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    (Original post by Windowswind123)
    How about lines that cross over each other?
    Personally I'd avoid that if possible, as your diagram could become very messy very quickly and you want it to be as easy to mark as possible. :P

    Does anyone know of any particularly difficult questions/papers?
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    For scheduling and Gantt charts, why is it that sometimes the activities are added to the next row like this on Q7(e) on June 2013
    https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

    and then sometimes they immediately follow another activity like on Q7(d) on june 2015
    https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

    Is it because the question on june 2015 asks for the minimum number of workers to be used whereas the question on june 2013 just asks for the chart to be completed?
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    (Original post by kennz)
    For scheduling and Gantt charts, why is it that sometimes the activities are added to the next row like this on Q7(e) on June 2013
    https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

    and then sometimes they immediately follow another activity like on Q7(d) on june 2015
    https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

    Is it because the question on june 2015 asks for the minimum number of workers to be used whereas the question on june 2013 just asks for the chart to be completed?
    NO,
    the June 2013 is asking for a cascade/gantt chart. For this each activity MUST be on a new row (other than the critical activities which go side by side as they make the critical time)

    The June 2015 is asking for scheduling which means you have to complete it with a few as possible workers and each row represents 1 worker and which activities they should do. This also has to be done in the critical time.
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    (Original post by 4nonymous)
    NO,
    the June 2013 is asking for a cascade/gantt chart. For this each activity MUST be on a new row (other than the critical activities which go side by side as they make the critical time)

    The June 2015 is asking for scheduling which means you have to complete it with a few as possible workers and each row represents 1 worker and which activities they should do. This also has to be done in the critical time.
    Ok thank you!
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    Name:  ImageUploadedByStudent Room1465676272.656807.jpg
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    Considering G. From the route, it indicates that the latest start time for G is 12. However, since its duration is only 1 and its latest finish time is 14, surely it can start at 13?


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    (Original post by Nikhilm)
    Name:  ImageUploadedByStudent Room1465676272.656807.jpg
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    Considering G. From the route, it indicates that the latest start time for G is 12. However, since its duration is only 1 and its latest finish time is 14, surely it can start at 13?


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    Name:  ImageUploadedByStudent Room1465676364.396606.jpg
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    Sorry here's a better quality image


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    Also, could there ever be a case where, in matching, set x and set y don't have the same number of nodes?
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    (Original post by Nikhilm)
    Also, could there ever be a case where, in matching, set x and set y don't have the same number of nodes?
    Yes, but a complete matching wouldn't be possible.
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    (Original post by NotNotBatman)
    Yes, but a complete matching wouldn't be possible.
    Ah thanks. Do you mind also answering my question above when free ^
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    Please could anyone explain question 4b, how would you use F as the starting node? Thanks Name:  image.jpg
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    (Original post by Nikhilm)
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    Sorry here's a better quality image


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    I'm sure it should be 13.
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    (Original post by NotNotBatman)
    I'm sure it should be 13.
    It's not. You're welcome to try the question if need be
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    (Original post by Nikhilm)
    It's not. You're welcome to try the question if need be
    I see. The latest time you can leave the node is 12 because F is dependent on C. If you left the node at 13, then it would delay F.
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    (Original post by economicss)
    Please could anyone explain question 4b, how would you use F as the starting node? Thanks Name:  image.jpg
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    Shortest from F to T, then shortest from S to F.
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    https://d0247e65b1a6ead8e6286cd43803...%20Edexcel.pdf

    Hey guys, for q7b here, for the y less tahn or qual to 4x constraint. Is it ok to say, 'Rose must make 4 large baskets for every small basket she makes' or is this worng?
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    (Original post by Nikhilm)
    Name:  ImageUploadedByStudent Room1465676364.396606.jpg
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    Sorry here's a better quality image


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    (Original post by NotNotBatman)
    I'm sure it should be 13.
    You must use the dummy activity (dotty arrow) when working back through the late event times.
    i.e. take the minimum of 14 - 1 and 12 - 0
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    (Original post by NotNotBatman)
    I see. The latest time you can leave the node is 12 because F is dependent on C. If you left the node at 13, then it would delay F.
    Ah yeah. If you consider those boxes in terms of C's earliest and latest finish time then it must finish by 12, but when drawing a scheduling, surely it's still okay to start G at 13 since F is not dependent on that - i.e. this is just a notional thing bc it has to consider C's latest time
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    Could someone help me on edexcel january 2015 (IAL) question 6d, what is the objective line?
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    (Original post by NotNotBatman)
    Shortest from F to T, then shortest from S to F.
    Thank you!
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    (Original post by amy4121)
    Could someone help me on edexcel january 2015 (IAL) question 6d, what is the objective line?
    Let the cost of selling 1 unit of x be a .The total cost is the cost of multiplied by the number sold. Set the objective function as Minimise C = ax + 3ay for any value a (as it will have the same gradient). The 3ay bit is because it costs 3 times the amount to produce a unit of y than it does to produce a unit of x.
 
 
 
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