Week 4 — Drills

Exam format. Show all work. Core items are for everyone; stretch items for whoever finishes early. Solutions at the end, worked the way an examiner wants to see them.

Running case for weeks 4–5: Arno Bikes is considering a new assembly line. It costs €2.4M now and is expected to return €700k, €800k, €900k and €1,000k at the end of each of the next four years. The firm can borrow at 9%.

Round money to the cent, rates to two decimals in a percentage.

A. Simple and compound interest

D1 (core). €8,000 is invested for 3 years at 6% a year. (a) Simple interest: the final amount. (b) Compounded annually: the final amount. (c) The difference, and one sentence on where it comes from.

D2 (core). €5,000 for 4 years at 8% nominal, compounded quarterly. (a) The final amount. (b) The interest earned.

D2b (core — the rate moves). €10,000 for eight years at 5%, compounded annually. (a) The final amount. (b) Recompute at 4% and at 6%. In euros, what is one point of interest worth over eight years? And over twenty? (c) One sentence: why is the answer to (b) so much larger at twenty years than at eight?

D3 (core — resit form). Select the correct alternative and justify in one line. A bank quotes 6% nominal. Which is worth more to a depositor? (i) compounded annually (ii) compounded monthly (iii) they are the same

B. Effective annual rate

D4 (core). Compute the EAR for a 12% nominal rate compounded (a) semi-annually, (b) quarterly, (c) monthly, (d) continuously.

D5 (core). Bank A offers 7.8% compounded monthly. Bank B offers 8% compounded annually. Which do you borrow from, and why? Show the comparison you made.

D6 (stretch). A rate is quoted as 5% nominal compounded continuously. What annual compounding rate would leave a depositor equally well off?

C. Discounting, present value and growth

D7 (core). What is €10,000 receivable in 5 years worth today at 7%?

D8 (core). A supplier offers you two ways to pay for a €50,000 machine: (i) €50,000 today; (ii) €18,000 at the end of each of the next three years. Your cost of capital is 9%. Which do you choose? Show the present value of (ii).

D9 (core — annuities and perpetuities). (a) An annuity pays €12,000 at the end of each year for 10 years. At 6%, what is it worth today? (b) A perpetuity pays €12,000 at the end of every year for ever. At 6%, what is it worth today? (c) Explain the gap between (a) and (b) in one sentence.

D10 (core — growing perpetuity). A perpetuity pays €5,000 next year and grows at 2% a year for ever. At 7%, what is it worth today?

D11 (core — Board form). Three years on, the Board asks again: what is Arno Bikes worth? The new forecast is a free cash flow of €350k next year, growing 4% a year for ever; the cost of capital is still 9%. (a) The value of the firm today. (b) By how much does the value change if growth is one point lower, 3% instead of 4%? In euros and in per cent. (c) One sentence for the Board.

D12 (stretch — what the market assumes). A listed rival is valued by the market at €12M. Next year’s free cash flow is €600k and its cost of capital is 9%. (a) What growth rate for ever is the market pricing in? (b) One line: is that believable for a mature firm, and what would you check?

D. NPV

D13 (core — the case). For the assembly line above, at a cost of capital of 9%: (a) Write the NPV as a sum. (b) Compute each discounted flow and the NPV. (c) Accept or reject, and say the sentence you would tell the Board.

D13b (core — where the rate comes from). Arno Bikes is financed 60% by equity and 40% by debt. Shareholders expect 12%; the bank charges 6%; the corporate tax rate is 25%. (a) Compute the weighted average cost of capital. (b) One line: why does the tax rate appear at all, and only on the debt side? (c) If the firm took on more debt and the mix became 40/60, what would happen to the WACC and therefore to the NPV of the assembly line? No arithmetic needed.

D14 (stretch). Recompute the NPV at 16%. What does the change in sign tell you, and what is the name of the rate at which it happens?

E. Challenge — the seven steps, on money

Use the function-study procedure of week 3. Nothing here is new; the point is to see that these curves are functions like any other.

D15 (challenge). A(t) = 10,000 × 1.05^t, for t ≥ 0. (a) Domain, A(0), and the limit as t → ∞. (b) A'(t). Its sign, and what it says. (Hint: d/dt aᵗ = aᵗ ln a.) (c) A''(t). Convex or concave? Is there an inflection? (d) Compare with simple interest S(t) = 10,000(1 + 0.05t): what is S''? Sketch both on the same axes, t from 0 to 30, marking t = 0 and t ≈ 14.2 where A = 20,000.

D16 (challenge — the NPV curve as a function). For the assembly line, NPV(r) = −2,400 + 700/(1+r) + 800/(1+r)² + 900/(1+r)³ + 1,000/(1+r)⁴, for r ≥ 0. (a) NPV(0). What does that number mean, and who said it this morning? (b) The limit as r → ∞, and what it means. Is there a horizontal asymptote? (c) Without computing it, say why NPV'(r) < 0 for every r. (Each term CF/(1+r)ᵗ has derivative −t·CF/(1+r)^(t+1).) (d) Sign of NPV(r): you know NPV(0.09) = 318.94 and NPV(0.16) = −73.14. Where is the zero, roughly, and what is it called? (e) Sketch: NPV against r, with the intercept, the zero and the asymptote. Then one sentence for the Board on what the sketch says about risk.

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Solutions

D1. (a) Simple: 8000(1 + 0.06 × 3) = 8000 × 1.18 = **€9,440.00**. (b) Compound: 8000(1.06)³ = 8000 × 1.191016 = **€9,528.13**. (c) Difference €88.13. Simple interest pays on the original €8,000 only; compounding pays interest on the interest already earned.

D2. (a) 5000(1 + 0.08/4)^(4×4) = 5000(1.02)^16 = 5000 × 1.372786 = **€6,863.93**. (b) Interest = **€1,863.93**.

D2b. (a) 10,000 × 1.05⁸ = 10,000 × 1.477455 = **€14,774.55**. (b) At 4%: 10,000 × 1.04⁸ = **€13,685.69**; at 6%: 10,000 × 1.06⁸ = **€15,938.48**. One point is worth about €1,100 over eight years. Over twenty: 4% → €21,911.23, 5% → €26,532.98, 6% → €32,071.35: about €4,600 to €5,500. (c) The rate sits in the base of a power with exponent t: a point of rate is multiplied by itself twenty times instead of eight. Small differences in the rate become large differences in the amount, and the longer the horizon the larger.

D3. (ii) compounded monthly. More frequent compounding earns interest on interest sooner, so the effective rate is higher: (1 + 0.06/12)^12 − 1 = 6.17% > 6%.

D4. EAR = (1 + i/m)^m − 1, and e^i − 1 in the limit. (a) (1.06)² − 1 = **12.36%** (b) (1.03)⁴ − 1 = **12.55%** (c) (1.01)^12 − 1 = **12.68%** (d) e^0.12 − 1 = **12.75%** The gaps shrink: that ceiling is where e comes from.

D5. Compare like with like, on EAR. Bank A: (1 + 0.078/12)^12 − 1 = (1.0065)^12 − 1 = 0.080850 = **8.085%**. Bank B: 8.00%. Borrow from B. A’s lower quoted rate hides more frequent compounding, and once both are put on the same basis A is the dearer of the two. The nominal rate is not the price.

D6. e^0.05 − 1 = **5.13%**.

D7. PV = 10,000 / (1.07)⁵ = 10,000 / 1.402552 = **€7,129.86**.

D8. PV(ii) = 18,000 × [1 − (1.09)⁻³]/0.09 = 18,000 × 2.5312955 = **€45,563.30**. 45,563.30 < 50,000, so pay in instalments: it costs about €4,437 less in today’s money.

D9. (a) PV = 12,000 × [1 − (1.06)⁻¹⁰]/0.06 = 12,000 × 7.360087 = **€88,321.04**. (b) PV = 12,000 / 0.06 = **€200,000.00**. (c) The perpetuity is worth more, but not infinitely more: the payments after year 10 are worth about €111,679 in total today, because discounting shrinks distant money towards nothing.

D10. Growing perpetuity: PV = C₁/(r − g) = 5,000/(0.07 − 0.02) = 5,000/0.05 = **€100,000.00**. With no growth the same €5,000 would be worth 5,000/0.07 = €71,428.57: two points of growth cut the denominator from 7% to 5% and add 40% to the value.

D11. (a) V = CF₁/(r − g) = 350/(0.09 − 0.04) = 350/0.05 = **€7,000.00k**, i.e. €7M. (b) At 3%: 350/(0.09 − 0.03) = 350/0.06 = **€5,833.33k**. The value falls by €1,166.67k, −16.67%. One point of growth moved the price by a sixth, because it made the denominator a fifth larger, from 5% to 6%, and 1/1.2 = 0.8333. (c) “On our forecast Arno Bikes is worth about €7M; if growth is one point lower it is worth €5.8M, so the price we accept should rest on the growth we can defend, not on the one we hope for.”

D12. (a) P = CF₁/(r − g) ⇒ g = r − CF₁/P = 0.09 − 600/12,000 = 0.09 − 0.05 = **4.00%**. (b) 4% a year for ever is at the top of what a mature firm can sustain — roughly the economy’s long-run nominal growth. Check whether recent cash-flow growth supports it, and remember the market price also embeds its own r: a lower r would need less growth to justify €12M.

D13. (a) NPV = −2,400 + 700/(1.09) + 800/(1.09)² + 900/(1.09)³ + 1,000/(1.09)⁴ (€ thousand). (b)

Year	Flow (€k)	Factor	Present value (€k)
0	−2,400	1.000000	−2,400.00
1	700	0.917431	642.20
2	800	0.841680	673.34
3	900	0.772183	694.97
4	1,000	0.708425	708.43
		NPV	+318.94

3. Accept. “At our 9% cost of capital the line adds about €319,000 of value in today’s money; the cash it returns covers the outlay by the end of year three.”

D13b. (a) WACC = 0.6(12%) + 0.4(6%)(1 − 0.25) = 7.2% + 1.8% = **9.0%** — the 9% we were handed all along. (b) Interest on debt is deductible from taxable profit, so a 6% loan costs the firm only 6%(1 − 0.25) = 4.5% after tax. Dividends are not deductible, so no such factor appears on the equity side. (c) More weight on the cheaper source → WACC falls → future cash flows are discounted less → the NPV rises. Worth saying out loud that this is not free: more debt raises the risk, and in time both r_e and r_d rise with it.

D14. At 16%: −2,400 + 603.45 + 594.53 + 576.59 + 552.29 = **−€73.14k**. Negative. The NPV falls as the discount rate rises and changes sign between 9% and 16%; the rate where it is exactly zero is the internal rate of return (about 14.6% here). Above it the project destroys value.

D15. (a) Domain [0, ∞). A(0) = 10,000 — the principal. lim (t→∞) A(t) = +∞: there is no ceiling. (b) A'(t) = 10,000 × 1.05^t × ln 1.05 = A(t) × 0.04879 > 0 for every t: always increasing, and the increase is proportional to the amount already there — that is compounding. (c) A''(t) = A(t) × (ln 1.05)² > 0: convex throughout, no inflection. The curve bends upward for ever: growth that keeps accelerating. (d) S''(t) = 0: a straight line. Both start at 10,000 with almost the same slope (S' = 500, A'(0) = 487.9); then A pulls away. A = 20,000 when 1.05^t = 2, t = ln 2 / ln 1.05 ≈ 14.2 years; at that point S = 17,100. At t = 30: A = 43,219, S = 25,000.

D16. (a) NPV(0) = −2,400 + 700 + 800 + 900 + 1,000 = **+1,000**: the “million of profit” — it is what you get by adding the flows with no discounting at all, i.e. at a zero cost of money. The Board member was computing NPV at r = 0. (b) lim (r→∞) NPV(r) = **−2,400**: every future flow is discounted to nothing and only the outlay remains. y = −2,400 is a horizontal asymptote. (c) Each term CF/(1+r)ᵗ falls as r rises (its derivative −t·CF/(1+r)^(t+1) is negative, since CF > 0); the constant −2,400 differentiates to zero. A sum of negatives is negative: NPV is decreasing for every r, so it crosses zero exactly once. (d) NPV is positive at 9% and negative at 16%, so the zero lies between; it is at r ≈ 14.6%, the internal rate of return. Sign: NPV > 0 for r < 14.6%, NPV < 0 above. (e) A falling, convex curve from (0, 1,000) through (0.09, 318.94) and (0.146, 0), flattening towards −2,400. Board sentence: “The project stays worth doing for any cost of money up to 14.6%; our 9% leaves about five and a half points of room, and the curve is steepest exactly where we are, so each point of rate costs us roughly €80k of value.”