jmc

By Cauchy-Schwarz, $$(a+2b)\left(\frac{2}{a}+\frac{1}{b}\right)\ge (\sqrt{2}+\sqrt{2}{)}^2=8.$$For equality to occur, we must have $a^2=4b^2,$ or $a=2b.$ Then $4b=1,$ so $b=\frac{1}{4},$ and $a=\frac{1}{2}.$

Hence, the minimum value is $\boxed{8}.$