OCR D1 (10th June) Discussion

I saw that someone already posted an unofficial mark scheme. What is the page and post number please?

I saw that someone already posted an unofficial mark scheme. What is the page and post number please?

Can't really add much to the great solutions already posted, but can provide you with marks for each Qn here. Sorry don't know how to do the spoiler thing.

1 Van packing
i) see last post (3marks)
ii) Van 1 250 250 100, Van 2 200 200 150, Van 3 150, 120, 120, 120 Van 4 120 (4 marks)
iii) see last post (2 marks)

2 Graph theory
ia) see last post, with graph drawn orders 2,2,3,3 (3 marks)
ib) sum of vertices = 10, min order 1(for connected), max order 3(for simple)
possibilities that fit these criteria are 3,3,3,1 or 3,3,2,2.
3,3,3,1 doesn't work because to be simple each of the arcs of order 3 must connect to every other node, so cannot have node of order 1 (5 marks)

ii) 5 graphs drawn (see post 290) (3marks)

3 Sum of squares algorithm
i) 4 runs through algorithm, print 4,22 (4 marks)
ii) 2 runs through algorithm, print FAIL (2 marks)
iii)so -ve sqrt doesn't occur and create error (to stop algorithm may also be credited) (1 mark)
iv) 0.7 x rt4 = 1.4s (2 marks)

4 treasure trail
i) B200m C100m D600m E300m F400m G800m H700m correct temporary labels should be shown on graph with no extras (5marks)
ii) odd nodes are A and G, repeat AG 800m, total = 4200m+800m = 5000m (3 marks)
iii) A-C-B-E-H (1 mark)
iv)need to make G even and H odd, repeat GH 100m, total = 4200m+100m = 4300m (2 marks)

5 treasure trail again
ia) G-H-F-E-B-C-A-D-G (must return to G) = 2000m (3marks)
ib) A-D-G-H-F-E-B-C-A (must return to A, could be reversed) (1 mark)
ii) AC, CB, BE, EF, FD, FH, HG (must show order in which arcs added) = 1000m (3marks)
iiia) remove FH,GH, add DG = 1000m (1 mark)
iiib) tour improvement algorithm gives routes given in last post 5vi. It is possible that there are others but I don't think so. (2 marks)
v) see 5vii from last post (2 marks)
vi) see 5viii from last post, (2 marks)

6 Sandie tanning lotions (!)
i) see last post (2marks)
ii) see last post (2 marks)
iii) 2a+4b+c<=176, 5a+b+3c<=180 (2 marks)
iv) £8 profit per bottle of amber (1 mark)
v) pivot on column a, Row 5; include operations for how each row calculated ( 5 marks)
vi) no negatives in objective row (1 mark)
vii) see last post (4 marks)
viii) 9x1 +35x0.8 = 37litres (1 mark)


In summary, IMO this is a 'safe' paper with all marks accessible, but not particularly any easier in terms of content. The fact that the questions are clear and straightforward means that the marking criteria will be kept tight, but it shouldn't have any effect on grade boundaries.

Most students will probably lose some marks in Q2b through poor communication of what they mean and also may have struggled to find all 5 graphs - many students just miss out this type of Qn altogether.
Similarly, although you might have found Q3 easy, I'm expecting quite a few abandoned attempts.
Lots of students bodge Djikstra's and the mark scheme is always tight on the working being shown. As these forum discussions show, the x100 will have caught out many students, but I wouldn't expect this to be heavily penalised - just 1 or maybe 2 marks total loss. Simplex is usually a mess and takes hours to mark, although this seemed nicer than some we've had. Many students just don't understand the structure and importance of the basic columns.
From what I've read on here, you lot sound like you've got great grades in the bag! D2 here we come!!

For the graph theory question, I have slightly different reasoning to you, however, I personally don't see what mine lacks, so will you GUESS/JUDGE, if mine is LIKELY to be accepted as 5 marks, or equivalent to yours?

"1) 5 arcs, therefore total order of 10.
2) We could have 4+2+2+2, but this would result in the graph not being simple, as one node is either directly connected to itself, or connected to another more than once.
3) Hence the maximum order of any node is 3.
4) This node then connects to the other 3, giving orders 3+1+1+1=6, so we have 3 arcs so far.
5) We add an arc joining two of the nodes of order 1, as we cannot add anything to the order of 3; giving us 3+2+2+1=8.
6) Finally, we must join an order of 2 with an order of 1 - we cannot join the order of 2's together because it would result in the graph not being simple, hence the only possible orders are 3+3+2+2=10. "

Also, for question 6i, this is the only mistake I made on the paper, I had a stupid moment and wrote each variable = 0.1L of whatever in order to fit the constraint. So when it said show this constraint, would I still get the method mark for doing it because they are my variables? As obviously 400*0.1=40L etc. And on the following question I also got the correct inequality but due to me defining the variables slightly differently, would I lose the method mark? Or would it be carried forward from my part (i).

In other words: would you predict I would lose 1 mark or 3?

I'm sorry to ask so much, just you know how it can be after an exam! So are you expecting a fairly average grade boundary of an A being approx. 55?

The questions that you are asking are the fine details that will be dealt with by the chief examiners and these decisions won't have been made yet! Therefore I can legitimately give you my best guesses. I guess definitely 4, possibly 5 for your graph theory as you have implied minimum order 1 for connectivity and clearly know what you are talking about (always good for a BOD mark!). For 6i, I think you might get 1 mark if you have clearly defined your variables. Using your variables and logic for part b would still give 2b+5c<=70 wouldn't it? Any multiple of this constraint should be fine also.

As for a boundary mark, absolutely no idea, but I repeat my earlier comments, I don't think that the content of this exam was significantly easier than previous series.

Oh yes I got the constraint in part ii correct but it was worth 2 marks (method + answer), so I'm wondering whether I would lose a mark for having the variables defined wrongly? Or if I could get both as ECF?

Edit: I will elaborate for clarification:
As I said I wrongly defined the variables as 0.1L of whatever
So for part ii, it was pretty simple for me to just straight put 2b+5c <=70; instead of 0.2b + 0.5c <=7 and then multiplying by 10. So do you think I would lose or keep the method mark?

No ECF. You got the correct answer, you will get 2 marks.

Oh really? So correct answer from the wrong method (due to having the wrong variables) always gets full marks? That is interesting to know!