Multi-choice Linear Programming in Fuzzy Random Hybrid Uncertainty Environment and Their Application in Multi-commodity Transportation Problem

Figures & data

Table 1. Node point for multi-choice parameter ${\tilde{\overline{c}}}_j$.

$u_j$	0	1	2	…	$k_j-1$
$f_{{\tilde{\overline{c}}}_j}\left(u_j\right)$	${\tilde{\overline{c}}}_j^{\left(1\right)}$	${\tilde{\overline{c}}}_j^{\left(2\right)}$	${\tilde{\overline{c}}}_j^{\left(3\right)}$	$\dots$	${\tilde{\overline{c}}}_j^{\left(k_j\right)}$

Table 2. Node point for multi-choice parameter ${\tilde{\overline{a}}}_{ij}$.

$w_{ij}$	0	1	2	…	$p_{ij}-1$
$f_{{\tilde{\overline{a}}}_{ij}}\left(w_{ij}\right)$	${\tilde{\overline{a}}}_{ij}^{\left(1\right)}$	${\tilde{\overline{a}}}_{ij}^{\left(2\right)}$	${\tilde{\overline{a}}}_{ij}^{\left(3\right)}$	$\dots$	${\tilde{\overline{a}}}_{ij}^{\left(p_{ij}\right)}$

Table 3. Node point for multi-choice parameter ${\tilde{\overline{b}}}_i$.

$v_i$	0	1	2	…	$r_i-1$
$f_{{\tilde{\overline{b}}}_i}\left(v_i\right)$	${\tilde{\overline{b}}}_i^{\left(1\right)}$	${\tilde{\overline{b}}}_i^{\left(2\right)}$	${\tilde{\overline{b}}}_i^{\left(3\right)}$	$\dots$	${\tilde{\overline{b}}}_i^{\left(r_i\right)}$

Table 4. Uncertain parameters of the example.

Source	Customer	Fuzzy Random Capacity	Fuzzy Random Multi-choice cost (commodity 1)	Fuzzy Random Multi-choice cost (commodity 1)
1	1	$\left(N\sim \left(1.5,0.1\right),0.5,2.5\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{11}^{\left(1\right)}=\left(N\sim \left(2,0.5\right),5.3,1.3\right)\\ {\tilde{\overline{c}}}_{11}^{\left(2\right)}=\left(N\sim \left(3,0.1\right),6.7,2.7\right)\\ {\tilde{\overline{c}}}_{11}^{\left(3\right)}=\left(N\sim \left(4,0.2\right),4.5,0.5\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{11}^{\left(1\right)}=\left(N\sim \left(3.5,0.2\right),2.3,0.3\right)\\ {\tilde{\overline{c}}}_{11}^{\left(2\right)}=\left(N\sim \left(3.7,0.5\right),0.5,1.7\right)\\ {\tilde{\overline{c}}}_{11}^{\left(3\right)}=\left(N\sim \left(4.1,0.7\right),1.2,4.8\right)\end{array}$
1	2	$\left(N\sim \left(3.3,0.2\right),2.8,5.6\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{12}^{\left(1\right)}=\left(N\sim \left(6.5,0.5\right),2.5,4.5\right)\\ {\tilde{\overline{c}}}_{12}^{\left(2\right)}=\left(N\sim \left(8.5,0.4\right),3.4,1.4\right)\\ {\tilde{\overline{c}}}_{12}^{\left(3\right)}=\left(N\sim \left(8.8,0.7\right),0.8,1.8\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{12}^{\left(1\right)}=\left(N\sim \left(1.5,0.2\right),2.6,4.6\right)\\ {\tilde{\overline{c}}}_{12}^{\left(2\right)}=\left(N\sim \left(2.5,0.3\right),3.1,5.1\right)\\ {\tilde{\overline{c}}}_{12}^{\left(3\right)}=\left(N\sim \left(3.5,0.7\right),4.2,6.2\right)\end{array}$
1	3	$\left(N\sim \left(2.7,0.5\right),0.2,1.4\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{13}^{\left(1\right)}=\left(N\sim \left(7.2,0.3\right),2.7,5.9\right)\\ {\tilde{\overline{c}}}_{13}^{\left(2\right)}=\left(N\sim \left(8.7,0.5\right),2.3,3.5\right)\\ {\tilde{\overline{c}}}_{13}^{\left(3\right)}=\left(N\sim \left(10.6,0.7\right),4.5,2.1\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{13}^{\left(1\right)}=\left(N\sim \left(7.5,0.3\right),3.1,1.1\right)\\ {\tilde{\overline{c}}}_{13}^{\left(2\right)}=\left(N\sim \left(7.6,0.1\right),0.5,2.1\right)\\ {\tilde{\overline{c}}}_{13}^{\left(3\right)}=\left(N\sim \left(8.5,0.7\right),4.3,6.3\right)\end{array}$
2	1	$\left(N\sim \left(3.5,0.2\right),3.7,1.7\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{21}^{\left(1\right)}=\left(N\sim \left(8.5,0.6\right),2.1,4.1\right)\\ {\tilde{\overline{c}}}_{21}^{\left(2\right)}=\left(N\sim \left(10.3,0.3\right),2.5,1.3\right)\\ {\tilde{\overline{c}}}_{21}^{\left(3\right)}=\left(N\sim \left(10.1,0.5\right),2.7,6.3\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{21}^{\left(1\right)}=\left(N\sim \left(2.5,0.5\right),4.6,2.6\right)\\ {\tilde{\overline{c}}}_{21}^{\left(2\right)}=\left(N\sim \left(3.5,0.2\right),3.7,1.7\right)\\ {\tilde{\overline{c}}}_{21}^{\left(3\right)}=\left(N\sim \left(3.6,0.3\right),1.4,3\right)\end{array}$
2	2	$\left(N\sim \left(2,0.75\right),0.7,4.7\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{22}^{\left(1\right)}=\left(N\sim \left(1.5,0.2\right),3.1,1.1\right)\\ {\tilde{\overline{c}}}_{22}^{\left(2\right)}=\left(N\sim \left(1,0.25\right),0.8,4.8\right)\\ {\tilde{\overline{c}}}_{22}^{\left(3\right)}=\left(N\sim \left(2.2,0.1\right),1.4,4.6\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{22}^{\left(1\right)}=\left(N\sim \left(1.7,0.2\right),3.3,4.5\right)\\ {\tilde{\overline{c}}}_{22}^{\left(2\right)}=\left(N\sim \left(2.5,0.1\right),2.5,4.5\right)\\ {\tilde{\overline{c}}}_{22}^{\left(3\right)}=\left(N\sim \left(3.5,0.2\right),1.8,3.8\right)\end{array}$
2	3	$\left(N\sim \left(4,0.25\right),5.7,1.7\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{23}^{\left(1\right)}=\left(N\sim \left(3.1,0.2\right),1.7,1.3\right)\\ {\tilde{\overline{c}}}_{23}^{\left(2\right)}=\left(N\sim \left(3.5,0.5\right),1.2,3.2\right)\\ {\tilde{\overline{c}}}_{23}^{\left(3\right)}=\left(N\sim \left(5.3,0.4\right),1.8,0.6\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{23}^{\left(1\right)}=\left(N\sim \left(1.5,0.5\right),2.3,0.3\right)\\ {\tilde{\overline{c}}}_{23}^{\left(2\right)}=\left(N\sim \left(1.8,0.2\right),0.8,1.6\right)\\ {\tilde{\overline{c}}}_{23}^{\left(3\right)}=\left(N\sim \left(2.5,0.5\right),0.3,2.3\right)\end{array}$
3	1	$\left(N\sim \left(3.4,1.0\right),0.4,2.8\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{31}^{\left(1\right)}=\left(N\sim \left(2,0.65\right),0.4,4.4\right)\\ {\tilde{\overline{c}}}_{31}^{\left(2\right)}=\left(N\sim \left(4.5,0.2\right),3.3,1.3\right)\\ {\tilde{\overline{c}}}_{31}^{\left(3\right)}=\left(N\sim \left(4.7,0.1\right),0.5,1.7\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{31}^{\left(1\right)}=\left(N\sim \left(16.4,0.6\right),2.4,4.8\right)\\ {\tilde{\overline{c}}}_{31}^{\left(2\right)}=\left(N\sim \left(17.5,0.7\right),3.5,5.5\right)\\ {\tilde{\overline{c}}}_{31}^{\left(3\right)}=\left(N\sim \left(18.5,0.8\right),2.8,4.8\right)\end{array}$
3	2	$\left(N\sim \left(2.2,0.2\right),3.2,6.4\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{32}^{\left(1\right)}=\left(N\sim \left(8.5,0.2\right),2.5,4.5\right)\\ {\tilde{\overline{c}}}_{32}^{\left(2\right)}=\left(N\sim \left(11,0.55\right),5.7,1.7\right)\\ {\tilde{\overline{c}}}_{32}^{\left(3\right)}=\left(N\sim \left(10.3,0.7\right),0.8,3.6\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{32}^{\left(1\right)}=\left(N\sim \left(2.3,0.2\right),2.8,5.6\right)\\ {\tilde{\overline{c}}}_{32}^{\left(2\right)}=\left(N\sim \left(3,0.55\right),0.3,4.3\right)\\ {\tilde{\overline{c}}}_{32}^{\left(3\right)}=\left(N\sim \left(5.5,0.4\right),3.3,1.3\right)\end{array}$
3	3	$\left(N\sim \left(2.4,0.6\right),1.3,3.7\right)$	$\begin{array}{l}{\tilde{\overline{c}}}_{33}^{\left(1\right)}=\left(N\sim \left(2.8,0.1\right),0.7,1.5\right)\\ {\tilde{\overline{c}}}_{33}^{\left(2\right)}=\left(N\sim \left(3.7,0.1\right),1.6,2.8\right)\\ {\tilde{\overline{c}}}_{33}^{\left(3\right)}=\left(N\sim \left(4.3,0.2\right),0.2,3\right)\end{array}$	$\begin{array}{l}{\tilde{\overline{c}}}_{33}^{\left(1\right)}=\left(N\sim \left(1.5,0.5\right),1.6,3.6\right)\\ {\tilde{\overline{c}}}_{33}^{\left(2\right)}=\left(N\sim \left(2.6,0.3\right),1.6,3.2\right)\\ {\tilde{\overline{c}}}_{33}^{\left(3\right)}=\left(N\sim \left(4.8,0.6\right),3.3,0.1\right)\end{array}$

Table 5. The optimal solution for the mathematical model.

$x_{ij}$	$x_{11}$	$x_{12}$	$x_{13}$	$x_{21}$	$x_{22}$	$x_{23}$	$x_{31}$	$x_{32}$	$x_{33}$
Commodity 1	1.5	0.0	0.5	0.0	2.0	0.0	0.5	0.0	1.5
Commodity 2	0.5	1.5	0.0	1.5	0.0	0.5	0.0	0.5	1.5