jmc

Since the measure of each interior angle of the octagon is the same, each measures $(8-2)(180^{\circ })/8=135^{\circ }$. We extend sides $\overline{AB},\overline{CD},\overline{EF}$, and $\overline{GH}$ to form a rectangle: let $X$ be the intersection of lines $GH$ and $AB$; $Y$ that of $AB$ and $CD$; $Z$ that of $CD$ and $EF$; and $W$ that of $EF$ and $GH$.



As $BC=2$, we have $BY=YC=\sqrt{2}$. As $DE=4$, we have $DZ=ZE=2\sqrt{2}$. As $FG=2$, we have $FW=WG=\sqrt{2}$.

We can compute the dimensions of the rectangle: $WX=YZ=YC+CD+DZ=3+3\sqrt{2}$, and $XY=ZW=ZE+EF+FW=2+3\sqrt{2}$. Thus, $HX=XA=XY-AB-BY=1+2\sqrt{2}$, and so $AH=\sqrt{2}HX=4+\sqrt{2}$, and $GH=WX-WG-HX=2.$ The perimeter of the octagon can now be computed by adding up all its sides, which turns out to be $\boxed{20+\sqrt{2}}$.