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Kuhn-Tucker Optimization Problem

  1. Sep 30, 2012 #1
    Well, I've been working on this problem, but I can't get the right path to the solution.

    1. The problem statement, all variables and given/known data
    "Consider the following version of the Ehrlich economic model of crime where an individual has the expected utility function:

    U = p ln(Iu) + (1 - p) ln (Is).

    p = objective probability of being caught
    Iu= Criminal's income if caught
    Is= Criminal's income if not caught

    The criminal's initial endowment of time is T hours a day and should be divided into "time for crime" (tc) and "time for legal work" (tl). So T = tc + tl.
    If the criminal choose to work legally, he gets an income of wl> 0 per unit of time; while engaged in crime he gets an income of wc> wl per unit of time.
    If the individual is caught committing crimes, he get a penalty f > wc - wl per unit of time.

    So, the income of an individual who commits a crime but is not arrested is Is = wltl + wctc while if he get arrested
    Iu = wltl + wctc - f tc .
    2. Relevant equations
    Then, the criminal divides his time between crime and work legally, an his problem is given by:

    max U = p ln(Iu) + (1 - p) ln (Is). with tc, tl[itex]\geq[/itex]0

    s.t. Is= wltl + wctc
    Iu = wltl + wctc - f tc
    T = tc + tl

    a) For solving the problem in ( Iu,Is), rewrite the budget constraint as a linear equation: Is = a - bIu . Find a and b.
    b) Graph the problem in ( Iu,Is). Solve the problem in ( Iu,Is) when wl=1 , wc= 2, f= 1.5 , T= 3. Then find the corresponding solutions for (tc,tl)."
    3. The attempt at a solution
    Using differentiation and the Kuhn - Tucker conditions I've concluded that

    a= ((1- p) + p (Is))/(1-p)
    b= p (Is)/(1-p)Iu

    But I don't believe that's correct.

    Hope you could help. Thank you for reading.
     
  2. jcsd
  3. Oct 1, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your expressions for a and b are incorrect.

    I hate this question's notation, so I re-cast it as
    max p*log(x1) + (1-p)*log(x2),
    s.t.
    x1 = w1*t1+w2*t2-f*t2
    x2 = w1*t1 + w2*t2
    T = t1+t2,
    all vars >= 0.
    You want to use the three constraints to eliminate x1, and so express x2, t1 and t2 in terms of x1. That is a simple linear-equation-solving exercise.

    RGV
     
  4. Oct 9, 2012 #3
    Thank you for your reply. It was very helpful .
     
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