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Math Help - linear programming

  1. #1
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    linear programming

    OK I have attempted 2 Linear Programming questions and i would like to know if i have got all the equations correct for the first question, and some help on the second one would be much appreciated.

    Q1:
    A bank is attempting to determine where its assets should be invested during the current year. At present $500 million is available for investment in bonds, home loans, car loans, and personal loans. The annual rate of return on each type of investment is known to be: bonds, 7%; home loans, 8%; car loans, 12%; personal loans, 11%. In order to ensure that the bank’s portfolio is not too risky, the bank’s investment manager has placed the following three restrictions on the bank’s portfolio:
    (a) The amount invested in personal loans cannot exceed the amount invested in bonds.
    (b) The amount invested in home loans cannot exceed the amount invested in car loans.
    (c) No more than 25% of the total amount invested may be in personal loans.
    The bank’s objective is to maximize the annual return on its investment portfolio. Formulate an LP (in standard form) that will enable the bank to meet this goal. Assume interest is calculated annually.

    My Answer:
    Let b=bonds h=home loans c=car loans and p=personal loans
    maximise z=0.07b+0.08h+0.12c+0.11p
    subject to:
    b+h+c+p<=500e6
    p<=b
    h<=c
    p<=125e6
    b, h, c, p>=0

    (note that 500e6 is 500,000,000 and p<=b means that p i s less than or equal to b)

    Q2:
    Bulldust Inc. blends silicon and nitrogen to produce two types of fertiliser. Fertiliser 1 must be at least 40% nitrogen and sells for $20/kg. Fertiliser 2 must be at least 70% silicon and sells for $18/kg. Bulldust can purchase up to 800 kg of nitrogen at $10/kg and up to 1000 kg of silicon at $8/kg. Assuming that all fertiliser produced can be sold, formulate an LP to help Bulldust maximize profit.

    HELP please =P
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  2. #2
    A riddle wrapped in an enigma
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    I'll give it a shot.

    Let \ x \ = \ no.\ of \ kg \ of \ fertilizer \ 1

    Let \ y \ = \ no. \ of \ kg \ of \ fertilizer \ 2

    .40x + .30y \leq800
    .60x + .70y \leq1000

    Maximum profit=20x+18y - overhead

    Overhead = cost of purchasing nitrogen (800 kg @ $10/kg) and silicon (1000kg @ $8/kg) which totals $16,000 in overhead.
    Last edited by masters; May 24th 2008 at 01:18 PM.
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  3. #3
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    Ok I think you are on the right track..but where did u get .30y and .60x from? are you jsut assuming that all silicon and nitrogen is used up?
    Also what about my first answer?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by masters View Post
    I'll give it a shot.

    Let \ x \ = \ no.\ of \ kg \ of \ fertilizer \ 1

    Let \ y \ = \ no. \ of \ kg \ of \ fertilizer \ 2

    .40x + .30y \leq800
    .60x + .70y \leq1000

    Maximum profit=20x+18y - overhead

    Overhead = cost of purchasing nitrogen (800 kg @ $10/kg) and silicon (1000kg @ $8/kg) which totals $16,000 in overhead.
    You have assumed that 40/60 and 30/70 mixes are those that maximise profit, if this is true you need to justify/show/prove it.

    RonL
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