Confusion with voltage drop equation
The 1.8V is a given for the LED; if current flows through a LED it will show a more or less constant voltage across it, just like a common silicon diode will drop about 0.7V at low currents. The anode will always be 1.8V higher than the cathode for your type of LED (the actual voltage mainly depends on the LED's color). The fact that this voltage is constant allows you to calculate the appropriate current limiting series resistor:  


  \$ R = \dfrac{V_+ - V_{LED}}{I_{LED}} \$  


or the current for a given resistor value:  


  \$ I_{LED} = \dfrac{V_+ - V_{LED}}{R} = \dfrac{5V -1.8V}{220 \Omega} = 15mA \$