A sequential bargaining protocol for land rental arrangements

Abstract

We consider land rental between a single tenant and several lessors. The tenant should negotiate sequentially with each lessor for the available land. In each stage, we apply the Nash bargaining solution, as a short-cut to solving non-cooperative bargaining games. Our results imply that, when all land is necessary, a uniform price per unit is more favorable for the tenant than a lessor-dependent price. Furthermore, a lessor is better off with a lessor-dependent price only when negotiating first. For the tenant, lessors’ merging is relevant with lessor-dependent price but not with uniform price.

Appendix

Proof of Theorem 4.1

We prove the following (stronger) result:

Given $\left( a_i\right) _{i=1}^{s-1}\in A^{s-1}$ where $a_i = (p_i,x_i)$ for all $i<s$ in stage s, and $\beta ^s = K - \left( \sum _{i<s}p_ic_i + r\sum _{i \ge s}^{}c_i \right)$, as $\rho \rightarrow 1$, the final payoff allocation approaches

$$\begin{aligned} \left( \frac{\beta ^s }{2^{n-s+1}}, (p_1-r )c_1,\dots ,(p_{s-1}-r )c_{s-1}, \frac{\beta ^s }{2},\frac{\beta ^s }{2^2}, \dots , \frac{\beta ^s }{2^{n-s+1}} \right) \end{aligned}$$

if $\beta ^s > 0$ and $x_i=c_i$ for all $i<s$, and $(0,\dots ,0)$ if $\beta ^s <0$ or $x_i<c_i$ for some $i<s$.

We proceed by backward induction on s.

Assume $s=n$. Let $\left( a_i\right) _{i=1}^{n-1}\in A^{n-1}$ with $a_i=(x_i,p_i)$ for all $i<n$. In case $\sum _{i<n}x_i < E-c_n$, then there is not enough land left and the final payoff is zero for everyone. Since $E=c(N)$, $\sum _{i<n}x_i < E-c_n$ implies $x_i<c_i$ for some $i<n$. For the same reason, $\sum _{i<n}x_i \ge E-c_n$ implies $x_i=c_i$ for all $i<n$. Hence, $\sum _{i<n}x_i \ge E-c_n$ implies $\sum _{i<n}x_i = E-c_n$.

If $\beta ^n < 0$, then the prices are too high and agreement is not possible, so the final payoff is zero for everyone.

Assume now $\beta ^n > 0$ and $x_i=c_i$ for all $i<n$. Then, the tenant and lessor n face the bargaining problem $(D^n,d^n)$ with $d^n=(0,0)$.

An efficient agreement implies $x_n=c_n$. So, the Pareto frontier of $D^n$ is:

$$\begin{aligned} \partial D^n=\left\{ \left( u_0\left( a^n \right) ,u_n \left( a^n \right) \right) \in {\mathbb {R}}^2 : p_n\ge 0, x_n = c_n\right\} . \end{aligned}$$

The Nash solution is obtained by the maximization problem

$$\begin{aligned} \max \left\{ u_0 u_n:(u_0,u_n)\in \partial D^n \right\}&= \max \left\{ \left( K-\sum _{i\in N}p_ic_i\right) (p_n-r)c_n:p_n\ge 0\right\} \\&= \max \left\{ \left( \beta ^n + rc_n - p_n c_n\right) (p_n-r)c_n:p_n\ge 0\right\} \end{aligned}$$

where the unique maximum is reached at $p^*_n=\frac{\beta ^n+2r c_n}{2c_n}$. Given this, it is straightforward to check that the final payoff allocation (as $\rho \rightarrow 1$) becomes:

$$\begin{aligned} \left( \frac{\beta ^n}{2}, (p_1-r)c_1,\dots ,(p_{n-1}-r)c_{n-1}, \frac{\beta ^n}{2} \right) . \end{aligned}$$

So, the hypothesis is satisfied for stage $s=n$.

We now consider stage s.

Assume that the result is true in stage $s+1$ for $s<n$. Let $\left( a_i\right) _{i=1}^{s-1}\in A^{s-1}$ with $a_i=(x_i,p_i)$ for all $i<s$. By analogous reasoning as in stage n, we deduce that in case $\sum _{i<s}x_i < E-\sum _{i\ge s}c_i$, there is not enough land left and the final payoff is zero for everyone, and $\sum _{i<s}x_i \ge E-\sum _{i \ge s}c_i$ implies $\sum _{i<s}x_i = E-\sum _{i \ge s}c_i$.

If $\beta ^s < 0$, then the prices are too high and agreement is not possible, so the final payoff is zero for everyone.

Assume now $\beta ^s > 0$ and $x_i=c_i$ for all $i<s$. The tenant and lessor s face the bargaining problem $(D^s,d^s)$ with $d^s=(0,0)$.

An efficient agreement implies $x_s=c_s$. So, the Pareto frontier of $D^s$ is:

$$\begin{aligned} \partial D^s=\left\{ \left( u_0 \left( a^s \right) ,u_s\left( a^s \right) \right) \in {\mathbb {R}}^2: p_s\ge 0, x_{s} = c_{s} \right\} . \end{aligned}$$

The Nash solution is obtained by the maximization problem, which is given as follows. By induction hypothesis, given that the payoff for the tenant in stage $s+1$ is $\frac{\beta ^{s+1}}{2^{n-s}}$, $\max \{u_0 u_s:(u_0,u_s)\in \partial D^s\}$ is equal to

$$\begin{aligned} \max \left\{ \frac{\beta ^{s+1}}{ 2^{n-s}}(p_s-r)c_s:p_s\ge 0\right\} \end{aligned}$$

where the unique maximum is reached at $p^*_s=\frac{\beta ^s+2r c_s}{2c_s}$. Given this, the final payoff allocation (as $\rho \rightarrow 1$) is:

$$\begin{aligned} \left( \frac{\beta ^s }{2^{n-s+1}}, (p_1-r )c_1,\dots ,(p_{s-1}-r )c_{s-1}, \frac{\beta ^s }{2},\frac{\beta ^s }{2^2},\dots , \frac{\beta ^s }{2^{n-s+1}} \right) . \end{aligned}$$

$\square$

Proof of Theorem 4.2

We define $\pi ^*:{\mathbb {R}}_+ \rightarrow {\mathbb {R}}_+$ as $\pi ^{*}(p) = \max \left\{ p,\frac{K+rE}{2E}\right\}$. We prove the following (stronger) result:

Given $\left( a_i\right) _{i=1}^{s-1}\in A^{s-1}$ with $a_i=(p_i,x_i)$ for all $i<s$ at stage s, and $p^{\prime }_{s-1}=\max _{i<s}\{p_i\}$ (for notational convenience, we assume $p^{\prime }_{0}=0$ when $s=1$), the final price is $\pi ^*\left( p^{\prime }_{s-1}\right)$ and the final payoff allocation (as $\rho \rightarrow 1$) becomes

$$\begin{aligned} \left( \frac{K-rE}{2}, \frac{K-rE}{2E}c_1, \dots , \frac{K-rE}{2E}c_n \right) \end{aligned}$$

(6)

if $p^{\prime }_{s-1}< \frac{K+rE}{2E} < \frac{K}{E}$ and $x_i=c_i$ for all $i<s$,

$$\begin{aligned} \left( K- p^{\prime }_{s-1} E, \left( p^{\prime }_{s-1}-r \right) c_1,\dots ,\left( p^{\prime }_{s-1}-r\right) c_n \right) \end{aligned}$$

(7)

if $\frac{K+rE}{2E} \le p^{\prime }_{s-1} \le \frac{K}{E}$ and $x_i=c_i$ for all $i<s$, and

$$\begin{aligned} \left( 0,\dots ,0 \right) \end{aligned}$$

(8)

if $p^{\prime }_{s-1} > \frac{K}{E}$ or $x_i<c_i$ for some $i<s$.

For any stage $s\in N$, in case $p^{\prime }_{s-1} > \frac{K}{E}$ or $x_i<c_i$ for any $i<s$, there is no possible agreement, so the final payoff is zero for everyone, as stated in (8). Hence, from now on, we assume $p^{\prime }_{s-1} \le \frac{K}{E}$ and $x_i=c_i$ for all $i<s$.

We proceed by backward induction on s.

Assume $s=n$. Let $\left( a_i \right) _{i=1}^{n-1}\in A^{n-1}$ with $a_i=(p_i,x_i)$ for all $i<n$. The tenant and lessor n face the bargaining problem $({\hat{D}}^n,{\hat{d}}^n)$ with ${\hat{d}}^n=(0,0)$. An efficient agreement is only possible when $x_n=c_n$ and $p_n\le \frac{K}{E}$, so the Pareto frontier of ${\hat{D}}^n$ is:

$$\begin{aligned} \partial {\hat{D}}^n&=\left\{ \left( u_0 \left( {\hat{a}}^n\right) ,u_0 \left( {\hat{a}}^n\right) \right) \in {\mathbb {R}}^{2}: p_n\le \frac{K}{E}, x_n=c_n \right\} \\&= \left\{ \left( K-p_{n}^{\prime }E,(p_{n}^{\prime }-r)x_n\right) : p_n\le \frac{K}{E}, x_n=c_n \right\} \\&= \left\{ \left( K-p_{n}E,(p_{n}-r)c_n\right) : p^{\prime }_{n-1} \le p_n\le \frac{K}{E} \right\} . \end{aligned}$$

The Nash solution is obtained by the maximization problem:

$$\begin{aligned} \max \left\{ u_0 u_n:(u_0,u_n)\in \partial {\hat{D}}^n \right\} =\max \left\{ (K-p_{n}E)(p_{n}-r)c_n: p^{\prime }_{n-1} \le p_n\le \frac{K}{E} \right\} . \end{aligned}$$

The product $u_0 u_n$ determines a concave parabola whose vertex is at $\frac{K+rE}{2E}$.

We have the following three cases:

First Case: If $p^{\prime }_{n-1}< \frac{K+rE}{2E} < \frac{K}{E}$, the maximum is reached at $p_n^*=\frac{K+rE}{2E}$. By definition, $\pi ^*(p_{n-1}^{\prime })=\max \{p_{n-1}^{\prime },\frac{K+rE}{2E}\}=\frac{K+rE}{2E}$. Then, the final price is $\pi ^*(p_{n-1}^{\prime })=\frac{K+rE}{2E}$. From this, it is straightforward to check that the final payoff allocation is given as in (6).

Second Case: If $\frac{K+rE}{2E} \le p^{\prime }_{n-1}\le \frac{K}{E}$, the maximum is reached at $p_{n}^*=p^{\prime }_{n-1}$. The final price is $\pi ^*(p_{n-1}^{\prime })=p_{n-1}^{\prime }$. From this, it is straightforward to check that the final payoff allocation is given as in (7).

Third Case: If $\frac{K+rE}{2E} \ge \frac{K}{E}$, we deduce that $rE \ge K$ which is a contradiction because of the condition $rE < K$. Therefore, this case is not possible.

We now consider stage s.

Assume that the result is true in stage $s+1$ for $s<n$. Let $\left( a_i \right) _{i=1}^{s-1}\in A^{s-1}$ with $a_i=(p_i,x_i)$ for all $i<s$. The tenant and lessor s face the bargaining problem $({\hat{D}}^s,{\hat{d}}^s)$ with ${\hat{d}}^s=(0,0)$. An efficient agreement is only possible when $x_s=c_s$ and $p_s \le \frac{K}{E}$. By induction hypothesis, the price is $\pi _{}^*(p_{s}^{\prime })$. So the Pareto frontier of ${\hat{D}}^s$ is:

$$\begin{aligned} \partial {\hat{D}}^s&= \left\{ \left( u_0({\hat{a}}^s), u_s({\hat{a}}^s) \right) \in {\mathbb {R}}^{2}: p_s \le \frac{K}{E}, x_s=c_s\right\} \\&=\left\{ \left( K-\pi _{}^*(p_{s}^{\prime })E,\left( \pi _{}^*(p_{s}^{\prime })-r\right) x_s\right) : p_s \le \frac{K}{E}, x_s=c_s \right\} \\&=\left\{ \left( K-\pi _{}^*(p_{s})E,\left( \pi _{}^*(p_{s})-r\right) c_s\right) : p^{\prime }_{s-1} \le p_s \le \frac{K}{E} \right\} . \end{aligned}$$

The Nash solution is obtained by the maximization problem $\max \{ u_0 u_s:(u_0,u_s)\in \partial {\hat{D}}^s \}$, or equivalently,

$$\begin{aligned} \max \left\{ (K-\pi _{}^*(p_{s})E)(\pi _{}^*(p_{s})-r)c_s: p^{\prime }_{s-1} \le p_s \le \frac{K}{E} \right\} . \end{aligned}$$

Since $\pi _{}^*(p_{s})$ is constant for $p_{s}^{}\le \frac{K+rE}{2E}$, the maximization problem can be rewritten as

$$\begin{aligned} \max \left\{ (K-p_{s}E)(p_{s}-r)c_s: \pi ^*\left( p^{\prime }_{s-1}\right) \le p_s \le \frac{K}{E} \right\} . \end{aligned}$$

The product $(K-p_{s}E)(p_{s}-r)c_s$ determines a concave parabola whose vertex is at $\frac{K+rE}{2E}$.

Now, we have the following three cases:

First Case: If $p^{\prime }_{s-1}< \frac{K+rE}{2E} < \frac{K}{E}$, since $\pi ^*(p_{s-1}^{\prime })=\max \{p_{s-1}^{\prime },\frac{K+rE}{2E}\}$, we deduce $\pi ^*(p_{s-1}^{\prime })=\frac{K+rE}{2E}$. Hence, $\pi ^*(p_{s-1}^{\prime })\le \frac{K+rE}{2E} \le \frac{K}{E}$, and thus the maximum is reached at $p_s^*=\frac{K+rE}{2E}$. Then, the final price is $\frac{K+rE}{2E}=\pi ^*(p_{s-1}^{\prime })$. From this, it is straightforward to check that the final payoff allocation is given as in (6).

Second Case: If $\frac{K+rE}{2E} \le p^{\prime }_{s-1}\le \frac{K}{E}$, since $\pi ^*(p_{s-1}^{\prime })=\max \{p_{s-1}^{\prime },\frac{K+rE}{2E}\}$, we deduce $\pi ^*(p_{s-1}^{\prime })= p^{\prime }_{s-1}$. Hence, $\frac{K+rE}{2E} \le \pi ^*(p_{s-1}^{\prime })\le \frac{K}{E}$, and thus the maximum is reached at $p_s^*= \pi ^*(p_{s-1}^{\prime })$. Then, the final price is $p_{s-1}^{\prime }=\pi ^*(p_{s-1}^{\prime })$. From this, it is straightforward to check that the final payoff allocation is given as in (7).

Third Case: If $\frac{K+rE}{2E}\ge \frac{K}{E}$, we deduce that $rE \ge K$ which is a contradiction, because of the hypothesis condition $rE < K$. Therefore, this case is not possible. $\square$

Proof of Theorem 5.1

Assume first $E\le c_{1},c_{2}$. The tenant, in case she does not reach an agreement with lessor 1 in stage 1, will face a bargaining problem $\left( D^{2},d^{2}\right)$ in stage 2 with lessor 2 with $d^{2}=\left( 0,0\right)$ and $D^{2}=\left\{ \left( u_{0},u_{2}\right) \in \mathbb {R}^{2}:u_{0}+u_{2}\le K-rE\right\}$. The Nash solution is $\left( \frac{K-rE}{2},\frac{K-rE}{2}\right)$. Hence, the expected final payoff for the tenant, provided there is disagreement in stage 1, is $\frac{K-rE}{2}$. From this, in stage 1, the tenant and lessor 1 face a bargaining problem $\left( D^{1},d^{1}\right)$ with $d^{1}=\left( \frac{K-rE}{2},0\right)$ and

$$\begin{aligned} D^{1}=\left\{ \left( K-p_{1}x_{1},\left( p_{1}-r\right) x_{1}\right) :\left( p_{1},x_{1}\right) \in \Delta ^0 \right\} \end{aligned}$$

where

$$\begin{aligned} \Delta ^0 =\left\{ (p_1,x_1)\in [r,\infty [\times \left[ E,c_{1}\right] : p_1x_1\le K \right\} . \end{aligned}$$

Now, any agreement $(p_1,x_1)\in \Delta ^0$ with $x_1>E$ is Pareto dominated by $(p'_1,x'_1)=\left( \frac{p_1x_1}{E},E\right)$. Hence, we can rewrite $D^{1}$ as

$$\begin{aligned} D^{1}=\left\{ \left( K-p_1E,(p_1-r)E\right) :p_1\in \left[ r,\frac{K}{E}\right] \right\} . \end{aligned}$$

The Nash solution for $\left( D{}^{1},d^{1}\right)$ maximizes $\left( K-p_1E-\frac{K-rE}{2}\right) (p_1-r)E$. This maximum is at $p^*_1=\frac{3r}{4}+\frac{K}{4E}$, so that the final payoff allocation, when $\rho \rightarrow 1$, becomes $\left( 3\frac{K-rE}{4},\frac{K-rE}{4},0\right)$.

Assume now $c_{2}<E\le c_{1}$ or $c_{2}<E\le c_{1}$. The smallest lessor is a dummy, so the tenant and the other lessor will equally share $K - rE$, and the dummy receives zero.

Finally, assume $E>c_{1},c_{2}$. Assume that the tenant and lessor 1 agree on some $\left( p_{1},x_{1}\right) \in \left[ 0,\infty \right. \left[ \times \left[ 0,c_{1}\right] \right.$ in stage 1. In order for agreement to be possible in stage 2, suppose $x_{1}\ge E-c_{2}$ and $\left( p_{1}-r\right) x_{1}\le K-rE$. Now, in stage 2, any individually rational choice $\left( p_{2},x_{2}\right)$ with $x_{2}>E-x_{1}$ is Pareto dominated by $\left( p_{2}',x_{2}'\right)$ with $p_{2}'=\frac{x_{2}}{E-x_{1}}p_{2}$ and $x_{2}'=E-x_{1}$. Hence, the tenant and lessor 2 face a bargaining problem $\left( D^{2},d^{2}\right)$ with $d^{2}=\left( 0,0\right)$ and

$$\begin{aligned} D^{2}=\left\{ \left( K-p_{1}x_{1}-\left( E-x_{1}\right) p_{2},\left( p_{2}-r\right) \left( E-x_{1}\right) \right) :p_{2}\in \Delta ^2 \right\} \end{aligned}$$

where $\Delta ^2=\left[ r,\frac{K-p_{1}x_{1}}{E-x_{1}}\right]$. Taking $\alpha =\left( E-x_{1}\right) p_{2}$, we can rewrite $D^2$ as

$$\begin{aligned} D^{2}&=\left\{ \left( K-p_{1}x_{1}-\alpha ,\alpha -\left( E-x_{1}\right) r\right) :\alpha \in \left[ \left( E-x_{1}\right) r,K-p_{1}x_{1}\right] \right\} \end{aligned}$$

equivalently (see Fig. 1, Right),

$$\begin{aligned} D^{2}&=\left\{ \left( u_0,u_2\right) \in {\mathbb {R}}^{2}:u_0+u_2\le K-rE-(p_1-r)x_1\right\} . \end{aligned}$$

By symmetry and efficiency, the Nash solution gives both players the same utility $\frac{K-rE-(p_1-r)x_1}{2}$ and it is uniquely determined with $x_{2}=E-x_{1}$ and $p_{2}=\frac{K-p_{1}x_{1}}{2\left( E-x_{1}\right) }+\frac{r}{2}$.

Given this, the bargaining problem in stage 1 is $\left( D^{1},d^{1}\right)$ given by $d^{1}=\left( 0,0\right)$ and

$$\begin{aligned} D^{1}=\left\{ \left( \frac{K-rE-(p_1-r)x_1}{2},\left( p_{1}-r\right) x_{1}\right) :\left( p_{1},x_{1}\right) \in \Delta ^1 \right\} \end{aligned}$$

where

$$\begin{aligned} \Delta ^1 =\left\{ (p_1,x_2)\in [0,\infty [\times \left[ E-c_{2},c_{1}\right] :\left( p_{1}-r\right) x_{1}\le K-rE\right\} . \end{aligned}$$

Taking $\beta =\left( p_{1}-r\right) x_{1}$, we can rewrite $D^{1}$ as

$$\begin{aligned} D^{1}&=\left\{ \left( \frac{K-rE-\beta }{2},\beta \right) :\beta \in \left[ -rc_{1},K-rE\right] \right\} \end{aligned}$$

equivalently (Fig. 2),

$$\begin{aligned} D^{1}&=\left\{ \left( u_0,u_2\right) :2u_0+u_2\le K-rE \right\} . \end{aligned}$$

The Nash solution maximizes $\left( K-rE-\beta \right) \beta$. This maximium is at $\beta ^{o}=\frac{K-rE}{2}$. Hence the final payoff allocation , when $\rho \rightarrow 1$, becomes

$$\begin{aligned} \left( \frac{K-rE-\beta ^{o}}{2},\beta ^{o},\frac{K-rE-\beta ^{o}}{2}\right) =\left( \frac{K-rE}{4},\frac{K-rE}{2},\frac{K-rE}{4}\right) . \end{aligned}$$

$\square$

Proof of Theorem 5.2

We focus on the case $c_{1},c_{2}<E$. The proof for the other cases is analogous to the proof of Theorem 5.1.

Assume that the tenant and lessor 1 agree on some $\left( p_{1},x_{1}\right) \in \left[ 0,\infty \right. \left[ \times \left[ 0,c_{1}\right] \right.$ in stage 1. In order for agreement to be possible in stage 2, we also assume $x_{1}\ge E-c_{2}$ and $p_{1}\le \frac{K}{E}$. Given this, the tenant and lessor 2 face the bargaining problem $\left( {\hat{D}}^{2},{\hat{d}}^{2}\right)$ with ${\hat{d}}^{2}=\left( 0,0\right)$ and

$$\begin{aligned} {\hat{D}}^{2}=\left\{ \left( K-\left( x_{1}+x_{2}\right) p_{2},\left( p_{2}-r\right) x_{2}\right) :\left( p_{2},x_{2}\right) \in \Delta ^2\right\} \end{aligned}$$

where

$$\begin{aligned} \Delta ^2=\left[ p_{1},\infty \right. \left[ \times \left[ E-x_{1},c_{2}\right] .\right. \end{aligned}$$

The individually rational Pareto frontier of ${\hat{D}}^{2}$ are the points $\left( u_0,g_{2}\left( u_{0}\right) \right)$, where $u_{0}\in \left[ 0,K- p_{1} E\right]$ and $g_{2}\left( u_{0}\right)$ is a function defined as follows: $g_2(u_0)$ the maximum of $\left( p_{2}-r\right) x_{2}$ subject to $p_{2}\ge p_{1}$, $p_{2}\ge r$, $E-x_{1}\le x_{2}\le c_{2}$ and $K-\left( x_{1}+x_{2}\right) p_{2}=u_{0}$.

Assume first $r=0$, this means that $g_2(u_0)$ reaches its maximum at $x_2=c_2$. Thus, we have that $g_2(u_0)=\frac{K-u_0}{x_1+c_2}c_2$.

Assume now $r>0$. The maximization is equivalent to maximizing the function

$$\begin{aligned} f\left( x_{2}\right) :=\left( \frac{K-u_{0}}{x_{1}+x_{2}}-r\right) x_{2} \end{aligned}$$

on the interval $x_{2}\in \left[ E-x_{1}, m_2 \right] ,$ where $m_2 = \min \left\{ c_{2}, \frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}\right\} .$

The derivative of f is given by $f'(x_2) = \frac{(K - u_0)x_1}{(x_1+x_2)^2} - r$, whose unique positive root is

$$\begin{aligned} x_{2}^{o}:=\sqrt{\frac{x_{1}}{r}\left( K-u_{0}\right) }-x_{1} \end{aligned}$$

associated to the price $p_{2}^{o}=\sqrt{\frac{\left( K-u_{0}\right) r}{x_{1}}}$. The second derivate is $f''(x_2) = -\frac{2(K-u_0)x_1}{(x_1 + x_2)^3}.$ Since $f''(x_2) < 0,$ we deduce that the maximum is unique, it is located at $x_2^o$ when it belongs to $[E - x_1, m_2]$, at $E - x_1$ when $x_2^o < E - x_1,$ and at $m_2$ when $m_2 < x_2^o.$

Hence, we have three cases:

• Case 1 If $x_{2}^{o}<E-x_{1}$ or, equivalently, $u_{0}>K-\frac{rE^{2}}{x_{1}}$, the unique maximum is at $x_{2}=E-x_{1}$ (with $p_{2}=\frac{K-u_{0}}{E}$), and it gives

  $$\begin{aligned} g_{2}\left( u_{0}\right) =f\left( E-x_{1}\right) =\left( \frac{K-u_{0}}{E}-r\right) \left( E-x_{1}\right) =\frac{E-x_{1}}{E}\left( K-rE-u_{0}\right) \end{aligned}$$

  which implies that, for $u_{0}>K-\frac{rE^{2}}{x_{1}}$, the frontier of ${\hat{D}}^{2}$ is a line with slope $-\frac{E-x_{1}}{E}$.

• Case 2 If $x_{2}^{o} \ge m_2,$ we have two subcases:

  • Case 2a If $c_{2}\le \frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}$ and $x_{2}^{o}\ge c_{2}$, or, equivalently, $u_{0}\le K - \left( x_{1} + c_{2}\right) \max \left\{ \max \{r, p_{1}\}, \frac{x_{1}+c_{2}}{x_{1}} r\right\}$, the maximum is at $x_{2}=c_{2}$ (with $p_{2}=\frac{K-u_{0}}{x_{1}+c_{2}}$), and it gives

    $$\begin{aligned} g_{2}\left( u_{0}\right) =f\left( c_{2}\right) =\frac{c_{2}}{x_{1}+c_{2}}\left( K-\left( x_{1}+c_{2}\right) r-u_{0}\right) \end{aligned}$$

    which implies that, for $u_{0}\le K-\left( x_{1}+c_{2}\right) \max \left\{ p_{1},\frac{x_{1}+c_{2}}{x_{1}}r\right\}$, the frontier of ${\hat{D}}^{2}$ is a line with slope $-\frac{c_{2}}{x_{1}+c_{2}}$.

  • Case 2b If $c_{2}\ge \frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}$ and $x_{2}^{o}\ge \frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}$, or, equivalently, $u_{0}\ge K - \frac{\max \{r, p_1\}x_1}{r} \min \left\{ \max \{r,p_{1}\}, \frac{x_{1} + c_2}{x_1}r \right\}$, the maximum is at $x_{2}=\frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}$ (with $p_{2}=\max \{r,p_{1}\}$), and it gives

    $$\begin{aligned} g_{2}\left( u_{0}\right)&=f\left( \frac{K-u_{0}}{\max \{r,p_{1}\}}-x_{1}\right) \\&=\frac{\max \{r,p_{1}\}-r}{\max \{r,p_{1}\}}\left( K-\max \{r,p_{1}\}x_{1}-u_{0}\right) \end{aligned}$$

    which implies that, for

    $$\begin{aligned} u_{0}\ge K - \frac{\max \{r, p_1\}x_1}{r} \min \left\{ \max \{r,p_{1}\}, \frac{x_{1} + c_2}{x_1}r \right\} , \end{aligned}$$

    the frontier of ${\hat{D}}^{2}$ is a line with slope $-\frac{\max \{r,p_{1}\}-r}{\max \{r,p_{1}\}}$.

• Case 3 In any other case, the maximum is at $x_{2}=x_{2}^{o}$ (with $p_{2}=p_{2}^{o}$), and it is

  $$\begin{aligned} g_{2}\left( u_{0}\right) =f\left( \sqrt{\frac{x_{1}}{r}\left( K-u_{0}\right) }-x_{1}\right) =\left( \sqrt{K - u_0} - \sqrt{rx_1}\right) ^2 \end{aligned}$$

  which implies that, in some cases, the frontier of ${\hat{D}}^2$ is a convex function.

From these cases allow us to define $g_{2}\left( u_{0}\right)$, for any $u_{0}\in \left[ 0,K-\max \left\{ p_{1},r\right\} E\right]$, as follows:

$$\begin{aligned} g_{2}\left( u_{0}\right) ={\left\{ \begin{array}{ll} \frac{E-x_{1}}{E}\left( K-rE-u_{0}\right) &{} \text { if } u_{0}>K-\frac{rE^{2}}{x_{1}}\\ \frac{c_{2}}{x_{1}+c_{2}}\left( K-\left( x_{1}+c_{2}\right) r-u_{0}\right) &{} \text { if } u_{0}\le K-\left( x_{1}+c_{2}\right) \max \left\{ \max \{r,p_{1}\},\frac{x_{1}+c_{2}}{x_{1}}r\right\} \\ \frac{\max \{r,p_{1}\}-r}{\max \{r,p_{1}\}}\left( K-\max \{r,p_{1}\}x_{1}-u_{0}\right) &{} \text { if } u_{0}\ge K-\frac{\max \{r,p_{1}\}x_{1}}{r}\min \left\{ \frac{x_{1}+c_{2}}{x_{1}}r,\max \{r,p_{1}\}\right\} \\ \left( \sqrt{K-u_{0}}-\sqrt{rx_{1}}\right) ^{2} &{} \text { otherwise} \end{array}\right. } \end{aligned}$$

for any $u_{0}\in \left[ 0,K-\max \left\{ p_{1},r\right\} E\right]$. Notice that Case 1 only applies when $K - \frac{rE^{2}}{x_{1}}<K-p_{1}E$ or, equivalently, $p_{1}x_{1}<rE$. In that case, and since $E\le x_{1}+c_{2}$, we have $\max \{r,p_{1}\} < \frac{x_{1} + c_{2}}{x_{1}}r$ and hence case 2b reduces to $u_{0}\ge K - \frac{\max \{r,p_{1}\}^{2}x_{1}}{r}$ which implies $u_{0}\ge K - \max \{r,p_{1}\}E$, which is impossible. Hence, case 1 and case 2b cannot happen simultaneously, and so $g_{2}$ is well-defined on the interval $\left[ 0,K - \max \left\{ p_{1}, r\right\} E\right]$.

Since $E\le x_{1}+c_{2}$, we have $\frac{rE}{x_{1}} \le \frac{x_{1}+c_{2}}{x_{1}}r$. There are three possibilities depending on $p_{1}x_{1}$:

• Small$p_{1}x_{1}$ If $p_{1}x_{1}<rE$, then case 2b vanishes:

  $$\begin{aligned} g_{2}\left( u_{0}\right) ={\left\{ \begin{array}{ll} \frac{E-x_{1}}{E}\left( K-rE-u_{0}\right) &{} \text{ if } u_{0}\ge K-\frac{rE^{2}}{x_{1}}\\ \frac{c_{2}}{x_{1}+c_{2}}\left( K-\left( x_{1}+c_{2}\right) r-u_{0}\right) &{} \text{ if } u_{0}\le K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r\\ \left( \sqrt{K-u_{0}}-\sqrt{rx_{1}}\right) ^{2} &{} \text{ if } u_{0}\in \left[ K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r,K-\frac{rE^{2}}{x_{1}}\right] . \end{array}\right. } \end{aligned}$$

  In this case, $g_{2}$ is convex. There are three candidates for the Nash solution:

  • Case 1$u_{0}^{s1}=\frac{K-rE}{2}$ only if $u_{0}^{s1}\ge K-\frac{rE^{2}}{x_{1}}$, but this case is not possible when K is large enough ($K>K^{s1}:=\frac{E+c_2}{E-c_{2}}rE$);

  • Case 2a$u_{0}^{s2}=\frac{K-\left( x_{1}+c_{2}\right) r}{2}$ only if $u_{0}^{s2}\le K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r$, which always holds for K large enough ($K>K^{s2}:=\frac{E+c_{2}}{E-c_{2}}\left( c_{1}+c_{2}\right) r$); and

  • Case 3$u_{0}^{s3}=\frac{K-r x_{1}}{8}\left( 1-\sqrt{1-\frac{16K}{K-r x_{1}}}\right)$ only if $K-rx_{1}\ge 16K$, which is not possible.

  Hence, for $K=\min \{K^{s1},K^{s2}\}=K^{s1}$ large enough, and for each pair $\left( x_{1},p_{1}\right)$ with $x_{1}p_{1}<rE$, the tenant and lessor 2 will agree on some $\left( p_{2}^{*},x_{2}^{*}\right)$ such that the tenant’s final payoff is $u_{0}^{2a}$ (case 2a). This implies $p_{2}^{*}=r+\frac{K}{\left( x_{1}+c_{2}\right) }$ and $x_{2}^{*}=c_{2}$.

• Medium$p_{1}x_{1}$ If $p_{1}x_{1}\in \left[ rE,\left( x_{1}+c_{2}\right) r\right]$, then case 1 vanishes:

  $$\begin{aligned} g_{2}\left( u_{0}\right) ={\left\{ \begin{array}{ll} \frac{c_{2}}{x_{1}+c_{2}}\left( K-\left( x_{1}+c_{2}\right) r-u_{0}\right) &{} \text{ if } u_{0}\le K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r\\ \frac{p_{1}-r}{p_{1}}\left( K-p_{1}x_{1}-u_{0}\right) &{} \text{ if } u_{0}\ge K-\frac{x_{1}}{r}p_{1}^{2}\\ \left( \sqrt{K-u_{0}}-\sqrt{rx_{1}}\right) ^{2} &{} \text{ if } u_{0}\in \left[ K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r,K-\frac{x_{1}}{r}p_{1}^{2}\right] . \end{array}\right. } \end{aligned}$$

  In this case, $g_{2}$ is again convex and there are three candidates for the Nash solution:

  • Case 2a$u_{0}^{2a}=\frac{K-\left( x_{1}+c_{2}\right) r}{2}$ only if $u_{0}^{2a}\le K-\frac{\left( x_{1}+c_{2}\right) ^{2}}{x_{1}}r$, which always holds for K large enough ($K>K^{s3}=\frac{E+c_{2}}{E-c_{2}}\left( c_{1}+c_{2}\right) r$);

  • Case 2b$u_{0}^{2b}=\frac{K-p_{1}x_{1}}{2}$ only if $u_{0}^{2b}\ge K-\frac{x_{1}}{r}p_{1}^{2}$, which does not hold for K large enough ($K>K^{s4}=\left( c_{1}+2c_{2}\right) \frac{rE}{E-c_{2}}$); and

  • Case 3$u_{0}^{3}=\frac{K-r{ x_{1}}}{8}\left( 1-\sqrt{1-\frac{16K}{K-r{ x_{1}}}}\right)$ only if $K-rx_{1}\ge 16K$, which is not possible.

  Hence, for $K=\min \{K^{s3},K^{s4}\}=K^{s4}$ large enough, and for each pair $\left( p_{1},x_{1}\right)$ with $p_{1}x_{1}\in \left[ rE,\left( x_{1}+c_{2}\right) r\right]$, the tenant and lessor 2 will agree on some $\left( p_{2}^{*},x_{2}^{*}\right)$ such that the tenant’s final payoff is $u_{0}^{2a}$ (case 2a). This implies again that $p_{2}^{*}=r+\frac{K}{\left( x_{1}+c_{2}\right) }$ and $x_{2}^{*}=c_{2}$.

• Large$x_{1}p_{1}$ If $x_{1}p_{1}\ge \left( x_{1}+c_{2}\right) r$, then cases 1 and 3 vanish

  $$\begin{aligned} g_{2}\left( u_{0}\right) ={\left\{ \begin{array}{ll} \frac{c_{2}}{x_{1}+c_{2}}\left( K-\left( x_{1}+c_{2}\right) r-u_{0}\right) &{} \text{ if } u_{0}\le K-\left( x_{1}+c_{2}\right) p_{1}\\ \frac{p_{1}-r}{p_{1}}\left( K-p_{1}x_{1}-u_{0}\right) &{} \text{ if } u_{0}\ge K-\left( x_{1}+c_{2}\right) p_{1}. \end{array}\right. } \end{aligned}$$

  In this case, $g_{2}$ is concave and the unique (generalized) Nash solution if given by:

  $$\begin{aligned} u_{0}^{*}={\left\{ \begin{array}{ll} u_{0}^{2a} &{} \text{ if } u_{0}^{2a}\le K-\left( x_{1}+c_{2}\right) p_{1}\\ u_{0}^{2b} &{} \text{ if } u_{0}^{2b}\ge K-\left( x_{1}+c_{2}\right) p_{1}\\ K-\left( c_{1}+c_{2}\right) p_{1} &{} \text{ otherwise } \end{array}\right. } \end{aligned}$$

  which is equivalent to:

  $$\begin{aligned} u_{0}^{*}={\left\{ \begin{array}{ll} \frac{K-\left( x_{1}+c_{2}\right) r}{2} &{} \text{ if } p_{1}\le \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}\\ \frac{K-p_{1}x_{1}}{2} &{} \text{ if } p_{1}\ge \frac{K}{x_{1}+2c_{2}}\\ K-\left( c_{1}+c_{2}\right) p_{1} &{} \text{ if } p_{1}\in \left[ \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2},\frac{K}{x_{1}+2c_{2}}\right] . \end{array}\right. } \end{aligned}$$

  For K large enough ($K>K^{s5}=\left( c_{1}+c_{2}\right) \left( 1+2\frac{c_{2}}{E-c_{1}}\right) r$), case $p_{1}\le \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}$ is compatible with $x_{1}p_{1}\ge \left( x_{1}+c_{2}\right) r$ and hence the three cases are nondegenerate. Hence, for each pair $\left( p_{1},x_{1}\right)$ with $p_{1}x_{1}\ge \left( x_{1}+c_{2}\right) r$, the tenant and lessor 2 will agree on some $\left( p_{2}^{*},x_{2}^{*}\right)$ such that the tenant’s final payoff is $u_{0}^{*}$ given as before.

See Fig. 3 for three examples of $({\hat{D}}^2,{\hat{d}}^2)$ for three possible choices of $(p_1,x_1)$.

Assume now we are in stage 1 and K is large enough ($K>K^{s6}=\frac{\left( c_{1}+2c_{2}\right) \left( c_{1}+c_{2}\right) r}{E-c_{2}}$). For each possible agreement $\left( p_{1},x_{1}\right)$, the above cases allow us to anticipate the agreement $\left( p_{2}^{\left( p_{1},x_{1}\right) },x_{2}^{\left( p_{1},x_{1}\right) }\right)$ in stage 2:

$$\begin{aligned} \left( p_{2}^{\left( p_{1},x_{1}\right) },x_{2}^{\left( p_{1},x_{1}\right) }\right) ={\left\{ \begin{array}{ll} \left( \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2},c_{2}\right) &{} \text{ if } p_{1}\le \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}\\ \left( p_{1},\frac{K}{2p_{1}}-\frac{x_{1}}{2}\right) &{} \text{ if } p_{1}\ge \frac{K}{x_{1}+2c_{2}}\\ \left( p_{1},c_{2}\right) &{} \text{ if } p_{1}\in \left[ \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2},\frac{K}{x_{1}+2c_{2}}\right] . \end{array}\right. } \end{aligned}$$

(9)

Therefore, we have a bargaining problem $\left( {\hat{D}}^{1},{\hat{d}}^{1}\right)$ with ${\hat{d}}^{1}=\left( 0,0\right)$ and

$$\begin{aligned} {\hat{D}}^{1}=\left\{ \left( K-\left( x_{1}+x_{2}^{\left( p_{1},x_{1}\right) }\right) p_{2}^{\left( p_{1},x_{1}\right) },\left( p_{2}^{\left( p_{1},x_{1}\right) }-r\right) x_{1}\right) :\left( p_{1},x_{1}\right) \in {\hat{\Delta }}^1 \right\} \end{aligned}$$

where ${\hat{\Delta }}^1 = \left[ 0,\frac{K}{E}\right] \times \left[ E-c_{2},c_{1}\right]$.

In particular, given an agreement $\left( p_{1},x_{1}\right) \in {\hat{\Delta }}^1$, the final payoff for the tenant and lessor 1, as $\rho \rightarrow 1$, is given by $\left( u^*_{0},u^*_{1}\right) =$

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \frac{K-\left( x_{1}+c_{2}\right) r}{2},\frac{K-\left( x_{1}+c_{2}\right) r}{2\left( x_{1}+c_{2}\right) } x_1\right) &{} \text{ if } p_{1}\le \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}\\ \left( \frac{K-p_{1}x_{1}}{2},\left( p_{1}-r\right) x_{1}\right) &{} \text{ if } p_{1}\ge \frac{K}{x_{1}+2c_{2}}\\ \left( K-\left( x_{1}+c_{2}\right) p_{1},\left( p_{1}-r\right) x_{1}\right) &{} \text{ if } p_{1}\in \left[ \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2},\frac{K}{x_{1}+2c_{2}}\right] . \end{array}\right. } \end{aligned}$$

(10)

The generalized Nash solution is given by a pair $(p_1,x_1)$ that maximizes $u^*_0 u^*_1$ (when this maximization problem has a unique solution). Let $u_{0}\in \left[ 0,K-rE\right]$ be the utility that the tenant can get. The Pareto frontier of ${\hat{D}}^{1}$ is determined by a function $g_{1}\left( u_{0}\right)$ which gives the maximum that lessor 1 can get when the tenant gets $u_{0}$, i.e.

$$\begin{aligned} \partial {\hat{D}}^{1}=\left\{ \left( u_0,g_{1}(u_0)\right) \in {\mathbb {R}}^{2}:u_{0}\in [0,K-rE]\right\} \end{aligned}$$

with $g_1(u_0)$ the maximum of $u^*_1$ subject to $u^*_0 = u_0$. From (10), we have three cases depending on $p_1$. If $p_{1}\le \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}$, we maximize $\frac{K-\left( x_{1}+c_{2}\right) r}{2\left( x_{1}+c_{2}\right) }x_{1}$ subject to $\frac{K-\left( x_{1}+c_{2}\right) r}{2}=u_{0}$, which is equivalent to maximize $\frac{ru_{0}x_{1}}{K-2u_{0}}$. Since it is increasing on $x_{1}$, we deduce that the optimal $x_1$ satisfies $p_{1}\ge \frac{K}{2\left( x_{1}+c_{2}\right) }+\frac{r}{2}$. If $p_{1}\ge \frac{K}{x_{1}+2c_{2}}$, we maximize $\left( p_{1}-r\right) x_{1}$ subject to $\frac{K-p_{1}x_{1}}{2}=u_{0}$, which is equivalent to maximize $K-2u_{0}-rx_{1}$. Since it is decreasing on $x_{1}$, we deduce that the optimal $x_1$ satisfies $p_{1}\le \frac{K}{x_{1}+2c_{2}}$.

We then maximize $\left( p_{1}-r\right) x_{1}$ subject to $K-\left( x_{1}+c_{2}\right) p_{1}=u_{0}$, which is equivalent to maximizing $\left( \frac{K-u_{0}}{x_{1}+c_{2}}-r\right) x_{1},$ equivalent to function f. Hence, by an analogous reasoning as before, we deduce that the maximum is at

$$\begin{aligned} x_{1}^{o}=\sqrt{\frac{\left( K-u_{0}\right) c_{2}}{r}}-c_{2}. \end{aligned}$$

Hence, we have three cases depending on wherever $x_{1}^{o}\ge c_{1}$ (equivalently, $u_{0}\le K-\frac{\left( c_{1}+c_{2}\right) ^{2}r}{c_{2}}$), $x_{1}^{o}\in \left[ E-c_{2},c_{1}\right]$ (equivalently, $u_{0}\in \left[ K-\frac{\left( c_{1}+c_{2}\right) ^{2}r}{c_{2}},K-\frac{rE^{2}}{c_{2}}\right]$), or $x_{1}^{o}\le E-c_{2}$ (equivalently, $u_{0}\ge K-\frac{rE^{2}}{c_{2}}$). The maximum is obtained, respectively, with $(p_1,x_1)=\left( \frac{K-u_{0}}{c_{1}+c_{2}},c_1\right)$, $(p_1,x_1)=\left( \sqrt{\frac{\left( K-u_{0}\right) r}{c_{2}}},x_{1}^{0}\right)$, and $(p_1,x_1)=\left( \frac{K-u_{0}}{E},E-c_{2}\right)$. From this, we have

$$\begin{aligned} g_1(u_0)={\left\{ \begin{array}{ll} \frac{c_{1}}{c_{1}+c_{2}}\left( K-\left( c_{1}+c_{2}\right) r-u_{0}\right) &{}\text { if }u_{0}\le K-\frac{\left( c_{1}+c_{2}\right) ^{2}r}{c_{2}}\\ \left( \sqrt{K-u_{0}}-\sqrt{rc_{2}}\right) ^{2}&{}\text { if }u_{0}\in \left[ K-\frac{\left( c_{1}+c_{2}\right) ^{2}r}{c_{2}},K-\frac{rE^{2}}{c_{2}}\right] \\ \frac{1}{\left( E-c_{2}\right) E}\left( K-rE-u_{0}\right) &{}\text { if }u_{0}\ge K-\frac{rE^{2}}{c_{2}}\\ \end{array}\right. } \end{aligned}$$

which determines the bargaining problem in stage 1 (see Fig. 4). For K large enough ($K>K^{s7}=\frac{2c_{1}+c_{2}}{c_{2}}\left( c_{1}+c_{2}\right) r$), the generalized Nash bargaining solution determines $u_{0}=\frac{K-\left( c_{1}+c_{2}\right) r}{2}$ as final payoff for the tenant, with $(p_1,x_1)=\left( \frac{K}{2(c_1+c_2)}+\frac{r}{2},c_1\right)$. From (9), we deduce that, at stage 2, the final agreement, as $\rho \rightarrow 1$, becomes $(p^*_2,x^*_2)=\left( \frac{K}{2(c_1+c_2)}+\frac{r}{2},c_2\right)$ and so the final payoff for each lessor $i\in \{1,2\}$ becomes $u_i = (p_2-r)x_i = \frac{c_iK}{2\left( c_{1} + c_{2}\right) } - \frac{c_ir}{2}$. $\square$