Exam FM — Loans (Amortization & Sinking Funds) Flashcards

The borrower makes a series of payments; each payment is split into an **interest** portion (on the outstanding balance) and a **principal** portion that reduces the balance. The outstanding balance falls to exactly $0$ with the last payment. The loan amount equals the present value of all payments at the loan rate.

The loan equals the present value of the payment stream: $L=P\cdot a_{\overline{n|}}$, where $a_{\overline{n|}}=\dfrac{1-v^{n}}{i}$ is the annuity-immediate factor and $v=(1+i)^{-1}$. So $P=\dfrac{L}{a_{\overline{n|}}}=\dfrac{L\,i}{1-v^{n}}$.

The balance equals the present value of the **remaining** payments: $B_{t}=P\cdot a_{\overline{n-t|}}$. This is usually the quickest method when the payment $P$ and the number of remaining payments $n-t$ are both known.

The balance equals the accumulated loan less the accumulated payments: $B_{t}=L\,(1+i)^{t}-P\cdot s_{\overline{t|}}$, where $s_{\overline{t|}}=\dfrac{(1+i)^{t}-1}{i}$. Use this when you know the original loan $L$ and the payments made so far but not the total number of payments $n$.

For a loan whose payments exactly amortize it at the stated rate, both methods give **identical** balances — pick whichever inputs you have. Prospective ($P\cdot a_{\overline{n-t|}}$) is easier when the remaining term is clean; retrospective ($L(1+i)^{t}-P\,s_{\overline{t|}}$) is easier when $n$ is unknown or the final payment is irregular. With a drop/balloon payment, retrospective is safer.

Interest is charged on the balance at the **start** of the period — the balance right after the previous payment: $I_{t+1}=i\cdot B_{t}$. A common error is to use $B_{t+1}$ (the end-of-period balance) instead of $B_t$.

$PR_{t+1}=P-I_{t+1}=P-i\,B_{t}$. For a level-payment loan there is a clean closed form: $PR_{t+1}=P\,v^{\,n-t}$ (the principal in a payment is $P$ discounted from the loan's end back to that payment).

The principal portions form a **geometric** sequence growing by the factor $(1+i)$: $PR_{t+1}=PR_{1}\,(1+i)^{t}$. Equivalently, the interest portions **decline** each period. Early payments are mostly interest; later payments are mostly principal. (This shortcut holds only for **level** payments.)

The principal portions must repay the whole loan, so $\sum PR_{k}=L$. The total interest paid is the total of all payments minus the principal: for level payments, $\text{Total interest}=n\,P-L$. There is no single rate to multiply by — interest is the residual after principal.

Columns: payment number, payment amount $P$, interest paid $I_{t}=i\,B_{t-1}$, principal repaid $PR_{t}=P-I_{t}$, and outstanding balance $B_{t}=B_{t-1}-PR_{t}$. Each row's balance feeds the next row's interest. The interest column shrinks, the principal column grows by $(1+i)$, and the final balance is $0$.

$P=\dfrac{L}{a_{\overline{10|}}}$ at $i=0.08$. $a_{\overline{10|}}=\dfrac{1-1.08^{-10}}{0.08}$. Since $1.08^{-10}\approx 0.463193$, $a_{\overline{10|}}\approx \dfrac{0.536807}{0.08}\approx 6.710081$. $P=\dfrac{10000}{6.710081}\approx \$1{,}490.29$.

$B_{4}=P\cdot a_{\overline{6|}}$ at $i=0.08$. $a_{\overline{6|}}=\dfrac{1-1.08^{-6}}{0.08}\approx \dfrac{1-0.630170}{0.08}\approx 4.622880$. $B_{4}=1490.29\times 4.622880\approx \$6{,}889.45$. (The retrospective formula $10000(1.08)^{4}-1490.29\,s_{\overline{4|}}$ gives the same $\$6{,}889.45$.)

Interest is on the prior balance: $I_{5}=i\,B_{4}=0.08\times 6889.45\approx \$551.16$. Principal: $PR_{5}=P-I_{5}=1490.29-551.16\approx \$939.14$. Check via the geometric shortcut: $PR_{1}=P-i\,L=1490.29-800=690.29$, and $690.29\times 1.08^{4}\approx 939.14$. ✓

Use the general closed form $PR_{t}=P\,v^{\,n-t+1}$ (principal in payment number $t$). With $n=12$, $t=5$: $PR_{5}=1500\,(1.06)^{-8}$. $1.06^{-8}\approx 0.627412$, so $PR_{5}\approx 1500\times 0.627412\approx \$941.12$. Interest: $I_{5}=P-PR_{5}=1500-941.12\approx \$558.88$.

$P=\dfrac{200000}{a_{\overline{30|}}}$, $a_{\overline{30|}}=\dfrac{1-1.06^{-30}}{0.06}\approx 13.764831$, so $P\approx \$14{,}529.78$. Prospective balance: $B_{10}=P\cdot a_{\overline{20|}}$, $a_{\overline{20|}}=\dfrac{1-1.06^{-20}}{0.06}\approx 11.469921$. $B_{10}\approx 14529.78\times 11.469921\approx \$166{,}655.46$.

Telescoping: the total principal between payments equals the **drop in the outstanding balance**: $\sum_{k=t_1}^{t_2}PR_{k}=B_{t_1-1}-B_{t_2}$. Compute the two balances (prospective is easiest) and subtract.

Total principal $=B_{2}-B_{6}$ (prospective). $B_{2}=P\,a_{\overline{8|}}=1490.29\times 5.746639\approx \$8{,}564.16$. $B_{6}=P\,a_{\overline{4|}}=1490.29\times 3.312127\approx \$4{,}936.03$. Principal in payments 3–6 $=8564.16-4936.03\approx \$3{,}628.13$.

Total interest $=$ (sum of payments) $-$ (principal) $=n\,P-L$. $=10\times 1490.29-10000=14902.90-10000\approx \$4{,}902.95$. (Do **not** multiply the loan by a rate — interest is the residual after principal is repaid.)

The principal portion is **largest in the last payment** (it grows geometrically by $(1+i)$). For an $n$-payment loan, the principal in the final payment is $PR_{n}=P\,v=\dfrac{P}{1+i}$, and its interest is $I_{n}=P-PR_{n}=P\dfrac{i}{1+i}=Pd$.

Final-payment principal: $PR_{5}=P\,v=1000\,(1.10)^{-1}\approx \$909.09$. Final-payment interest: $I_{5}=P-PR_{5}=1000-909.09\approx \$90.91$ (equivalently $Pd=1000\times\tfrac{0.10}{1.10}$). The last payment is almost all principal, as expected.

Fall back to the definitions via the **prospective** outstanding balance. For payment number $t+1$: $I_{t+1}=i\,B_{t}$ and $PR_{t+1}=P_{t+1}-I_{t+1}$, where $P_{t+1}$ is that period's (varying) payment. The only extra work is getting $B_{t}$: discount the **remaining** payments back to time $t$, $B_{t}=\sum_{k=t+1}^{n}P_{k}\,v^{\,k-t}$. The closed forms $PR_{t+1}=P\,v^{\,n-t}$ and $PR_{t+1}=PR_1(1+i)^{t}$ are **level-payment only** — never use them here.

Loan $=900\,a_{\overline{10|}}+100\,(Ia)_{\overline{10|}}$ at $6\%$ (writing payment $k$ as $900+100k$). With $a_{\overline{10|}}\approx 7.360087$ and $(Ia)_{\overline{10|}}=\dfrac{\ddot a_{\overline{10|}}-10v^{10}}{i}\approx 36.962408$: $L\approx 900(7.360087)+100(36.962408)\approx \$10{,}320.32$. Split the 4th payment via the prospective balance after payment $3$ — the PV of the remaining payments $\$1{,}300,\dots,\$1{,}900$ (first $\$1{,}300$, increment $\$100$, $7$ left): $B_{3}=1200\,a_{\overline{7|}}+100\,(Ia)_{\overline{7|}}$. With $a_{\overline{7|}}\approx 5.582381$, $(Ia)_{\overline{7|}}\approx 21.032076$: $B_{3}\approx 1200(5.582381)+100(21.032076)\approx 6698.86+2103.21\approx \$8{,}802.07$. Interest: $I_{4}=i\,B_{3}=0.06\times 8802.07\approx \$528.12$. Principal: $PR_{4}=P_{4}-I_{4}=1300-528.12\approx \$771.88$ (and $B_{4}=B_{3}-PR_{4}\approx \$8{,}030.19$).

Refinancing replaces the remaining obligation with a new loan equal to the **current outstanding balance** $B_t$, repaid at a new rate $i'$ over a new term $m$. The new payment is $P'=\dfrac{B_{t}}{a_{\overline{m|}\,i'}}$. Critical: the new loan is the balance at refinance, **not** the original principal, and you reset $n$ to the new remaining term.

New payment on the balance: $P'=\dfrac{166655.46}{a_{\overline{20|}\,0.045}}$. $a_{\overline{20|}}=\dfrac{1-1.045^{-20}}{0.045}\approx 13.007936$, so $P'\approx \$12{,}811.83$. Annual saving $=14529.78-12811.83\approx \$1{,}717.95$.

Solving $a_{\overline{n|}}=L/P$ for $n$ rarely yields a whole number, so a fractional final period remains. Two conventions: a **balloon** payment (a larger-than-$P$ amount paid with the last regular payment, at the earlier date) or a **drop** payment (a smaller final payment one period later). Both clear the same outstanding balance, accumulated to the relevant date.

Solve $a_{\overline{n|}}=\tfrac{50000}{8000}=6.25$ at $9\%$: $n\approx 9.59$, so there are **$9$ full payments**, with a smaller payment at time $10$. Balance after $9$: $B_{9}=50000(1.09)^{9}-8000\,s_{\overline{9|}}\approx \$4{,}426.37$. Drop payment $=B_{9}(1.09)\approx \$4{,}824.75$ at time $10$.

A balloon clears the loan at time $9$ in one larger payment. Balance after $8$ payments: $B_{8}=50000(1.09)^{8}-8000\,s_{\overline{8|}}\approx \$11{,}400.34$. Balloon $=B_{8}(1.09)\approx \$12{,}426.37$ at time $9$ (this single payment replaces the $9$th regular $\$8{,}000$ and clears the loan).

The borrower keeps the full loan $L$ outstanding and (1) pays the lender **interest only** each period, $i\cdot L$, and (2) makes level deposits into a separate **sinking fund** earning rate $j$, sized so the fund accumulates to exactly $L$ at maturity to repay the principal in one lump sum.

Deposit: $D=\dfrac{L}{s_{\overline{n|}\,j}}$, since $D\,s_{\overline{n|}\,j}=L$ at maturity. Total periodic outlay $=i\,L+\dfrac{L}{s_{\overline{n|}\,j}}$ — the first term is the interest paid to the lender, the second is the fund deposit. Key point: $i$ (paid to the lender) and $j$ (earned in the fund) are generally **different** rates.

Interest to lender: $0.10\times 20000=\$2{,}000$. Deposit: $D=\dfrac{20000}{s_{\overline{8|}\,0.06}}$, $s_{\overline{8|}\,0.06}=\dfrac{1.06^{8}-1}{0.06}\approx 9.897468$, so $D\approx \$2{,}020.72$. Total annual outlay $=2000+2020.72\approx \$4{,}020.72$.

The face loan $L$ stays outstanding to the lender, but the borrower's **net** liability is $L$ minus the accumulated sinking fund. Writing the net balance at time $t$ as $NB_{t}$: $NB_{t}=L-D\,s_{\overline{t|}\,j}$. This net balance plays the role that the outstanding balance does under amortization, and reaches $0$ at maturity when $D\,s_{\overline{n|}\,j}=L$.

$D=\dfrac{30000}{s_{\overline{10|}\,0.06}}$, $s_{\overline{10|}\,0.06}\approx 13.180795$, so $D\approx \$2{,}276.04$. Fund after $4$ deposits: $D\,s_{\overline{4|}\,0.06}=2276.04\times 4.374616\approx \$9{,}956.80$. Net loan balance $=30000-9956.80\approx \$20{,}043.20$.

Each period the fund earns $j$ on its prior balance; this interest is **retained inside the fund**, accelerating its growth. The borrower's level deposit $D$ is the *new money* added; together with reinvested fund interest, the fund reaches $L$ at maturity. The borrower's deposit alone (without fund interest) sums to only $nD<L$.

Under **amortization**, each payment immediately reduces the balance, so the lender's exposure falls every period and interest is charged on a shrinking balance. Under the **sinking fund** method, the full $L$ stays owed to the lender the entire term (interest is paid on the full $L$ every period), while a side fund grows to repay it at the end.

They are **equal**. Algebraically $\dfrac{1}{a_{\overline{n|}}}=\dfrac{1}{s_{\overline{n|}}}+i$ (the deposit-plus-interest identity), so $i\,L+\dfrac{L}{s_{\overline{n|}\,i}}=\dfrac{L}{a_{\overline{n|}\,i}}$, which is exactly the amortization payment. The two methods are financially identical only when $i=j$.

The **sinking fund** method costs **more**. The borrower pays interest on the full $L$ all term, yet the fund accumulates at the slower rate $j<i$, so it needs larger deposits to reach $L$. The amortization payment $L/a_{\overline{n|}\,i}$ is cheaper. Only when $j\geq i$ can the sinking fund match or beat amortization.

(A) Amortization: $P=\dfrac{100000}{a_{\overline{15|}\,0.07}}=\dfrac{100000}{9.107914}\approx \$10{,}979.46$; total paid $\approx \$164{,}691.94$. (B) Sinking fund: deposit $=\dfrac{100000}{s_{\overline{15|}\,0.05}}=\dfrac{100000}{21.578564}\approx \$4{,}634.23$; annual outlay $=7000+4634.23\approx \$11{,}634.23$; total $\approx \$174{,}513.43$. Sinking fund costs $\approx\$9{,}821$ more because $j=5\%<i=7\%$.

Set the sinking-fund total outlay equal to an amortization payment and solve for the rate $i^{*}$: $i\,L+\dfrac{L}{s_{\overline{n|}\,j}}=\dfrac{L}{a_{\overline{n|}\,i^{*}}}$, i.e. $a_{\overline{n|}\,i^{*}}$ equals $L$ divided by the total periodic outlay. Solve for $i^{*}$ by calculator/iteration. When $j<i$, the equivalent rate $i^{*}>i$.

Solve $a_{\overline{8|}\,i^{*}}=\dfrac{20000}{4020.72}\approx 4.974232$. Finding the rate with $a_{\overline{8|}}=4.97423$ (e.g. BA II Plus: $N=8$, $PV=-4.97423$, $PMT=1$, $FV=0$, solve $I/Y$) gives $i^{*}\approx 11.96\%$. As expected, $i^{*}>10\%$ because the fund earns only $6\%$.

Original loan $=2000\cdot a_{\overline{12|}\,0.05}=2000\times 8.863252\approx \$17{,}726.50$. Total payments $=12\times 2000=\$24{,}000$. Total interest $=24000-17726.50\approx \$6{,}273.50$.