Cost allocation for less-than-truckload collaboration among perishable product retailers

Abstract

We study cost allocation problem arising from less-than-truckload collaboration among perishable product retailers. The relevant costs we consider include fixed transportation cost, variable transportation cost, and decay loss of perishable products. Cooperative game theory is applied to study this cost allocation problem. The corresponding cooperative game, called transportation facility choice game, is established. First, we show that the core of the transportation facility choice game is non-empty. Then, we identify some conditions for concavity and quasi-concavity of the transportation facility choice game with the linear decay and negative exponential decay functions, respectively. Finally, simulation is conducted to analyze how optimal solutions differ under the linear decay and exponential decay functions, and intuitive cost allocation schemes are proposed and compared with the $\tau$-value and the Shapley value of the corresponding game. Simulation results show that the optimal solution under linear decay function tends to choose facilities with higher fixed cost than that under exponential decay function. Additionally, among all the cost allocation schemes compared, the simple cost allocation scheme called A-IM, the $\tau$-value, and the Shapley value have better performance in terms of the percentage of allocations lying in the core.

Appendices

Appendix 1: Proof of Proposition 3

Proof

For notational convenience, in this proof we use c(S) to denote $c_{L}(S)$ (that is, the subscript are omitted). Other notions are simplified similarly. If $c(S)=\underset{k\in F}{\min }c(S)_{k}$, then $c(S)=\underset{k\in F}{\min }\left\{ f_{k}+\underset{j\in S}{\sum }a_{k} d_jq_{j}+\underset{j\in S}{\sum }t_{k}d_j\theta _{j}q_{j}\right\}$. Let $f^{S}$, $a^{S}$ and $t^{S}$ be the fixed cost, unit variable cost and traveling time per unit traveling distance corresponding to c(S) , respectively. Then,

$$\begin{aligned} c(S)=f^{S}+\underset{j\in S}{\sum }a^{S}d_jq_{j}+\underset{j\in S}{\sum }t^{S}d_j\theta _{j}q_{j}. \end{aligned}$$

(1) For $i,j\in N$, we have $c(\{i\})=\underset{k\in F}{\min }\left\{ f_{k}+a_{k}d_iq_{i}+t_{k}d_i\theta _{i}q_{i}\right\} =f^{\{i\}}+a^{\{i\}}d_iq_{i} +t^{\{i\}}d_i\theta _{i}q_{i}$. Similarly $c(\{j\})=f^{\{j\}}+a^{\{j\}}d_jq_{j}+t^{\{j\}}d_j\theta _{j}q_{j}$. Let $c(S)_{W}$ be the total cost for coalition S when W is the optimal set of facilities used by coalition S. Then

$$\begin{aligned} c(\{i,j\})_{\{i\}}=f^{\{i\}}+\sum _{k\in \{i,j\}}a^{\{i\}}d_k q_{k}+\sum _{k\in \{i,j\}}t^{\{i\}}d_k\theta _{k}q_{k}, \end{aligned}$$

and

$$\begin{aligned} c(\{i,j\})_{^{\{j\}}}=f^{\{j\}}+\sum _{k\in \{i,j\}}a^{\{j\}}d_k q_{k}+\sum _{k\in \{i,j\}}t^{\{j\}}d_k\theta _{k}q_{k}. \end{aligned}$$

Thus,

$$\begin{aligned} c(\{i,j\})_{\{i\}}-c(\{i\})-c(\{j\})&=-f^{\{j\}}+\big (a^{\{i\}}-a^{\{j\}} \big )d_jq_{j}+\big (t^{\{i\}}-t^{\{j\}}\big )d_j\theta _{j}q_{j}\\ {}&=-f^{\{j\}}+\big (f^{\{j\}} -f^{\{i\}}\big )\left( \delta d_j\theta _{j}q_{j}-a^{\{j\}}d_jq_{j}/f^{\{j\}}\right) , \end{aligned}$$

$$\begin{aligned} c(\{i,j\})_{^{\{j\}}}-c(\{i\})-c(\{j\})&=-f^{\{i\}}+\big (a^{\{j\}}-a^{\{i\}} \big )d_iq_{i}+\big (t^{\{j\}}-t^{\{i\}}\big )d_i\theta _{i}q_{i}\\ {}&=-f^{\{i\}}+\big (f^{\{i\}}-f^{\{j\}}\big )\left( \delta d_i\theta _{i}q_{i}-a^{\{i\}}d_iq_{i}/f^{\{i\}}\right) . \end{aligned}$$

If $f_i\theta _{\min }\delta \ge a_i$, then $\delta d_j\theta _{j}q_{j}-a^{\{j\}}d_jq_{j}/f^{\{j\}}\ge 0$, and $\delta d_i\theta _{i}q_{i}-a^{\{i\}}d_iq_{i}/f^{\{i\}}\ge 0$.

If $f^{\{i\}}\ge f^{\{j\}}$, we have $c(\{i,j\})_{^{\{i\}}}-c(\{i\})-c(\{j\})\le -f^{\{j\}}<0$; otherwise $c(\{i,j\})_{\{j\}}-c(\{i\})-c(\{j\})\le -f^{\{i\}}<0$.

If $f_i\theta _{\max }\delta \le a_i$, then $\delta d_j\theta _{j}q_{j} -a^{\{j\}}d_jq_{j}/f^{\{j\}}\le 0$, and $\delta d_i\theta _{i}q_{i}-a^{\{i\}}d_iq_{i}/f^{\{i\}}\le 0$.

If $f^{\{i\}}\le f^{\{j\}}$, then $c(\{i,j\})_{\{i\}}-c(\{i\})-c(\{j\})\le -f^{\{j\}}<0$; otherwise $c(\{i,j\})_{^{\{j\}}}-c(\{i\})-c(\{j\})\le -f^{\{i\}}<0$.

Given that at least one inequality is less than 0, $\min \left\{ c(\{i,j\})_{\{i\}},c(\{i,j\})_{\{j\}}\right\} <c(\{i\})+c(\{j\})$, so $c(\{i,j\})=\underset{_{k\in F}}{\min }c(\{i,j\})_{k}$.

(2) For $S,T\in N$, and $S\cap T=\emptyset$, let $S\cup T=ST$, and suppose that $c(S)=\underset{k\in F}{\min }c(S)_{k}$ on the set S and T. We then arrive at

$$\begin{aligned} c(S)= & {} f^{S}+t^{S}\sum _{j\in S}d_j\theta _{j}q_{j}+a^{S}\sum _{j\in S}d_jq_{j}, c(T)=f^{T}+t^{T}\sum _{j\in T}d_j\theta _{j} q_{j} \\&+ \, a^{T}\sum _{j\in T}d_jq_{j}. \end{aligned}$$

Note that $\underset{_{k\in F}}{\min }~c(S\cup T)_{k}\le f^{S}+t^{S}\underset{j\in S\cup T}{\sum }d_j\theta _{j}q_{j}+a^{S}\underset{j\in S\cup T}{\sum }d_jq_{j}$, and $\underset{_{k\in F} }{\min }~c(S\cup T)_{k}\le f^{T}+t^{T}\underset{j\in S\cup T}{\sum }d_j\theta _{j}q_{j} +a^{T}\underset{j\in S\cup T}{\sum }d_jq_{j}$. Let $c(S,T)=\underset{_{k\in F}}{\min }~c(S\cup T)_{k}-c(S)-c(T)$. Thus, we have

$$\begin{aligned} c(S,T)\le -f^{S}+\big (t^{T}-t^{S}\big )\underset{j\in S}{\sum }d_j\theta _{j}q_{j}+\big (a^{T}-a^{S}\big )\underset{j\in S}{\sum }d_jq_{j} , \end{aligned}$$

(A.1)

$$\begin{aligned} c(S,T)\le -f^{T}+\big (t^{S}-t^{T}\big )\underset{j\in T}{\sum }d_j\theta _{j}q_{j}+\big (a^{S}-a^{T}\big )\underset{j\in T}{\sum }d_jq_{j}. \end{aligned}$$

(A.2)

Combining Condition A and Condition B with (A.1) and (A.2), respectively, we can obtain

$$\begin{aligned} c(S,T)\le \big (f^{S}-f^{T}\big )\left( \underset{j\in S}{\sum }\delta d_j\theta _{j} q_{j}-\sum _{j\in S}\frac{a^Sd_jq_{j}}{f^S}\right) -f^{S}. \end{aligned}$$

(A.3)

$$\begin{aligned} c(S,T)\le \big (f^{T}-f^{S}\big )\left( \underset{j\in T}{\sum }\delta d_j\theta _{j} q_{j}-\sum _{j\in T}\frac{a^Td_jq_{j}}{f^T}\right) -f^{T}, \end{aligned}$$

(A.4)

When $f_i\theta _{\min }\delta \ge a_i$, we have $\underset{j\in S}{\sum } \delta d_j\theta _{j}q_{j}-\underset{j\in S}{\sum }\frac{a^Sd_jq_{j}}{f^S}\ge 0$ and $\underset{j\in T}{\sum }\delta d_j\theta _{j}q_{j}-\underset{j\in T}{\sum }\frac{a^Td_jq_{j}}{f^T}\ge 0$. Furthermore, if $f^{S}\ge f^{T}$, then it follows from (A.4) that $c(S,T)\le -f^{T}<0$; otherwise, $c(S,T)\le -f^{S}<0$. Thus, $c(S,T)<0$.

When $f_i\theta _{\max }\delta \le a_i$, we have $\underset{j\in S}{\sum }\delta d_j\theta _{j}q_{j}-\underset{j\in S}{\sum } \frac{a^Sd_jq_{j}}{f^S}\le 0$ and $\underset{j\in S}{\sum }\delta d_j\theta _{j}q_{j}-\underset{j\in S}{\sum }\frac{a^Sd_jq_{j}}{f^S}\le 0$. Furthermore, if $f^{S}\ge f^{T}$, then it follows from (A.3) that $c(S,T)\le -f^{S}<0$; otherwise, $c(S,T)\le -f^{T}<0$. Thus, $c(S,T)<0$.

From the analysis above, we have $\underset{_{k\in F}}{\min }c(S\cup T)_{k}<c(S)+c(T)$.

Given that

$$\begin{aligned} c(S\cup T)=\underset{\Pi _{S\cup T}}{\min }\underset{k\in F}{\sum }\left( f_{k}+\underset{j\in S\cup T}{\sum }a_{k}d_jq_{j}+\underset{j\in S\cup T}{\sum }t_{k}d_j\theta _{j}q_{j}\right) <c(S)+c(T), \end{aligned}$$

we have $c(S\cup T)=\underset{k\in F}{\min }c(S\cup T)_{k}$.

(3) From (1), we know for $S,T\in N$, and $S\cap T=\emptyset$, $\left| S\cup T\right| =2$, and so $c(S\cup T)=\underset{k\in F}{\min }c(S\cup T)_{k}$. From (2), we know if $\left| S\cup T\right| \le 3$, then $c(S\cup T)=\underset{k\in F}{\min }c(S\cup T)_{k}$. It can be derived similarly that $c(S\cup T)=\underset{k\in F}{\min }c(S\cup T)_{k}$, for all $S\in N$. $\square$

Appendix 2: Proof of Proposition 5

Proof

For $l\notin S\subset T\subseteq N$, let $h(S,\{l\})=c(S\cup \{l\})-c(S)$. Then,

$$\begin{aligned} h(S,\{l\})= & {} f^{S\cup \{l\}}+a^{S\cup \{l\}}\underset{j\in S\cup \{l\}}{\sum }d_jq_{j}+t^{S\cup \{l\}}\underset{j\in S\cup \{l\}}{\sum } d_j\theta _{j}q_{j}-f^{S}-a^{S}\underset{j\in S}{\sum }d_jq_{j}\\&\,- \, t^{S}\underset{j\in S}{\sum }d_j\theta _{j}q_{j}. \end{aligned}$$

Since

$$\begin{aligned} c(S\cup \{l\})\le f^{S}+a^{S}\underset{j\in S\cup \{l\}}{\sum }d_jq_{j}+t^{S}\underset{j\in S\cup \{l\}}{\sum }d_j\theta _{j}q_{j}, \end{aligned}$$

and

$$\begin{aligned} c(S)\le f^{S\cup \{l\}}+a^{S\cup \{l\}}\underset{j\in S}{\sum } d_jq_{j}+t^{S\cup \{l\}}\underset{j\in S}{\sum }d_j\theta _{j}q_{j}, \end{aligned}$$

we have

$$\begin{aligned} t^{S\cup \{l\}}d_l\theta _{l}q_{l}+a^{S\cup \{l\}}d_lq_{l}\le h(S,\{l\})\le t^{S}d_l\theta _{l}q_{l}+a^{S}d_lq_{l}. \end{aligned}$$

Similarly, we can show that

$$\begin{aligned} t^{T\cup \{l\}}d_l\theta _{l}q_{l}+a^{T\cup \{l\}}d_lq_{l}\le h(T,\{l\})\le t^{T}d_l\theta _{l}q_{l} +a^{T}d_lq_{l}. \end{aligned}$$

Therefore,

$$\begin{aligned} h(S,\{l\})-h(T,\{l\})\ge \big (t^{S\cup \{l\}}-t^{T}\big )d_l\theta _{l} q_{l}+\big (a^{S\cup \{l\}}-a^{T}\big )d_lq_{l}. \end{aligned}$$

(A.5)

Furthermore,

$$\begin{aligned} h(S,\{l\})-h(T,\{l\})&=f^{S\cup \{l\}}+a^{S\cup \{l\}} \underset{j\in S\cup \{l\}}{\sum }d_jq_{j}+t^{S\cup \{l\}}\underset{j\in S\cup \{l\}}{\sum }d_j\theta _{j}q_{j}\\&\quad -f^{S}-a^{S}\underset{j\in S}{\sum }d_jq_{j} -t^{S}\underset{j\in S}{\sum }d_j\theta _{j}q_{j}-f^{T\cup \{l\}}\\&\quad -a^{T\cup \{l\}}\underset{j\in T\cup \{l\}}{\sum }d_jq_{j}-t^{T\cup \{l\}}\underset{j\in T\cup \{l\}}{\sum }d_j\theta _{j}q_{j}\\&\quad +f^{T}+a^{T}\underset{j\in T}{\sum }d_jq_{j} +t^{T}\underset{j\in T}{\sum }d_j\theta _{j}q_{j}. \end{aligned}$$

Given that

$$\begin{aligned} c(T\cup \{l\})\le f^{S\cup \{l\}}+a^{S\cup \{l\}}\underset{j\in T\cup \{l\}}{\sum }d_jq_{j}+t^{S\cup \{l\}}\underset{j\in T\cup \{l\}}{\sum } d_j\theta _{j}q_{j}, \end{aligned}$$

and

$$\begin{aligned} c(S)\le f^{T}+a^{T}\underset{j\in S}{\sum }d_jq_{j} +t^{T}\underset{j\in S}{\sum }d_j\theta _{j}q_{j}, \end{aligned}$$

we have

$$\begin{aligned} h(S,\{l\})-h(T,\{l\})\ge \big (t^{T}-t^{S\cup \{l\}}\big )\underset{j\in T-S}{\sum }d_j\theta _{j}q_{j}+\big (\ a^{T}-a^{S\cup \{l\}}\big )\underset{j\in T-S}{\sum }d_jq_{j}. \end{aligned}$$

(A.6)

Combining Condition A and Condition B with (A.5) and (A.6), respectively, we obtain

$$\begin{aligned} h(S,l)-h(T,l)\ge \big (f^{T}-f^{S\cup \{l\}}\big )\left( \delta d_l\theta _{l}q_{l} -\frac{a^Td_lq_{l}}{f^T}\right) , \end{aligned}$$

(A.7)

$$\begin{aligned} h(S,l)-h(T,l)\ge \big (f^{S\cup \{l\}}-f^{T}\big )\left( \underset{j\in T-S}{\sum }\delta d_j\theta _{j}q_{j}-\underset{j\in T-S}{\sum }\frac{a^{S\cup \{l\}}d_jq_{j}}{f^{S\cup \{l\}}}\right) . \end{aligned}$$

(A.8)

When $f_i\theta _{\min }\delta \ge a_i$, we have $\delta d_l\theta _{l}q_{l}-\frac{a^Td_lq_{l}}{f^T}\ge 0$ and $\underset{j\in T-S}{\sum }\delta d_j\theta _{j}q_{j}-\underset{j\in T-S}{\sum }\frac{a^{S\cup \{l\}}d_jq_{j}}{f^{S\cup \{l\}}}\ge 0$. Furthermore if $f^{S\cup \{l\}}\le f^{T}$, then from (A.7) we have $h(S,l)-h(T,l)\ge 0$; otherwise from (A.8) we get $h(S,l)-h(T,l)\ge 0$.

When $f_i\theta _{\max }\delta \le a_i$, we have $\delta d_l\theta _{l}q_{l} -\frac{a^Td_lq_{l}}{f^T}\le 0$ and $\underset{j\in T-S}{\sum }\delta d_j\theta _{j} q_{j}-\underset{j\in T-S}{\sum }\frac{a^{S\cup \{l\}}d_jq_{j}}{f^{S\cup \{l\}}}\le 0$. Furthermore if $f^{S\cup \{l\}}\le f^{T}$, from (A.8) we have $h(S,l)-h(T,l)\ge 0$; otherwise from (A.7) we have $h(S,l)-h(T,l)\ge 0$.

The analysis above shows that $c(S\cup \{l\})-c(S)\ge c(T\cup \{l\})-c(T)$. This establishes the concavity of the game $(N,c_{L})$. $\square$

Appendix 3: Proof of Proposition 7

Proof

Suppose that $c_{E}(S)=\underset{k\in F}{\min }c_{E}(S)_{k}$ on the set S, T ($S,T\subset N$). For notational convenience, in this proof use c(S) to denote $c_{E}(S)$ (other notations are simplified in a similar way). For $S,T\subset N$, and $S\cap T=\emptyset$, we have

$$\begin{aligned} c(S)=f^{S}+\underset{j\in S}{\sum }q_{j}\big (1-e^{-\theta _{j}d_jt^{S}}\big )+a^{S}\underset{j\in S}{\sum }d_jq_{j}, \end{aligned}$$

and

$$\begin{aligned} c(T)=f^{T}+\underset{j\in T}{\sum }q_{j}\big (1-e^{-\theta _{j}d_jt^{T}}\big )+a^{T} \underset{j\in T}{\sum }d_jq_{j}. \end{aligned}$$

Given that

$$\begin{aligned} c(S\cup T)\le f^{S}+\underset{j\in S\cup T}{\sum }q_{j}\big (1-e^{-\theta _{j}d_jt^{S}}\big )+a^{S}\underset{j\in S\cup T}{\sum }d_jq_{j}, \end{aligned}$$

and

$$\begin{aligned} c(S\cup T)\le f^{T}+\underset{j\in S\cup T}{\sum }q_{j} \big (1-e^{-\theta _{j}d_jt^{T}}\big )+a^{T}\underset{j\in S\cup T}{\sum }d_jq_{j}, \end{aligned}$$

then

$$\begin{aligned} c(S,T)\le -f^{S}+\underset{j\in S}{\sum }q_{j}\big (e^{-\theta _{j}d_jt^{S}} -e^{-\theta _{j}d_jt^{T}}\big )+\big (a^{T}-a^{S}\big )\underset{j\in S}{\sum }d_jq_{j}, \end{aligned}$$

(A.9)

and

$$\begin{aligned} c(S,T)\le -f^{T}+\underset{j\in T}{\sum }q_{j}\big (e^{-\theta _{j}d_jt^{T}} -e^{-\theta _{j}d_jt^{S}}\big )+\big (a^{S}-a^{T}\big )\underset{j\in T}{\sum }d_jq_{j}. \end{aligned}$$

(A.10)

If $t^{T}\ge t^{S}$, then it follows from Lemma 1 that

$$\begin{aligned} \left( e^{-\theta _{j}d_jt^{S} }-e^{-\theta _{j}d_jt^{T}}\right) \le \theta _{j}d_j\big (t^{T}-t^{S}\big ). \end{aligned}$$

From (A.9), we have

$$\begin{aligned} c(S,T)\le -f^{S}+\big (t^{T}-t^{S}\big )\underset{j\in S}{\sum }d_j\theta _{j}q_{j} +\big (a^{T}-a^{S}\big )\underset{j\in S}{\sum }d_jq_{j}. \end{aligned}$$

Combining with Condition B, we have

$$\begin{aligned} c(S,T)\le -f^{S}+\big (t^{T}-t^{S}\big )\left( \underset{j\in S}{\sum }d_j\theta _{j} q_{j}-\frac{1}{\delta }\underset{j\in S}{\sum }\frac{a^Sd_jq_{j}}{f^S}\right) . \end{aligned}$$

(A.11)

If $t^{T}<t^{S}$, then it follows from Lemma 1 that

$$\begin{aligned} (e^{-\theta _{j}d_jt^{T} }-e^{-\theta _{j}d_jt^{S}})\le \theta _{j}d_j(t^{S}-t^{T}). \end{aligned}$$

From (A.10), we have

$$\begin{aligned} c(S,T)\le -f^{T}+(t^{S}-t^{T})\underset{j\in T}{\sum }d_j\theta _{j}q_{j} +(a^{S}-a^{T})\underset{j\in T}{\sum }d_jq_{j}. \end{aligned}$$

Combining with Condition B, we have

$$\begin{aligned} c(S,T)\le -f^{T}+(t^{S}-t^{T})\left( \underset{j\in T}{\sum }d_j\theta _{j} q_{j}-\frac{1}{\delta }\underset{j\in T}{\sum }\frac{a^Td_jq_{j}}{f^T}\right) . \end{aligned}$$

(A.12)

Given that

$$\begin{aligned} \theta _{\max }\le \frac{a_j}{f_j\delta }, \end{aligned}$$

then

$$\begin{aligned} \underset{j\in S}{\sum }d_j\theta _{j}q_{j}-\frac{1}{\delta }\underset{j\in S}{\sum }\frac{a^Sd_jq_{j}}{f^S}\le 0, \end{aligned}$$

and

$$\begin{aligned} \underset{j\in T}{\sum }d_j\theta _{j}q_{j}-\frac{1}{\delta }\underset{j\in T}{\sum }\frac{a^Td_jq_{j}}{f^T}\le 0. \end{aligned}$$

Therefore $c(S,T)<0$. When $t^{T}\ge t^{S}$, from (A.11), we have $c(S,T)<0$. Otherwise, (A.12) yields $c(S,T)\le -f^{T}<0$. Similar to the proof of Proposition 3, we can prove $c(S)=\underset{k\in F}{\min }c(S)_{k}$, for all $S\in N$. $\square$