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    (Original post by theseeker)
    I did May 2009 paper and just wanted to ask a question..for question 3, Bipartitate graph..the c part..starting from E, does it have to go all the way to 6 like they did in the mark scheme or can I do

    E-5=r-4 then cs

    then from h-1=s-3=c-6
    then you get the same answer...?

    http://www.mathsgeeks.co.uk/pdf/C2Ed...elJune09MS.pdf
    Would have helped if the link works

    I think your going wrong because your looking at figure 2 & not at what you got in part a)
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    (Original post by otrivine)
    you mean the solomon, the questions the way they are structured is quite different in our text book questions and edexcel papers/
    Yh, but you know that question we had in January... the first question was horrible, do you think that flow stuff could come up again ?? :confused:
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    (Original post by posthumus)
    Yh, but you know that question we had in January... the first question was horrible, do you think that flow stuff could come up again ?? :confused:
    I doubt flow chart could come ,cause they asked in Jan 2013, most probably the first question will be on bubble/quick sort/ bin packing or kruscals/prims
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    (Original post by posthumus)
    Would have helped if the link works

    I think your going wrong because your looking at figure 2 & not at what you got in part a)
    Just realised. Thanks
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    (Original post by theseeker)
    Just realised. Thanks
    Thought so... because I think I did the same thing in January I was sitting there puzzled !! & so didn't do the question
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    (Original post by Epic Flawless)
    Of course, they are!

    I'm worried about making a mistake in the exam, because I know that would be deadly in decision since the following questions always stem from the first (e.g inspection, CPA, Dijkstra's). Especially because I was making silly mistakes in the mechanics exam, but fortunately, I had enough time to check, idenify and correct them.
    Not sure if this is what you mean but I heard if you misread the questions the examiners knock off 2 marks and then mark the question as normal??


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    I'm stuck on part e) Why wouldn't reducing activity "I" not make it profitable?? Name:  ImageUploadedByStudent Room1368627074.993265.jpg
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Size:  166.3 KBName:  ImageUploadedByStudent Room1368627089.564207.jpg
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    Thanks in advance


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    (Original post by posthumus)
    I'm stuck on part e) Why wouldn't reducing activity "I" not make it profitable?? Name:  ImageUploadedByStudent Room1368627074.993265.jpg
Views: 87
Size:  166.3 KBName:  ImageUploadedByStudent Room1368627089.564207.jpg
Views: 80
Size:  160.4 KB

    Thanks in advance


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    can you explain to me part d) ?
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    (Original post by otrivine)
    can you explain to me part d) ?
    For each day day the project is delayed you get £50,000 fine

    now activity F is delayed by 3 days... since its critical, the whole project is delayed by 3 days

    therefore fine = 3 x 50,000 = £150,000

    If you reduce N by 1 day, since it's critical you'll save 1 day on whole project, so overall it will only be 2 days delayed... so your fine would be 2 x 50,000 = £100,000

    saving you £50,000

    however to reduce activity N by 1 day it costs £90,000

    therefore your savings are outweighed, and you make a net loss of 90,000 - 50,000 = £40,000

    So your £40,000 better off if you just delay the project by 3 days
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    (Original post by posthumus)
    For each day day the project is delayed you get £50,000 fine

    now activity F is delayed by 3 days... since its critical, the whole project is delayed by 3 days

    therefore fine = 3 x 50,000 = £150,000

    If you reduce N by 1 day, since it's critical you'll save 1 day on whole project, so overall it will only be 2 days delayed... so your fine would be 2 x 50,000 = £100,000

    saving you £50,000

    however to reduce activity N by 1 day it costs £90,000

    therefore your savings are outweighed, and you make a net loss of 90,000 - 50,000 = £40,000

    So your £40,000 better off if you just delay the project by 3 days
    Thanks but how did activity N become 2 days , dont quite follow that please
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    when you are doing doing a binary search for names, it is important to write out the full names of the people involved? for example could you write...:

    1.) A
    2.) C
    3.) E

    etc rather than

    1.) Amy
    2.) Charlie
    3.) Elliot

    because writing it out in full would be a waste of time but then i dont want to run the risk of dropping any marks :/
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    (Original post by mbisababe)
    when you are doing doing a binary search for names, it is important to write out the full names of the people involved? for example could you write...:

    1.) A
    2.) C
    3.) E

    etc rather than

    1.) Amy
    2.) Charlie
    3.) Elliot

    because writing it out in full would be a waste of time but then i dont want to run the risk of dropping any marks :/
    Yes that is fine. or you can use numbers such as reject 1-3 ,
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    (Original post by otrivine)
    Thanks but how did activity N become 2 days , dont quite follow that please
    Oh sorry it will only reduce the project by 1 day, since it's critical path.....

    .... so all other days being reduced is pointless, because it's not reducing overall project time

    I don't think we should worry about the kind of Solomon Questions in all honesty

    The Linear programming one was worse though, in our spec does it suggest we will always get linear programming question with only 2 variables? (as apposed to 3 x,y,z we got in solomon B)
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    (Original post by otrivine)
    Yes that is fine. or you can use numbers such as reject 1-3 ,
    thats great thankyou!
    you know when you find the middle value, does it matter how you do it? everyone i know tends to do it a different way....
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    (Original post by mbisababe)
    when you are doing doing a binary search for names, it is important to write out the full names of the people involved? for example could you write...:

    1.) A
    2.) C
    3.) E

    etc rather than

    1.) Amy
    2.) Charlie
    3.) Elliot

    because writing it out in full would be a waste of time but then i dont want to run the risk of dropping any marks :/
    No they don't expect you to.... that's why you'll never find more than one person with a name starting with the same letter
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    I'm resitting D1 from last year.. but still don't understand linear programming at all! Its probably easy but i can't get my head round it at all.. how do you find out where they objective line is if you use that method? in the mark schemes they use specific points but i don't understand how to get them?
    And also when you are asked to find interger optimal points how do you do that? Is it just from looking at the graph?
    Thanks!
    Good luck in the exam everyone!
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    (Original post by a_m_y)
    I'm resitting D1 from last year.. but still don't understand linear programming at all! Its probably easy but i can't get my head round it at all.. how do you find out where they objective line is if you use that method? in the mark schemes they use specific points but i don't understand how to get them?
    And also when you are asked to find interger optimal points how do you do that? Is it just from looking at the graph?
    Thanks!
    Good luck in the exam everyone!
    When you have the objective function make it equal to something appropriate so it can be drawn onto your graph the reason you can choose what it's equal to, is because you only want the gradient

    Say you have 3x + 2y

    and your scale goes in 10s' all the way up to 50 on the x & y axis
    You could go for 60 which is a nice number to choose

    x=20 & y=30

    mark those on the axis and draw a line through em, you have your objective line. Align your rule with this and move it across:

    The first point your ruler touches is the minimum point, it's possible to read this off the graph, but usually you have to do simultaneous equations to find the point's co-ordinates.

    The latest point your ruler touches is the maximum point

    For integer value - it's again possible to see why is in the feasible region. But you can test various points you have and put them in your 2 inequality equations to see if they make sense.

    So say I am testing x=2.4 & y=6.2 ... in the inequalities:
    2x + y < 0 & 0< x + y

    Though the second one gives me 0<8.6, the first inequality gives me 15<0 ..which obviously doesn't make sense, therefore the point is not in the feasible region and so on.....

    I find this tedious and so long... so I hope we don't have to resort to this method in the exam
    • Welcome Squad
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    Welcome Squad
    (Original post by a_m_y)
    I'm resitting D1 from last year.. but still don't understand linear programming at all! Its probably easy but i can't get my head round it at all.. how do you find out where they objective line is if you use that method? in the mark schemes they use specific points but i don't understand how to get them?
    And also when you are asked to find interger optimal points how do you do that? Is it just from looking at the graph?
    Thanks!
    Good luck in the exam everyone!
    Hi, to plot the obj line: use the obj function so for eg if it's 3x + 2y, just choose any suitable number that goes into both 3 and 2 so here it could be 6, 60, 18, 600, 1200, just any number so lets choose 600 now 3x + 2y = 600 therefore you'd plot ( 0, 300 ) & (200, 0) & join to make a straight line. Obv the no you choose will depend on the scale on the axis as it would be no good choosing 600 if the axis only goes up to 50. So yeah that's just it because even though everyone's numbers will be diff they will still be multiples of each other. Now with finding the optimal point, the best way is to use simultaneous equations as the OP will always be the intersection point of 2 lines so use simultaneous equations to work out x & y as that will be more accurate than reading off your graph. In terms of integer solutions, I'm not entirely sure myself, I first thought that you just rounded the numbers but don't think that's right.
    • Welcome Squad
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    Welcome Squad
    (Original post by posthumus)
    When you have the objective function make it equal to something appropriate so it can be drawn onto your graph the reason you can choose what it's equal to, is because you only want the gradient

    Say you have 3x + 2y

    and your scale goes in 10s' all the way up to 50 on the x & y axis
    You could go for 60 which is a nice number to choose

    x=20 & y=30

    mark those on the axis and draw a line through em, you have your objective line. Align your rule with this and more it across:

    The first point your rule touches is the minimum point, it's possible to read this off the graph, but usually you have to do simultaneous equations to find the points co-ordinates.

    The latest point your ruler touches is the maximum point

    For interger value - it's again possible to see why is in the feasible region. But you can test various points you have and put them in your 2 inequality equations to see if they make sense.

    So say I am testing x=2.4 & y=6.2 ... in the inequalities:
    2x + y < 0 & 0< x + y

    Though the second one gives me 0<8.6, the first inequality gives me 15<0 ..which obviously doesn't make sense, therefore the point is not in the feasible region and so on.....

    I find this tedious and so long... so I hope we don't have to resort to this method in the exam
    Ha ha your way of explaining was way better than mine & I chose 3x + 2y as well but for integer solutions there must be a quicker way, there has to be!
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    (Original post by posthumus)
    Oh sorry it will only reduce the project by 1 day, since it's critical path.....

    .... so all other days being reduced is pointless, because it's not reducing overall project time

    I don't think we should worry about the kind of Solomon Questions in all honesty

    The Linear programming one was worse though, in our spec does it suggest we will always get linear programming question with only 2 variables? (as apposed to 3 x,y,z we got in solomon B)
    yay
    this is our syllabus for linear
    5 Linear programming
    What students need to learn:
    Formulation of problems as linear programs.
    Graphical solution of two variable problems using
    ruler and vertex methods
    Consideration of problems where solutions must
    have integer values

    we only have 2 variables not 3
 
 
 
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