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D1 (Decision 1) 17 May 2013 Official Thread

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Original post by Fortitude
I've always thought it was first fit.


Huh, why do you have to do 2 types? I'm so confused :confused:
Original post by posthumus
I don't quite get what you mean ?


He meant that after you've found some full bins what do you do with the rest of the numbers, do you apply first-fit or first-fit decreasing. :smile:
Original post by posthumus
Huh, why do you have to do 2 types? I'm so confused :confused:


You don't do two types, usually or as we've learnt you just apply the first-fit but in 1 paper it seems like they did first-fit decreasing but if you see one of my previous posts I don't think it matters. I'd just stick w/ first-fit though. :smile:
Original post by Fortitude
He meant that after you've found some full bins what do you do with the rest of the numbers, do you apply first-fit or first-fit decreasing. :smile:


Oh how sillly of me :tongue: damn it, I haven't come across full bin packing yet in past papers... but would have thought first fit

Oh and thanks !
Reply 404
Original post by Fortitude
I've looked at the mark scheme again & I think we will get the mark because 2 marks are for the 2 full bins & then it states "2A1: A 4 bin solution found." so yeah I don't think it really matters or else they would have specified. :smile:


Perfect cheers.

Just got 68/75 on June 2009 so very happy 1 more A*! Only just but right now I am being super strict with my Marking, Removing ALL EASY ERRORS!!!!

Hahaha looking forward to this Exam.

Then onto revision for my other Exams.
Reply 405
Original post by Fortitude
You don't do two types, usually or as we've learnt you just apply the first-fit but in 1 paper it seems like they did first-fit decreasing but if you see one of my previous posts I don't think it matters. I'd just stick w/ first-fit though. :smile:


Fortitude one last question.

What is the definition of a Graph ironically it's the only definition I'm confused about now.

2009 Papers are odd. I just want to get it confirmed.
Reply 406
I find it funny how suddenly all the Crammers have jumped to the Forum Loooool.

There is no way you can Cram D1. C3 and C4 if you are super focused. But not D1.

D1 Boundaries.PNG
D1 Boundaries.PNG
And I'm not saying that just to be an idiot. See the D1 June Grade Boundaries, they are on average lower than the January Exams.

People who sit it in January are usually more prepared.

There is at times a 5-6 Mark Difference which is huge.
(edited 10 years ago)
Original post by Better
I find it funny how suddenly all the Crammers have jumped to the Forum Loooool.

There is no way you can Cram D1. C3 and C4 if you are super focused. But not D1.


What do you mean by a crammer? Also definition of a graph I use:
A finite amount of verticies/nodes which are connected by edges/arcs


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Reply 408
Original post by Westeros
What do you mean by a crammer? Also definition of a graph I use:
A finite amount of verticies/nodes which are connected by edges/arcs


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That's what I do too but it's not correct in the Jan 2009 Markscheme.
(edited 10 years ago)
Original post by Better
That's what I do too but it's not correct in the Jan 2009 Markscheme.


I don't see a question asking the definition of a graph in the Jan 09 Paper? Can you help me with this linear programming question
ImageUploadedByStudent Room1368649595.970097.jpg
I need to Minimise C = 500x+800y
how do I do this?
(Summer 2010 Paper Q7)


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Reply 410
Original post by Westeros
I don't see a question asking the definition of a graph in the Jan 09 Paper? Can you help me with this linear programming question
ImageUploadedByStudent Room1368649595.970097.jpg
I need to Minimise C = 500x+800y
how do I do this?
(Summer 2010 Paper Q7)


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I could be wrong because I haven't had done that Paper in a while now; but I don't remember any of my Feasible Regions looking like that.

Secondly your best of writing in Pencil.

Thirdly just use your Objective Line ---- and stop at the earliest point you hit.

But anyway time for me to do Physics. Don't think I'll go to school tomorrow, they go way to slowly in Chemistry and Physics. It is annoying. Will stay in and do D1 and Chemistry revision!
Reply 411



How do you do part b? I tried log6/log2 but that doesnt round to 15 :/
Reply 412
Original post by Westeros
I don't see a question asking the definition of a graph in the Jan 09 Paper? Can you help me with this linear programming question
ImageUploadedByStudent Room1368649595.970097.jpg
I need to Minimise C = 500x+800y
how do I do this?
(Summer 2010 Paper Q7)


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You are minimising so you are looking for the first intersection if you do the ruler method.
Original post by Better
I could be wrong because I haven't had done that Paper in a while now; but I don't remember any of my Feasible Regions looking like that.

Secondly your best of writing in Pencil.

Thirdly just use your Objective Line ---- and stop at the earliest point you hit.

But anyway time for me to do Physics. Don't think I'll go to school tomorrow, they go way to slowly in Chemistry and Physics. It is annoying. Will stay in and do D1 and Chemistry revision!


I checked the mark scheme and my graph is correct, so if i use the ruler method, the first point I hit is (11.2,6) and this works out £10400, but the mark scheme says (11,7) with a cost of £11100
AH! I hate Linear Programming.
Good Luck with your revision :frown: If you can look at this question tomorrow it would be much appreciated :colondollar:
Original post by Westeros
I don't see a question asking the definition of a graph in the Jan 09 Paper? Can you help me with this linear programming question
ImageUploadedByStudent Room1368649595.970097.jpg
I need to Minimise C = 500x+800y
how do I do this?
(Summer 2010 Paper Q7)


Posted from TSR Mobile


You have to move the objective line across and see which point it hits first (minimum)... you can achieve this by first aligning your ruler with the objective line then moving it across, the idea is to keep the gradient the same :smile: There's a technique where you can also use a set square to help you keep the rule consistent with objective lines gradient.

However like in most cases I think here it's obvious that the "minimum" point is where the lines y=6 & 5x+4y=80 intersect :smile:

You can find the x & y values by using simultaneous equations technique. When you have the co-ordinates, x & y .... sub them into your objective function :smile:
Original post by QwertyG



How do you do part b? I tried log6/log2 but that doesnt round to 15 :/


No you use the log thingy for finding how many iterations in binary search :smile:

In this case use this: n/2 (n-1)

therefore 3 x 5 = 15
Original post by posthumus
No you use the log thingy for finding how many iterations in binary search :smile:

In this case use this: n/2 (n-1)

therefore 3 x 5 = 15


Ive never been taught the log thingy, what is it? :tongue:
Original post by Nitrogen
Ive never been taught the log thingy, what is it? :tongue:


To find the number of iterations in binary search, it came up in January (had no bloody clue how to do it), so I really don't think it will come up again.

the most common way is actually to half it, and see how many times you half it... :smile:

Another way is to use this

2n = x

where n is the number of iterations and x is the number of variables :smile: In QwertyG's case if it were binary search:

2n = 6

therefore : n = log26
which equals = 2.6

therefore 3 iterations are possible... I'll use the common method to confirm this:

6/2 = 3
3/2 = [1.5] = 2
2/2= 1
Can I just clarify that Hamilton cycles are not in this spec?????


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Original post by Flounder1
Can I just clarify that Hamilton cycles are not in this spec?????


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Nope, it's not ! :smile:

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