Loan Amortization and EMI Calculations

When you take a car loan or a house loan from a bank, you usually repay it through an equated monthly instalment (EMI): the same amount every month until the debt is cleared. Behind that single number is a neat piece of mathematics that splits each payment into interest and principal. Understanding it helps students solve finance assignments and helps anyone judge whether a loan offer is fair.

What Amortization Means

To amortize a loan is to pay it off gradually through regular payments. Each EMI does two jobs:

1. It pays the interest that has built up on the outstanding balance during the month.
2. Whatever is left over reduces the principal, the amount still owed.

Early in the loan the balance is large, so most of the EMI goes to interest. As the balance falls, the interest portion shrinks and the principal portion grows. The EMI itself never changes, which makes budgeting easy for the borrower.

Deriving the EMI Formula

Let \( P \) be the amount borrowed, \( i \) the interest rate per month (annual rate divided by 12), and \( n \) the number of monthly payments. The bank's view is that the present value of all the EMIs, discounted at rate \( i \), must equal the amount lent:

$$ P = \frac{E}{1+i} + \frac{E}{(1+i)^2} + \cdots + \frac{E}{(1+i)^n} $$

This is a geometric series. Summing it gives

$$ P = E \cdot \frac{1-(1+i)^{-n}}{i} $$

Solving for \( E \):

$$ E = \frac{P\,i\,(1+i)^n}{(1+i)^n - 1} $$

This is the formula every bank's EMI calculator uses.

Worked Example 1: A Car Loan

Hamza borrows PKR 2,000,000 to buy a car. The bank charges 18% per year, and the loan is repaid monthly over 3 years. Find the EMI, the total amount repaid and the total interest.

Step 1: Convert to monthly terms.

$$ i = \frac{0.18}{12} = 0.015, \qquad n = 3 \times 12 = 36 $$

Step 2: Compute the growth factor.

$$ (1.015)^{36} = 1.709140 $$

Step 3: Apply the formula.

$$ E = \frac{2{,}000{,}000 \times 0.015 \times 1.709140}{1.709140 - 1} = \frac{51{,}274.19}{0.709140} $$

$$ E = \text{PKR } 72{,}304.79 $$

Step 4: Total repaid and total interest.

$$ \text{Total repaid} = 72{,}304.79 \times 36 = \text{PKR } 2{,}602{,}972.48 $$

$$ \text{Total interest} = 2{,}602{,}972.48 - 2{,}000{,}000 = \text{PKR } 602{,}972.48 $$

Sanity check. At 18% simple interest on the full PKR 2,000,000 for three years, interest would be PKR 1,080,000. Our answer is a little over half that, which makes sense because the balance falls steadily and averages roughly half the original amount.

Building the Amortization Schedule

An amortization schedule lists, month by month, how each EMI is split. The rules are:

• Interest for the month = opening balance \( \times i \)
• Principal repaid = EMI \( - \) interest
• Closing balance = opening balance \( - \) principal repaid

Here are the first three months of Hamza's loan:

Check month 2: interest is \( 1{,}957{,}695.21 \times 0.015 = 29{,}365.43 \), and principal is \( 72{,}304.79 - 29{,}365.43 = 42{,}939.36 \). Each month the interest falls and the principal rises. In the final month almost the whole EMI is principal, and the closing balance reaches zero.

Worked Example 2: Outstanding Balance After One Year

Hamza wants to sell the car after 12 payments and needs to know how much he still owes. Instead of building twelve rows, use the outstanding balance formula. The balance after \( k \) payments is the future value of the loan minus the future value of the payments made:

$$ B_k = P(1+i)^k - E\cdot\frac{(1+i)^k - 1}{i} $$

With \( k = 12 \): first, \( (1.015)^{12} = 1.19561817 \).

$$ P(1.015)^{12} = 2{,}000{,}000 \times 1.19561817 = 2{,}391{,}236.34 $$

$$ E \cdot \frac{0.19561817}{0.015} = 72{,}304.79 \times 13.041211 = 942{,}942.07 $$

$$ B_{12} = 2{,}391{,}236.34 - 942{,}942.07 = \text{PKR } 1{,}448{,}294.27 $$

An equivalent method: the balance equals the present value of the 24 EMIs still to come. Either way, after paying PKR 867,657.49 in the first year, Hamza has reduced the debt by only about PKR 551,706, because the rest went on interest.

Worked Example 3: How the Term Changes the Cost

A family takes a PKR 10,000,000 home loan at 15% per year. Compare a 10-year and a 20-year term. The monthly rate is \( 0.15/12 = 0.0125 \).

10 years (120 months):

$$ E = \frac{10{,}000{,}000 \times 0.0125 \times (1.0125)^{120}}{(1.0125)^{120} - 1} = \text{PKR } 161{,}334.96 $$

Total interest \( = 161{,}334.96 \times 120 - 10{,}000{,}000 = \text{PKR } 9{,}360{,}194.85 \).

20 years (240 months):

$$ E = \frac{10{,}000{,}000 \times 0.0125 \times (1.0125)^{240}}{(1.0125)^{240} - 1} = \text{PKR } 131{,}678.96 $$

Total interest \( = 131{,}678.96 \times 240 - 10{,}000{,}000 = \text{PKR } 21{,}602{,}949.98 \).

Doubling the term lowers the EMI by only about 18%, but more than doubles the total interest. This is the central trade-off of every long loan: affordability now versus cost overall.

Flat Rate Versus Reducing Balance

Some lenders advertise a flat rate, where interest is charged on the original amount for the whole term. At 18% flat, Hamza's car loan would cost \( 2{,}000{,}000 \times 0.18 \times 3 = 1{,}080{,}000 \) in interest, giving a monthly payment of

$$ \frac{2{,}000{,}000 + 1{,}080{,}000}{36} = \text{PKR } 85{,}555.56 $$

That is over PKR 13,000 a month more than the reducing-balance EMI at the same headline rate. Always ask which method a lender uses.

Islamic Home and Car Financing

Islamic banks in Pakistan typically offer car financing through ijarah (leasing) and home financing through diminishing musharakah, in which the bank and customer jointly own the asset and the customer gradually buys the bank's share while paying rent on it. The contracts are different from conventional loans, but the monthly payment is usually worked out with the same annuity mathematics, with rental rates linked to a benchmark such as KIBOR.

Did You Know?

• In the first month of a 20-year loan at 15%, about 95% of the EMI goes to interest: PKR 125,000 out of PKR 131,679.
• Making one extra principal payment early in a loan saves far more interest than the same payment made near the end.
• If the interest rate were zero, the EMI would simply be \( P/n \); the formula gives this as a limit when \( i \to 0 \).

Key Takeaways

• EMI formula: \( E = \dfrac{P\,i\,(1+i)^n}{(1+i)^n - 1} \), with \( i \) and \( n \) in months.
• Each EMI pays the month's interest first; the rest reduces the principal.
• The outstanding balance can be found directly without building the full schedule.
• Longer terms lower the EMI but greatly increase the total interest paid.
• Flat-rate quotes are much more expensive than reducing-balance quotes at the same headline rate.

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