How to increase unbounding and dominating numbers for $(\kappa^\lambda,\leq^*)$ Let $\kappa$ be regular and $\lambda\geq\kappa$. For $f, g\in\kappa^\lambda$ say that $f\le^* g$ if the set $\{\gamma<\lambda:f(\gamma)>g(\gamma)\}$ has
size less than  $\kappa$. Set
$\mathfrak{b}_\kappa^\lambda:=\min\{|F|:F\subseteq \kappa^\lambda\text{ and }\neg\exists y\in \kappa^\lambda\forall x\in F(x\leq^* y)\}$,
$\mathfrak{d}_\kappa^\lambda:=\min\{|D|:D\subseteq \kappa^\lambda\text{ and }\forall x\in \kappa^\lambda\exists y\in D(x\leq^* y)\}$. 
I'm studying the cardinals $\mathfrak{b}_\kappa^\lambda$ and $\mathfrak{d}_\kappa^\lambda$. My question what forcing should I use to use to increase $\mathfrak{b}_\kappa^\lambda$ and $\mathfrak{d}_\kappa^\lambda$?
For example: If $\lambda\geq\kappa$, $\mu>\lambda$ and $\mathrm{cf}(\mu)>\lambda$. What forcing can I use for $\mathfrak{d}_\kappa^\lambda\geq\mu$?
 A: Short version:  an appropriate version of Cohen forcing will increase $\mathfrak d^\lambda_\kappa$. 
With more details: Consider the forcing $Q=Q(\mu,\kappa,\mathord<\kappa)$, the set of all partial functions from $\mu$ (equivalently, from $\mu\times \lambda$) into $\kappa$ of size $<\kappa$.  The generic function $g:\mu\times\lambda\to \kappa$ induces a family $(g_i:i<\mu)$, each $g_i$ a function from $\lambda $ to $\kappa$. 
Assuming $\kappa^{<\kappa} = \kappa$, the forcing notion $Q$ has the $\kappa^+$-cc, and adds no bounded sets to $\kappa$, hence preserves all cardinals. 
I claim that $Q$ forces   $ {\mathfrak d^\kappa_\lambda \ge \mu}$. Indeed, any function $f\colon \lambda\to \kappa$ in the extension already lives in a $Q(A, \kappa, \mathord<\kappa)$-extension, for some $A$ of size $\lambda$, as the values of $f$ are decided by a family of $\lambda$ many (labelled) antichains. Such a function $f$ cannot be forced to dominate any $g_i$ for $i\notin A$.  Hence if you have fewer than $\mu$ many functions, you will find an index $i<\mu$ such that these functions cannot dominate $g_i$. 
A: A partial answer concerning $\mathfrak{b}_\kappa^\lambda$:
Assume $\lambda > \kappa$, regular. I claim $\mathfrak{b}_\kappa^\lambda =\kappa$:
It is obvious, why $\mathfrak{b}_\kappa^\lambda <\kappa$ is impossible.
The family $(f_\alpha \equiv \alpha)_{\alpha < \kappa}$ is unbounded. Assume $g$ dominates all $f_\alpha$. As $\lambda$ is regular, there exists $\beta < \kappa$ such that $\{i < \lambda \colon g(i) = \beta\}$  has size $\lambda$. So $g$ does not dominate $f_{\beta +1}$. 
