jmc

By AM-GM, $$\begin{array}{rl}\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}& \ge 4\sqrt[4]{\frac{r_1}{2}\cdot \frac{r_2}{4}\cdot \frac{r_3}{5}\cdot \frac{r_4}{8}}\\ & =4\sqrt[4]{\frac{r_1{r}_2{r}_3{r}_4}{320}}.\end{array}$$Since $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1,$ this gives us $$r_1{r}_2{r}_3{r}_4\le \frac{320}{4^4}=\frac{5}{4}.$$By Vieta's formulas, $r_1{r}_2{r}_3{r}_4=\frac{5}{4},$ so by the equality condition in AM-GM, $$\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac{1}{4}.$$Then $r_1=\frac{4}{2}=\frac{1}{2},$ $r_2=1,$ $r_3=\frac{5}{4},$ and $r_4=2,$ so $$r_1+r_2+r_3+r_4=\frac{1}{2}+1+\frac{5}{4}+2=\frac{19}{4}.$$So by Vieta's formulas, $a=\boxed{19}.$