As suggested by @RobPratt, here is the new opened question. I have these decomposed master and subproblems:

\begin{align} &\text{minimize} &\sum_t \sum_s \text{slack}_{ts} \\ &\text{subject to} &\sum_i \sum_r \text{motivation}_{its}^r \lambda_{ir} + \text{slack}_{ts} & = \text{demand}_{ts} &&\forall t,s &&\\ &&\sum_r \lambda_{ir} &= 1 &&\forall i &&\\ &&\lambda_{ir} &\in\mathbb{Z}^+ &&\forall i,r\\ &&\text{slack}_{ts} &\ge 0 &&\forall t,s \end{align} and subproblems($$i$$): \begin{align} &\text{minimize} &0-\sum_{t,s} \pi_{ts} \text{motivation}_{its} - \mu_i \\ &\text{subject to} &-M(1-x_{its}) \le \text{motivation}_{its} - \text{mood}_{it} &\le M(1-x_{its}) &&\forall t,s \\ &&\text{motivation}_{its} &\le x_{its} && \forall t,s \\\ &&\alpha_{it} \sum_s x_{its} + \text{mood}_{it} &= 1 &&\forall t\\ &&\text{motivation}_{its} &\in[0,1] &&\forall t,s \\ &&\text{mood}_{it} &\in[0,1] &&\forall t \\ &&x_{its}&\in \{0,1\} &&\forall t,s\\ \end{align}

Now i want to additionally impose the constraint: $$\sum_i x_{its} \geq 1 \quad \forall t, s$$

One approach would be to add a new single problem that includes all $$i$$ and impose the new constraint there. How would that look like?

Your original compact problem is: \begin{align} &\text{minimize} &\sum_t \sum_s \text{slack}_{ts} \\ &\text{subject to} &\sum_i \text{motivation}_{its} + \text{slack}_{ts} &= \text{demand}_{ts} &&\forall t,s \\ &&-M(1-x_{its}) \le \text{motivation}_{its} - \text{mood}_{it} &\le M(1-x_{its}) &&\forall i,t,s \\ &&\text{motivation}_{its} &\le x_{its} && \forall i,t,s \\ &&\alpha_{it} \sum_s x_{its} + \text{mood}_{it} &= 1 &&\forall i,t\\ &&\sum_i x_{its} &\ge 1 &&\forall t,s\\ &&\text{motivation}_{its} &\in[0,1] &&\forall i,t,s \\ &&\text{slack}_{ts} &\ge 0 &&\forall t,s \\ &&\text{mood}_{it} &\in[0,1] &&\forall i,t \\ &&x_{its}&\in \{0,1\} &&\forall i,t,s\\ \end{align}

For the Dantzig-Wolfe reformulation, introduce $$\lambda_r \ge 0$$ and substitute $$\text{motivation}_{its} = \sum_r \text{motivation}_{its}^r \lambda_r$$ in the (objective and) complicating constraints to obtain the master problem: \begin{align} &\text{minimize} &\sum_t \sum_s \text{slack}_{ts} \\ &\text{subject to} &\sum_i \sum_r \text{motivation}_{its}^r \lambda_r + \text{slack}_{ts} & = \text{demand}_{ts} &&\forall t,s &&\text(\text{\pi_{ts} free})\\ &&\sum_r \lambda_r &= 1 &&&&\text{(\mu free)}\\ &&\lambda_r &\in\mathbb{Z}^+ &&\forall r\\ &&\text{slack}_{ts} &\ge 0 &&\forall t,s \end{align} The reduced cost of $$\lambda_r$$ is $$0-\sum_{i,t,s} \pi_{ts} \text{motivation}_{its}^r - \mu$$, so the subproblem is: \begin{align} &\text{minimize} &0-\sum_{i,t,s} \pi_{ts} \text{motivation}_{its} - \mu \\ &\text{subject to} &-M(1-x_{its}) \le \text{motivation}_{its} - \text{mood}_{it} &\le M(1-x_{its}) &&\forall i,t,s \\ &&\text{motivation}_{its} &\le x_{its} && \forall i,t,s \\\ &&\alpha_{it} \sum_s x_{its} + \text{mood}_{it} &= 1 &&\forall i,t\\ &&\sum_i x_{its} &\ge 1 &&\forall t,s\\ &&\text{motivation}_{its} &\in[0,1] &&\forall i,t,s \\ &&\text{mood}_{it} &\in[0,1] &&\forall i,t \\ &&x_{its}&\in \{0,1\} &&\forall i,t,s\\ \end{align}

• Isn't there a separate subproblem for each $i$? If so, how can you sum over $i$ in a subproblem?
– prubin
Commented May 28 at 18:13
• @prubin Yes, decomposing by $i$ is the more natural approach. But in another comment the OP asked for details about instead putting the new constraints in a single subproblem that includes all $i$: or.stackexchange.com/questions/11608/… Commented May 28 at 18:30
• So is your answer above replacing the various subproblems by a single one? This version of the question still mentions "subproblems($i$)".
– prubin
Commented May 28 at 18:33
• Yes, after imposing the new constraints, the subproblems are no longer independent and instead form a single block. Commented May 28 at 18:35