mathematics assignment loans & interest

Module 9: Finance

Annuities and loans, learning outcomes.

• Calculate the balance on an annuity after a specific amount of time
• Discern between compound interest, annuity, and payout annuity given a finance scenario
• Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan
• Determine which equation to use for a given scenario
• Solve a financial application for time

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. In this section, we will explore the math behind specific kinds of accounts that gain interest over time, like retirement accounts. We will also explore how mortgages and car loans, called installment loans, are calculated.

Savings 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

[latex]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}[/latex]

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

[latex]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}+d[/latex]

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. Write an explicit formula that represents this scenario.

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

[latex]{{P}_{m}}=\left(1+\frac{0.06}{12}\right){{P}_{m-1}}+100=\left(1.005\right){{P}_{m-1}}+100[/latex]

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

[latex]P_0=0[/latex]

[latex]P_1=(1.005)P_0+100=100[/latex]

[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[/latex]

[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[/latex]

Continuing this pattern, after m deposits, we’d have saved:

[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[/latex]

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. The last month’s deposit (L) would have earned only one month’s 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:

[latex]1.005{{P}_{m}}=1.005\left(100{{\left(1.005\right)}^{m-1}}+100{{\left(1.005\right)}^{m-2}}+\cdots+100(1.005)+100\right)[/latex]

Distributing on the right side of the equation gives

[latex]1.005{{P}_{m}}=100{{\left(1.005\right)}^{m}}+100{{\left(1.005\right)}^{m-1}}+\cdots+100{{(1.005)}^{2}}+100(1.005)[/latex]

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

[latex]\begin{align}&\begin{matrix}1.005{{P}_{m}}&=&100{{\left(1.005\right)}^{m}}+&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&{}\\{{P}_{m}}&=&{}&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&+100\\\end{matrix}\\&\\\end{align}[/latex]

Almost all the terms cancel on the right hand side when we subtract, leaving

[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex]

Factor [latex]P_m[/latex] out of the terms on the left side.

[latex]\begin{array}{c}P_m(1.005-1)=100{{\left(1.005\right)}^{m}}-100\\(0.005)P_m=100{{\left(1.005\right)}^{m}}-100\end{array}[/latex]

Solve for P m

[latex]\begin{align}&0.005{{P}_{m}}=100\left({{\left(1.005\right)}^{m}}-1\right)\\&\\&{{P}_{m}}=\frac{100\left({{\left(1.005\right)}^{m}}-1\right)}{0.005}\\\end{align}[/latex]

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

[latex]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-1\right)}{0.005}[/latex]

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 savings annuity formula.

Annuity Formula

[latex]P_{N}=\frac{d\left(\left(1+\frac{r}{k}\right)^{Nk}-1\right)}{\left(\frac{r}{k}\right)}[/latex]

• P N 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.

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, k = 12.
• If you make your deposits every year, use yearly compounding, k = 1.
• If you make your deposits every quarter, use quarterly compounding, k = 4.

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.

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,

Putting this into the equation:

[latex]\begin{align}&{{P}_{20}}=\frac{100\left({{\left(1+\frac{0.06}{12}\right)}^{20(12)}}-1\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{20}}=\frac{100\left({{\left(1.005\right)}^{240}}-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(3.310-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(2.310\right)}{\left(0.005\right)}=\$46200 \\\end{align}[/latex]

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 $46,200 – $24,000 = $22,200.

This example is explained in detail here.

A 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?

d = $5              the daily deposit

r = 0.03           3% annual rate

k = 365            since we’re doing daily deposits, we’ll compound daily

N = 10             we want the amount after 10 years

[latex]P_{10}=\frac{5\left(\left(1+\frac{0.03}{365}\right)^{365*10}-1\right)}{\frac{0.03}{365}}=21,282.07[/latex]

 Financial planners typically recommend that you have a certain amount of savings upon retirement.  If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.

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 .

In this case, we’re going to have to set up the equation, and solve for d .

[latex]\begin{align}&200,000=\frac{d\left({{\left(1+\frac{0.08}{12}\right)}^{30(12)}}-1\right)}{\left(\frac{0.08}{12}\right)}\\&200,000=\frac{d\left({{\left(1.00667\right)}^{360}}-1\right)}{\left(0.00667\right)}\\&200,000=d(1491.57)\\&d=\frac{200,000}{1491.57}=\$134.09 \\\end{align}[/latex]

So you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.

View the solving of this problem in the following video.

Solving For Time

We can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.

If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?

This is a savings annuity problem since we are making regular deposits into the account.

We don’t know N , but we want P N to be $10,000.

[latex]10,000=\frac{100\left({{\left(1+\frac{0.03}{12}\right)}^{N(12)}}-1\right)}{\left(\frac{0.03}{12}\right)}[/latex]                        Simplifying the fractions a bit

[latex]10,000=\frac{100\left({{\left(1.0025\right)}^{12N}}-1\right)}{0.0025}[/latex]

We want to isolate the exponential term, 1.002512 N , so multiply both sides by 0.0025

[latex]25=100\left({{\left(1.0025\right)}^{12N}}-1\right)[/latex]                             Divide both sides by 100

[latex]0.25={{\left(1.0025\right)}^{12N}}-1[/latex]                                     Add 1 to both sides

[latex]1.25={{\left(1.0025\right)}^{12N}}[/latex]                                        Now take the log of both sides

[latex]\log\left(1.25\right)=\log\left({{\left(1.0025\right)}^{12N}}\right)[/latex]                              Use the exponent property of logs

[latex]\log\left(1.25\right)=12N\log\left(1.0025\right)[/latex]                          Divide by 12log(1.0025)

[latex]\frac{\log\left(1.25\right)}{12\log\left(1.0025\right)}=N[/latex]                                               Approximating to a decimal

N = 7.447 years

It will take about 7.447 years to grow the account to $10,000.

This example is demonstrated here:

Payout Annuities

Removing money from annuities.

In the last section you learned about annuities. In an 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 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

[latex]P_{0}=\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}[/latex]

• P 0 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)
• 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.

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.

• Payout Annuity: Many withdrawals

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?

We’re looking for P 0:  how much money needs to be in the account at the beginning.

[latex]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-20(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-{{\left(1.005\right)}^{-240}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-0.302\right)}{\left(0.005\right)}=\$139,600 \\\end{align}[/latex]

You will need to have $139,600 in your account when you retire.

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 – $139,600 = $100,400 in interest.

View more about this problem in this video.

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

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?

[latex]\begin{align}&500,000=\frac{d\left(1-{{\left(1+\frac{0.08}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.08}{12}\right)}\\&500,000=\frac{d\left(1-{{\left(1.00667\right)}^{-360}}\right)}{\left(0.00667\right)}\\&500,000=d(136.232)\\&d=\frac{500,000}{136.232}=\$3670.21 \\\end{align}[/latex]

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

A detailed walkthrough of this example can be viewed here.

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?

d = unknown

r = 0.04                       4% annual rate

k = 1                           since we’re doing annual scholarships

N = 20                         20 years

P0 = 100,000               we’re starting with $100,000

[latex]100,000=\frac{d\left(1-\left(1+\frac{0.04}{1}\right)^{-20*1}\right)}{\frac{0.04}{1}}[/latex]

Solving for d gives $7,358.18 each year that they can give in scholarships.

It is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely.   If this donor had specified that, $100,000(0.04) = $4,000 a year would have been available.

Conventional 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

• P 0 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)
• 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.

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.

• Annuity: Many deposits
• Loans: Many payments

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?

We’re looking for P 0 , the starting amount of the loan.

[latex]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\\end{align}[/latex]

You can afford a $11,120 loan.

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-$11,120 = $880 interest total.

Details of this example are examined in this video.

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?

[latex]\begin{align}&140,000=\frac{d\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&140,000=\frac{d\left(1-{{\left(1.005\right)}^{-360}}\right)}{\left(0.005\right)}\\&140,000=d(166.792)\\&d=\frac{140,000}{166.792}=\$839.37 \\\end{align}[/latex]

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.

View more about this example here.

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?

d =                               unknown

r = 0.16                       16% annual rate

k = 12                         since we’re making monthly payments

N = 2                           2 years to repay

P0 = 3,000                   we’re starting with a $3,000 loan

[latex]\begin{array}{c}3000=\frac{{d}\left(1-\left(1+\frac{0.06}{12}\right)^{-2*12}\right)}{\frac{0.16}{12}}\\\\3000=20.42d\end{array}[/latex]

Solve for d to get monthly payments of $146.89

Two years to repay means $146.89(24) = $3525.36 in total payments.  This means Janine will pay $3525.36 – $3000 = $525.36 in interest.

Calculating the 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?”

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 P 0 when

[latex]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-10(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\left(1-{{\left(1.005\right)}^{-120}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\left(1-0.5496\right)}{\left(0.005\right)}=\$90,073.45 \\\end{align}[/latex]

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

This example is explained in the following video:

Oftentimes answering remaining balance questions requires two steps:

• Calculating the monthly payments on the loan
• Calculating the remaining loan balance based on the remaining time on the loan

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 .

We set up the equation and solve for d .

[latex]\begin{align}&180,000=\frac{d\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&180,000=\frac{d\left(1-{{\left(1.00333\right)}^{-360}}\right)}{\left(0.00333\right)}\\&180,000=d(209.562)\\&d=\frac{180,000}{209.562}=\$858.93 \\\end{align}[/latex]

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.

[latex]\begin{align}&{{P}_{0}}=\frac{858.93\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-25(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&{{P}_{0}}=\frac{858.93\left(1-{{\left(1.00333\right)}^{-300}}\right)}{\left(0.00333\right)}\\&{{P}_{0}}=\frac{858.93\left(1-0.369\right)}{\left(0.00333\right)}=\$162,794.53 \\\end{align}[/latex]

The loan balance after 5 years, with 25 years remaining on the loan, will be $162,794.53.

Over that 5 years, the couple has paid off $180,000 – $162,794.53 = $17,205.47 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 – $17,205.47 = $34,330.33 of what they have paid so far has been interest.

Solving for Time

Recall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.

Joel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?

d = $30                        The monthly payments

r = 0.12                       12% annual rate

P0 = 1,000                   we’re starting with a $1,000 loan

We are solving for N , the time to pay off the loan

[latex]1000=\frac{30\left(1-\left(1+\frac{0.12}{12}\right)^{-N*12}\right)}{\frac{0.12}{12}}[/latex]

Solving for N gives 3.396. It will take about 3.4 years to pay off the purchase.

Home loans are typically paid off through an amortization process, amortization refers to paying off a debt (often from a loan or mortgage) over time through regular payments. An amortization schedule is a table detailing each periodic payment on an  amortizing loan as generated by an amortization calculator .

If you want to know more, click on the link below to view the website “How is an Amortization Schedule Calculated?” by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.

• How is an Amortization Schedule Calculated?

Which Formula to Use?

Now that we have surveyed the basic kinds of finance calculations that are used, it may not always be obvious which one to use when you are given a problem to solve. Here are some hints on deciding which equation to use, based on the wording of the problem.

The easiest types of problems to identify are loans.  Loan problems almost always include words like loan , amortize (the fancy word for loans), finance (i.e. a car), or mortgage (a home loan). Look for words like monthly or annual payment.

Interest-Bearing Accounts

Accounts that gain interest fall into two main categories.  The first is on where you put money in an account once and let it sit, the other is where you make regular payments or withdrawals from the account as in a retirement account.

• If you’re letting the money sit in the account with nothing but interest changing the balance, then you’re looking at a compound interest problem. Look for words like compounded, or APY. Compound interest assumes that you put money in the account once and let it sit there earning interest.

COMPOUND INTEREST

[latex]P_{N}=P_{0}\left(1+\frac{r}{k}\right)^{Nk}[/latex]

• P 0 is the starting balance of the account (also called initial deposit, or principal)
• r is the annual interest rate in decimal form
• If the compounding is done annually (once a year), k = 1.
• If the compounding is done quarterly, k = 4.
• If the compounding is done monthly, k = 12.
• If the compounding is done daily, k = 365.
• The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest .

SIMPLE INTEREST OVER TIME

[latex]\begin{align}&I={{P}_{0}}rt\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\end{align}[/latex]

• I is the interest
• A is the end amount: principal plus interest
• [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
• r is the interest rate in decimal form

The units of measurement (years, months, etc.) for the time should match the time period for the interest rate.

• If you’re putting money into the account on a regular basis (monthly/annually/quarterly) then you’re looking at a basic annuity problem.  Basic annuities are when you are saving money.  Usually in an annuity problem, your account starts empty, and has money in the future. 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.

ANNUITY FORMULA

• If you’re pulling money out of the account on a regular basis, then you’re looking at a payout annuity problem.  Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty in the future. 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.

PAYOUT ANNUITY FORMULA

Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and then determine what approach will best allow you to solve the problem.

For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.

• Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left? Show Solution This is a payout annuity problem. She can pull out $1833.60 a quarter.
• Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years? Show Solution This is a savings annuity problem. He will have saved up $7,524.11
• Keisha is managing investments for a non-profit company.       They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?  Show Solution This is compound interest problem. She would need to deposit $22,386.46.
• Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?  Show Solution This is a loans problem. She can buy $4,609.33 of new equipment
• How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years? Show Solution This is a savings annuity problem. You would need to save $200.46 each month

In the following video, we present more examples of how to use the language in the question to determine which type of equation to use to solve a finance problem.

In the next video example, we show how to solve a finance problem that has two stages, the first stage is a savings problem, and the second stage is a withdrawal problem.

• Outcomes and Introduction. Provided by : Lumen Learning. License : CC BY: Attribution
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Bad Credit? We can help!. Authored by : BookMama. Located at : https://www.flickr.com/photos/myloonyland/430367107 . License : CC BY: Attribution . License Terms : 2.0
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• Question ID 6691, 6688. Authored by : Lippman, David. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• Determining The Value of an Annuity. Authored by : Sousa, James (Mathispower4u.com). Located at : https://youtu.be/DWFezRwYp0I . License : CC BY: Attribution
• Determining The Value of an Annuity on the TI84. Authored by : Sousa, James (Mathispower4u.com). Located at : https://youtu.be/OKCZRZ-hWH8 . License : CC BY: Attribution
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• Payout Annuities. Authored by : OCLPhase2's channel. Located at : https://youtu.be/HK2eRFH6-0U . License : CC BY: Attribution
• Payout annuity - solve for withdrawal. Authored by : OCLPhase2's channel. Located at : https://youtu.be/XK7rA6pD4cI . License : CC BY: Attribution
• Question ID 6681, 6687. Authored by : David Lippman. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• Payout Annuity Formula - Part 1, Part 2. Authored by : Sousa, James (Mathispower4u.com). Located at : https://youtu.be/zxRFWEj5v5w,%20https://youtu.be/h3CWS9Rs7-s . License : CC BY: Attribution
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• Calculating payment on a home loan. Authored by : OCLPhase2's channel. Located at : https://youtu.be/BYCECTyUc68 . License : CC BY: Attribution
• Question ID 6684, 6685. Authored by : Lippman, David. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• Identifying type of finance problem. Authored by : OCLPhase2's channel. Located at : https://youtu.be/V5oG7lLTECs . License : CC BY: Attribution
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Loan Amortization

4.1 Loan Amortization: Calculator Approach

A. amortization definitions.

Amortization of a loan is the process of the gradual reduction in the loan amount through periodic repayments, usually of equal size, over a predetermined length of time. The amortization period is the time period set to repay the loan. An amortization schedule is a detailed table, similar to what is shown in Table 4.1.1, that breaks down each loan payment amount into its interest (INT) and principal portions (PRN). It also shows the outstanding principal balance (BAL) after each payment is made.

Table 4.1.1 An Amortization Schedule

Note that the size of payment (PMT) is the same for every period except the last payment as it needs to be adjusted to fully repay the loan. In banking, all actual monetary values are rounded to the nearest cent (two decimal places). Because of this rounding, the final payment needs to be adjusted and thus is almost always different from other payments.

B. Amortization Schedules for Loans 

For each payment (row) in the schedule, you can use the Amortization (AMORT) Worksheet in a financial calculator to calculate the principal portion (PRN), interest portion (INT), and the balance (BAL) after that payment. Given all the other information about the loan has already been entered, set P1 and P2 to that payment number to obtain the principal portion and interest portion of a particular payment number. For example, when P1 = 5 and P2 = 5, PRN displays the principal portion of the 5th payment and INT displays the interest portion of the 5th payment. BAL displays the principal balance after the 5th payment.

Figure 4.1.1 The keys on the AMORT worksheet on a financial calculator

How to Construct an Amortization Schedule: Calculator Approach

1. Enter the given information about the loan annuity in the TVM worksheet and compute the unknown variable, which is usually either the size of payment (PMT) or the number of payments in the term (N).

• Unknown PMT : compute PMT first using the TVM worksheet, round it off to the nearest cent (if needed), re-enter the rounded value (ensure to keep the same sign) to the PMT button in the calculator, and then use the Amortization Worksheet.
• Unknown N : compute N first using the TVM worksheet, round it up to the nearest whole payment, re-enter it to the N button in the calculator (optional), and then use the Amortization Worksheet.

2. In the first row of the schedule, enter zero for the payment number and the loan amount ( PV ) under the balance.

To construct the other rows except for the last two rows, complete steps 3-7:

3. Enter the size of payment ( PMT) against its respective payment number. The periodic payment amount is usually a rounded amount (to the nearest cent).

4. In the AMORT worksheet, set both P1 and P2 to the payment number to obtain the principal portion and interest portion of that particular payment number. Use the down arrow key to move to BAL.

5. Enter the outstanding principal balance ( BAL) in the 5th column of the schedule. Use the down arrow key to move to PRN.

6. Enter the principal portion ( PRN) of the payment in the 4th column. Use the down arrow key to move to INT.

7. Enter the interest portion (INT) of the payment in the 3rd column.

To construct the row of last payment, complete steps 8-11:

8. For the final payment, we only take the interest portion from the AMORT worksheet, and we will calculate the BAL and the size of the payment as below.

9. In the row of the final payment number, the principal balance will be zero because the loan will be fully paid off.

10. For the final payment, the principal portion equals the remaining balance before the last payment ( [asciimath]PRN_N=[/asciimath]   [asciimath]BAL_(N-1)[/asciimath] )

11. Since PMT or N might be rounded, the final payment may be different than PMT. The final payment amount has to be calculated using any of the two methods:

  [asciimath]PMT_N=PRN_(N)+INT_N[/asciimath]   where [asciimath]PRN_N=BAL_(N-1)[/asciimath]

•   [asciimath]PMT_N=PMT-"Overpayment"[/asciimath] where the overpayment is the absolute value of BAL on the last payment ( [asciimath]BAL_(N)[/asciimath]  ).

12.   The last row lists all the totals, which can be used to cross-check the calculations:

• Total Principal Portion = Original loan amount
• Total Amount Paid = Total Interest Portion + Total Principal Portion

Note that for clear presentation all values in the table should be entered as nonnegative values.

Pearline took out a loan of $10,000 from TD Bank to buy office supplies. The loan has an annual interest rate of 10%, compounded annually, and is to be repaid over four years. a) Determine the amount of her payments due at the end of each year. b) Create an amortization schedule for her loan.

Given information:

• Interest is compounded annually so [asciimath]C//Y = 1[/asciimath]
• Deposits are made at the end of every year so [asciimath]P//Y = 1[/asciimath]

  [asciimath]C//Y = P//Y[/asciimath]   [asciimath]=>[/asciimath]    Ordinary Simple Annuity

• Loan amortization period: [asciimath]t = 4[/asciimath] years
• Number of payments in the term: [asciimath]N =P//Y*t=1 (4) =4[/asciimath]
• Nominal interest rate: [asciimath]I//Y = 10%[/asciimath]
• Loan amount is cash inflow: [asciimath]PV = $10,000[/asciimath]
• Loan is fully repaid, so  [asciimath]FV=0[/asciimath]

a) First, we use the TVM worksheet to compute PMT.

Thus, the size of the payment is $3,154.71.

Note that the PMT value should be rounded to the nearest cent. After rounding, this adjusted value should then be re-entered into the PMT key with the same negative sign.

b) Next, we follow the steps given in the “How To – Calculator Approach” to fill in the amortization schedule. Steps 1 to 3 are already done. We now use the AMORT worksheet to obtain BAL, PRN, and INT for each payment.

It is important to note that when entering values into the amortization schedule, they should be rounded to two decimal places. Also, notice that the balance after payment number 4 is negative, which indicates overpayment.

Therefore, we need to compute the final PMT separately (Steps 8-11):

  [asciimath]PMT_4=BAL_3+INT_4[/asciimath] [asciimath]=2867.91+286.79=$3154.70[/asciimath]

  [asciimath]PMT_N=PMT-"Overpayment"[/asciimath]

Here the overpayment is the absolute value of BAL on the last payment ( [asciimath]BAL_(4)[/asciimath] ).

  [asciimath]PMT_4=3154.71-0.01=$3154.70[/asciimath]

Step 12: Finally, we check that all totals add up as we did before using the formula approach.

The figure above features a bar chart that shows the interest and principal portions of each payment for this example. Additionally, it includes a line chart that tracks the remaining principal balance after each payment is made. An important observation is that as the outstanding balance reduces over time, the interest portion of each payment also decreases. Since the payment amount (PMT) is constant for each period (except for the last payment), the diminishing interest portion results in an increased proportion of each payment being allocated to reduce the loan principal.

A $15,000 loan is settled by end-of-quarter payments of $4500. The interest is 6.8% compounded semi-annually. a) Find the number of payments needed to settle the loan. b) construct an amortization schedule for the loan.

• Interest is compounded semi-annually so [asciimath]C//Y = 2[/asciimath]
• Payments are made at the end of every quarter so [asciimath]P//Y = 4[/asciimath]

  [asciimath]C//Y != P//Y[/asciimath]   [asciimath]=>[/asciimath]    Ordinary General Annuity

• Nominal interest rate: [asciimath]I//Y = 6.8%[/asciimath]
• Loan principal is cash inflow: [asciimath]PV = $15,000[/asciimath]
• Periodic payments are cash outflow: [asciimath]PMT = -$4500[/asciimath]
• The loan is fully paid:  [asciimath]FV=0[/asciimath]

a)  [asciimath]N=?[/asciimath]  

First, we use the TVM worksheet to compute N.

Note that N is rounded up to 4 payments. The rounded value is then re-entered in the N key.

b) Next, we follow the steps given in the “How To – Calculator Approach” to fill in the amortization schedule. We now use the AMORT worksheet to obtain BAL, PRN, and INT for each payment.

Note that when entering values into the amortization schedule, they should be rounded to two decimal places. Additionally, observe that the balance after payment number 4 is negative, which indicates overpayment. Therefore, we need to adjust the final payment using the methods discussed in Steps 8-11:

  [asciimath]PMT_4=BAL_3+INT_4[/asciimath] [asciimath]=2042.61+34.43=$2077.04[/asciimath]

  [asciimath]PMT_4=4500-2422.96=$2077.04[/asciimath]

Step 12: Finally, we check that all totals add up:

The below figure features a bar chart that shows the interest and principal portions of each payment for this example. It also includes a line chart that depicts the remaining principal balance after each payment is made. It is important to note that as the outstanding balance reduces over time, the interest portion of each payment also decreases. Since the payment amount (PMT) is constant for each period (except for the last payment), the decreasing interest portion leads to an increased proportion of each payment being allocated to reduce the loan principal.

C. Calculating The Principal Balance

At times, we may need to determine the principal balance of a loan before the loan term ends, such as when considering an early settlement or making a partial repayment. This can be achieved by constructing an amortization schedule to find the outstanding principal at the desired time, or by utilizing the calculator approach outlined below.

Calculator Approach:

• Enter the loan details into the Time Value of Money (TVM) worksheet of the financial calculator.
• Calculate any variable (if any) that is unknown, for example, the payment size (PMT) or the number of payments (N).
• Access the Amortization (AMORT) worksheet on the calculator. Input the range of payment periods for which the amortization needs to be calculated. For instance, to determine the balance at the end of the first year of a loan with monthly payments, set P2 to 12. P1 can be any value less than or equal to 12 since the calculator primarily considers P2 to calculate the remaining balance.
• The calculator will provide the balance at the end of the specified range, which represents the principal balance at that date.

Megan took out a $20,000 loan at an interest rate of 4% compounded quarterly. The loan is scheduled to be fully repaid over 8 years, with payments due at the end of each quarter. Calculate the remaining balance on the loan after the first year.

• Interest is compounded annually so  [asciimath]C//Y = 4[/asciimath]
• Deposits are made at the end of every year so  [asciimath]P//Y = 4[/asciimath]

[asciimath]C//Y = P//Y[/asciimath]    [asciimath]=>[/asciimath]    Ordinary Simple Annuity

• Loan amortization period:  [asciimath]t = 8[/asciimath]  years
• Number of payments in the term:  [asciimath]N =P//Y*t=4 (8) =32[/asciimath]
• Nominal interest rate: [asciimath]I//Y = 4%[/asciimath]
• Loan principal is cash inflow: [asciimath]PV = $20,000[/asciimath]

a)  [asciimath]PMT=?[/asciimath]  

First, we need to compute the size of the payment (PMT) using the TVM worksheet.

Thus, the size of the payment is $733.42. Note that PMT is rounded to the nearest cents, and then the rounded value is re-entered in the PMT key.

b)  The number of payments up to the focal date of [asciimath]t=1[/asciimath] year is [asciimath]N= 4(1)=4[/asciimath] .

To find the outstanding balance at the end of the first year, we use the AMORT worksheet and set P1 = P2 = 4 (P1 can be any value less than or equal to 4 but make sure P2 = 4).

Therefore, the outstanding principal balance at the end of the first year is $17,834.10.

D. Calculating the Interest and Principal Portions

Sometimes, it may be necessary to determine the interest or principal portion of a specific payment or a specific period of time within the loan term.

For a Specific Payment: To find the interest and principal portion of a specific payment, in the AMORT worksheet, set both P1 and P2 to that specific payment number. Then, review the details provided to see the breakdown of interest and principal for that specific payment.

For a Specific Period: To determine the total principal or interest paid during a certain period of a loan, such as the 5th year, first, identify the payment numbers that correspond to the start and end of this period. In the AMORT worksheet, set P1 to the payment number marking the start of the period and P2 to the payment number at the end of the period. The calculator will then display the cumulative interest and principal portions paid between these two points.

Samuel took out a $308,000 mortgage to buy an apartment. The mortgage is structured to be repaid with monthly payments of $2,375.11 at the end of each month. The interest rate on the mortgage is 4.62%, compounded monthly, and the loan is amortized over 15 years. a) Calculate the interest portion of the 21st payment. b) Calculate the principal portion of the 21st payment.

• Interest is compounded monthly so [asciimath]C//Y = 12[/asciimath]
• Deposits are made at the end of every year so  [asciimath]P//Y = 12[/asciimath]

  [asciimath]C//Y = P//Y[/asciimath]    [asciimath]=>[/asciimath]    Ordinary Simple Annuity

• Loan amortization period:  [asciimath]t = 15[/asciimath]  years
• Number of payments in the term:  [asciimath]N =P//Y*t=12 (15) =180[/asciimath]
• Nominal interest rate: [asciimath]I//Y = 4.62%[/asciimath]
• Loan principal is cash inflow: [asciimath]PV = $308,000[/asciimath]
• Periodic payments are cash outflow: [asciimath]PMT=-$2,375.11[/asciimath]

To find the interest and principal portions of the 21st payment, start by entering the loan information into the TVM worksheet. Next, access the AMORT worksheet and set both P1 and P2 to 21, which represents the 21st payment. Then, review the details provided to see the breakdown of interest and principal for that specific payment.

a) Therefore, the interest portion of the 21st payment is $1090.80.

b) The principal portion of the 21st payment is $1284.31.

Samuel took out a $308,000 mortgage to buy an apartment. The mortgage is structured to be repaid with monthly payments of $2,375.11 at the end of each month. The interest rate on the mortgage is 4.62%, compounded monthly, and the loan is amortized over 15 years. a) Calculate the total principal amount repaid in the 8th year. b) Calculate the total amount of interest paid in the 8th year.

To determine the total principal or interest paid during a certain year of a loan, such as the 8th year, we need to identify the specific payment numbers that mark the beginning and end of that year. We then use the AMORT worksheet on a financial calculator to calculate those amounts:

i) To find the first payment of the 8th year, we calculate the number of payments made by the end of the 7th year, which is 12 payments per year times 7 years, equaling 84 payments. Therefore, the first payment of the 8th year is payment number 85 (84 from the previous years plus 1).

ii) The last payment of the 8th year is determined by the total number of payments made in 8 years, which is 12 payments per year times 8 years, equaling 96 payments.

iii) To calculate the interest and principal portions for the 8th year, you input the loan details into the TVM worksheet. Then, in the AMORT worksheet, set P1 to 85 and P2 to 96. The calculator will then display the cumulative interest and principal portions paid between these two points in time.

a) Therefore, the total principal amount repaid in the 8th year is $20,131.41.

b) The total interest amount paid in the 8th year is $8369.91.

Section 4.1 Exercises

a) PMT = $811.45

b) BAL = $73,724.15

a)  PMT = $372.80

b) INT = $1,048.36

c) PRN = $3,425.24

a)  PMT = $241.83

b) INT = $24.91

c) PRN = $216.92

Mathematics of Finance Copyright © 2024 by Amir Tavangar is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Mathematics Advanced: Financial Mathematics

Module 2 for Mathematics Advanced (HSC)

Simple and Compound Interest

• $P$ is the principal (initial) sum,
• $R$ is the rate of interest per unit of time
• $n$ is the number of time intervals which have passed
• If the question asks for the total amount, add $P$ to $I$ at the end
• $A_n$ is the amount of interest after $n$ units of time
• To find the interest (without the initial amount), subtract $P$ from $A_n$
• Depreciation is a form of compound in terest, where the value decreases over time
• $R$ is the rate of depreciation per unit time
• Annuities are compound interest investments, from which equal payments are recieved on a regular basis, for a fixed period of time

Practice Question

After 1 month, the account has $200(1+0.005)$ dollars

After 2 months, $200(1.005)^{2}+200(1.005)$

After $n$ months, we have $200(1.005^{n}+1.005^{n-1}+…+1.005)$

The geometric progression in the brackets is: $$S_{(20\times 12)}=\frac{1.005(1.005^{240})-1}{1.005-1}=464.3511$$

Therefore, $464.3511\times 200=92870.22$ $ after 20 years

Present and Future Values

• The Future value $(FV)$ is the total value of an investment at the end of its term, including all interest
• The Present value $(PV)$ is the single lump of money that could be initially invested to yield a given future value over a given period
• Present values are calculated using the compound interest formula
• Future value is calculated using a variant of the compound interest formula: $$\color{orange}FV=PV(1+r)^{n}$$

Loan Repayments

• Loans are usually repaid through regular installments, with compound interest charged on the balance owed
• The loan is paid off when $A_{n}=0$

Michael takes out $10000 to buy a car. He will repay the loan in 5 years, paying 60 equal monthly instalments, beginning 1 month after he takes out the loan. Interest is 6% p.a. compounded monthly. How much is the monthly installment?

Let $M$ be the monthly installment:

• $A_{1}=10000(1.005)-M$
• $A_{2}=(10000(1.005)-M)(1.005)-M$
• $\therefore A_{2}=10000(1.005)^{2}-1.005M-M$
• $A_{60}=0=10000(1.005)^{60}-M(1+1.005+…+1.005^{59})$

GP inside the brackets is $\frac{10000(1.005^{60})}{\frac{1.005^{60}-1}{0.005}}=\$193.33$

Method 2 (Speed Hack):

• $A_{n}=10000(1.005)^{n}-M(1+1.005+…+1.005^{n-1})$
• $10000(1.005)^{60}=M(1+1.005+…+1.005^{59})$

GP inside brackets is $S_{60}=\frac{1.005^{60}-1}{0.005}=69.77$

• $\therefore M=\frac{10000(1.005)^{60}}{69.77}$
• $=\$193.33 $

Jackson Taylor

Post writer.

2021 Graduate, UNSW Medicine first year.

Pranav Sharma

UNSW Student, site owner and developer.

6.9 Understanding Student Loans

Learning objectives.

After completing this section, you should be able to:

• Describe how to obtain a student loan.
• Distinguish between federal and private student loans and state distinctions.
• Understand the limits on student loans.
• Summarize the standard prepayment plan.
• Understand student loan consolidation.
• Summarize and describe benefits or drawbacks of other repayment plans.
• Summarize possible courses of action if a student loan defaults.

Obtaining a Student Loan

All college students are eligible to apply for a loan regardless of their financial situation or credit rating. Federal student loans do not require a co-signer or a credit check. Most students do not have a credit history when they begin college, and the federal government is aware of this. However, private loans will generally require a co-signer as well as a credit check. The co-signer will assume responsibility for paying off the loan if the student cannot make the payments.

The first step in applying for student loans is to fill out the FAFSA (Free Application for Student Aid). FAFSA determines financial need and what type of loan the student is qualified to obtain. For students who are still dependents on their parent’s taxes, the parents also fill out the FAFSA, as their wealth and income impacts what the dependent student is eligible for. Students who cannot demonstrate financial need will also be helped by applying with FAFSA, as it will help guide them to the type of loan most appropriate. The FAFSA must be submitted each year.

The FAFSA deadline is the spring of the student’s next academic year. The deadline is often in March. Do not allow this deadline to pass.

As soon as an offer letter from the college is received, the student should start the application process. The college will determine the loan amount needed. Also, there are limits on the amount a student can borrow. There are both yearly limits and aggregate limits. See the table later in this section that outlines the loan limits per school year and in the aggregate.

If the student receives a direct subsidized loan, there is a limit on the eligibility period. The time limit on eligibility depends on the college program into which the student enrolls. The school publishes how long a program is expected to take. The eligibility period is 150% of that published time. For example, if a student is enrolled in a 4-year program, such as a bachelor’s degree program, their eligibility period is 6 years, as 1.50(4) = 6. Therefore, the student may receive direct subsidized loans for a period of 6 years.

Types and Features of Student Loans

Once a tuition statement is received, and all the non-loan awards are analyzed that are applicable to the costs of college (such as scholarships and grants), there still may be quite of bit of an expense to attend college. This difference between what college will cost (including tuition, room and board, books, computers) and the non-loan awards received is the college funding gap.

Example 6.81

College funding gap.

Ishraq receives her award and tuition letter from the college she wants to attend. Her tuition, fees, books, and room and board all come to $24,845 for the year. Her non-loan awards include an instant scholarship from the school for $7,500, a scholarship she earned for enrolling in a STEM program for $3,750, and a $1,000 scholarship from her church. What is Ishraq’s college funding gap?

Her awards total to $12,250. Her cost to attend is $24,845. Her college funding gap is then $ 24,845 − $ 12,250 = $ 12,595 $ 24,845 − $ 12,250 = $ 12,595 . She will need to find $12,595 in funding.

Your Turn 6.81

There are several loan types, which basically break down into four broad categories: subsidized loans, unsubsidized loans, PLUS loans, and private loans. These loans are meant to fill the college funding gap.

Federal subsidized loans are backed by the U.S. Department of Education. These loans are intended for undergraduate students who can demonstrate financial need. Subsidized federal loans, including Stafford loans, defer payments until the student has graduated. During the deferment, the government pays the interest while the student is enrolled at least half-time. These loans are generally made directly to students. However, there are restrictions on how the money can be used. It can only be used for tuition, room and board, computers, books, fees, and college-related expenses. Interest rates are not based on the financial markets but determined by Congress. Federal loans are backed by the Department of Education.

Federal unsubsidized loans , including unsubsidized Stafford loans, are available for undergraduate and graduate students who cannot demonstrate financial need. If the student meets the program requirements, they are automatically approved. The student is not required to pay these loans during their time in college (enrolled at least half-time). However, the interest rate is generally higher and there is no deferment period, as with subsidized loans. Interest begins accruing as soon as the money is disbursed.

The immediate accrual of interest means the balance of the loan grows as the student attends school. A loan that was for $10,000 can grow past $13,000 over five years of college. Some advisors tell students to pay the interest portion of the loan while it is deferred to prevent this growth of debt.

Parent Loans for Undergraduate Students (PLUS) are federal loans made directly to parents. They are available even if parents are not deemed financially needy. A credit check is performed and approval is not automatic. The limit to what parents can borrow from a PLUS loan each year is still the college funding gap, but the aggregate of the PLUS loans does not have a limit. This means the PLUS loan can cover whatever is left in the funding gap once all other aid and loans are applied. Payments do not begin until the student is out of school, but interest begins to accrue the moment funds are disbursed. Because the parents take out the loan, the parents are responsible for paying back the loan.

Private student loans are backed by a bank or credit institution and require a credit check, and interest rates are variable. As private loans are not subsidized by the government, no one pays the interest but the borrower. The student does not have to start repaying the loan until after graduation, but interest starts to accrue immediately. This loan has fewer repayment options, more fees and penalties, and the loan cannot be discharged through bankruptcy. Many students need a co-signer to acquire a private loan. Like PLUS loans, private student loans can cover whatever is left in the funding gap once all other aid and loans are applied

Student loans, in general, have a term of 10 years, that is, the loans are paid back over 10 years. This can vary, but 10 years is the standard.

School-Channel Loans and Direct-to-Consumer Loans

Private loans can fall into one of two categories: school-channel loans and direct to consumer loans. School-channel loans are disbursed directly to the school. The school verifies the loan does not exceed the cost to attend school. Direct–to-consumer loans do not have the verification process. Those proceeds are sent directly to the borrower. They are processed more quickly, but often have higher interest rates.

Types of Student Loans

Limits on Student Loans

As mentioned earlier, there are limits to how much a student can borrow, per year and in total. The following table shows a general breakdown of the amounts the federal government and private lenders will lend. Amounts are based on level of need and whether the student is a dependent or an independent student. Independent students include those who are at least 24 years old, married, a professional, a graduate student, a veteran, a member of the armed forces, an emancipated minor, or an orphan. The amounts shown are as of this writing in 2022.

Check out this Edvisors page on the limits of student borrowing to learn more!

Example 6.82

Loan for year 5 of college.

Efraim is a dependent undergraduate student enrolled in a biology program. He’s about to attend for the fifth year. In year 1 he took out $5,000 in federal subsidized and unsubsidized loans, in year 2 he took out $6,400 in federal subsidized and unsubsidized loans, in years 3 and 4, he took out the maximum federal subsidized and unsubsidized loans amounts. He needs federal subsidized and unsubsidized loans for his fifth year of school. How much can he obtain in federal subsidized and unsubsidized student loans?

The sum of his previous loans is $ 5,000 + $ 6,400 + $ 7,500 + $ 7,500 = $ 26,400 $ 5,000 + $ 6,400 + $ 7,500 + $ 7,500 = $ 26,400 . The limit for federal subsidized and unsubsidized loans is $31,000, so in year 5 he can get student loans in the amount of $ 31,000 − $ 26,400 = $ 4,600 $ 31,000 − $ 26,400 = $ 4,600 .

Your Turn 6.82

Putting this all together, we have a way to determine the student loans needed for a student to attend college.

• First, determine the funding gap. If the student or family can cover the gap, then no loans are necessary.
• Second, determine how much in federal subsidized and unsubsidized student loans can be taken out. If the total federal loans available is more than the funding gap, no other loans are needed.
• Third, if the federal subsidized and unsubsidized loans do not cover the gap, PLUS and private student loans can be taken out to cover the remainder of the gap.

At each step, if the student and family can cover some or all of the gap, they can do so without taking out a loan.

Example 6.83

College funding gap and plus and private student loans.

Olivia receives her award and tuition letter from the college she wants to attend. Her tuition, fees, books, and room and board all come to $44,845 for her second year. Her non-loan awards include an instant scholarship from the school for $13,500, a scholarship she earned for enrolling in an engineering program for $5,750, and a $2,000 scholarship from her parent’s workplace. For her first year, what is Olivia’s college funding gap? How much can Olivia borrow in federal subsidized and unsubsidized student loans? Once Olivia takes out her maximum subsidized and unsubsidized federal student loans, how much will have to be paid for using PLUS and private student loans?

Her awards total to $21,250. Her cost to attend is $44,845. Her college funding gap is then $ 44,845 − $ 21,250 = $ 23,595 $ 44,845 − $ 21,250 = $ 23,595 . The maximum in federal student loans that Olivia can borrow is $6,500 in year 2. The remaining funding gap is $ 23,595 − $ 6,500 = $ 17,095 $ 23,595 − $ 6,500 = $ 17,095 . Private student loans, PLUS loans, or other sources must be used to cover this gap.

Your Turn 6.83

Student loan interest rates.

Student loans are first and foremost loans. Students will pay them back and will pay interest. In the fall of 2022, the federal student loan interest rate was 4.99%. Private student loans rates ranged between 3.22% and 13.95%. Finding the lowest interest rate you can helps with the payments, and especially helps if the loan is not federally subsidized. Remember, if the loan is not federally subsidized, the student is on the hook for the interest that is accumulating with the loan.

Interest Accrual

The interest on student loans begins as soon as the loan is disbursed (paid to the borrower). When the loan is federally subsidized, the government pays that interest for the student. This means the loan for a subsidized loan of $3,000 is still a loan for $3,000 when the student graduates. However, if the loan is not federally subsidized, the student is responsible for the interest that accrues on the loan. The $3,000 loan from year 1 of college is now a loan for more due to that added interest. The interest on that loan grew while the student was in college. The formula for growth of the loan’s balance is the same as compound interest formula from Compound Interest , A = P ( 1 + r n ) n t A = P ( 1 + r n ) n t .

Example 6.84

Denise takes out unsubsidized student loan, in August, in her first year of college for $2,000. She manages an interest rate of 8%. She graduates after her fifth year of college, in May. She does not pay the interest on the loan during her time in college. What is the balance of her first year loan in May of her graduation year?

The principal of the loan is $2,000. Her interest rate is 8%. Since student loans are typically paid monthly, there are 12 periods per year. Since the time she has had the loan is not in years, we will use the number of months for the value of nt nt in the formula. She has had the loan for 4 years and 9 months, meaning 57 period have passed. Substituting those values into the formula and calculating, we find her balance in May of her graduating year is A = P ( 1 + r n ) n t = $ 2,000 ( 1 + 0.08 12 ) 57 = $ 2,000 ( 1.0 6 ¯ ) 57 = $ 2,920.89 A = P ( 1 + r n ) n t = $ 2,000 ( 1 + 0.08 12 ) 57 = $ 2,000 ( 1.0 6 ¯ ) 57 = $ 2,920.89 .

Your Turn 6.84

The standard repayment plan.

There are various repayment plans available. The one most likely to apply to a student loan is the standard repayment plan, which is available to everyone. Borrowers pay a fixed amount monthly so the loan is paid in full within 10 years. Consolidated loans, discussed later in this section, also qualify for the standard repayment plan, and may allow the payoff period to range from 10 to 30 years. Direct subsidized and unsubsidized loans, PLUS loans, and federal Stafford loans are eligible.

Since these are loans, they are paid back with interest. As with most installment loans, their payments are due monthly. The formula for paying back these loans is the same as the formula used for paying loans in The Basics of Loans :

Using that formula, we can calculate how much the payment is for a student loan. Remember that all loan payments are rounded up to the next penny.

Example 6.85

Standard repayment plan.

Find the payment for the following student loans using the standard repayment plan:

• Loan is $3,500, interest is 4.99%
• Loan is for $6,200, interest is 6.75%
• The principal is P P = $3,500 and the rate is r r = 0.0499. Since this is the standard repayment plan, there are n n = 12 payments per year for 10 years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 3,500 × ( 0.0499 / 12 ) × ( 1 + 0.0499 / 12 ) 12 × 10 ( 1 + 0.0499 / 12 ) 12 × 10 − 1 = $ 3,500 × ( 0.004158 3 ¯ ) × ( 1.004158 3 ¯ ) 120 ( 1.004158 3 ¯ ) 120 − 1 = $ 23.9469911276 0.645370131872 = $ 37.11 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 3,500 × ( 0.0499 / 12 ) × ( 1 + 0.0499 / 12 ) 12 × 10 ( 1 + 0.0499 / 12 ) 12 × 10 − 1 = $ 3,500 × ( 0.004158 3 ¯ ) × ( 1.004158 3 ¯ ) 120 ( 1.004158 3 ¯ ) 120 − 1 = $ 23.9469911276 0.645370131872 = $ 37.11
• The principal is P P = $6,200 and the rate is r r = 0.0675. Since this is the standard repayment plan, there are n n = 12 payments per year for 10 years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 6,200 × ( 0.0675 / 12 ) × ( 1 + 0.0675 / 12 ) 12 × 10 ( 1 + 0.0675 / 12 ) 12 × 10 − 1 = $ 6,200 × ( 0.005625 ) × ( 1.005625 ) 120 ( 1.005625 ) 120 − 1 = 68.3662233426 0.960321816275 = $ 71.20 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 = $ 6,200 × ( 0.0675 / 12 ) × ( 1 + 0.0675 / 12 ) 12 × 10 ( 1 + 0.0675 / 12 ) 12 × 10 − 1 = $ 6,200 × ( 0.005625 ) × ( 1.005625 ) 120 ( 1.005625 ) 120 − 1 = 68.3662233426 0.960321816275 = $ 71.20

Your Turn 6.85

Example 6.86, standard repayment plan for an unsubsidized loan.

Erson has a balance of $8,132.55 when he starts paying off the 8.6% unsubsidized student loan he took out in his third year. How much are his payments if the term for his loan is the standard 10 years?

Using the payment formula, p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 , with P P = $8,132.55, r r = 0.086, t t = 10 and n n = 12, we calculate that his monthly payment will be

Your Turn 6.86

Student loan consolidation.

When a student graduates, they may have multiple different student loans. Keeping track of them and paying them off separately can be a burden. Instead, these loans can be consolidated into a single loan. If they are federal loans the combination is called federal consolidation. Combining private loans is often referred to as refinancing. Refinancing, or private consolidation, can be used to combine both private and federal student loans. Be aware that consolidated federal loans may still be subject to the rules and protections that govern subsidized loans. Refinancing loans, private or federal, are no longer subject to those rules and guidelines. Check out this Experian article about consolidation and refinancing for more deatil.

In consolidation of federal direct student loans, the interest rate is the weighted average of the interest rates on the subsidized loans. This means the interest rate remains the same. However, if the term is extended, then the student will pay back more over time than if they did not extend the loan term.

In refinancing, it is possible to obtain a lower interest rate on the student loans, which may lower how much is paid per month and lower the total paid back over time. These monthly payments are calculated using the same formula as for any other loan payment, p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 . The term of the refinanced loan may also be changed, which would also impact the payment per month.

In either case, consolidating or refinancing, the monthly financial burden on the student can decrease. However, if the term is extended, the total amount repaid may increase.

Example 6.87

Federal loan consolidation and interest rates.

Ernest has four federal student loans that he wants to consolidate. He combines them into one loan. What is the maximum Ernest can reduce the interest rate by?

Consolidating subsidized loans has no impact on the interest rate of the loans, so the maximum that the interest rate can be reduced is 0%.

Your Turn 6.87

Example 6.88, payments for consolidated student loans.

Brianna consolidates her student loans, some federal and some private, into a single refinanced student loan with a principal of $27,800. The interest rate that Brianna received was 8.375%. If Brianna’s new term is 15 years, how much are her payments per month?

The principal is P P = $27,800 and the rate is r r = 0.08375. Since the payments are monthly, n n =12. The loan term is for 15 years, so t t = 15. years. Substituting those values into p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 p m t = P × ( r / n ) × ( 1 + r / n ) n × t ( 1 + r / n ) n × t − 1 and calculating gives a monthly payment of

Your Turn 6.88

Other repayment plans.

There are various other repayment plans available to students. Plans other than the standard repayment plan typically require the student to meet certain criteria. The following plans are independent of student income, but may make early payments easier.

• Graduated repayment plans are plans where the amount of payments gradually increases so that the loan is paid off in 10 years, or within 10 to 30 years for consolidated loans. Payments start off small and increase approximately every 2 years. Almost all loan types are eligible, including direct subsidized and unsubsidized loans, Stafford loans, PLUS loans, and consolidated loans.
• Extended repayment plans are available to the direct loan borrower if the outstanding direct loans are over $30,000. The payments, fixed or graduated, are designed so that the loans are satisfied within 25 years. Eligible loans include both direct subsidized or unsubsidized loans, Stafford loans, PLUS loans, and consolidated loans.

If student earnings are such that the standard, graduated, or extended repayment plans are unaffordable, then one can make payments that are based on their discretionary income. Discretionary income is federally defined to be the difference between (adjusted) gross income and 150% of the poverty guideline for location and family size. This discretionary income then depends on where one lives (contiguous United States or Hawaii or Alaska) and how many dependents one has. If married, a spouse’s income will be included in the adjusted gross income. Understanding discretionary income is necessary to understand how income driven payments plans work.

Example 6.89

Discretionary income.

• The poverty guideline for a single person living in Arkansas, is $12,000. If Harriet is a single person in Arkansas with a (adjusted) gross income of $23,500, what is her discretionary income?
• For California, the poverty guideline for a person with four people in the household is $27,750. If such a person has a (adjusted) gross income of $48,600, what is their discretionary income?
• The poverty guideline for a single person in Arkansas is $12,000. 150% of that guideline value is 1.5 × $ 12,000 = $ 18,000 1.5 × $ 12,000 = $ 18,000 . The gross income that Harriet makes over that $18,000 is her discretionary income. That gross income is $23,500, so her discretionary income is $ 23,500 - $ 18,000 = $ 5,500 $ 23,500 - $ 18,000 = $ 5,500 .
• The poverty guideline for a household of four in California is $27,750. 150% of that guideline value is 1.5 × $ 27,750 = $ 41,625 1.5 × $ 27,750 = $ 41,625 . The gross income that Harriet makes over that $41,625 is her discretionary income. That gross income is $48,600, so her discretionary income is $ 48,600 − $ 41,625 = $ 6,975 $ 48,600 − $ 41,625 = $ 6,975 .

Your Turn 6.89

The following plans all depend on discretionary income.

• Pay As You Earn (PAYE) repayment plans have monthly payments that are 10% of discretionary income based on a student’s updated income and family size. If a borrower files a joint tax return, their spouse’s income and debt may also be considered. Eligible loans include direct subsidized and unsubsidized loans, PLUS loans made to students, and some consolidated loans. These loans are forgiven (student does not pay the remaining balance) after 20 years of monthly payments if they were direct federal student loans.
• Revised Pay As You Earn (REPAYE) repayment plans have payment amounts that are based on income and family size and calculated as 10% of discretionary income. Eligible loans include direct subsidized and unsubsidized loans, direct PLUS loans, and some consolidated loans. These loans are forgiven, that is, the student does not pay more, after 20 or 25 years, provided they were direct federal student loans.
• Income-Based Repayment (IBR) plans sound like a few of the others and there are similarities in all of them. The payments are either 10% or 15% of discretionary income, but this plan is meant for those with a relatively high debt. Every year, income and family size must be updated, and payments are calculated based on those figures. Eligible loans include direct subsidized and unsubsidized loans, Stafford loans, and PLUS loans made to students, but not PLUS loans made to parents.
• Income-Contingent Repayment (ICR) plans have payments that are either 20% of discretionary income or whatever would be paid if a student were on a fixed payment plan for more than 12 years, whichever is less. Eligible loans include direct subsidized and unsubsidized, PLUS loans to students, and consolidated loans.

There are many similarities among these repayment plans, and it is easy to misunderstand the nuances of each. Therefore, be careful entering into any type of repayment contract until you fully understand all the details and repercussions of the plan you choose. For more detail, see this nerdwallet article "Income-Driven Repayment: Is It Right for You?" to learn more !

Example 6.90

Payments for a repaye program.

Warren qualifies for a REPAYE payment plan. His gross income is $32,700. He is single and live in Montana, so the federal poverty guideline for Warren is $12,000.

• What is Warren’s discretionary income?
• Under the REPAYE plan, he pays 10% of his discretionary income, but monthly. How much are Warren’s REPAYE payments?
• The poverty guideline for Warren is $12,000. 150% of that guideline value is 1.5 * $ 12,000 = $ 18,000 1.5 * $ 12,000 = $ 18,000 . The gross income that Warren makes over that $18,000 is his discretionary income. That gross income is $32,700, so his discretionary income is $ 32,700 − $ 18,000 = $ 14,700 $ 32,700 − $ 18,000 = $ 14,700 .
• 10% of Warren’s discretionary income is 0.1 × $ 14,700 = $ 1,470 0.1 × $ 14,700 = $ 1,470 . He pays monthly, so his monthly payments are $1,470 divided by 12, or $122.50 per month.

Your Turn 6.90

Using income to determine payments initially seems excellent. However, if there is no forgiveness at the end of the loan, then the income driven payment plans can cause problems. For one, your payment may not be sufficient to cover the interest rate of your loans. In that case, your loan balance actually increases as you make your payments. Eventually, you are paying for not only your original loan balance, but interest that has been growing and compounding over time. Also, if the loan term is extended, you may pay more, perhaps a lot more, money over time. You may find yourself in the position of paying these loans for decades.

With those possible drawbacks, great care must be taken to avoid large problems down the line.

Repayment Plans

Student Loan Default and Consequences

The first day a payment is late, the account becomes delinquent . After 90 days, this delinquency is reported to the credit bureaus, and goes into default . This is serious, as now a credit score is affected, meaning that it will be harder to buy a car, a home, get a credit card, or a cell phone. Even renting an apartment may be a task not easily overcome. The default rate for students who do not complete their degree is three times higher than for students who do.

Further, defaulting on a student loan may mean that the borrower loses eligibility for repayment plans, as the balance and any unpaid interest may become due immediately, and any tax refunds may be withheld and applied to the loan, and wages may be garnished. One should immediately contact your loan servicer and try to make other arrangements for repayment if this situation becomes apparent, as different repayment plans are available, if actions are taken quickly.

There are several options that may be open to avoid defaulting. One is called rehabilitation, or is the process in which a borrower may bring a student loan out of default by adhering to specified repayment requirements, and the other is consolidation. Certain criteria must be met to enter these programs.

Both of these options are detailed, including the criteria required for eligibility, on the studentaid.gov loan management page .

Professionals advise hiring an attorney if one of these paths is chosen.

Check Your Understanding

Section 6.9 exercises.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/contemporary-mathematics/pages/1-introduction

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• Book title: Contemporary Mathematics
• Publication date: Mar 22, 2023
• Location: Houston, Texas
• Book URL: https://openstax.org/books/contemporary-mathematics/pages/1-introduction
• Section URL: https://openstax.org/books/contemporary-mathematics/pages/6-9-understanding-student-loans

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How to calculate interest on a loan, what is the loan interest formula, how to use the loan interest calculator.

The loan interest calculator (or interest calculator on loan) is a simple tool that helps you estimate the interest on your loan . In addition, you can check the loan's balance including periodic interest and principal payments in the loan amortization schedule .

In the following article, we show you how to calculate interest on a loan, and you can read some interesting details in our FAQ .

To learn more about loan amortization, check our amortization calculator .

The best way to understand how interest is calculated on a loan is to introduce it with a real-life example.

Let's assume you are considering obtaining a loan for a car purchase, so you decide to turn to a bank that offers you a personal loan of 10,000 dollars with 6% interest, repaid monthly in 10 years with the same compounding frequency. You can easily insert this data into our loan interest calculator :

• Loan balance: $10,000
• Loan term: 10
• Interest rate: 6%
• Payment frequency: Monthly
• Compounding frequency: Monthly

As a first step, you need to compute the equivalent rate , which is adjusted for compounding frequency. Since, in the present case, the payment frequency and the compounding frequency coincide, the equivalent rate equals the given interest rate. If you want to check the formula for this calculation, visit our equivalent rate calculator .

Calculate the periodic rate ( i i i ) by dividing the annual interest rate by the number of payments in a year. In our case, it is 0.06 / 12 = 0.005 .

Compute the total number of payments (or periods, n n n ) required to repay the loan principal. In our case, it is 12 × 10 = 120 .

Apply the below formula for calculating the periodic payment.

• P P P — Periodic payment;
• A A A — Loan balance or principal;
• i i i — Periodic rate; and
• n n n — Number of payments or periods.

In our case, the periodic (monthly) payment is $111.02 .

Calculate the total payment by multiplying the periodic payment by the number of payments. Therefore, the total payment is 111.02 × 120 = $13,322.46 .

The interest payment is the difference between the total payment and the principal balance (or loan amount). That is, the interest on the above loan is 13,322.46 – 10,000 = $3,322.46 .

The loan interest formula can be formulated in the following way.

Interest = A - (i × A × n)/(1 – (1 + i) -n )

• A is the loan balance or principal;
• i is the periodic equivalent rate; and
• n is the number of payments or periods.

You need to follow the below simple steps to apply for our loan interest calculator :

• Input the loan balance you plan to borrow. It will be the principal, which needs to be repaid.
• Provide the loan term .
• Provide the interest rate .
• Set the payment frequency.
• Choose the compounding frequency, which will be the timing of capitalization of the interest (the unpaid amount of interest added to the loan's principal balance).

You can also follow the accumulation of the total interest on the chart of balances and the periodic or annual interest payments in the amortization schedule displayed below the main results.

What is the interest of a $10,000 loan with a 6% rate?

The interest of a $10,000 loan with a 6% rate with ten years loan term repaid monthly is $3,322.46 .

How can I calculate loan interest?

Follow the below steps to calculate loan interest.

Calculate the periodic rate ( i ) by dividing the annual interest rate by the number of payments in a year ( n ).

Calculate the total payment ( P ) by multiplying the periodic rate ( i ) with the loan amount ( A ) and the number of payment ( n ) and then divide it by the factor of 1 – (1 + i) -n .

Finally, calculate the loan interest by subtracting the loan amount from the total payment ( interest = P - A ).

Why are interest rates are increasing?

Interest rates are increasing due to monetary policy intervention responding to high inflation rates. The higher interest rates reduce aggregate demand as fewer consumers take a loan, which eventually can lead to disinflation and lower inflation expectations.

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8.5: Amortized Loans

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• Page ID 84634

• Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
• Coconino Community College

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In Section 8.2, we learned how to find the future value of a lump sum, and in Section 8.4, we learned how to find the future value of an annuity. With these two concepts in hand, we will now learn how to amortize a loan , and how to find the present value of an annuity .

Amortize a Loan

If a person or business needs to buy or pay for something now (a car, a home, college tuition, equipment for a business) but does not have the money, they can borrow the money as a loan.

They receive the loan amount now, called the principal, P , (or present value), and are obligated to pay back the principal in the future over a stated amount of time (term of the loan), as regular periodic payments, PMT , plus interest.

Example \(\PageIndex{1}\)

Find the monthly payment for a car costing $15,000 if the loan is amortized over five years at an interest rate of 9%.

Consider the following scenario:

Two people, Mr. Cash and Mr. Credit, go to buy the same car that costs $15,000. Mr. Cash pays cash and drives away, but Mr. Credit wants to make monthly payments for five years.

Our job is to determine the amount that Mr. Credit needs to pay each month for 5 years. We reason as follows:

If Mr. Credit pays PMT dollars per month, then the PMT dollar payment deposited each month at 9% for 5 years should yield the same amount as the $15,000 lump sum deposited in an annuity for 5 years.

Again, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like them to be the same.

Since Mr. Cash is paying a lump sum of $15,000, the future value \(F\) is given by the lump sum formula (8.2), and it is

\[F =\ 15,000\left(1+\dfrac{.09}{12}\right)^{60} \nonumber\]

Mr. Credit wishes to make a sequence of payments of \(PMT\) dollars per month, and the future value is given by the annuity formula (8.4), and this value is

\[F = PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \nonumber\]

We set the two future amounts equal to eachother and solve for the unknown value, PMT .

\[\begin{array}{l} \ 15,000\left(1+\dfrac{.09}{12}\right)^{60}=PMT\frac{\left[\left(1+\dfrac{.09}{12}\right)^{60}-1\right]}{\dfrac{.09}{12}} \\ \ 15,000(1.5657)=PMT(75.4241) \\ \ 311.3792=PMT \end{array} \nonumber\]

Therefore, the monthly payment needed to repay the loan is $311.38 for five years.

The formula used above (and restated here), for finding payments on an amortized loan, can appear cumbersome.

\(\begin{array}{l} \ P\left(1+\dfrac{r}{m}\right)^{m t}=PMT\frac{\left[\left(1+\dfrac{r}{m}\right)^{m t}-1\right]}{\dfrac{r}{m}} \end{array} \nonumber\)

If we do the necessary algebra to solve this equation for PMT , we can use the new formula to find the payments. The algebra has been omitted and the new formula is stated in the box below.

Amortization Formula

\(PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\)

\(P\) is the balance in the account at the beginning (the principal, or amount of the loan)

\(r\) is the annual interest rate in decimal form

\(t\) is the length of the loan, in years

\(m\) is the number of compounding periods in one year

\(PMT\) is the loan payment (monthly payment, annual payment, etc.)

*Notes: The compounding frequency is not always explicitly given, but is determined by how often you make payments. We will round payments on a loan up to the next cent.

Example \(\PageIndex{2}\)

You want to take out a $340,000 mortgage (home loan). The interest rate on the loan is 3.5%, and the loan is for 30 years. How much will your monthly payments be? How much interest will you pay over the life of the loan?

We’re looking for \(PMT\).

\(\begin{array}{ll} P = \$340,000 & \text{the starting loan amount} \\ r = 0.035 & 3.5\% \text{ annual rate} \\ t = 30 & \text{since we’re making monthly payments for 30 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l} \ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\ \ PMT=340000\cdot \dfrac{\left(\dfrac{0.035}{12}\right)}{\left[1-\left( 1+\dfrac{0.035}{12}\right)^{-12(30)}\right]} \\ \ PMT=1526.7519 \end{array} \nonumber\]

You will make payments of $1526.76 per month for 30 years. (Remember to round up for payments.)

You will pay a total of \(\$ 1526.76\) per month for 360 months which equals \(\$ 549,633.60\) to the loan company. The total paid over the life of the loan is \(\$ 549,633.60 - \$ 340,000=\$ 209,633.60\).

Example \(\PageIndex{3}\)

Jack goes to a car dealer to buy a new car for $18,000 at 2% APR with a five-year loan. The dealer quotes him a monthly payment of $425. Verify that this is the correct monthly payment.

P = $18,000, r = 0.02, t = 5, m = 12

\[\begin{array}{l} \ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\ \ PMT=18000\cdot \dfrac{\left(\dfrac{0.02}{12}\right)}{\left[1-\left( 1+\dfrac{0.02}{12}\right)^{-12(5)}\right]} \\ \ PMT=315.4996 \end{array} \nonumber\]

Jack should have a monthly payment of $315.50, not $425.

What should the total principal and interest be with the $315.50 monthly payment? \(\$ 315.50(12)(5) =\$ 18,930\) The $315.50 payment per month has a total of $930 in interest paid over the 2-year loan period.

However, the dealer is trying to get Jack to pay $425 per month. This equates to \(\$ 425(12)(5) =\$ 25,500\) which is significantly more than the calculation above. And notice the difference in the the total interest charges; $25,500 - $18,000 = $7,500. This means that the quoted rate of 2% APR is not accurate, or the quoted price of $18,000 is not accurate, or both.

Try it Now 1

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? What is the total interest charged?

\(\begin{array}{ll} P = 3,000 & \text{the starting loan amount \$3,000 loan} & \\ r = 0.16 & 16\% \text{ annual rate} \\ t = 2 & \text{2 year to repay} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l} \ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\ \ PMT=3000\cdot \dfrac{\left(\dfrac{0.16}{12}\right)}{\left[1-\left( 1+\dfrac{0.16}{12}\right)^{-12(2)}\right]} \\ \ PMT=146.8893 \end{array} \nonumber\]

Janine will need to make monthly payments of \(\$ 146.89\).

In total, she will pay \(\$ 3,525.36\) to the store, meaning she will pay \(\$ 525.36\) in interest over the two years.

Example \(\PageIndex{4}\)

With a fixed rate mortgage, you are guaranteed that the interest rate will not change over the life of the loan. Suppose you need $250,000 to buy a new home. The mortgage company offers you two choices: a 30-year loan with an APR of 6% or a 15-year loan with an APR of 5.5%. Compare your monthly payments and total loan cost to decide which loan you should take. Assume no difference in closing costs.

Option 1 : First calculate the monthly payment:

The monthly payment for a 30-year loan at 6% interest is $1498.88.

Now calculate the total cost of the loan over the 30 years:

The monthly payments are $1498.88 and the total cost of the loan is $539,596.80.

Option 2 : First calculate the monthly payment:

The monthly payment for a 15-year loan at 5.5% interest is $2042.71.

Now calculate the total cost of the loan over the 15 years:

The monthly payments are $2042.71 and the total cost of the loan is $367,687.80.

Therefore, the monthly payments are higher with the 15-year loan, but you spend a lot less money overall.

Present Value of an Annuity

The present valu e of an annuity is the amount of money we would need now in order to be able to make the annuity payments in the future. Often, we know how much we can afford to pay for each regular payment, so we need to find how much money we can borrow.

Example \(\PageIndex{5}\)

Jordan can afford $400 per month as a car payment. The car dealership offers an auto loan at 12% interest for 4 years. What is the present value of the car? In other words, what loan amount can Jordan afford at $400 per month?

In this example,

\(\begin{array}{ll} PMT = \$400 & \text{the monthly loan payment} \\ r = 0.12 & 12\% \text{ annual rate} \\ t = 4 & \text{since we’re making monthly payments for 4 years} \\ m = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \end{array}\)

\[\begin{array}{l} \ PMT=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]} \\ \ 400=P\cdot \dfrac{\left(\dfrac{0.12}{12}\right)}{\left[1-\left( 1+\dfrac{0.12}{12}\right)^{-12(4)}\right]} \\ \ P=15189.5838 \end{array} \nonumber\]

Jordan will pay a total of $15,189.58 ($400 per month for 48 months) to the car dealership. The difference between the amount paid and the amount of the loan is the interest paid . In this case, Jordan is paying \(\$ 19,200 -\$ 15,189.58=\$ 4,010.42\) interest total.

Solve the Amortization Formula for P, Present Value

In the example above, notice we were given the monthly payment and asked to find the the present value, P . It would be helpful to solve the amortization formula for the present value first.

---Since P is multiplied by a fraction, solve for P by multiplying both sides of the equation by the reciprocal of that fraction.

\(PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}=P\cdot \dfrac{\left(\dfrac{r}{m}\right)}{\left[1-\left( 1+\dfrac{r}{m}\right)^{-m t}\right]}\cdot {\dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}}\)

---Simply the right side of the equation, then rewrite as,

\(P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)}\)

Present Value of an Annuity Formula

Use this formula when you know the payment and you want to find the present value, P .

*Note: The compounding frequency is not always explicitly given, but is determined by how often you make payments.

Example \(\PageIndex{6}\)

Grace buys an iPad from a rent-to-own business on credit with payments of $30 a month for four years at 14.5% APR compounded monthly. If Grace had bought the iPad from Best Buy or Amazon it would have cost $500. What is the price that Grace paid for the iPad at the rent-to-own business? How much interest was paid over the life of the loan? What is the better option?

PMT = $30, r = 0.145, t = 4, m = 12

\[\begin{array}{l} \ P=PMT\cdot \dfrac{\left[1-\left(1+\dfrac{r}{m}\right)^{-m(t)}\right]}{\left(\dfrac{r}{m}\right)} \\ \ P=30\cdot \dfrac{\left[1-\left(1+\dfrac{0.145}{12}\right)^{-12(4)}\right]}{\left(\dfrac{0.145}{12}\right)} \\ \ P=1087.8254 \end{array} \nonumber\]

The price Grace paid for the iPad was $1,087.83.

That’s a lot more that $500!

Example \(\PageIndex{7}\)

Suppose you have won a lottery that pays $1,000 per month for the next 20 years. But, you prefer to have the entire amount now. If the interest rate is 8%, how much will you accept?

Consider, for argument purposes, that two people Mr. Cash, and Mr. Credit have won the same lottery of $1,000 per month for the next 20 years. Mr. Credit is happy with his $1,000 monthly payment, but Mr. Cash wants to have the entire amount now.

Our job is to determine how much Mr. Cash should get. We reason as follows:

If Mr. Cash accepts P dollars, then the P dollars deposited at 8% for 20 years should yield the same amount as the $1,000 monthly payments for 20 years. In other words, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like the future values to equal.

Since Mr. Cash is receiving a lump sum of \(x\) dollars, its future value is given by the lump sum formula we studied in Section 6.2, and it is

\[\mathrm{A}=\mathrm{P}(1+.08 / 12)^{240} \nonumber\]

Since Mr. Credit is receiving a sequence of payments, or an annuity, of $1,000 per month, its future value is given by the annuity formula we learned in Section 6.3. This value is

\[\mathrm{A}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \nonumber\]

The only way Mr. Cash will agree to the amount he receives is if these two future values are equal. So we set them equal and solve for the unknown.

\[\begin{array}{l} \mathrm{P}(1+.08 / 12)^{240}=\frac{\$ 1000\left[(1+.08 / 12)^{240}-1\right]}{.08 / 12} \\ \mathrm{P}(4.9268)=\$ 1000(589.02041) \\ \mathrm{P}(4.9268)=\$ 589020.41 \\ \mathrm{P}=\$ 119,554.36 \end{array} \nonumber\]

The present value of an ordinary annuity of $1,000 each month for 20 years at 8% is $119,554.36

The reader should also note that if Mr. Cash takes his lump sum of \(\mathrm{P}\) = $119,554.36 and invests it at 8% compounded monthly, he will have an accumulated value of \(\mathrm{A}\)=$589,020.41 in 20 years.

Which equation to use?

When presented with a finance problem (on an exam or in real life), you're usually not told what type of problem it is or which equation to use. Here are some hints on deciding which equation to use based on the wording of the problem.

The easiest types of problem to identify are loans. Loan problems almost always include words like: "loan", "amortize" (the fancy word for loans), "finance (a car)", or "mortgage" (a home loan). Look for these words. If they're there, you're probably looking at a loan problem. To make sure, see if you're given what your monthly (or annual) payment is, or if you're trying to find a monthly payment.

If the problem is not a loan, the next question you want to ask is: "Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?" If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a compound interest problem. The exception would be bonds and other investments where the interest is not reinvested; in those cases you’re looking at simple interest.

If you're making regular payments or withdrawals, the next questions is: "Am I putting money into the account, or am I pulling money out?" If you're putting money into the account on a regular basis (monthly/annually/quarterly) then you're looking at a basic Annuity problem. Basic annuities are when you are saving money. Usually in an annuity problem, your account starts empty, and has money in the future.

If you're pulling money out of the account on a regular basis, then you're looking at a Payout Annuity problem. Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty in the future.

Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and to determine what approach will best allow you to solve the problem.

Try it Now 5

For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem.

• Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?
• Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?
• Keisha is managing investments for a non-profit company. They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?
• Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?
• How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?
• This is a payout annuity problem. She can pull out $1833.60 a quarter.
• This is a savings annuity problem. He will have saved up $7,524.11.
• This is compound interest problem. She would need to deposit $22,386.46.
• This is a loans problem. She can buy $4,609.33 of new equipment.
• This is a savings annuity problem. You would need to save $200.46 each month

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Tver city, Russia

The capital city of Tver oblast .

Tver - Overview

Tver is a city in Russia located on the banks of the Volga River about 180 km north-west of Moscow, the administrative center of Tver Oblast. It is a large industrial, scientific, and cultural center, a major transport hub at the intersection of the St. Petersburg - Moscow railway line and the federal highway M10 “Russia” with the Upper Volga.

The population of Tver is about 424,900 (2022), the area - 152 sq. km.

The phone code - +7 4822, the postal codes - 170000-170043.

Tver city flag

Tver city coat of arms.

Tver city map, Russia

Tver city latest news and posts from our blog:.

27 December, 2023 / Amazing Church of the Transfiguration in Krasnoye in Tver Oblast .

25 October, 2020 / Rzhev Memorial to the Soviet Soldier .

4 April, 2019 / Cities of Russia at Night - the Views from Space .

16 October, 2018 / The Oldest Monastery in the Tver Region .

30 September, 2018 / Church of the Transfiguration of the Savior in Krasnoye .

More posts..

History of Tver

Tver in the 12th-16th centuries.

Tver has a rich history dating back to the first half of the 12th century. Officially, the year of foundation of the city is 1135. According to archaeological data, a settlement on the cape at the confluence of the T’maka River with the Volga existed already in the 9th-10th centuries. In the 12th century, it was a small trading settlement. The first birch bark letter, found in Tver on the territory of the old Tver Kremlin, dates back to 1200-1220.

In the first third of the 13th century, Tver was part of the Principality of Pereyaslavl. In 1238, the town was ravaged by the Mongols, but quickly recovered from the defeat. In 1247, Tver became the capital of the Principality of Tver.

The geographical position of Tver on an important trade route connecting Novgorod with northeastern Russia, and the relative remoteness from the Golden Horde contributed to the influx of population from other Russian lands into the region. The growth of Tver was also caused by the change in its political role. In 1264, Prince Yaroslav of Tver became the Grand Duke of Vladimir, but stayed to live in Tver.

Chronicles and stone construction were resumed in Tver for the first time in Rus after a 50-year hiatus. Tver developed its own original art school: architecture, icon painting, jewelry, decorative and applied arts. Its own coin was also minted here.

More Historical Facts…

In the summer of 1327, after the arrival of the khan’s ambassador Shevkal in Tver, rumors spread about the imminent conversion of the residents of Tver to Islam. A powerful uprising broke out in the town. With the help of Ivan Kalita, Grand Duke of Moscow, it was brutally suppressed; Tver was ruined. The suppression of the uprising marked the beginning of the decline of Tver.

Until the second half of the 15th century, Tver was subjected to repeated attacks by the Mongols and Moscow. In this struggle, Tver gradually lost its leading position among the old principalities of North-Eastern Russia. Moscow became the unifier of the Russian lands. This struggle undermined the strength of Tver. However, in the 14th-15th centuries, it remained a large trade, craft, and cultural center, one of the most developed Russian towns. In 1485, Moscow troops occupied Tver, the Tver Principality ceased to exist and became part of the emerging Russian centralized state.

In the 16th century, the oldest surviving church in Tver, the Trinity Church, known as the White Trinity, was built. In December 1569, Tver was plundered by the oprichnina army of Ivan IV the Terrible during his campaign against Novgorod.

Tver in the 17th-19th centuries

At the beginning of the 17th century, about 10,000 people lived in Tver. In 1612, during the Time of Troubles, it was completely devastated by the Polish-Lithuanian troops. In 1629, the town had 11 abandoned churches and monasteries, 1,450 empty houses. In 1685, its population was about 4,600 residents.

In the 18th century, after the founding of St. Petersburg, Tver became an important communication point between the new capital of Russia and Moscow. In 1701, by order of Peter I, a floating bridge on rafts was built in Tver, which existed until 1900. By 1709, the construction of the Vyshnevolotsk water system was completed, which connected the Volga with the Baltic Sea.

A lot of civil buildings were constructed in Tver. Many of them have survived to this day. The oldest of them is considered to be the house of the merchant Arefiev, which houses the museum of Tver everyday life now.

In 1763, a fire destroyed the central part of Tver. The main features of the new regular planning of the town were the long axial Millionnaya (Sovetskaya) Street, named so because a million rubles were allocated for the construction of stone houses in the town center; as well as the “Versailles trident” - a three-rayed composition of streets converging at one point, modeled on a similar urban planning technique used in St. Petersburg.

In 1764-1766, the main attraction of Tver was constructed - the Travel Palace of Empress Catherine in the style of classicism with baroque elements, designed by M.F. Kazakov. The palace was intended for the rest of the members of the imperial family on the way from St. Petersburg to Moscow. In 1787, the population of Tver was about 15,100 people.

In 1851, Tver was connected with St. Petersburg and Moscow by the Nikolaev railway. In the second half of the 19th century, a steamship society, a weaving manufactory, a mechanical plant for the production of parts for textile machines, sawmills, and other enterprises were opened in Tver. In 1850-1860, three textile factories were founded in Tver.

Tver in the 20th century and beyond

In 1900, a permanent bridge across the Volga River was built in Tver, designed by the Czech engineer L. Mashek. In 1914, the population of Tver was 63,900 people. With the outbreak of the First World War, the Russian Empire experienced problems in exchanging information with its allies - France and the British Empire, since most of the European land lines passed through the German Empire. The key role in the exchange of information between the allies was played by the Tver special purpose radio station of the Russian military department. During the war, the Russian-Baltic Carriage Works were evacuated to Tver from Riga.

In the 1920s-1930s, dozens of churches, architectural monuments of the 17th-19th centuries, were closed and destroyed in the city. On November 20, 1931, Tver was renamed Kalinin, in honor of M.I. Kalinin, the Soviet party and statesman, a native of the Tver Governorate. In 1935, Kalinin became the administrative center of Kalinin Oblast. The population of the city was 216,131 people in 1939.

On October 17, 1941, during the Second World War, the city was captured by the Germans, however, their further advance was delayed, and in the northwest direction it was completely stopped. It was under German occupation for about two months. During the occupation and heavy fighting, the city was badly damaged, more than half of the residential buildings and about 70 enterprises were destroyed. Until the liberation of Rzhev (March 3, 1943), Kalinin was subjected to systematic raids by German aircraft.

On November 1, 1945, a decree was adopted on the priority restoration of 15 most important cities of Russia destroyed during the war, including the city of Kalinin. In the 1950s-1960s, magnificent buildings in the style of Stalinist classicism were erected: the drama theater (1951), the library named after M. Gorky, the technical school on Lenin Avenue (1957), the ensembles of Novopromyshlennaya Square, Peace Square, Gagarin Square.

In the second half of the 20th century, an electrical equipment plant, a chemical fiber plant, a silicate brick plant, a silk-weaving factory, a meat processing plant, a pharmaceutical factory, a fiberglass plant, etc. were opened in Kalinin. It became not only a large industrial city, but also an important transport hub. In 1961, a river port and a circular highway were opened here, connecting highways to Moscow, Leningrad, Rzhev, and Volokolamsk.

In 1989, the population of Tver was 450,941 people. By the early 1990s, there were about 80 industrial enterprises of 28 industries, 5 universities, 3 theaters, 12 movie theaters, 8 palaces of culture in Kalinin. On July 17, 1990, the historical name and coat of arms were returned to the city.

Due to the negative economic consequences caused by the collapse of the Soviet Union, active development in the city temporarily ceased. The only new microdistrict built in the 1990s-2000s was the microdistrict Mamulino, built for the withdrawal of a contingent of Soviet troops from East Germany in 1993.

The new economic reality turned out to be disastrous for the local industry, especially the light one, traditional for the city, as a result of which the largest enterprises of Tver stopped their work, which caused an increase in the level of unemployment and social tension.

In the mid-2000s, large-scale housing construction resumed in Tver, active construction of public buildings, shopping and entertainment centers began. At the same time, new industrial enterprises were commissioned.

Architecture of Tver

Winter in Tver

Author: Andrey Dmitriev

Tver architecture

Wooden architecture in Tver

Author: Alexandra Kilanova

Tver - Features

There are several versions about the etymology of the name of Tver. According to one of them, it was named after the Tvertsa River, and Tvertsa, in turn, is derived from the Finnish word “tiort” (“fast”). According to another version, the name of the city comes from the Slavic “tverd” (“fortress”). In addition, this name is associated with the Polish word “twierdza” (“fortress”) and the Lithuanian words “tvora” (“fence”) and “tverti” (“to enclose”).

Tver is located at the intersection of the railway and highway connecting Moscow and St. Petersburg with the Volga in its upper reaches; about 180 km north-west of Moscow and 520 km south-east of St. Petersburg. The City Day of Tver is celebrated on the last Saturday of June.

The climate in Tver is moderately continental, with cold and rather long winters and warm, humid summers. The average temperature in February is minus 8.3 degrees Celsius, in July - plus 19.8 degrees Celsius.

The coat of arms of Tver corresponds to the historical coat of arms of Tver approved by Empress Catherine II in 1780 with the following description: “In a red field, on a green pillow, a golden crown”.

Tver is one of the largest industrial centers in Russia. The main industrial sectors of the city are mechanical engineering, printing, and food processing. A characteristic feature of local machine building is its concentration in one enterprise: Tver Carriage Works (the largest employer in Tver with more than 8 thousand employees).

Today, Tver is an important tourist center of Russia. The city has preserved a lot of historical mansions of the 18th century, with which almost the entire Volga embankment is built up. The restored churches of the 17th century are also of great interest.

The Volga River is worth a separate mention. Within the limits of Tver, its banks are dressed in picturesque embankments, where comfortable areas for recreation and walking have been created.

Main Attractions of Tver

Tver Imperial Travel Palace (1764-1766) - an architectural monument in the classicism style with baroque elements, which is located on Cathedral Square in the historical center of Tver. The palace was intended for the rest of the members of the imperial family on the way from St. Petersburg to Moscow. Today, this palace is not only an object of historical and cultural heritage, but also one of the oldest buildings in Tver.

Today, the Tver Imperial Travel Palace houses the Tver Picture Gallery. The art collection of the gallery numbers about 32,000 exhibits: collections of cult (“Old Russian”) art of the 14th-20th centuries, Russian painting, graphics and sculpture of the 18th-20th centuries, Western European art of the 15th-20th centuries, decorative and applied art of Russia, Europe, East.

Stepan Razin Embankment - one of the most interesting and popular walking places for locals and tourists in Tver, named after the famous Don Cossack. It is located on the right bank of the Volga River and stretches from the city garden to the stadium. In addition to the former merchant buildings on this embankment, there are noble mansions, “The House of Voroshilov Riflemen” - an architectural monument of the Stalinist era, and “Zvezda” (“Star”) movie theater, built in the style of late constructivism.

Afanasy Nikitin Embankment . This embankment stretches along the left bank of the Volga. There are a lot of stylized restaurants and open-air verandas here, where tourists love to drop in in the warm season. On the embankment, there are several monuments in honor of famous historical figures, as well as a city amusement park.

Trekhsvyatskaya Street - a 1.3 km long pedestrian street known as “Tver Arbat”. This is one of the most picturesque streets in Tver, built up with old houses and mansions. Almost all of the surviving buildings date from the late 19th - early 20th centuries. This beautiful and well-maintained place attracts musicians, artists, and souvenir sellers. The first floors of the buildings are given over to restaurants, cafes, and shops. All major city holidays are held here.

Sovetskaya Street - the main street of Tver stretching through the historic center of the city for 2.2 km. A lot of buildings on this street are recognized as monuments of architecture and history. Among them are the Ascension Cathedral (1751-1760) and the Tver Imperial Travel Palace, the House of the Nobility Assembly, and the building of the Drama Theater. Of the modern buildings, the Catholic Church of the Transfiguration of the Lord, which combines elements of Gothic and modernism, is of interest.

Tver State United Museum . The exposition of the museum tells about the nature of this region, its history, and archaeological finds. Here you can see antiques, learn about the history of the Tver Principality, the religion of the Slavs, get acquainted with the architecture and life of the nobles, who lived on the territory of the Tver region in the 17th-19th centuries. Sovetskaya Street, 5.

Museum of Tver Everyday Life - a small museum located in the complex of buildings of the estate of the Arefiev merchants. The exposition of the museum is devoted to the everyday life of various segments of people of the Tver Governorate and works of applied art created by urban artisans and peasants. Gor’kogo Street, 19/4.

White Trinity Church (1564) - the oldest building on the territory of Tver, constructed of white stone and brick, a monument of history and architecture. Inside, an old iconostasis has been preserved, the wall painting belongs to a later period - the 19th-20th centuries. Troitskaya Street, 38.

Resurrection Cathedral (1912-1913) - a beautiful building in the neo-Russian style, built for the 300th anniversary of the reign of the House of Romanov. The decoration of the cathedral is stylized to resemble the forms of Pskov and Novgorod architecture. Barrikadnaya Street, 1.

Tver Mosque (1906). Due to its bright and very unusual appearance, this mosque is deservedly one of the main architectural attractions of Tver. The building was constructed in neo-Moorish style with semicircular arches on the ground floor and a scalloped ridge on the roof. Sovetskaya Street, 63.

Not far from the mosque, there is an observation deck “Panorama” (Smolensky Lane, 29), where you can enjoy the views of the old Tver and the Volga River from the height of the 24th floor.

Starovolzhsky Bridge (1900) - the first permanent bridge across the Volga River in Tver, connecting Afanasy Nikitin Embankment on one side with the City Garden on the other. The openwork spans of this bridge look like the famous Liberty Bridge in Budapest. It is one of the main sights of the city.

Tver city of Russia photos

Monuments in tver.

Monument to Afanasy Nikitin in Tver

Author: Inna Zyuganova

Monument to Mikhail Kalinin in Tver

Monument to Vladimir Lenin in Tver

Author: Cherepanov Timofey

Sights of Tver

Tver River Station

Monument Agreement of Thousands in Tver

Tver Regional Drama Theater

Churches of Tver

Church of the Life-Giving Trinity beyond the Volga in Tver

Author: Igor Stolyarov

Cathedral of the Ascension of the Lord in Tver

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Simple Interest Calculator

The Simple Interest Calculator calculates the interest and end balance based on the simple interest formula. Click the tabs to calculate the different parameters of the simple interest formula. In real life, most interest calculations involve compound Interest. To calculate compound interest, use the Interest Calculator .

Related Interest Calculator | Compound Interest Calculator

What is Simple Interest?

Interest is the cost you pay to borrow money or the compensation you receive for lending money. You might pay interest on an auto loan or credit card, or receive interest on cash deposits in interest-bearing accounts, like savings accounts or certificates of deposit (CDs).

Simple interest is interest that is only calculated on the initial sum (the "principal") borrowed or deposited. Generally, simple interest is set as a fixed percentage for the duration of a loan. No matter how often simple interest is calculated, it only applies to this original principal amount. In other words, future interest payments won't be affected by previously accrued interest.

Simple Interest Formula

The basic simple interest formula looks like this:

Simple Interest = Principal Amount × Interest Rate × Time

Our calculator will compute any of these variables given the other inputs.

Simple Interest Calculated Using Years

You may also see the simple interest formula written as:

In this formula:

• I = Total simple interest
• P = Principal amount or the original balance
• r = Annual interest rate
• t = Loan term in years

Under this formula, you can manipulate "t" to calculate interest according to the actual period. For instance, if you wanted to calculate interest over six months, your "t" value would equal 0.5.

Simple Interest for Different Frequencies

• I = total interest
• P = Principal amount
• r = interest rate per period
• n = number of periods

Under this formula, you can calculate simple interest taken over different frequencies, like daily or monthly. For instance, if you wanted to calculate monthly interest taken on a monthly basis, then you would input the monthly interest rate as "r" and multiply by the "n" number of periods.

Simple Interest Examples

Let's review a quick example of both I=Prt and I=Prn.

For example, let's say you take out a $10,000 loan at 5% annual simple interest to repay over five years. You want to know your total interest payment for the entire loan.

To start, you'd multiply your principal by your annual interest rate, or $10,000 × 0.05 = $500.

Then, you'd multiply this value by the number of years on the loan, or $500 × 5 = $2,500.

Now that you know your total interest, you can use this value to determine your total loan repayment required. ($10,000 + $2,500 = $12,500.) You can also divide the value to determine how much interest you'd pay daily or monthly.

Alternatively, you can use the simple interest formula I=Prn if you have the interest rate per month.

If you had a monthly rate of 5% and you'd like to calculate the interest for one year, your total interest would be $10,000 × 0.05 × 12 = $6,000. The total loan repayment required would be $10,000 + $6,000 = $16,000.

What Financial Instruments Use Simple Interest?

Simple interest works in your favor as a borrower, since you're only paying interest on the original balance. That contrasts with compound interest, where you also pay interest on any accumulated interest. You may see simple interest on short-term loans.

For this same reason, simple interest does not work in your favor as a lender or investor. Investing in assets that don't offer compound growth means you may miss out on potential growth.

However, some assets use simple interest for simplicity — for example bonds that pay an interest coupon. Investments may also offer a simple interest return as a dividend. To take advantage of compounding you would need to reinvest the dividends as added principal.

By contrast, most checking and savings accounts, as well as credit cards, operate using compound interest.

Simple Interest Versus Compound Interest

Compound interest is another method of assessing interest. Unlike simple interest, compound interest accrues interest on both an initial sum as well as any interest that accumulates and adds onto the loan. (In other words, on a compounding schedule, you pay interest not just on the original balance, but on interest, too.)

Over the long run, compound interest can cost you more as a borrower (or earn you more as an investor). Most credit cards and loans use compound interest. Savings accounts also offer compounding interest schedules. You can check with your bank on the compounding frequency of your accounts.

Compound Interest Formula

The basic formula for compound interest is:

A = P × (1 +

r n A = ending balance P = Principal balance r = the interest rate (expressed as a decimal) n = the number of times interest compounds in a year t = time (expressed in years) Note that interest can compound on different schedules – most commonly monthly or annually. The more often interest compounds, the more interest you pay (or earn). If your interest compounds daily, you'd enter 365 for the number of time interest compounds annually. If it compounds monthly, you'd input 12 instead. Learn More About Compound Interest Compound interest calculations can get complex quickly because it requires recalculating the starting balance every compounding period. For more information on how compound interest works, we recommend visiting our compound interest calculator . Which is Better for You: Simple or Compound Interest? As a borrower, paying simple interest works in your favor, as you'll pay less over time. Conversely, earning compound interest means you'll net larger returns over time, be it on a loan, investment, or your regular savings account. For a quick example, consider a $10,000 loan at 5% interest repaid over five years. As established above, a loan this size would total $12,500 after five years. That's $10,000 on the original principal plus $2,500 in interest payments. Now consider the same loan compounded monthly. Over five years, you'd repay a total of $12,833.59. That's $10,000 of your original principal, plus $2,833.59 in interest. Over time, the difference between a simple interest and compound interest loan builds up exponentially. Search Tver Oblast Overview Map Directions Satellite Photo Map Overview Map Directions Satellite Photo Map Tap on the map to travel Popular Destinations Type: State with 1,280,000 residents Description: administrative division (oblast) in central Russia Neighbors: Moscow Oblast , Novgorod Oblast , Pskov Oblast , Smolensk Oblast , Vologda Oblast and Yaroslavl Oblast Categories: oblast of Russia and locality Location: Central Russia , Russia , Eastern Europe , Europe View on Open­Street­Map Tver Oblast Satellite Map Popular Destinations in Central Russia Escape to a random place. Tver Oblast 2 Other destinations 3 Understand 6 Get around 11 Stay safe <a href=\"https://tools.wmflabs.org/wikivoyage/w/poi2gpx.php?print=gpx&amp;lang=en&amp;name=Tver_Oblast\" title=\"Download GPX file for this article\" data-parsoid=\"{}\"><img alt=\"Download GPX file for this article\" resource=\"./File:GPX_Document_rev3-20x20.png\" src=\"//upload.wikimedia.org/wikipedia/commons/f/f7/GPX_Document_rev3-20x20.png\" decoding=\"async\" data-file-width=\"20\" data-file-height=\"20\" data-file-type=\"bitmap\" height=\"20\" width=\"20\" class=\"mw-file-element\" data-parsoid='{\"a\":{\"resource\":\"./File:GPX_Document_rev3-20x20.png\",\"height\":\"20\",\"width\":\"20\"},\"sa\":{\"resource\":\"File:GPX Document rev3-20x20.png\"}}'/></a></span>"}'/> Tver Oblast is a region in Central Russia , which borders Smolensk Oblast to the southwest, Pskov Oblast to the west, Novgorod Oblast to the north, Vologda Oblast to the northeast, Yaroslavl Oblast to the east, and Moscow Oblast to the southeast. 56.857828 35.921928 1 Tver — the capital and only major city of the region; an ancient city with a prestigious past that once contended with Moscow for control of Russia, that is now a shell of its former self following widespread destruction from cultural vandalism at the hands of the Soviet government and the Nazi Wehrmacht Other destinations Nilov Monastery — a fantastically beautiful and enormous monastery on Stolbny Island in Lake Seliger ; served as a gulag for many of the Polish prisoners of war who were massacred at Katyn 57.251242 32.473697 2 Volgoverkhovie Russian is the principal language in all aspects of life in the region. However, English is now taught regularly in most Russian schools, and more and more Russians nationals can speak English, some quite fluently. There is a still vivid Karelian minority (1   %, 6.7   % in 1897), who originally moved here following the Treaty of Stolbova 1617. People in the conceded areas would have to convert to Lutheranism, which was the only legal religion in Sweden – and pay heavy taxes – and many chose to move to land still Russian. There is also a Ukrainian minority (1,5   %). Tver does have an international airport (Migalovo), but most travelers arrive via the well-traveled railroad running between Saint Petersburg and nearby Moscow . Within the city limits of Tver, there are several small shuttle vans available and a dated, but generally reliable, electric trolley system, at nominal cost. Commercial taxicabs should be avoided, however, as they can be very expensive, especially if the driver discovers that the passengers are foreigners. There is also a bus service connecting Tver with Moscow. There is also rail service as Tver is a major stop on the regular trains that operate between Moscow and St Petersburg. Fast food restaurants have taken root in Tver, including fast-food chicken and hamburger outlets. Prices are reasonable in most cases. Some of the independent restaurants tend to be a bit pricey and intentional overcharges often occur, especially if foreigners are the customers. There are a few nightclubs and restaurants that serve liquor in Tver; however, some of them has cover charges. Novgorod is an easy and rewarding destination to get to from Tver Oblast, which could be done as a day trip. This travel guide to is an and may need more content. It has a , but there is not enough present. If there are and listed, they may not all be at status or there may not be a and a "Get in" section describing all of the typical ways to get here. Please and ! Has custom banner Has mapframe Has map markers Outline regions Outline articles Region articles Bottom-level regions Has Geo parameter Central Russia All destination articles Pages using the Kartographer extension Navigation menu 494 COMMENTS 6.1: Simple and Compound Interest Interest rates are usually given as an annual percentage rate (APR) - the total interest that will be paid in the year. If the interest is paid in smaller time increments, the APR will be divided up. For example, a 6% 6 % APR paid monthly would be divided into twelve 0.5% 0.5 % payments. A 4% 4 % annual rate paid quarterly would be divided ... PDF GRADE 11 SUBJECT Mathematical Literacy WEEK 1 TOPIC Finance ... •Lender - the person/institution who lends money to the borrower. The money must be paid back. • Loan term - the period over which the money may be paid back • Interest is the amount of money paid in return for the use of someone else's money. • Interest rate is the rate at which interest is paid by a borrower for the use of money that they borrow from a lender. 6.8 The Basics of Loans Deposits at the bank are used by the bank to generate loans. The bank has to pay those depositors interest. The bank must charge more for loans they give than they pay to people with deposits in the bank. Banks also borrow money from each other. These loans have an interest rate, and once more, the bank making a loan must make a profit. 6.4 Compound Interest This module covers the mathematics of compound interest. Understand and Compute Compound Interest. As we saw in Simple Interest, an account that pays simple interest only pays based on the original principal and the term of the loan. Accounts offering compound interest pay interest at regular intervals. After each interval, the interest is ... PDF Finance Amortized Loans MAT Example 1. Find the monthly payment and total interest paid for a simple interest amor-tized loan of $15,000 at an annual interest rate of 63 8 % for 8 years. Solution: For this problem we are given the loan amount (15000), the interest rate (0.06375 in decimal form), the compounding period (monthly or 12 periods per year), and nally the time ... 4: Mathematics of Finance 4.1: Simple Interest and Discount. It costs to borrow money. The rent one pays for the use of money is called the interest. The amount of money that is being borrowed or loaned is called the principal or present value. Simple interest is paid only on the original amount borrowed. Annuities and Loans 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. 4.1 Loan Amortization: Calculator Approach Samuel took out a $308,000 mortgage to buy an apartment. The mortgage is structured to be repaid with monthly payments of $2,375.11 at the end of each month. The interest rate on the mortgage is 4.62%, compounded monthly, and the loan is amortized over 15 years. a) Calculate the total principal amount repaid in the 8th year. PDF Essential Mathematics 2019 v1 181238. Essential Mathematics 2019 v1.1. IA3 high-level annotated sample response. September 2018. Problem-solving and modelling task. This sample has been compiled by the QCAA to assist and support teachers to match evidence in student responses to the characteristics described in the instrument -specific standards. Assessment objectives. 6.4: Loans You can afford a $11, 120 $ 11, 120 loan. 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, 120 = $880 $ 12, 000 − $ 11, 120 = $ 880 interest total. Mathematics Advanced: Financial Mathematics Compound interest is found by A n = P ( 1 + R) n, a Geometric Progression. A n is the amount of interest after n units of time. To find the interest (without the initial amount), subtract P from A n. Depreciation is a form of compound in terest, where the value decreases over time. Depreciation is expressed as A n = P ( 1 − R) n (also a ... PDF WORKSHEET ­ General 2 Mathematics WORKSHEET General 2 Mathematics. Topic Areas: Financial Mathematics. FM4 - Credit and Borrowing. Teacher: PETER HARGRAVES. nutes Worked Solutions: IncludedNote: Each. question has designated marks. Use this information as both a guide to the question's difficulty and as a timing indicator, whereby each mark should equate to 1.5 minutes. 6.9 Understanding Student Loans Students will pay them back and will pay interest. In the fall of 2022, the federal student loan interest rate was 4.99%. Private student loans rates ranged between 3.22% and 13.95%. Finding the lowest interest rate you can helps with the payments, and especially helps if the loan is not federally subsidized. ... Contemporary Mathematics ... Amortization calculator What is the original value of a mortgage if it is amortized over 10 months at 7% interest and $25.3 monthly payments? example 3: ex 3: You wish to take out a $50000 mortgage with monthly payments of 4.5% , and you can afford $550 per month. financial mathematics Study with Quizlet and memorize flashcards containing terms like lesson 21, rachel's income for july is $3300. she spent $1000 for rent, $150 on student loans, $100 for a new tablet, and $450 on groceries. how much of her monthly income is left?, the expression 12x + 4y + 365z represents irma's yearly expenses. she pays for rent every month, a student loan every three months, and bus service ... Loan Interest Calculator Follow the below steps to calculate loan interest. Calculate the periodic rate ( i) by dividing the annual interest rate by the number of payments in a year ( n ). Calculate the total payment ( P) by multiplying the periodic rate ( i) with the loan amount ( A) and the number of payment ( n) and then divide it by the factor of 1 - (1 + i)-n. 8.5: Amortized Loans Math for Liberal Arts 8: Consumer Mathematics 8.5: Amortized Loans Expand/collapse global location ... The monthly payment for a 30-year loan at 6% interest is $1498.88. Now calculate the total cost of the loan over the 30 years: The monthly payments are $1498.88 and the total cost of the loan is $539,596.80. Khan Academy Khan Academy Tver city, Russia travel guide Tver - Overview. Tver is a city in Russia located on the banks of the Volga River about 180 km north-west of Moscow, the administrative center of Tver Oblast. It is a large industrial, scientific, and cultural center, a major transport hub at the intersection of the St. Petersburg - Moscow railway line and the federal highway M10 "Russia" with the Upper Volga. Simple Interest Calculator Alternatively, you can use the simple interest formula I=Prn if you have the interest rate per month. If you had a monthly rate of 5% and you'd like to calculate the interest for one year, your total interest would be $10,000 × 0.05 × 12 = $6,000. The total loan repayment required would be $10,000 + $6,000 = $16,000. Tver Oblast Map Tver Oblast. Tver Oblast is a region in Central Russia, which borders Smolensk Oblast to the southwest, Pskov Oblast to the west, Novgorod Oblast to the north, Vologda Oblast to the northeast, Yaroslavl Oblast to the east, and Moscow Oblast to the southeast. Photo: Belliy, CC BY-SA 4.0. Photo: Florstein, CC BY-SA 3.0. Tver Oblast Map of Tver Oblast. 56.857828 35.921928. 1 Tver — the capital and only major city of the region; an ancient city with a prestigious past that once contended with Moscow for control of Russia, that is now a shell of its former self following widespread destruction from cultural vandalism at the hands of the Soviet government and the Nazi ... assignment essay homework phd report resume review speech thesis