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Linear programming

  1. Nov 11, 2004 #1
    All-Easy manufactures three products whose unit profits are $1, $9 and $5, respectively. The company has budgeted 70 hrs. of labor time
    and 45 hours of machine time for the production of three products.
    The labor requirements per unit of products A,B C are 2, 3 and 5 hours, respectively. The corresponding machine time requirements per unit are 1, 4 and 5 hour.

    All-Easy regards the budgeted labor and machine hours as goals that must be exceeded, if necessary,but at the additional cost of $15 per labor hour and $5 per machine hour. Formulate the problem as an LP model.

    Doubts w/ solutions:

    I let x = no. of units of product A, y = no. of units of product B, z = no. of units of product C.

    Maximize: z = x + 9y + 5z (profit)
    subject to:
    2x + 3y + 5z <= 70 (labor hrs.)
    x + 4y + 5z <= 45 (machine hrs.)
    x,y,z >= 0

    "All-Easy regards the budgeted labor and machine hours as goals that must be exceeded, if necessary, but at the additional cost of $15 per labor hour and $5 per machine hour." - if I were to make mathematical model out of these, am i going to adjust my objective function or my constraints or both? How?
     
  2. jcsd
  3. Nov 11, 2004 #2
    Let u = additional labor hours, and v = additional machine hours. How do they affect your profit, and how do they affect your constraint functions? Express it algebraically.
     
  4. Nov 12, 2004 #3
    I got this idea... so at least, I can show you where am I... got stuck

    We have a constraint on Labor: 2x + 3y + 5z <= 70,
    but it can be exceeded ... at extra cost.

    If 2x + 3y + 5z is greater than 70, it costs an additional $15/hour.

    The excess is: (2x + 3y + 5z - 70) hours which costs $15/hr.
    The extra labor cost is: 15(2x + 3y + 5z- 70) dollars,
    which, of course, reduces the profit.


    Similarly, we have a constraint on Machine time: x + 4y + 5z <= 45
    which can be exceeded ... at extra cost.

    The excess is (x + 4y + 5z - 45) hours which costs $5/hr.
    The extra machine cost is: 5(x + 4y + 5z - 45) dollars,
    which also reduces the profit.

    is this correct? Can I relate it to what you've replied... and um, that's it, how will I re-formulate my LP model?
     
  5. Nov 12, 2004 #4
    OK, but remember that your labor hours are no longer limited to 70.
    I would also not use "z" to represent profit, since you are already using it for product C. :)

    How about:
    Maximize: P = x + 9y + 5z - 5u - 15v (profit)
    subject to:
    2x + 3y + 5z <= 70 + u (labor hrs.)
    x + 4y + 5z <= 45 + v (machine hrs.)
    x,y,z,u,v >= 0

    And if the goals must be exceeded, you have actual equality:
    2x + 3y + 5z = 70 + u (labor hrs.)
    x + 4y + 5z = 45 + v (machine hrs.)
    Solving for u and v in those get you the two relationships you mention in your post.
     
  6. Nov 12, 2004 #5
    Um... I did understand about it.... but if I were to write the final part as my model for the LP, it seems that it is "unstable"... bec. I am following the standard form of a LP model...
     
  7. Nov 12, 2004 #6
    The final part? you mean the two equalities?
     
  8. Nov 13, 2004 #7
    yes.. bec. I don't feel that my model is a formal one yet... =)
     
  9. Nov 13, 2004 #8
    The equalities are the bounding surfaces, and the solution is found on the surface--the vertices, in fact.
     
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