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    (Original post by Fortitude)
    Hmmm, well here I am thinking there must be a quicker way I should seriously give up finding quicker ways to do D1 ...because so far there aren't any
    It's just practice really ! I need to time myself tonight with at least 2 past papers
    But I can't see how anyone can finish more than 5 minutes early in that exam regardless of how much they've prepared :confused: Like in M1 on Monday I finished 30 minutes early and so there wasn't so much pressure.... D1 there'll probably be a point where I'm rushing/writing desperately due to time can't wait until it's over !
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    If your doing Prims from a table, do you list the smallest first and consider them in that order?
    Because on the january 10 paper it says to start with London and then they put the S-P one second last on their list. Can someone explain to me why this is?


    http://www.edexcel.com/migrationdocu...c_20100219.pdf

    http://www.edexcel.com/migrationdocu...e_20100115.pdf
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    Any last minute tips? I've only got the may 2012 to do out of actual past papers- probs should go over definitions- sigh
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    do we need to bother learning about the triangle inequality (just seen a bit in the textbook on it)
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    in the mark schemes you know for like prims and kruscals algorithms when your writing the arcs out, in the mark scheme some arcs are like in brackets

    e.g. [ (AC) (FG) ] does it mean u must have it both or either one of them
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    (Original post by Priya08)
    in the mark schemes you know for like prims and kruscals algorithms when your writing the arcs out, in the mark scheme some arcs are like in brackets

    e.g. [ (AC) (FG) ] does it mean u must have it both or either one of them
    either one
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    okay cheers
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    How many marks do you think will be awarded for the Linear Programming question? I find it kinda confusing. Also i REALLY hope scheduling and drawing graphs from precedence tables doesnt come up. How likely do you reckon these topics will come up? (obviously LP will come up im talking about the others)
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    Welcome Squad
    (Original post by dh1995)
    If your doing Prims from a table, do you list the smallest first and consider them in that order?
    Because on the january 10 paper it says to start with London and then they put the S-P one second last on their list. Can someone explain to me why this is?


    http://www.edexcel.com/migrationdocu...c_20100219.pdf

    http://www.edexcel.com/migrationdocu...e_20100115.pdf
    Hi, what you do is you cross out the row with London in & label the London column w/ 1 & then you see which is the smallest no in that column which is 56 so you circle that & draw that arc so draw OL this is your first vertex, then cross the Oxford row out & label the Oxford column 2. Find the next smallest vertex in the uncrossed rows, circle ...repeat until all rows are crossed
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    Can someone explain to me how to do scheduling please?
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    For route inspection do I have to show various possible routes I can take and then mark with a * which one is the shortest ? What if I write 1 or 2 & I already know which one is going to be shorter

    If so then how many possible routes do I need to show ?

    Thanks in advance
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    (Original post by posthumus)
    You don't need to

    Just find the amount of time all tasks will take

    then divide by the time allocated for the project... which looking at your gantt chart I assume it 22

    & I think it's pretty neat
    That would be 59/22 which works out 3 workers, but the mark scheme have chosen a point, e.g at time X.. so many activities must be happening (and the answer was 4 if I remember correctly) is there a proper way of doing this?
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    (Original post by posthumus)
    For route inspection do I have to show various possible routes I can take and then mark with a * which one is the shortest ? What if I write 1 or 2 & I already know which one is going to be shorter

    If so then how many possible routes do I need to show ?

    Thanks in advance
    For Route Inspection, once you've found all your odd nodes, you have to write down ALL possible pairings - since you get 'A1' marks which are accuracy marks, normally it'll be 4 odd nodes, 3 possible pairings, so even if you know which one is going to be shortest, I'd still write down the others!
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    Good Luck everyone!!
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    for part b why does the mark scheme only start at J and not D?
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    If you try starting with D, some possible solutions would go like:
    D-2=1-5=M-2
    Here we end up back at 2, and we used 2 previously, so you can't use that solution.

    Then try again for example:
    D-5=M-2=A-5 For the last part, A could go to 5 or 2, both which we have used in the previous part. So therefore we must start at J!
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    (Original post by QwertyG)


    for part b why does the mark scheme only start at J and not D?
    start anywhere, it doesn't matter
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    I forgot to quote you! Hope you understand my explanation :-)

    (Original post by Westeros)
    If you try starting with D, some possible solutions would go like:
    D-2=1-5=M-2
    Here we end up back at 2, and we used 2 previously, so you can't use that solution.

    Then try again for example:
    D-5=M-2=A-5 For the last part, A could go to 5 or 2, both which we have used in the previous part. So therefore we must start at J!
    (Original post by QwertyG)


    for part b why does the mark scheme only start at J and not D?
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    yhhh i started from J and did the whole question and i got the whole of par c right but in the markscheme for some weird reason it doesnt have any suggestions from starting from D
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    (Original post by Westeros)
    That would be 59/22 which works out 3 workers, but the mark scheme have chosen a point, e.g at time X.. so many activities must be happening (and the answer was 4 if I remember correctly) is there a proper way of doing this?
    I couldn't find the question in summer 2010, but I think I know what you mean....

    That isn't the lower bound then The lower bound is 3 workers.

    The way you can figure out how many workers are needed at a specific time is by possibly drawing a line (to make things easier). It's similar to the question where they ask how many activities should be happening at a particular time.

    Once you've found that you will need 4 at this particular time, this show you that your going to need more workers than the lower bound. The value 4 is not the lower bound itself.
 
 
 
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