Discrete and Financial Mathematics

Time Value of Money and Compound Interest

Money has a time value. This means that a dollar today is worth more than a dollar received in the future, because money today can be invested and can earn interest.

This idea is one of the most important parts of financial mathematics. It helps us compare savings plans, loans, investments, and payments made at different times.

In this lesson, you will learn how to:

• understand the time value of money,
• calculate future value,
• calculate present value,
• work with interest compounded at discrete time intervals, and
• work with continuous compounding.

1. Simple idea behind the time value of money

If you invest money in a bank account, the bank pays you interest. Because of this, the amount of money grows over time.

For example, if you invest \(1000\) dollars at \(5\%\) interest for one year, you will have more than \(1000\) dollars at the end of the year. So receiving \(1000\) dollars now is better than receiving \(1000\) dollars one year from now.

This is the key idea:

Money available now can grow, so it is worth more than the same amount received later.

2. Compound interest

With compound interest, interest is added to the original amount, and then future interest is calculated on the new total. This means you earn interest on both:

• the original principal, and
• the interest already earned.

This is why compound interest causes money to grow faster over time than simple interest.

Important terms

• Principal: the starting amount of money, usually written as \(P\)
• Interest rate: the annual percentage rate, usually written as \(r\) in decimal form
• Time: how long the money is invested or borrowed, usually written as \(t\) in years
• Compounding frequency: how many times per year interest is added, usually written as \(n\)
• Amount or future value: the total value after time has passed, usually written as \(A\) or \(FV\)

3. Formula for compound interest with discrete compounding

When interest is compounded a fixed number of times per year, we use:

where:

• \(A\) = future value
• \(P\) = principal
• \(r\) = annual interest rate as a decimal
• \(n\) = number of compounding periods per year
• \(t\) = time in years

Common compounding frequencies

• Annually: \(n=1\)
• Semiannually: \(n=2\)
• Quarterly: \(n=4\)
• Monthly: \(n=12\)
• Daily: \(n=365\) (sometimes \(360\) is used in finance)

4. Future value

The future value tells us how much an investment will grow to after earning compound interest.

If you know the starting amount and the interest details, use the compound interest formula directly.

Worked Example 1: Future value with annual compounding

Suppose \(P=2000\), the annual interest rate is \(6\%\), compounded annually, for \(3\) years. Find the future value.

Step 1: Write the known values

• \(P=2000\)
• \(r=0.06\)
• \(n=1\)
• \(t=3\)

Step 2: Substitute into the formula

Answer: The future value is approximately \(\$2382.03\).

Worked Example 2: Future value with monthly compounding

Find the future value of \(\$5000\) invested at \(4.8\%\) annual interest compounded monthly for \(5\) years.

Step 1: Identify the values

• \(P=5000\)
• \(r=0.048\)
• \(n=12\)
• \(t=5\)

Step 2: Use the formula

Answer: The investment grows to approximately \(\$6352.45\).

Notice that monthly compounding means interest is added more often, so the amount grows a little more than if it were compounded only once per year.

5. Present value

The present value is the amount you would need now to reach a certain future amount later.

This is like working backward through compound interest.

Starting from

we solve for \(P\):

This formula is very useful when comparing offers or deciding how much money a future payment is worth today.

Worked Example 3: Present value with quarterly compounding

You want to have \(\$10{,}000\) in \(4\) years. If the account pays \(5\%\) interest compounded quarterly, how much should you invest now?

Step 1: Identify the values

• \(A=10000\)
• \(r=0.05\)
• \(n=4\)
• \(t=4\)

Step 2: Substitute into the present value formula

Answer: You should invest approximately \(\$8200.80\) now.

This makes sense: if your money grows over time, you need less than \(\$10{,}000\) today to end up with \(\$10{,}000\) in the future.

6. Continuous compounding

Sometimes interest is compounded continuously. This means the compounding happens constantly, rather than at separate times like monthly or quarterly.

For continuous compounding, the formula becomes:

where \(e\) is a mathematical constant approximately equal to \(2.718\).

If you need present value with continuous compounding, solve for \(P\):

Worked Example 4: Continuous compounding

Find the future value of \(\$3000\) invested at \(7\%\) annual interest compounded continuously for \(6\) years.

Step 1: Identify the values

• \(P=3000\)
• \(r=0.07\)
• \(t=6\)

Step 2: Use the continuous compounding formula

Answer: The future value is approximately \(\$4565.89\).

7. Comparing compounding methods

If the interest rate and time are the same, then more frequent compounding gives a larger future value.

In general, the order from smallest growth to greatest growth is:

• annual compounding,
• semiannual compounding,
• quarterly compounding,
• monthly compounding,
• daily compounding,
• continuous compounding.

However, the differences become smaller and smaller as compounding becomes more frequent. Continuous compounding gives the maximum possible growth for a fixed annual rate.

8. How to decide whether to use future value or present value

• Use future value when you know the amount now and want to know how much it will become later.
• Use present value when you know a future amount and want to know what it is worth today.

A useful question to ask is:

• Am I moving forward in time? Use future value.
• Am I moving backward in time? Use present value.

9. Common mistakes to avoid

• Not changing the percent to a decimal: \(5\% = 0.05\), not \(5\).
• Using the wrong value of \(n\): monthly means \(12\), quarterly means \(4\).
• Forgetting that \(t\) is in years: convert months to years if needed.
• Mixing up present value and future value: check whether you are solving for the amount now or later.
• Using the discrete formula for continuous compounding: if compounding is continuous, use \(A=Pe^{rt}\).

10. Quick problem-solving steps

1. Read the question carefully.
2. Identify whether it asks for future value or present value.
3. Write down the known values: \(P\), \(A\), \(r\), \(n\), and \(t\).
4. Choose the correct formula.
5. Substitute carefully.
6. Use correct order of operations.
7. Round money answers to the nearest cent unless told otherwise.

11. Why this matters in real life

Time value of money appears in many real financial decisions:

• savings accounts,
• retirement planning,
• investments,
• loans,
• credit cards, and
• comparing payment options.

For example, if someone offers you \(\$900\) today or \(\$1000\) in two years, the better choice depends on interest rates. Present value helps you compare those options fairly.

Brief Summary

The time value of money means money today is worth more than the same amount in the future because it can earn interest.

For discrete compound interest, use

and for present value use

For continuous compounding, use

and

Once you can identify what each variable means and whether the question moves forward or backward in time, you can solve many financial mathematics problems with confidence.

Effective Annual Rate (APY) helps us compare savings accounts, investments, and loans fairly when interest is compounded at different times.

Two financial products may both advertise an annual interest rate, but if one compounds monthly and the other compounds daily, they do not actually grow by the same amount over a year. APY fixes this by showing the actual yearly percentage growth after compounding is included.

In this lesson, you will learn what APY means, how to calculate it, why it matters, and how to use it to compare financial choices.

1. Review: What is compound interest?

Compound interest means interest is added to the balance, and then future interest is calculated on the new, larger balance.

For example, if money compounds monthly, then each month interest is added. The next month, interest is earned on both the original amount and the interest already added.

This is different from simple interest, where interest is only calculated on the original principal.

The compound interest formula is

$$A = P\left(1+\frac{r}{n}\right)^{nt}$$

where:

• 
  A = final amount
• 
  P = principal (starting amount)
• 
  r = nominal annual interest rate written as a decimal
• 
  n = number of compounding periods per year
• 
  t = time in years

2. What is APY?

APY stands for Annual Percentage Yield. It is also called the effective annual rate.

APY tells you the real percent increase in one year when compounding is taken into account.

If a bank says the nominal rate is 6% compounded monthly, the APY will be slightly more than 6% because interest is being added during the year.

The formula for APY is

$$\text{APY} = \left(1+\frac{r}{n}\right)^n - 1$$

where:

• 
  r = nominal annual rate as a decimal
• 
  n = number of compounding periods per year

To write APY as a percent, multiply the decimal result by 100.

3. Why APY matters

APY allows you to compare financial products equitably, meaning fairly, even when they use different compounding schedules.

For example:

• Account A: 5.8% compounded daily
• Account B: 5.85% compounded quarterly

You cannot just compare 5.8% and 5.85% directly and assume B is better. The compounding frequency changes the actual yearly growth.

APY puts both accounts into the same form: actual growth over one year.

4. Common compounding periods

• Annually: \(n=1\)
• Semiannually: \(n=2\)
• Quarterly: \(n=4\)
• Monthly: \(n=12\)
• Daily: \(n=365\)

In general, the more often interest is compounded, the higher the APY, assuming the same nominal rate.

5. Important idea: nominal rate vs. APY

The nominal rate is the stated yearly rate before compounding is taken into account.

The APY is the actual yearly yield after compounding is included.

So if the nominal rate is 8% compounded monthly, the APY is a little more than 8%.

If interest is compounded only once per year, then the nominal rate and APY are the same.

6. Worked Example 1: Find APY from a nominal rate compounded quarterly

A savings account advertises 6% annual interest compounded quarterly. Find the APY.

Step 1: Identify the values.

• \(r = 0.06\)
• \(n = 4\)

Step 2: Use the APY formula.

$$\text{APY} = \left(1+\frac{0.06}{4}\right)^4 - 1$$

$$\text{APY} = (1.015)^4 - 1$$

$$\text{APY} \approx 1.06136 - 1 = 0.06136$$

Step 3: Convert to a percent.

$$0.06136 = 6.136\%$$

Answer: The APY is approximately 6.14%.

This means the account really grows by about 6.14% over one year.

7. Worked Example 2: Compare two accounts fairly

Which account gives a better yearly return?

• Account A: 5.9% compounded monthly
• Account B: 6.0% compounded semiannually

Find APY for Account A.

$$\text{APY}_A = \left(1+\frac{0.059}{12}\right)^{12}-1$$

$$\text{APY}_A \approx (1.0049167)^{12}-1 \approx 0.06061$$

$$\text{APY}_A \approx 6.061\%$$

Find APY for Account B.

$$\text{APY}_B = \left(1+\frac{0.06}{2}\right)^2-1$$

$$\text{APY}_B = (1.03)^2-1 = 1.0609-1 = 0.0609$$

$$\text{APY}_B = 6.09\%$$

Compare.

Account B has an APY of 6.09%, which is slightly higher than Account As 6.061%.

Answer: Account B gives the better yearly return, even though the two advertised rates are very close.

8. Worked Example 3: Use APY to find actual balance after one year

You deposit \(\$2{,}500\) into an account with a nominal rate of 4.8% compounded monthly. Find:

1. the APY
2. the balance after one year

Step 1: Find the APY.

$$\text{APY} = \left(1+\frac{0.048}{12}\right)^{12}-1$$

$$\text{APY} = (1.004)^{12}-1 \approx 0.04907$$

$$\text{APY} \approx 4.907\%$$

Step 2: Use APY to find one-year growth.

Since APY is the actual one-year growth rate,

$$A = P(1+\text{APY})$$

$$A = 2500(1.04907)$$

$$A \approx 2622.68$$

Answer:

• APY \(\approx 4.91\%\)
• Balance after one year \(\approx \$2622.68\)

You could also find the balance directly with the compound interest formula. Both methods should agree for one year.

9. Worked Example 4: Comparing a monthly account and a daily account

Which is better for one year?

• Option 1: 5.4% compounded monthly
• Option 2: 5.35% compounded daily

Option 1:

$$\text{APY}_1 = \left(1+\frac{0.054}{12}\right)^{12}-1$$

$$\text{APY}_1 = (1.0045)^{12}-1 \approx 0.05533$$

$$\text{APY}_1 \approx 5.533\%$$

Option 2:

$$\text{APY}_2 = \left(1+\frac{0.0535}{365}\right)^{365}-1$$

$$\text{APY}_2 \approx 0.05496$$

$$\text{APY}_2 \approx 5.496\%$$

Compare.

Even though Option 2 compounds more often, Option 1 has the higher APY because its nominal rate is high enough to make up for the less frequent compounding.

Answer: Option 1 is better.

10. How to solve APY questions step by step

1. Write the nominal rate as a decimal.
2. Identify the compounding frequency \(n\).
3. Substitute into $$\text{APY} = \left(1+\frac{r}{n}\right)^n - 1$$
4. Calculate carefully.
5. Convert the result to a percent.
6. Use APY values to compare accounts fairly.

11. APY and decision-making

In financial mathematics, you often have to choose between different options. APY is useful because it gives a standard measure for yearly growth.

When comparing savings accounts, certificates of deposit, or investment products with different compounding schedules, APY helps you make a better decision.

For borrowing, a related idea is also used to understand the true yearly cost of interest. In this lesson, we focus on APY as a way to measure actual yearly yield.

12. Common mistakes to avoid

• Using the percent instead of the decimal. For example, use \(0.07\), not \(7\).
• Forgetting to subtract 1. The formula gives a growth factor first, so you must subtract 1 to get the rate.
• Comparing nominal rates only. Always compare APY values if compounding frequencies differ.
• Using the wrong value for \(n\). Monthly is 12, quarterly is 4, daily is usually 365.

13. Quick check questions

Try these on your own:

• What is the APY for 7% compounded monthly?
• Which is better: 6.2% compounded quarterly or 6.15% compounded monthly?
• If an account has an APY of 5.12%, what does that mean in words?

14. Brief summary

APY, or effective annual rate, tells you the actual percent growth of money in one year after compounding is included.

The formula is

$$\text{APY} = \left(1+\frac{r}{n}\right)^n - 1$$

It is important because it lets you compare different financial products fairly, even if they compound interest at different rates.

Whenever you are choosing between accounts or investments, compare the APY, not just the advertised nominal rate.

Lesson: Annuities and Sinking Funds

When money is deposited or paid regularly, and interest is added over time, we use the ideas of annuities and sinking funds to model what happens.

These ideas are important in financial mathematics because they help answer questions like:

• How much will I have if I save a fixed amount every month?
• How much should I deposit regularly to reach a future goal?
• How do retirement savings grow over time?
• How can a business or person build up a target fund for a future expense?

This lesson focuses on the geometric growth created by regular deposits and compound interest.

1. What is an annuity?

An annuity is a sequence of equal payments or deposits made at regular time intervals.

Examples include:

• Depositing \(R500\) every month into a savings account
• Paying a loan installment every month
• Contributing regularly to a retirement account

In this lesson, we will mainly look at an annuity where regular deposits are made into an account that earns compound interest.

2. What is a sinking fund?

A sinking fund is a fund built up by making regular deposits so that a specific amount of money is available in the future.

For example, if you want \(R100\,000\) in 10 years for a car, university fees, or a business cost, you can make regular deposits into a sinking fund.

So, the difference in focus is:

• Annuity: often asks, “How much will my regular deposits grow to?”
• Sinking fund: often asks, “What regular deposit is needed to reach a target amount?”

3. Why is this geometric?

Each deposit earns compound interest for a different amount of time.

The first deposit stays in the account the longest, so it earns the most interest. The last deposit stays in the account for the shortest time, so it earns the least interest.

This creates a geometric sequence of values at the end of the investment period.

Suppose you deposit \(P\) at the end of every period, and the interest rate per period is \(i\). After \(n\) deposits, the future value is:

This is a geometric series with:

• first term \(1\)
• common ratio \((1+i)\)
• \(n\) terms

Using the sum of a geometric series, we get the formula:

This is the future value of an ordinary annuity.

4. Ordinary annuity

An ordinary annuity means deposits are made at the end of each payment period.

This is the most common form used in school financial mathematics.

The formula is:

where:

• \(S\) = future value of the annuity
• \(P\) = deposit made each period
• \(i\) = interest rate per period
• \(n\) = total number of deposits

Important: The deposit frequency and interest compounding period must match.

For example:

• monthly deposits require a monthly interest rate
• quarterly deposits require a quarterly interest rate

If the annual nominal rate is \(r\), compounded \(m\) times per year, then:

and if the money is deposited for \(t\) years, then:

5. Sinking fund formula

If you know the target future amount \(S\) and want to find the regular deposit \(P\), rearrange the annuity formula:

So:

This is the main sinking fund formula.

It tells you how much to deposit each period to reach a future goal.

6. Steps for solving annuity and sinking fund problems

1. Identify what is given: regular deposit, future value, annual rate, compounding frequency, and time.
2. Convert the annual rate to the rate per period: \(i = \frac{r}{m}\).
3. Find the total number of deposits: \(n = mt\).
4. Choose the correct formula:

• For future value of regular deposits: \(S = P\left(\frac{(1+i)^n - 1}{i}\right)\)
• For required deposit to reach a target: \(P = S\left(\frac{i}{(1+i)^n - 1}\right)\)

5. Substitute carefully.
6. Use brackets on your calculator.
7. Round money values appropriately, usually to the nearest cent.

7. Worked Example 1: Finding the future value of an annuity

Question: A student deposits \(R800\) at the end of every month into a savings account that earns \(9\%\) per year, compounded monthly. How much will be in the account after 5 years?

Step 1: Identify values

• \(P = 800\)
• \(r = 0.09\)
• \(m = 12\)
• \(t = 5\)

Step 2: Find rate per month

Step 3: Find number of deposits

Step 4: Use the annuity formula

Answer: The account will contain approximately \(R60\,339.20\).

Meaning: Even though the student deposited \(60 \times 800 = R48\,000\), the final amount is larger because of compound interest.

8. Worked Example 2: Sinking fund deposit needed

Question: A person wants to have \(R150\,000\) in 8 years. The account earns \(10.8\%\) per year, compounded monthly. How much must be deposited at the end of each month?

Step 1: Identify values

• \(S = 150\,000\)
• \(r = 0.108\)
• \(m = 12\)
• \(t = 8\)

Step 2: Find monthly interest rate

Step 3: Find number of monthly deposits

Step 4: Use the sinking fund formula

Now calculate:

Answer: The person must deposit approximately \(R989.70\) per month.

9. Worked Example 3: Quarterly deposits

Question: A company places \(R5\,000\) into a maintenance fund at the end of each quarter. The fund earns \(8\%\) per year, compounded quarterly. Find the value of the fund after 6 years.

Step 1: Identify values

• \(P = 5\,000\)
• \(r = 0.08\)
• \(m = 4\)
• \(t = 6\)

Step 2: Find rate per quarter

Step 3: Find number of quarters

Step 4: Apply formula

Answer: The fund will be worth approximately \(R152\,110.00\).

10. Worked Example 4: Retirement savings model

Question: A worker saves \(R2\,500\) at the end of every month for 20 years in a retirement account earning \(12\%\) per year, compounded monthly. Find the future value.

Step 1: Values

• \(P = 2\,500\)
• \(r = 0.12\)
• \(m = 12\)
• \(t = 20\)

Step 2: Monthly rate and total deposits

Step 3: Apply formula

Answer: The retirement account will grow to about \(R2\,473\,075.00\).

Interpretation: Long-term saving is powerful because many deposits are made, and each deposit earns compound interest. Time makes a very big difference.

11. Comparing total deposits and future value

It is useful to compare:

• Total amount deposited = \(P \times n\)
• Future value = amount in the fund after interest

In retirement or long-term savings problems, the future value can be much larger than the total deposits because of compound interest.

12. Common mistakes to avoid

• Using the annual rate directly instead of converting to the rate per payment period
• Using the wrong number of periods, for example using years instead of months
• Mixing deposit intervals and compounding intervals
• Forgetting that this formula is for end-of-period deposits
• Calculator errors from missing brackets

13. Annuities and decision-making

These calculations help people make financial decisions.

For example, a learner or family may compare:

• saving \(R1\,000\) per month for 10 years
• saving \(R1\,500\) per month for 7 years

By calculating the future value of both annuities, they can decide which savings plan better meets a financial goal.

Similarly, if a target amount is fixed, the sinking fund formula can show whether the required monthly deposit is affordable.

14. Key formulas

Future value of an ordinary annuity:

Sinking fund deposit:

Rate per period:

Number of periods:

15. Brief summary

An annuity involves equal deposits made at regular intervals, with compound interest causing the deposits to grow as a geometric series.

A sinking fund is a savings plan designed to reach a specific future amount through regular deposits.

To solve these problems, always match the deposit period with the interest period, calculate \(i\) and \(n\) correctly, and then use the correct formula.

These ideas are especially useful for understanding retirement savings, education funds, emergency funds, and other long-term financial goals.

Amortization Schedules help us understand exactly how a loan is paid off over time. When you borrow money for something like a car, education, or a house, you usually repay it in equal installments. Even though the payment amount may stay the same, the part that goes to interest and the part that goes to principal change each time.

An amortization schedule is a table that shows this breakdown for every payment. It tells you:

• the payment number,
• the total payment amount,
• how much of that payment is interest,
• how much reduces the principal, and
• the remaining loan balance after the payment.

This topic is important because it connects compound interest to real-life borrowing decisions. It helps you see how loans really work and why the total amount paid can be much more than the amount borrowed.

Key vocabulary:

• Principal: the amount originally borrowed.
• Interest: the cost of borrowing money.
• Installment: each regular payment.
• Balance: the amount still owed.
• Amortization: paying off a loan gradually through regular payments.

Suppose a loan has principal amount \(P\), interest rate per payment period \(r\), and total number of payments \(n\). If the loan is fully amortized with equal payments, the payment amount \(A\) is:

This formula is used when interest is charged once per payment period. For example:

• monthly payments use a monthly interest rate,
• quarterly payments use a quarterly interest rate.

If the annual interest rate is \(i\) and payments are made monthly, then the periodic rate is usually:

To build an amortization schedule, we repeat the same steps for each payment.

1. Start with the current balance.
2. Calculate the interest for the period:
   \(\text{Interest} = \text{Balance} \times r\)
3. Calculate the principal repaid:
   \(\text{Principal repaid} = \text{Payment} - \text{Interest}\)
4. Update the balance:
   \(\text{New balance} = \text{Old balance} - \text{Principal repaid}\)
5. Repeat until the balance reaches 0.

A very important pattern appears in amortized loans:

• At the beginning, a larger part of each payment goes to interest.
• Later, as the balance becomes smaller, the interest part gets smaller.
• So a larger part of each later payment goes to principal.

This happens because interest is always calculated from the remaining balance. As the balance drops, the interest charged also drops.

Let us work through some examples.

Worked Example 1: One-step breakdown of a loan payment

A student borrows \(\$10{,}000\) at an annual interest rate of \(6\%\), with monthly payments over 5 years. Find the monthly payment and break down the first payment into interest and principal.

Step 1: Identify the values.

• \(P = 10{,}000\)
• Annual rate \(= 6\% = 0.06\)
• Monthly rate \(r = 0.06/12 = 0.005\)
• Number of payments \(n = 5 \times 12 = 60\)

Step 2: Use the payment formula.

Now calculate:

So the monthly payment is about \(\$193.33\).

Step 3: Find the interest in the first payment.

Step 4: Find the principal repaid in the first payment.

Step 5: Find the new balance.

So in the first payment:

• Payment = \(\$193.33\)
• Interest = \(\$50.00\)
• Principal = \(\$143.33\)
• Remaining balance = \(\$9,856.67\)

Worked Example 2: First three rows of an amortization schedule

Using the same loan from Example 1, create the first three rows of the amortization schedule.

We already found the payment amount: \(\$193.33\).

Payment 1

• Starting balance: \(\$10,000.00\)
• Interest: \(10{,}000.00 \times 0.005 = 50.00\)
• Principal: \(193.33 - 50.00 = 143.33\)
• New balance: \(10{,}000.00 - 143.33 = 9{,}856.67\)

Payment 2

• Starting balance: \(\$9,856.67\)
• Interest: \(9{,}856.67 \times 0.005 \approx 49.28\)
• Principal: \(193.33 - 49.28 = 144.05\)
• New balance: \(9{,}856.67 - 144.05 = 9{,}712.62\)

Payment 3

• Starting balance: \(\$9,712.62\)
• Interest: \(9{,}712.62 \times 0.005 \approx 48.56\)
• Principal: \(193.33 - 48.56 = 144.77\)
• New balance: \(9{,}712.62 - 144.77 = 9{,}567.85\)

These rows can be shown in a table:

Notice the pattern:

• interest goes down: \(50.00, 49.28, 48.56\)
• principal goes up: \(143.33, 144.05, 144.77\)

This is the main idea of amortization.

Worked Example 3: Total interest paid over the whole loan

For the same \(\$10,000\) loan, how much total interest is paid over 5 years?

Step 1: Find total amount paid.

Step 2: Subtract the amount borrowed.

So the borrower pays about \(\$1,599.80\) in interest over the life of the loan.

This is why a longer repayment time can make a loan more expensive, even if the monthly payment feels easier to manage.

Worked Example 4: Comparing two loan choices

A borrower needs \(\$8,000\). They are considering these two options:

• Option A: 4 years at \(5\%\) annual interest, monthly payments
• Option B: 6 years at \(5\%\) annual interest, monthly payments

Which option has the lower monthly payment, and which option costs more overall?

Option A

• \(P = 8{,}000\)
• \(r = 0.05/12 \approx 0.004167\)
• \(n = 48\)

Monthly payment for Option A is about \(\$184.21\).

Total paid:

Total interest:

Option B

• \(P = 8{,}000\)
• \(r = 0.05/12 \approx 0.004167\)
• \(n = 72\)

Monthly payment for Option B is about \(\$128.85\).

Total paid:

Total interest:

Conclusion:

• Option B has the lower monthly payment.
• Option B also has the higher total interest.
• Option A costs less overall, but the monthly payment is larger.

This kind of comparison is one of the most useful reasons for studying amortization schedules. They help people make informed financial choices.

Important ideas to remember

• The payment in a fully amortized loan is usually constant.
• The interest part changes every time because it depends on the remaining balance.
• Early payments mostly cover interest.
• Later payments mostly reduce principal.
• A longer loan term lowers the payment but increases total interest.

Common mistakes students make

• Using the annual rate instead of the periodic rate.
  You must match the interest rate to the payment period.
• Forgetting to update the balance after each payment.
  Each new interest amount must be based on the new balance, not the original principal.
• Mixing up interest and principal.
  Remember: \(\text{Principal repaid} = \text{Payment} - \text{Interest}\).
• Assuming equal payments mean equal principal amounts.
  They do not. The principal part changes over time.

How to read an amortization schedule

If you are given a schedule, you should be able to answer questions like:

• How much interest is paid in the 10th payment?
• What is the balance after 2 years?
• How much principal has been repaid after 12 payments?
• How much total interest will be paid by the end of the loan?

To find principal repaid after several payments, subtract the current balance from the original principal:

To find total interest paid so far, add the interest entries up to that point, or use:

Final Summary

An amortization schedule shows how each loan payment is divided into interest and principal over time. The payment amount often stays the same, but the interest part gets smaller and the principal part gets larger as the balance decreases.

To create a schedule, calculate interest from the current balance, subtract that from the payment to find principal repaid, and then update the balance. This process repeats for every payment until the loan is fully paid off.

Understanding amortization schedules helps you compare loans, predict total interest paid, and make better financial decisions.

Introduction to Graph Theory

Graph theory is a branch of mathematics used to study networks and relationships. In graph theory, we model a situation using points and connections. This helps us understand systems such as road maps, airline routes, social networks, computer networks, and delivery paths.

Even though the word graph may make you think of coordinate axes, graph theory is different. Here, a graph is a collection of vertices and edges. A vertex represents an object or location, and an edge represents a connection between two vertices.

This topic is important in discrete mathematics because many real-life problems can be turned into network problems. Once a situation is represented as a graph, we can study routes, connections, loops, and efficient ways to move through the network.

1. Basic Ideas and Vocabulary

A graph consists of:

• Vertices (also called nodes): the points in the graph
• Edges: the lines or links joining pairs of vertices

We often write a graph as:

$$G = (V, E)$$

where:

• \(V\) is the set of vertices
• \(E\) is the set of edges

For example, if a graph has vertices \(A, B, C\) and edges joining \(A\) to \(B\) and \(B\) to \(C\), then:

$$V = \{A, B, C\}, \qquad E = \{AB, BC\}$$

2. Vertices and Edges in Context

To build a graph, we first decide what the vertices and edges represent.

• In a road network, vertices may be towns and edges may be roads.
• In a friendship network, vertices may be people and edges may show friendships.
• In a flight network, vertices may be airports and edges may be direct flights.

This is called modelling. A graph is a mathematical model of a real situation.

3. Adjacent Vertices

Two vertices are called adjacent if an edge joins them directly. For example, if there is an edge from \(A\) to \(B\), then \(A\) and \(B\) are adjacent.

Similarly, two edges are adjacent if they share a common vertex.

4. Degree of a Vertex

The degree of a vertex is the number of edges connected to it. It tells us how many direct connections that vertex has.

We write the degree of a vertex \(A\) as \(\deg(A)\).

For example, if vertex \(A\) is connected to \(B\), \(C\), and \(D\), then:

$$\deg(A) = 3$$

A vertex with degree 0 is called an isolated vertex because it has no connections.

5. Path

A path is a sequence of vertices connected by edges. It shows a way to move through the graph.

For example, if edges \(AB\), \(BC\), and \(CD\) exist, then:

$$A \to B \to C \to D$$

is a path from \(A\) to \(D\).

The length of a path is the number of edges used. The path \(A \to B \to C \to D\) has length 3.

6. Closed Circuit

A closed circuit is a path that starts and ends at the same vertex. In simple language, it is a loop in the graph.

For example, if the edges \(AB\), \(BC\), and \(CA\) are present, then:

$$A \to B \to C \to A$$

is a closed circuit.

Closed circuits are useful when studying delivery routes, inspection paths, or systems where you return to your starting point.

7. Connected and Disconnected Graphs

A graph is connected if there is a path between every pair of vertices. This means every part of the graph can be reached from every other part.

A graph is disconnected if some vertices cannot be reached from others.

In a transport network, a connected graph means all locations are reachable. In a disconnected graph, some places are cut off from the rest.

8. Why Graph Theory Matters

Graph theory helps us answer important questions such as:

• Can one place be reached from another?
• What route connects several locations?
• Does a network contain a loop?
• Which points have the most direct connections?
• Is the whole network connected?

These ideas are used in transport, communication, business planning, and technology.

Worked Example 1: Identifying Vertices and Edges

A school wants to model direct bus routes between four stops: \(A, B, C, D\). The direct routes are:

• \(A\) to \(B\)
• \(A\) to \(C\)
• \(B\) to \(D\)
• \(C\) to \(D\)

Step 1: List the vertices.

These are the bus stops:

$$V = \{A, B, C, D\}$$

Step 2: List the edges.

These are the direct routes:

$$E = \{AB, AC, BD, CD\}$$

Step 3: Interpret the graph.

This graph shows which stops are directly connected. For example, \(A\) is directly connected to \(B\) and \(C\), but not directly to \(D\).

Answer: The graph is defined by \(V = \{A, B, C, D\}\) and \(E = \{AB, AC, BD, CD\}\).

Worked Example 2: Finding Degrees

Use the graph from Example 1.

$$E = \{AB, AC, BD, CD\}$$

Find the degree of each vertex.

Step 1: Count edges touching each vertex.

• Vertex \(A\) touches \(AB\) and \(AC\), so \(\deg(A) = 2\)
• Vertex \(B\) touches \(AB\) and \(BD\), so \(\deg(B) = 2\)
• Vertex \(C\) touches \(AC\) and \(CD\), so \(\deg(C) = 2\)
• Vertex \(D\) touches \(BD\) and \(CD\), so \(\deg(D) = 2\)

Answer:

$$\deg(A)=2, \quad \deg(B)=2, \quad \deg(C)=2, \quad \deg(D)=2$$

This tells us each stop has exactly two direct connections.

Worked Example 3: Finding a Path and Its Length

Consider the graph with:

$$V = \{A, B, C, D, E\}$$

$$E = \{AB, BC, CD, DE, AC\}$$

Find one path from \(A\) to \(E\), and state its length.

Step 1: Look for connected edges from \(A\) to \(E\).

One possible path is:

$$A \to C \to D \to E$$

Step 2: Count the edges in the path.

• \(A \to C\)
• \(C \to D\)
• \(D \to E\)

So the path uses 3 edges.

Answer: One path is \(A \to C \to D \to E\), and its length is \(3\).

Worked Example 4: Identifying a Closed Circuit and Checking Connectivity

Consider the graph with vertices \(A, B, C, D\) and edges:

$$E = \{AB, BC, CA, CD\}$$

Part A: Is there a closed circuit?

The edges \(AB\), \(BC\), and \(CA\) form:

$$A \to B \to C \to A$$

This starts and ends at \(A\), so it is a closed circuit.

Part B: Is the graph connected?

Check whether every vertex can be reached from every other vertex.

• \(A\), \(B\), and \(C\) are connected in a triangle.
• \(D\) is connected to \(C\) by edge \(CD\).

Since \(D\) can be reached through \(C\), all vertices are connected.

Answer: The graph contains the closed circuit \(A \to B \to C \to A\), and the graph is connected.

9. Steps for Modelling a Real Situation as a Graph

When solving a graph theory problem, it helps to follow a clear process:

1. Decide what the vertices represent.
2. Decide what the edges represent.
3. Draw or describe the graph clearly.
4. Look for direct connections, paths, and closed circuits.
5. Use the graph to answer questions about the network.

10. Common Mistakes to Avoid

• Confusing graph theory with coordinate graphs: graph theory studies networks, not lines on axes.
• Forgetting what the vertices and edges represent: always connect the graph back to the real situation.
• Counting vertices instead of edges in a path: the length of a path is the number of edges.
• Assuming a graph is connected: always check whether every vertex can be reached.
• Missing a closed circuit: look for a route that returns to its starting point.

11. Quick Check Questions

Try these on your own:

• If a graph has vertices \(P, Q, R\) and edges \(PQ\) and \(QR\), is there a path from \(P\) to \(R\)?
• What is the degree of \(Q\) in that graph?
• Does the graph contain a closed circuit?

Answers:

• Yes, the path is \(P \to Q \to R\).
• \(\deg(Q) = 2\).
• No, because there is no loop returning to the starting point.

Brief Summary

Graph theory is used to model networks and relationships. A graph is made of vertices and edges. Important ideas include adjacent vertices, degree, paths, path length, closed circuits, and whether a graph is connected.

When you see a real-world network, think about what the points and connections represent. Then use graph theory to describe the structure clearly and answer questions about movement, links, and loops in the network.

Eulerian and Hamiltonian Paths are two important ideas in graph theory. They help us study routes through a network, such as roads between towns, delivery routes, or connections in a computer network.

A graph is made of vertices (points or nodes) and edges (lines connecting the points). In this lesson, we will learn how to tell whether a graph has a route that uses every edge exactly once, or a route that visits every vertex exactly once.

Although Eulerian and Hamiltonian paths sound similar, they focus on different things:

• Eulerian path: uses every edge exactly once.
• Hamiltonian path: visits every vertex exactly once.

Understanding the difference is the key to solving many graph theory questions.

1. Basic vocabulary

Before we begin, here are some terms you need to know.

• Vertex: a point in a graph.
• Edge: a line joining two vertices.
• Degree of a vertex: the number of edges connected to that vertex.
• Path: a route through a graph.
• Circuit or cycle: a path that starts and ends at the same vertex.

For Eulerian questions, the degree of each vertex is very important. For Hamiltonian questions, we focus more on whether a route can pass through every vertex once without repeating a vertex.

2. Eulerian paths and circuits

An Eulerian path is a path that travels along every edge exactly once. It does not have to end where it started.

An Eulerian circuit is an Eulerian path that starts and ends at the same vertex.

The main test for Eulerian paths depends on the number of vertices with odd degree.

Eulerian Path Rule

• If a connected graph has exactly 0 odd-degree vertices, then it has an Eulerian circuit.
• If a connected graph has exactly 2 odd-degree vertices, then it has an Eulerian path but not an Eulerian circuit.
• If a connected graph has more than 2 odd-degree vertices, then it has no Eulerian path.

So, to decide whether a graph is Eulerian, count the degree of each vertex and see how many are odd.

Why does odd degree matter?

Every time you enter a vertex along one edge, you usually need another edge to leave it. That uses edges in pairs. This means most vertices on an Eulerian route must have even degree.

If the path starts and ends at different vertices, then those two vertices are special. One has one extra “leaving” edge, and the other has one extra “arriving” edge. That is why exactly two odd-degree vertices allow an Eulerian path.

3. Hamiltonian paths and cycles

A Hamiltonian path visits every vertex exactly once.

A Hamiltonian cycle visits every vertex exactly once and returns to the starting vertex.

Unlike Eulerian paths, there is no simple degree-counting rule that always tells us whether a graph has a Hamiltonian path or cycle at this level. Usually, we test the graph carefully and try to build a route.

For Hamiltonian questions, ask yourself:

• Can I visit each vertex once without getting stuck?
• Are there vertices that force my route in a certain way?
• If I use one edge, will I block access to another vertex later?

Important difference:

• Eulerian problems are about using every edge once.
• Hamiltonian problems are about visiting every vertex once.

4. Comparing Eulerian and Hamiltonian ideas

It is possible for a graph to have:

• an Eulerian path but no Hamiltonian path,
• a Hamiltonian path but no Eulerian path,
• both,
• or neither.

This is because the two ideas measure different kinds of routes.

For example, a graph with many repeated connections might let you use every edge once, but visiting each vertex only once may be impossible. On the other hand, a graph may let you visit each vertex once, but some vertex degrees may stop it from being Eulerian.

5. Steps for solving Eulerian questions

1. Check that the graph is connected.
2. Find the degree of each vertex.
3. Count how many vertices have odd degree.
4. Use the Eulerian Path Rule to decide whether there is an Eulerian path or circuit.

6. Steps for solving Hamiltonian questions

1. List the vertices.
2. Try to create a route that visits each vertex once.
3. Watch for vertices with very limited connections, because they often determine where the path must go.
4. If you can return to the start after visiting every vertex once, then you have a Hamiltonian cycle.

Worked Example 1: Deciding if a graph is Eulerian

Suppose a graph has vertices A, B, C, and D with edges:

AB, BC, CD, DA

This forms a square.

First, find the degree of each vertex:

• \(\deg(A)=2\)
• \(\deg(B)=2\)
• \(\deg(C)=2\)
• \(\deg(D)=2\)

All vertices have even degree. There are 0 odd-degree vertices.

Therefore, the graph has an Eulerian circuit.

One Eulerian circuit is:

\(A \to B \to C \to D \to A\)

This route uses every edge exactly once and returns to the start.

Worked Example 2: Eulerian path but not circuit

Consider a graph with vertices A, B, C, D and edges:

AB, BC, CD

This is just a chain of 4 vertices.

Now find the degrees:

• \(\deg(A)=1\)
• \(\deg(B)=2\)
• \(\deg(C)=2\)
• \(\deg(D)=1\)

The odd-degree vertices are A and D. There are exactly 2 odd-degree vertices.

Therefore, the graph has an Eulerian path but not an Eulerian circuit.

One Eulerian path is:

\(A \to B \to C \to D\)

This uses every edge exactly once, but it does not end where it started.

Worked Example 3: Testing for a Hamiltonian path

Consider the same square graph with vertices A, B, C, D and edges:

AB, BC, CD, DA

Can we find a Hamiltonian path?

Yes. A Hamiltonian path must visit each vertex exactly once. One example is:

\(A \to B \to C \to D\)

This visits A, B, C, and D once each.

Can we find a Hamiltonian cycle?

Yes. Since the graph also has edge DA, we can return to the start:

\(A \to B \to C \to D \to A\)

This is a Hamiltonian cycle because every vertex is visited exactly once before returning to A.

Worked Example 4: A graph that is Hamiltonian but not Eulerian

Consider a graph with vertices A, B, C, D, E and edges:

AB, BC, CD, DE, EA, AC

First, test for a Hamiltonian cycle. The route

\(A \to B \to C \to D \to E \to A\)

visits every vertex once and returns to the start, so the graph is Hamiltonian.

Now test whether it is Eulerian by checking degrees:

• \(\deg(A)=3\) because A is connected to B, E, and C
• \(\deg(B)=2\)
• \(\deg(C)=3\)
• \(\deg(D)=2\)
• \(\deg(E)=2\)

The odd-degree vertices are A and C. There are exactly 2 odd-degree vertices.

So the graph has an Eulerian path, but not an Eulerian circuit.

This example shows that a graph can be Hamiltonian and also have only an Eulerian path, not an Eulerian circuit.

7. Common mistakes to avoid

• Mixing up edges and vertices. Eulerian means every edge once. Hamiltonian means every vertex once.
• Ignoring connectivity. If the graph is not connected, you cannot travel through the whole graph in one path.
• Repeating a vertex in a Hamiltonian path. A Hamiltonian path must visit each vertex exactly once.
• Using degree rules for Hamiltonian paths. The odd/even degree rule applies to Eulerian paths, not Hamiltonian paths.

8. Real-life meaning

Eulerian paths can model problems where every connection must be used once, such as inspecting every road in a neighborhood or tracing every wire in a network.

Hamiltonian paths can model problems where every location must be visited once, such as planning a trip through several cities without revisiting any city.

In decision-making and network optimization, these ideas help us choose efficient routes depending on whether the goal is to cover all connections or to visit all locations.

9. Quick check

When you see a graph, ask:

• Am I being asked about edges or vertices?
• If it is about edges, count odd-degree vertices.
• If it is about vertices, try building a route through each vertex once.

Summary

An Eulerian path uses every edge exactly once, and an Eulerian circuit also returns to the starting point. In a connected graph, 0 odd-degree vertices means an Eulerian circuit, and 2 odd-degree vertices means an Eulerian path.

A Hamiltonian path visits every vertex exactly once, and a Hamiltonian cycle returns to the starting point. There is no single easy rule for Hamiltonian paths, so you usually test possible routes carefully.

If you remember that Eulerian = edges and Hamiltonian = vertices, you will be able to start most problems correctly.

Trees and Minimum Spanning Trees

In graph theory, we often use a graph to model a network. A graph is made of vertices (also called nodes) and edges (the connections between nodes).

For example, vertices could represent towns, computers, or buildings, and edges could represent roads, cables, or pipelines. In many real-life problems, each edge has a weight, such as distance, cost, or time.

A common goal is to connect all points in the network as cheaply as possible. This is where trees and minimum spanning trees are useful.

1. What is a tree?

A tree is a connected graph with no circuits (no loops you can travel around and return to the starting point without repeating edges).

A graph is a tree if:

• every vertex is connected to the others, and
• there are no closed paths.

Important properties of a tree with \(n\) vertices are:

• it has exactly \(n-1\) edges,
• there is exactly one path between any two vertices,
• removing any edge disconnects the graph.

So if a graph has 6 vertices and is a tree, then it must have:

$$6-1=5 \text{ edges}$$

2. What is a spanning tree?

A spanning tree of a connected graph is a tree that includes all the vertices of the graph.

This means a spanning tree:

• connects every vertex,
• uses only some of the original edges if necessary,
• contains no circuits.

A single graph can have many different spanning trees.

3. What is a minimum spanning tree?

If the edges in a graph have weights, then different spanning trees can have different total weights.

A minimum spanning tree (MST) is the spanning tree with the smallest possible total edge weight.

If the weights represent cost, then the MST gives the cheapest way to connect the entire network without creating unnecessary circuits.

The total weight of a spanning tree is found by adding the weights of its edges.

For example, if a spanning tree uses edges with weights \(2, 3, 5, 6\), then its total weight is:

$$2+3+5+6=16$$

4. Why no circuits?

If a network contains a circuit, then at least one edge in that circuit can be removed and the graph will still remain connected. That means the extra edge is unnecessary.

Since we want the lowest total weight, we avoid circuits and keep only the edges needed to connect all vertices.

5. How to find a minimum spanning tree

Two common methods are Kruskal's algorithm and Prim's algorithm.

Both methods aim to connect all vertices with no circuits and the smallest total weight.

6. Kruskal's algorithm

Kruskal's algorithm builds the MST by choosing edges from smallest weight to largest weight.

Steps:

1. List all edges in order from smallest weight to largest weight.
2. Choose the smallest edge.
3. Keep adding the next smallest edge unless it creates a circuit.
4. Stop when all vertices are connected.

Key idea: Choose cheap edges first, but never form a circuit.

7. Prim's algorithm

Prim's algorithm starts at one vertex and grows one connected tree.

Steps:

1. Start at any vertex.
2. Add the smallest edge that connects a vertex already in the tree to a vertex not yet in the tree.
3. Repeat until all vertices are included.

Key idea: Grow the tree outward, always taking the cheapest possible next connection.

8. Worked Example 1: Is this graph a tree?

A graph has 5 vertices and 4 edges. It is connected and has no circuits.

Is it a tree?

Solution:

A graph is a tree if it is connected and has no circuits. This graph satisfies both conditions.

Also, for \(n=5\) vertices, a tree should have:

$$n-1=5-1=4 \text{ edges}$$

The graph has 4 edges, so this matches as well.

Conclusion: Yes, it is a tree.

9. Worked Example 2: Find the total weight of a spanning tree

Suppose a spanning tree uses these edges:

• \(AB=4\)
• \(BC=2\)
• \(CD=7\)
• \(DE=3\)

Find the total weight.

Solution:

Add the edge weights:

$$4+2+7+3=16$$

Conclusion: The total weight is \(16\).

10. Worked Example 3: Minimum spanning tree using Kruskal's algorithm

Consider a graph with vertices \(A, B, C, D\) and weighted edges:

• \(AB=2\)
• \(AC=5\)
• \(AD=6\)
• \(BC=1\)
• \(BD=4\)
• \(CD=3\)

Find a minimum spanning tree using Kruskal's algorithm.

Step 1: Order the edges from smallest to largest

• \(BC=1\)
• \(AB=2\)
• \(CD=3\)
• \(BD=4\)
• \(AC=5\)
• \(AD=6\)

Step 2: Add edges one at a time

Add \(BC=1\).

No circuit is formed.

Add \(AB=2\).

Still no circuit.

Add \(CD=3\).

This connects vertex \(D\) without making a circuit.

Now all 4 vertices are connected: \(A, B, C, D\).

Since a tree with 4 vertices needs \(4-1=3\) edges, we stop here.

MST edges:

• \(BC=1\)
• \(AB=2\)
• \(CD=3\)

Total weight:

$$1+2+3=6$$

Conclusion: The minimum spanning tree has total weight \(6\).

11. Worked Example 4: Minimum spanning tree using Prim's algorithm

Use the same graph:

• \(AB=2\)
• \(AC=5\)
• \(AD=6\)
• \(BC=1\)
• \(BD=4\)
• \(CD=3\)

Find an MST using Prim's algorithm, starting at \(A\).

Step 1: Start at \(A\)

The edges from \(A\) are:

• \(AB=2\)
• \(AC=5\)
• \(AD=6\)

Choose the smallest: \(AB=2\).

Tree so far: \(A, B\)

Step 2: Look for the smallest edge leaving the current tree

Available edges from \(A\) or \(B\) to outside vertices are:

• \(AC=5\)
• \(AD=6\)
• \(BC=1\)
• \(BD=4\)

Choose the smallest: \(BC=1\).

Tree so far: \(A, B, C\)

Step 3: Continue

Available edges from the tree to outside vertices are:

• \(AD=6\)
• \(BD=4\)
• \(CD=3\)

Choose the smallest: \(CD=3\).

Now all vertices are included.

MST edges:

• \(AB=2\)
• \(BC=1\)
• \(CD=3\)

Total weight:

$$2+1+3=6$$

Conclusion: Prim's algorithm also gives an MST with total weight \(6\).

12. Comparing Kruskal's and Prim's algorithms

• Kruskal's algorithm focuses on the smallest edges in the whole graph.
• Prim's algorithm focuses on growing one connected tree from a starting vertex.
• Both produce a minimum spanning tree when used correctly.

In some graphs, there may be more than one MST if different spanning trees have the same total weight.

13. Common mistakes to avoid

• Including a circuit: An MST must not contain any circuit.
• Stopping too early: All vertices must be connected.
• Using too many edges: For \(n\) vertices, a spanning tree must have exactly \(n-1\) edges.
• Choosing the smallest edge incorrectly in Prim's algorithm: The edge must connect the current tree to a new vertex.
• Ignoring weights: The goal is to minimize total weight, not just to connect the graph somehow.

14. Real-life applications

Minimum spanning trees are useful in many network design problems, such as:

• laying the least amount of cable to connect computers,
• building roads to connect towns at minimum cost,
• connecting power lines or water pipes efficiently,
• designing transport or communication networks.

These applications match an important idea in discrete mathematics: making the best decision when there are many possible paths or connections.

15. Quick check for understanding

• What two conditions must a graph satisfy to be a tree?
• How many edges does a tree with 8 vertices have?
• What is the difference between a spanning tree and a minimum spanning tree?
• In Kruskal's algorithm, what must you check before adding an edge?
• In Prim's algorithm, what kind of edge can be added next?

Answers:

• It must be connected and have no circuits.
• \(8-1=7\) edges.
• A spanning tree connects all vertices with no circuits; a minimum spanning tree is the spanning tree with the smallest total weight.
• Check that the edge does not create a circuit.
• An edge that connects the current tree to a vertex not yet included.

Summary

A tree is a connected graph with no circuits, and a tree with \(n\) vertices has exactly \(n-1\) edges. A spanning tree connects all vertices of a graph without forming circuits. A minimum spanning tree is the spanning tree with the smallest total weight.

You can find an MST using Kruskal's algorithm, which adds the smallest edges without creating circuits, or Prim's algorithm, which grows a tree from one starting vertex by always choosing the cheapest edge to a new vertex. These methods are important for solving real-world network optimization problems efficiently.

Fair Division and Apportionment is about sharing things in a way that is as fair as possible.

In everyday life, this might mean dividing money, tasks, land, or prizes among people. In government, it often means deciding how many representatives or seats each state, region, or group should receive based on population.

This topic is part of discrete mathematics because we are usually dividing whole items, not fractions. For example, a state cannot receive 7.4 seats in a legislature. It must receive either 7 or 8 seats. That makes fair division challenging.

In this lesson, you will learn what fair division means, how apportionment works, and how to apply common methods to distribute seats fairly and logically.

1. What is fair division?

Fair division is the process of dividing goods, resources, or responsibilities among several parties in a way that follows a clear rule of fairness.

There are two common situations:

• Divisible goods: items that can be split, such as money, land, or time.
• Indivisible goods: items that must stay whole, such as seats in Congress, scholarships, or committee positions.

In this lesson, we focus mainly on indivisible goods, especially the division of seats. This process is called apportionment.

2. What is apportionment?

Apportionment is the assignment of a fixed number of indivisible items among groups according to their size.

A common example is assigning legislative seats to states based on population. If one state has twice the population of another, it should usually receive about twice as many seats. But because seats must be whole numbers, exact proportionality is often impossible.

The goal is to make the distribution as close as possible to each group's fair share.

3. Standard divisor and standard quota

Most apportionment methods begin with two important quantities: the standard divisor and the standard quota.

Suppose:

• The total population is \(P\)
• The total number of seats is \(S\)
• A group's population is \(p\)

Then the standard divisor is

This tells us how many people should correspond to one seat.

The standard quota for a group is

This gives the exact fractional number of seats the group deserves before rounding.

The standard quota is often not a whole number. That is why we need apportionment methods.

4. Lower quota and upper quota

If a group's standard quota is not a whole number, it lies between two whole numbers.

For example, if a quota is \(8.37\):

• The lower quota is \(8\)
• The upper quota is \(9\)

A fair apportionment method often tries to give each group either its lower quota or upper quota.

5. Hamilton's Method

One of the most common apportionment methods taught at this level is Hamilton's Method, also called the Method of Largest Remainders.

Steps in Hamilton's Method:

1. Find the standard divisor.
2. Find each group's standard quota.
3. Give each group its lower quota.
4. Add the assigned seats.
5. If there are still seats left, give them one at a time to the groups with the largest fractional parts of their quotas.

This method is easy to compute and tries to stay close to proportional fairness.

Worked Example 1: Finding standard divisor and quotas

A school district has 4 regions that must share 20 council seats. Their student populations are:

• North: 4,200
• South: 3,100
• East: 7,400
• West: 5,300

Find the standard divisor and each standard quota.

Step 1: Find total population

Step 2: Find standard divisor

This means 1 seat corresponds to 1,000 students.

Step 3: Find standard quotas

So the standard quotas are:

• North: \(4.2\)
• South: \(3.1\)
• East: \(7.4\)
• West: \(5.3\)

These quotas total:

Now we are ready to assign whole-number seats.

Worked Example 2: Using Hamilton's Method

Use the quotas from Example 1 to assign the 20 seats.

Step 1: Assign lower quotas

• North: \(4\)
• South: \(3\)
• East: \(7\)
• West: \(5\)

Total assigned so far:

There is 1 seat left.

Step 2: Compare fractional parts

• North: \(0.2\)
• South: \(0.1\)
• East: \(0.4\)
• West: \(0.3\)

The largest fractional part is \(0.4\), for East.

Step 3: Give the remaining seat to East

Final apportionment:

• North: \(4\)
• South: \(3\)
• East: \(8\)
• West: \(5\)

Check:

So the 20 seats are fairly apportioned using Hamilton's Method.

6. Why is apportionment difficult?

Apportionment is difficult because fairness can mean different things.

We want a method that is:

• Proportional to population
• Based on whole numbers
• Consistent
• Reasonable for both large and small groups

But these goals can conflict. A method may seem fair in one situation and less fair in another.

For example, if two groups have very close quotas, deciding who gets the extra seat can be difficult. Small differences in population can change the result.

7. Jefferson's Method

Another important apportionment method is Jefferson's Method.

Instead of using the standard divisor exactly, Jefferson's Method adjusts the divisor until the total number of assigned seats is correct.

How Jefferson's Method works:

1. Start with the standard divisor.
2. Compute each quota using that divisor.
3. Round down every quota.
4. If too few seats are assigned, use a smaller divisor and try again.
5. Continue until the total number of assigned seats equals the required total.

Because Jefferson's Method always rounds down, reducing the divisor makes quotas larger and can increase the total number of seats assigned.

Worked Example 3: Jefferson's Method

A committee has 10 seats to divide among three departments with employee counts:

• A: 480
• B: 260
• C: 160

Step 1: Total population

Step 2: Standard divisor

Step 3: Compute quotas using divisor 90

Round down:

• A: \(5\)
• B: \(2\)
• C: \(1\)

Total:

That is too few, so we need a smaller divisor.

Try divisor 75

Round down:

• A: \(6\)
• B: \(3\)
• C: \(2\)

Total:

Now there are too many seats, so the correct divisor must be between 75 and 90.

Try divisor 80

Round down:

• A: \(6\)
• B: \(3\)
• C: \(2\)

Total:

Still too many. Try a slightly larger divisor.

Try divisor 82

Round down:

• A: \(5\)
• B: \(3\)
• C: \(1\)

Total:

Too few. So we need a divisor between 80 and 82.

Try divisor 81

Round down:

• A: \(5\)
• B: \(3\)
• C: \(1\)

Total:

Still too few. We need a divisor just under 80.

Try divisor 79

Round down:

• A: \(6\)
• B: \(3\)
• C: \(2\)

Total:

This example shows that Jefferson's Method can require testing divisors carefully. In practice, a divisor is chosen so that the rounded-down totals equal the required number of seats. The important idea is the process: adjust divisor, compute quotas, round down, and repeat.

8. Comparing Hamilton's and Jefferson's methods

These methods can produce different results.

• Hamilton's Method uses the standard divisor, gives each group its lower quota, then awards remaining seats by largest fractional parts.
• Jefferson's Method adjusts the divisor and always rounds down.

Hamilton's Method often feels natural because it starts from the exact quotas. Jefferson's Method can sometimes favor larger groups slightly because of repeated rounding down and divisor adjustment.

9. Fair share in general fair division

Outside politics, fair division also means each person should receive a fair share.

If a resource is worth a total of \(V\) and there are \(n\) people, then one person's fair share is often thought of as

For example, if \(\$600\) must be shared equally among 4 people, each fair share is

But when the resource is made of indivisible items, exact equality may not be possible. Then we use rules or algorithms to get as close as possible.

Worked Example 4: Apportionment with larger numbers

A country must assign 25 seats to 5 provinces based on population:

• P: 12,500
• Q: 8,300
• R: 6,700
• S: 5,900
• T: 3,600

Use Hamilton's Method.

Step 1: Find total population

Step 2: Find standard divisor

Step 3: Find standard quotas

Step 4: Assign lower quotas

• P: \(8\)
• Q: \(5\)
• R: \(4\)
• S: \(3\)
• T: \(2\)

Total assigned:

There are 3 seats left.

Step 5: Compare fractional parts

• P: \(0.45\)
• Q: \(0.61\)
• R: \(0.53\)
• S: \(0.99\)
• T: \(0.43\)

The three largest fractional parts are:

• S: \(0.99\)
• Q: \(0.61\)
• R: \(0.53\)

So each of those provinces gets one additional seat.

Final apportionment:

• P: \(8\)
• Q: \(6\)
• R: \(5\)
• S: \(4\)
• T: \(2\)

Check:

10. Important ideas to remember

• The total number of seats is fixed.
• Populations determine each group's share.
• The standard divisor is total population divided by total seats.
• The standard quota is a group's population divided by the standard divisor.
• Hamilton's Method uses lower quotas first, then assigns extra seats by largest fractional parts.
• Jefferson's Method changes the divisor and rounds down until the total number of seats is correct.

11. Common mistakes

• Using the wrong total population: Always add all groups carefully.
• Rounding too early: Keep decimals until the final step.
• Forgetting lower quotas in Hamilton's Method: Start by assigning all lower quotas first.
• Comparing whole quotas instead of fractional parts: Extra seats go to the largest decimal remainders.
• Not checking the final total: The assigned seats must add up exactly to the required number.

12. How to decide which method to use

In school problems, the method is usually stated in the question. If it is not, look for clues.

• If the problem says largest remainders, use Hamilton's Method.
• If it says modified divisor and round down, use Jefferson's Method.

Be sure to write your work clearly, since most of the credit often comes from showing the process.

Brief Summary

Fair division and apportionment are ways to distribute resources or seats fairly among groups. In apportionment, the challenge is that seats are whole numbers, so exact proportional division is usually impossible.

The key tools are the standard divisor, standard quota, and methods such as Hamilton's Method and Jefferson's Method. If you understand how quotas are found and how each method turns quotas into whole numbers, you can solve most apportionment problems confidently.

Voting methods are mathematical ways to decide a winner when a group of people has different preferences. In many elections, voters do not all agree on the same candidate, so the way votes are counted can change the result.

In this lesson, you will learn three common voting methods: plurality, Borda count, and ranked-choice voting. You will also learn about paradoxes, which are surprising situations where a voting system gives a result that seems unfair or inconsistent.

This topic is part of discrete mathematics because it deals with counting, ranking, and decision-making using finite choices. It also connects to real-world problems such as elections, awards, surveys, and group decisions.

Main idea: a voting method is a rule for turning voter preferences into a final result. Different rules can lead to different winners, even when the same people vote.

1. Preference ballots

Many voting systems use a preference ballot. Instead of choosing only one candidate, each voter ranks the candidates from most preferred to least preferred.

For example, if there are three candidates: A, B, and C, then one voter might rank them:

A > C > B

This means the voter prefers A most, then C, then B.

When many voters submit rankings, we can organize them into a preference schedule.

Here is an example:

• 12 voters: A > B > C
• 9 voters: B > C > A
• 7 voters: C > A > B

This tells us how many voters chose each ranking.

2. Plurality method

The plurality method is the simplest voting method. Each voter's first-choice candidate gets one vote. The candidate with the most first-place votes wins.

This is often called “most votes wins”, even if the winner does not get more than half of the total votes.

How to use plurality:

1. Count only the number of first-place votes for each candidate.
2. Ignore second and third choices.
3. The candidate with the highest total wins.

Worked Example 1: Plurality

Suppose 30 voters rank three candidates A, B, and C as follows:

• 14 voters: A > B > C
• 10 voters: B > C > A
• 6 voters: C > B > A

Count the first-place votes:

• A: 14
• B: 10
• C: 6

So the plurality winner is A.

Notice that A has only 14 out of 30 votes. Since

$$\frac{14}{30} < \frac{1}{2},$$

A does not have a majority. A won only because A had more first-place votes than anyone else.

Strength of plurality: easy to calculate.

Weakness of plurality: it ignores most of the information on the ballot. A candidate disliked by most voters can still win.

3. Borda count

The Borda count uses all rankings on the ballot. Candidates receive points based on their position in each ranking.

If there are 3 candidates, a common point system is:

• 1st place = 2 points
• 2nd place = 1 point
• 3rd place = 0 points

In general, if there are \(n\) candidates, then a common Borda system gives:

$$n-1,\; n-2,\; n-3,\; \dots,\; 0$$

points from top rank to bottom rank.

How to use Borda count:

1. Assign points to each ranking position.
2. Multiply the number of voters by the points earned in each group.
3. Add the total points for each candidate.
4. The candidate with the highest point total wins.

Worked Example 2: Borda Count

Use the same election:

• 14 voters: A > B > C
• 10 voters: B > C > A
• 6 voters: C > B > A

With 3 candidates, use:

• 1st place = 2 points
• 2nd place = 1 point
• 3rd place = 0 points

Candidate A

• From 14 voters ranking A first: \(14 \times 2 = 28\)
• From 10 voters ranking A third: \(10 \times 0 = 0\)
• From 6 voters ranking A third: \(6 \times 0 = 0\)

Total for A:

$$28 + 0 + 0 = 28$$

Candidate B

• From 14 voters ranking B second: \(14 \times 1 = 14\)
• From 10 voters ranking B first: \(10 \times 2 = 20\)
• From 6 voters ranking B second: \(6 \times 1 = 6\)

Total for B:

$$14 + 20 + 6 = 40$$

Candidate C

• From 14 voters ranking C third: \(14 \times 0 = 0\)
• From 10 voters ranking C second: \(10 \times 1 = 10\)
• From 6 voters ranking C first: \(6 \times 2 = 12\)

Total for C:

$$0 + 10 + 12 = 22$$

The Borda count winner is B.

This is important: plurality chose A, but Borda count chose B. The same votes produced different winners because the methods value ballots differently.

Strength of Borda count: it uses more information from each ballot.

Weakness of Borda count: a candidate who is many voters’ second choice can beat a candidate with the most first-place votes.

4. Ranked-choice voting (instant runoff)

Ranked-choice voting, often called instant runoff voting, also uses ranked ballots. The idea is to simulate a runoff election without holding multiple rounds of voting.

How it works:

1. Count all first-place votes.
2. If a candidate has a majority, meaning more than half the votes, that candidate wins.
3. If no one has a majority, eliminate the candidate with the fewest first-place votes.
4. Transfer those ballots to the next preferred candidate still in the race.
5. Repeat until one candidate has a majority.

A majority means:

$$\text{majority} > \frac{\text{total number of votes}}{2}$$

Worked Example 3: Ranked-Choice Voting

Suppose 25 voters rank candidates A, B, and C like this:

• 10 voters: A > C > B
• 8 voters: B > C > A
• 7 voters: C > B > A

Round 1: Count first-place votes

• A: 10
• B: 8
• C: 7

To win, a candidate needs more than

$$\frac{25}{2} = 12.5,$$

so a majority is at least 13 votes.

No candidate has 13 votes, so eliminate the candidate with the fewest first-place votes: C.

Transfer C’s ballots

The 7 voters who chose C first ranked B second, so those 7 votes now go to B.

Round 2 totals

• A: 10
• B: \(8 + 7 = 15\)

Now B has 15 votes, which is a majority. So the ranked-choice winner is B.

Again, notice something interesting:

• Plurality winner: A
• Ranked-choice winner: B

Why? Because ranked-choice considers backup preferences after the weakest candidate is removed.

5. Comparing the three methods

Here is a quick comparison:

• Plurality: only first-place votes matter.
• Borda count: every ranking position matters through points.
• Ranked-choice: first-place votes matter first, but lower choices matter when candidates are eliminated.

No system is perfect. Each method tries to answer the question “Who should win?” in a different way.

6. What is a voting paradox?

A paradox is a result that seems strange or unfair, even though the voting method was followed correctly. Voting paradoxes show that fair decision-making can be more complicated than it first appears.

In voting theory, paradoxes happen because different ideas of fairness can conflict with one another.

Some important paradoxes for this course are:

• Majority paradox
• Condorcet paradox
• Different-winner paradox across methods

7. Majority winner and majority paradox

A majority winner is a candidate ranked first by more than half of the voters.

It may seem natural that if a candidate has a majority of first-place votes, that candidate should win. Many people consider this an important fairness idea.

A majority paradox happens when a voting method fails to choose the majority winner.

This can happen in methods like Borda count.

Worked Example 4: Majority Paradox with Borda Count

Suppose 11 voters rank candidates A, B, and C as follows:

• 6 voters: A > B > C
• 5 voters: B > C > A

First, check for a majority winner.

A has 6 first-place votes out of 11. Since

$$6 > \frac{11}{2} = 5.5,$$

A is the majority winner.

Now apply Borda count using 2, 1, 0 points.

Candidate A

• From 6 voters ranking A first: \(6 \times 2 = 12\)
• From 5 voters ranking A third: \(5 \times 0 = 0\)

Total:

$$12$$

Candidate B

• From 6 voters ranking B second: \(6 \times 1 = 6\)
• From 5 voters ranking B first: \(5 \times 2 = 10\)

Total:

$$6 + 10 = 16$$

Candidate C

• From 6 voters ranking C third: \(6 \times 0 = 0\)
• From 5 voters ranking C second: \(5 \times 1 = 5\)

Total:

$$5$$

Borda count chooses B, even though A had a majority of first-place votes.

This is a majority paradox.

8. Condorcet winner and Condorcet paradox

A Condorcet winner is a candidate who would beat every other candidate in a head-to-head comparison.

To compare two candidates head-to-head, count how many voters prefer one over the other.

For example, if more voters prefer A over B than prefer B over A, then A beats B head-to-head.

Sometimes there is no Condorcet winner because the preferences form a cycle. This is called the Condorcet paradox.

The cycle looks like this:

• A beats B
• B beats C
• C beats A

This seems impossible at first, but it can happen when different groups of voters have different rankings.

Worked Example 5: Condorcet Paradox

Suppose 9 voters have these preferences:

• 3 voters: A > B > C
• 3 voters: B > C > A
• 3 voters: C > A > B

Now compare candidates two at a time.

A vs. B

• The 3 voters with A > B > C prefer A over B.
• The 3 voters with C > A > B also prefer A over B.
• The 3 voters with B > C > A prefer B over A.

So:

• A over B: \(3 + 3 = 6\)
• B over A: 3

A beats B.

B vs. C

• The 3 voters with A > B > C prefer B over C.
• The 3 voters with B > C > A prefer B over C.
• The 3 voters with C > A > B prefer C over B.

So:

• B over C: \(3 + 3 = 6\)
• C over B: 3

B beats C.

C vs. A

• The 3 voters with B > C > A prefer C over A.
• The 3 voters with C > A > B prefer C over A.
• The 3 voters with A > B > C prefer A over C.

So:

• C over A: \(3 + 3 = 6\)
• A over C: 3

C beats A.

We get the cycle:

$$A \text{ beats } B, \quad B \text{ beats } C, \quad C \text{ beats } A$$

There is no Condorcet winner. This is the Condorcet paradox.

This shows that even if every voter has a clear ranking, the group as a whole may not produce a perfectly consistent order.

9. Why different methods can disagree

Different methods disagree because they answer different fairness questions.

• Plurality asks: Who has the most first-choice support?
• Borda count asks: Who is ranked well overall?
• Ranked-choice asks: Who can build majority support after weaker candidates are removed?

Because these questions are different, the winners can be different too.

This does not mean one method is always wrong. It means that designing a perfectly fair voting system is difficult.

10. Steps for solving voting method problems

When you are given a preference schedule, use this process:

1. Read the schedule carefully. Make sure you know how many voters chose each ranking.
2. For plurality, count only first-place votes.
3. For Borda count, assign points for each rank and total them.
4. For ranked-choice, count first-place votes, eliminate the lowest, and transfer those ballots.
5. Check for paradoxes. Ask whether a majority winner lost, or whether pairwise comparisons form a cycle.

11. Common mistakes to avoid

• Counting all rankings in plurality. In plurality, only first-place votes matter.
• Using the wrong point values in Borda count. For 3 candidates, the usual values are 2, 1, 0.
• Forgetting to transfer votes correctly in ranked-choice. A ballot transfers to the next candidate still not eliminated.
• Mixing up majority and plurality. A plurality is the largest number of votes; a majority is more than half.
• Ignoring ties. In some problems, ties may need a special rule, so read carefully.

12. Final summary

Voting methods are mathematical systems for making decisions from ranked preferences. The plurality method uses only first-place votes, the Borda count gives points to all rankings, and ranked-choice voting eliminates the weakest candidate and redistributes votes until someone has a majority.

These systems can produce different winners from the same set of ballots. That is why voting paradoxes matter. The majority paradox shows that a majority winner can lose under some methods, and the Condorcet paradox shows that group preferences can form a cycle with no clear head-to-head champion.

Understanding these methods helps you analyze elections logically and see that mathematics plays an important role in decision-making and fairness.