The “Forever” Assumption
When the same payment is to be received each time period (e.g., year)
determining the price or discount rate can be a tedious business. If one knows the
discount rate to be used, then it is just a matter of a tedious calculation to find the price.
On the other hand, if one knows the price, solving for the discount rate becomes a game
of trial and error. Be clear, I am using the term ‘discount rate’ so that any type of future
payment can be addressed whether interest payments on a bond or dividend payments on
a stock (or, profit from an investment in a physical asset).
Suppose a bond promises to pay $100 each year for the next 5 years. Further
suppose that given the default risk of the issuer (i.e., borrower) you decide on a 10%
discount rate in order to price the bond (i.e., what you would be willing to pay for the
bond). We have the following calculation to make.
P
$100
$100
$100
$100
$100




2
3
4
(1  .10) (1  .10)
(1  .10)
(1  .10)
(1  .10) 5
Thus, we are only attempting to find the present value of the future payments. We have
taken the risk characteristics of the bond into account with the discount rate – in the case
of a bond this discount rate is normally called the ‘yield to maturity’ or just yield for
short. Now, the calculations are easy enough to do, just a bit tedious.
P  $90.91  $82.64  $75.13  $68.30  $62.09  $379.08
You can imagine the dread you would feel in doing this for a 20 year bond. Of course,
you could hopefully just plug it into a spreadsheet and have the answer in a split second.
Suppose the same bond was currently selling at a price of $400. Now, you would
like to know what the yield to maturity (i.e., the interest rate) is on this bond assuming
you bought it at the current market price.
$400 
$100
$100
$100
$100
$100




2
3
4
(1  k ) (1  k )
(1  k )
(1  k )
(1  .k ) 5
Here, you wish to solve for the k. You can rearrange this equation in different ways, but
there will not be an easy formula that results for us to use. For this type of problem, one
is forced to use a trial and error method to find the k that works.
So how does assuming that a payment is made ‘forever’ help us? The reason it
helps is that the present value formula reduces to something very simple. For example,
suppose a bond pays $100 each year, but this time assume it pays this amount forever.
We have the following present value calculation.
(1)
P
100
100
100
100
100





2
3
4
(1  k ) (1  k )
(1  k )
(1  k )
(1  .k ) 5
First, let us factor out the $100 from each term.
(2)
P  100(
1
1
1
1
1




 )
2
3
4
(1  k ) (1  k )
(1  k )
(1  k )
(1  .k ) 5
Now, we need to figure out how to sum the amount within the parentheses. Let’s begin
by defining that sum as S.
(3)
S
1
1
1
1
1





2
3
4
(1  k ) (1  k )
(1  k )
(1  k )
(1  .k ) 5
We can employ a couple of little tricks to find the solution. First, multiply both sides of
(3) by the first term on the right hand side.
(4)
1
1
1
1
1
1
S





2
3
4
5
(1  k )
(1  k )
(1  k )
(1  k )
(1  k )
(1  .k ) 6
Now, simply subtract (4) from (3).
(5)
S
1
1
S
(1  k )
(1  k )
Factor the S out of the left hand side and simplify.

1 
1
S 1 


 (1  k )  (1  k )
1  k  1 
1
S


 (1  k )  (1  k )
 k 
1
S


 (1  k )  (1  k )
Now, we cancel out the denominators and divide both sides by k (or, just cross multiply).
(6)
S
1
k
Now, we can return to our original question about the bond that pays $100 each year
forever. We simply substitute (6) back into (2).
(7)
 1  100
P  100  
k
k
We can even solve for the discount rate (or, yield to maturity) if the bond price is already
known.
(8)
k
100
P
This is all much simpler than attempting to solve for a 10 year bond, etc. You may want
to see if you can now use a similar approach to solving for the simple formula of the
Gordon Model. Recall, the Gordon Model assumes that a dividend gets paid forever. If
the dividend does not grow, then the formula simplifies to something like (7). However,
the Gordon Model allows for a constant growth rate of the dividend as well (the big
assumption is that the growth rate continues forever). See if you can derive the simple
solution.
Start from
V0E 
D0 (1  g ) D0 (1  g ) 2 D0 (1  g ) 3 D0 (1  g ) 4 D0 (1  g ) 5





(1  k )
(1  k ) 2
(1  k ) 3
(1  k ) 4
(1  k ) 5
And arrive at
V0E 
D1
kg