A-level COMPUTER SCIENCE 7517/1 Paper 1 Mark scheme June 2022 Version: 1.0 Final

A-level COMPUTER SCIENCE 7517/1 Paper 1 Mark scheme June 2022 Version: 1.0 Final *226A7517/1/MS* 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 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. 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. A-level Computer Science Paper 1 (7517/1) – applicable to all programming languages A, B, C, D and E June 2022 The following annotation is used in the mark scheme: ; – means a single mark // – means an alternative response / – means an alternative word or sub-phrase A. – means an acceptable creditworthy answer R. – means reject answer as not creditworthy NE. – means not enough I. – means ignore DPT. – means ‘Don't penalise twice’. 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. Examiners are required to assign each of the candidate’s responses to the most appropriate level according to its overall quality, and 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 question 05.1, the marks available for the AO3 elements are as follows: AO3 (design) 4 marks AO3 (programming) 8 marks Where a candidate’s answer only reflects one element of the AO, the maximum mark they can receive will be restricted accordingly. Question Marks 01 All marks AO1 (knowledge) A. O(n x log n) NE. O(log n) I. missing brackets I. missing O Mark as follows: 1 mark: 1st row correct 1 mark: 2nd row correct 1 mark: 3rd row correct 3 Question Marks 02 1 All marks for AO1 (understanding) 1. Check the queue is (not already) empty; 2. Compare the value of the front pointer with the maximum size of the array; 3. If equal then front pointer becomes one; A. index of the first position in the array instead of one 4. Otherwise, add one to the front pointer; Alternative answer 1 1. Check the queue is (not already) empty; 2. Compare the value of the front pointer with the maximum size of the array minus one; 3. If equal then front pointer becomes zero; A. index of the first position in the array instead of zero 4. Otherwise, add one to the front pointer; Alternative answer 2 1. Check the queue is (not already) empty; 2. Add one to the front pointer; 3. Compare the value of the front pointer with the maximum size of the array; 4. If equal then front pointer becomes zero; A. index of the first position in the array instead of zero Alternative answer 3 1. Check the queue is (not already) empty; 2. Add one to the front pointer; 3. Compare the value of the front pointer with the maximum size of the array plus one; 4. If equal then front pointer becomes one; A. index of the first position in the array instead of one Alternative answer 4 1. Check the queue is (not already) empty; 2. Add one to the front pointer; 3. Use modulus/modulo operator/function with new value of front pointer; 4. Use modulus/modulo operator/function with maximum size of array; Max 3 if any errors 4 Question Marks 02 2 All marks for AO1 (understanding) Static data structures have storage size determined at compile-time / before program is run / when program code is translated / before the data structure is first used // dynamic data structures can grow / shrink during execution / at run-time // static data structures have fixed (maximum) size // size of dynamic data structures can change; Static data structures can waste storage space / memory if the number of data items stored is small relative to the size of the structure // dynamic data structures only take up the amount of storage space required for the actual data; Dynamic data structures require (memory to store) pointers to the next item(s) // static data structures (typically) do not need (memory to store) pointers; Static data structures (typically) store data in consecutive memory locations // dynamic data structures (typically) do not store data in consecutive memory locations; Max 3 3 02 3 Mark is for AO2 (apply) Jib; 1 02 4 Mark is for AO2 (apply) Jib; 1 02 5 All marks for AO1 (understanding) (Until the queue is empty) repeatedly remove / delete (the front item) from the queue and push it on to the stack; (Until the stack is empty) repeatedly pop items from the stack and add them to the (rear of the) queue; 2 Question Marks 03 1 Mark is for AO2 (analyse) Statement 1 can’t be correct because it means Statement 5 / Statement 6 is true which means Statement 1 is false; Statement 1 can’t be correct because it would mean Statement 2 is correct which would mean all of the other statements have to be both correct and incorrect; Statement 1 can’t be correct because it would mean Statement 4 is correct which means that Statements 2 and 3 have to be both correct and incorrect; Questions says only one of the statements is true so Statement 1 can’t be true as that means more than one statement would be true; Max 1 1 03 2 Mark is for AO2 (analyse) (Statement) 5; 1 03 3 All marks AO2 (analyse) Statement 3 can’t be correct because Statement 1 is false; Statement 3 can’t be correct because the question says only one of the statements is correct; Statement 3 can’t be correct because that would mean Statement 2 would be a contradiction as this would mean Statement 3 would have to be incorrect; If Statement 2 is true then Statement 4 has to be false. As Statements 1 and 3 are false for Statement 4 to be false Statement 2 has to be false as well (otherwise one of the above would be true). This is a contradiction so Statement 2 can’t be true; Statements 1, 2 and 3 are false so Statement 4 is false; If Statement 6 is true then 5 has to be false implying at least one of Statements 1 to 4 have to be true but they are all false so Statement 6 has to be false; Max 2 2