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# Edexcel A2 D2 Mathematics June 2016 - Official Thread watch

1. (Original post by grannyG)
I got a -3, can anyone agree?
In last iteration it was -2 iirc.

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2. (Original post by ltbitters)
on 6 how do u make the simplex tableau??? i just left it blank because i didn't know how to do it lmao
use your constraints from the previous part
3. (Original post by ltbitters)
on 6 how do u make the simplex tableau??? i just left it blank because i didn't know how to do it lmao
You bring all variables on side of V and add slack variables r s t u to the 4 equations to make them equal too. Then for top row in table use V p1 p2 p3 r s t u and basic values r s t u. The values in the table are the coefficients of V p1 p2 p3 r s t u in there respective equations.
4. Ok so:
Travelling salesman went well ish : somehow messed up on nearest neighbour and got length 127 for both!! No idea how hahah but some else got it - I thought I was good at this topic!
But I then got 113 and represented the inequality correctly, just got the wrong upperbound so let's say I dropped 4 (being harsh) - probably not 4 as I would get error carried forward and must have done the method slightly correct.
Question 2: Didn't find the cut but new what to do + think I got one of the capacities as 64 and not 72 so dropped 4 harshly.
Question 3: 9/9 - Hungarian was nice, awkward numbers though.
Question 4 : 8/8 - Simplex is nice, just confusing as hell with so many awkward numbers. I got 11/2 appearing each now and then.
Question 5: I got theta as 3 then 11 but I forgot to do a full second iteration of my indices again. So I had negative improvement indices values and said it wasn't optimal. Being harsh, dropped 6.
Question 6: 13/13 did so much practise on this one and new it would come up.
Question 7: I got a value of 1325 some how, however I can remember having 1700 in my last box too. Hopefully, even though I must have made errors to get to 1325, the fact that 1700 appeared could net me 4/5 marks? Maybe more. Being harsh then, I dropped 10! :O
That equates to my minimum mark as being 50 ish. But I'll go very harsh and say 45.
I found my upper mark as being around 60, but it's not realistic.
Therefore my marks are 45<my marks<60 (travelling salesman reference)
Hopefully gotten a B, which id be content with and I'd say realistically around 53 ish. If I got last question, I'd be on track to an A but didn't so oh well, no point in worrying. Not the best exam but not as bad as D1 haha, kind of on par with my fp1. Just meh exams, but it's over now so I'm going to just relax and see what happens! Worst comes to worst I resit FP1 and maybe D2 or something next year, can't worry about it. Enjoy your summer all
5. (Original post by physicsmaths)
In last iteration it was -2 iirc.

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I got -2 as well. Think a -1 one and a -2 or something? Definitely -2
6. (Original post by TH3-FL45H)

1. Classical - each vertex only visited once, practical - vertices can be revisited
Nearest neighbour lengths - 158 and 149
Lower bound = 86 (RMST) + 27 (Shortest arcs from A) = 113
113<optimal<=149

2. Initial flow = 59
C1 = 72
C2 = 86
New flow = 62, Cut = 62 so maximum due to minimum cut = maximum flow theorem

3. Alexa - 2, Ewan - 4, Faith - 1, cant remember - 3
maximum score = 248

4. Pivot row = R1
P+13....-5s=27
P=5s......
increasing s would increase P so not optimal

5. ....

6. Row minimums = ..... Row maximin = 0
Column maximums = ...... Column minimax = 4
0=/=4 so Row maximin =/= Column minimax so not stable

Let p1 be the probability with which A plays 1, p2 be the probability with which A plays 2 and p3 be the probability with which A plays 3 with p1,p2,p3>=0

Let v be the value of the original game to A
Let V=v+5 be the value of the new game to A

Maximise P-V=0

Subject to V - ...... + r = 0
V - ...... + r = 0
V - ...... + s = 0
V - ...... + t = 0
p1 + p2 + p3 + u = 1

7. Cost = 1700, 3,3,5,5,3
Profit = 5550
In question 6 wouldn't P=V+5?
7. (Original post by anonymous2410)
In question 6 wouldn't P=V+5?
No V is defined as value of game+5 so P=v

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8. (Original post by TH3-FL45H)

1. Classical - each vertex only visited once, practical - vertices can be revisited
Nearest neighbour lengths - 158 and 149
Lower bound = 86 (RMST) + 27 (Shortest arcs from A) = 113
113<optimal<=149

2. Initial flow = 59
C1 = 72
C2 = 86
New flow = 62, Cut = 62 so maximum due to minimum cut = maximum flow theorem

3. Alexa - 2, Ewan - 4, Faith - 1, cant remember - 3
maximum score = 248

4. Pivot row = R1
P+13....-5s=27
P=5s......
increasing s would increase P so not optimal

5. ....

6. Row minimums = ..... Row maximin = 0
Column maximums = ...... Column minimax = 4
0=/=4 so Row maximin =/= Column minimax so not stable

Let p1 be the probability with which A plays 1, p2 be the probability with which A plays 2 and p3 be the probability with which A plays 3 with p1,p2,p3>=0

Let v be the value of the original game to A
Let V=v+5 be the value of the new game to A

Maximise P-V=0

Subject to V - ...... + r = 0
V - ...... + r = 0
V - ...... + s = 0
V - ...... + t = 0
p1 + p2 + p3 + u = 1

7. Cost = 1700, 3,3,5,5,3
Profit = 5550
What was question 5?
9. (Original post by edmundh52)
You bring all variables on side of V and add slack variables r s t u to the 4 equations to make them equal too. Then for top row in table use V p1 p2 p3 r s t u and basic values r s t u. The values in the table are the coefficients of V p1 p2 p3 r s t u in there respective equations.
ahhh thank you !!! i left the last part as inequalities, so thats probably why i was confused :///
10. How many marks would you lose if you wrote down the answer to game theory as inequalities instead of equations? I didn't realise the question said explicitly to write them down as equalities
11. (Original post by ReeceFraser)
What was question 5?
The transportation problem. I couldnt remember my answer so i didnt put it up
12. (Original post by TH3-FL45H)

1. Classical - each vertex only visited once, practical - vertices can be revisited
Nearest neighbour lengths - 158 and 149
Lower bound = 86 (RMST) + 27 (Shortest arcs from A) = 113
113<optimal<=149

2. Initial flow = 59
C1 = 72
C2 = 86
New flow = 62, Cut = 62 so maximum due to minimum cut = maximum flow theorem

3. Alexa - 2, Ewan - 4, Faith - 1, cant remember - 3
maximum score = 248

4. Pivot row = R1
P+13....-5s=27
P=5s......
increasing s would increase P so not optimal

5. ....

6. Row minimums = ..... Row maximin = 0
Column maximums = ...... Column minimax = 4
0=/=4 so Row maximin =/= Column minimax so not stable

Let p1 be the probability with which A plays 1, p2 be the probability with which A plays 2 and p3 be the probability with which A plays 3 with p1,p2,p3>=0

Let v be the value of the original game to A
Let V=v+5 be the value of the new game to A

Maximise P-V=0

Subject to V - ...... + r = 0
V - ...... + r = 0
V - ...... + s = 0
V - ...... + t = 0
p1 + p2 + p3 + u = 1

7. Cost = 1700, 3,3,5,5,3
Profit = 5550
I got every thing the same.

For forming linear programing of game theory.
I put -v+......-r=0
-v+.....-s=0 and so on, is that fine, I basically moved V and (r,s,t) rather than p1 p2 p3
is that fine?
13. Q5-
add dummy cuz supply need to equal demand

dummy demand was 14

theta were 3 the first time and 11 the second time.

it was optimal as all improvement indices were positive
14. (Original post by dragozox)
Q5-
add dummy cuz supply need to equal demand

dummy demand was 14

theta were 3 the first time and 11 the second time.

it was optimal as all improvement indices were positive
I got that it wasnt optimal -2 for the lowest Improvement Indice in B3. This question seems to be the most split from what ive seen.
15. (Original post by dragozox)
I got every thing the same.

For forming linear programing of game theory.
I put -v+......-r=0
-v+.....-s=0 and so on, is that fine, I basically moved V and (r,s,t) rather than p1 p2 p3
is that fine?
that still makes sense right so you should get the marks

I just put V - ...... + r = 0 because thats what it was in my book

Yours should be fine
16. (Original post by dragozox)
Q5-
add dummy cuz supply need to equal demand

dummy demand was 14

theta were 3 the first time and 11 the second time.

it was optimal as all improvement indices were positive
I checked all the improvement indices in my calculator and there weren't any negative ones, but i didn't wrote them down as i was running out of space. I said it was optimal, do you i will lose marks? And how many marks was that part in total?
17. (Original post by Ramil Ahmadov)
I checked all the improvement indices in my calculator and there weren't any negative ones, but i didn't wrote them down as i was running out of space. I said it was optimal, do you i will lose marks? And how many marks was that part in total?
It was 3 marks so maybe lose 2. If you said that there are no negative improvement indices then it should only be 1 mark lost...
18. (Original post by js.int)
It was 3 marks so maybe lose 2. If you said that there are no negative improvement indices then it should only be 1 mark lost...
I am sure i said it is optimal, because there isn't any negative improvement indices.
19. (Original post by Ramil Ahmadov)
I checked all the improvement indices in my calculator and there weren't any negative ones, but i didn't wrote them down as i was running out of space. I said it was optimal, do you i will lose marks? And how many marks was that part in total?
You could have asked for extra paper
20. (Original post by Ramil Ahmadov)
I am sure i said it is optimal, because there isn't any negative improvement indices.
Then surely it's just 1 mark lost

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