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I need help with the following question.

Take a two-agent bargaining problem. d=(0,0). Pick a strictly concave function for the bargaining frontier. Fix the intersection of the frontier with the y-axis to be at 1. Vary the intersection of the frontier with the x-axis between 0.1 and 10 with increments of 0.1. Graph the points proposed by the following solution concepts.

So y^2=1-x^2/0.1 can be the function. Does anyone know how can it be computed in matlab or a similar program?

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  • $\begingroup$ Not sure if there's an existing MATLAB program for it, but it should be pretty easy to code one yourself. Have you tried this at all? $\endgroup$ – Herr K. Mar 11 at 3:43
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The answer depends on the "following solution concepts", which you did not specify. I guess the Nash solution and the Kalai-Smorodinsky solution will be among them.

The frontier functions you seem to propose are of the form $y_n(x)=\sqrt{1-\frac{100x^2}{n^2}}$, where $n\in\{1,2,\ldots,100\}$.

For the Nash solution you maximize $xy_n(x)$ and after substituting for $y_n(x)$ and setting the derivative to zero you will find $x_n=\frac{n}{\sqrt{200}}$ and $y_n=\frac{1}{\sqrt{2}}$. For the KS solution you set $y_n(x)=\frac{10x}{n}$ and after substituting you will again find $x_n=\frac{n}{\sqrt{200}}$ and $y_n=\frac{1}{\sqrt{2}}$. Graphing these solutions is easy.

Remark: That the solutions coincide is not a coincidence. The different frontier functions are constructed in such a way that they are all just horizontally "compressed" or "stretched" versions of the frontier function $y_{10}(x)=\sqrt{1-x^2}$. This means that they are really all equivalent in the sense that they result from each other after some rescaling of $x$ (the utility of agent 1). But the frontier function $y_{10}$ just describes the quarter unit circle in the positive orthant, which is symmetric w.r.t. the agents. In symmetric feasible sets the Nash and the KM solutions are known to coincide.

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