Loan Repayment Methods: Level Payment and Non-Level
Compare level payment, non-level, and other loan repayment methods for Exam FM.
Level Payment Loans
The most common repayment structure is level payments, where the borrower pays a constant amount R each period. For a loan of L repaid over n periods at rate i: R = L / a-angle-n. Early payments are mostly interest; later payments are mostly principal. The outstanding balance, interest, and principal in any period can be found using the amortization formulas covered in the amortization schedule topic.
The level payment formula is derived from the equation of value: L = R * a-angle-n, which states that the present value of all payments equals the loan amount. Solving for R gives R = L * i / (1 - v^n). This can also be written as R = L * (1/a-angle-n) = L * (i + 1/s-angle-n), using the identity 1/a-angle-n = i + 1/s-angle-n.
Example: A $200,000 mortgage at 6% annual effective rate, repaid monthly over 30 years. The monthly rate is j = (1.06)^(1/12) - 1 = 0.004868. The monthly payment is R = 200,000 / a-angle-360 at j = 200,000 * 0.004868 / (1 - (1+0.004868)^(-360)) = $1,182.07.
Level Principal Repayment
A variation is level principal repayment, where the principal portion is constant at L/n each period. The total payment decreases over time as the outstanding balance (and thus interest) falls. In period t, the outstanding balance before payment is L * (n - t + 1)/n, the interest is i * L * (n - t + 1)/n, and the total payment is L/n + i * L * (n - t + 1)/n.
Example: A $10,000 loan at 5% repaid over 5 years with level principal. The principal each year is $2,000. Year 1 payment: 2,000 + 0.05*10,000 = $2,500. Year 2: 2,000 + 0.05*8,000 = $2,400. Year 3: 2,000 + 0.05*6,000 = $2,300. The total interest paid is 0.05*(10,000 + 8,000 + 6,000 + 4,000 + 2,000) = 0.05*30,000 = $1,500. Compare this to the level payment method, where total interest is 5*R - 10,000 = 5*2,309.75 - 10,000 = $1,548.75. Level principal repayment costs less total interest because the principal is repaid faster.
Arithmetically Increasing and Decreasing Payments
Payments that increase by a constant amount Q each period can be decomposed into a level component and an increasing component. If the first payment is P and each subsequent payment increases by Q, the equation of value is: L = P * a-angle-n + Q * (Ia)-angle-n - Q * a-angle-n. Alternatively, L = (P - Q) * a-angle-n + Q * (Ia)-angle-n + Q * a-angle-n. Use whichever form is more convenient.
Decreasing payments work similarly but with a negative increment. If payments start at P and decrease by Q each period, ensure that all payments remain positive (P - (n-1)*Q > 0). The present value uses the same general arithmetic annuity formula with a negative Q.
Geometrically Increasing Payments
Payments that grow by a constant percentage g each period are common in practice (for example, when payments are tied to salary growth or inflation). If the first payment is R_1, the payment in period t is R_1 * (1+g)^(t-1). The present value is: L = R_1 * (1 - ((1+g)/(1+i))^n) / (i - g) when i is not equal to g.
Solving for R_1: R_1 = L * (i - g) / (1 - ((1+g)/(1+i))^n). This formula directly gives the first payment. Each subsequent payment is (1+g) times the previous one. On Exam FM, verify whether the growth rate applies to each payment relative to the previous one (which is the standard assumption) or relative to the first payment.
Balloon and Drop Payments
A balloon payment structure involves level payments for n-1 periods followed by a larger final payment. This occurs when the level payment is calculated for a longer term than the actual loan duration, or when the borrower agrees to a partially amortizing loan. The balloon payment equals the outstanding balance after n-1 level payments, plus the interest on that balance: Balloon = B_{n-1} * (1+i).
A drop payment is a smaller final payment. This arises when the level payment is rounded up, causing the loan to be overpaid slightly. The drop payment equals the outstanding balance after n-1 payments, accumulated with interest. On Exam FM, problems may ask you to find the balloon or drop payment amount, or to determine how many level payments are needed before the final adjusted payment.
Refinancing and Rate Changes
When a loan is refinanced at a new rate, compute the outstanding balance at the refinancing date using the original terms, then use that balance as a new loan at the new rate. The outstanding balance at the changeover point is the same regardless of whether you use the prospective or retrospective formula.
Refinancing is beneficial when the new rate is lower than the original rate and the savings exceed any fees or penalties. On Exam FM, refinancing problems may ask you to find the new payment, the savings per period, or the break-even point (the number of periods needed for the cumulative savings to exceed the refinancing costs).
For adjustable-rate loans where the rate changes at a specified time, recompute the payment for the remaining term using the outstanding balance and the new rate. The key insight is that the outstanding balance depends only on past payments and the original rate, while the new payment depends on the remaining balance, the new rate, and the remaining term.
Study Tips and Common Pitfalls
The fundamental principle underlying all loan repayment methods is the equation of value: the present value of all payments at the loan rate must equal the loan amount. When you encounter an unfamiliar payment structure, set up this equation and solve for the unknown.
Common mistakes include using the wrong interest rate for each component of a split-rate problem, forgetting to account for a final balloon or drop payment, and confusing the payment timing (beginning vs. end of period). Practice at least 3 problems for each payment type: level, level principal, arithmetic, geometric, and balloon/drop. This variety ensures you are prepared for any loan repayment question on Exam FM.