### 3.7.101.3. Flow models for $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$, $\mathrm{𝚜𝚊𝚖𝚎}$

Figure 3.7.24 presents flow models for the $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$ and the $\mathrm{𝚜𝚊𝚖𝚎}$ constraints. Blue arcs represent feasible flows respectively corresponding to the solutions $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$$\left(〈{x}_{1}=2,{x}_{2}=4,{x}_{3}=6〉,〈{y}_{1}=2,{y}_{2}=4〉\right)$ and $\mathrm{𝚜𝚊𝚖𝚎}$$\left(〈{x}_{1}=2,{x}_{2}=4,{x}_{3}=5〉,〈{y}_{1}=2,{y}_{2}=4,{y}_{3}=5〉\right)$, while pink arcs correspond to arcs that cannot carry any flow if the constraint has a solution. Within the context of the $\mathrm{𝚜𝚊𝚖𝚎}$ constraint, the assignment ${x}_{1}=1$ is forbidden since $1\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$. Consequently ${x}_{1}=2$ and, since ${y}_{1}$ is the only variable of $\left\{{y}_{1},{y}_{2},{y}_{3}\right\}$ that can be assigned value 2, the assignment ${y}_{1}=3$ is forbidden. Now since $3\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$ the assignment ${x}_{2}=3$ is also forbidden. Finally ${x}_{3}=6$ is forbidden since $6\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$.

##### Table 3.7.24. Domains of the variables for the $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$ constraint of Figure 3.7.24.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({y}_{i}\right)$
1$\left\{1,2\right\}$1$\left\{2,3\right\}$
2$\left\{3,4\right\}$2$\left\{4,5\right\}$
3$\left\{4,5,6\right\}$
##### Table 3.7.24. Domains of the variables for the $\mathrm{𝚜𝚊𝚖𝚎}$ constraint of Figure 3.7.24.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({y}_{i}\right)$
1$\left\{1,2\right\}$1$\left\{2,3\right\}$
2$\left\{3,4\right\}$2$\left\{4,5\right\}$
3$\left\{4,5,6\right\}$3$\left\{4,5\right\}$