Doesn't this mean that the implication is wrong then?(Original post by otherdan)
Attachment 546233
If you have a look at this truth table, the only time where the implies falls down is when A and B are both 1 and C is 0. I believe this was the case for current flow in the circuit too.
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OCR MEI Decision 2 (D2) 4772: 8 June 2016
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 08062016 20:28

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 08062016 20:32
(Original post by daaa)
Doesn't this mean that the implication is wrong then? 
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 08062016 20:34
(Original post by otherdan)
I don't think so, if the circuit's output is the same as the implication for the given input combinations, then it has been shown to represent that proposition. 
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 08062016 20:36
(Original post by daaa)
But if A and B are up then there is current flowing through the circuit? So the implication is 0 and the output is 1? I don't know I might just be completely wrong. 
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 08062016 20:47
(Original post by otherdan)
I think you had to make up and down equal to 0/1 a specific way around for it to work. Because I don't have the diagram I can't be sure which. 
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 08062016 20:48
(Original post by daaa)
Ohh. Well you can prove that (~An~B) => ~C. But I thought in the question up was X and down was ~X  if I missread that well then I guess that's where I went wrong but I don't know if I did. 
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 09062016 07:23
Surely (AnB)=>c can be written as ~(AnB)uC and hence can be written as (~A)u(~b)uC.

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 09062016 10:55
(Original post by otherdan)
The ones I can remember:
Decision Analysis
 Initially, EMV minimised with not getting vaccinated (18?).
 Value of questionnaire was £5.69 (1812.31).
 To minimise unpleasantness, get vaccinated (EMV depends on how you modelled unpleasantness).
Logic
 We couldn't tell whether she'd go outside. Even if it was warm, we don't know how dry it was. Even then, she might just not feel like going.
 We can deduce that headlights are off and dashboard lights are not dimmed.
 Prove using truth table/Boolean algebra.
 Show how the switching circuit represented (AnB) => C.
 We can deduce that both the ganged switches and D must be up.
LP
 yx=0 ensured that there would be at least as much expensive paint as there was less expensive paint.
 x+y>= 350 not included since the maximisation will make it as high as it possibly can anyway.
 LP solved to get x = y = 500/3, therefore total area painted = 1000/3 = 333.3.
 When increased budget to £450, there would be x = y = 187.5 therefore total area of 375.
 Not optimal since more area than needed, and he would rather use more expensive paint where possible.
 For two stage, change objective to maximise P = x, introduce x+y=350 using a <= and >= constraint combined. Using artificial and surplus variable. Minimise A to zero and then proceed to maximise P as usual.
Networks
 Route inspection = 109?
 Floyd's  do 4th iteration and show no change on 5th.
 Nearest neighbour and minimum connector  49(?) and 53(?) upper and lower bounds?
 Interpretation of Hamilton cycle in original network required going via one more node than in complete network.
It said to "prove your deduction" and so I'm wondering if my approach was correct.
I defined the variables (c=car headlights on, l = dashboard lights on, d = it is daylight, r = it is raining)
Then my next part was to turn the two statements given in the question into expressions, like this:
(d /\ c) => r (1)
l => c (2)
Then found the contrapositive of both:
(1) ~r => ~(d /\ c)
therefore (1) is ~r => ~d v ~c
(2) ~c => ~l
Using (1)  Since we knew it was not raining, ~r is true. Therefore ~d or ~c is true.
Since we know d is true, ~d is false, therefore ~c is true
Using (2)  And then of course since ~c => ~l, ~l is true.
Would that have gotten be 6/6 marks? My answer doesn't feel very mark schemelike if you know what I mean 
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 09062016 10:57
(Original post by dinobot)
Surely (AnB)=>c can be written as ~(AnB)uC and hence can be written as (~A)u(~b)uC. 
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 09062016 10:59
(Original post by ComputerMaths97)
Yo mate firstly I appreciate you doing this! Secondly, for the logic question where we must prove the headlights where not on and the dashboard lights where not dimmed:
It said to "prove your deduction" and so I'm wondering if my approach was correct.
I defined the variables (c=car headlights on, l = dashboard lights on, d = it is daylight, r = it is raining)
Then my next part was to turn the two statements given in the question into expressions, like this:
(d /\ c) => r (1)
l => c (2)
Then found the contrapositive of both:
(1) ~r => ~(d /\ c)
therefore (1) is ~r => ~d v ~c
(2) ~c => ~l
Using (1)  Since we knew it was not raining, ~r is true. Therefore ~d or ~c is true.
Since we know d is true, ~d is false, therefore ~c is true
Using (2)  And then of course since ~c => ~l, ~l is true.
Would that have gotten be 6/6 marks? My answer doesn't feel very mark schemelike if you know what I mean
How did you find the exam overall? 
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 09062016 11:01
(Original post by dinobot)
Surely (AnB)=>c can be written as ~(AnB)uC and hence can be written as (~A)u(~b)uC.(Original post by ComputerMaths97)
That's what I did! And then I said if switches are set such that down = current through, diagram shows this (or something like that). 
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 09062016 11:11
(Original post by otherdan)
That's exactly what I did (underneath a possibly useless truth table, I don't know). I know what you mean, seemed a bit too wordy to me too, but I think the logic is fine, so hopefully should be good for the marks. I'm assuming that your definition of l = dashboard lights dimmed.
How did you find the exam overall?
The EMV question I got it all, except the last 3 marker I didn't draw a decision tree, just wrote some nonsense then concluded vaccination would be better to minimise unpleasantness with no backing so 0/3 there.
The Logic question I think I got perfect, maybe lost 1 mark for the answer to the 4 marker circuit one, as I got to ~a v ~b v c and then said about the fact that the switches could be set up such that down allows current through or visa versa  but that seems a little weird to me.
The network question I'm hoping I got it all, only marks I would've lost would've been to mistakes.
But, the simplex question, I messed up. Got the first mark, definitely didn't get the next two.
For the first 7 marker I did it all well, but I think I calculated an integer solution as well as the normal solution, which will probably lose me marks. The next 4 marks of writing I hope I got 3 of.
And the last 7 marker I got completely wrong. I instead said add a x+y>=350 constraint by itself, and then minimise P=x+y. In my mind that made sense  make sure it's over 350 but minimise the waste. But I definitely got 0/7 for that as they clearly didn't want that.
So overall I'm hoping I got around 60  which tends to be a solid A which I guess will have to do.
How did you find it? 
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 09062016 11:27
(Original post by ComputerMaths97)
Yeah I'm hoping I wrote l = dashboard lights dimmed but I can't remember
The EMV question I got it all, except the last 3 marker I didn't draw a decision tree, just wrote some nonsense then concluded vaccination would be better to minimise unpleasantness with no backing so 0/3 there.
The Logic question I think I got perfect, maybe lost 1 mark for the answer to the 4 marker circuit one, as I got to ~a v ~b v c and then said about the fact that the switches could be set up such that down allows current through or visa versa  but that seems a little weird to me.
The network question I'm hoping I got it all, only marks I would've lost would've been to mistakes.
But, the simplex question, I messed up. Got the first mark, definitely didn't get the next two.
For the first 7 marker I did it all well, but I think I calculated an integer solution as well as the normal solution, which will probably lose me marks. The next 4 marks of writing I hope I got 3 of.
And the last 7 marker I got completely wrong. I instead said add a x+y>=350 constraint by itself, and then minimise P=x+y. In my mind that made sense  make sure it's over 350 but minimise the waste. But I definitely got 0/7 for that as they clearly didn't want that.
So overall I'm hoping I got around 60  which tends to be a solid A which I guess will have to do.
How did you find it?
I think I lost quite a few on logic. Perhaps got 3/6 on the 'prove your deductions' for being unclear about whether I was using a truth table/Boolean algebra. Perhaps lost all 4 for using a truth table not algebra for the switching question.
Simplex, I think all okay, except I somehow had a negative in one of my final objective rows. I knew I had a feasible solution though, and I think it's right, so probably just a small arithmetic error. Lose one or two probably.
Networks, not sure what I got wrong. Probably lost one for my comment on the last question.
So probably around 60 for me too. Looking at the A* boundaries from 2005 to 2015:
52, 65, 59, 58, 63, 53, 63, 62, 64, 51, 59
How do you think this paper ranked in terms of difficulty? 
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 09062016 11:36
(Original post by otherdan)
I think I lost one on Decision Analysis for not subtracting the EMV of the questionnaire from the EMV of not using it.
I think I lost quite a few on logic. Perhaps got 3/6 on the 'prove your deductions' for being unclear about whether I was using a truth table/Boolean algebra. Perhaps lost all 4 for using a truth table not algebra for the switching question.
Simplex, I think all okay, except I somehow had a negative in one of my final objective rows. I knew I had a feasible solution though, and I think it's right, so probably just a small arithmetic error. Lose one or two probably.
Networks, not sure what I got wrong. Probably lost one for my comment on the last question.
So probably around 60 for me too. Looking at the A* boundaries from 2005 to 2015:
52, 65, 59, 58, 63, 53, 63, 62, 64, 51, 59
How do you think this paper ranked in terms of difficulty? 
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 09062016 11:40
(Original post by ComputerMaths97)
I think it was a pretty easy paper but the 100 UMS boundary will be like 67 as there were a few marks I reckon very very few people got, so I reckon it will be A*,A,B at 62,57,52
Yeah, I was worried that people found it on the whole fairly easy too. I'm borderline A*/A there, which means I need to iron out the mistakes in S2 if I want to forget about FP2! 
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 09062016 11:42
(Original post by otherdan)
Yeah, I was worried that people found it on the whole fairly easy too. I'm borderline A*/A there, which means I need to iron out the mistakes in S2 if I want to forget about FP2!
I got 91 in M2 last year so I'm allowed a slip up this year technically, but I thought FP2 had to be one of the 3 used for the A*? 
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 09062016 11:46
(Original post by ComputerMaths97)
Doesn't FP2 have to be one of the 3 for the A* tho?
I got 91 in M2 last year so I'm allowed a slip up this year technically, but I thought FP2 had to be one of the 3 used for the A*?
Not that I'm aware of, it's just any 3 A2 units (ie something with a '2' or higher that's not C24). If I'm wrong, I'm in a little trouble. 
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 09062016 11:50
(Original post by otherdan)
Not that I'm aware of, it's just any 3 A2 units (ie something with a '2' or higher that's not C24). If I'm wrong, I'm in a little trouble. 
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 09062016 12:00
(Original post by otherdan)
Not that I'm aware of, it's just any 3 A2 units (ie something with a '2' or higher that's not C24). If I'm wrong, I'm in a little trouble.
It seems that you have to of taken FP2, but your best 3 A2 units are used. So you could have M2:90, D2:90, S2:90 and FP2:80 and still get an A* it seems?! 
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 09062016 12:05
(Original post by ComputerMaths97)
Ah it does seem like it.
It seems that you have to of taken FP2, but your best 3 A2 units are used. So you could have M2:90, D2:90, S2:90 and FP2:80 and still get an A* it seems?!
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Updated: June 15, 2016
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