The complete supplied teaching material, including deeper learning and source qualifications. Reflection questions here are not automatically graded.
Week 2 — Financial Mathematics
Monastery Complete Masterclass · Academic English · HS2026 Mathematics with Statistics A
How to use this course
Start with a short lesson, attempt its problems before consulting the solution, and move on as soon as your explanation and calculations are correct. An arithmetic slip does not by itself justify restarting a lesson. Use the mixed set at the end without formula cues.
Unified financial notation
Let time 0 be the valuation origin. Write $i$ for the effective interest rate per payment period, $n$ for an integer number of periods, $P$ for value at time 0, $A$ for a later accumulated value, and $C$ for an equal periodic cash flow. For an annual nominal rate compounded $m$ times annually use $j$, and for signed dated net inflows use $B_k$. A period is not automatically a year.
Cash-flow signs must be unambiguous: inflows positive, outflows negative. An initial cost $K_0>0$ appears as $B_0=-K_0$.
1. Geometric series and payment streams
Classification: prerequisite. Learning objective: Sum finite geometric progressions; identify time indices; relate a cash-flow timeline to a sum.
Intuition
A stream is not a single amount: each payment occupies a distinct date. Multiplying by a fixed discount factor makes the resulting values a geometric progression.
Definition and primary equation
\[\sum_{k=0}^{n-1}aq^k=a\frac{1-q^n}{1-q}\]
Derivation
Start with S=a+aq+...+aq^{n-1}. Multiply by q and subtract: (1-q)S=a(1-q^n). Dividing requires q≠1; if q=1 the sum is na.
Conditions and boundary cases
q≠1 for the fraction; n a nonnegative integer; for infinite sums require |q|<1.
Worked examples: increasing difficulty
1. Introductory — W2-C01-E01
Compute 100+110+121.
Worked solution. 100(1+1.1+1.1²)=331.
Numerical check: 331 (round only at the end).
2. Standard — W2-C01-E02
Find the sum of six annual cash flows 200, 210, ... growing at 5%.
Worked solution. 200((1.05^6−1)/0.05).
Numerical check: 1,360.3826 (round only at the end).
3. Transfer — W2-C01-E03
Calculate the infinite sum 30+24+19.2+....
Worked solution. 30/(1−0.8)=150.
Numerical check: 150 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for geometric series and payment streams. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C01-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
2. Compound accumulation and inverse problems
Classification: core. Learning objective: Compute future amounts, recover initial principal, rates and integer holding periods.
Intuition
Interest is credited on both the original principal and all earlier interest. Each period multiplies the balance by the same factor.
Definition and primary equation
\[A_n=P(1+i)^n\]
Derivation
A_1=P(1+i), A_2=P(1+i)^2, hence by induction A_n=P(1+i)^n. For P,A>0 and i>0, solving for time gives n=ln(A/P)/ln(1+i).
Conditions and boundary cases
Discrete integer periods n≥0; i>−1, positive bases for real logarithms; when i=0, A_n=P and time cannot be recovered from growth.
Worked examples: increasing difficulty
1. Introductory — W2-C02-E01
Invest CHF 2,000 for three years at 4% effective annually.
Worked solution. 2000(1.04)^3.
Numerical check: 2,249.7280 (round only at the end).
2. Standard — W2-C02-E02
What is the minimum full number of years for CHF 5,000 to exceed CHF 6,000 at 3%?
Worked solution. n>ln(1.2)/ln(1.03); choose the smallest integer satisfying strict exceedance..
Numerical check: 7 (round only at the end).
3. Transfer — W2-C02-E03
CHF 8,000 earns 2% for two years then 5% for three more.
Worked solution. 8000(1.02)^2(1.05)^3.
Numerical check: 9,635.1444 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for compound accumulation and inverse problems. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C02-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
3. Nominal rates, effective rates and continuous compounding
Classification: core. Learning objective: Convert compounding conventions; compare bank rates on equivalent horizons.
Intuition
A quoted annual rate is not necessarily the actual percentage earned in one year. Interest frequency changes the multiplier.
Definition and primary equation
\[A=P(1+j/m)^{mt};\quad i_{\mathrm{eff}}=(1+j/m)^m-1\]
Derivation
Each of mt subperiods has rate j/m. Over one year the multiplier is (1+j/m)^m. Taking the m→∞ limit gives e^j and A=Pe^{jt}.
Conditions and boundary cases
m positive integer; mt integer for literal discrete payments; 1+j/m>0; continuous rate j is a force of interest, not an effective annual rate.
Worked examples: increasing difficulty
1. Introductory — W2-C03-E01
Find the effective annual rate of 6% nominal compounded quarterly.
Worked solution. (1+0.06/4)^4−1.
Numerical check: 0.0614 (round only at the end).
2. Standard — W2-C03-E02
Compare 4% continuously compounded and 4.1% effective for one year.
Worked solution. e^0.04−1 vs 0.041; the 4.1% effective offer is higher..
Numerical check: 0.0408 (round only at the end).
3. Transfer — W2-C03-E03
What nominal rate compounded monthly is equivalent to 5% effective annually?
Worked solution. j=12((1.05)^(1/12)−1).
Numerical check: 0.0489 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for nominal rates, effective rates and continuous compounding. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C03-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
4. Present value, future value and valuation dates
Classification: core. Learning objective: Bring all amounts to one common date; recognise delayed cash flows.
Intuition
Money at different dates cannot be added as if it had the same value. The comparison date is part of the problem.
Definition and primary equation
\[PV=\frac{F_t}{(1+i)^t};\quad V_s=\sum_k B_k(1+i)^{s-k}\]
Derivation
An amount X at t becomes X(1+i)^{s-t} at s. When s<t, the exponent is negative and the operation is discounting.
Conditions and boundary cases
Consistent periodic discount rate and period length; i>−1; one agreed valuation date.
Worked examples: increasing difficulty
1. Introductory — W2-C04-E01
Present value of CHF 1,500 received after four years at 5%.
Worked solution. 1500/(1.05)^4.
Numerical check: 1,234.0537 (round only at the end).
2. Standard — W2-C04-E02
At time 0 compare CHF 2,000 now with CHF 2,200 in two years, using 4%.
Worked solution. 2200/(1.04)^2 vs 2000; future offer has higher PV..
Numerical check: 2,034.0237 (round only at the end).
3. Transfer — W2-C04-E03
Value CHF 300 at time 1 and CHF 500 at time 4 at time 2, using 5%.
Worked solution. 300(1.05)+500/(1.05)^2.
Numerical check: 768.5147 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for present value, future value and valuation dates. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C04-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
5. Project valuation and net present value
Classification: core. Learning objective: Build net cash flow timelines; calculate NPV and rank mutually exclusive choices.
Intuition
A project is worthwhile relative to the opportunity cost of capital if the discounted value of its net benefits is positive.
Definition and primary equation
\[NPV(i)=\sum_{k=0}^{n}\frac{B_k}{(1+i)^k}\]
Derivation
Write cash flow B_k=R_k−K_k where R and K are nonnegative magnitudes. Discount separately, then use linearity to combine into one NPV sum.
Conditions and boundary cases
Deterministic cash flows; same currency and aligned periodic discount rate; risk adjustment and capital rationing require further models.
Worked examples: increasing difficulty
1. Introductory — W2-C05-E01
Pay CHF 1,000 now; receive CHF 1,150 in one year; discount at 5%.
Worked solution. −1000+1150/1.05.
Numerical check: 95.2381 (round only at the end).
2. Standard — W2-C05-E02
A project costs CHF 4,000 now and yields CHF 1,500 each year for three years at 6%.
Worked solution. −4000+1500[1−(1.06)^−3]/.06.
Numerical check: 9.5179 (round only at the end).
3. Transfer — W2-C05-E03
At 5%, compare project A: −2000 now, +2300 at t=1 with project B: −2000 now, +2500 at t=2.
Worked solution. Compute both NPVs and choose the larger..
Numerical check: 267.5737 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for project valuation and net present value. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C05-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
6. Internal rate of return and decision limitations
Classification: core. Learning objective: Solve NPV(r)=0; assess when IRR rankings or existence claims fail.
Intuition
IRR asks which discount rate makes a project exactly break even. It is a root, not an additional cash flow.
Definition and primary equation
\[\sum_{k=0}^{n}\frac{B_k}{(1+r)^k}=0\]
Derivation
For a one-outflow/one-inflow project −P+F/(1+r)^n=0 implies r=(F/P)^{1/n}−1. For unconventional signs, NPV can cross zero more than once.
Conditions and boundary cases
Solve on r>−1 mathematically; restrict r>0 when following the HSG introductory presentation. IRR>i decision rule only for conventional cash flows with an appropriate unique root.
Worked examples: increasing difficulty
1. Introductory — W2-C06-E01
A project costs CHF 1,000 now and pays CHF 1,100 in one year. Find IRR.
Worked solution. −1000+1100/(1+r)=0 gives r=10%..
Numerical check: 0.1000 (round only at the end).
2. Standard — W2-C06-E02
Invest CHF 900 now for CHF 1,089 in two years. Find IRR.
Worked solution. r=sqrt(1089/900)−1..
Numerical check: 0.1000 (round only at the end).
3. Transfer — W2-C06-E03
Consider cash flows (−100, +230, −132). Find all positive IRRs.
Worked solution. −100+230/(1+r)−132/(1+r)^2=0; multiply by (1+r)^2: roots r=0.1 and r=0.2..
Numerical check: 0.2000 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for internal rate of return and decision limitations. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C06-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
7. Annuity-immediate
Classification: core. Learning objective: Identify end-of-period payments; derive and apply PV and FV annuity factors.
Intuition
An ordinary annuity pays at t=1,…,n. Each payment is discounted a different number of times.
Definition and primary equation
\[P_I=C\frac{1-(1+i)^{-n}}{i};\quad A_I=C\frac{(1+i)^n-1}{i}\]
Derivation
PV=CΣ_{k=1}^n(1+i)^{-k}. Apply the geometric sum. Multiply PV by (1+i)^n for terminal value.
Conditions and boundary cases
Equal end-of-period payments; constant periodic i>−1 and integer n≥1; factor formulas with /i require i≠0; at i=0 both values equal nC.
Worked examples: increasing difficulty
1. Introductory — W2-C07-E01
Find the PV of three end-year payments of CHF 500 at 4%.
Worked solution. 500[1−1.04^−3]/.04.
Numerical check: 1,387.5455 (round only at the end).
2. Standard — W2-C07-E02
Accumulate six annual end-year CHF 200 deposits at 5%; value at t=6.
Worked solution. 200[(1.05)^6−1]/.05.
Numerical check: 1,360.3826 (round only at the end).
3. Transfer — W2-C07-E03
Value CHF 400 paid at the end of every month for 24 months, given an effective annual rate of 6%.
Worked solution. Convert i_month=(1.06)^(1/12)−1, then compute immediate PV..
Numerical check: 9,039.7467 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for annuity-immediate. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C07-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
8. Annuity-due
Classification: core. Learning objective: Identify beginning-of-period payments and translate ordinary-annuity factors.
Intuition
At t=0,…,n−1 each payment arrives one period sooner than an otherwise identical annuity-immediate.
Definition and primary equation
\[P_D=(1+i)P_I;\quad A_D=(1+i)A_I\]
Derivation
Shift every immediate payment back by one period: CΣ_{k=0}^{n−1}(1+i)^{-k}=(1+i)P_I. The same factor applies to the value at time n.
Conditions and boundary cases
Same C,n,i in both streams; exact timing definitions; special case i=0 produces equal values.
Worked examples: increasing difficulty
1. Introductory — W2-C08-E01
PV of three CHF 500 beginning-year payments at 4%.
Worked solution. 500[1+1/1.04+1/1.04²].
Numerical check: 1,443.0473 (round only at the end).
2. Standard — W2-C08-E02
Value at t=5 of five CHF 1,000 beginning-year deposits at 3%.
Worked solution. 1000(1.03)[((1.03)^5−1)/.03].
Numerical check: 5,468.4099 (round only at the end).
3. Transfer — W2-C08-E03
A five-payment stream begins immediately, C=CHF 700, i=5%. What is its advantage over end-year timing?
Worked solution. Difference P_D−P_I=i P_I..
Numerical check: 151.5317 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for annuity-due. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C08-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
9. Payment, term and deferred-annuity inverse problems
Classification: core. Learning objective: Solve for payment or term; correctly shift deferred streams.
Intuition
Pricing a payment stream can be reversed: initial capital, instalment size, and number of payments are different unknowns of the same valuation equation.
Definition and primary equation
\[C=\frac{Pi}{1-(1+i)^{-n}};\quad n=\frac{-\ln(1-Pi/C)}{\ln(1+i)}\]
Derivation
Rearrange the immediate PV formula. For a deferred stream first value the ordinary annuity just before its first payment, then discount to time 0.
Conditions and boundary cases
For positive i,C,P require 0<Pi/C<1 for finite positive real n; for integer payments, verify feasibility and rounding; i=0 gives C=P/n.
Worked examples: increasing difficulty
1. Introductory — W2-C09-E01
Find end-year payments amortising a CHF 10,000 PV over five years at 5%.
Worked solution. C=10000×.05/(1−1.05^−5).
Numerical check: 2,309.7480 (round only at the end).
2. Standard — W2-C09-E02
For CHF 5,000 capital, CHF 800 end-year withdrawal, i=4%, find real-valued duration.
Worked solution. n=−ln(1−5000×.04/800)/ln(1.04)..
Numerical check: 7.3350 (round only at the end).
3. Transfer — W2-C09-E03
Find PV of CHF 600 annually at t=4,5,6,7 with 5% interest.
Worked solution. Value four-payment annuity at t=3, then discount three years..
Numerical check: 1,837.8752 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for payment, term and deferred-annuity inverse problems. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C09-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
10. Loan repayment and outstanding balances
Classification: core. Learning objective: Compute fixed loan payments and residual principal after a payment date.
Intuition
A loan is the present value of its promised repayments. Outstanding debt grows with interest and shrinks with every instalment.
Definition and primary equation
\[L=C\frac{1-(1+i)^{-n}}{i};\quad L_k=L(1+i)^k-C\frac{(1+i)^k-1}{i}\]
Derivation
Recurrence L_k=(1+i)L_{k−1}−C for payments at the ends of periods. Unroll the recurrence into a geometric sum; alternatively price remaining payments at t=k.
Conditions and boundary cases
Ordinary amortising-loan timing, constant periodic rate and payments; balance L_k defined immediately after payment k.
Worked examples: increasing difficulty
1. Introductory — W2-C10-E01
Repay CHF 12,000 through four equal end-year payments at 5%. Find C.
Worked solution. 12000×.05/(1−1.05^−4).
Numerical check: 3,384.1420 (round only at the end).
2. Standard — W2-C10-E02
A CHF 10,000 loan at 4% is repaid in five equal end-year payments. Find balance immediately after payment 2.
Worked solution. C=10000×.04/(1−1.04^−5); L₂=10000×1.04²−C(1.04²−1)/.04..
Numerical check: 6,233.6069 (round only at the end).
3. Transfer — W2-C10-E03
A CHF 20,000 loan accrues 6% yearly. A payment of CHF 3,000 is made after year 1. Find balance then, and interest charged in year 2.
Worked solution. L₁=20000(1.06)−3000=18200; interest next year=0.06×18200..
Numerical check: 1,092.0000 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for loan repayment and outstanding balances. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C10-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
11. Mixed valuation, deferred payments and perpetuities
Classification: extension. Learning objective: Combine irregular flows; distinguish infinite streams from finite annuities; transfer timing methods.
Intuition
A perpetuity is an infinite annuity, not a large but finite one. Mixed problems are solved by decomposing streams and shifting values.
Definition and primary equation
\[P_{\infty}=\frac{C}{i}\quad (i>0)\]
Derivation
As n→∞ in P_I=C[1−(1+i)^{−n}]/i, the final term vanishes for i>0, yielding C/i. A perpetuity with first payment at t=d is worth (C/i)(1+i)^{−(d−1)}.
Conditions and boundary cases
Constant C, first payment convention explicit; i>0 for positive-payment perpetuity; classification as extension pending HS26 lecture-source verification.
Worked examples: increasing difficulty
1. Introductory — W2-C11-E01
Value a perpetuity paying CHF 300 every year starting at t=1, i=5%.
Worked solution. 300/.05.
Numerical check: 6,000.0000 (round only at the end).
2. Standard — W2-C11-E02
Value the same perpetuity starting at t=4.
Worked solution. (300/.05)/(1.05)^3.
Numerical check: 5,183.0256 (round only at the end).
3. Transfer — W2-C11-E03
Compute the time-0 PV of CHF 200 at t=1,2,3, plus CHF 500 every year from t=4 onward, i=5%.
Worked solution. 200(1−1.05^−3)/.05+(500/.05)/1.05³.
Numerical check: 9,183.0256 (round only at the end).
Stop-and-retrieve prompt
Explain aloud why the defining equation is appropriate for mixed valuation, deferred payments and perpetuities. Include payment dates and the rate’s time unit.
Practice and transfer
Attempt W2-C11-Q01, then Q02, followed by Q03 when ready. The solution and targeted feedback are in questions.json.
Cross-topic synthesis: a decision protocol
- Draw an explicit timeline. 2. Set cash-flow signs. 3. Convert rates to the payment period. 4. Choose a common valuation date. 5. Calculate the value of each independent stream. 6. Compare on the same date, explaining the economic criterion. 7. Check exceptional cases such as $i=0$, non-integer terms, and unusual cash-flow signs.
How to distinguish NPV and IRR
NPV is measured in money at a specified discount rate. IRR is measured as a rate satisfying an equation. Mutually exclusive projects should be compared by NPV at the stipulated opportunity cost; an IRR ranking can fail. Unconventional cash-flow signs can produce multiple admissible IRRs.
Deferred flows and perpetuity caveat
The original earlier masterclass included both. Their mathematics follows from geometric-series valuation, but their exact inclusion in Lecture 2 cannot be certified using the available locked-slide snapshot. They remain clearly labelled extensions rather than claimed HSG lecture requirements.
Mixed exam training — no advertised method
X01
You can receive CHF 5,000 today or CHF 5,900 after four years. A safe account earns 4% effective annually. Which option is financially preferable?
Answer key (consult only after attempting): Compare present values; 5900/1.04^4 versus 5000.
X02
A machine costs CHF 18,000 now and produces net cash inflows of CHF 7,000 at the end of each of the next three years. At 7%, should it be purchased?
Answer key (consult only after attempting): Compute −18000+7000(1−1.07^−3)/.07 and test its sign.
X03
A borrower receives CHF 15,000 and pays CHF 4,000 at the end of each year for four years. At 5% annual effective interest, what residual liability remains immediately after year 4?
Answer key (consult only after attempting): Evaluate 15000(1.05)^4−4000[(1.05)^4−1]/.05.
X04
An account provides 5% nominal annually, compounded every month. A pension promises CHF 250 at the beginning of each month for 36 months. Find its present value.
Answer key (consult only after attempting): Convert to periodic 0.05/12, then value annuity due with n=36.
X05
Project A costs CHF 1,000 and pays CHF 1,300 next year; B costs CHF 10,000 and pays CHF 12,000 next year. At 6% choose the project creating more present-value wealth.
Answer key (consult only after attempting): NPV A=−1000+1300/1.06; B=−10000+12000/1.06. Select B.
X06
A contract pays CHF 1,000 at t=2, CHF 1,000 at t=3, and CHF 1,000 at t=4. Another pays CHF 2,700 at t=1. At 5%, which is more valuable at t=0?
Answer key (consult only after attempting): Compute both present values. Do not presume the larger undiscounted sum is preferable.
Mastery and remediation
For each concept, a mathematically correct explanation plus appropriate correct transfer work permits immediate progress. If the wrong answer exposes a conceptual error, use the matching M-tag in questions.json, read its targeted correction and solve the new problem. Do not invent an unlock threshold or repeat a lesson solely because of one arithmetic error.
Anki recommendation policy
Suggest only durable definitions, timing distinctions, conditions and frequently recurring conceptual confusions, with a stable ID. Do not automatically add every computational question. Prioritise concepts the learner demonstrably misunderstood.
Advanced conceptual development and proofs
This section deepens the core material beyond formula application. It also supplies the mathematical safeguards that distinguish reliable financial modelling from mere pattern recognition.
1. Unifying accumulation and discounting with a value operator
For a constant effective rate $i>-1$, define the accumulation function $a(t)=(1+i)^t$ for integral $t$. The value at time $s$ of one payment $X$ at time $t$ is
\[
V_s(X,t)=X\frac{a(s)}{a(t)}=X(1+i)^{s-t}.
\]
This one equation contains compounding ($s>t$), present-value discounting ($s<t$), and direct valuation ($s=t$). It also ensures valuation-date invariance: valuing a complete project at any two dates changes its amount by the common positive factor $(1+i)^{s-r}$ and therefore preserves the sign of NPV. For a project with balances $B_k$,
\[
V_s=\sum_{k=0}^n B_k(1+i)^{s-k}=(1+i)^s NPV_0.
\]
Thus $V_s>0$ if and only if $NPV_0>0$ when $i>-1$. This identity gives a useful diagnostic: if you calculate a positive NPV at time 0 and a negative equivalent value at time 5, you have made a timing, sign, or interest-rate error.
A cash-flow comparison is meaningless unless all alternatives are valued using the same rate convention, currency, and valuation date. Treat the rate as part of the model, not as an interchangeable percentage.
2. Nominal versus effective rates: equivalent quotations
Suppose interest is credited $m$ times annually at an annual nominal rate $j^{(m)}$. The periodic crediting factor is $1+j^{(m)}/m$. A year's effective factor is
\[
1+i_{\rm annual}=\left(1+\frac{j^{(m)}}m\right)^m.
\]
For a target effective annual rate $i_{\rm annual}>-1$, an equivalent nominal rate is
\[
j^{(m)}=m\left[(1+i_{\rm annual})^{1/m}-1\right].
\]
The corresponding effective monthly rate is $(1+i_{\rm annual})^{1/12}-1$, not $i_{\rm annual}/12$ unless the quoted annual interest rate is nominal with monthly conversion. The distinction survives even when differences appear numerically small: a tiny conversion error compounds across a 30-year mortgage.
The continuously compounded, or log, rate equivalent to annual effective $i_{\rm annual}$ is
\[
\delta=\ln(1+i_{\rm annual}), \qquad 1+i_{\rm annual}=e^{\delta}.
\]
For constant positive $j$, more frequent nominal compounding increases the effective annual yield, approaching $e^j-1$ as $m\to\infty$. This limiting result depends on holding the nominal annual rate fixed; it is not a statement that every continuously compounded quotation is superior to every discrete quotation.
3. Net present value: cash-flow algebra and economic reasoning
Use $R_k\ge0$ for benefits, $K_k\ge0$ for costs, and $B_k=R_k-K_k$. The project's present value is
\[
NPV(i)=\sum_{k=0}^n\frac{R_k-K_k}{(1+i)^k}.
\]
An initial outlay of CHF 10,000 at $t=0$ is represented by $B_0=-10,000$, not by an additional positive return. Importantly, positive NPV is relative to the assumed opportunity cost. An investor can make a nominal gain and still have a negative NPV when a safe alternative offers a higher return.
Mutually exclusive projects: accepting a positive-NPV project is not enough when only one project can be chosen. Compare the projects' NPVs calculated at the same discount rate, including their different scales and dates. The project with the larger NPV produces the larger incremental present-value wealth, subject to the financial assumptions and non-financial constraints of the model.
Worked diagnostic. Project A costs CHF 100 now and pays CHF 130 after one year; Project B costs CHF 1,000 and pays CHF 1,200 at that time. At 5%,
\[
NPV_A=-100+\frac{130}{1.05}\approx23.81,
\qquad
NPV_B=-1000+\frac{1200}{1.05}\approx142.86.
\]
A has IRR 30%, whereas B has IRR 20%. Yet B creates about CHF 119.05 more present-value wealth. Choosing solely by the bigger IRR would answer a different question from the one asked.
4. When the internal rate of return misleads
Let $x=1/(1+r)$. Finding an IRR is equivalent to finding positive roots of the polynomial $\sum_{k=0}^n B_k x^k=0$, restricted to $r>-1$ (and possibly $r>0$ if that is the required economic domain). A polynomial can have zero, one, or several such roots. The introductory acceptance test $IRR>i$ can be defended for conventional projects with one initial outflow followed by strictly nonnegative inflows, at least one of which is positive, because the NPV function is strictly decreasing with respect to the discount rate for $r>-1$:
\[
\frac{d}{dr}NPV(r)=-\sum_{k=1}^n \frac{kB_k}{(1+r)^{k+1}}<0
\]
whenever $B_k\ge0$ for $k\ge1$ with at least one strictly positive later flow. This monotonicity need not hold when later outflows occur.
Counterexample with two positive IRRs: for cash flows $(B_0,B_1,B_2)=(-100,230,-132)$,
\[
-100+\frac{230}{1+r}-\frac{132}{(1+r)^2}=0.
\]
Multiplying through by $(1+r)^2$ gives $100(1+r)^2-230(1+r)+132=0$, whose roots are $1+r=1.1$ and $1.2$. Both $r=10\%$ and $r=20\%$ satisfy the definition. Asking for the IRR without discussing the multiplicity is incomplete. NPV at the specified discount rate is the appropriate direct decision criterion.
5. Annuity geometry: why the factor $1+i$ appears
Place $n$ payments $C$ on a timeline. For an annuity-immediate, payment dates are $1,2,\dots,n$; for an annuity-due, they are $0,1,\dots,n-1$. Write the present values explicitly:
\[
P_I=C\bigl[(1+i)^{-1}+(1+i)^{-2}+\cdots+(1+i)^{-n}\bigr],
\]
\[
P_D=C\bigl[1+(1+i)^{-1}+\cdots+(1+i)^{-(n-1)}\bigr].
\]
Each due term is the corresponding immediate term multiplied by $1+i$. Therefore $P_D=(1+i)P_I$. This is not because annuity-due has one additional payment: the streams have the same $n$ payments at different dates. Likewise, at time $n$ the due payments have each accumulated for one extra period. This is a shift theorem rather than a separate formula to memorise.
When $i=0$, both present values equal $nC$, established directly from the finite sum. The quotient formulas contain a removable singularity, not an economic discontinuity.
6. Inverse annuity equations and feasibility
With payments of size $C>0$ at the end of each period, $i>0$, and present capital $P>0$, the annuity equation implies
\[
1-\frac{Pi}{C}=(1+i)^{-n}>0.
\]
Consequently, a finite positive real repayment horizon with constant payments requires $C>Pi$. If $C=Pi$, each payment exactly equals periodic interest and principal never falls; the horizon tends to infinity. If $C<Pi$, the initial period's interest exceeds the payment and the loan grows. This is a conceptual feasibility check that should be performed before attempting a logarithm.
For $C>Pi$, logarithms give
\[
n=\frac{-\ln(1-Pi/C)}{\ln(1+i)}.
\]
If $n$ is noninteger, this is an algebraic solution for a formal exponent, not a literal schedule with fractional instalment dates. A practical loan contract needs an integer number of payments with a potentially different final payment. Distinguish these two interpretations explicitly in exam reasoning.
7. Loan balance: prospective versus retrospective valuation
For a loan $L$ with $n$ equal end-of-period payments $C$, $i>0$, let $L_k$ be the debt immediately after payment $k$. Two independent methods must agree.
Retrospective method: compound the original loan to $k$ and subtract the accumulated value of payments already made:
\[
L_k=L(1+i)^k-C\frac{(1+i)^k-1}{i}.
\]
Prospective method: just after payment $k$, the only obligations are the $n-k$ remaining payments, so
\[
L_k=C\frac{1-(1+i)^{-(n-k)}}{i},\qquad 0\le k\le n.
\]
These expressions are identical whenever $L$ and $C$ satisfy the original loan's annuity PV equation. A student can use the agreement to self-check both timing and arithmetic. The balance immediately before payment $k$ is $L_k+C$ (under the end-of-period convention), not $L_k$.
For $k=0$, the prospective formula equals the initial principal. For $k=n$, it equals zero exactly if the payments were chosen to amortise the loan fully. Any nonzero terminal balance in that idealised model exposes an error.
8. Deferred annuities and indefinite cash flows (extension)
Suppose the first of $n$ end-of-period payments is at $t=d$, not $t=1$. The last payment is at $t=d+n-1$. A standard annuity-immediate formula values that stream one period before the first payment, namely at $t=d-1$:
\[
V_{d-1}=C\frac{1-(1+i)^{-n}}{i},\qquad
V_0=\frac{C[1-(1+i)^{-n}]}{i(1+i)^{d-1}}.
\]
Students frequently discount by $d$ instead of $d-1$. List the first date on the timeline before doing any algebra.
For an infinite end-year stream with constant positive $C$ and $i>0$,
\[
\sum_{k=1}^{\infty}\frac{C}{(1+i)^k}=\frac{C}{i}.
\]
If the first payment occurs at $t=d$, delay the stream by discounting $C/i$ through $d-1$ periods. This is a rigorous consequence of the geometric-series theorem; it does not mean perpetuities were verified as explicit content of the accessible Week 2 slides.
9. Integrated worked case: investment financed with debt
A machine costs CHF 12,000 today and yields CHF 4,600 at the end of each of years 1, 2 and 3. The discount rate is 6% effective yearly. The purchase can be financed through a three-year loan at 5% effective yearly with equal end-year payments. Determine whether the machine itself has positive NPV, and determine the contractual annual loan payment. Do not confuse the cash flows of the investment appraisal with those of the financing contract.
The machine's project NPV is
\[
NPV_{machine}=-12{,}000+4{,}600\frac{1-1.06^{-3}}{0.06}.
\]
The loan's contractual payment is separately
\[
C_{loan}=\frac{12{,}000(0.05)}{1-1.05^{-3}}.
\]
The investment discount rate and the loan's contractual interest rate serve different roles; substituting 5% into the machine NPV without an explicit economic justification conflates them. An exam response should state this distinction, and never subtract loan instalments from project cash flows while also leaving the entire purchase price as an initial project expense unless the chosen valuation framework deliberately and consistently includes financing.
10. Self-examination without formula hints
The following prompts are additional, not replications of HSG old exams. Attempt each cold, recording the timeline and the precise reason for your chosen method.
- A retirement contract promises CHF 800 at the beginning of each quarter for five years. The only quoted market rate is 4.5% effective annually. Determine the time-0 value and explain why division by four is not the correct rate conversion.
- A two-year project produces one negative initial cash flow followed by one positive payment and a later environmental remediation expense. Explain why a simple IRR acceptance criterion may fail even if you find a positive root.
- An investor deposits CHF 300 at the end of each month for twelve months and receives a lump sum at month 18. Show how to value both streams at time 12 before comparing them.
- Explain why equal payment size and equal number of instalments do not imply equal present value when the first payment dates differ.
- Construct two cash-flow streams with the same undiscounted total but different present values at any given $i>0$.
Each question tests conceptual selection; correct answers must justify timing, sign, rate conversion and domain restrictions, rather than displaying a formula alone.