2017 AQA A Level Computer Science Paper 2 Unofficial Markscheme

Tell me the answers and I will edit this markscheme. Also suggest some grade boundary predictions.
1.1 Fetch stage -
Program counter passes info to MAR. Using address bus to to relay address in main memory. Main memory returns data and passes to MBR via data bus, At the same time the program counter increments by one.

1.2 Why not MBR - MBR merely holds the instruction to be executed, CIR is needed as it decodes the instruction before it is executable, and MBR does not have this capability. Therefore MBR is not suitable for the purpose listed in the question?
1.3 Harvard - Harvard, parallel aquasition of data and instruction as memories are independent. Harvard memories can also be different IE: instructions are read only and data is read/write. Simultaneous access to data and instruction memories.

2.1 KAITLEN
2.2 Vernam - I said only use key once and make sure key is not just the same letter repeated more than once
2.3 Symmetric (same key) asymmetric (different keys)

3.1 No. of bits - 3
3.2 1500 bps
3.3 Which line - B
3.4 Serial communication - signal doesn't degrade over long distances, cheaper to manufacture since it only requires one wire.

4.1 AND with A and B and OR C and D followed by ANDing these 2 results
4.2 (A.B).(C+D)
4.3 Boolean algebra - B + C?

5.1 Direct addressing (operand is memory location) immediate (operand is datum)
5.2 10 40 50 and 1 were the numbers, don't remember what specific registers they went into (and memory locations)
5.3 -
Using low level language is appropriate for:

6.1 10
6.2 [1,9,25], [1,5], 18
6.3 Higher-order function - Takes a function as an argument and/or returns a function as a result

7 - consequences, wifi - WPA/WPA2, SSID broadcast disabled, white list - legal issues e.g. data protection act.
Camera - (how it works in capturing data)
Machine learning algorithms to blur out faces and car registration
Don't transmit data till the car is back at a company hub.
I said introduce Code of Conduct, more thorough testing of systems.

8.1 1200 kilobytes
8.2 Nyquists theorem, should have been >= 29000Hz
8.3 MIDI
MIDI is an alternative method for storing and communicating sound digitally that does not use sound waves; instead, information about each musical note is stored. This information includes:

9.1 something.something.something.x x != 0 or 255
9.2 Logical (conceptually how it works) and physical topologies (how it is actually laid out). A physical star topology can work logically as a bus topology.
9.3 peer to peer (equal status), client server (one or more nominated as servers, majority are clients that request services from the servers)

Databases
10.1 - Composite primary key with CarRegNo and Date
10.2 - Relations were to be in 3NF, no non-key dependencies, eliminate data redundancy and inconsistencies.
10.3 - ER diagram - Some possibilities: part to job is many to many; PartUsedForJob to part is many to one; car to job is one to many.
10.4 - UPDATE, SET, WHERE
10.5 -
INSERT INTO PartUsedForJob
VALUES ([jobID], [partID], 2)
10.6 -
SELECT [parameters in question]
FROM PartUsedForJob, Part
WHERE PartID = 106 ' number given in question; may not have been 106
AND Part.PartID = PartUsedForJob.PartID
ORDER BY PartID
10.7 - Nasty! any ideas?

Floating Point
11.1 - C?
11.2 - B?
11.3 - -5
11.4 - exponent was 6 which is 0110
11.5 - 0.05
11.6 - 0.36%

Mark allocations per question:

1.1 - 4
1.2 - 2
1.3 - 2
2.1 - 1
2.2 - 2
2.3 - 1
3.1 - 1
3.2 - 1
3.3 - 1
3.4 - 2
4.1 - 3
4.2 - 2
4.3 - 4
4.4 - 1
4.5 - 2
5.1 - 2
5.2 - 4
5.3 - 1
5.4 - 4
6.1 - 1
6.2 - 3
7 - 12
8.1 - 2
8.2 - 2
8.3 - 4
9.1 - 2
9.2 - 2
9.3 - 4
10.1 - 1
10.2 - 2
10.3 - 2
10.4 - 3
10.5 - 2
10.6 - 5
10.7 - 3
11.1 - 1
11.2 - 1
11.3 - 2
11.4 - 3
11.5 - 1
11.6 - 1

I'm still pretty sure that both sql questions asked to update the database.

-> Update the time taken
-> update the number of parts used for that job (job was already a booked entry therefore it wouldn't make sense to insert values).

Might be wrong^tm

I think you're wrong. First question was update, 2nd one was insert into. In the question it said the job has already been added (so update) and a new record needs to be created for number of parts.
First bit was 3 marks (UPDATE, SET, WHERE) and the next bit was 2 marks (INSERT INTO, VALUES).

THE BOOLEAN algebra i got was A + B + C. Did anyone get this or is this completely rwrong?

I also got A + B + C. Is this wrong??????


Posted from TSR Mobile

Bro You are the same person lol

I know, but that was the only question i used logic to answer, so please

Tell me im right!!!!!!!!! 😭😭😭😭


Posted from TSR Mobile

I know, but that was the only question i used logic to answer, so please

Tell me im right!!!!!!!!! 😭😭😭😭


Posted from TSR Mobile

Pretty sure that if the instruction was decoded in MBR then if there is a direct address operand, the MAR would get the address of the datum and bring it back into the MBR which would overwrite the contents of the MBR and hence the current instruction. Therefore the program would not run any further.

Grade Boundaries:
85 for A* for this paper, not going to too low since the paper was very standard aside from Boolean Algebra

FOR 10.7

I said to create a new link table for the models and parts

impossible it must be minimum of 96 for A*

impossible it must be minimum of 96 for A*

I said MBR is needed to be freed up for it to process the next instruction coming from PC after it had been incremented. Therefore we use CIR to process the current instruction

Is this wrong?


Posted from TSR Mobile

No way, it will be somewhere in the 80s for sure