Finance-Theory · Unit 3 · Video 3 · Interactive Practice

The NPV Rule at Work: A Lighting-System Investment and the CNOOC Loan Subsidy

IKey Formulas

Formula	Name	What it settles
$\dfrac{1}{(1+r)^t}$	Discount factor for date $t$	What one dollar arriving in year $t$ is worth today: thirty years out, about 31 cents at 4%, 10 cents at 8%, 3 cents at 12%
$V_0 = \dfrac{C_1}{1+r} + \dfrac{C_2}{(1+r)^2} + \cdots = \sum_t \dfrac{C_t}{(1+r)^t}$	Value of a cash-flow stream	The asset is the sequence $C_1, C_2, \ldots$ with its dates; the value is the one number the rate produces from it
$\text{NPV} = -\text{cost} + V_0$	Net present value	The outlay is a cash flow like any other — dated 0, entered negative
$\text{NPV} > 0 \Rightarrow \text{take}, \qquad \text{NPV} < 0 \Rightarrow \text{reject}$	The NPV rule	Usable only with an $r$ obtainable in the open market, never a rate chosen to suit the project

Key Insight: An asset is a sequence of cash flows; its value is one number that sequence and the rate produce together. The same $230,000 lighting system is worth +$19,758.19 at 4% and −$6,183.32 at 10% — identical savings, opposite verdicts — so the rate must be one the market will actually give you.

IIWhat a Later Dollar Is Worth Today

A dollar arriving later is worth less today, and the higher the rate, the steeper the fall.

Year t 30

IIIThe Lighting System at Any Rate

Same savings, same cost: the rate alone decides whether the $230,000 lighting system is worth buying.

Rate r 4.00%

💡 The rate at which the net present value crosses zero — just under 8.5% for this project — is its internal rate of return, which a later unit takes up.

IVThirty Years of Interest Saved

The cheap loans save interest every year; discounting at 8% turns thirty of those savings into one number.

Through year 30

💡 Each year's saving is worth the same fraction, $1/1.08$ , of the one before, which is how a payment that never stops can still be worth a finite amount — the perpetuity, next.

VQuiz Questions

Problem 1 · One Cash Flow, Brought Back

Given: a certain saving of $50,000 arrives at the end of year 4, and the market rate is $r = 6\%$ . What is that saving worth today?

✅ Correct! Dividing by

(1.06)^4 = 1.26247696

carries the saving back four years:

50{,}000 / 1.26247696 = 39{,}604.68

❌ That is simple interest. Dividing by

1 + 0.06 \times 4 = 1.24

credits four years of interest on the original amount only. Discounting uses the compounded factor

(1.06)^4 = 1.26247696

, because interest earns interest.

❌ One year short. $41,980.96 is

50{,}000/(1.06)^3

. The exponent is the date the cash flow arrives, so a year-4 amount is divided by

(1.06)^4

❌ That runs the clock the wrong way. Multiplying by

(1.06)^4

carries money forward; a cash flow that arrives later has to be brought back, so you divide.

Show solution

A cash flow $C_t$ dated $t$ is worth $C_t/(1+r)^t$ today. Here $C_4 = 50{,}000$ , $r = 0.06$ and $t = 4$ :

(1.06)^4 = 1.26247696, \qquad \frac{50{,}000}{1.26247696} = 39{,}604.68

Equivalently, one dollar arriving in year 4 is worth $1/(1.06)^4 \approx 0.7921$ today, and the saving is 50,000 of those dollars.

Problem 2 · Which Cash Flows Are the Project

Given: a plant pays $120,000 a year for fuel. A $40,000 retrofit, installed today, would cut the fuel bill to $105,000 a year for each of the next four years. Which cash flows belong on the retrofit's timeline?

✅ Correct! The project is the change it causes: $40,000 out today, and a bill lighter by

120{,}000 - 105{,}000 = 15{,}000

dollars in each of four years.

❌ The fuel bill is paid either way. It runs with or without the retrofit, so it is not a cash flow of this project. Only the $15,000 reduction each year is.

❌ $105,000 is the new bill, not an inflow. The plant still pays it; what the retrofit delivers is the $15,000 difference between the old bill and the new one.

❌ The outlay belongs on the timeline. $40,000 leaves the firm today, so it enters as a negative cash flow at date 0 — that subtraction is exactly what makes the sum a net present value.

Show solution

Define the asset before valuing it: the asset is the sequence of cash flows the decision changes.

Date 0: the retrofit costs $40,000, entered as $C_0 = -40{,}000$ .
Dates 1–4: the bill falls from $120,000 to $105,000, so $C_t = +15{,}000$ .

The $120,000 bill itself never appears: it is paid in both worlds, so it cancels. Only differences between the two worlds are the project's cash flows.

\text{NPV} = -40{,}000 + \frac{15{,}000}{1+r} + \frac{15{,}000}{(1+r)^2} + \frac{15{,}000}{(1+r)^3} + \frac{15{,}000}{(1+r)^4}

Problem 3 · The Same Retrofit at 20%

Given: the retrofit above — $40,000 today, then $15,000 saved at the end of each of years 1 through 4 — but now the firm's market rate is $r = 20\%$ . At 8% its NPV is $+9{,}681.90$ dollars. What is its NPV at 20%?

✅ Correct! Discounted at 20%, the four savings are worth

V_0 = 38{,}831.02

dollars today, which falls $1,168.98 short of the $40,000 cost. Same cash flows as at 8%, opposite verdict.

❌ That is $V_0$ , not the NPV. $38,831.02 is what the four savings are worth today; the $40,000 outlay still has to be subtracted.

❌ Nothing was discounted.

4 \times 15{,}000 - 40{,}000 = 20{,}000

adds cash from four different dates as though it all arrived today. Each saving must be divided by

(1.20)^t

first.

❌ That is the 8% answer. At 20% every saving is divided by a much larger factor —

(1.20)^4 = 2.0736

against

(1.08)^4 = 1.3605

— so the savings are worth far less today.

Show solution

Discount each saving at 20%:

\frac{15{,}000}{1.20} = 12{,}500.00, \qquad \frac{15{,}000}{(1.20)^2} = 10{,}416.67

\frac{15{,}000}{(1.20)^3} = 8{,}680.56, \qquad \frac{15{,}000}{(1.20)^4} = 7{,}233.80

Their sum, taken before the terms are rounded to the cent, is $V_0 = 38{,}831.02$ dollars, so

\text{NPV} = -40{,}000 + 38{,}831.02 = -1{,}168.98

Negative: reject. Nothing about the machine changed between 8% and 20% — the savings are the same four amounts on the same four dates. What changed is the opportunity cost of the $40,000: at 20% that money has somewhere better to be.

Problem 4 · Pricing a Loan Subsidy

Given: a parent company lends its subsidiary $6 billion for 10 years at 2%, when the subsidiary's own market borrowing rate is 8%. The principal is repaid in full either way, and interest is paid at the end of each of the ten years.

What does the subsidy save the subsidiary each year?

What is that stream of ten yearly savings worth today at 8%?

✅ Correct! The gap is

0.08 - 0.02 = 0.06

, so

6 \times 0.06 = 0.36

billion a year; the ten discount factors at 8% add to

6.7101

, and

0.36 \times 6.7101 = 2.4156

billion.

❌ That is the whole market interest bill.

6 \times 0.08 = 0.48

billion is what the subsidiary would pay at 8%, but it still pays 2%, so only the gap

0.08 - 0.02

is saved.

❌ That is the interest actually paid.

6 \times 0.02 = 0.12

billion leaves the firm each year; the saving is what it does not pay,

6 \times (0.08 - 0.02)

❌ That is all ten years added together, and added without discounting. The question asks what one year saves:

6 \times 0.06 = 0.36

billion.

❌ The principal is repaid in full either way, so it cancels out of the comparison and never enters the subsidy at all. The subsidy is the interest saved.

❌ That adds ten dates as if they were one. Each $360 million arrives a year further out than the last, so the value today must be well below $3.6 billion.

❌ That discounts the whole sum as if it arrived in year 10:

3.6/(1.08)^{10} = 1.6675

. Nine of the ten savings arrive earlier than that, so they are discounted less.

Show solution

Step 1 — the asset. The principal is repaid either way, so the only difference the cheap loan makes is interest:

6 \times (0.08 - 0.02) = 6 \times 0.06 = 0.36 \text{ billion per year, years 1 through 10}

Step 2 — the value. Discount each year's saving at the rate the borrower would otherwise face, 8%:

\text{PV} = \frac{0.36}{1.08} + \frac{0.36}{(1.08)^2} + \cdots + \frac{0.36}{(1.08)^{10}}

The first term is $0.3333$ , the second $0.3086$ , the tenth $0.1667$ . The ten discount factors add to $6.7101$ , so

\text{PV} = 0.36 \times 6.7101 = 2.4156 \text{ billion}

Check it against the bounds. Every saving is discounted, so the value has to be below the undiscounted $3.6 billion; and only the last one is discounted a full ten years, so it has to be well above $3.6/(1.08)^{10} = 1.67$ billion. Only $2.42 billion sits between them.

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