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    (Original post by iMacJack)
    Yeah I'm sure about them all apart from Dijkstras and Algorithms, they may have been the other way around, not sure
    For the flow chart, was it 135 as the ouput, something like 80+55?
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    (Original post by Pablo Picasso)
    For the flow chart, was it 135 as the ouput, something like 80+55?
    output t = 135

    algorithm multiplies x + y together
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    (Original post by James10090)
    Does anyone know the breakdown for the linear programme question?
    (a) 2 marks (for the cost inequality)
    (b) 4 marks (for the graph)
    (c) 1 mark (for the objective function)
    (d) 3 marks (for the objective line and optimal vertex's location)
    (e) 2 marks (for finding the exact co-ordinates of V)
    (f) 3 marks (for finding integer solutions and the corresponding cost)

    Total: 14 marks.

    I hope that this helps!

    (By the way, I don't even know how I managed to remember that!)

    (Wait, that can't possibly be right: that comes to 15 marks! Please excuse me whilst I go and time-travel back to yesterday's paper, and try to remember exactly where I went wrong just there...)
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    (Original post by iMacJack)
    output t = 135

    algorithm multiplies x + y together
    What multiplies what to what?
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    (Original post by Pablo Picasso)
    What multiplies what to what?
    The algorithm's output was simply the product of x and y, so yes the output was 135
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    (Original post by Pablo Picasso)
    What multiplies what to what?
    Overall, it was x and y: as 27 X 5 = 135.

    (The trace-table had you doing 80 + 55 - so, that was definitely right!)

    You then had to use this multiplication in part (b)(ii): 122 X (1/2) = 61.
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    (Original post by iMacJack)
    The algorithm's output was simply the product of x and y, so yes the output was 135
    What was the value of x and y?
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    (Original post by Pablo Picasso)
    What was the value of x and y?
    x = 27, y = 5
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    (Original post by mupsman2312)
    Overall, it was x and y: as 27 X 5 = 135.

    (The trace-table had you doing 80 + 55 - so, that was definitely right!)

    You then had to use this multiplication in part (b)(ii): 122 X (1/2) = 61.
    Can remember get 135 from 80+55 = 135, but cant remember doing 27*5. I'm screwed
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    (Original post by iMacJack)
    x = 27, y = 5
    I can remember doing 80 + 55 and outputting 135 as the answer, would that mean i must have gotten x and y as 27 and 5? Vaguely remember what i wrote now
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    (Original post by mupsman2312)
    (a) 2 marks (for the cost inequality)
    (b) 4 marks (for the graph)
    (c) 1 mark (for the objective function)
    (d) 3 marks (for the objective line and optimal vertex's location)
    (e) 2 marks (for finding the exact co-ordinates of V)
    (f) 3 marks (for finding integer solutions and the corresponding cost)

    Total: 14 marks.

    I hope that this helps!

    (By the way, I don't even know how I managed to remember that!)
    That adds up to 15 mate, i think (a) was 1 mark
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    Do curved arcs matter when drawing the activity network in question 2?
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    (Original post by iMacJack)
    The algorithm's output was simply the product of x and y, so yes the output was 135
    Do you know how many marks that was worth?
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    (Original post by Jporter1399)
    Do you know how many marks that was worth?
    I think the first part might have been 5 and then the second part 3

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    (Original post by James10090)
    Does anyone know the breakdown for the linear programme question? Or just the marks breakdown in general. Would be helpful for everyone
    I was something like this (I'm pretty sure, at least!):


    Matchings:
    (a) 2 marks (definition of a bipartite graph);
    (b) 3 marks (alternating path).
    {Total: 5 marks}

    Activity network:
    5 marks (drawing only).
    {Total: 5 marks}

    List of numbers:
    (a) 4 marks (quick sort);
    (b) 3 marks (bin-packing);
    (c) 2 marks ("Is this solution optimal?").
    {Total: 9 marks}

    Flow Chart:
    (a) 4 marks (trace-table);
    (b) 3 marks - (i): choice and explanation (presumably 2 marks);
    (ii): new output (presumably 1 mark);
    {Total: 7 marks}

    Shortest route:
    (a) 6 marks (Dijkstra's algorithm);
    (b) 2 marks (A to F, via J);
    (c) 3 marks (Prim's algorithm);
    (d) 1 mark (weight of MST).
    {Total: 12 marks}

    Route inspection:
    (a) 5 marks (repetition and length);
    (b) 2 marks (explanation: starting at a vertex with an odd degree);
    (c) 2 marks (determining the finish point);
    (d) 2 marks (route and new length).
    {Total: 11 marks}

    Critical path analysis:
    (a) 3 marks (finding the values of w, x, y and z);
    (b) 4 marks (cascade/Gantt chart);
    (c) 2 marks (lower bound);
    (d) 3 marks ([that really annoying] schedule!).
    {Total: 12 marks}

    Linear programming:
    (a) 1-2 marks [I'm not quite sure...] (finding a cost inequality);
    (b) 3-4 marks [the same thing again!] (plotting [and shading] the graph);
    (c) 1 mark (stating the objective function);
    (d) 3 marks (drawing an objective line, and locating the optimal vertex);
    (e) 2 marks (calculating the exact co-ordinates of V);
    (f) 3 marks (seeking for integer values, and then calculating the corresponding cost).
    {Total: 14 marks}*

    (*I know that that comes to 15 marks, but I can't quite remember exactly where I seemed to just add an extra one in... Was it only 1 mark for the cost inequality? 3 for the graph? It was something like that...)

    {Grand total for paper: 75 marks} [Obviously!]


    Wow - that took a pretty while to do! After all, that was all from memory! I sincerely hope that this can help!
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    (Original post by frances98)
    I think the first part might have been 5 and then the second part 3

    Posted from TSR Mobile
    Unfortunately, they were really harsh - and only gave us 4 marks for completing such a long-winded trace-table!
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    (Original post by Pablo Picasso)
    Can remember get 135 from 80+55 = 135, but cant remember doing 27*5. I'm screwed
    Don't worry - you never did have to do 27 X 5; this was only what the algorithm achieved, overall. This and that addition both gave 135. This is how I knew to then do 122 X (1/2) in part (b)(ii), in order to give 61.
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    Thanks a lot
    (Original post by mupsman2312)
    I was something like this (I'm pretty sure, at least!):


    Matchings:
    (a) 2 marks (definition of a bipartite graph);
    (b) 3 marks (alternating path).
    {Total: 5 marks}

    Activity network:
    5 marks (drawing only).
    {Total: 5 marks}

    List of numbers:
    (a) 4 marks (quick sort);
    (b) 3 marks (bin-packing);
    (c) 2 marks ("Is this solution optimal?".
    {Total: 9 marks}

    Flow Chart:
    (a) 4 marks (trace-table);
    (b) 3 marks - (i): choice and explanation (presumably 2 marks);
    (ii): new output (presumably 1 mark);
    {Total: 7 marks}

    Shortest route:
    (a) 6 marks (Dijkstra's algorithm);
    (b) 2 marks (A to F, via J);
    (c) 3 marks (Prim's algorithm);
    (d) 1 mark (weight of MST).
    {Total: 12 marks}

    Route inspection:
    (a) 5 marks (repetition and length);
    (b) 2 marks (explanation: starting at a vertex with an odd degree);
    (c) 2 marks (determining the finish point);
    (d) 2 marks (route and new length).
    {Total: 11 marks}

    Critical path analysis:
    (a) 3 marks (finding the values of w, x, y and z);
    (b) 4 marks (cascade/Gantt chart);
    (c) 2 marks (lower bound);
    (d) 3 marks ([that really annoying] schedule!).
    {Total: 12 marks}

    Linear programming:
    (a) 1-2 marks [I'm not quite sure...] (finding a cost inequality);
    (b) 3-4 marks [the same thing again!] (plotting [and shading] the graph);
    (c) 1 mark (stating the objective function);
    (d) 3 marks (drawing an objective line, and locating the optimal vertex);
    (e) 2 marks (calculating the exact co-ordinates of V);
    (f) 3 marks (seeking for integer values, and then calculating the corresponding cost).
    {Total: 14 marks}*

    (*I know that that comes to 15 marks, but I can't quite remember exactly where I seemed to just add an extra one in... Was it only 1 mark for the cost inequality? 3 for the graph? It was something like that...)

    {Grand total for paper: 75 marks} [Obviously!]


    Wow - that took a pretty while to do! After all, that was all from memory! I sincerely hope that this can help!
    (Original post by mupsman2312)
    I was something like this (I'm pretty sure, at least!):


    Matchings:
    (a) 2 marks (definition of a bipartite graph);
    (b) 3 marks (alternating path).
    {Total: 5 marks}

    Activity network:
    5 marks (drawing only).
    {Total: 5 marks}

    List of numbers:
    (a) 4 marks (quick sort);
    (b) 3 marks (bin-packing);
    (c) 2 marks ("Is this solution optimal?".
    {Total: 9 marks}

    Flow Chart:
    (a) 4 marks (trace-table);
    (b) 3 marks - (i): choice and explanation (presumably 2 marks);
    (ii): new output (presumably 1 mark);
    {Total: 7 marks}

    Shortest route:
    (a) 6 marks (Dijkstra's algorithm);
    (b) 2 marks (A to F, via J);
    (c) 3 marks (Prim's algorithm);
    (d) 1 mark (weight of MST).
    {Total: 12 marks}

    Route inspection:
    (a) 5 marks (repetition and length);
    (b) 2 marks (explanation: starting at a vertex with an odd degree);
    (c) 2 marks (determining the finish point);
    (d) 2 marks (route and new length).
    {Total: 11 marks}

    Critical path analysis:
    (a) 3 marks (finding the values of w, x, y and z);
    (b) 4 marks (cascade/Gantt chart);
    (c) 2 marks (lower bound);
    (d) 3 marks ([that really annoying] schedule!).
    {Total: 12 marks}

    Linear programming:
    (a) 1-2 marks [I'm not quite sure...] (finding a cost inequality);
    (b) 3-4 marks [the same thing again!] (plotting [and shading] the graph);
    (c) 1 mark (stating the objective function);
    (d) 3 marks (drawing an objective line, and locating the optimal vertex);
    (e) 2 marks (calculating the exact co-ordinates of V);
    (f) 3 marks (seeking for integer values, and then calculating the corresponding cost).
    {Total: 14 marks}*

    (*I know that that comes to 15 marks, but I can't quite remember exactly where I seemed to just add an extra one in... Was it only 1 mark for the cost inequality? 3 for the graph? It was something like that...)

    {Grand total for paper: 75 marks} [Obviously!]


    Wow - that took a pretty while to do! After all, that was all from memory! I sincerely hope that this can help!
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    (Original post by mupsman2312)
    Don't worry - you never did have to do 27 X 5; this was only what the algorithm achieved, overall. This and that addition both gave 135. This is how I knew to then do 122 X (1/2) in part (b)(ii), in order to give 61.
    So at the end 80+55 = 135 would get you the full marks provided u didnt make any mistakes?
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    (Original post by Pablo Picasso)
    So at the end 80+55 = 135 would get you the full marks provided u didnt make any mistakes?
    Yes, that algorithm was to find the product of the two values x and y! You didn't have to say that 27*5 was 135, because your t value 80 + 55 showed it anyway
 
 
 
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