# Homework Help: Kuhn-Tucker Optimization Problem

1. Sep 30, 2012

### Dev06

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$\geq$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. Oct 1, 2012

### Ray Vickson

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

3. Oct 9, 2012