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    (Original post by alow)
    I've got the paper, don't know when I'm allowed to upload it though, dont you have to wait a day or something?
    You can upload it now!, its only edexcel i think or international qualifications you have to wait a day
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    (Original post by alow)
    I've got the paper, don't know when I'm allowed to upload it though, dont you have to wait a day or something?
    There's no time restriction for OCR (posting it would be a breach of their copyright but I think they are unlikely to go after you legally).
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    (Original post by alow)
    I've got the paper, don't know when I'm allowed to upload it though, dont you have to wait a day or something?
    Please could you post it. If anybody posts it I will try and do an unofficial mark scheme so everybody has an idea of how they've done.
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    (Original post by alow)
    I've got the paper, don't know when I'm allowed to upload it though, dont you have to wait a day or something?
    It would be great if you could share the paper! If you don't want to, could to type out all the questions with the number of marks for each question so we can get an idea about how many marks we got?
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    (Original post by alow)
    I've got the paper, don't know when I'm allowed to upload it though, dont you have to wait a day or something?
    Can you please upload it again at least for around half an hour so I can finish the mark scheme. Thanks.

    How did you manage to get the paper btw, did you ask a teacher or just strolled out with it.
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    For the linear programmin question the one were u had to draw it out, where did you guys get the feasible region ? Was it between the centre and 2 other points?
    because i got two triangles but i found the feasible regoin to be between the centre and 2 other points, but i wasnt sure if its right?
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    Work so far, will be continued if paper re-uploaded:

    Unofficial Mark Scheme

    1i)  24 \ 57 \ 9 \ 31 \ 16 \ 4

     24 \ 9 \ 57 \ 31 \ 16 \ 4

     24 \ 9 \ 31 \ 57 \ 15 \ 4

     24 \ 9 \ 31 \ 15 \ 57 \ 4

     24 \ 9 \ 31 \ 15 \ 4 \ 57

    Circles where necessary to indicate comparisons. (2 marks)

    ii)  Pass \ 2 \ : \ 9 \ 24 \ 15 \ 4 \ 31 \ 57

     Pass \ 3 \ : \ 9 \ 15 \ 4 \ 24 \ 31 \ 57

    (2 marks)

    iii) Pass 1 - 4 swaps, Pass 2 - 3 swaps, Pass 3 - 2 swaps , Pass 4 - 1 swap, Pass 5 -1 swap (2 marks)

    2i)a) Square plus a loop at one vertex for example (1 mark)

    b) Max vertex order for simple graph with 4 vertices is 3, but is Eulerian so order has to be 2 (as 0 means one vertex is not connected). Hence total order = 4x2 = 8. 8/2 = 4, so 4 arcs, not 5 arcs. (2 marks)

    ii)a) Total order = 2 x no. of arcs = 2 x 10 = 20 (1 mark)

    b) Max - 6, min - 2, and correct graph drawn (3 marks)

    c) Two vertices of order 4, six of order 2. (1 mark)

    3i) Correct trace through the algorithm, yielding outputs 6 and 90. (6 marks)

    ii) After you fill in the space in the answer booklet, you will get F D and E the same as in part i). From this it follows that G = 18 /6 = 3 and M = 3 x 30 = 90. The same as in part i). (2 marks)

    iii) F = 4 and M = 24. (2 marks)

    4i)  A - B -C-D-E-F-G
    Weight = 51 km
    82km - 51km = 31km

    so:  A-B-C-D-E-F-G-F-B-A

    with weight 82km. (4 marks)

    ii) 51 - 5 - 7 + 8 = 47km.
    Go back through D so:

     A-B-C-E-F-G-D-A (2 marks)

    iii) NN starts  A - B-C-D-E-F-G

    Stalls as can't return to A.

     B - A - C - D - E -F-G-B

    Weight = 76km.

     B - A - C - D - E - G - F - B

    Weight = 69km. (6 marks)

    5i)  A - B - F - G

    Weight = 31km. (5 marks)

    ii) 25 hours (2 marks).

    iii)  11 + 15 + CE \leq 31

     CE \leq 5

    8 - 5 = 3km (2 marks)

    iv) Odd nodes A,B,C and F.
    All repeated arc lengths are 32km.
    32 + 224 - 11 - 8 = 237km. (6 marks)

    6i) Attached using Wolfram (4 marks)

    ii)  (6,3) ,  \ (4,0), \  (0,3) \ and \  (4,4.5)

    At (6,3) P = 54
    At (4,0) P = 20
    At (0,3) P = 24
    At (4,4.5) P = 56.

    So at optimal point x = 4, y = 4.5, P = 56. (4 marks)

    iii) When x = 0, max y = 3, P = 24.
    When x = 1, max y = 3, P = 29
    When x = 2, max y = 3, P =34
    When x = 3, max y = 4, P = 47
    When x = 4, max y = 4, P = 52
    When x = 5, max y = 3, P = 49
    When x = 6, max y = 3, P = 54.

    So optimal point is when x = 6, y = 3 and P = 54. (4 marks)

    iv) To make the RHS non-negative. (1 mark)

    v) Correct Simplex Tableau (2 marks)

    vi) First pivot 8 in y-column.
    Show how rows calculated.
    P = 24, x = 0, y = 3.
    Pivot is 4.5 in x-column.
    Verify that optimum point is when x =4, y = 4.5 and P = 56.
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    (Original post by metaltron)
    Work so far, will be continued if paper re-uploaded:

    Unofficial Mark Scheme

    1i)  24 \ 57 \ 9 \ 31 \ 16 \ 4

     24 \ 9 \ 57 \ 31 \ 16 \ 4

     24 \ 9 \ 31 \ 57 \ 15 \ 4

     24 \ 9 \ 31 \ 15 \ 57 \ 4

     24 \ 9 \ 31 \ 15 \ 4 \ 57

    Circles where necessary to indicate comparisons. (2 marks)

    ii)  Pass \ 2 \ : \ 9 \ 24 \ 15 \ 4 \ 31 \ 57

     Pass \ 3 \ : \ 9 \ 15 \ 4 \ 24 \ 31 \ 57

    (2 marks)

    iii) Pass 1 - 4 swaps, Pass 2 - 3 swaps, Pass 3 - 2 swaps , Pass 4 - 1 swap, Pass 5 -1 swap (2 marks)

    2i)a) Square plus a loop at one vertex for example (1 mark)

    b) Max vertex order for simple graph with 4 vertices is 3, but is Eulerian so order has to be 2 (as 0 means one vertex is not connected). Hence total order = 4x2 = 8. 8/2 = 4, so 4 arcs, not 5 arcs. (2 marks)

    ii)a) Total order = 2 x no. of arcs = 2 x 10 = 20 (1 mark)

    b) Max - 6, min - 2, and correct graph drawn (3 marks)

    c) Two vertices of order 4, six of order 2. (1 mark)

    3i) Correct trace through the algorithm, yielding outputs 6 and 90. (6 marks)
    Saved the files before they were deleted, knew that would happen! But I'm one step ahead...
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    (Original post by metaltron)
    Can you please upload it again at least for around half an hour so I can finish the mark scheme. Thanks.

    How did you manage to get the paper btw, did you ask a teacher or just strolled out with it.
    Just put it in my pocket and walked out
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    (Original post by alow)
    Just put it in my pocket and walked out
    LOL but at the end don't the invigilators collect them in? they do at my college they collect everything off the desks that's not ours then tell us to leave.
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    (Original post by DingDong!)
    LOL but at the end don't the invigilators collect them in? they do at my college they collect everything off the desks that's not ours then tell us to leave.
    They come round and get all of the papers and formulae books, I just handed the invigilator the formula book and he didn't care. Loads of people at my college always take maths papers out
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    (Original post by alow)
    They come round and get all of the papers and formulae books, I just handed the invigilator the formula book and he didn't care. Loads of people at my college always take maths papers out
    Oh, well it's easier when they do that. At my college one invigilator collects collects formula books, one question papers and the other answer booklets! I asked if I could take it once but he said no. :mad:
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    (Original post by DingDong!)
    Oh, well it's easier when they do that. At my college one invigilator collects collects formula books, one question papers and the other answer booklets! I asked if I could take it once but he said no. :mad:
    Yeah you cant really dodge it if they collect separarately. Do you have any other maths exams in may/june?
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    Thanks alow , Much Appreciated
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    (Original post by metaltron)
    What I can remember:

    Bubble sort:
    1i) First pass correct with correct comparisons and swaps.
    ii) Second and third pass correct.
    iii) Pass 1 - 4 swaps
    Pass 2 - 3 swaps
    Pass 3 - 2 swaps
    Pass 4 - 1 swap
    Pass 5 - 1 swap

    Eulerian Graphs:

    2

    i) Maximum = 6 arcs
    Minimum = 2 arcs

    ii) Correct drawing with one vertex order 6.

    iii) Order = no. of arcs = 10 x 2 = 20

    iv) Correct drawing, two vertices with order 4 and other order 2.


    3.i) Trace the algorithm, outputs 6 and 90.
    ii) Fill in table, values the same as in part i) until it comes to dividing by A. Nevertheless outputs are still 6 and 90.
    iii) 4 and 24

    4i) Kruskal's's
    ii) Algorithm cannot return to A.
    Start from B to get a cycle of weight of
    Alteration to obtain a length of 69 km.

    5i) Dijkstra's
    ii) Chinese Postman on reduced network.

    6i) Correct drawing.
    ii) x = 4, y= 4.5 , P= 56
    iii) Try all integer values of x with their respective y-values. Obtain P = 54.
    iv) To get constraint in the form ax+by <= c, as algorithm can not be used for greater than constraints.
    v) Initial Simplex tableau
    ii) Can't remember the first values, but you will then verify for non-integer values that P = 56, when x = 4 and y =4.5
    For 6(iii). wasn't (0,7) in the feasible region so therefore that gives 56 (P=5x+8y)
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    (Original post by missypink<3)
    For 6(iii). wasn't (0,7) in the feasible region so therefore that gives 56 (P=5x+8y)
    I think I got (0,3) for my highest y, due to one of the inequalities being y<=3+(3/8)x
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    Can anyone draw 6i and upload it please?
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    (Original post by DarkPeople)
    Yeah I got 88 too after the second iteration I think. Yep got 69km for that! yeah all 3 pairs for me were 32k each as well ;]

    Simplex should have given you (4,4.5) which corresponded to P=56. This was the same as in 6(i) and (ii).
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    Full UO ms there now.
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    (Original post by metaltron)
    Work so far, will be continued if paper re-uploaded:

    Unofficial Mark Scheme

    1i)  24 \ 57 \ 9 \ 31 \ 16 \ 4

     24 \ 9 \ 57 \ 31 \ 16 \ 4

     24 \ 9 \ 31 \ 57 \ 15 \ 4

     24 \ 9 \ 31 \ 15 \ 57 \ 4

     24 \ 9 \ 31 \ 15 \ 4 \ 57

    Circles where necessary to indicate comparisons. (2 marks)

    ii)  Pass \ 2 \ : \ 9 \ 24 \ 15 \ 4 \ 31 \ 57

     Pass \ 3 \ : \ 9 \ 15 \ 4 \ 24 \ 31 \ 57

    (2 marks)

    iii) Pass 1 - 4 swaps, Pass 2 - 3 swaps, Pass 3 - 2 swaps , Pass 4 - 1 swap, Pass 5 -1 swap (2 marks)

    2i)a) Square plus a loop at one vertex for example (1 mark)

    b) Max vertex order for simple graph with 4 vertices is 3, but is Eulerian so order has to be 2 (as 0 means one vertex is not connected). Hence total order = 4x2 = 8. 8/2 = 4, so 4 arcs, not 5 arcs. (2 marks)

    ii)a) Total order = 2 x no. of arcs = 2 x 10 = 20 (1 mark)

    b) Max - 6, min - 2, and correct graph drawn (3 marks)

    c) Two vertices of order 4, six of order 2. (1 mark)

    3i) Correct trace through the algorithm, yielding outputs 6 and 90. (6 marks)

    ii) After you fill in the space in the answer booklet, you will get F D and E the same as in part i). From this it follows that G = 18 /6 = 3 and M = 3 x 30 = 90. The same as in part i). (2 marks)

    iii) F = 4 and M = 24. (2 marks)

    4i)  A - B -C-D-E-F-G
    Weight = 51 km
    82km - 51km = 31km

    so:  A-B-C-D-E-F-G-F-B-A

    with weight 82km. (4 marks)

    ii) 51 - 5 - 7 + 8 = 47km.
    Go back through D so:

     A-B-C-E-F-G-D-A (2 marks)

    iii) NN starts  A - B-C-D-E-F-G

    Stalls as can't return to A.

     B - A - C - D - E -F-G-B

    Weight = 76km.

     B - A - C - D - E - G - F - B

    Weight = 69km. (6 marks)

    5i)  A - B - F - G

    Weight = 31km. (5 marks)

    ii) 25 hours (2 marks).

    iii)  11 + 15 + CE \leq 31

     CE \leq 5

    8 - 5 = 3km (2 marks)

    iv) Odd nodes A,B,C and F.
    All repeated arc lengths are 32km.
    32 + 224 - 11 - 8 = 237km. (6 marks)

    6i) Attached using Wolfram (4 marks)

    ii)  (6,3) ,  \ (4,0), \  (3,0) \ and \  (4,4.5)

    At (6,3) P = 54
    At (4,0) P = 20
    At (0,3) P = 24
    At (4,4.5) P = 56.

    So at optimal point x = 4, y = 4.5, P = 56. (4 marks)

    iii) When x = 0, max y = 3, P = 24.
    When x = 1, max y = 3, P = 29
    When x = 2, max y = 3, P =34
    When x = 3, max y = 4, P = 47
    When x = 4, max y = 4, P = 52
    When x = 5, max y = 3, P = 49
    When x = 6, max y = 3, P = 54.

    So optimal point is when x = 6, y = 3 and P = 54. (4 marks)

    iv) Simplex is designed for inequalities with a less than or equals sign. (1 mark)

    v) Correct Simplex Tableau (2 marks)

    vi) First pivot 8 in y-column.
    Show how rows calculated.
    P = 24, x = 0, y = 3.
    Pivot is 4.5 in x-column.
    Verify that optimum point is when x =4, y = 4.5 and P = 56.
    For 5 part 3, should be a less than sign, not less than or equal to surely? The question how much shorter should it be for it to be the SHORTEST as in the only short one, that's how I interpreted it.
 
 
 
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