2.2a. Annuities

Annuities (Future Value Annuities)

When periodic payments are made into an account in order to increase the value of the account, we call this a future value annuity.

When periodic payments are paid from an account (or paid on a loan) in order to decrease the value of the account, we call this a present value annuity.

We will first discuss future value annuities.

For most of us, we aren't able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a

savings annuity. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.

An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship

Pm=(1+rk)Pm1\displaystyle{P}_{{m}}={\left({1}+\frac{{r}}{{k}}\right)}{P}_{{{m}-{1}}}


For a savings annuity, we simply need to add a deposit, d, to the account with each compounding period:

Pm=(1+rk)Pm1+d\displaystyle{P}_{{m}}={\left({1}+\frac{{r}}{{k}}\right)}{P}_{{{m}-{1}}}+{d}


Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general.

Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. In this example:

r = 0.06 (6%)

k = 12 (12 compounds/deposits per year)

d = $100 (our deposit per month)

Writing out the recursive equation gives

Pm=(1+0.0612)Pm1+100=(1.005)Pm1+100\displaystyle{P}_{{m}}={\left({1}+\frac{{0.06}}{{12}}\right)}{P}_{{{m}-{1}}}+{100}={\left({1.005}\right)}{P}_{{{m}-{1}}}+{100}


Assuming we start with an empty account, we can begin using this relationship:

P0 = 0

P1 = (1.005)P0 + 100 = 100

P2 = (1.005)P1 + 100 = (1.005)(100) + 100 = 100(1.005) + 100

P3 = (1.005)P2 + 100 = (1.005)(100(1.005) + 100) + 100 = 100(1.005)2 + 100(1.005) + 100

Continuing this pattern, after

m deposits, we'd have saved:

Pm = 100(1.005)m – 1 + 100(1.005)m – 2 + L + 100(1.005) + 100

In other words, after

m months, the first deposit will have earned compound interest for m – 1 months. The second deposit will have earned interest for m – 2 months. Last months deposit would have earned only one month worth of interest. The most recent deposit will have earned no interest yet.

This equation leaves a lot to be desired, though—it doesn't make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:

1.005

Pm = 1.005(100(1.005)m – 1 + 100(1.005)m – 1 + L + 100(1.005) +100)

Distributing on the right side of the equation gives

1.005

Pm = 100(1.005)m + 100(1.005)m – 1 + L + 100(1.005)2 + 100(1.005)

Now we'll line this up with like terms from our original equation, and subtract each side

1.005Pm\displaystyle{1.005}{P}_{{m}}
=
100(1.005)m\displaystyle{100}{\left({1.005}\right)}^{{m}}
+
100(1.005)m1\displaystyle{100}{\left({1.005}\right)}^{{{m}-{1}}}
+
L\displaystyle{L}
+
100(1.005)\displaystyle{100}{\left({1.005}\right)}
Pm\displaystyle{P}_{{m}}
=
100(1.005)m1\displaystyle{100}{\left({1.005}\right)}^{{{m}-{1}}}
+
L\displaystyle{L}
+
100(1.005)\displaystyle{100}{\left({1.005}\right)}
+
Almost all the terms cancel on the right hand side when we subtract, leaving

1.005PmPm = 100(1.005)m – 100

Solving for Pm
0.005Pm=100((1.005)m1)\displaystyle{0.005}{P}_{{m}}={100}{\left({\left({1.005}\right)}^{{m}}-{1}\right)}


Pm=100((1.005)m1)0.005\displaystyle{P}_{{m}}=\frac{{{100}{\left({\left({1.005}\right)}^{{m}}-{1}\right)}}}{{0.005}}


Replacing

m months with 12N, where N is measured in years, gives

PN=100((1.005)12N1)0.005\displaystyle{P}_{{N}}=\frac{{{100}{\left({\left({1.005}\right)}^{{{12}{N}}}-{1}\right)}}}{{0.005}}


Recall 0.005 was

r/k and 100 was the deposit d. 12 was k, the number of deposit each year. Generalizing this result, we get the saving annuity formula.

Annuity Formula

PN=d((1+rk)Nk1)(rk)\displaystyle{P}_{{N}}=\frac{{{d}{\left({\left({1}+\frac{{r}}{{k}}\right)}^{{{N}{k}}}-{1}\right)}}}{{{\left(\frac{{r}}{{k}}\right)}}}


  • PN is the balance in the account after N years.
  • d is the regular deposit (the amount you deposit each year, each month, etc.)
  • r is the annual interest rate in decimal form.
  • k is the number of compounding periods in one year.


Insert By Professor P: The above formula actually describes the future value (FV) of an ordinary annuity. I typically use this formula for the Future Value of an ordinary annuity.
FV=PMT((1+rm)n1)(r/m)\displaystyle{FV}=\frac{{{PMT}{\left({\left({1}+\frac{{r}}{{m}}\right)}^{{{n}}}-{1}\right)}}}{{{\left({{r/m}}{{}}\right)}}}


  • FV= future value of the annuity
  • PMT= amount of the periodic payment
  • r= annual interest rate written in decimal form
  • m=number of compounding periods per year.
  • n=total number of compounding periods. {n=mt, where t=number of years}


Note: Some texts prefer to replace the rate per period (r/m) with i. The above formula would look like this.
FV=PMT((1+i)n1)(i)\displaystyle{FV}=\frac{{{PMT}{\left({\left({1}+{i}\right)}^{{{n}}}-{1}\right)}}}{{{\left({{i}}{{}}\right)}}}


If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.

For example, if the compounding frequency isn't stated:

If you make your deposits every month, use monthly compounding

If you make your deposits every year, use yearly compounding.

If you make your deposits every quarter, use quarterly compounding

When Do You Use This?

Annuities assume that you put money in the account

on a regular schedule (every month, year, quarter, etc.) and let it sit there earning interest.

Compound interest assumes that you put money in the account

once and let it sit there earning interest.

Compound interest: One deposit Annuity: Many deposits.

Example 7

A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?

In this example,

The monthly deposit
d=$100\displaystyle{d}=\${100}
6% annual rate
r=0.06\displaystyle{r}={0.06}
12 months in 1 year
k=12\displaystyle{k}={12}
We want the amount after 20 years
N=20\displaystyle{N}={20}
Putting this into the equation:

P20=100((1+0.0612)20(12)1)(0.0612)\displaystyle{P}_{{20}}=\frac{{{100}{\left({\left({1}+\frac{{0.06}}{{12}}\right)}^{{{20}{\left({12}\right)}}}-{1}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}


P20=100((1.005)2401)(0.005)\displaystyle{P}_{{20}}=\frac{{{100}{\left({\left({1.005}\right)}^{{240}}-{1}\right)}}}{{{\left({0.005}\right)}}}


P20=100(3.3101)(0.005)\displaystyle{P}_{{20}}=\frac{{{100}{\left({3.310}-{1}\right)}}}{{{\left({0.005}\right)}}}


P20=100(2.310)(0.005)=$46200\displaystyle{P}_{{20}}=\frac{{{100}{\left({2.310}\right)}}}{{{\left({0.005}\right)}}}=\${46200}


Professor P's method: We know the following: PMT=100, i=0.06/12=0.005, n=12(20)=240

FV=PMT((1+0.005)2401)(0.005)\displaystyle{FV}=\frac{{{PMT}{\left({\left({1}+{0.005}\right)}^{{{240}}}-{1}\right)}}}{{{\left({{0.005}}{{}}\right)}}}
=$46,204.09

The account will grow to $46,200 after 20 years.

Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned.

In this case it is $46,204.09 - $24,000 = $22,204.09.

Example 8

You want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?

In this example,

We're looking for d.

8% annual rate
r=0.08\displaystyle{r}={0.08}
12 months in a year
k=12\displaystyle{k}={12}
30 years
N=30\displaystyle{N}={30}
The amount we want to have in 30 years
P30=$200,000\displaystyle{P}_{{30}}=\${200},{000}
In this case, we're going to have to set up the equation, and solve for d.

200,000=d((1+0.0812)30(12)1)(0.0812)\displaystyle{200},{000}=\frac{{{d}{\left({\left({1}+\frac{{0.08}}{{12}}\right)}^{{{30}{\left({12}\right)}}}-{1}\right)}}}{{{\left(\frac{{0.08}}{{12}}\right)}}}


200,000=d((1.00667)3601)0.00667\displaystyle{200},{000}=\frac{{{d}{\left({\left({1.00667}\right)}^{{360}}-{1}\right)}}}{{{0.00667}}}


200,000=d(1491.57)\displaystyle{200},{000}={d}{\left({1491.57}\right)}


d=200,0001491.57=$134.09\displaystyle{d}=\frac{{{200},{000}}}{{{1491.57}}}=\${134.09}


The above method is in error due to rounding.

Professor P's method: We know the following: FV=200000, i=0.08/12 (do not round), n=12(30)=360

I will use the previous formula to get a sinking fund formula.

PMT=FV(r/m)((1+r/m)n1)\displaystyle{PMT}={FV}\frac{(r/m)}{((1+r/m)^{n}-1)}


PMT=200000(0.08/12)((1+0/08/12)3601)\displaystyle{PMT}={200000}\frac{(0.08/12)}{((1+0/08/12)^{360}-1)}
=$134.20

A monthly deposit of $134.20 will grow to $200,000.  

Try it Now 2

A more conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest? Note: Assume 365 days/year (no leap years)









Payout Annuities (Present Value Annuities)

In the last section you learned about (future value) annuities. In an (future value) annuity, you start with nothing, put money into an account on a regular basis, and end up with money in your account.

In this section, we will learn about a variation called a Payout (present value) Annuity. With a payout annuity, you start with money in the account, and pull money out of the account on a regular basis. Any remaining money in the account earns interest. After a fixed amount of time, the account will end up empty.

Payout annuities are typically used after retirement. Perhaps you have saved $500,000 for retirement, and want to take money out of the account each month to live on. You want the money to last you 20 years. This is a payout annuity. The formula is derived in a similar way as we did for savings annuities. The details are omitted here.

Payout Annuity Formula

P0=d(1(1+rk)Nk)(rk)\displaystyle{P}_{{0}}=\frac{{{d}{\left({1}-{\left({1}+\frac{{r}}{{k}}\right)}^{{-{N}{k}}}\right)}}}{{{\left(\frac{{r}}{{k}}\right)}}}


P0 is the balance in the account at the beginning (starting amount, or principal).

d is the regular withdrawal (the amount you take out each year, each month, etc.)

r is the annual interest rate (in decimal form. Example: 5% = 0.05)

k is the number of compounding periods in one year.

N is the number of years we plan to take withdrawals

Like with annuities, the compounding frequency is not always explicitly given, but is determined by how often you take the withdrawals.

Professor P's Payout (PV) Annuity Formula

PV=PMT(1(1+rm)n)(rm)\displaystyle{PV}=\frac{{{PMT}{\left({1}-{\left({1}+\frac{{r}}{{m}}\right)}^{{-{n}}}\right)}}}{{{\left(\frac{{r}}{{m}}\right)}}}


PV is the balance in the account at the beginning (starting amount, or principal).

PMT is the regular withdrawal (the amount you take out each year, each month, etc.)

r is the annual interest rate (in decimal form. Example: 5% = 0.05)

m is the number of compounding periods in one year.

n is the total number of payouts (n=mt)

When Do You Use This?

Payout annuities assume that you take money from the account on a regular schedule (every month, year, quarter, etc.) and let the rest sit there earning interest.

Compound interest: One deposit Annuity: Many deposits. Payout Annuity: Many withdrawals

Example 9

After retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement account. The account earns 6% interest. How much will you need in your account when you retire?

In this example,

The monthly withdrawal
d=$1000\displaystyle{d}=\${1000}
6% annual rate
r=0.06\displaystyle{r}={0.06}
12 months in a year
k=12\displaystyle{k}={12}
We're taking withdrawals for 20 years
N=20\displaystyle{N}={20}
We're looking for

P0; how much money needs to be in the account at the beginning.

Putting this into the equation:

P0=1000(1(1+0.0612)20(12))(0.0612)\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{20}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}


P0=1000(1(1.005)240)(0.005)\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1.005}\right)}^{{-{{240}}}}\right)}}}{{{\left({0.005}\right)}}}


P0=1000(10.302)(0.005)=$139,600\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{0.302}\right)}}}{{{\left({0.005}\right)}}}=\${139},{600}


You will need to have $139,580.77 in your account when you retire.

Professor P's Method: PMT=1000, i=0.06/12=0.005, n=12(20)=240
PV=1000(1(1+0.005)240)(0.005)\displaystyle{PV}={1000}\frac{(1-(1+0.005)^{-{240}})}{(0.005)}
=$139580.77

Notice that you withdrew a total of $240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is $240,000 – $139580.77 = $100,419.23 in interest.

Evaluating Negative Exponents on Your Calculator

With these problems, you need to raise numbers to negative powers. Most calculators have a separate button for negating a number that is different than the subtraction button. Some calculators label this (–) , some with +/–. The button is often near the = key or the decimal point.

If your calculator displays operations on it (typically a calculator with multiline display), to calculate 1.005–240 you'd type something like: 1.005 ^ (–) 240

If your calculator only shows one value at a time, then usually you hit the (–) key after a number to negate it, so you'd hit: 1.005 yx 240 (–) =

Give it a try — you should get 1.005–240 = 0.302096

Example 10

You know you will have $500,000 in your account when you retire. You want to be able to take monthly withdrawals from the account for a total of 30 years. Your retirement account earns 8% interest. How much will you be able to withdraw each month?

In this example,

We're looking for d.

8% annual rate
r=0.08\displaystyle{r}={0.08}
12 months in a year
k=12\displaystyle{k}={12}
30 years
N=30\displaystyle{N}={30}
We are beginning with $500,000
P0=$500,000\displaystyle{P}_{{0}}=\${500},{000}
In this case, we're going to have to set up the equation, and solve for

d.

500,000=d(1(1+0.0812)30(12))(0.0812)\displaystyle{500},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.08}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.08}}{{12}}\right)}}}


500,000=d(1(1.00667)360)(0.00667)\displaystyle{500},{000}=\frac{{{d}{\left({1}-{\left({1.00667}\right)}^{{-{360}}}\right)}}}{{{\left({0.00667}\right)}}}


500,000=d(136.232)\displaystyle{500},{000}={d}{\left({136.232}\right)}


d=500,000136.232=$3670.21\displaystyle{d}=\frac{{{500},{000}}}{{{136.232}}}=\${3670.21}


You would be able to withdraw $3,670.21 each month for 30 years.

(The above answer is wrong because they rounded too much.)

Professor P's Method: For a Present Value annuity the formula for PMT can easily be derived.

PMT=PV(r/m)(1(1+r/m)n)\displaystyle{PMT}={PV}\frac{(r/m)}{(1-(1+r/m)^{-{n}})}


PV=500000, i=0.08/12 (don't round), n=12(30)=360
PMT=500000(0.08/12)(1(1+0.08/12)360)\displaystyle{PMT}={500000}\frac{(0.08/12)}{(1-(1+0.08/12)^{-{360}})}
=$3668.82

My method, just requires you to enter one formula into the calculator then when you tap enter, you got the answer.

Try it Now 3

A donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year? Note: In this case, i and r are both 4% because it is annual compounding.





Loans

In the last section, you learned about payout annuities.

In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.

One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you're acting as the bank, you pay interest. The car lender takes payments until the balance is zero.

Loans Formula

P0=d(1(1+rk)Nk)(rk)\displaystyle{P}_{{0}}=\frac{{{d}{\left({1}-{\left({1}+\frac{{r}}{{k}}\right)}^{{-{N}{k}}}\right)}}}{{{\left(\frac{{r}}{{k}}\right)}}}


P0 is the balance in the account at the beginning (the principal, or amount of the loan).

d is your loan payment (your monthly payment, annual payment, etc)

r is the annual interest rate in decimal form.

k is the number of compounding periods in one year.

N is the length of the loan, in years

Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.

When Do You Use This?

The loan formula assumes that you make loan payments

on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.

Compound interest:

One deposit

Annuity:

Many deposits.

Payout Annuity:

Many withdrawals

Loans:

Many payments

Example 11

You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?

In this example,

The monthly loan payment
d=$200\displaystyle{d}=\${200}
3% annual rate
r=0.03\displaystyle{r}={0.03}
12 months in a year
k=12\displaystyle{k}={12}
We're making monthly payments for 5 years
N=5\displaystyle{N}={5}
We're looking for P0, the starting amount of the loan.

P0=200(1(1+0.0312)5(12))(0.0312)\displaystyle{P}_{{0}}=\frac{{{200}{\left({1}-{\left({1}+\frac{{0.03}}{{12}}\right)}^{{-{5}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.03}}{{12}}\right)}}}


P0=200(1(1.0025)60)(0.0025)\displaystyle{P}_{{0}}=\frac{{{200}{\left({1}-{\left({1.0025}\right)}^{{-{60}}}\right)}}}{{{\left({0.0025}\right)}}}


P0=200(10.861)(0.0025)=$11,120\displaystyle{P}_{{0}}=\frac{{{200}{\left({1}-{0.861}\right)}}}{{{\left({0.0025}\right)}}}=\${11},{120}


You can afford a $11,120 loan. (Should be $11,130.47) You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you're paying $12,000-$11,130 = $870 interest total (estimate).

Professor P's Method
PV=200(1(1+0.03/12)60)(0.03/12)\displaystyle{PV}={200}\frac{(1-(1+0.03/12)^{-{60}})}{(0.03/12)}
=$11,130.47

Example 12

You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?

In this example,

We're looking for d.

6% annual rate
r=0.06\displaystyle{r}={0.06}
12 months in a year
k=12\displaystyle{k}={12}
30 years
N=30\displaystyle{N}={30}
Starting loan amount
P0=$140,000\displaystyle{P}_{{0}}=\${140},{000}
In this case, we're going to have to set up the equation, and solve for

d.

140,000=d(1(1+0.0612)30(12))(0.0612)\displaystyle{140},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}


140,000=d(1(1.005)360)(0.005)\displaystyle{140},{000}=\frac{{{d}{\left({1}-{\left({1.005}\right)}^{{-{360}}}\right)}}}{{{\left({0.005}\right)}}}


140,000=d(166.792)\displaystyle{140},{000}={d}{\left({166.792}\right)}


d=140,000166.792=$839.37\displaystyle{d}=\frac{{{140},{000}}}{{{166.792}}}=\${839.37}


You will make payments of $839.37 per month for 30 years.

You're paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 – $140,000 = $162,173.20 in interest over the life of the loan.

Professor P's Method: PV=140000, i=0.06/12 , n=12(30)=360
PV=140000(0.06/12)(1(1+0.06/12)360)\displaystyle{PV}={140000}\frac{(0.06/12)}{(1-(1+0.06/12)^{-{360}})}
=$839.37

My method, just requires you to enter one formula into the calculator then when you tap enter, you got the answer.

Try it Now 4

Janine bought $3,000 of new furniture on credit. Because her credit score isn't very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?





Remaining Loan Balance

With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.

To determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don't already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will

not have paid off $12,000 of the loan balance.

To determine the remaining loan balance, we can think "How much loan will these loan payments be able to pay off in the remaining time on the loan?"

Example 13

If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?

To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we're looking for P0 when

The monthly loan payment
d=$1,000\displaystyle{d}=\${1},{000}
6% annual rate
r=0.06\displaystyle{r}={0.06}
12 months in a year
k=12\displaystyle{k}={12}
10 years
N=10\displaystyle{N}={10}
P0=1000(1(1+0.0612)10(12))(0.0612)\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1}+\frac{{0.06}}{{12}}\right)}^{{-{10}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.06}}{{12}}\right)}}}


P0=1000(1(1.005)120)(0.005)\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{\left({1.005}\right)}^{{-{{120}}}}\right)}}}{{{\left({0.005}\right)}}}


P0=1000(10.5496)(0.005)=$90,073.45\displaystyle{P}_{{0}}=\frac{{{1000}{\left({1}-{0.5496}\right)}}}{{{\left({0.005}\right)}}}=\${90},{073.45}


The loan balance with 10 years remaining on the loan will be $90,073.45

Often times answering remaining balance questions requires two steps:

  1. Calculating the monthly payments on the loan
  2. Calculating the remaining loan balance based on the remaining time on the loan


Example 14

A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?

First we will calculate their monthly payments.

We're looking for d.

4% annual rate
r=0.04\displaystyle{r}={0.04}
12 months in a year
k=12\displaystyle{k}={12}
30 years
N=30\displaystyle{N}={30}
Starting loan amount
P0=$180,000\displaystyle{P}_{{0}}=\${180},{000}
We set up the equation and solve for d.

180,000=d(1(1+0.0412)30(12))(0.0412)\displaystyle{180},{000}=\frac{{{d}{\left({1}-{\left({1}+\frac{{0.04}}{{12}}\right)}^{{-{30}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.04}}{{12}}\right)}}}


180,000=d(1(1.00333)360)(0.00333)\displaystyle{180},{000}=\frac{{{d}{\left({1}-{\left({1.00333}\right)}^{{-{360}}}\right)}}}{{{\left({0.00333}\right)}}}


180,000=d(209.562)\displaystyle{180},{000}={d}{\left({209.562}\right)}


d=180,000209.562=$858.93\displaystyle{d}=\frac{{{180},{000}}}{{{209.562}}}=\${858.93}


Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.

Monthly loan payment
d=$858.93\displaystyle{d}=\${858.93}
4% annual rate
r=0.04\displaystyle{r}={0.04}
12 months in a year
k=12\displaystyle{k}={12}
25 years
N=25\displaystyle{N}={25}
P0=858.93(1(1+0.0412)25(12))(0.0412)\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{\left({1}+\frac{{0.04}}{{12}}\right)}^{{-{25}{\left({12}\right)}}}\right)}}}{{{\left(\frac{{0.04}}{{12}}\right)}}}


P0=858.93(1(1.00333)300)(0.00333)\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{\left({1.00333}\right)}^{{-{300}}}\right)}}}{{{\left({0.00333}\right)}}}


P0=858.93(10.369)(0.00333)=$155,793.91\displaystyle{P}_{{0}}=\frac{{{858.93}{\left({1}-{0.369}\right)}}}{{{\left({0.00333}\right)}}}=\${155},{793.91}


The loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91

Over that 5 years, the couple has paid off $180,000 – $155,793.91 = $24,206.09 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 – $24,206.09 = $27,329.71 of what they have paid so far has been interest.


David Lippman, Math in Society, "Finance," licensed under a CC BY-SA 3.0 license.

More Annuity Examples

Any retirement-age adult with a mandatory full-time job will likely tell you that the one thing they would change is to have planned for retirement. Does that mean a person falling into this category forgot to put away one large deposit a long time ago? Perhaps, but we often plan for retirement by making smaller payments at regular intervals. This is known as an annuity.

On a smaller scale, let's suppose you decide to put away $100 for each of the next 12 months, at which time you'll take that money to make a purchase. Additionally, suppose that you find a rate of 12% compounded monthly. The time diagram will look like this:

You'll notice that the first payment is actually not made until the

end of month 0 and that the last payment is made at the time of withdrawal. This is known as an annuity-immediate. For simplicity, we assume this is the case. Alternatively, it is possible to have payments made at the beginning of each month, which is known as an annuity-due.

To compute the future, 12-month value of each of these 12 payments, we note that the first payment will earn interest for 11 months, since it wasn't made until the end of the initial month. Thus, for this payment, the future value is:

A = 100(1.01)11

We note that each additional payment has one month fewer of interest-earning capabilities thant the previous payment. In other words, the payment made at the end of month 1 has 10 months to go, or

A = 100(1.01)10

We have to do this for each of the remaining payments, which we show in the table below:

End of Month Balance
0 100(1.01)11 = 111.57
1 100(1.01)10 = 110.46
2 100(1.01)9 = 109.37
3 100(1.01)8 = 108.29
4 100(1.01)7 = 107.21
5 100(1.01)6 = 106.15
6 100(1.01)5 = 105.1
7 100(1.01)4 = 104.06
8 100(1.01)3 = 103.03
9 100(1.01)2 = 102.01
10 100(1.01)1 = 101
11 100(1.01)0 = 100
Importantly, notice that the final payment made at the end of month 11 does not earn any interest, since the account is closed out and the balance is withdrawn. To find the total, we sum these values up and get $1,268.25. Since 12 payments of $100 were made, the total comes out to be $1,200 in outflows. The remainder is earned interest, which comes out to be $1,268.25 – $1,200 = $68.25.

Clearly, making these computations is no pleasant matter. Let's do a little bit of manipulation. The sum can be written in the following way:

A=100(1.01)11+100(1.10)10+100(1.01)9+100(1.01)8++100(1.01)0\displaystyle{A}={100}{\left({1.01}\right)}^{{11}}+{100}{\left({1.10}\right)}^{{10}}+{100}{\left({1.01}\right)}^{{9}}+{100}{\left({1.01}\right)}^{{8}}+\cdot\cdot\cdot+{100}{\left({1.01}\right)}^{{0}}


Is there a quick way to write this all out? Here's a trick: multiply both sides of the equation by 1.01, so that we get:

1.01A=1.01×[100(1.01)11+100(1.10)10+100(1.01)9+100(1.01)8++100(1.01)0]\displaystyle{1.01}{A}={1.01}\times{\left[{100}{\left({1.01}\right)}^{{11}}+{100}{\left({1.10}\right)}^{{10}}+{100}{\left({1.01}\right)}^{{9}}+{100}{\left({1.01}\right)}^{{8}}+\cdot\cdot\cdot+{100}{\left({1.01}\right)}^{{0}}\right]}


Distributing 1.01 results in increasing all exponents by 1:

1.01A=100(1.01)12+100(1.01)11+100(1.01)10+100(1.01)9++100(1.01)1\displaystyle{1.01}{A}={100}{\left({1.01}\right)}^{{12}}+{100}{\left({1.01}\right)}^{{11}}+{100}{\left({1.01}\right)}^{{10}}+{100}{\left({1.01}\right)}^{{9}}+\cdot\cdot\cdot+{100}{\left({1.01}\right)}^{{1}}


Notice that

A and 1.01A have right sides that are similar. The only terms that are different are that 1.01A has an extra term with a power of 12 and that A has a term with a power of 0 that 1.01A is missing. Thus, if we take the difference between 1.01A and A, we get:

1.01AA=100(1.01)12100(1.01)0\displaystyle{1.01}{A}-{A}={100}{\left({1.01}\right)}^{{12}}-{100}{\left({1.01}\right)}^{{0}}


We need not write out the terms with powers between 1 and 11, since they zero each other out by the operation of subtraction. We now need to solve for A, which is the total future value of all payments:

.01A=100(1.01)12100(1)\displaystyle{.01}{A}={100}{\left({1.01}\right)}^{{12}}-{100}{\left({1}\right)}


.01A=100[(1.01)121]\displaystyle{.01}{A}={100}{\left[{\left({1.01}\right)}^{{12}}-{1}\right]}


A=100[(1.01)121].01\displaystyle{A}=\frac{{{100}{\left[{\left({1.01}\right)}^{{12}}-{1}\right]}}}{{.01}}


Since the payment is constant, and the duration of time is irrelevant, since all but the

power and the 0 power terms will cancel out, we can generalize this result:

Formula for Future Value of an Annuity with Regular Payments

For an annuity with APR

r, compounded n times per year for t years, lasting n × t periods, and with a constant payment, PMT, that is made each compounding period, then the resulting balance, A, is given by

A=PMT[(1+rn)n×t1]rn\displaystyle{A}=\frac{{{P}{M}{T}{\left[{\left({1}+\frac{{r}}{{n}}\right)}^{{{n}\times{t}}}-{1}\right]}}}{{\frac{{r}}{{n}}}}


Example

A company offers its employees a 401(k) plan and offers to match its employees' contributions at 10% for up to a $600 employer contribution per year. An employee decides to use her 401(k). She makes payments of $90 per month to an account that pays an average of 7%, compounded monthly. If she retires in 30 years, how much can she retire on from this account?

Solution

The employee's contributions will be matched by

.10($90) = $9 per month, which will only be $108 per year and so will be well below the allowable contribution. Thus, the employee will have $99 going to her 401(k) fund every month at a monthly rate of
.0712.00583\displaystyle\frac{{.07}}{{12}}\approx{.00583}
months. On a time diagram, we have:

Using the annuity formula, we have:

A=99[(1.00583)3601].00583$120,681.85\displaystyle{A}=\frac{{{99}{\left[{\left({1.00583}\right)}^{{360}}-{1}\right]}}}{{.00583}}\approx\${120},{681.85}


She put in a total of

out-of-pocket, so she earned $88,281.85 in interest.

Thus, the future value according to the calculator is $120,777.13. Why the difference? This occurs simply because we rounded the interest rate in our calculations by hand. The calculator keeps track of more significant digits. Our error in calculating by hand is $95.28, which is a relative error of

95.281220777.13.07%\displaystyle\frac{{95.28}}{{1220777.13}}\approx{.07}\%
, which is insignificant.

Example

When planning for retirement and opening up a 401(k) or 403(b), which are simply names for annuities based on corresponding IRS tax codes, you have the option of choosing how aggressive you want your fund to be. (A 401(k) is typically seen within private sector companies. A 401(b) is specific to government employers, such as the military or public educational institutions.) The riskier you are willing to be, the more of the potential return, but the higher the risk of loss and major fluctuations. Suppose you open a 401(k) with your current employer and decide to contribute $300 each month. For the first 5 months, the average return is 8%, and then jumps to 15% for the next 15 months. For the 20 years following, the rate remains fairly stable at about 7%. Assume the first two rates are compounded monthly and that the last is compounded daily. How much will be in your 401(k) at that time?

Solution

We must evaluate three separate time periods—one for each of the three rate periods. Following is the time diagram we consider:

Notice that we indicate regular payments in the boxes, but that we additionally have question marks in the inflows and outflows. This is because when the rates change, we have already accumulated an overall balance for the given period. For the first 5 months, we calculate the ending balance:

We have an accumulated balance of $1,520.13 after 5 months. Note that this amount is principle and interest.

Even though the rate changes, we proceed to reinvest the $1,520.13, along with recurring payments of $300 at the new interest rate. This is where it becomes handy to use PV for lump sum amounts and payments for recurring values. First, we re-label the time diagram:

We calculate with these values in TVM Solver:

Relabeling the time diagram gives:

Similarly to before, the current reinvested balance at the new rate of 7% for 240 months is $6,747.40 with regular recurring payments of $300. We adjust P/Y and C/Y to be 365 for daily compounding, and must multiply 365 by 20 to get the number of daily compound periods in 20 years. This gives:

Updating the time diagram, we have that the ending balance of the retirement account will be $4,805,712.62.

CAUTION: It is simple to overlook the fact that payments are based on deposits per compounding period. In the last 240 months, it was the case that compounding was taking place daily. Thus, the $300 payment is also assumed to be made on a daily basis! So, don't assume that it is this easy to be a multi-millionaire!

It is often useful or necessary to determine the amount of interest earned on an investment account. Since money accumulates with each additional payment for

each payment made prior, the amount of earned interest grows exponentially. Since every payment is assumed to have been made at the end of the month, we know that only the accumulation of payments and interest prior to that most recent payment earns interest.

For example, in the last example we noticed that after 5 months the balance was $1,520.13. This means that interest for the next month, month 6, will be

.1512\displaystyle\frac{{.15}}{{12}}
, multiplied by the account balance to get in interest dollars.

We can use the TVM Solver to expedite this. Since, month 6 is actually month 1 of a new calculation, we will have to reference this as month 1. We first enter the FV calculation:

This step is CRUCIAL! Without it, the interest reported will not be valid because the calculator relies on the current financial computation. Now, we exit out of TVM Solver by pressing 2nd, then MODE.

Calculator Clinic

Calculating interest on an annuity between time

t1 and t2:

  1. Press .
  2. Go to 1: Finance
  3. Scroll down to A: ΣInt and press ENTER. (Σ is the Greek letter sigma. In mathematics, this symbol means "sum" or "add up.")
  4. A parenthesis is automatically entered. In the parenthesis, we specify the two points in time during which we would like the interest. The format of your inputs will be ΣInt(t1,t2). If you want to find the interest during a single period, t1 and t2 will be the same value. Note that the first number must be 1 or greater. Think of these as the payment number. There is no such thing as the 0th payment!


This confirms our hand-calculation. Again, note that we did

not enter month 6, since the investment resets each time something changes and we need to reinvest.

Example

Suppose you decide today that you will put away $400 per month for the next 35 years, at which point you will retire. You are confident you will be able to find a rate that averages around 6.5% for the duration of the investment. At the time you retire, you transfer your money to a safe account that is invested namely in bonds and savings. This account returns about 3% per year, compounded monthly. How much money will you be able to withdraw each month so that you never touch the accumulation of money you made over the 35 years?

Solution

The only way to never touch the balance of an account is to survive entirely off interest. We first calculate the future value of this investment:

Putting $640,126.48 into the safe account will yield

.0312\displaystyle\frac{{.03}}{{12}}
. Thus, each month you will earn .0025(640126.48) = $1,600.32, assuming you remove the interest each month, thus leaving only the $640,126.48 that you had at the close of the retirement account.

Practice Problems

    1. Give possible stories for the following scenarios. Be sure to clearly describe the investment taking place.







    1. Suppose you open an IRA. You have an automatic withdrawal of $250 each month from your checking account into the IRA, which pays an average of 9.1% compounded monthly. Ignoring fees and taxes, answer the following questions.

      1. How much will you have saved up at retirement in 35 years?
      2. How long would it take you to save up $3 million?
      3. Suppose you were able to find a rate that earns you $3 million in 35 years. What would the interest rate have to be to earn this amount exactly?






    1. 3.Chevy is thinking about purchasing a car next year. He prefers to pay in cash whenever possible so as not to take out a loan and incur interest. The car he is considering is currently running for $26,995 (with a sunroof for his Jack Russell Terrier!). Having done his homework, he knows that this particular make and model depreciates by 24% over the first year. He is able to find a 12-month CD at an APY of 3.4%, which is the result of monthly compounding. What would he have to put away each month in order to pay cash for the depreciated car next year?




    1. Suppose you open a 401(k) through your employer. You decide to use a recent bonus of $1000 given to you to get the account started and you plan on making regular deposits of $150 per month. Suppose you are able to get a rate of 7.88% compounded monthly.

      1. What will you be able to accumulate in 20 years?
      2. How much interest will be accumulated during the last year?
      3. How much larger would your payment need to be in order to reach a goal of $1,000,000 by retirement? Assume that you will still make the lump-sum deposit of $1000.
      4. Suppose you are set on making the $150 monthly payment along with your initial $1000 lump-sum deposit.

        1. What would your interest rate need to be to accomplish your $1,000,000 goal at retirement?


      5. Keeping the original rate, how much longer would you need to make $150 payments in order to accomplish your goal of $1,000,000 by retirement?






    1. 5.Homer wants to retire by the time he is 50. Today he turned 18. Homer wants to be able to withdraw $3,000 each month (for as long as he lives) once he closes his annuity at the time of retirement. Currently, he is earning an average of 8.05% compounded monthly. When he retires, he puts the balance into a safer, less volatile investment account that only earns 2% compounded monthly. He does not make additional payments to the account after retirement.

      1. What accumulation of money should he have in his annuity at retirement for the above scenario to be possible? (Keep in mind that in order to withdraw payments as long as he lives, he must live off interest alone.)
      2. What would be his required monthly annuity payment for the accumulation of money to occur?
      3. How much would he have put in out of his pocket over the duration of his annuity?
      4. How much interest will he have earned on the annuity?






    1. As people say, "hindsight is always 20/20." This is to mean that looking at the past makes it seem as if the events that occurred should have been predictable. Suppose that two twin sisters, Darla and Thelma, are of retirement age. Both took different routes throughout their annuity investing careers. They each put away $450 per month since they were 20 and are now both 61. Darla started out with an account that paid 11.06% for the first 18 months, 1% for the next 30 months, and then averaged 6.01% for the remaining time. Thelma was a bit more conservative and went with an account that paid 5.2% for the first 10 months, but then jumped to 16.1% for the next 50 months, and then dropped to 5.4% for the remainder of the time. Who is better off now that they're both retired, and by how much?





Milos Podmanik, By the Numbers, "The Mathematics of Finances," licensed under a CC BY-NC-SA 3.0 license.

MathIsGreatFun, "4 2 P1 MAT217," licensed under a Standard YouTube license.

MathIsGreatFun, "4 2 P2 MAT217," licensed under a Standard YouTube license.

MathIsGreatFun, "4 2 P3 MAT217," licensed under a Standard YouTube license.

MathIsGreatFun, "4 2 P4 MAT217," licensed under a Standard YouTube license.



MathIsGreatFun, "4 2 P5 MAT217," licensed under a Standard YouTube license.



MathIsGreatFun, "4 2 P6 MAT217," licensed under a Standard YouTube license.