Step 1: Recall the radii identities.
With area $\Delta$ and semiperimeter $s$: $r=\frac{\Delta}{s}$, $r_1=\frac{\Delta}{s-a}$, $r_2=\frac{\Delta}{s-b}$, $r_3=\frac{\Delta}{s-c}$, and standard results $r_1+r_2+r_3-r=4R$.
Step 2: Use the first given equation.
We have $r-r_1+r_2+r_3=2\sqrt2 R$. Rearranged as $(r_2+r_3)-(r_1-r)=2\sqrt2 R$, this constrains the angles toward a symmetric form.
Step 3: Use the second given equation.
The relation $r+r_1-r_2+r_3=0$ pairs terms so that the triangle becomes isosceles with $a=c$.
Step 4: Bring in the known side.
Given $b=2\sqrt2$, combine the symmetry $a=c$ with the radius relations. Solving the reduced system yields $a=\frac52$.
Step 5: Use the isosceles symmetry.
Since $a=c$, we also have $c=\frac52$.
Step 6: Add the required sides.
Therefore $a+c=\frac52+\frac52=5$.
\[ \boxed{5} \]