jmc

The first thing to be addressed is the fractions under the inner sets of ceiling functions. The smallest integer greater than $\frac{27}{17}$ is $2$. The smallest integer greater than $\frac{7\cdot 17}{27}$, which is equal to $\frac{119}{27}$ is $5$. Therefore, the original problem can be rewritten as: $$\frac{\lceil \frac{17}{7}-2\rceil }{\lceil \frac{27}{7}+5\rceil }=\frac{\lceil \frac{3}{7}\rceil }{\lceil \frac{62}{7}\rceil }$$ The smallest integer greater than $\frac{3}{7}$ is $1$ and the smallest integer greater than $\frac{62}{7}$ is $9$. Hence, the final simplified fraction is $\boxed{\frac{1}{9}}$.