Certain lengths and angles are marked in the diagram. First, notice that $ a+b=120^\circ$, and $ \angle E=a-30>0$. Similarly, $ b-30>0$. Thus, $ 0<a-30<60$. Let us consider the triangle that has the smaller perimeter. WLOG let it be $ \triangle CDF$. We shall consider the case where the two triangles are congruent. Since $ \angle BCE=\angle DCF=60^\circ$, congruency is achieved when $ a=b=60^\circ$. In this case, $ \triangle CDF$ is 30-60-90 and has perimeter $ 1+2+\sqrt{3}<5$. Notice that this case results in the maximum possible perimeter for$ \triangle CDF$. Since the perimeter must be an integer, we conclude that the perimeter is less than 5, i.e. $ 1+x+y\le4$.\ Now we consider the case where the perimeter of $ \triangle CDF$ is the smallest. This occurs when $ b-30$ is maximized, or when $ b-30=60$, making $ \triangle CDF$ an equilateral triangle with perimeter 3. This can never be achieved because it would imply $ \angle E=a-30=0$, so we conclude that the perimeter of $ \triangle CDF$ must be greater than 3, i.e. $ 1+x+y\ge 4$. Combined with the previous inequality $ 1+x+y\le4$ we discover that the perimeter of $ \triangle CDF$ must be 4, or $ x+y=3$. Now we find $ x+y$.\ From the law of sines, \[ \frac{1}{\sin (b-30)}=\frac{y}{\sin 60} \Longrightarrow y=\frac{\sin 60}{\sin (b-30)}\\\\ \frac{1}{\sin (b-30)}=\frac{x}{\sin \angle CDF}=\frac{x}{\sin (150-b)} \Longrightarrow x=\frac{\sin (150-b)}{\sin (b-30)}\\\\ x+y=\frac{\sin 60 +\sin (150-b)}{\sin (b-30)}=\frac{\sqrt{3}+\cos b + \sqrt{3}\sin b}{\sqrt{3}\sin b - \cos b}=3 \Longrightarrow \\\\ \sqrt{3}+\cos b + \sqrt{3}\sin b=3\sqrt{3}\sin b - 3\cos b \Longrightarrow \\\\ \sqrt{3}+4\cos b=2\sqrt{3}\sin b =2\sqrt{3}\sqrt{1-\cos^2 b} \Longrightarrow \\\\ 3+8\sqrt{3}\cos b+ 16\cos^2 b=12-12\cos^2 b \Longrightarrow \\\\ 28\cos^2 b+8\sqrt{3}\cos b -9=0 \Longrightarrow \\\\ \cos b=\frac{-8\sqrt{3}+ \sqrt{192+1008}}{56}=\boxed{\frac{3\sqrt{3}}{14}} \Longrightarrow \\\\ \sin b=\sqrt{1-\cos^2 b}=\boxed{\frac{13}{14}}\] With one last application of the law of sines on triangle ABD, we have \[ \frac{AD}{\sin a}=\frac{BD}{\sin 60}=2 \Longrightarrow AD=2\sin a\] Similarly $ AB=2\sin b=\boxed{\bf{\frac{13}{7}}}$. Since $ a+b=120$, then $ \sin a=\sin (120-b)= \frac{\sqrt{3}\cos b + \sin b}{2}=\frac{9}{28}+\frac{13}{28}=\frac{11}{14}$. Thus, $ AD=2\sin a= \boxed{\bf{\frac{11}{7}}}$, so the perimeter of quadrilateral $ ABCD$ is $ \frac{13}{7}+\frac{11}{7}+2=$ \[ \boxed{\frac{38}{7}}\]