On horizontal cooperation in linear production processes with a supplier that controls a limited resource

Abstract

In this paper we consider a two-echelon supply chain with one supplier that controls a limited resource and a finite set of manufacturers who need to purchase this resource. We analyze the effect of the limited resource on the horizontal cooperation of manufacturers. To this end, we use cooperative game theory and the existence of stable distributions of the total profit among the manufacturers as a measure of the possibilities of cooperation. The game theoretical model that describes the horizontal cooperation involves externalities, which arise because of the possible scarcity of the limited resource and the possible coalition structures that can be formed. Furthermore, manufacturers do not know how the supplier will allocate the limited resource, therefore, how much of this resource they will obtain is uncertain for all concerned. Nevertheless, when the limited resource is not scarce for the grand coalition, the existence of stable distributions of the total profit is guaranteed and consequently the collaboration among the manufacturers is profitable for them all. In the event that the limited resource is insufficient for the grand coalition, we introduce a new cooperative game that assesses the expectations of each coalition of manufacturers regarding the amount of the limited resource they can obtain. We analyze two extreme expectations: the optimistic and the pessimistic. In the optimistic case, we cannot reach a conclusion regarding the full cooperation of the manufacturers. In the pessimistic case, with one reasonable assumption, the existence of stable distributions of the total profit is guaranteed and as a result the collaboration among manufacturers is a win–win deal.

Appendix

In this appendix we include all proofs of the results given throughout the paper. This is done in order to improve the readability of the paper for those readers not interested in the proofs, but only in the results and insights provided on horizontal cooperation of manufacturers when there is a limited resource.

1.1 Proof of Proposition 1

Given $P\in \mathcal {P}(N),V\left( \left. S\right| P\right) =value\left( S;z_{S}(P)\right) ,\forall S\in P$ such that ${\displaystyle \sum \nolimits _{S\in P}}z_{S}(P)\le r.$ Let be $\left( x^{S};z_{S}(P)\right)$ an optimal plan for each coalition $S\in P$. Thus, $Ax^{S}\le \left( \begin{array}{c} b^{S}\\ z_{S}(P) \end{array}\right)$ and

$$\begin{aligned} A\left( \sum _{S\in P}x^{S}\right) \le \left( \begin{array}{c} {\displaystyle \sum _{S\in P}}b^{S}\\ {\displaystyle \sum _{S\in P}}z_{S}(P) \end{array}\right) \le \left( \begin{array}{c} b^{N}\\ r \end{array}\right) . \end{aligned}$$

Then, $\left( {\displaystyle \sum \nolimits _{S\in P}}x^{S};{\displaystyle \sum \nolimits _{S\in P}}z_{S}(P)\right)$ is a feasible production plan for N and

$$\begin{aligned} \sum _{S\in P}value\left( S;z_{S}(P)\right) \le value\left( N;\sum _{S\in P}z_{S}(P)\right) \le V\left( \left. N\right| \left\{ N\right\} \right) . \end{aligned}$$

$\square$

1.2 Proof of Proposition 2

(i):

We know that $z_{S}(P)=d_{S}$ for all $S\subseteq N,$ due to $d\left( P\right) \le r$ for all P. Therefore, for each $S\subseteq N, V\left( \left. S\right| P\right) =V\left( \left. S\right| P^{\prime }\right)$ for all $P,P^{^{\prime }}\in \mathcal {P}(N)$ such that $S\in P$, i.e. for each coalition the value does not depend on the coalitions formed by other players.

(ii):

Since $d_{N}>r$ and $d\left( P\right) \le r,$ for all $P\in \mathcal {P}(N)$ , $P\ne \left\{ N\right\} ,$ similarly to (i), for each $S\subseteq N,V\left( \left. S\right| P\right) =V\left( \left. S\right| P^{\prime }\right)$ for all $P,P^{^{\prime }}\in \mathcal {P}(N),$ i.e. the value of coalition S does not depend on the coalitions formed by other players. On the other hand, the value of the grand coalition

$$\begin{aligned} \begin{array}{ll} \max &{} \sum \limits _{j=1}^{g}p_{j}x_{j}-cr\\ \text {s.t:} &{} Ax\le \left( \begin{array}{c} b^{N}\\ r \end{array}\right) \\ &{} x\ge \mathbf {0}_{g}. \end{array} \end{aligned}$$

(3)

only depends on its own, since there is no partition including N as a proper subset. Therefore, the related game has no externalities.

$\square$

1.3 Proof of Theorem 1

The dual problem of (1)³ for the grand coalition, N, is

$$\begin{aligned} \begin{array}{ll} \min &{} \sum \limits _{t=1}^{q}b_{t}^{N}y_{t}+0y_{q+1}\\ \text {s.t:} &{} A^{t}y\ge p\\ &{} y_{q+1}\le c\\ &{} y\ge \mathbf {0}_{q+1}. \end{array} \end{aligned}$$

(4)

An optimal solution of (1) for the grand coalition N is given by $\left( x^{N};d_{N}\right)$ with $d_{N}\le r,$ and the related dual optimal solution is $\left( y_{q}^{N};y_{q+1}^{N}\right)$, where with an abuse of notation from now on, we represent by $y_{q}^{N}$ the vector $\left( y_{1}^{N},\ldots ,y_{q}^{N}\right)$. From duality, it is known that

$$\begin{aligned} \sum _{j=1}^{g}p_{j}x_{j}^{N}-cd_{N}=\sum _{t=1}^{q}b_{t}^{N}y_{t}^{N}+0y_{q+1}^{N}=v\left( N\right) . \end{aligned}$$

Therefore, somehow, the cost of the common resource is charged to (discounted from) the value of the resources. It is easy to check that $\left( y_{q}^{N};y_{q+1}^{N}\right)$ is feasible in the dual problem of (1) for every coalition $S\subset N.$ Moreover, we have that for a dual optimal solution $\left( y_{q}^{S};y_{q+1}^{S}\right)$ associated with the optimal solution $\left( x^{S};d_{S}\right) ,$ it holds that

$$\begin{aligned} \sum _{t=1}^{q}b_{t}^{S}y_{t}^{N}+0y_{q+1}^{N}\ge \sum _{t=1}^{q}b_{t}^{S}y_{t}^{S}+0y_{q+1}^{S}=v\left( S\right) . \end{aligned}$$

Thus, $\sum _{i\in S}\left( \sum _{t=1}^{q}b_{t}^{i}y_{t}^{N}+0y_{q+1}^{N}\right) \ge v\left( S\right) ,\forall S\subset N,$ and this implies that $\left( b^{i}y^{N}\right) _{i\in N}\in C\left( v\right) .$$\square$

1.4 Proof of Theorem 2

Before giving the proof of Theorem 2 we need the following technical lemma, which is given without a proof because it is easy to derive. It give us two linear programs that, although they have different optimal solution sets, also have the same optimal values, i.e. they are optimally equivalents. Note that an optimal solution of the second one is the optimal demand of the limited resource for each coalition S,  $d_{S}.$ We should highlight that they only differ in a redundant constraint, $z\le d_{S}$, however, this is the key with which to prove the next theorem.

Lemma 1

Let $\left( A,B,p,r,c\right)$ be an LPLR situation. For all S, the following linear programs are optimally equivalents ,

$$\begin{aligned}&\begin{array}{ll} \max &{} \sum \limits _{j=1}^{g}p_{j}x_{j}-cz\\ \text {s.t:} &{} Ax\le \left( \begin{array}{c} b^{S}\\ z \end{array}\right) \\ &{} z\le d_{S}\\ &{} x\ge \mathbf {0}_{g},z\ge 0. \end{array} \end{aligned}$$

(5)

$$\begin{aligned}&\begin{array}{ll} \max &{} \sum \limits _{j=1}^{g}p_{j}x_{j}-cz\\ \text {s.t:} &{} Ax\le \left( \begin{array}{c} b^{S}\\ z \end{array}\right) \\ &{} x\ge \mathbf {0}_{g},z\ge 0. \end{array} \end{aligned}$$

(6)

The previous result provides us with two different, but equivalent, ways in which to tackle the linear programs. In the proof of the following theorem, we use one or the other depending on which will be more helpful.

Proof of Theorem 2

Since $d_{N}\le r$, consider the linear program (5) for the grand coalition,

$$\begin{aligned} \begin{array}{ll} \max &{} \sum \limits _{j=1}^{g}p_{j}x_{j}-cz\\ \text {s.t:} &{} Ax\le \left( \begin{array}{c} b^{N}\\ z \end{array}\right) \\ &{} z\le d_{N}\\ &{} x\ge \mathbf {0}_{g},z\ge 0. \end{array} \end{aligned}$$

(7)

its dual is given by

$$\begin{aligned} \begin{array}{ll} \min &{} \sum \limits _{t=1}^{q}b_{t}^{N}y_{t}+0y_{q+1}+d_{N}y_{q+2}\\ \text {s.t:} &{} A^{t}y\ge p\\ &{} y_{q+1}-y_{q+2}\le c\\ &{} y\ge \mathbf {0}_{q+2}. \end{array} \end{aligned}$$

(8)

Let $\left( x^{N};d_{N}\right)$ and $\left( y_{q}^{N},y_{q+1}^{N},0\right)$ be primal and dual optimal solutions for (7) and (8), respectively with $d_{N}\le r$ and $y_{q+2}^{N}=0$. Effectively, by the lemma we know that linear problems (1) and (5) are optimally equivalent, and therefore their dual problems will also be optimally equivalent. Now if $\left( y_{q}^{N},y_{q+1}^{N}\right)$ is an optimal solution of the dual problem (4) associated with the linear problem (1), then $\left( y_{q}^{N},y_{q+1}^{N},0\right)$ is an optimal solution of the dual problem (5). Therefore, we can take a dual optimal solution with $y_{q+2}^{N}=0$ associated with the optimal solution $\left( x^{N};d_{N}\right)$.

Now, it is easy to check that $\left( y_{q}^{N},y_{q+1}^{N},0\right)$ is a feasible solution for the dual problem of (5) for every coalition S. If $\left( y_{q}^{S},y_{q+1}^{S},y_{q+2}^{S}\right)$ is an optimal dual solution associated with $\left( x^{S};d_{S}\right)$, it holds that

$$\begin{aligned} \sum _{t=1}^{q}b_{t}^{S}y_{t}^{N}+0y_{q+1}^{N}+d_{S}y_{q+2}^{N}\ge & {} \sum _{t=1}^{q}b_{t}^{S}y_{t}^{S}+0y_{q+1}^{S}+d_{S}y_{q+2}^{S}\\= & {} value(S;d_{S})\ge V(S\mid P), \end{aligned}$$

for all $P\in \mathcal {P}(N)$. Therefore, $\sum \nolimits _{i\in S}\left( \sum \nolimits _{t=1}^{q}b_{t}^{i}y_{t}^{N}\right) \ge V(S\mid P),$$\forall S\subseteq N$, and for all $P\in \mathcal {P}(N)$ such that $(S\mid P)$ is an embedded coalition, and this implies that $\left( b^{i}y^{N}\right) _{i\in N}\in \overline{C}\left( V\right) .$$\square$

1.5 Proof of Theorem 3

Since $C\left( R\right) \ne \varnothing ,$ there is $u\in \mathbb {R}^{N}$ such that $u(S)=\sum _{i\in S}u_{i}\ge R\left( S\right) ,$ for all S, and $u(N)=r.$ Let $y^{*}$ be an optimal solution of the dual problem of (3). From duality theory, we know that $\sum _{t=1}^{q}b_{t}^{N}y_{t}^{*}+ry_{q+1}^{*}-cr=v^{R}(N).$ On the other hand, $\forall S\subseteq N$

$$\begin{aligned} \sum _{t=1}^{q}b_{t}^{S}y_{t}^{*}+u\left( S\right) y_{q+1}^{*}-cu\left( S\right) \ge \sum _{t=1}^{q}b_{t}^{S}y_{t}^{*}+R\left( S\right) (y_{q+1}^{*}-c)\ge v^{R}(S), \end{aligned}$$

where the last inequality holds because $y^{*}$ is feasible for the dual problem of coalition S and $y_{q+1}^{*}>c$ since $d_{N}>r$. In effect, assume that $y_{q+1}^{*}\le c$. Consider a variation $\varDelta r$ such that the basic feasible solution does not change and $r+\varDelta r<d_{N}$. Then,

$$\begin{aligned} value\left( N;r+\varDelta r\right) =value\left( N;r\right) +\left( y_{q+1}^{*}-c\right) \varDelta r\text {.} \end{aligned}$$

We distinguish two cases:

• 1. If $\varDelta r>0$, then

  $$\begin{aligned}&value\,\left( N;r+\varDelta r\right) \le value\,\left( N;r\right) \\&\quad r<r+\varDelta r<d_{N}\Rightarrow \exists \alpha \in \left( 0,1\right) \text { such that }r+\varDelta r=\alpha r+\left( 1-\alpha \right) d_{N}. \end{aligned}$$

  Since $\alpha x^{r}+\left( 1-\alpha \right) x^{d_{N}}$ is a feasible solution for the problem with $r+\varDelta r$, $\left( x^{r},x^{d_{N}}\right)$ is an optimal solution for r and $d_{N}$, we have

  $$\begin{aligned} value\,\left( N;r+\varDelta r\right) \ge \alpha value\,\left( N;r\right) +\left( 1-\alpha \right) value\,\left( N;d_{N}\right) . \end{aligned}$$

  However, $value\,\left( N;d_{N}\right) >value\,\left( N;r+\varDelta r\right)$ by definition of $d_{N}$ and $value\,\left( N;r\right) \ge value\,\left( N;r+\varDelta r\right)$. Therefore, we have a contradiction.

• 2. If $\varDelta r<0,$ then

  $$\begin{aligned}&value\,\left( N;r+\varDelta r\right) =value\,\left( N;r\right) +\left( y_{q+1}^{*}-c\right) \varDelta r\\&\quad \Rightarrow value\,\left( N;r+\varDelta r\right) \ge value\,\left( N;r\right) \text {.}\\&\quad r+\varDelta r<r<d_{N}\Rightarrow \exists \alpha \in \left( 0,1\right) \text { such that }\alpha \left( r+\varDelta r\right) +\left( 1-\alpha \right) d_{N}=r \end{aligned}$$

  Being that $\alpha x^{r+\varDelta r}+\left( 1-\alpha \right) x^{d_{N}}$ is a feasible solution for the problem with r, $\left( x^{r+\varDelta r},x^{d_{N}}\right)$ is an optimal solution with $r+\varDelta r$ and $d_{N}$, respectively. We obtain

  $$\begin{aligned} value\,\left( N;r\right) \ge \alpha value\,\left( N;r+\varDelta r\right) +\left( 1-\alpha \right) value\,\left( N;d_{N}\right) . \end{aligned}$$

  But $value\,\left( N;d_{N}\right) >value\,\left( N;r\right)$ by definition of $d_{N}$ and $value\,\left( N;r+\varDelta r\right) \ge value\,\left( N;r\right)$. Thus, there is a contradiction.

Thereby we can conclude that $y_{q+1}^{*}>c$.

Therefore, $\left( b^{i}y^{*}+u_{i}\left( y_{q+1}^{*}-c\right) \right) _{i\in N}\in C\left( v^{R}\right) \ne \varnothing .$$\square$

1.6 Proof of Theorem 4

Let $\left( d_{i}\right) _{i\in N}$ be the individual demands of agents in N. Consider the resource game (N, w), where $w\left( S\right) =\left( r-\sum \nolimits _{i\notin S}d_{i}\right) _{+},\forall S\subsetneq N,$ and $w\left( N\right) =r.$

We will distinguish two cases:

• (a) If $\sum \limits _{i\in N}d_{i}>r$, (N, w) is a standard bankruptcy game (O’Neill 1982) and, therefore, it has a non empty core. Then, an $u\in \mathbb {R}^{N}$ such that $u\left( N\right) =r$ exists and

  $$\begin{aligned} u\left( S\right)\ge & {} \left( r-\sum \nolimits _{i\notin S}d_{i}\right) _{+}\ge \min \limits _{P:S\in P}\left\{ \left( r-\sum \nolimits _{\begin{array}{c} T\in P\\ T\ne S \end{array} }d_{T}\right) _{+}\right\} \\\ge & {} \min \left\{ \min \limits _{P:S\in P}\left\{ \left( r-\sum \nolimits _{\begin{array}{c} T\in P\\ T\ne S \end{array} }d_{T}\right) _{+}\right\} ,d_{S}\right\} =R^{pes}\left( S\right) . \end{aligned}$$

  Therefore, $C\left( R^{pes}\right) \ne \varnothing$ and using the same arguments as in Theorem 3, the result holds.

• (b) If $\sum \limits _{i\in N}d_{i}=r$,

  $$\begin{aligned} d\left( S\right)= & {} \sum \limits _{i\in S}d_{i}\ge \left( r-\sum \nolimits _{i\notin S}d_{i}\right) _{+}\ge \min \limits _{P:S\in P}\left\{ \left( r-\sum \nolimits _{\begin{array}{c} T\in P\\ T\ne S \end{array} }d_{T}\right) _{+}\right\} \ge \\\ge & {} \min \left\{ \min \limits _{P:S\in P}\left\{ \left( r-\sum \nolimits _{\begin{array}{c} T\in P\\ T\ne S \end{array} }d_{T}\right) _{+}\right\} ,d_{S}\right\} =R^{pes}\left( S\right) . \end{aligned}$$

  Thus, $C\left( R^{pes}\right) \ne \varnothing$ and, then, from Theorem 3$C\left( v^{pes}\right) \ne \varnothing$. $\square$

1.7 Proof of Theorem 5

If $d_N \le r$, the result immediately follows by using a similar argument as in Theorem 2.

If $d_N > r$, we first note the following:

• (i) Since the manufacturers are symmetric $d_i=d \in \mathbb {R}_+, \forall i \in N$

• (ii) If (x; z) is an optimal plan for just one manufacturer, then by the linearity of the production system, (sx; sz) is an optimal plan for s manufacturers and, consequently, $d_S=|S|d$. Therefore, $d >\frac{r}{n}$.

• (iii) $R^{pes}(S)=\left( r-(n-|S|)d\right) _{+}$

• (iv) $r-(n-s)d \le r-(n-s)\frac{r}{n} = \frac{sr}{n}$.

Taking into account $(i)-(iv)$, it is not difficult to prove that $\left( \frac{r}{n}\right) _{i \in N}$ belongs to $C(R^{pes})$, therefore $C(R^{pes}) \ne \varnothing$. Now by Theorem 3, we have that $C(v^{pes}) \ne \varnothing$. $\square$

1.8 Proof of Theorem 6

Before giving the proof of Theorem 6 we need the following technical lemma.

Lemma 2

Let $\left( A,B,p,r,c\right)$ be an LPLR situation and $(N,R^{pes})$ the associated pessimistic resource game. Each coalition configuration $\{S_1,\ldots ,S_k\}$ is compatible.

Proof

Given $\{S_1,\ldots ,\S _k\}$, we have to prove that

$$\begin{aligned} \sum _{j=1}^{k}R^{pes}(S_j) \le r. \end{aligned}$$

First we recall that

$$\begin{aligned} R^{pes}\left( S\right) =\min \left\{ \min \limits _{P:S\in P}\left\{ \left( r-\sum \limits _{\begin{array}{c} T\in P\\ T\ne S \end{array} }d_{T}\right) _{+}\right\} ,d_{S}\right\} . \end{aligned}$$

We now distinguish two cases:

• 1. $\sum _{j=1}^{k}d_{S_j} \le r$. In this situation, the result immediately holds.

• 2. $\sum _{j=1}^{k}d_{S_j} > r$. First, note that this relationship implies that

  $$\begin{aligned} d_{S_h}>r - \sum _{j:j \ne h}d_{S_j} > r. \end{aligned}$$

  Now we have the following chain of inequalities:

  $$\begin{aligned} \sum _{j=1}^{k}R^{pes}(S_j)\le & {} \sum _{j=1}^{k}\min \left\{ \min \limits _{P:S_j\in P}\left\{ \left( r-\sum \limits _{\begin{array}{c} T\in P\\ T\ne S_j \end{array}}d_{T}\right) _{+}\right\} ,d_{S_j}\right\} \\\le & {} \sum _{j=1}^{k} \left( r - \sum _{h:h \ne j}d_{S_h}\right) _+ \le r, \end{aligned}$$

  where the last inequality holds because the $\sum _{j=1}^{k} \left( r - \sum _{h:h \ne j}d_{S_h}\right) _+$ is a non decreasing function in r and when $r = \sum _{j=1}^{k}d_{S_j}$ the left hand side equals r.

$\square$

Proof of Theorem 6

First, we take a maximally stable coalition $S_1 \subseteq N$. This is always possible because in the worst case a singleton is maximally stable. Next we take a maximally stable coalition $S_2 \subseteq N\backslash S_1$ and so on so forth. This procedure ends in a finite number of steps k and it is well defined. The coalition configuration obtained $\{S_1,\ldots ,S_k\}$ is by construction cc-stable compatible. Indeed, Conditions (1) and (2) in Definition 2 hold by definition of maximally stable coalition. Finally, by Lemma 2 the constructed coalition configuration is compatible. Therefore, the result holds. $\square$