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Homework Help: Shipping problem (algebra)

  1. Dec 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A friend of mine passed along this problem he's having trouble with in a business class, and I have to admit, I'm a bit stumped coming up with the proper mathematical framework for it. Here's the problem:

    A company manufactures loudspeakers in plants I and II. Output at plant I is at most 800 units per month, output at plant II is at most 600 units per month. The units are shipped seperately to three warehouses, A, B, and C, whose minimum monthly requirements are 500, 400, and 400 units respectively. Shipping costs for plant I to warehouses A, B, and C are $16, $20, and $22 per unit, repectively, and shipping costs for plant II to each warehouse are $18, $16, and $14, respectively. What shipping schedule will enable the company to meet the warehouses' requirements and at the same time minimize shipping costs?

    The attempt at a solution
    Basically, I looked at minimizing the cost for the shipping. Since plant I is cheaper to send to warehouse A, and plant II is cheaper to send to B and C, I then looked at which warehouse would minimize costs. So, I had plant II send 400 units to C and 200 units to B. This leaves plant I sending 200 units to B, and 500 units to A.

    This looks right to me, but I'm curious, is there a way to set this up with equations to get an exact and correct answer? I tried setting this up with a system of equations like this:

    (Plant I's production I'm calling x, y, and z for shipping to the three warehouses, respectively. Plant II's production I'm calling A, B, and C for shipping to the three warehouses)

    x + A = 500
    y + B = 400
    z + C = 400
    A + B + C = 600
    x + y + z = 700

    However, this is six unknowns with only five equations... Also, I'm having to assume that plant II (A + B + C) will be maxed out in it's production. Is there any way to approach this where I don't have to make those sorts of assumptions?

    Thanks for your help!

  2. jcsd
  3. Dec 21, 2009 #2
    This is no ordinary algebra question.

    For the number of units ship from Manufacturer I to Factory A we represent as [tex]A_{I}[/tex] and the rest of other combination will follow the similar notation.

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    Last edited: Dec 21, 2009
  4. Dec 21, 2009 #3
    Right. So are you suggesting this is a combinatorics problem? If so, I'll have to start reading up on that-- the highest I've taken is linear algebra and ODE.
  5. Dec 21, 2009 #4
    Its not combinatoric problem.
    I have no ideal how to get started too .
    Lets wait for the homework helpers to assist.(=
  6. Dec 21, 2009 #5
    {500,200,0,0,200,400} is the best arrangement you can get if i have no mistake.
    My solution is derived mainly by logic , if you need the proof i have give it to you but i'm afraid i can't translate it into a good explaination.
  7. Dec 21, 2009 #6


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    Staff: Mentor

    Simplex algorithm?

    This is a linear programming problem.
  8. Dec 21, 2009 #7


    Staff: Mentor

    This is indeed a linear programming problem. The main point here is that the total cost of shipping speakers from the two plants to the three warehouses is to be minimized, subject to all of the constraints. icystrike is on the right path except that his first equation should be an inequality. That is, AI + BI + CI + AII + BII + CII [itex]\geq[/itex] 1300.

    I'm using the same variable names as icystrike used, where AI represents the number of speakers sent to warehouse A from plant I. The other five variables have similar names.

    There should be only one equation - the one for shipping cost. All other relationships are inequalities.
  9. Dec 21, 2009 #8
    It's easy to see that shipping more than the required minimum to any warehouse can only
    increase costs. So the relation can be an equality and you could use it to eliminate one of
    the variables.

    Icystrike also combined the equations for the 3 warehouses. I can see no reason to do this
    If you have AI+AII = 500 and B{SUB]I[/SUB]+BII = 400 and
    CI+CII = 400 it implies AI + BI + CI + AII + BII + CII = 1300, but the converse need not be true.

    The 3 equations can be used to eliminate 3 variables (but you wouldn't do that if you're going to use the simplex method)
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