Payout annuities:
Start with P dollars, e.g., P = 100, 000. Over a 30 year period you receive equal payments of A dollars at the end of each month. The amount of money left in the account, the balance, earns an interest rate of 8.25% (compounded monthly). At the end of 30 years the amount of money left is 0.
In our example i = r/12 = .0825/12
At the end of the first month after your payment is received your amount in the account, the balance, is P (1 + i)− A
At the end of the second month after your payment is received the balance is (P (1 + i)− A)(1 + i) − A
At the end of the third month after your payment is received the balance is ((P (1 + i)− A)(1 + i) − A)(1 + i) − A To simplify notation, denote x = 1 + i. After three months, the balance is now
P x3− Ax2 − Ax − A = P x3 − A(x2+ x + 1) = P x3 − Ax 3 − 1 x− 1
by the geometric series formula from last time.
(Recall the formula for a geometric series: 1 + x + x2+· · · + xn= x
n+1
− 1 x− 1 . ) Continuing in this fashion, after n payments, the balance is
P xn− Ax
n− 1
x− 1 Substituting x = 1 + i, after n payments, the amount is
P (1 + i)n− A(1 + i)
n− 1
i
Your monthly payment A is the quantity so that in 30 years, i.e. when n = 30× 12 = 360, the amount left in the account is 0 so that with n = 30× 12 = 360,
P (1 + i)n− A(1 + i)
n− 1
Then solving for A,
A = P i(1 + i)
n
(1 + i)n− 1
Summarizing, if P is the original principal, i is the interest per period, and n is the number of periods, then the payment per period is
A = P i(1 + i)
n
(1 + i)n− 1
In our example, i = .0825/12, P = 100000, and n = 12× 30 = 360. Plug those numbers in and get
A = P i(1 + i)
n
(1 + i)n− 1 = $751.27
So this is the monthly payment you receive each month, so that the original principal reduces to 0 in 30 years.
During the payout annuity, after n payments, the balance is P (1 + i)n− A(1 + i)
n− 1
i
The principal paid up to this point would be P− the amount owed. The total amount of money paid to you up to this point would be n× A.
The interest paid to you up to this point would be total paid− the principal paid. Summary: after n payments:
Balance P (1 + i)n− A(1 + i)
n− 1
i Principal paid P− amount owed
Total paid n× A
Interest paid Total paid− principal paid
Balance $61,251.57 Principal paid $ 38,748.43 Total paid $ 180,304.80 Interest paid $ 141,556.37
If you look over the notes for the section on amortized loans, you will see that a payout annuity is the same thing as an amortized loan if you just change your point of view. The simple observation is the following: Start with a principal P at the beginning of a period (a month).
With an amortized loan at the end of the period you owe P (1 + i) and then after the loan payment of A is received you owe P(1+i)-A.
With a payout annuity at the end of the period your new balance is P (1 + i) because your principal earned interest, and then after your payment is sent your balance is P (1 + i)− A.
methods to compute the monthly payment, etc. We demonstrate with our example of P = 100, 000 and annual interest rate of 8.25% compounded monthly, with the payout-annuity to expire in thirty years. What will the monthly payment be?
Using Excel: Open a worksheet. In a cell enter =pmt(.0825/12,360,100000)
If you forget the formula, click on a cell and then click on the paste function f∗ Then choose Financial, then click on PMT and click on that. You will fill in the values for rate (the interest rate per period=.0825/12), the number of periods (360), the present value (100,000). The last two quantities, the future value and payment type are assumed to be zero unless entered otherwise.
Using a TI-83: The quickest and easiest way would be to just use the formula A = P i(1 + i)
n
(1 + i)n− 1 = $751.27
and use the values P = 100000, i = .0825/12, n = 360.
If you want to calculate the balance after 20 years, just enter these quantities into the formula for the balance after n payments.
We know that after n payments, the balance is
P (1 + i)n− A(1 + i)
n− 1
i
You should get $61, 251.57
Example problem: Pam is going to deposit a principal that will be used for a payout annuity. She will receive twelve monthly payments of $1000, and after the twelve payments the annuity expires. The balance earns 10% compounded monthly. How much should she deposit into the annuity?
You start with an unknown principal P . The monthly interest rate i = .1/12, the monthly payment A = 1000, and the number of periods n = 12. So
P (1 + i)n− A(1 + i)
n
− 1
i = 0
with n = 12, A = 1000. Then solving for P , P = A(1 + i)
12
− 1
i× (1 + i)12 = $11, 374.51
rounded to the nearest penny.
To avoid lots of parenthesis, I first store (1 + i)12 in ANSWER in the calculator: (1 + .1/12)12ENTER
Then I take 1000*(ANSWER -1)/((.1/12)*ANSWER)
Notice that this should seem reasonable. If no interest were paid, you would need to deposit $12, 000 to receive twelve payments of $1, 000. Since the balance earns interest, you can deposit less than $12, 000. Cost of Living Adjustments:
Cost of Living Adjustments:
In order to keep the formulas simpler, we pretend that there is just one payment per year, and for the first year the payment is A, for the second year the payment is (1 + c)A, for the third year the payment is (1 + c)2A , ..., and for the kthyear the payment is (1 + c)k−1A. The quantity c represents an annual
increase in the payment to compensate for inflation. The quantity c is called the annual C.O.L.A. rate. Denote r as the annual interest rate. Let P denote the original amount.
At the end of the first year, after your payment A is sent to you, the amount in the account is P (1 + r)− A
At the end of the second year, after your new payment A(1 + c) is sent, the amount in the account is (P (1 + r)− A)(1 + r) − A(1 + c).
At the end of the third year, after your payment A(1 + c)2is sent, the amount in the account is
At the end of the fourth year, after your payment A(1 + c)3is sent, the amount in the account is
(((P (1 + r)− A)(1 + r) − A(1 + c))(1 + r) − A(1 + c)2)(1 + r)
− A(1 + c)3.
Expanding, at the end of the fourth year, after the payment is sent, the amount is P (1 + r)4− A[(1 + r)3+ (1 + c)(1 + r)2+ (1 + c)2(1 + r) + (1 + c)3] = P (1 + r)4− A[(1 + r)3+ (1 + c 1 + r)(1 + r) 3+ (1 + c 1 + r) 2(1 + r)3+ (1 + c 1 + r) 3(1 + r)3] = P (1 + r)4− A(1 + r)3[1 + x + x2+ x3]
where we factored out (1 + r)3 and to simplify notation, write 1 + c
1 + r as x. We use the geometric series formula
(1 + x + x2+ x3) = (1− x4)/(1
− x) so after the fourth year the amount is
P (1 + r)4− A(1 + r)4[1 − (1 + c 1 + r) 4]/(r − c) after substituting 1+c
1+r for x and a little algebra.
The above represents the amount of money left in the account after 4 years. After t years, the amount of money left in the account is
P (1 + r)t− A(1 + r)t[1− (1 + c 1 + r)
t]/(r
− c)
The annual payments A, A(1 + c), A(1 + c)2, etc. are such that after n years, with n=30 in this example,
the amount of money left in the account is 0. So we set P (1 + r)t− A(1 + r)t[1 − (1 + c 1 + r) t]/(r − c) = 0 with t=30, and we get
P = A[1− (1 + c 1 + r) 30]/(r − c) and A = P (r− c)/[1 − (1 + c 1 + r) 30]
In our example, with P=100,000 and an annual interest rate of 8.25% and a C.O.L.A. rate of 2%, the first annual payment
A = 100000(.0825− .02)/[1 − ((1 + .02)/(1 + .0825))30] = $7511.53
rounded to the nearest penny.
The second annual payment would be (1.02)×the first annual payment = $7661.76 rounded to the nearest penny, etc.
——————————————————————————– Summary
For a payout annuity with a beginning principal P , payment A per period, interest rate i per period, after k periods the balance is
P (1 + i)k− A(1 + i)
k
− 1 i
For a payout annuity with COLA rate c, annual interest rate r, beginning principal P , first annual payment A, second annual payment (1 + c)A, third annual payment (1 + c)2A, etc., after k years the
balance is P (1 + r)k− A (1 + r) k(1− (1+c 1+r) k) r− c
You can distribute the term (1 + r)k in the numerator and rewrite the expression for the balance as
P (1 + r)k− A (1 + r)
k− (1 + c)k
r− c
Notice that if each period is a year, and if there is no COLA adjustment (c = 0), then the formula simplifies to that of a payout annuity.
