed to write anything credit worthy. Insert a hyphen when a candidate has not attempted a question, so that eventually the Principal Examiner will be able to distinguish between the two (not attempted / nothing credit worthy) in any statistics. • This mark scheme contains the correct responses which we believe that candidates are most likely to give. Other valid responses are possible to some questions and should be credited. Examiners should refer responses that are not covered by the mark scheme, but which they deem creditworthy, to a Team Leader. The following annotation is used in the mark scheme: ; - means a single mark // - means alternative response / - means an alternative word or sub-phrase A. - means acceptable creditworthy answer R. - means reject answer as not creditworthy NE. - means not enough I. - means ignore DPT. - in some questions a specific error made by a candidate, if repeated, could result in the loss of more than one mark. The DPT label indicates that this mistake should only result in a candidate losing one mark on the first occasion that the error is made. Provided that the answer remains understandable, subsequent marks should be awarded as if the error was not being repeated. TO. - a mark that would otherwise have been awarded has not be awarded because another part of the candidate’s response indicates that they did not understand the point they had made. For example, they might have made a contradictory point. MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 5 Examiners are required to assign each of the candidates’ responses to the most appropriate level according to its overall quality, then allocate a single mark within the level. When deciding upon a mark in a level examiners should bear in mind the relative weightings of the assessment objectives. eg In the following questions, the marks available are as follows: Question 07.4 (max 7 marks) AO2 (analyse) – 5 marks AO3 (programming) – 2 marks Question 08.3 (max 3 marks) AO1 (knowledge) – 1 mark AO1 (understanding) – 2 marks MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 6 Question Marks 01 1 All marks AO1 (knowledge) Put the bits into groups of four/nibbles (starting at the right); Convert each group of bits/nibble into a hexadecimal digit; A. group/section/chunk of bits, even if number of bits not stated or incorrect for this mark A. “value” or “number” for “digit” A. convert each group of bits into decimal for 0 to 9, and letter A-F for 10 to 15 Do not award marks if the conversion has been done but has not been described, although it is acceptable for the conversion to be performed as part of the description. 2 Question Marks 01 2 Mark is for AO1 (understanding) More compact when displayed // can be displayed using fewer digits; Easier (for people) to understand / remember; A. read, write Lower likelihood of an error when typing in data; Saves (the programmer) time writing / typing in data; NE. takes up less space R. if stated that hexadecimal uses less memory / storage Max 1 1 Question Marks 02 1 All marks AO1 (knowledge) Multiple bits transmitted simultaneously / at same time; NE. data, values etc for bits Each (simultaneously transmitted) bit is sent down a different wire / cable / path / line; A. multiple wires / cables / paths / lines used for transmission 2 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 7 Question Marks 02 2 Mark is AO1 (understanding) The hardware / wiring required for serial data transmission is cheaper; NE. cheaper without reference to hardware or wiring Serial transmission does not suffer from crosstalk // (two) bits cannot interfere with each other because they are not sent simultaneously; Serial transmission does not suffer from data skewing // bits transmitted are guaranteed to arrive in the order they were sent; NE. more reliable, lower probability of interference / corruption Serial transmission can be used over longer distances; Max 1 1 Question Marks 02 3 Mark is AO1 (understanding) B Latency is the rate at which signals on a wire or line can change; R. if more than one lozenge shaded 1 Question Marks 02 4 Mark is AO1 (knowledge) Start the receiver clock ticking; A. to wake up the receiver Synchronise the clock in the receiver to the transmitter clock // bring the clock in the receiver into phase with the clock in the transmitter; A. to synchronise the receiver and transmitter clocks A. synchronise the clocks in the devices NE. synchronise the (two) clocks R. indicates start of transmission Max 1 1 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 8 Question Marks 02 5 Mark is AO1 (knowledge) Provides time for the receiver to process / transfer the received data; NE. indicates that the received data can be processed Allows the (next) start bit to be recognised; R. indicates end of transmission R. indicates clocks no longer need to be synchronised Max 1 1 Question Marks 03 1 Mark is AO2 (apply) ? ? ? + ? ?̅ ?̅ ?̅ ∙ ?̅ ? ̅̅̅̅ ∙ ̅̅?̅̅ 0 0 0 1 1 1 0 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 0 1 1 mark: Table correctly completed 1 Question Marks 03 2 Mark is AO1 (understanding) De Morgan’s (Law); 1 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 9 Question Marks 03 3 All marks AO2 (apply) Simplification of the two sub-expressions A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ and C ⋅ (A + A̅ ∙ (B + 1)) should be marked independently. Stop awarding marks for a sub-expression as soon as a mistake has been made in that subexpression, but continue to award marks for simplifying the other subexpression. Marking guidance for examiners: • award marks for working out until an incorrect step has been made • ignore missing steps from the example solutions, as long as the jumps between steps are logically correct • if, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. For example, if the expression P.P.(P+Q) + P.P.1 was changed to P.(P+Q)+P.0, the candidate would get one mark for simplifying the first part to P.(P+Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P.0 (ie to 0). 1 mark for final answer: A ∙ B̅ + C Max 3 for working. Award up to three marks for applying each one of the three techniques (one mark per application): • a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression • applying an identity other than cancelling NOTs that produces a simpler expression • successfully putting terms into brackets • successfully expanding brackets • successfully using the distributive law. Note: A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators. Max 3 overall if any working is incorrect 4 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 10 Example Solution (1) A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅ ∙ (B + 1)) A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅ ∙ 1) By ? + 1 = 1 A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅) By ? ∙ 1 = ? A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ 1 By ? + ?̅ = 1 A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C By ? ∙ 1 = ? A̅ + B ∙ (C + C̅) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ + C Put into brackets A̅ + B ∙ 1 ̅̅̅̅̅̅̅̅̅̅̅ + C By ? + ?̅ = 1 A̅ + B ̅̅̅̅̅̅̅ + C By ? ∙ 1 = ? A ∙ B̅ + C Application of De Morgan Example Solution (2) A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅ ∙ (B + 1)) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ (A̅ + B ∙ C + B ⋅ C̅) ⋅ C ⋅ (A + A̅ ∙ (B + 1)) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ Application of De Morgan (A̅ + B ∙ (C + C̅)) ⋅ C ⋅ (A + A̅ ∙ (B + 1)) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ Put into brackets (A̅ + B ∙ (1)) ⋅ C ⋅ (A + A̅ ∙ (B + 1)) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ By ? + ?̅ = 1 (A̅ + B) ⋅ C ⋅ (A + A̅ ∙ (B + 1)) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ By ? ∙ 1 = ? (A̅ + B) ⋅ C ⋅ (A + A̅ ∙ 1) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ By ? + 1 = 1 (A̅ + B) ⋅ C ⋅ (A + A̅) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ By ? ∙ 1 = ? (A̅ + B) ⋅ C ⋅ (1) ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ By ? + ?̅ = 1 ( ̅̅A̅̅̅+̅̅̅̅B ̅ ) ̅̅ ⋅ ̅̅C̅̅ By ? ∙ 1 = ? A̅ + B ̅̅̅̅̅̅̅ + C Application of De Morgan A ∙ B + C Application of De Morgan MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 11 Example Solution (3) A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅ ∙ (B + 1)) A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅ ∙ 1) By ? + 1 = 1 A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ (A + A̅) By ? ∙ 1 = ? A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C ⋅ 1 By ? + ?̅ = 1 A ̅̅̅̅+̅̅̅B ̅̅̅ ∙ ̅C ̅̅̅+̅̅̅B ̅̅ ⋅ ̅̅C̅̅ + C By ? ∙ 1 = ? (A̅ + B ∙ C) ⋅ B ⋅ C̅ + C Application of De Morgan (A̅ + B ∙ C) ⋅ (B + C) + C Application of De Morgan (A ∙ B ∙ C) ⋅ (B + C) + C Application of De Morgan A ∙ (B + C) ⋅ (B + C) + C Application of De Morgan A ∙ B ∙ B + A ∙ C ∙ B + A ∙ B ∙ C + A ∙ C ∙ C + C Expand Brackets A ∙ B + A ∙ C ∙ B + C By ? + any term with ? = ? / distributive law A ∙ B + C By distributive law MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 12 Question Marks 04 1 Mark is AO1 (knowledge) Software used in the management of a computer system // layer(s) of software that abstract the user from how the computer works // software that provides a platform for other software to use; A. software used to run the computer A. software that provides a virtual machine NE. software that maintains a computer 1 Question Marks 04 2 Mark is AO1 (knowledge) B Bitmap image editors; R. if more than one lozenge shaded 1 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 13 Question Marks 04 3 All marks AO1 (knowledge) To hide the complexities of the hardware from the user; NE. virtual machine without description R. user interface To handle interrupts // to call appropriate interrupt handler (A. ISR) when an interrupt occurs; To allocate processors/cores to processes // schedule processes // decide which process to carry out when // manage the execution of multiple processes; NE. processor management To allocate memory/RAM to processes // to determine what areas of memory are used for what purpose // moving data into and out of RAM / to a paging file for virtual memory // ensuring processes can only write to memory that they have been allocated; NE. memory management To allocate I/O devices to processes // manages communication between processes and I/O devices // automatic installation of drivers for new I/O devices; A. examples of devices (but no more than one mark) NE. manages I/O devices To allocate space on a storage device to files // organising files into directories // determines where on a device to save a file // recognising storage devices when they are connected; A. defragmentation of disks NE. saving a file Installation of new software // automatic / managing updating of software; A. “programs” or “tasks” for “processes” Manage power consumption / use of battery; A. examples of this eg controlling clock speed, brightness of screen Note: Students must describe – phrases such as “processor management”, “allocating memory” etc are not enough. Max 2 2 MARK SCHEME – A-LEVEL COMPUTER SCIENCE – 7517/2 – JUNE 2022 14 Question Marks 05 1 All marks AO2 (apply) Award 2 marks for correct answer: –19 5/16 // –309/16 // –19.3125 If answer is incorrect then award 1