Optimizing Payoffs for Renting an Apartment

[brojesus111]

Homework Statement

A boyfriend (B) and girlfriend (G) are going to rent an apartment. They have narrowed the offer to between $300,000 and $400,000. Their respective ordinal payoff functions for the amount spent are:

u_B(x)= -2x+7 when 3 <= x <= 3.5 and 0 when 3.5 <= x <= 4
u_G(x)= 0 when 3 <= x <= 3.5 and 3x-10.5 when 3.5 <= x <= 4

a) The following optimization problem will give the best compromise solution:

min { max { | u_B(x) - u*_B| , | u_G(x)-u*_G| } }
subject to 3.5 <= x <= 4

where u*_B and u*_G are the best possible payoff for the boyfriend and girlfriend respectively. Will this point ever be Pareto efficient? Explain.

The Attempt at a Solution

a) I realize that u*_B and u*_G are fixed numbers and that we are only considering the interval between 3.5 and 4. Our professor has given us the hint that the above statement can be simplified to max(1, 12 - 3x).

I don't understand why the term |u_B(x)-u*_B| is a constant 1. From 3.5 to 4, u_B(x) is 0. How do we get 1 from |0 - u*_B|? I'm also not sure where 12 - 3x comes from. From 3.5 to 4 u_G(x) is 3x - 10.5, so we have |3x - 10.5 - u*_G|.

In regards to Pareto efficient, I know that it has to do with the idea that it is impossible to increase one of the payoffs without hurting the other decision maker's payoff. I guess if one of the terms is constant, then this point will be Pareto efficient.

Any help is appreciated. Thank you.

 

[subha_iit]

The domain of the x is 3<=x<=4;
In the Domain of x, the value of u*_B= -(2*3)+7 = 1; [As u*_B= max value of u_B(x)]
In the same way, the value of u*_G= (3*4)-10.5 = 1.5.

Now find out function |u_B(x)-u*_b| and |u_G(x)-u*_G| in the domain of interest, i.e, 3.5<=x<=4

So, in the doamin of 3.5<=x<=4,

|u_B(x)-u*_b|=|0-1|=1

And |u_G(x)-u*_G|= |3x-10.5-1.5| = |3x-12| = (12-3x)

So now the problem is min{max(1, 12-3x)}

I think now you can figure it out.

 

[brojesus111]

The domain of the x is 3<=x<=4;
In the Domain of x, the value of u*_B= -(2*3)+7 = 1; [As u*_B= max value of u_B(x)]
In the same way, the value of u*_G= (3*4)-10.5 = 1.5.

Now find out function |u_B(x)-u*_b| and |u_G(x)-u*_G| in the domain of interest, i.e, 3.5<=x<=4

So, in the doamin of 3.5<=x<=4,

|u_B(x)-u*_b|=|0-1|=1

And |u_G(x)-u*_G|= |3x-10.5-1.5| = |3x-12| = (12-3x)

So now the problem is min{max(1, 12-3x)}

I think now you can figure it out.

Got it. I don't know why I didn't think about actually finding the values for the u_B* and u_G*.

And since the boyfriend has a constant payoff in this optimization problem, the point will be Pareto efficient since the girlfriend can seek the biggest payoff without hurting the payoff of her boyfriend.

If we graph the function with respect to x, we can see that from 3.667 to 4 we get the best possible solution.

Thanks!