Quick question:
Does the persistance factor (1+r-ω) basically replace the (r-g) in the standard Terminal Value formula? Then we discount the Terminal Value back to find the Present Value today. Will “g” always be 0 in the persistance factor? Maybe that is my question.
Let’s do some algebra:
1 + r – ω = r – ω + 1 = r – (ω – 1)
So:
g = ω – 1
Thus, if, for example, ω = 0.8, then g = ω – 1 = 0.8 – 1 = –0.2: the growth rate of residual income is –20% (or, next year’s residual income is 80% of this year’s residual income).
Wow. You truly are a magician. Thank you sir.
You’re quite welcome.
How should one treat the 1+g in the numerator ? Should we take the g value as calculated above ?
In some questions the growth factor with the mentioned g rate has been multiplied with in the numerator and in some it hasn’t. What’s the difference?
The way I understand it is, you would always use the 1+g (with g being the long-term sustainable growth rate of residual income) depending on the time period you’re asked to calculate. If you’re given expected yr. 4 RI and are asked to calculate continuing RI starting in year 5, you would take RIYr. 4 x (1+g) / (1+r-ω). However, if you’re given yr. 5 RI and are asked to calculate continuing RI from that, you won’t need to include (1+g) since you already have the relevant RI.
S2000Magician - did I get that right?
Looks good to me.
Thanks a lot, tuchsford and s2000magician. I don’t know why this had to confuse me.
My pleasure.