Algebras derived equivalent to quasi-hereditary algebras Let an algebra always be finite dimensional over a field and connected.
It is well known that a quasi-hereditary algebra with $n$ simple modules has global dimension at most $2n-2$.

Questions:
  1. Do algebras derived equivalent to quasi-hereditary algebras with $n$ simple modules also have global dimension at most $2n-2$?

  
*Is there a good discrete test (for example with a computer) for checking whether an algebra of finite global dimension is derived equivalent to a quasi-hereditary algebra?
  
*Is every Nakayama algebra of finite global dimension derived equivalent to a quasi-hereditary algebra?

For example the Nakayama algebras with $n$ simples and Kupisch series $[n,n+1,...,n+1]$ are not quasi-hereditary for $n \geq 3$, but they are derived equivalent to the representation-finite block of a Schur algebra, which is quasi-hereditary.
Here are some Kupisch series of Nakayama algebras of finite global dimension that are not quasi-hereditary, but Im not sure whether they are derived equivalent to a quasi-hereditary algebra:
[ [ 3, 3, 3, 4 ], [ 3, 5, 5, 4 ], [ 4, 5, 6, 5 ], 
  [ 4, 6, 5, 5 ], [ 4, 6, 6, 5 ] ]
It would be interesting to see how to deal with this problem even in small examples.
 A: The answer to question 1 is "no": there are algebras with two simple modules, derived equivalent to quasi-hereditary algebras, but with global dimension three.
This can probably be extracted from one or both of the papers
Dubnov, Dmitry, On derived categories of modules over 2-vertex basic algebras, Commun. Algebra 28, No. 9, 4355-4374 (2000). ZBL0983.16009.
Volkov, Y. On the derived category of quasi-hereditary algebras with two simple modules,   arXiv:1812.00351,
but I'll give an explicit example.
Let $A$ be the algebra given by a quiver with two vertices $1$ and $2$, one arrow $\alpha$ from vertex $1$ to vertex $2$, two arrows $\beta_1$, $\beta_2$ from vertex $2$ to vertex $1$, and relations $\beta_1\alpha=0=\beta_2\alpha$.
Denoting by $S(i)$ and $P(i)$ the simple and indecomposable projective modules corresponding to vertex $i$, $P(2)$ is $3$-dimensional with head $S(2)$ and radical $S(1)\oplus S(1)$, and $P(1)$ is $4$-dimensional with head $S(1)$ and radical isomorphic to $P(2)$. So $A$ is quasihereditary with standard modules $\Delta(1)=S(1)$ and $\Delta(2)=P(2)$.
There is a tilting complex
$$T:=\dots\to0\to P(1)^3\to P(2)\to0\to\dots,$$
with $P(2)$ in degree zero, where one copy of $P(1)$ is in the kernel of the differential, and the other two copies map onto $\text{rad }P(2)$.
Let $B=\text{End}_{D^b(A)}(T)$, so there is an equivalence of triangulated categories $F:D^b(A)\to D^b(B)$ sending $T$ to $B$.
Since $\text{Hom}_{D^b(A)}(T,S(2)[i])=0$ for $i\neq0$, $FS(2)$ has homology only in degree zero, and so is a $B$-module $M$.
If $X$ is the quotient of $P(2)$ by a $1$-dimensional submodule of its socle, then $\text{Hom}_{D^b(A)}(T,X[i])=0$ for $i\neq1$, and so $FX[1]$ is a $B$-module $N$.
Now
$$\text{Ext}^3_B(N,M)\cong\text{Hom}_{D^b(B)}(N,M[3])
\cong\text{Hom}_{D^b(A)}(X,S(2)[2])\cong\text{Ext}^2_A(X,S(2)),$$
which is easily calculated to be non-zero.
Hence $B$ has global dimension at least three (in fact, it's equal to three).
