Financing a Balloon With a Replacement Loan

Replacing a Balloon With Level Installments

A balloon is a single large payment due at a specified date. If the borrower cannot pay it from available funds, a replacement loan can provide the money at that date and repay it through installments. The essential calculation is to treat the balloon amount being financed as the new loan’s principal, then equate that principal to the value of the replacement loan’s scheduled payments.

The date matters. The replacement loan begins when the balloon is due, and its principal is measured at that date. Unless the problem states otherwise, assume the replacement loan has no fees, its level installments are made at the end of each payment period, and it has no further balloon payment. The rate used in the payment formula must be the effective rate for one payment period.

Definition: A replacement loan is a new loan used to provide some or all of the funds needed to discharge an obligation, such as a balloon payment, at a specified date. If the replacement loan advances principal $L$ and has $n$ equal end-of-period payments at rate $i$ per period, its payments have present value $L$ at the loan date. When the balloon amount $B$ is financed in full and there are no additional amounts, $L=B$.

For a positive rate $i$ per payment period and $n$ level payments in arrears, the present value of the payments is $R a_{\overline{n}|}$, where $R$ is the payment and $a_{\overline{n}|}=(1-v^n)/i$, with $v=1/(1+i)$. Thus a replacement loan of principal $L$ has payment $R=L/a_{\overline{n}|}$. If a quoted annual rate is nominal and payments are more frequent than annual, first convert it to the rate for one payment period.

Why the Payment Formula Holds

At the replacement loan date, the lender advances the principal. In exchange, the lender is entitled to the agreed installments. Under the assumed rate and timing, the value at that date of each installment is its amount discounted for the number of periods until it is paid. Equating the advance to the total discounted value gives the payment formula.

Write the value equation.
For a loan of principal $L$, with $n$ equal payments $R$ at the ends of periods, $L=Rv+Rv^2+\cdots+Rv^n$.

Recognize the annuity factor.
The discounted payments sum to $R a_{\overline{n}|}$, since $a_{\overline{n}|}=v+v^2+\cdots+v^n$.

Solve for the installment.
Rearranging $L=R a_{\overline{n}|}$ gives $R=L/a_{\overline{n}|}$. Using $a_{\overline{n}|}=(1-v^n)/i$ gives the equivalent expression below.

R=\frac{L}{a_{\overline{n}|}} =\frac{Li}{1-(1+i)^{-n}} =\frac{Li(1+i)^n}{(1+i)^n-1}

This equation assumes a positive periodic rate, a fixed term of $n$ periods, and level payments at the ends of those periods. If the rate is zero, the payment is $L/n$. If payments are due at the beginning of each period, their present value is an annuity-due value instead; the ordinary-annuity formula above does not apply unchanged.

Worked Example: Finding Two Annual Payments for a Balloon

A balloon of $2,100 is due today. A replacement loan advances exactly $2,100 and is repaid with two equal annual payments, each made at year-end. The effective annual interest rate is 10%. Find the annual payment and verify the repayment schedule.

Method. Set the present value of the two end-of-year payments equal to the $2,100 principal. The rate per payment period is $i=0.10$, and the term is $n=2$.

R=\frac{2{,}100(0.10)(1.10)^2}{(1.10)^2-1} =\frac{2{,}100(0.10)(1.21)}{0.21} =\frac{254.10}{0.21} =1{,}210

The annual payment is $1,210. To verify, in year 1 the interest is $2{,}100(0.10)=210$. The first payment reduces principal by $1{,}210-210=1{,}000$, leaving a balance of $2{,}100-1{,}000=1{,}100$. In year 2, interest is $1{,}100(0.10)=110$, so the final payment of $1,210 repays $1,100 of principal and $110 of interest. The balance after that payment is zero.

Common mistake avoided. Do not omit the factor $i$ in the numerator of the payment formula. Here, $2{,}100(1.21)/0.21$ is $12,100, not $1,210, and does not solve the value equation.

Rate and Payment-Period Consistency

The time unit for the interest rate must match the time unit for the installments. For annual payments, an effective annual rate is used directly. For monthly payments, the periodic rate is monthly. For example, if a nominal annual rate $i^{(12)}$ is convertible monthly, the monthly rate is $i^{(12)}/12$; the number of periods is the number of months in the term. Using an annual rate with a monthly number of payments overstates the interest applied in each period.

The replacement loan’s principal need not always equal the entire balloon. If the balloon is $B$ and the new loan advances $L$, the difference $B-L$ must be supplied from another source when $L<B$. When the problem says that the balloon is fully financed and gives no other amounts, use $L=B$. Do not silently change the loan principal to include an unstated fee or cash payment.

Worked Example: Correcting the Payment Period

A $3,000 balloon is due today and is financed in full by a replacement loan with two equal annual payments at the ends of years 1 and 2. The effective annual rate is 5%. Find the payment to the nearest cent.

Method. Use the annuity-immediate present value with an annual rate and two annual payments. The present-value equation is $3{,}000=R(1/1.05+1/1.05^2)$.

\frac{1}{1.05}=0.95238095,\qquad \frac{1}{1.05^2}=\frac{1}{1.1025}=0.90702948

R=\frac{3{,}000}{0.95238095+0.90702948} =\frac{3{,}000}{1.85941043} =1{,}613.4146\ldots

Rounded to the nearest cent, the annual payment is $1,613.41. As a check, the unrounded payment discounted for one and two years has value $3,000 at the loan date, by construction.

Common mistake avoided. There are two payments, not one payment in each half-year. The loan’s specified annual payment timing determines both the number of periods and the discounting. Rounding the installment only after solving also avoids accumulating rounding error.

Finding a Balance After Replacement-Loan Payments

At a payment date, the outstanding balance can be found prospectively: it is the present value, at that date, of the payments that remain. If $k$ payments have been made, then $n-k$ payments remain. Under the same fixed-rate, level-payment assumptions, the balance immediately after the $k$th payment is $R a_{\overline{n-k}|}$. Equivalently, it is the amount of the original loan accumulated to that date less the accumulated value of the payments already made.

The phrase immediately after is important. At a payment date, the scheduled installment has just been applied, so it is not included among the remaining payments. A balance immediately before that installment would include its value as a payment still due.

Worked Example: Balance After the First Replacement-Loan Payment

A $4,200 balloon is financed in full by a replacement loan with two equal annual payments at the ends of years 1 and 2. The effective annual rate is 10%. Find the payment and the balance immediately after the first payment.

Method. First calculate the level payment from the original principal. Then find the balance prospectively as the value of the one remaining payment.

R=\frac{4{,}200(0.10)(1.10)^2}{(1.10)^2-1} =\frac{4{,}200(0.10)(1.21)}{0.21} =\frac{508.20}{0.21} =2{,}420

Immediately after the first payment, one payment of $2,420 remains, due one year later. Its value at the first payment date is

B_1=\frac{2{,}420}{1.10}=2{,}200

The balance is therefore $2,200. A retrospective check gives the same result: first-year interest is $4{,}200(0.10)=420$, so the first $2,420 payment reduces principal by $2{,}420-420=2{,}000$. The balance is $4{,}200-2{,}000=2{,}200$. In year 2, interest of $2{,}200(0.10)=220$ plus principal of $2,200 equals the final $2,420 payment.

Common mistake avoided. Do not discount the remaining payment for two years. The balance is measured at the end of year 1, so the one remaining payment is discounted for one year.

Checks Before Finalizing a Calculation

A correct-looking payment can still come from a mismatched rate, term, or payment timing. Before using the result, confirm that the value equation describes the actual replacement loan. In particular, check these items:

Principal: Is the loan amount the balloon being financed, or does the problem state a different advance?
Period rate: Does the rate represent one payment period?
Number of payments: Does $n$ count installments, rather than years when payments are more frequent?
Timing: Are installments at period-end, as assumed by $a_{\overline{n}|}$, or at another time?
Balance date: Is the requested balance just before or just after a scheduled installment?
Final payment: Does the schedule reduce the balance to zero after the stated final payment?

This method is useful whenever a single obligation at a known date is replaced with a fixed schedule of payments. It converts the balloon into a principal at the replacement-loan date and then applies ordinary loan valuation. It does not, by itself, determine whether refinancing is economically preferable; that would require comparing the relevant cash flows and costs. The calculations here focus on the loan payment or balance once the replacement principal, rate, term, and payment timing are specified. This treatment is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Key takeaway: At the date a balloon is financed, equate the replacement loan’s principal to the present value of its installments. Use a rate and term measured in the payment period, and value only the payments remaining when finding a later balance.

Check Your Understanding

For each problem, assume no fees and end-of-period level payments unless stated otherwise.

A $5,000 balloon is fully financed by a replacement loan with two annual payments at an effective annual rate of 6%. Find the annual payment to the nearest cent. (Answer: $2,727.18)
A $10,000 balloon is fully financed by a two-year replacement loan with two annual payments at an effective annual rate of 10%. Find the annual payment. (Answer: $5,761.90)
A $2,100 balloon is replaced by a two-year loan with annual payments at an effective annual rate of 10%. Find the balance immediately after the first payment. (Answer: $1,100.00)
A $12,000 balloon is financed by a loan with two monthly payments. The nominal annual rate is 12% convertible monthly. Find the monthly payment to the nearest cent. (Answer: $6,090.15)
A $25,000 balloon is due today. A replacement loan advances $24,000 and is repaid by two annual payments at an effective annual rate of 5%. Find the annual payment and the amount the borrower must supply separately at the balloon date. (Answer: annual payment $12,907.32; separate funds required $1,000.00)

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