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# OCR (not MEI) D1 24th May 2013 Watch

1. (Original post by kabutsu12)
how many marks were the sections of 6?
IIRC it was (4) for the graph, (4) for the non-integer value, (4) for the integer value, (1) for why they changed a constraint, (2) for representing the simplex table and (6) for solving 2 iterations.
2. (Original post by JimmyA*)
Yep i did 30/6 aswell. We must have made a mistake somewhere. It meant we couldn't do the third part as i had no clue it was the LCM and HCF . I think i did awful!
Ah I see now. Don't worry mate, just focus on your upcoming exams if you have any! Hopefully you do well and the grade boundaries helps us all out a bit.
3. I spent 15 mins redoing dijkstras as I messed it up so had to redraw it on seperate paper!!
4. ****, you know what I think the 88 I was spurting on about was from the travelling salesman question. Just redrawn out the inequalities and am positive I drew out the 3x+4y equality correctly. Did others get 88 on the tsp one?
5. Answer was 56 to the simplex.Your linear programming optimal should be 56 too. As both methods are just different techniques the constraints remained the same so you should have the same answer for both, I asked my maths teacher.
6. I found the paper very pushed for time! I think the grade boundaries will probably be 51/72 for A so fingers crossed I've scraped an A!

For the algorithm question, I think I got 90 and 5. Is that right?
7. Oh, and if anyone has the paper, it would be great if you could upload it!

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8. i got 56 for LP, if you didnt know what region to use, as i couldn't see it for a while, the next question said "without the origin" so i guessed it was in that quad shape by the origin. also then got 56 for the simplex. x = 4, y =4.5

a harder paper ,yes. if you hadn't read the questions carefully it was easy to make mistakes. i think the grade boundaries will be about 52/53 for an A.

D1 (and D2) is one of those papers that if you are not well practiced will devour you.
9. that could well be where those of us who got 88 messed up. I distinctly rememember finding the vertices of a triangle, not a quad near the origin. That's really pissed me off, need to do well on this and have probs lost at least 10 marks through that mistake :'(
10. (Original post by ben494)
Yeah i got 56 as well but everyone else on here seemed to be getting 88. Glad somebody else got 56!

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I got 56 at (6,3) too!
11. (Original post by benitashrestha)
OMG THANKYOU! I THOUGHT IT WAS AN OKAY PAPER IN TERMS OF CONTENT BUT **** TIME FLEW, I ACTUALLY LOST ABOUT 10 MARKS JUST BECAUSE OF TIME BUT APPARENTLY GRADE BOUNDARIES ARE ALWAYS LOW? (sorry for caps)

Same, I am really annoyed. Honestly! I mean, I missed out the second iteration on the last question, I missed out 3 marks for one of the explain questions, I missed out drawing a cycle as I thought I'd come back to it, but I never did. Honestly, I thought I could get back to them. D1 should be 2 hours long. not one hour and a half. It's absurd because it's not truly testing mathematical ability by trying to squeeze in so much. I'm sure I could've answered it if I was actually able to pace myself without being frenzied from trying to stay within the damn time limits. I mean why do they have to make questions so long-winded.

I really needed a good mark on this, and I think I've blown it majorly.
12. A few of my answers I can remember, do you agree?

Max. order = 6, min order = 2
Eulerian Graph with 2 vertices of order 4.
Arc has to be decreased by 3km.
Also I remember getting 3 32km's for Chinese Postman.
13. (Original post by JASApplications)
I found the paper very pushed for time! I think the grade boundaries will probably be 51/72 for A so fingers crossed I've scraped an A!

For the algorithm question, I think I got 90 and 5. Is that right?

At my rate, I'll need something like 15/72 for an A.
14. (Original post by kabutsu12)
It being only 6 marks was a blessing in disguise for me!
Same - I messed up choice of pivot with a stupid plus/minus error ending up choosing 4 instead then at the last minute changed it and tried to get as much down as possible to maybe get a method mark. If grade boundaries are about the 52 mark I reckon I should be okay for an A.

5 and 90 for the algorithm question that everyone hated.
3 x 32 for the route inspection one worth 6 marks (thought that was over-generous for the amount of working unless all of us have missed a trick, compared with miserly six marks for Simplex).
The thing with 82/81 was upper and lower bounds for TSP, which wasn't too bad.
Max 6 and min 2 for Eulerian graph question. 7 nodes of 2 order and 1 node of order 6 connected to all the others except one.Stuffed up the part (c) though because I had vertices of order 1. Lolz.
ABFG (I think?) for Djikstra's, nice easy one that was I thought. Answer was 31km I'm pretty sure.

Anyone know when Mr. M or someone else will come up with an unofficial mark scheme for D1?

Best wishes for results day y'all!
15. What was the Route Inspection/Chinese postman question again, was this the landslide one? Because I swear after pairing up the odd nodes, you then had to go on and calculate the minimum distance required to walk?
16. (Original post by sj_1995)
What was the Route Inspection/Chinese postman question again, was this the landslide one? Because I swear after pairing up the odd nodes, you then had to go on and calculate the minimum distance required to walk?
Yeah it was something like 220 + 32 - (weight of deleted arcs)
17. (Original post by metaltron)
Yeah it was something like 220 + 32 - (weight of deleted arcs)
That just jogged my memory; it was 224 + 32 = 256km

18. (Original post by metaltron)
Yeah it was something like 220 + 32 - (weight of deleted arcs)
Yeah that's right! Phew, thought I had done some extra unnecessary working and potentially lost even more marks!
19. (Original post by Amos36)
That just jogged my memory; it was 224 + 32 = 256km

I think you had to delete the arc that had now been destroyed by the landslide (since the total weight would no longer be 224)
20. What did people get for that little 2 marker where you had to find by how much CE would need to be shorter for that route to be used?
4?

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