OCR MEI Decision 2 (D2) 4772: 8 June 2016

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.

Doesn't this mean that the implication is wrong then?

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.

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.

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.

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.

The ones I can remember:

Decision Analysis
- Initially, EMV minimised with not getting vaccinated (18?).
- Value of questionnaire was £5.69 (18-12.31).
- 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
- y-x=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.

Surely (AnB)=>c can be written as ~(AnB)uC and hence can be written as (~A)u(~b)uC.

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 scheme-like if you know what I mean

Surely (AnB)=>c can be written as ~(AnB)uC and hence can be written as (~A)u(~b)uC.

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).

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?

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 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?

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!

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 C2-4). If I'm wrong, I'm in a little trouble.

Not that I'm aware of, it's just any 3 A2 units (ie something with a '2' or higher that's not C2-4). If I'm wrong, I'm in a little trouble.

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?!