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    couldnt you do

    D-2 = a - 5 = m - 1 (alternate)

    D=2 A=5 M=1 (changed)
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    (Original post by QwertyG)


    for part b why does the mark scheme only start at J and not D?
    Dont worry, in the exam they have lots of choices for matchings , you can start at either J or D they are both correct because you start with an unmatched node
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    (Original post by Fortitude)
    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
    thank you for replying
    I still don't get why S-P wouldn't be the next smallest vertex in the uncrossed rows after i've labelled the oxford column 2?
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    (Original post by Priya08)
    couldnt you do

    D-2 = a - 5 = m - 1 (alternate)

    D=2 A=5 M=1 (changed)
    M can't do activity 1?

    (Original post by posthumus)
    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.


    http://www.examsolutions.net/a-level...estions/q8.gif

    Here is the question, it specifically asks for the lower bound and the mark scheme says 4 :mad:

    http://www.edexcel.com/migrationdocu...c_20100618.pdf
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    (Original post by Priya08)
    couldnt you do

    D-2 = a - 5 = m - 1 (alternate)

    D=2 A=5 M=1 (changed)
    that is correct but be careful with your changed status

    it should be D=2-A=5-M=1

    remember the - sign
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    (Original post by QwertyG)


    for part b why does the mark scheme only start at J and not D?
    Is this January 2013?

    If it is... then damn this question got me!


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    (Original post by otrivine)
    Dont worry, in the exam they have lots of choices for matchings , you can start at either J or D they are both correct because you start with an unmatched node
    All the alternative paths in the mark scheme don't start with D They must start with J because if you try starting with D, you end up back where you started
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    (Original post by posthumus)
    Is this January 2013?

    If it is... then damn this question got me!


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    Jan 11
    • Welcome Squad
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    Welcome Squad
    (Original post by dh1995)
    thank you for replying
    I still don't get why S-P wouldn't be the next smallest vertex in the uncrossed rows after i've labelled the oxford column 2?
    Sorry I forgot to mention when looking for the next smallest vertex, you have to look in the numbered columns.

    So the next smallest would be 60 (CL)
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    (Original post by Westeros)
    All the alternative paths in the mark scheme don't start with D They must start with J because if you try starting with D, you end up back where you started
    oh I see are you trying to say if you used D you would get a complete mathcing for part b) because we dont want a complete matching in part b we use j to get D to be unmathced so a complete mathcing is not possible right?
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    (Original post by Fortitude)
    Sorry I forgot to mention when looking for the next smallest vertex, you have to look in the numbered columns.

    So the next smallest would be 60 (CL)
    thank you very much!
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    (Original post by otrivine)
    oh I see are you trying to say if you used D you would get a complete mathcing for part b) because we dont want a complete matching in part b we use j to get D to be unmathced so a complete mathcing is not possible right?
    Nope!
    If you use D, you end up returning back to a number/activity you have already used, like I said in my earlier post, try attempting using D for part b) you'd either get

    D-2=A-5=M-2 <<<<<<<returned back to 2, when we used D-2 at start
    or
    D-5=M-2=A-5 <<<<<<<returned back to 5, when we used D-5 at start
    you understand?
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    (Original post by Westeros)
    Nope!
    If you use D, you end up returning back to a number/activity you have already used, like I said in my earlier post, try attempting using D for part b) you'd either get

    D-2=A-5=M-2 <<<<<<<returned back to 2, when we used D-2 at start
    or
    D-5=M-2=A-5 <<<<<<<returned back to 5, when we used D-5 at start
    you understand?
    But D-2 = a - 5 = m - 1

    the person Priya she made a mistake above because it should be 2 not 1 , oh and yes I get it forms like a cycle and you cannot get out of the cycle to arrive to the unmathced node on the RHS
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    (Original post by otrivine)
    But D-2 = a - 5 = m - 1

    the person Priya she made a mistake above because it should be 2 not 1 , oh and yes I get it forms like a cycle and you cannot get out of the cycle to arrive to the unmathced node on the RHS
    Yeah, M cannot be matched to 1, so therefore M would be matched to 2, but we already used 2 at the start :-) that's why for all possible matchings in part b, we must start at J
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    (Original post by Westeros)
    Yeah, M cannot be matched to 1, so therefore M would be matched to 2, but we already used 2 at the start :-) that's why for all possible matchings in part b, we must start at J
    I see very clever question
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    also has anyone managed to undertstand this question in the book , question 12 on page 165 about the oxford and york
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    I usually skip all the part b's of the chinese postman questions - terrible habit i know!

    i need to learn how to do it now but are they usually the same (repeated questions)
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    (Original post by Westeros)
    M can't do activity 1?



    [/B]http://www.examsolutions.net/a-level...estions/q8.gif

    Here is the question, it specifically asks for the lower bound and the mark scheme says 4 :mad:

    http://www.edexcel.com/migrationdocu...c_20100618.pdf
    Sorry I'm going to have a look now !
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    (Original post by otrivine)
    I crossed it out to show the examiner that I am choosing the smaller value but very weirdly in the mark scheme they leave it, I would suggest you not to show any cancelling to be on the safe side

    (Original post by tobywalsh)
    My teacher tells me that you should always write every working value you obtain - even if it is bigger than your lowest working value, just put it in brackets... .

    Imagine you get 26 then 23 then 24, you would write 26 [space] 23 [space] then (24) - note the brackets... .

    Don't give them an excuse to take a mark off you!!

    (Original post by otrivine)
    if its bigger i usually show a cross that this is not the smallest

    (Original post by tobywalsh)
    Thank you! And also, you're right about the cross for the larger working values, I have done that before in a mock and it worked out! As long as you make it stand out I doubt it'll be a problem including it... .

    Toby

    (Original post by IWantSomeMushu)
    Huh, I was taught not to put them in.

    I'll do it from now on then and put them in brackets.
    I teach my students not to put them in at all. Having just checked up on several students who did D1 last summer, they all got full marks on the Dijkstra question, so not putting the larger values in is fine.
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    quick question when drawing the activity network, where you have 2 activties sharing the same start vertex and end vertex because of uniquely represented, you use a dummy but does it matter which one out of those 2 activities to put the dummies?
 
 
 
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