Calculate a discount rate (am learning share market investing)

[rnikey]

Hi everyone. I just joined and am hoping for some help that is now unfortunately beyond me. I'm learning share market investing and have been going over discounting formula. I've developed a formula that includes growth. It's basically the sum of a geometric series....

V is the value today of the future payments.
G is the growth = (g+1) if g is a percentage.
D is the discount rate = (d+1) if d is a percentage.
a is the last dividend, assumed to be just gone. We are discounting back future payments which (may) grow.

My formula...
V = a(G/D + (G/D)^2 + (G/D)^3 + ... + (G/D)^n)

This solves to
V= a.G/(D-G) (1-(G/D)^n)

This is great and it fits into my excel spreadsheet nicely.

The way most people use it is to calculate V using your personal discount rate (commonly 10% but could be anything), and then compare that to the share price. You would then buy or sell the share accordingly.

I now want to solve for D (or d) rather than solve for V. So V will be the share price currently offered, I want to calculate the discount rate I'm getting for that price.

I am good at math but unfortunately this is now beyond me.

Please help.

 

[Deleted member 4993]

Hi everyone. I just joined and am hoping for some help that is now unfortunately beyond me. I'm learning share market investing and have been going over discounting formula. I've developed a formula that includes growth. It's basically the sum of a geometric series....

V is the value today of the future payments.
G is the growth = (g+1) if g is a percentage.
D is the discount rate = (d+1) if d is a percentage.
a is the last dividend, assumed to be just gone. We are discounting back future payments which (may) grow.

My formula...
V = a(G/D + (G/D)^2 + (G/D)^3 + ... + (G/D)^n)

This solves to
V= a.G/(D-G) (1-(G/D)^n) ...............................................................(1)

This is great and it fits into my excel spreadsheet nicely.

The way most people use it is to calculate V using your personal discount rate (commonly 10% but could be anything), and then compare that to the share price. You would then buy or sell the share accordingly.

I now want to solve for D (or d) rather than solve for V. So V will be the share price currently offered, I want to calculate the discount rate I'm getting for that price.

I am good at math but unfortunately this is now beyond me.

Please help.

Substitute u = G/D in equation (1)

Rewrite equation above in terms of 'u'.

Can you isolate 'u'? Please show us your results.

 

[rnikey]

Substitute u = G/D in equation (1)

Rewrite equation above in terms of 'u'.

Can you isolate 'u'? Please show us your results.

Replacing u = G/D and solving for u still leaves other G's and D's around. I have tried to solve for D or G or even D/G, but I have failed.

 

[rnikey]

That means V= a.G/(D-G) * (1-(G/D)^n)

Or did you mean V= a.G/[(D-G) (1-(G/D)^n)] ?

Yes, the (1-(G/D)^n) is above the line. You could write...
V = a * G * (1-(G/D)^n)
--------------------
(D - G)

And do you mean something like this simple example:

YEAR  PAYMENT  INTEREST  BALANCE
  0                          .00 (PV = ?)
  1   1000.00       .00  1000.00
  2   1100.00    120.00  2220.00
  3   1210.00    266.40  3696.40
  4   1331.00    443.57  5470.97

Received over 4 years is an annual flow of $1000 at end of 1st year,
increasing by 10% each year.
The interest rate paid is 12% annually.

So the PV = 5470.97 / 1.12^4 = 3476.90

And you are trying to solve for the payment increase percentage of 10% ?

If not, please supply an example.

d	g	The full growth and discount formula	P1	P2	V1	V2	V
0.1	0.1	#DIV/0!	110	121	100	100	200
0.2	0.1	175.694444	110	121	91.66667	84.02778	175.6944
0.3	0.1	156.213018	110	121	84.61538	71.59763	156.213
0.4	0.1	140.306122	110	121	78.57143	61.73469	140.3061
0.5	0.1	127.111111	110	121	73.33333	53.77778	127.1111

In the example above, a = $100 and n = 2.
You can see that the full formula, which is the one I posted above, does indeed evaluate to the sum of the next two future discounted values, except when d = g.

The #DIV/0 is another annoying quirk of my formula. When d = g, the formula gets a 0/0, even though in reality there is a real value, here = $200. If you can change my formula to remove this quirk I would also be grateful, however this is not the problem here.

Discounting when applied to share trading means you are deciding what future payments are worth today. You decide that you would go without your money today to have d% more tomorrow. It's a personal rate. You may want to google it and see a tutorial on discounting. I do recommend that you learn share market investing to improve your wealth in the future.

I have the formula that solves for V, or the amount I think future payments are worth. Now I want to change the formula to solve for d, the discount rate, after plugging in the current share price for V.

 

[stapel]

My formula:

V = a(G/D + (G/D)^2 + (G/D)^3 + ... + (G/D)^n)

Replacing u = G/D and solving for u still leaves other G's and D's around. I have tried to solve for D or G or even D/G, but I have failed.

Please reply showing your work, so we can see what's going on. For instance, I am not understanding how you're starting from "My formula", plugging "u" in for each instance of "G/D":

. . . . .\(\displaystyle V\, =\, a\, \bigg(\dfrac{G}{D}\, +\, \left(\dfrac{G}{D}\right)^2\, +\, \left(\dfrac{G}{D}\right)^3\, +\, ...\, \left(\dfrac{G}{D}\right)^n\bigg)\)

. . . . .\(\displaystyle V\, =\, a\, (u\, +\, u^2\, +\, u^3\, +\, ...\, +\, u^n)\)

...and then somehow getting G's and D's back into the formula. We'll need to see your steps (because we can't guess from a mere reference to a result) in order to figure out what's going wrong.

Please be complete. Thank you!

 

[rnikey]

Please reply showing your work, so we can see what's going on. For instance, I am not understanding how you're starting from "My formula", plugging "u" in for each instance of "G/D":

. . . . .\(\displaystyle V\, =\, a\, \bigg(\dfrac{G}{D}\, +\, \left(\dfrac{G}{D}\right)^2\, +\, \left(\dfrac{G}{D}\right)^3\, +\, ...\, \left(\dfrac{G}{D}\right)^n\bigg)\)

. . . . .\(\displaystyle V\, =\, a\, (u\, +\, u^2\, +\, u^3\, +\, ...\, +\, u^n)\)

...and then somehow getting G's and D's back into the formula. We'll need to see your steps (because we can't guess from a mere reference to a result) in order to figure out what's going wrong.

Please be complete. Thank you!

I didn't ever say something was going wrong. Rather, everything is going right. I only ever said that the problem is that I can't make d the subject of the formula.
My formula is correct by definition. It's the one I tested and decided I'm going to use. Whether or not there is a perceived mistake in it's derivation is immaterial. I will say again, THAT MY PROBLEM IS I CAN'T MAKE d THE SUBJECT OF THE FORMULA. I don't have that skill.

However, for completeness, I will do the derivation for my formula using u = D/G. However, I don't see the point. All I'm doing is teaching you how to do math, and I thought I could assume you could do it when I came here in the first place.

Equation 1: V = a (u + u^2 + u^3 + ... + u^n)
Divide both sides by u...
Equation 2: V/u = a(1 + u + u^2 + ... + u^(n-1))
Divide both sides of both equations by a, and then subtract Equation 2 from Equation 1. Note that all the terms cancel except for the u^n in equation 1 and the 1 in equation 2.
V/a - V/a.u = u^n - 1
Multiply both sides by a.u...
u.V - V = a.u.(u^n - 1)
V(u-1) = a.u.(u^n - 1)
V = a.u.(u^n - 1)/(u-1)
Multiply top and bottom by -1...
V = a.u(1 - u^n) / (1-u)

But u = G/D...
V = a.(G/D)(1-(G/D)^n) / (1 - (G/D))
There's a division by D on the bottom line which ends up on the top line, cancelling another D out and leaving the expression (G-D) on the bottom.
V = a.G.(1 - (G/D)^n) / (G-D)
This is the formula that originally posted.

 

[rnikey]

That's all very nice...BUT your problem is to solve that equation for d;
no background required...just solve the darn thing for d!

Here's a shorter/simpler standard financial equation:
f = a[(1 + i)^n - 1] / i

Can you solve that for i?

Try it. The solution to yours will become apparent...

Thank you Denis, yes, thank you, problem is to solve for d.

i = a[(1 + i)^n - 1] / f

Whilst i is by itself on the left, there is still an i on the RHS. So the equation is up itself.
I can't solve it. I need something without i on the RHS that I can put into my excel spreadsheet.
I did learn logarithms at high school, but as far as I remember, they are only good for multiplication and powers. As soon as there is a subtraction in the formula then one can't take logarithms. But I may be wrong.

 

[rnikey]

That's RIGHT: can't be solved directly; numeric methods required.
Now your problem is the SAME: can't be solved directly...get my drift?

No.

I don't know what you mean by "numeric methods required".

I did also put up the derivation using u of my formula. However it's gone to moderation. It looks like my 3rd post went to moderator and my 4th post got posted directly.

In my original post I did say that it's also the sum of the series.
V = a(G/D + (G/D)^2 + (G/D)^3 + ... + (G/D)^n).
If this can be solved for G/D then I will accept it and work with it. So we know the result (V), but not the geometric factor (G/D).

Actually, I just saw a solution if n = infinite and d>g, which it always will be.

I've been resisting solutions with n = infinite until now for some reasons I might explain later. But it might be good enough for my purposes of calculating d for now.

 

[JeffM]

I have not been following this thread so I don't know whether or not this is helpful. My recollection is that excel will do a numeric approximation for you on simple discount problems without your having a clue what they are. If you can transform your problem into a traditional discount problem, excel will take it from there.

 

[rnikey]

I have not been following this thread so I don't know whether or not this is helpful. My recollection is that excel will do a numeric approximation for you on simple discount problems without your having a clue what they are. If you can transform your problem into a traditional discount problem, excel will take it from there.

Thanks, I'll have a look for it.

 

[Deleted member 4993]

Thank you Denis, yes, thank you, problem is to solve for d.

i = a[(1 + i)^n - 1] / f

Whilst i is by itself on the left, there is still an i on the RHS. So the equation is up itself.
I can't solve it. I need something without i on the RHS that I can put into my excel spreadsheet.
I did learn logarithms at high school, but as far as I remember, they are only good for multiplication and powers. As soon as there is a subtraction in the formula then one can't take logarithms. But I may be wrong.

i = a[(1 + i)^n - 1] / f

f*i = a[(1 + i)^n - 1]

f*i/a + 1 = (1 + i)^n

This is a non-linear equation in 'i'. Do you know Newton's method for calculating roots of non-linear equations?

Do a google-search on roots of non-linear equations to familiarize yourself in this subject.