Finitistic dimension equal to the dominant dimension

Given a finite-dimensional self-injective algebra $A$ and an indecomposable non-projective module $N$, let $M:=A \oplus N$ and $B:=End(M)$. Does $B$ always have dominant dimension equal to the finitistic dimension?

The problem can be restated as follows: let $d:=\inf \{ i \geq 1 | Ext^{i}(M,M) \neq 0 \} +1$ (this is the dominant dimension of $B$) and $M^{\perp d-2} := \{ X | Ext^{i}(M,X)=0$ for $i=1,...,d-2 \}$. The question can be restated whether every module in $M^{\perp d-2}$ but not in $add(M)$ has infinite $add(M)$-resolution dimension. It is true in the following cases:

1. $A$ local and $M$ with $Ext^{1}(M,M) \neq 0$, a proof can be found here Finite add(N)-resolution as an answer. (For example this includes all local Hopf algebras.)

2. $A$ serial and $N$ a simple module or a radical of a projective indecomposable module (I only have a highly complicated proof of that at the moment). It is also true for various other representation-finite symmetric algebras when choosing certain simple modules.

3. $N$ with $\tau(N) \cong N$.

This would be highly surprising if always true, but the calculation of finitistic dimension is highly complicated and my computer can just test it for some small cases where $B$ is representation-finite. Had no luck with a counterexample yet.