Effective Yield, and Converting Between Interest Rates
For this handout, keep in mind the following two interest formulas from the class on 4.2 and 4.3:
r nt
n compounds: A = P 1 +
Continuous compounding: A = P ert
n
In these, P is the principal (initial amount of money), A is the final amount of money, t is the number of interest
periods (like years), and r is the interest rate measured as a decimal.
Effective Yield
Suppose you have some interest-paying plan. The effective yield reff of this plan is the simple interest rate
that produces the same money in the same time.
Example: Suppose we invest one dollar for one year, compounded monthly at an annual rate of 10%. That
means P = 1, t = 1 (measured in years), r = 0.1, and n = 12. The amount of money I earn after that year is
12·1
0.1
≈ 1.10471 i.e. about $1.10
A=1 1+
12
In contrast, if we use a simple interest rate (n = 1) of r = 0.10471 (i.e. 10.471%) for that same year, we earn
A = 1(1 + 0.10471)1 = 1.10471
Thus, we get the same amount whether we use 10% monthly or ≈ 10.471% annually. We say that 10.471% is
the effective yield of the 10% monthly plan!
FINDING EFFECTIVE YIELD:
1. Write the amount of money your plan makes.
(I recommend leaving P and t unknown... they’ll cancel later.)
2. Write the amount you make with simple interest using an unknown rate reff .
3. Set your two values equal and solve for reff . Any values like P and t should cancel.
Example revisited: Let’s revisit that earlier example more thoroughly. This time, however, let’s not assume
any values of P or t. The 10% plan compounded monthly produces
12t
0.1
A=P 1+
12
If we use a simple interest plan with interest rate reff , we get
A = P (1 + reff )t
We set these equal:
P
1+
0.1
12
12t
= P (1 + reff )t
The P ’s cancel (we can divide by P ). The t’s cancel (i.e. you can take powers of 1/t to cancel them). You’re
left with
12
0.1
1+
= 1 + reff
12
In this form, though, we just need to subtract 1!
reff =
1+
0.1
12
12
− 1 ≈ 0.10471
You try it: Find the effective yield when 5% interest is applied: (a) Quarterly, (b) Continuously.
NOTE: Some people, instead of leaving P and t unknown to cancel, prefer to plug in simple values, like
P = 1 and t = 1. Those values cause the effects of P and t to disappear anyway, sort of like the example at the
top of the page.
Converting Between Any Two Rates
The procedure you see above works more generally of converting any interest plan to any other! In this
scenario, let’s say you have two different r values (like r1 and r2 ) and two different n values (n1 and n2 )
representing two different compounding strategies. However, both plans will use the same prinicipal P and the
same time t, and they should produce the same final amount A.
Example: Suppose one plan of interest makes money by taking a 5% annual rate and compounding it
monthly. If we were to use a daily plan instead (365 days), what would be the equivalent rate?
Let’s say r1 = 0.05, n1 = 12 (monthly), r2 is not yet known, and n2 = 365. When we write our formulas
and set them equal, we get
12t
0.05
r2 365t
P 1+
=P 1+
12
365
As with effective yield, P and t cancels. We have
12 0.05
r2 365
1+
= 1+
12
365
The first thing to cancel, to solve for r2 , is the 365th power. We apply a 365th root:
1+
r2
=
365
1+
0.05
12
12/365
Next, subtract 1, and then multiply 365:
!
12/365
0.05
1+
r2 = 365
− 1 ≈ 0.498995
12
Thus, the daily plan would use a 4.98995% rate (only slightly smaller!) to produce the same amount of money.
Another example: What simple interest rate is equivalent to continuous compounding of 1% a year?
This is basically asking for the effective yield of 1% continuous. Say r is the unknown simple rate (so n = 1).
We set the plans equal:
P (1 + r)t = P e0.01t
Cancel P and t, and we get 1 + r = e0.01 . Therefore, r = e0.01 − 1 , approximately 1.00502%.
You try it:
(a) Convert a simple interest rate of 7% to a monthly compounding rate.
(b) Convert a continuously compounded rate of 10% to a semiannual rate.
Warning about Equation Solving
In all the examples you see, the solution process involved using roots... i.e. our unknowns were in the BASE
of a power. We say this is a polynomial equation type when the equation has its unknown in the base. Sections
4.2 and 4.3 do not introduce enough technique for solving most equations with the unknown in the exponent,
which are exponential equations. (After Section 4.4, covering logarithms, you can return to these problems!)