Refinancing

Study guide: Exam FM-2

Refinancing is paying off an existing loan with the proceeds of a new one on different terms — usually a lower interest rate, sometimes a different term or payment. The new loan's principal is the old loan's outstanding balance at the refinancing date, plus any fees or prepayment penalty the borrower finances.

$L' = OB_k + \text{fees}$

$OB_k = P\,a_{\overline{n-k}|\,i}$

$P' = \frac{L'}{a_{\overline{n'}|\,i'}}$

$P$ , $n$ and $i$ are the old loan's payment, term and rate; $k$ is the number of payments made before refinancing; $i'$ and $n'$ are the new rate and term. Unless a question says otherwise, the balance is computed at the old loan's rate — that is the amount the lender is owed.
After refinancing, the new loan is amortized from scratch: interest in its first payment is $i' L'$ , and the Amortization Schedule restarts from $L'$ . Nothing about the payments already made changes.
There are two usual ways to take the benefit. Lower payment: keep the remaining term and solve for $P'$ . Shorter term: keep the old payment and solve for $n'$ , with a smaller final Drop Payment if $n'$ is not a whole number.
Whether it is worth doing is an Equation of Value question: compare the present value of the payment savings with the fees. A fee paid in cash up front can be set directly against that present value.

Refinancing to a Lower Rate A $250,000 loan is repaid by 20 level annual payments at 7%. Just after the 8th payment, the borrower refinances the balance at 5% over the remaining 12 years, adding a $2,000 fee to the new loan. Find the new payment, the annual saving, and the split of the first new payment.
Answer Old payment and the balance after 8 payments (12 remain):
 $\begin{align*} P &= \frac{250{,}000}{a_{\overline{20}|0.07}} \\ &= \frac{250{,}000}{10.5940} \\ &= 23{,}598.23 \\ OB_8 &= 23{,}598.23\,a_{\overline{12}|0.07} \\ &= 187{,}433.35 \end{align*}$ 
New loan and payment:
 $\begin{align*} L' &= 187{,}433.35 + 2{,}000 \\ &= 189{,}433.35 \\ P' &= \frac{189{,}433.35}{a_{\overline{12}|0.05}} \\ &= \frac{189{,}433.35}{8.86325} \\ &= 21{,}372.90 \end{align*}$ 
The payment falls by $2,225.34 a year. The first new payment contains $0.05 \times 189{,}433.35 = 9{,}471.67$ of interest and $21{,}372.90 - 9{,}471.67 = 11{,}901.23$ of principal — the old loan's next payment would have carried $0.07 \times 187{,}433.35 = 13{,}120.33$ of interest.

Keeping the Old Payment to Shorten the Term In the example above, the borrower instead keeps paying $23,598.23 a year on the new $189,433.35 loan at 5%. How many payments are needed, and what is the final one?
Answer
 $\begin{align*} a_{\overline{n'}|0.05} &= \frac{189{,}433.35}{23{,}598.23} \\ &= 8.02744 \\ n' &= \frac{-\ln(1 - 0.05 \times 8.02744)}{\ln 1.05} \\ &= 10.52 \end{align*}$ 
So 10 full payments and a smaller 11th. The balance after 10 payments, retrospectively:
 $\begin{align*} OB_{10} &= 189{,}433.35(1.05)^{10} - 23{,}598.23\,s_{\overline{10}|0.05} \\ &= 11{,}750.95 \\ \text{Final payment} &= 11{,}750.95 \times 1.05 \\ &= 12{,}338.49 \end{align*}$ 
The loan is paid off in 11 years instead of 12, with a final payment of $12,338.49.