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    (Original post by NotNotBatman)
    The tagging system isn't working properly for me.

    For this question you would just have to look at the area of the Gantt chart with the most overlaps. If I were doing this question, I would look around the place where there's a lot of activities occurring with minimal floats, as there's not a lot of room for manoeuvre and then consider an interval of time, where activities must be taking place.
    Thanks!! 👍


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    For exercise F question 2b https://86495a24d9b181f09087f4168bce...VJZTW8/CH1.pdf dos anyone know whether there's a formula or anything that can be used please? Thanks
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    (Original post by economicss)
    For exercise F question 2b https://86495a24d9b181f09087f4168bce...VJZTW8/CH1.pdf dos anyone know whether there's a formula or anything that can be used please? Thanks
    \frac{n(n-1)}{2}
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    (Original post by NotNotBatman)
    \frac{n(n-1)}{2}
    Brilliant, thanks so much!
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    (Original post by NotNotBatman)
    \frac{n(n-1)}{2}
    what about minimum?
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    (Original post by 4nonymous)
    what about minimum?
    0 if the list is already in order.
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    does anyone have access to the jan 2016 paper?
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    also am I right in thinking with filling in trace tables, if we use a separate row for every individual box in the algorithm we would get full marks even though the mark scheme has it condensed?
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    (Original post by Saywhatyoumean)
    also am I right in thinking with filling in trace tables, if we use a separate row for every individual box in the algorithm we would get full marks even though the mark scheme has it condensed?
    Yes, you would get full marks either way, as long as the information is correct.
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    (Original post by NotNotBatman)
    Yes, you would get full marks either way, as long as the information is correct.
    Thanks again

    Sorry I've been teaching myself D1 so have a few questions, with scheduling diagrams as long as you meet the dependencies and have the right length of events you'd get full marks right?

    I know the book says if you have a choice of scheduling events you should choose the one with the lowest latest start time first but sometimes that makes it hard to schedule the other activities so was just wondering if other alternatives were allowed?


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    (Original post by Saywhatyoumean)
    Thanks again

    Sorry I've been teaching myself D1 so have a few questions, with scheduling diagrams as long as you meet the dependencies and have the right length of events you'd get full marks right?

    I know the book says if you have a choice of scheduling events you should choose the one with the lowest latest start time first but sometimes that makes it hard to schedule the other activities so was just wondering if other alternatives were allowed?


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    You should get all the marks as long as there is a minimum number of workers, the activities finish in the critical time (unless stated otherwise, where you need to restrict the number of workers) and the dependencies are fulfilled.

    Choosing the activity with the lowest latest finish time makes it easier, so that activities with larger floats that have more room for manoeuvre can be considered later, but it doesn't have to be done like this and there is not a unique solution.
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    What would you make M equal to? x is red hats and y is green hats.
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    (Original post by target21859)
    What would you make M equal to? x is red hats and y is green hats.
    The gradient would be -\frac{1}{3}

    economicss
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    (Original post by NotNotBatman)
    The gradient would be -\frac{1}{3}

    economicss
    Could you explain why? Sorry I'm not very good at D1.
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    (Original post by target21859)
    Could you explain why? Sorry I'm not very good at D1.
    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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    (Original post by NotNotBatman)
    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.
    Ah thanks that makes sense
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    (Original post by NotNotBatman)
    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.
    But if it costs 3 times a much to produce a green hat (y) doesnt that mean y=3x
    so isnt it C=3x+y?
    I hate these type of questions so annoying
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    (Original post by 4nonymous)
    But if it costs 3 times a much to produce a green hat (y) doesnt that mean y=3x
    so isnt it C=3x+y?
    I hate these type of questions so annoying
    Remember x is the number of red hats and y is the number of green hats, not the cost.

    No, y is the number of green hats and x is the number of red hats, it has nothing to do with the cost What you've written, y=3x, essentially says "The number of green hats equals 3 times the number of green hats," this is not what the question says. If x had a cost of a, then y has a cost of 3a. So the total cost of red hats is ax and the cost of green hats is 3ay, so the total cost is ax+3ay.

    Or you could use numbers, if red hats had a cost of 5 each and you wanted the total cost of red hats, that would be 5 times the number of red hats, 5x. And do the same for yellow hats the total cost of the yellow hats would be 15y. So overall the total Cost would be 5x+15y. This could be used as the objective line as it has the same gradient.
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    (Original post by NotNotBatman)
    Remember x is the number of red hats and y is the number of green hats, not the cost.

    No, y is the number of green hats and x is the number of red hats, it has nothing to do with the cost What you've written, y=3x, essentially says "The number of green hats equals 3 times the number of green hats," this is not what the question says. If x had a cost of a, then y has a cost of 3a. So the total cost of red hats is ax and the cost of yellow hats is 3ay, so the total cost is ax+3ay.

    Or you could use numbers, if red hats had a cost of 5 each and you wanted the total cost of red hats, that would be 5 times the number of red hats, 5x. And do the same for yellow hats the total cost of the yellow hats would be 15y. So overall the total Cost would be 5x+15y. This could be used as the objective line as it has the same gradient.
    oh thanks. I knew you'd prove me wrong.
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    In an activity network, do I start from 0 or 1 for the first node?
 
 
 
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