Finding the Break-Even Date for Refinancing

What Does “Break-Even” Mean?

A refinancing plan can reduce the next installment but extend repayment, change the interest rate, or require costs at closing. As a result, the date when installment savings first equal the refinancing cost may not be the date when refinancing has actually caught up: the replacement loan may still have a larger balance outstanding.

This tutorial uses a balance-adjusted, simple cash-payback measure at installment dates. It adds the payments made under each option through that date and includes the amount still owed under each loan. The measure does not discount past payments or include investment earnings on savings. It is a payback comparison, not a present-value comparison.

Definition: Let $C$ be the refinancing cost paid upfront, $R_j^{(o)}$ and $R_j^{(n)}$ the old and new loan payments at date $j$, and $B_k^{(o)}$ and $B_k^{(n)}$ the respective balances immediately after the payment at date $k$. Define the refinancing advantage through date $k$ as $$ G_k=\sum_{j=1}^{k}\left(R_j^{(o)}-R_j^{(n)}\right)-C+B_k^{(o)}-B_k^{(n)}. $$ The first balance-adjusted break-even date is the earliest installment date $k$ for which $G_k\geq 0$. If there is no such date in the comparison period, there is no break-even within that period.

A positive payment difference, $R_j^{(o)}-R_j^{(n)}$, is a saving for that period. The balance difference $B_k^{(o)}-B_k^{(n)}$ adjusts for how much each option still requires the borrower to repay. A larger balance on the replacement loan reduces the refinancing advantage; a smaller balance increases it.

Calculating the Balances and the Break-Even Measure

Assume payments are made at the end of each period, interest is effective for that period, and there are no additional fees or transactions beyond the stated refinancing cost. For a loan with balance $B_{k-1}$, periodic effective interest rate $i$, and end-of-period payment $R_k$, the balance immediately after that payment is

B_k=B_{k-1}(1+i)-R_k.

Use the appropriate rate and payment schedule for each loan. If a payment pays off the loan, the balance after that payment is zero. If the loan has already been repaid, its later payments and balances are also zero.

Why the balance adjustment is necessary: Through date $k$, the old-loan option has had total outflows of $\sum_{j=1}^k R_j^{(o)}$ and still has liability $B_k^{(o)}$. The refinancing option has had outflows $C+\sum_{j=1}^k R_j^{(n)}$ and still has liability $B_k^{(n)}$. Subtracting the second total from the first gives $$ \left(\sum_{j=1}^k R_j^{(o)}+B_k^{(o)}\right) -\left(C+\sum_{j=1}^k R_j^{(n)}+B_k^{(n)}\right)=G_k. $$ Thus $G_k\geq0$ means that, under this simple cash-payback measure, refinancing has no greater combined outflow and remaining liability through that date.

The following process helps avoid a common timing error: comparing cumulative payments but forgetting that the two loan balances may differ.

Set the comparison dates.
Use matching installment dates for both options and state the comparison horizon.

Calculate both schedules.
At each date, apply the balance recurrence using that loan’s own rate and payment. Record zero balances after payoff.

Calculate $G_k$ date by date.
Add cumulative old-minus-new payments, subtract the upfront cost, and add the old-minus-new balance.

Find the first crossing.
The first date with $G_k\geq0$ is the break-even date under this measure. Continue checking later dates if you need to know whether the advantage persists.

Worked Refinancing Comparisons

Worked Example: Lower Payments but a Larger Replacement Balance

A borrower owes $10,000. The old loan charges 2% per month and requires payments of $5,200 at the end of month 1 and $5,100 at the end of month 2. The replacement loan advances the same $10,000, charges 1% per month, and requires $3,500 at the end of each of months 1 and 2. It is paid off at the end of month 3 with its remaining balance. Refinancing costs $120, paid upfront. Find the first balance-adjusted break-even date and check whether the advantage persists through payoff.

Method and assumptions. Compare end-of-month balances and use the balance-adjusted simple cash-payback measure. Both loans have monthly effective rates as stated; payments occur at month-end, and the $120 cost is paid at refinancing.

Old-loan balances. Starting from $10,000, after month 1:

B_1^{(o)}=10{,}000(1.02)-5{,}200=10{,}200-5{,}200=5{,}000.

After month 2, the amount due is $5{,}000(1.02)=5{,}100$, exactly the scheduled payment. Therefore $B_2^{(o)}=0$, and there are no later old-loan payments.

Replacement-loan balances. After month 1:

B_1^{(n)}=10{,}000(1.01)-3{,}500=10{,}100-3{,}500=6{,}600.

After month 2:

B_2^{(n)}=6{,}600(1.01)-3{,}500=6{,}666-3{,}500=3{,}166.

The payoff at month 3 is $3{,}166(1.01)=\$3{,}197.66$. The replacement loan is then fully repaid.

Month 1 comparison. The cumulative payment saving is $5{,}200-3{,}500=\$1{,}700$, but the replacement balance exceeds the old balance by $6{,}600-5{,}000=\$1{,}600$. Hence

G_1=1{,}700-120+5{,}000-6{,}600=-\$20.

Month 2 comparison. Cumulative old payments are $5{,}200+5{,}100=\$10{,}300$, and cumulative replacement payments are $3{,}500+3{,}500=\$7{,}000$. The cumulative payment saving is therefore $3,300. Including the balances:

G_2=3{,}300-120+0-3{,}166=\$14.

The first balance-adjusted break-even date is month 2. The payment savings alone exceeded the $120 cost in month 1, but the replacement loan still had a substantially larger balance, so month 1 was not balance-adjusted break-even.

Check through payoff. By month 3, total old payments are $10,300, while total replacement payments are $3{,}500+3{,}500+3{,}197.66=\$10{,}197.66$. Both balances are zero, so

G_3=10{,}300-10{,}197.66-120=-\$17.66.

The advantage at month 2 is temporary; it does not persist through payoff. The trap avoided is treating the first break-even crossing as proof that refinancing remains advantageous thereafter.

Worked Example: A Longer Term Can Prevent Break-Even

A borrower owes $8,000 on a loan charging 1% per month. The old loan requires a $4,100 payment at month 1 and is paid off at month 2 with the remaining amount due. A replacement loan advances $8,000 at the same 1% monthly rate and requires payments of $2,800 at months 1 and 2, with the remaining balance paid at month 3. The upfront refinancing cost is $100. Determine whether the replacement loan breaks even by month 3.

Method and assumptions. Apply the balance recurrence separately to both contracts, then evaluate $G_k$ at each month-end. Payments are at month-end and the cost is paid upfront.

The old balance after month 1 is

B_1^{(o)}=8{,}000(1.01)-4{,}100=8{,}080-4{,}100=3{,}980.

The final old-loan payment is $3{,}980(1.01)=\$4{,}019.80$, leaving $B_2^{(o)}=0$.

For the replacement loan,

B_1^{(n)}=8{,}000(1.01)-2{,}800=5{,}280, \qquad B_2^{(n)}=5{,}280(1.01)-2{,}800=2{,}532.80.

The month 3 payoff is $2{,}532.80(1.01)=\$2{,}558.128$, which rounds to $2,558.13 if reported to cents.

At month 1, the payment saving is $4{,}100-2{,}800=\$1{,}300$. The balance difference is $3{,}980-5{,}280=-\$1{,}300$, so

G_1=1{,}300-100-1{,}300=-\$100.

At month 2, cumulative old payments are $4{,}100+4{,}019.80=\$8{,}119.80$, and cumulative replacement payments are $5,600. Thus the cumulative payment difference is $2,519.80. The old balance is zero and the replacement balance is $2,532.80:

G_2=2{,}519.80-100+0-2{,}532.80=-\$113.00.

At month 3, cumulative replacement payments are $5{,}600+2{,}558.128=\$8{,}158.128$. Both balances are zero, giving

G_3=8{,}119.80-8{,}158.128-100=-\$138.328.

Rounded to cents, the month 3 advantage is $-\$138.33$. There is no break-even by month 3. The common mistake is to count the lower monthly installments as savings without accounting for the larger balance carried into month 2 and the additional replacement-loan payment.

Worked Example: Unequal Payments and a Break-Even at Month 2

A hypothetical borrower owes $15,000. The old loan charges 1.5% per month and requires payments of $5,000 at the ends of months 1 and 2, then a final payment at month 3. A replacement loan charges 1% per month and requires $4,000 at the ends of months 1 and 2, then a final payment at month 3. The upfront cost is $100. Find the first break-even date and assess the result through payoff.

Method and assumptions. These are monthly effective rates, all payments occur at month-end, and the stated cost is paid immediately. The final payment for each loan equals the balance due at month 3.

The old-loan balances are

B_1^{(o)}=15{,}000(1.015)-5{,}000=10{,}225, $$ $$ B_2^{(o)}=10{,}225(1.015)-5{,}000=10{,}378.375-5{,}000=5{,}378.375, $$ $$ B_3^{(o)}=5{,}378.375(1.015)=5{,}459.050625.

The old loan’s month 3 payoff is therefore $5,459.050625. The replacement balances are

B_1^{(n)}=15{,}000(1.01)-4{,}000=11{,}150, $$ $$ B_2^{(n)}=11{,}150(1.01)-4{,}000=11{,}261.50-4{,}000=7{,}261.50, $$ $$ B_3^{(n)}=7{,}261.50(1.01)=7{,}334.115.

At month 1, the payment saving is $1,000, but the old-minus-new balance difference is $10{,}225-11{,}150=-\$925$. Thus $G_1=1{,}000-100-925=-\$25$.

At month 2, cumulative old payments are $10,000 and cumulative replacement payments are $8,000, a $2,000 saving. The balance difference is $5{,}378.375-7{,}261.50=-\$1{,}883.125$, giving

G_2=2{,}000-100-1{,}883.125=\$16.875.

The first break-even date is month 2. At month 3, the total old payments are $5{,}000+5{,}000+5{,}459.050625=\$15{,}459.050625$; the total replacement payments are $4{,}000+4{,}000+7{,}334.115=\$15{,}334.115$. The balances are then zero, so $G_3=15{,}459.050625-15{,}334.115-100=\$24.935625$, or $24.94 to the nearest cent. Here the break-even advantage persists through payoff.

Common Traps and Interpretation

The break-even date depends on the comparison measure and the dates included. Under the measure defined here, a first crossing means $G_k\geq0$ at that date; it does not guarantee $G_j\geq0$ at every later date. A longer replacement term, a balloon payment, or uneven payment amounts can cause the measure to change direction.

Ignoring the remaining balance. Cumulative installment savings alone can make a longer replacement loan look attractive even while it leaves more debt outstanding.
Using one rate for both schedules. Each balance must be advanced using its own loan’s effective rate for the payment period.
Comparing different dates. Calculate both balances immediately after payments at the same date. Do not compare one loan before its installment with the other after its installment.
Forgetting payoff dates. Once a loan is paid off, set its later balance and payments to zero; do not continue its former installment schedule.
Confusing simple payback with present value. This measure does not discount dated cash flows. If the question asks for a present-value comparison, use a present-value method instead of interpreting $G_k$ as a discounted value.

A clear exam solution identifies the payment dates, rate per period, upfront cost, and comparison horizon before calculating. Report the first date $G_k$ reaches zero or more, and state whether the result remains nonnegative at the later dates that matter.

Check Your Understanding

For each problem, use the balance-adjusted simple cash-payback measure and assume payments occur at period-end. Treat stated rates as effective rates per period.

A borrower owes $6,000. The old loan charges 1% per month, with a $3,100 payment in month 1 and a final payoff in month 2. The replacement loan charges 0.5% per month, with $2,000 payments in months 1 and 2 and a final payoff in month 3. The upfront cost is $50. Is there a break-even date by month 3? (Answer: No; $G_1=-\$20$, $G_2=-\$10.55$, and $G_3=-\$20.80$, to the nearest cent.)
At month 1, cumulative old-minus-new payments are $240 and old-minus-new balances are $-\$100$. At month 2, those amounts are $500 and $-\$250$. If the upfront cost is $180, what is the first break-even date? (Answer: Month 2; $G_1=-\$40$ and $G_2=\$70$.)
A $5,000 old loan charges 2% per month and has payments of $2,600 in month 1 and a final payoff in month 2. A replacement loan charges 1% per month, has payments of $1,800 in months 1 and 2, and is paid off in month 3. The upfront cost is $90. Is there a break-even date by month 3? (Answer: No; $G_1=-\$40$, $G_2=-\$22.50$, and $G_3=-\$37.33$, to the nearest cent.)
A refinancing comparison has an upfront cost of $300. At months 1, 2, and 3, respectively, cumulative old-minus-new payments are $500, $900, and $1,100; old-minus-new balances are $-\$250$, $-\$500$, and $-\$900$. Find the first break-even date and determine whether the advantage persists at month 3. (Answer: First break-even is month 2, where $G_2=\$100$; it does not persist to month 3, where $G_3=-\$100$.)

What You'll Learn