Purpose for Shunt Resistor on Opto Isolator Input?
The currents that you show appear to agree with a quick circuit simulation I did with LTSpice. Do note that the 1.2K and 200 ohm resistors are not acting like a normal voltage divider. Instead the input diode in the opto-coupler is tending to clamp the voltage drop across the 200 ohm resistor to about 1.1V. The current from the upper resistor gets split into one current going through the 220 resistor and the other going through the opto coupler input diode. The current split ratio will depend upon the forward voltage drop of the input diode in the opto coupler. 

If you remove the opto coupler from the circuit then the two resistor voltage divider will not be clamped and the divider node will be at over 3.7V at a current of about  16.9mA in the divider. 

The inclusion of the C1 capacitor in the circuit acts like a low pass filter to the input signal. The time constant is primarily due to the 1.2K resistor. The 220 ohm resistor helps to discharge the filter capacitor when the input signal goes low or open.