# Apply monotone convex interpolation to swap rate input data

I'm trying to apply Hagan & West's monotone convex interpolation to a 6m EURIBOR (forward) curve using ESTR (already bootstrapped) for discounting.

In their paper Hagan & West use discrete forward rates $$f_i^d$$ belonging to the interval $$[\tau_{i-1},\tau_i]$$ as input. They then try to identify and instantaneous forward rate function f that satisfies $$\frac{1}{\tau_i-\tau_{i-1}} \int_{\tau_{i-1}}^{\tau_i} f(t) dt = f_i^d$$ plus some other constraints.

The problem I'm struggling now with is how to get the discrete forwards from my input data, which are FRA rates in the short end and fix-float-swap rates form 2y onwards. In their paper the authors state that "we have (or can rearrange our inputs so that we have) discrete forwards for entire intervals". The FRA rates directly give me discrete forward rates. For the swaps it's a bit more complicated. For a m-year swap to be at par the following condition has to satisfied $$\sum_{j=1}^m S_{my} \cdot dcf(\tau_{j-1},\tau_j) \cdot DF_{ESTR}(\tau_j) = \sum_{i=1}^n f_i^d \cdot dcf(\tau_{i-1},\tau_i) \cdot DF_{ESTR}(\tau_i)$$ whith $$\tau_j \in [0,1y,2y,...]$$ and $$\tau_i \in [0,6m,1y,...]$$. Now the 3y formula can be rewritten using the 2y one as $$f_5^d \cdot dcf(\tau_4,\tau_5) \cdot DF_{ESTR}(\tau_5) + f_6^d \cdot dcf(\tau_5,\tau_6) \cdot DF_{ESTR}(\tau_6) = \underbrace{\left( \sum_{j=1}^3 S_{3y} \cdot dcf(\tau_{j-1},\tau_j) \cdot DF_{ESTR}(\tau_j) \right) - \left( \sum_{i=1}^4 S_{2y} \cdot dcf(\tau_{i-1},\tau_i) \cdot DF_{ESTR}(\tau_i) \right)}_{=:A}$$ Or in integral notation as $$\frac{1}{\tau_5-\tau_4} \int_{\tau_4}^{\tau_5} f(t) dt \cdot dcf(\tau_4,\tau_5) \cdot DF_{ESTR}(\tau_5) + \frac{1}{\tau_6-\tau_5} \int_{\tau_5}^{\tau_6} f(t) dt \cdot dcf(\tau_5,\tau_6) \cdot DF_{ESTR}(\tau_6) = A$$ This means I only have combined information about $$f_5^d$$ and $$f_6^d$$.

Can anybody give me a hint on how to continue? The comment of the authors imply that I can rewrite this as $$\frac{1}{\tau_6-\tau_4} \int_{\tau_4}^{\tau_6} f(t) dt = B$$. But I don't see how ...