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Thread: Modelling a maximising problem

  1. #1
    Junior Member
    Sep 2010

    Modelling a maximising problem

    Hi, I am completely lost with this problem...

    The UN wishes to determine the composition of 1000 identical food parcels, each of which is to be packed into a 20 litre container. For each possible food, its density, dietary value per kilogramme and availability are given below.

    Item 1:
    Density (kg/litre) 0.1
    Dietary value (units/kg): 10
    Availability (kg): 2000

    Item 2:
    Density (kg/litre) 0.2
    Dietary value (units/kg): 6
    Availability (kg): 3000

    Item 3:
    Density (kg/litre) 0.25
    Dietary value (units/kg): 3
    Availability (kg): 10000

    Model the problem of maximising the total dietary value of the food parcels as a linear programming problem (assuming that the foods can be packed without any air spaces).

    So I am guessing that the objective function will be:


    (1=10*01, so that it is units/litre, and so on)

    Is that correct? And how can I get the constraints? :-/

    Thanks a lot!
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  2. #2
    MHF Contributor

    Apr 2005
    I would recommend you specifically identify what your "variables" represent. I can see from you first "object function" that x1 is "the number of litres of Item 1", x2 is "the number of litres of item 2" and x3 is "the number of litres of item 3" but it is better to say that explicitely (and it will shock your teacher!).
    Yes, the food value in each package is the density in kg/litre times the "food value" per kg giving "food value per litre" and then multiply by the variables in litres, gives "total food value", the quantity you want to maximize.

    Now, there are several restrictions: you have to be able to put them all into a "30 litre container" so, since x1, x2, and x3 are measured in litres, you must have $\displaystyle x1+ x2+ x3\le 20$. You have available 200 kg of item 1 but since x1 is in litres, you need to multiply by its density: item 1 is .1 kg/litre so 200 kg is (200 kg)/(.1 kg/litre)= 2000 litres. A constraint on x1 is $\displaystyle x1\le 2000$.

    Of course, you have similar constraits on x2 and x3.

    Notice how I am keeping careful track of "units of measurement" to get the relations. I know I need to divide by .1 kg/litre instead of multiply because I need to cancel "kg" and get "litre" in the numerator, not the denominator.

    And don't forget the constraints that are often not mentioned: $\displaystyle x1\ge 0$, $\displaystyle x2\ge 0$, and $\displaystyle x3\ge 0$.
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  3. #3
    Junior Member
    Sep 2010
    Thank you so much, I understand this whole thing so much better now!
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