Consistency requirements and pattern methods in cost sharing problems with technological cooperation

Abstract

Using the discrete cost sharing model with technological cooperation, we investigate the implications of the requirement that demand manipulations must not affect the agents’ shares. In a context where the enforcing authority cannot prevent agents (who seek to reduce their cost shares) from splitting or merging their demands, the cost sharing methods used must make such artifices unprofitable. The paper introduces a family of rules that are immune to these demand manipulations, the pattern methods. Our main result is the characterization of these methods using the above requirement. For each one of these methods, the associated pattern indicates how to combine the technologies in order to meet the agents’ demands. Within this family, two rules stand out: the public Aumann–Shapley rule, which never rewards technological cooperation; and the private Aumann–Shapley rule, which always rewards technology providers. Fairness requirements imposing natural bounds (for the technological rent) allow to further differentiate these two rules.

Appendix

1.1 Figures

Fig. 1

An example of a pattern method for $m=3$

Fig. 2

Elementary flows $f^{\mathcal {S}_{{\tiny pub} }^{m}}$ and $f^{\mathcal {S}_{pr}^{m}}$ for $m=3$

Fig. 3

Remaining elementary flows for $m=3$

1.2 Additional computations for Example 3

Let us calculate the shares for agents 2 and 3. As previously discussed, $\partial _{2}C^{u*}(t)=0$ for all t such that $t_{2}=1$ and $\partial _{3}C^{u*}(t)=0$ for all t such that $t_{3}=2.$ The following table describes the marginal cost of the demand of agent 2 depending on the technologies and demands already introduced, as well as the flow associated to each of these marginal costs by the elementary flows $f^{ \mathcal {S}_{{\tiny pub}}^{4}}$ and $f^{\mathcal {S}_{{\tiny pr}}^{4}}$, associated respectively to the private and public Aumann–Shapley methods, as well as the elementary flow $f^{\mathcal {S}_{{\tiny *}}^{4}}.$

r	$\partial _{2}C^{u*}(r)$	$f_{2}^{\mathcal {S}_{{\tiny pub}}^{4}}(r)$	$f_{2}^{\mathcal {S}_{{\tiny pr}}^{4}}(r)$	$f_{2}^{\mathcal {S}_{{\tiny *}}^{4}}(r)$
(0, 0, 2, 0)	2	0	$\frac{1}{4}$	0
(1, 0, 2, 0)	2	0	0	$\frac{1}{12}$
(0, 1, 2, 0)	2	0	0	$\frac{1}{12}$
(0, 0, 2, 1)	1	0	0	$\frac{1}{12}$
(1, 1, 2, 0)	2	0	0	0
(1, 0, 2, 1)	1	0	0	0
(0, 1, 2, 1)	1	0	0	0
(1, 1, 2, 1)	1	$\frac{1}{4}$	0	0
(2, 0, 2, 0)	2	0	$\frac{1}{12}$	0
(0, 2, 2, 0)	2	0	$\frac{1}{12}$	0
(0, 0, 2, 2)	1	0	$\frac{1}{12}$	0
(2, 1, 2, 0)	2	0	0	$\frac{1}{24}$
(2, 0, 2, 1)	1	0	0	$\frac{1}{24}$
(1, 2, 2, 0)	2	0	0	$\frac{1}{24}$
(0, 2, 2, 1)	1	0	0	$\frac{1}{24}$
(1, 0, 2, 2)	1	0	0	$\frac{1}{24}$
(0, 1, 2, 2)	1	0	0	$\frac{1}{24}$
(2, 2, 2, 0)	2	0	$\frac{1}{12}$	0
(2, 0, 2, 2)	1	0	$\frac{1}{12}$	0
(0, 2, 2, 2)	1	0	$\frac{1}{12}$	0
(2, 1, 2, 1)	1	$\frac{1}{12}$	0	0
(1, 2, 2, 1)	1	$\frac{1}{12}$	0	0
(1, 1, 2, 2)	1	$\frac{1}{12}$	0	0
(2, 2, 2, 1)	1	$\frac{1}{12}$	0	$\frac{1}{12}$
(2, 1, 2, 2)	1	$\frac{1}{12}$	0	$\frac{1}{12}$
(1, 2, 2, 2)	1	$\frac{1}{12}$	0	$\frac{1}{12}$
(2, 2, 2, 2)	1	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$

We hence have $AS_{2}^{{\tiny pub}}(P)=AS_{2}^{{\tiny pub} }(P^{u})=\sum \nolimits _{r\in [e^{2},2e]}f_{2}^{\mathcal {S}_{{\tiny pub} }^{4}}(r)\partial _{2}C^{u*}(r)=1$, and $AS_{2}^{{\tiny pr}}(P)=AS_{2}^{ {\tiny pr}}(P^{u})=\sum \nolimits _{r\in [e^{2},2e]}f_{2}^{\mathcal {S}_{ {\tiny pr}}^{4}}(r)\partial _{2}C^{u*}(r)=\frac{3}{2}$. We also find $y_{2}^{*}(P)=y_{2}^{*}(P^{u})=\sum \nolimits _{r\in [e^{2},2e]}f_{2}^{\mathcal {S}_{{\tiny *}}^{4}}(r)\partial _{2}C^{u*}(r)=\frac{5}{4}.$

The following table describes the marginal cost of adding the technology of agent 3 depending on the technologies and demands already introduced, as well as the flow associated to each of these marginal costs by the elementary flows $f^{\mathcal {S}_{{pub}}^{4}}$ and $f^{\mathcal {S}_{{\small pr}}^{4}}$, associated respectively to the private and public Aumann–Shapley methods, as well as the elementary flow $f^{\mathcal {S}_{{\small *} }^{4}}.$

r	$\partial _{3}C^{u*}(r)$	$f_{3}^{\mathcal {S}_{{\tiny pub}}^{4}}(r)$	$f_{3}^{\mathcal {S}_{{\tiny pr}}^{4}}(r)$	$f_{3}^{\mathcal {S}_{{\tiny *}}^{4}}(r)$
(0, 0, 0, 1)	0	$\frac{1}{4}$	0	$\frac{1}{4}$
(1, 0, 0, 1)	0	$\frac{1}{12}$	0	$\frac{1}{12}$
(0, 1, 0, 1)	0	$\frac{1}{12}$	0	$\frac{1}{12}$
(0, 0, 1, 1)	0	$\frac{1}{12}$	0	$\frac{1}{12}$
(1, 1, 0, 1)	0	$\frac{1}{12}$	0	0
(1, 0, 1, 1)	0	$\frac{1}{12}$	0	0
(0, 1, 1, 1)	0	$\frac{1}{12}$	0	0
(1, 1, 1, 1)	0	$\frac{1}{4}$	0	0
(2, 0, 0, 1)	− 1	0	$\frac{1}{12}$	0
(0, 2, 0, 1)	− 1	0	$\frac{1}{12}$	0
(0, 0, 2, 1)	− 1	0	$\frac{1}{12}$	0
(2, 1, 0, 1)	− 1	0	0	$\frac{1}{24}$
(2, 0, 1, 1)	− 1	0	0	$\frac{1}{24}$
(1, 2, 0, 1)	− 1	0	0	$\frac{1}{24}$
(0, 2, 1, 1)	− 1	0	0	$\frac{1}{24}$
(1, 0, 2, 1)	− 1	0	0	$\frac{1}{24}$
(0, 1, 2, 1)	− 1	0	0	$\frac{1}{24}$
(2, 2, 0, 1)	− 4	0	$\frac{1}{12}$	0
(2, 0, 2, 1)	− 2	0	$\frac{1}{12}$	0
(0, 2, 2, 1)	− 2	0	$\frac{1}{12}$	0
(2, 1, 1, 1)	− 1	0	0	0
(1, 2, 1, 1)	− 1	0	0	0
(1, 1, 2, 1)	− 1	0	0	0
(2, 2, 1, 1)	− 4	0	0	$\frac{1}{12}$
(2, 1, 2, 1)	− 2	0	0	$\frac{1}{12}$
(1, 2, 2, 1)	− 2	0	0	$\frac{1}{12}$
(2, 2, 2, 1)	− 5	0	$\frac{1}{4}$	0

Taking the scalar product of column with the following ones, we obtain agent 3’s respective shares: $AS_{3}^{{\tiny pub}}(P)=AS_{3}^{{\tiny pub}}(P^{u})=\sum \nolimits _{r\in [e^{3},2e]}f_{3}^{\mathcal {S}_{{\tiny pub}}^{4}}(r)\partial _{3}C^{u*}(r)=0$, $AS_{3}^{{\tiny pr}}(P)=AS_{3}^{ {\tiny pr}}(P^{u})=\sum \nolimits _{r\in [e^{3},2e]}f_{3}^{\mathcal {S}_{ {\tiny pr}}^{4}}(r)\partial _{3}C^{u*}(r)=-\frac{13}{6}$ and $y_{3}^{*}(P)=y_{3}^{*}(P^{u})=\sum \nolimits _{r\in [e^{3},2e]}f_{3}^{\mathcal {S}_{*}^{4}}(r)\partial _{3}C^{u*}(r)=- \frac{11}{12}.$

1.3 Proofs of the Theorems

1.3.1 Theorem 2

Suppose y is a pattern method represented by a sequence of patterns $h=\left( h^{m}\right) _{m\ge 2}.$

Step 1  Showing that pattern methods satisfy NMS and ID

Substep 1a  Pattern methods satisfy NMS

Consider $P=(N,x,C)$, $P^{\prime }=(N^{\prime },C^{\prime },x^{\prime })\in P$ such that $P^{\prime }$ is a splitting manipulation of P available to some agent $i\in N$, with $N^{\prime }=(N{\setminus } i)\cup I$ . First, observe that $m=\sum _{j\in N\backslash i}x_{j}+x_{i}+\left| N_{0}(x)\right| =\sum _{j\in N\backslash i}x_{j}^{\prime }+\sum _{j\in I}x_{j}^{\prime }+\left| N_{0}(x^{\prime })\right| =m^{\prime }$ by the properties of a splitting manipulation. We thus have that,

$$\begin{aligned} y_{i}^{h_{m}}(N,x,C)=\sum _{k=1}^{x_{i}}y_{i_{k}}^{h_{m}}(N^{u},x^{u},C^{u})= \sum _{j\in I}\sum _{k=1}^{x_{j}^{\prime }}y_{j_{k}}^{h_{m^{\prime }}}=\sum _{j\in I}y_{j}^{h_{m}}(N^{\prime },C^{\prime },x^{\prime }) \end{aligned}$$

as $m=m^{\prime }$ and the fact that $P^{\prime }$ is a splitting manipulation of P.

We also need to show that it does not affect the shares of agents in $N\backslash i.$ Consider agent $j\in N\backslash i.$ Clearly, since $m=m^{\prime }$ and $x_{j}=x_{j}^{\prime },$ we have that $y_{j}^{h_{m}}(N,x,C)=y_{j}^{h_{m}}(N^{\prime },C^{\prime },x^{\prime }).$

Substep 1b  Pattern methods satisfy ID

Suppose that i is a demand dummy in problem $P=(N,x,C),$ with $x_{i}=1.$ Consider the problem $P^{\prime }=(N,C^{\prime },\left( 0_{i},x_{N\backslash i}\right) )$ where $C^{\prime }$ is the restriction of C to $\left[ 0,e^{N}+x_{N\backslash i}\right] .$ Since agent i is a demand dummy, we have that $(N^{u},x^{u},C^{u})=(N^{u}, \left( 0_{i},x_{N\backslash i}\right) ^{u},\left( C^{\prime }\right) ^{u})$ and thus that $y_{i}^{h_{m}}(N,x,C)= \sum _{k=1}^{x_{i}}y_{i_{k}}^{h_{m}}(N^{u},x^{u},C^{u})=y_{i_{1}}^{h_{m}}(N^{u}, x^{u},C^{u})$ while $y_{i}^{h_{m}}(N, \left( 0_{i},x_{N\backslash i}\right) , C)=y_{i_{1}}^{h_{m^{\prime }}}(N^{u},x^{u},C^{u}).$ Since $m=m^{\prime },$ we have that $y_{i}^{h_{m}}(N,x,C)=y_{i}^{h_{m}}(N,\left( 0_{i},x_{N\backslash i}\right) ,C )$ and $y_{j}^{h_{m}}(N,x,C)=y_{j}^{h_{m}}(N,\left( 0_{i},x_{N\backslash i}\right) ,C )$ for all $m\in N\backslash i.$

Step 2  Showing that a flow method that satisfies NMS and ID is a pattern method.

Consider a problem (N, x, C) where, without loss of generality, the first K agents in N have $x_{i}>0$ while the remaining ones have null demands.

Substep 2a

Since y meets no merging or splitting, we can write

$$\begin{aligned} y_{1}(P)=\sum _{i^{\prime }\in I_{1}}y_{i^{\prime }}(P^{(1)}),\quad \text { and } \quad y_{N{\setminus } 1}(P)=y_{N{\setminus } 1}(P^{(1)}), \end{aligned}$$

where $P^{(1)}$ is the problem obtained from P by splitting the demand of agent 1 into $x_{1}$ distinct (unitary) demands of 1. We have: $P^{(1)}=(N^{(1)},x^{(1)},C^{(1)})$, where

$$\begin{aligned} N^{(1)}= & {} (N{\setminus } 1)\cup I_{1}~~(\text {with}~|I_{1}|=x_{1}); \\ x^{(1)}= & {} (\underbrace{1,\ldots ,1}_{I_{1}},x_{N{\setminus } 1}); \\ C^{(1)*}(t_{I_{1}},t_{N{\setminus } 1})= & {} C^{*}\left( \hat{t} _{1},t_{N{\setminus } 1}\right) . \end{aligned}$$

By the same token, starting from the problem $P^{(1)}$ and splitting the demand of agent 2 into $x_{2}$ unitary demands gives:

$$\begin{aligned} y_{2}(P^{(1)})=\sum _{i^{\prime }\in I_{2}}y_{i^{\prime }}\left( P^{(2)}\right) ,\quad \text { and }\quad y_{N^{(1)}{\setminus } 2}\left( P^{(1)}\right) =y_{N^{(1)}{\setminus } 2}\left( P^{(2)}\right) . \end{aligned}$$

Combining, we obtain

$$\begin{aligned} y_{1}(P)=\sum _{i^{\prime }\in I_{1}}y_{i^{\prime }}\left( P^{(2)}\right) ,\quad y_{2}(P)=\sum _{i^{\prime }\in I_{2}}y_{i^{\prime }}\left( P^{(2)}\right) \quad \text { and }\quad y_{N{\setminus } \{1,2\}}(P)=y_{N{\setminus } \{1,2\}}\left( P^{(2)}\right) . \end{aligned}$$

Using the same argument repeatedly (K times, in fact), we get:

$$\begin{aligned} y_{1}(P)= & {} \sum _{i^{\prime }\in I_{1}}y_{i^{\prime }}\left( P^{(K)}\right) ,\ldots ,y_{K}(P)=\sum _{i^{\prime }\in I_{K}}y_{i^{\prime }}\left( P^{(K)}\right) \end{aligned}$$

(6)

$$\begin{aligned} y_{j}(P)= & {} y_{j}\left( P^{(K)}\right) ,\quad \forall j\in N{\setminus } \{1,\ldots ,K\}. \end{aligned}$$

(7)

Substep 2b

Since y meets irrelevant demand, we can write

$$\begin{aligned} y_{K+1}(P^{(K)})=y_{K+1}(\bar{P}^{(K+1)}),\quad \text { and }\quad y_{N^{(K)}{\setminus } K+1}(P^{(K)})=y_{N^{(K)}{\setminus } K+1}(\bar{P}^{(K+1)}), \end{aligned}$$

where $\bar{P}^{(K+1)}$ is the problem obtained from $P^{(K)}$ by adding a costless unit of demand to agent $K+1$. We have: $\bar{P}^{(K+1)}=(N^{(K)}, \bar{x}^{(K+1)},\bar{C}^{(K+1)})$, where $\bar{x}^{(K+1)}=\left( 1,x_{N^{(K)}{\setminus } K+1}^{(K)}\right)$ and $\bar{C}^{(K+1)}$ is such that for all $t\in \left[ 0,z^{N^{(K)}{\setminus } K+1}\right] ,$

$$\begin{aligned} \bar{C}^{(K+1)*}(t)= & {} C^{(K)*}(t); \\ \bar{C}^{(K+1)*}(t+e^{K+1})= & {} C^{(K)*}(t+e^{K+1}); \\ \bar{C}^{(K+1)*}(t+2e^{K+1})= & {} C^{(K)*}(t+e^{K+1}). \end{aligned}$$

Combining with the previous substep, we obtain

$$\begin{aligned} y_{1}(P)= & {} \sum _{i^{\prime }\in I_{1}}y_{i^{\prime }}(\bar{P} ^{(K+1)}),\ldots ,y_{K}(P)=\sum _{i^{\prime }\in I_{K}}y_{i^{\prime }}(\bar{P} ^{(K+1)}) \\ y_{j}(P)= & {} y_{j}(\bar{P}^{(K+1)}),\quad \forall j\in N{\setminus } \{1,\ldots ,K\}. \end{aligned}$$

Repeating the argument $n-\left| K\right|$ times, we obtain

$$\begin{aligned} y_{1}(P)= & {} \sum _{i^{\prime }\in I_{1}}y_{i^{\prime }}(\bar{P} ^{(n)}),\ldots ,y_{K}(P)=\sum _{i^{\prime }\in I_{K}}y_{i^{\prime }}(\bar{P} ^{(n)}) \\ y_{j}(P)= & {} y_{j}(\bar{P}^{(n)}),\quad \forall j\in N{\setminus } \{1,\ldots ,K\}. \end{aligned}$$

where $\bar{P}^{(n)}=(N^{(K)},\bar{x}^{(n)},\bar{C}^{(n)})$ and

$$\begin{aligned} N^{(K)}= & {} \cup _{i\in \{1,\ldots ,K\}}I_{i}\cup N{\setminus } \{1,\ldots ,K\}=N^{u} \\ \bar{x}^{(n)}= & {} e^{N^{(K)}}=e^{N^{u}}=x^{u} \\ \bar{C}^{(n)*}(t)= & {} C^{*}(\hat{t})=C^{u*}(t). \end{aligned}$$

Thus, $\bar{P}^{(n)}=(N^{u},x^{u},C^{u})$ and $\left\{ \begin{array}{ll} y_{i}(N,x,C)=\sum \limits _{k=1}^{x_{i}}y_{i_{k}}(N^{u},x^{u},C^{u}), &{}\quad \mathrm{if }\;x_{i}>0; \\ y_{i}(N,x,C)=y_{i}(N^{u},x^{u},C^{u}), &{}\quad \mathrm{if }\;x_{i}=0. \end{array} \right. ~~$ $\square$

1.3.2 Theorem 3

(i) $AS^{pub}$ satisfies Fair Rent

Consider a problem (N, x, C) s.t. $N_{0}(x)\equiv N{\setminus } N(x)\ne \emptyset$. By definition of $\mathcal {S}_{pub}^{m}$ we have $p(m)=e$, for $m=|N_{0}(x)|+x(N)$ and $p\in \mathcal {S}_{pub}^{m}$; and therefore each technological agent $i\in N_{0}(x)$ receives a null rent under the public Aumann–Shapley rule: $y_{i}(N,x,C)=0,\forall i\in N_{0}(x)$. Thus, we have by efficiency,

$$\begin{aligned} \sum _{j\in N}AS_{j}^{pub}(N,x,C)=\sum _{j\in N(x)}AS_{i}^{pub}(N,x,C)=C(N,x,C)\le C((N(x),x_{N(x)}); \end{aligned}$$

and hence Fair Rent is satisfied by $AS^{pub}$. Note that the inequality above comes from Property 1 of our cost functions with technological cooperation (see Sect. 2).

(ii) All other pattern methods violate Fair Rent

Suppose now that we have a pattern method $y\ne AS^{pub}$ that is represented by the symmetric flow $h^{m}$ (for each $m\in \mathrm {IN}$) on the box [0, 2e]. First, note from Definitions 5 and 6 that $y\ne AS^{pub}$ implies that there exists $m\ge 2$ and $z\in [0,2e]$ such that:⁷

$$\begin{aligned} h_{i}^{m}(2e,z)>0,z_{i}=1,z_{j}=2\quad \mathrm{for\;some }\;\;i,j\in N. \end{aligned}$$

(8)

Next, pick any problem (N, x, C) s.t. $|N_{0}(x)|+x(N)=m$, $i\in N_{0}(x)$ [$x_i=0$], $x_j=1$, and $\partial C^{u*}(z)<\frac{C^{*}(z^{x})-C^{*}((z^{x})^{N(x)})}{h_{i}^{m}(2e,z)}$. Then, since $i\in N_{0}(x)$ and $\partial _{i}C^{u*}(r)\le 0$ (from $x_{i}=0$), we have

$$\begin{aligned} \sum _{k\in N_{0}(x)}y_{k}(N,x,C)\le y_{i}(N,x,C)= & {} \sum \limits _{r\in \left[ e^{i},2e\right] }h_{i}^{m}(2e,r)\partial _{i}C^{u*}(r) \\= & {} h_{i}^{m}(2e,z)\partial _{i}C^{u*}(z)+\sum \limits _{r\in \left[ e^{i},2e\right] {\setminus } \{z\}}\overbrace{h_{i}^{m}(2e,r)\partial _{i}C^{u*}(r)}^{\le 0} \\< & {} h_{i}^{m}(2e,z)\frac{C^{*}(z^{x})-C^{*}\left( (z^{x})^{N(x)}\right) }{ h_{i}^{m}(2e,z)}=C^{*}(z^{x})-C^{*}((z^{x})^{N(x)}). \end{aligned}$$

That is to say,

$$\begin{aligned} \sum \limits _{k\in N(x)}y_{k}(N,x,C)=C^{*}(z^{x})-\sum _{k\in N_{0}(x)}y_{k}(N,x,C)>C^{*}\left( (z^{x})^{N(x)}\right) \equiv C\left( N(x),x_{N(x)}\right) . \end{aligned}$$

Therefore, Fair Rent is not met by y.  $\square$

1.3.3 Theorem 4

• (i) $AS^{pub}$ does not meet LBR. Fix $N\in \mathcal {N}$ and $x\in \mathrm {IN}^{N}$; and suppose that $i\in N_{0}(x)$, with $\left| C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \right| >0$. Since $AS_{i}^{pub}(N,x,C)=0$, we can claim that

  $$\begin{aligned} \left| AS_{i}^{pub}(N,x,C)\right| =0<k\left| C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \right| ,\quad \forall k\in (0,1). \end{aligned}$$

  Hence, $AS^{pub}$ violates LBR.

• (ii) $AS^{pr}$ satisfies $\frac{1}{m}$-LBR. Fix $N\in \mathcal {N}$, $x\in \mathrm {IN}^{N}$ and let $m\equiv |N_{0}(x)|+x(N)$. Next consider $i\in N_{0}(x)$ and, for any path $p:\{0,1,\ldots ,z_{1}^{x}+\cdots +z_{m}^{x}\}\rightarrow [0,2e]$, denote by $k_{i}^{p}\in \{0,1,\ldots ,z_{1}^{x}+\cdots +z_{m}^{x}\}$ the unique number such that $p_{i}(k_{i}^{p})=1$ and $p_{i}(k_{i}^{p}-1)=0$. Using this notation and the definition of $AS^{pr}$, we may then write:

$$\begin{aligned} AS_{i}^{pr}(N,x,C)= & {} \frac{1}{m!}\sum \limits _{p\in \mathcal {S}_{pr}^{m}} \overbrace{\partial _{i}C^{u*}(p(k_{p}^{i}))}^{\le 0} \end{aligned}$$

(9)

$$\begin{aligned}\le & {} \frac{1}{m!}\sum \limits _{{\tiny \begin{array}{c} p\in \mathcal {S}_{pr}^{m}: \\ p(k_{i}^{p})=2e-e^{i} \end{array}}}\partial _{i}C^{u*}(p(2e-e_{i})) \end{aligned}$$

(10)

Note that $p(k_{i}^{p})=2e-e^{i}$ means that the technology of i is activated last, when the demands of all other agents have already been produced. That is to say, $\partial _{i}C^{u*}(p(2e-e_{i}))=C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \le 0$. Substituting in (9), we get:

$$\begin{aligned} AS_{i}^{pr}(N,x,C)\le \frac{C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) }{m!}\sum \limits _{{\tiny \begin{array}{c} p\in \mathcal {S}_{pr}^{m}: \\ p(k_{i}^{p})=2e-e^{i} \end{array} }}1 \end{aligned}$$

(11)

A basic combinatorics exercise allows to see that we have exactly $(m-1)!$ paths $p:\{0,1,\ldots ,z_{1}^{x}+\cdots +z_{m}^{x}\}\rightarrow [0,2e]$ such that $p(k_{i}^{p})=2e-e^{i}$. Therefore, (11) becomes:

$$\begin{aligned} AS_{i}^{pr}(N,x,C)\le \frac{(m-1)!}{m!}\left[ C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \right] =\frac{1}{m}\left[ C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \right] . \end{aligned}$$

Since both the right and left hand sides are nonpositive, taking the absolute value gives $|AS_{i}^{pr}(N,x,C)|\ge \frac{1}{m}\left| C(N,x)-C\left( N{\setminus } i,x_{N{\setminus } i}\right) \right|$, which is the desired result.  $\square$

1.3.4 Theorem 6

Fix a separable problem $P=(N,x,C)$, with $C(S,t_{S})=M(|S|) \tilde{C} (t(S))$ and $N_{0}(x)\ne \emptyset$, and recall that $m\equiv |N^{u}|$. Using Theorem 2, it is sufficient to show that $| y^{f}_{i}(P)| \le | AS^{pr}_{i}(P)|$, for any symmetric flow f to $2e^{m}$ and any $i\in N_{0}(x)$. We proceed in 4 steps.

1.  It is well known that every flow f (to some $z\in \mathbb {R}^{m}$) can be written as a convex combination of paths (to z). In other words, denoting by $\mathcal {P}^{z}$ the set of paths to z, one can write

$$\begin{aligned} f= & {} \sum _{p\in \mathcal {P}^{z}}\beta ^{p}p, \end{aligned}$$

(12)

$$\begin{aligned} \mathrm{where }\;\beta _{p}\ge & {} 0\quad \mathrm{and }\quad \sum _{p\in \mathcal {P}^{z}}\beta ^{p}=1. \end{aligned}$$

(13)

Moreover, it is not difficult to see that, for every symmetric flow f, any two symmetric paths p and $p^{\prime }$ (i.e., p and $p^{\prime }$ are permutations of one another) will be given the same weight under the decomposition (12). More precisely, if we call $\mathcal {S}(p)$ the class of paths that can be obtained as a permutation of p, we have $\beta ^{p^{\prime }}=\beta ^{p}$, for any $p^{\prime }\in \mathcal {S}(p)$. Let then $\mathcal {S}_{1},\ldots ,\mathcal {S}_{L}$ be the list of equivalence classes for the equivalence relation R defined on $\mathcal {P}^{z}$ by:⁸ $pRp^{\prime }\Leftrightarrow p^{\prime }\in \mathcal {S}(p)$. Recall that we have already introduced $\mathcal {S}_{1}$ and $\mathcal {S}_{2}$ in Sect. 4, when defining the public and private Aumann–Shapley methods. Using this partition of $\mathcal {P}^{z}$ into equivalence classes and defining $\beta (\mathcal {S}_{l})\equiv \beta ^{p}$ (for any $p\in \mathcal {S}_{l}$), one can then rewrite (12) as

$$\begin{aligned} f=\sum _{p\in \mathcal {S}_{1}}\beta (\mathcal {S}_{1})p+\sum _{p\in \mathcal {S} _{2}}\beta (\mathcal {S}_{2})p+\cdots +\sum _{p\in \mathcal {S}_{L}}\beta ( \mathcal {S}_{L})p. \end{aligned}$$

(14)

In addition, since there are exactly m! possible permutations for any path to $z\in \mathbb {R}^{m}$, we deduce from (13) that

$$\begin{aligned} m!\left[ \beta (\mathcal {S}_{1})+\beta (\mathcal {S}_{2})+\cdots +\beta ( \mathcal {S}_{L})\right] =1. \end{aligned}$$

(15)

Hence, defining $\tilde{\beta }(\mathcal {S}_{l})=m!\beta (\mathcal {S}_{l})$, we may rewrite (14) as

$$\begin{aligned} f=\tilde{\beta }(\mathcal {S}_{1})\sum _{p\in \mathcal {S}_{1}}\frac{p}{m!}+ \tilde{\beta }(\mathcal {S}_{2})\sum _{p\in \mathcal {S}_{2}}\frac{p}{m!}+\cdots +\tilde{\beta }(\mathcal {S}_{L})\sum _{p\in \mathcal {S}_{L}}\frac{p}{m!}. \end{aligned}$$

(16)

For any $l=1,\ldots ,L$, observe that $f^{\mathcal {S}_{l}}\equiv \sum \nolimits _{p\in \mathcal {S}_{l}}\frac{p}{m!}$ is a symmetric flow to z since it is the average of all paths (to z) contained in $\mathcal {S}_{l}$ . We call this symmetric flow the elementary flow (to z) associated with $\mathcal {S}_{l}$. In other words, Eqs. (16) and (15) mean that any symmetric flow (to z) can be decomposed as a convex combination of elementary flows (to z).⁹ Thus, to prove Theorem 6, it suffices to show that, the private Aumann–Shapley flow ($f^{\mathcal {S}^{2}}$) minimizes the technological players’ shares within the class of elementary flows to $z=2e^{m}$.

2.  Fix $l=1,\ldots ,L$. We will now construct a one-to-one mapping $\hat{\cdot }:\mathcal {S}_{l}\rightarrow \mathcal {S}_{2}$ . Let then $p\in \mathcal {S}_{l}$ and consider the ordering $\sigma ^{p}=[i_{1},i_{2},\ldots ,i_{m}]$ according to which the technologies of the m unit-players are activated under the path p.¹⁰ Observe that the path

$$\begin{aligned} \hat{p}\equiv \left\{ 0,e_{i_{1}},2e_{i_{1}},2e_{i_{1}}+e_{i_{2}},2e_{i_{1}}+2e_{i_{2}},\ldots , \sum \limits _{k=1}^{m-1}2e_{i_{k}}+e_{i_{m}},\sum \limits _{k=1}^{m}2e_{i_{k}}\right\} \end{aligned}$$

(17)

is the unique path in $\mathcal {S}_{2}$ that activates technological units in the same order as p. In addition, for any distinct $p,p^{\prime }\in \mathcal {S}_{l}$, we have $\hat{p}\ne \hat{p}^{\prime }$. Therefore, $\hat{\cdot }$ is a one-to-one mapping (since $\mathcal {S}_{l}$ and $\mathcal {S}_{2}$ each have exactly m! elements).

3.  For any path p (to 2e) and any $r\in [0,2e]\subset \mathbb {R}^{m}$, define

$$\begin{aligned} t(r)\equiv & {} \{i\in N:\exists k\in I_{i}\mathrm{s.t. }r_{k}\ge 1\}; \end{aligned}$$

(18)

$$\begin{aligned} q(r)\equiv & {} \sum _{i\in N^{u}{\setminus } N_{0}(x)}\max (r_{i}-1,0). \end{aligned}$$

(19)

The set t(r) describes the (actual) technologies that are available when the path p reaches r,¹¹ whereas q(r) gives the number of costly units that have been produced when p reaches the profile r.¹² In addition, letting $k_{i}^{p}$ be the (unique) index $k\in \{1,\ldots ,2m\}$ s.t. $p_{i}(k)=1$ and $p_{i}(k-1)=0$, one can see from (17) to (19) that,

$$\begin{aligned} t\left( p\left( k_{i}^{p}\right) \right) =t\left( \hat{p}\left( k_{i}^{\hat{p}}\right) \right) \quad \mathrm{and }\quad q\left( p\left( k_{i}^{p}\right) \right) \le q\left( \hat{p}\left( k_{i}^{\hat{p}}\right) \right) ,\quad \forall i\in N^{u}. \end{aligned}$$

(20)

4.  We now show that, for any $i\in N_{0}(x)$, $l=1,\ldots ,L$ and $p\in \mathcal {S}_{l}$, we have $y^{p}_{i}(P)\ge y^{\hat{ p}}_{i}(P)$. That is to say, a technological agent i always receives under p a higher share than under $\hat{p}$.

Let us then fix $i\in N_{0}(x)$, $l=1,\ldots ,L$ and $p\in \mathcal {S}_{l}$. Using the notation above, Eq. (20) and the fact that P is a separable problem, one can compute the shares associated with the paths p and $\hat{p}$ (for any $i\in N_{0}(x)\ne \emptyset$):

$$\begin{aligned} y_{i}^{p}(P)= & {} \underbrace{\left[ M(t(p(k_{i}^{p})))-M(t(p(k_{i}^{p}) \backslash i))\right] }_{\le 0}\overbrace{C(q(p(k_{i}^{p})))}^{\le C(q(\hat{ p}(k_{i}^{\hat{p}})))} \\\ge & {} \left[ M\left( t\left( p\left( k_{i}^{p}\right) \right) \right) -M\left( t\left( p\left( k_{i}^{p}\right) \backslash i\right) \right) \right] C\left( q\left( \hat{p}\left( k_{i}^{\hat{p}}\right) \right) \right) \\= & {} \left[ M\left( t\left( \hat{p}\left( k_{i}^{\hat{p}}\right) \right) \right) -M\left( t\left( \hat{p} \left( k_{i}^{\hat{p}}\right) \backslash i\right) \right) \right] C\left( q\left( \hat{p}\left( k_{i}^{\hat{p}}\right) \right) \right) =y_{i}^{\hat{p} }(P). \end{aligned}$$

Hence, $y^{p}_{i}(P)\ge y^{\hat{p}}_{i}(P)$, for any $p\in \mathcal {S}_{l}$; and we can then write:

$$\begin{aligned} y^{f^{\mathcal {S}_{l}}}_{i}(P)=\frac{1}{m!}\sum _{p\in \mathcal {S} _{l}}y^{p}_{i}(P)\ge \frac{1}{m!}\sum _{p\in \mathcal {S}_{l}}y^{\hat{p} }_{i}(P)=\frac{1}{m!}\sum _{p^{\prime }\in \mathcal {S}_{2}}y^{p^{ \prime }}_{i}(P)=AS^{pr}_{i}(P).\nonumber \\ \end{aligned}$$

(21)

This proves that the private Aumann–Shapley method always minimizes (maximizes) the share (rent) of the technological agents among symmetric flow methods.

To conclude the proof, note that the inequality in (21) is strict whenever $\tilde{C}$ is increasing and $q(p(k_{i}^{p}))<q(\hat{p}(k_{i}^{\hat{p}})))$, which will happen—for at least one homogeneous problem (N, x, C), one equivalence class $\mathcal {S} _{l}$, and one path $p\in \mathcal {S}_{l}$—as soon as the considered method y differs from $AS^{pr}$. $\square$