OCR (not MEI) D1 24th May 2013

For 5 part 3, should be a less than sign, not less than or equal to surely? The question how much shorter should it be for it to be the SHORTEST as in the only short one, that's how I interpreted it.

I interpreted as needing less that equals to because it asks for it to be a shortest path not the shortest path. They done a dirty one like this in a past paper which I got wrong, so I'm hoping for better fortune this time round.

It says for "it to become part of a shortest path" , the word "it" referring to CE.

It says for "it to become part of a shortest path" , the word "it" referring to CE.

Just put it in my pocket and walked out

work so far, will be continued if paper re-uploaded:

unofficial mark scheme

1i) $24 \ 57 \ 9 \ 31 \ 16 \ 4$

$24 \ 9 \ 57 \ 31 \ 16 \ 4$

$24 \ 9 \ 31 \ 57 \ 15 \ 4$

$24 \ 9 \ 31 \ 15 \ 57 \ 4$

$24 \ 9 \ 31 \ 15 \ 4 \ 57$

circles where necessary to indicate comparisons. (2 marks)

ii) $pass \ 2 \ : \ 9 \ 24 \ 15 \ 4 \ 31 \ 57$

$pass \ 3 \ : \ 9 \ 15 \ 4 \ 24 \ 31 \ 57$

(2 marks)

iii) pass 1 - 4 swaps, pass 2 - 3 swaps, pass 3 - 2 swaps , pass 4 - 1 swap, pass 5 -1 swap (2 marks)

2i)a) square plus a loop at one vertex for example (1 mark)

b) max vertex order for simple graph with 4 vertices is 3, but is eulerian so order has to be 2 (as 0 means one vertex is not connected). Hence total order = 4x2 = 8. 8/2 = 4, so 4 arcs, not 5 arcs. (2 marks)

ii)a) total order = 2 x no. Of arcs = 2 x 10 = 20 (1 mark)

b) max - 6, min - 2, and correct graph drawn (3 marks)

c) two vertices of order 4, six of order 2. (1 mark)

3i) correct trace through the algorithm, yielding outputs 6 and 90. (6 marks)

ii) after you fill in the space in the answer booklet, you will get f d and e the same as in part i). From this it follows that g = 18 /6 = 3 and m = 3 x 30 = 90. The same as in part i). (2 marks)

iii) f = 4 and m = 24. (2 marks)

4i) $a - b -c-d-e-f-g$
weight = 51 km
82km - 51km = 31km

so: $a-b-c-d-e-f-g-f-b-a$

with weight 82km. (4 marks)

ii) 51 - 5 - 7 + 8 = 47km.
Go back through d so:

$a-b-c-e-f-g-d-a$ (2 marks)

iii) nn starts $a - b-c-d-e-f-g$

stalls as can't return to a.

$b - a - c - d - e -f-g-b$

weight = 76km.

$b - a - c - d - e - g - f - b$

weight = 69km. (6 marks)

5i) $a - b - f - g$

weight = 31km. (5 marks)

ii) 25 hours (2 marks).

Iii) $11 + 15 + ce \leq 31$

$ce \leq 5$

8 - 5 = 3km (2 marks)

iv) odd nodes a,b,c and f.
All repeated arc lengths are 32km.
32 + 224 - 11 - 8 = 237km. (6 marks)

6i) attached using wolfram (4 marks)

ii) $(6,3) , \ (4,0), \ (3,0) \ and \ (4,4.5)$

at (6,3) p = 54
at (4,0) p = 20
at (0,3) p = 24
at (4,4.5) p = 56.

So at optimal point x = 4, y = 4.5, p = 56. (4 marks)

iii) when x = 0, max y = 3, p = 24.
When x = 1, max y = 3, p = 29
when x = 2, max y = 3, p =34
when x = 3, max y = 4, p = 47
when x = 4, max y = 4, p = 52
when x = 5, max y = 3, p = 49
when x = 6, max y = 3, p = 54.

So optimal point is when x = 6, y = 3 and p = 54. (4 marks)

iv) simplex is designed for inequalities with a less than or equals sign. (1 mark)

v) correct simplex tableau (2 marks)

vi) first pivot 8 in y-column.
Show how rows calculated.
P = 24, x = 0, y = 3.
Pivot is 4.5 in x-column.
Verify that optimum point is when x =4, y = 4.5 and p = 56.

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.

Simplex should have given you (4,4.5) which corresponded to P=56. This was the same as in 6(i) and (ii).

For 5 part 3, should be a less than sign, not less than or equal to surely? The question how much shorter should it be for it to be the SHORTEST as in the only short one, that's how I interpreted it.

thought the final pass of bubble sort should have 0 swap?

Hi everyone. I also found the timing impossible for that exam. Q3 where you had to carry out the algorithm took forever. And how did everyone do Q4i/ii where you had to get 81 and 82? I knew you had to take away 51, but then what?...

For all the questions 4 and 5 I didn't put km i just wrote the numbers. Will i lose marks for that?

Pass 4: 9 4 15 24 ... 1 swap

Pass 5: 4 9 15 24 ... 1 swap

i must have misread the question and thought it meant the fifth pass was the final pass

will you get follow through marks for 6ii and 6iii if you draw the graph for 6i wrong? if not, i would have lost 12 marks on that question.... great

48/72 for 80 UMS I think

I think it wud be about 54/55 for an A