The "correct" formulas are for C charging through R1+R2 towards VCC from 1/3 VCC to 2/3 VCC (output high period), and for C discharging through R2 towards GND from 2/3 VCC to 1/3 VCC (output low period).

The time taken for the voltage across the resistor to drop to half its initial value when charging or discharging a capacitor is: t = R C -ln(0.5). -ln(0.5) is 0.69314718056. Therefore the ideal formulas are:

t(high) = 0.69314718056 * (R1 + R2) * C

t(low) = 0.69314718056 * R2 * C

The reciprocal 0.69314718056 is 1.44269504089. The value given in the Philips data sheet, 1.49, is 3.3% different. I don't know why this is.

Assuming R1, R2 and C are precisely right, and the frequency is relatively low, the biggest cause of frequency error in practice will be the saturation voltage of the discharge transistor inside the 555 between pin 7 and GND. This will increase the output low period, i.e. reduce the frequency, especially if R1 and/or R2 is a low value. This source of error is not taken into account in the formula that uses a constant 0.693... value but it probably will be simulated properly by Multisim. Therefore I would expect the simulation to have a slightly lower frequency than you calculated from that formula. That is what you're seeing, but the error, especially for t(low), is pretty high.

Re the capacitor on the control voltage pin (pin 5). In the standard 555, pin 5 is connected to the first (top) tap in a chain of three 5k resistors in series from VCC to GND. The two taps in the chain set the upper and lower trigger voltages. Adding a decoupling capacitor from pin 5 to ground should make no difference to the operating frequency. But you're right John; the TLC555 data sheet does say that the capacitor makes a 10% difference. They don't explain why, and I can't suggest a reason.

Temperature and supply voltage will make a pretty small difference to the frequency, I think. The differences he is seeing are 3% for t(high) and 10% for t(low). I think those are too high to be explained as temperature and voltage variations.

Here's an idea. If the difference in t(low) in the simulation is due to the saturation voltage of the transistor, the error will be less if you increase the resistor values. So multiply the resistors by 10 and divide the capacitor by 10, and re-run the simulation. According to the simple formula, this should give the same frequencies. My guess is that with the higher resistor values, t(low) will still be higher than the value you calculated with the simple formula, but not so much higher.