Hi,
It's a voltage divider.
R1 (10k or 1k)
R2 (10k or 100k)
Calculation: R2/(R1 + R2). e.g. 10k/(10k + 10k) = 0.5. Or, 100k/101k = 0.99
If you put say 10k/10k, you'll only get 2.5V out (of a 5V supply).
If you put 1k/100k, you'll theoretically get 4.95V out of a 5V supply at the resistive divider junction.
Then, subtract the diode forward voltage drop. Let's assume it's around 0.7V. So, 4.95V - 0.7V = 4.25V out. Theoretically, it will be a little different, especially depending on type of diode used, also, unlikely the BJT is ideal so it will leak a tiny bit when off ("off" state is not always truly "off" with transistors), I have no idea how much but maybe a few mV or so will be lost there, so let's make up a final value of e.g. 4.2V out of the resistive junction after the diode.
Hope I haven't missed anything pertinent to your calculation.