The picture shows an example of solving the integer problem with a decomposition method. However, what I am trying to ask is about the dual function part instead of the integer part. As you can see, the integer variables are relaxed. However, I cannot find the same dual function as the author shows.
The dual function should be as:
$$d(\lambda)=-b^T\lambda+min_{x\in X}\sum_{i\in I}(c_i^T+\lambda^TH_i)x_i \label{1},$$
where $b$ is the shared resouce ($11.1$ in the problem), $I$ is the set of agents. In this case, it should be:
$$d(\lambda)=-11.1\lambda+min_{x\in X}\sum_{i=1}^4(c_i^T+\lambda^TH_i)x_i.$$
Let us focus on the minimization part, which is:
$$\min_{x\in X}\sum_{i=1}^4(c_i^T+\lambda^TH_i)x_i.$$
Then for the first agent $i=1$, denote by $x_{11}, x_{21}$ its two variables, we have,
$$\min_{x_1\in X_1}(c_1^T+\lambda^TH_1)x_1=(x_{11}+x_{21}+\lambda^T(x_{11}+x_{21}))=(\lambda^T+1)x_{11}+(\lambda^T+1)x_{21}.$$
Similarly, we have for $i=2$
$$\min_{x_2\in X_2}(c_2^T+\lambda^TH_2)x_2=(-2x_{12}+x_{22}+\lambda^T(5x_{12}+x_{22}))=(5\lambda^T-2)x_{12}+(\lambda^T+1)x_{22},$$
for $i=3$
$$\min_{x_3\in X_3}(c_3^T+\lambda^TH_3)x_3=(0.5x_{13}-x_{23}+\lambda^T(x_{13}+x_{23}))=(\lambda^T+0.5)x_{13}+(\lambda^T-1)x_{23},$$
for $i=4$
$$\min_{x_4\in X_4}(c_4^T+\lambda^TH_4)x_4=(-3x_{14}+0.5x_{24}+\lambda^T(x_{14}+x_{24}))=(\lambda^T-3)x_{14}+(\lambda^T+0.5)x_{24}.$$
Now, we can write the piecewise function about $\lambda$. But here, I cannot get the same function as the author.
For example, if $\lambda \in [0,2/5]$, I think these variables should take the larger values, i.e. $x_{12}=2.1,x_{23}=1.1,x_{14}=1.2$, because the coefficients are negative, while the others take zero value, because the coefficients are positive. Hence, the dual function in $[0, 2/5]$ is
$d(\lambda)=-11.1\lambda+2.1(5\lambda-2)+1.1(\lambda-1)+1.2(\lambda-3)=(-11.1+10.5+1.1+1.2)\lambda-4.2-1.1-3.6=1.7\lambda-8.9$, which is totally different from the author.
I know there must be something wrong. My understanding of the dual function might be incorrect. Can someone please help me with this?
The intent in the given solution appears to be that we don't drop the integrality constraint on $x_1, x_2, x_3, x_4$. With this modification, though your overall approach is correct, the optimal choice of primal variables for $\lambda \in [0,2/5]$ is $$x_1 = (0,0), \quad x_2 = (2,0), \quad x_3 = (0,1), \quad x_4 = (1,0).$$ This changes $2.1(5\lambda-2)$ to $2(5\lambda-2)$, $1.1(\lambda-1)$ to $1(\lambda-1)$, and $1.2(\lambda-3)$ to $1(\lambda-3)$, and altogether we get $$-11.1\lambda + 2(5\lambda-2) + (\lambda-1) + (\lambda-3) = -8 + 0.9\lambda,$$ just as in the given solution.
There is another discrepancy that I believe is simply a calculation error. For higher values of $\lambda$, we will drop the $2(5\lambda-2)$ term, then the $(\lambda-1)$ term, then the $(\lambda-3)$ term as each one becomes positive, because we can always switch the corresponding $x_i$ to $(0,0)$. This gives us: $$d(\lambda) = \begin{cases}-8 + 0.9\lambda & 0 \le \lambda < 2/5 \\ -4 - 9.1\lambda & 2/5 \le \lambda < 1 \\ -3 - 10.1\lambda & 1 \le \lambda < 3 \\ -11.1\lambda & 3 \le \lambda\end{cases}$$ (It is very easy to subtract $10$ from $0.9$ and accidentally get $-8.9$ rather than the correct $-9.1$.)