D1 OCR - 19th January

the squirrel question was alright.. it was the bus question that did me in.. the question about the bus was so stupid.. this year they've gone for a cake

What was Cyril the Squirrel? Do tell!

Hmm.. it was interesting I guess, but not too bad I don't think!

1

a) managed to get 297.. with a minute to go.. mind my uselessness hehe - struggled adding 1 to C twice
b) got 0 with no problem
c) don't remember what i got!

2

a) a bit of a giveaway nice and easy hehe
b) semi-eulerian (1,2,2,3,4... 2 odds i think.. :P)
c) would form a cycle so it isnt a tree drew a pic for the examiner too!

3

long and bloody beautiful a very nice 1/3 of the paper free marks there!
100 for spanning tree
77 lower bound
126 upper bound (2 shortest from E + 77)
connect the 4 odd nodes up into 2 nice pairs of even ones, shortest being 42 and 9... thus 300 + 51 = 351 weight
other stuff, very nice question hehe

4

23 comparisons... 17 swaps... fairly 100% sure its right hehe, like a good shuttle sort
shuttle is better as its quicker (..due to less comparisons needed)

5

loved the first few parts i feel they were nice and easy for a last question
for 1 iteration i got something like P = 20, X = 5, rest = 0? it was along those lines, when discussing with peers it was right at the time anyway haha!

the graph part was interesting the 2nd hardest part after the 1st part of question 1 hehe! remembering y=2x as the last constraint, (x=4/3,y=8/3) where 4-x = 2x... as it isn't batch-able (like that new word hehe), find the most suitable (x,y) value along the top of the feasible region (y=4-x graph), giving (0,4) or (1,3)... work them out, giving £16 and £17 value of P respectively. Thus she should make a batch of plain cookies and 3 of the choc chips

...hope this helped anyone, tried my best with the questions posed here hehe

NB; these aren't right for sure, just the answers I got myself, and am fairly certain most of it is correct !

lower bound is 77 (MST without E) + 2 arcs of least weight from E (Can't remember what it was) so you forgot to add the 2 arcs of least weight from E? i got similiar for other answers apart from where i've made silly mistakes

lower bound is 77 (MST without E) + 2 arcs of least weight from E (Can't remember what it was) so you forgot to add the 2 arcs of least weight from E? i got similiar for other answers apart from where i've made silly mistakes

No problem mate.

Hopefully (if examiners mark it right) i'll get around 58 marks which hopefully should be around 80 UMS. (drop 3 on q1, 2 on q2, 0 on q3, 0 on q4, 9 on q5. = 58/72. After seeing Mr Ms mark schemes for C1/C2/C3 i think ill get around 275/300 UMS for them.. so 355/400 which means i need 125/200 on C4/S1 for A overall so pressure taken off a bit

Only 4 exams left now, 3 unimportant (A2 general studies 4,5,6), 1 important (Unit 4 physics)

I'm still confused about the last part of Q5, the graph, I am curious to see where (1,3) came from, I did think it should be that, but with little time left I couldn't work out where you could get it from, unless I drew the graph wrong of course?

I think this is what my graph sort of looked like.

x + y = 4 (0, 4) to (4, 0) with upper shaded
4x + 6y = 24 (0, 4) to (6, 0) with upper shaded
2x = y (0,0) through (1, 2), (2, 4), (3,6) etc..
and then below x-axis shaded and y-axis to the left shaded..
optimum pts, (0,0), (0,4) and (4/3, 8/3) - leaving max of them (0,4) = £16 profit?

4

23 comparisons... 17 swaps... fairly 100% sure its right hehe, like a good shuttle sort
shuttle is better as its quicker (..due to less comparisons needed)

5

loved the first few parts i feel they were nice and easy for a last question
for 1 iteration i got something like P = 20, X = 5, rest = 0? it was along those lines, when discussing with peers it was right at the time anyway haha!

the graph part was interesting the 2nd hardest part after the 1st part of question 1 hehe! remembering y=2x as the last constraint, (x=4/3,y=8/3) where 4-x = 2x... as it isn't batch-able (like that new word hehe), find the most suitable (x,y) value along the top of the feasible region (y=4-x graph), giving (0,4) or (1,3)... work them out, giving £16 and £17 value of P respectively. Thus she should make a batch of plain cookies and 3 of the choc chips

...hope this helped anyone, tried my best with the questions posed here hehe

NB; these aren't right for sure, just the answers I got myself, and am fairly certain most of it is correct !

Does anybody have any idea at all?

I'm still confused about the last part of Q5, the graph, I am curious to see where (1,3) came from, I did think it should be that, but with little time left I couldn't work out where you could get it from, unless I drew the graph wrong of course?

I think this is what my graph sort of looked like.

x + y = 4 (0, 4) to (4, 0) with upper shaded
4x + 6y = 24 (0, 4) to (6, 0) with upper shaded
2x = y (0,0) through (1, 2), (2, 4), (3,6) etc..
and then below x-axis shaded and y-axis to the left shaded..
optimum pts, (0,0), (0,4) and (4/3, 8/3) - leaving max of them (0,4) = £16 profit?

what you are missing is what was so easily misunderstood in the question. it said you were to use suitable values of x to find the max for P, when you have drawn the graph... leaving 0,4 or 1,3... trying both, it shows that 1,3 is optimum, giving £17 profit. if that helps