{ "deckName": "Exam FM — Loans (Amortization & Sinking Funds)", "examCode": "Exam FM", "cards": [ { "front": "Define the **amortization method** of repaying a loan.", "back": "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.", "tag": "Amortization basics" }, { "front": "For a level-payment amortized loan of amount $L$ with $n$ payments of $P$ at rate $i$ per period, how are $L$ and $P$ related?", "back": "The loan equals the present value of the payment stream:\n$L=P\\cdot a_{\\overline{n|}}$, where $a_{\\overline{n|}}=\\dfrac{1-v^{n}}{i}$ is the annuity-immediate factor and $v=(1+i)^{-1}$.\nSo $P=\\dfrac{L}{a_{\\overline{n|}}}=\\dfrac{L\\,i}{1-v^{n}}$.", "tag": "Amortization basics" }, { "front": "State the **prospective** formula for the outstanding loan balance $B_t$ immediately after the $t$-th payment (level payments).", "back": "The balance equals the present value of the **remaining** payments:\n$B_{t}=P\\cdot a_{\\overline{n-t|}}$.\nThis is usually the quickest method when the payment $P$ and the number of remaining payments $n-t$ are both known.", "tag": "Outstanding balance" }, { "front": "State the **retrospective** formula for the outstanding loan balance $B_t$ immediately after the $t$-th payment.", "back": "The balance equals the accumulated loan less the accumulated payments:\n$B_{t}=L\\,(1+i)^{t}-P\\cdot s_{\\overline{t|}}$, where $s_{\\overline{t|}}=\\dfrac{(1+i)^{t}-1}{i}$.\nUse this when you know the original loan $L$ and the payments made so far but not the total number of payments $n$.", "tag": "Outstanding balance" }, { "front": "When do the **prospective** and **retrospective** outstanding-balance formulas give the same answer, and when should you prefer one?", "back": "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.", "tag": "Outstanding balance" }, { "front": "How is the **interest portion** $I_{t+1}$ of payment number $t+1$ computed?", "back": "Interest is charged on the balance at the **start** of the period — the balance right after the previous payment:\n$I_{t+1}=i\\cdot B_{t}$.\nA common error is to use $B_{t+1}$ (the end-of-period balance) instead of $B_t$.", "tag": "Interest & principal split" }, { "front": "How is the **principal portion** $PR_{t+1}$ of payment $t+1$ computed for a level-payment loan, and what is its closed form?", "back": "$PR_{t+1}=P-I_{t+1}=P-i\\,B_{t}$.\nFor 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).", "tag": "Interest & principal split" }, { "front": "On a level-payment amortized loan, how does the **principal repaid** in successive payments behave over time?", "back": "The principal portions form a **geometric** sequence growing by the factor $(1+i)$:\n$PR_{t+1}=PR_{1}\\,(1+i)^{t}$.\nEquivalently, the interest portions **decline** each period. Early payments are mostly interest; later payments are mostly principal. (This shortcut holds only for **level** payments.)", "tag": "Interest & principal split" }, { "front": "What is the **sum of all principal repayments** over the life of an amortized loan, and the **sum of all interest payments**?", "back": "The principal portions must repay the whole loan, so $\\sum PR_{k}=L$.\nThe 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.", "tag": "Total interest" }, { "front": "Describe the columns of a standard **amortization schedule** and how each row is built.", "back": "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 $0$.", "tag": "Amortization basics" }, { "front": "A $\\$10{,}000$ loan is repaid with $10$ equal annual payments at an annual effective rate of $8\\%$. Find the level annual payment.", "back": "$P=\\dfrac{L}{a_{\\overline{10|}}}$ at $i=0.08$.\n$a_{\\overline{10|}}=\\dfrac{1-1.08^{-10}}{0.08}$. Since $1.08^{-10}\\approx 0.463193$, $a_{\\overline{10|}}\\approx \\dfrac{0.536807}{0.08}\\approx 6.710081$.\n$P=\\dfrac{10000}{6.710081}\\approx \\$1{,}490.29$.", "tag": "Calculation: payment" }, { "front": "For the $\\$10{,}000$, $10$-year, $8\\%$ loan with level payment $P\\approx\\$1{,}490.29$, find the outstanding balance immediately after the $4$th payment (prospective).", "back": "$B_{4}=P\\cdot a_{\\overline{6|}}$ at $i=0.08$.\n$a_{\\overline{6|}}=\\dfrac{1-1.08^{-6}}{0.08}\\approx \\dfrac{1-0.630170}{0.08}\\approx 4.622880$.\n$B_{4}=1490.29\\times 4.622880\\approx \\$6{,}889.45$.\n(The retrospective formula $10000(1.08)^{4}-1490.29\\,s_{\\overline{4|}}$ gives the same $\\$6{,}889.45$.)", "tag": "Calculation: balance" }, { "front": "For the same $\\$10{,}000$, $8\\%$, $10$-year loan ($P\\approx\\$1{,}490.29$, $B_{4}\\approx\\$6{,}889.45$), split the **5th** payment into interest and principal.", "back": "Interest is on the prior balance: $I_{5}=i\\,B_{4}=0.08\\times 6889.45\\approx \\$551.16$.\nPrincipal: $PR_{5}=P-I_{5}=1490.29-551.16\\approx \\$939.14$.\nCheck via the geometric shortcut: $PR_{1}=P-i\\,L=1490.29-800=690.29$, and $690.29\\times 1.08^{4}\\approx 939.14$. ✓", "tag": "Calculation: split" }, { "front": "A loan is repaid by $12$ level annual payments of $\\$1{,}500$ at $6\\%$. Without building a schedule, find the **interest** and **principal** in the $5$th payment.", "back": "Use the general closed form $PR_{t}=P\\,v^{\\,n-t+1}$ (principal in payment number $t$). With $n=12$, $t=5$: $PR_{5}=1500\\,(1.06)^{-8}$.\n$1.06^{-8}\\approx 0.627412$, so $PR_{5}\\approx 1500\\times 0.627412\\approx \\$941.12$.\nInterest: $I_{5}=P-PR_{5}=1500-941.12\\approx \\$558.88$.", "tag": "Calculation: split" }, { "front": "A loan of $\\$200{,}000$ is amortized with $30$ level annual payments at $6\\%$. Find the payment, then the outstanding balance just after the $10$th payment.", "back": "$P=\\dfrac{200000}{a_{\\overline{30|}}}$, $a_{\\overline{30|}}=\\dfrac{1-1.06^{-30}}{0.06}\\approx 13.764831$, so $P\\approx \\$14{,}529.78$.\nProspective balance: $B_{10}=P\\cdot a_{\\overline{20|}}$, $a_{\\overline{20|}}=\\dfrac{1-1.06^{-20}}{0.06}\\approx 11.469921$.\n$B_{10}\\approx 14529.78\\times 11.469921\\approx \\$166{,}655.46$.", "tag": "Calculation: balance" }, { "front": "How do you find the **total principal repaid between two payments** (say payments $t_1$ through $t_2$) without summing every row?", "back": "Telescoping: the total principal between payments equals the **drop in the outstanding balance**:\n$\\sum_{k=t_1}^{t_2}PR_{k}=B_{t_1-1}-B_{t_2}$.\nCompute the two balances (prospective is easiest) and subtract.", "tag": "Total interest" }, { "front": "For the $\\$10{,}000$, $8\\%$, $10$-year loan ($P\\approx\\$1{,}490.29$), find the total **principal** repaid in payments $3$ through $6$.", "back": "Total principal $=B_{2}-B_{6}$ (prospective).\n$B_{2}=P\\,a_{\\overline{8|}}=1490.29\\times 5.746639\\approx \\$8{,}564.16$.\n$B_{6}=P\\,a_{\\overline{4|}}=1490.29\\times 3.312127\\approx \\$4{,}936.03$.\nPrincipal in payments 3–6 $=8564.16-4936.03\\approx \\$3{,}628.13$.", "tag": "Calculation: split" }, { "front": "For the $\\$10{,}000$, $8\\%$, $10$-year loan ($P\\approx\\$1{,}490.29$), what is the **total interest** paid over the life of the loan?", "back": "Total interest $=$ (sum of payments) $-$ (principal) $=n\\,P-L$.\n$=10\\times 1490.29-10000=14902.90-10000\\approx \\$4{,}902.95$.\n(Do **not** multiply the loan by a rate — interest is the residual after principal is repaid.)", "tag": "Total interest" }, { "front": "On a level amortized loan, in which payment is the principal portion largest, and what is the principal in the **final** payment?", "back": "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 $I_{n}=P-PR_{n}=P\\dfrac{i}{1+i}=Pd$.", "tag": "Interest & principal split" }, { "front": "A $5$-payment loan at $10\\%$ has level annual payments of $\\$1{,}000$. Find the interest and principal contained in the **last** payment.", "back": "Final-payment principal: $PR_{5}=P\\,v=1000\\,(1.10)^{-1}\\approx \\$909.09$.\nFinal-payment interest: $I_{5}=P-PR_{5}=1000-909.09\\approx \\$90.91$ (equivalently $Pd=1000\\times\\tfrac{0.10}{1.10}$).\nThe last payment is almost all principal, as expected.", "tag": "Calculation: split" }, { "front": "For a loan repaid by **non-level** payments (e.g. an arithmetically increasing annuity), the geometric principal shortcut fails. How do you split a given payment into interest and principal?", "back": "Fall back to the definitions via the **prospective** outstanding balance. For payment number $t+1$:\n$I_{t+1}=i\\,B_{t}$ and $PR_{t+1}=P_{t+1}-I_{t+1}$, where $P_{t+1}$ is that period's (varying) payment.\nThe only extra work is getting $B_{t}$: discount the **remaining** payments back to time $t$,\n$B_{t}=\\sum_{k=t+1}^{n}P_{k}\\,v^{\\,k-t}$.\nThe 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.", "tag": "Interest & principal split" }, { "front": "A loan is repaid by $10$ annual payments at $6\\%$ that **increase** by $\\$100$ each year: $\\$1{,}000,\\,\\$1{,}100,\\,\\dots,\\,\\$1{,}900$. Find the loan amount, then split the **4th** payment ($\\$1{,}300$) into interest and principal.", "back": "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 $(Ia)_{\\overline{10|}}=\\dfrac{\\ddot a_{\\overline{10|}}-10v^{10}}{i}\\approx 36.962408$:\n$L\\approx 900(7.360087)+100(36.962408)\\approx \\$10{,}320.32$.\nSplit 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):\n$B_{3}=1200\\,a_{\\overline{7|}}+100\\,(Ia)_{\\overline{7|}}$. With $a_{\\overline{7|}}\\approx 5.582381$, $(Ia)_{\\overline{7|}}\\approx 21.032076$:\n$B_{3}\\approx 1200(5.582381)+100(21.032076)\\approx 6698.86+2103.21\\approx \\$8{,}802.07$.\nInterest: $I_{4}=i\\,B_{3}=0.06\\times 8802.07\\approx \\$528.12$.\nPrincipal: $PR_{4}=P_{4}-I_{4}=1300-528.12\\approx \\$771.88$ (and $B_{4}=B_{3}-PR_{4}\\approx \\$8{,}030.19$).", "tag": "Calculation: split" }, { "front": "What is **refinancing**, and how do you compute the new payment when a loan is refinanced after $t$ payments at a new rate?", "back": "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 $P'=\\dfrac{B_{t}}{a_{\\overline{m|}\\,i'}}$.\nCritical: the new loan is the balance at refinance, **not** the original principal, and you reset $n$ to the new remaining term.", "tag": "Refinancing" }, { "front": "A $\\$200{,}000$, $30$-year loan at $6\\%$ ($P\\approx\\$14{,}529.78$, $B_{10}\\approx\\$166{,}655.46$) is refinanced after $10$ payments at $4.5\\%$ over the remaining $20$ years. Find the new payment and annual saving.", "back": "New payment on the balance: $P'=\\dfrac{166655.46}{a_{\\overline{20|}\\,0.045}}$.\n$a_{\\overline{20|}}=\\dfrac{1-1.045^{-20}}{0.045}\\approx 13.007936$, so $P'\\approx \\$12{,}811.83$.\nAnnual saving $=14529.78-12811.83\\approx \\$1{,}717.95$.", "tag": "Refinancing" }, { "front": "Why does the number of payments $n$ sometimes come out **non-integer**, and what are the two ways to handle the final irregular amount?", "back": "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.", "tag": "Drop & balloon" }, { "front": "A $\\$50{,}000$ loan at $9\\%$ is repaid with annual payments of $\\$8{,}000$ until a final smaller (**drop**) payment clears it. How many full payments are made, and what is the drop payment?", "back": "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 $10$.\nBalance after $9$: $B_{9}=50000(1.09)^{9}-8000\\,s_{\\overline{9|}}\\approx \\$4{,}426.37$.\nDrop payment $=B_{9}(1.09)\\approx \\$4{,}824.75$ at time $10$.", "tag": "Drop & balloon" }, { "front": "For the same $\\$50{,}000$, $9\\%$ loan with $\\$8{,}000$ payments, suppose the borrower instead makes a **balloon** payment with the $9$th payment. Find the balloon amount.", "back": "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$.\nBalloon $=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).", "tag": "Drop & balloon" }, { "front": "Define the **sinking fund method** of loan repayment and the borrower's two periodic obligations.", "back": "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.", "tag": "Sinking fund" }, { "front": "Give the formulas for the periodic sinking-fund **deposit** and the borrower's **total periodic outlay**, with loan rate $i$ and fund rate $j$.", "back": "Deposit: $D=\\dfrac{L}{s_{\\overline{n|}\\,j}}$, since $D\\,s_{\\overline{n|}\\,j}=L$ at maturity.\nTotal 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.\nKey point: $i$ (paid to the lender) and $j$ (earned in the fund) are generally **different** rates.", "tag": "Sinking fund" }, { "front": "A $\\$20{,}000$ loan is repaid by the sinking fund method over $8$ years: interest at $10\\%$ is paid to the lender annually, and the sinking fund earns $6\\%$. Find the total annual outlay.", "back": "Interest to lender: $0.10\\times 20000=\\$2{,}000$.\nDeposit: $D=\\dfrac{20000}{s_{\\overline{8|}\\,0.06}}$, $s_{\\overline{8|}\\,0.06}=\\dfrac{1.06^{8}-1}{0.06}\\approx 9.897468$, so $D\\approx \\$2{,}020.72$.\nTotal annual outlay $=2000+2020.72\\approx \\$4{,}020.72$.", "tag": "Calculation: sinking fund" }, { "front": "Under the sinking fund method, what is the **net amount owed** (net loan balance) at an intermediate time, and how does it relate to the fund?", "back": "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 $NB_{t}$:\n$NB_{t}=L-D\\,s_{\\overline{t|}\\,j}$.\nThis net balance plays the role that the outstanding balance does under amortization, and reaches $0$ at maturity when $D\\,s_{\\overline{n|}\\,j}=L$.", "tag": "Sinking fund" }, { "front": "A $\\$30{,}000$ loan uses a sinking fund earning $6\\%$ over $10$ years. Find the annual deposit, the fund balance after $4$ deposits, and the net loan balance then.", "back": "$D=\\dfrac{30000}{s_{\\overline{10|}\\,0.06}}$, $s_{\\overline{10|}\\,0.06}\\approx 13.180795$, so $D\\approx \\$2{,}276.04$.\nFund after $4$ deposits: $D\\,s_{\\overline{4|}\\,0.06}=2276.04\\times 4.374616\\approx \\$9{,}956.80$.\nNet loan balance $=30000-9956.80\\approx \\$20{,}043.20$.", "tag": "Calculation: sinking fund" }, { "front": "Under the sinking fund method, how is the **interest earned by the fund** each period treated, and what does the borrower's deposit cover?", "back": "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 $nDi$.", "tag": "Comparison" }, { "front": "For the $\\$20{,}000$, $8$-year sinking fund (lender rate $10\\%$, fund rate $6\\%$, total outlay $\\approx\\$4{,}020.72$), find the equivalent level amortization rate.", "back": "Solve $a_{\\overline{8|}\\,i^{*}}=\\dfrac{20000}{4020.72}\\approx 4.974232$.\nFinding 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 $i^{*}\\approx 11.96\\%$.\nAs expected, $i^{*}>10\\%$ because the fund earns only $6\\%$.", "tag": "Calculation: comparison" }, { "front": "A loan is repaid with payments of $\\$2{,}000$ per year for $12$ years at $5\\%$. The borrower wants to know the original loan amount and the total interest paid.", "back": "Original loan $=2000\\cdot a_{\\overline{12|}\\,0.05}=2000\\times 8.863252\\approx \\$17{,}726.50$.\nTotal payments $=12\\times 2000=\\$24{,}000$.\nTotal interest $=24000-17726.50\\approx \\$6{,}273.50$.", "tag": "Calculation: payment" } ] }