How to fully discharge supercapacitor?
The energy stored in a capacitor is given by : 

$$E= \frac{CV^2}{2}$$

Fill in the numbers for both 2.7 V and 900mV:

$$E_{\text{full}} = \frac{100 \text{F} \cdot 2.7 \text{V}^2}{2}\approx365 \text{J}$$
$$E_{\text{end}} = \frac{100 \text{F} \cdot 0.9 \text{V}^2}{2}\approx41 \text{J}$$

In other words, we have \$41/365\approx 11\% \$ of the full capacity left when your converter dies. 

Making DC/DC converters that can extract (part of) that 11% capacity left in the capacitor, without losing it all to lower overall efficiency, is an active and challenging topic within research.