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    (Original post by Arsey)
    well technically = as you will deliver all your stock to meet the demand.

    However, you will get away with <= on all transportation questions.
    (Original post by Miken Moose)
    Okay... I've just noticed something glaringly obvious which should settle this. First, this is the solution laid out by example 12 page 27:

    Let x_i_j be the number of units transported from i to j.

    Minimise C = 3x_1_1 + 3x_1_2 + 2x_1_3 + 4x_2_1 + 2x_2_2 + 3x_2_3 + 3x_3_1 + 4x_3_2 + 3x_3_3

    Subject to:
    x_1_1 + x_1_2 + x_1_3 \leq 25
    x_2_1 + x_2_2 + x_2_3 \leq 40
    x_3_1 + x_3_2 + x_3_3 \leq 31
    x_1_1 + x_2_1 + x_3_1 \leq 30
    x_1_2 + x_2_2 + x_3_2 \leq 30
    x_1_3 + x_2_3 + x_3_3 \leq 36
    x_i_j \geq 0

    Okay? So, why don't I just set all the x_i_j = 0? This satisfies all constraints and clearly gives the minimum value of C.

    Unless of course, I'm missing something (which for me is quite likely).

    Thanks. Also, about the question above, say you had an unbalanced problem, then setting xij=0 seems to be the best solution when this obviously isn't the case. Do you know why that is?
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    (Original post by brittanna)
    Thanks. Also, about the question above, say you had an unbalanced problem, then setting xij=0 seems to be the best solution when this obviously isn't the case. Do you know why that is?
    Xij is the amount you are transporting from (factory) i to (shop) j

    You wouldn't have a very good business if you didn't transport anything!

    If the supply is enough you have to satisfy the demand and you are trying to minimise the costs of doing that. You cannot just say I am going to minimise my costs by not delivering anything to anyone.
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    (Original post by Mallika)
    Hi Arsey, could you please explain this?
    So say you have 3 lines on the well graph for each strategy for B.

    If the best mixed strategy for A is where B1 and B2 meet and if B knows that A
    is playing the best mixed strategy then there is no point B playing 3 as he will lose more (or gain less); therefore you can delete column B3 from the game which reduces it to a 2x2 matrix
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    (Original post by RYRK)
    ok , i got a quick question, when writing a maximizing allocation problem as a linear progamming problem, u first transform the matrix and then continue as normal, so would the objective be to maximise or minimise, i looked on the example of the book on page 55 , and this one is minimising and i am not sure it it is right?
    once transformed you are looking to minimise.
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    (Original post by Arsey)
    Xij is the amount you are transporting from (factory) i to (shop) j

    You wouldn't have a very good business if you didn't transport anything!

    If the supply is enough you have to satisfy the demand and you are trying to minimise the costs of doing that. You cannot just say I am going to minimise my costs by not delivering anything to anyone.
    But say you used simplex to solve it, would it not give Xij=0 as the optimal solution?
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    (Original post by Arsey)
    So say you have 3 lines on the well graph for each strategy for B.

    If the best mixed strategy for A is where B1 and B2 meet and if B knows that A
    is playing the best mixed strategy then there is no point B playing 3 as he will lose more (or gain less); therefore you can delete column B3 from the game which reduces it to a 2x2 matrix
    Thank you
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    (Original post by brittanna)
    But say you used simplex to solve it, would it not give Xij=0 as the optimal solution?
    you don't use simplex to solve transportation problems.

    you only solve max problems using simplex in current spec D2
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    (Original post by Lilmzbest)
    As far as i remember one of the arrows was in the wrong direction (which you can see by applying flow conservation) and this gave a completely wrong answer :confused: chances are that your answer is correct
    (Original post by Hamburglar)
    I'm not looking at it right now, but beware that mark schemes for mock papers and practice papers are generally very wrong. I do recall remembering that the mark scheme was wrong when labelling some arrows on one of those papers.



    Yes possibly, June 2008 was part of the current spec as far as I know. It is pretty much the same idea anyway, so I don't see why they couldn't ask. Although I can't say I remember being asked to use Prim's, usually they ask you to use Kruskal's but either way, you should be comfortable with both.



    As you're filling the table in, if supply and demand are satisfied at the same time, you can either go across one and put the 0 in the cell, then move down. Or, go down and put the 0 in that cell, and move along one. Basically in the diagonal you have, you want to put the zero in one of the spaces where normally you'd expect a number.

    Technically you can actually put it anywhere, but it's best to stick to the rule of moving along and downwards only.



    64 if I am not mistaken

    (Original post by Arsey)
    probably, there were quite a few errors on the mock paper.
    Thanks guys
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    (Original post by Arsey)
    Xij is the amount you are transporting from (factory) i to (shop) j

    You wouldn't have a very good business if you didn't transport anything!

    If the supply is enough you have to satisfy the demand and you are trying to minimise the costs of doing that. You cannot just say I am going to minimise my costs by not delivering anything to anyone.
    So if supply and demand are met then you use equalities in the constraints and if supply is greater than demand you use inequalities?
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    GOOD LUCK ANYONE WHO'S READING THIS!


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    Let the posts begin.

    Who managed to finish Dynamic Programming? If so; what value did you get for the end? and did you end up with keep keep keep replace?
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    arsey have you got a model answer for this exam yet?
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    someone please give me answers
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    the last question was a bugger, hated it, well there goes my A , that question was really different from the ones from the past papers , most people who finished it at my school got around 32000
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    I got 32000 but keep keep replace keep I think!
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    linear programming. How was it done?
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    (Original post by RYRK)
    the last question was a bugger, hated it, well there goes my A , that question was really different from the ones from the past papers , most people who finished it at my school got around 32000
    I got 29000
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    (Original post by bowersbros)
    I got 29000
    I got keep keep replace keep with 29000, but I think it is probably 32000.
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    I got keep keep replace keep, got £31,000

    I forgot to add the value when the bikes are sold at the end...
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    How was the linear programming question done where u had to maximise the values?
 
 
 
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