Time Value of Money & Economic Equivalence

The money you have now versus the money you will have later — are they the same? The answer is no. An earlier dollar is worth more than a later dollar, and for simplified thinking, two effects drive this. Interest increases your earning power: leave \$100 in a savings account at 3% and after a year you have \$103. Inflation decreases your buying power: start the year with \$100 under 3% inflation and it is only worth \$97 a year from now.

The combined effect is what matters. Suppose a refrigerator costs \$100 today and you have exactly \$100. If inflation runs at 8% while your savings earn 6%, after one year the refrigerator costs \$108 but you only have \$106, so you lost 2% of purchasing power by waiting. Flip the rates (6% earning, 4% inflation) and you have \$106 against a \$104 price tag: you gained. Simple rule: when the earning rate is above the inflation rate, holding and saving money works in your favor.

Keep in mind that these rates are not abstractions. The 1973 oil crisis, the 2008 housing bubble, the COVID pandemic: wars, trade wars, and financial crises all move interest and inflation, and with them the time value of money.

Think of interest as renting your money. Lend money to someone (or deposit it in a bank) and you are giving them the privilege of using it; they pay you for that privilege, just like rent on a room. Borrow, and you are the one paying rent. Interest is quoted as a percentage tied to a fixed time period, and you receive it only after the period has passed.

Single-period example: deposit \$100 at 5% annually and withdraw \$105 a year later. Borrow \$100 at 5% and repay \$105. Either way, \$100 now is equivalent to \$105 a year from now: \$100 is the present value $P$, \$105 the future value $F$, linked by the rate $i$ and the number of periods $N$.

Simple interest is applied only on the initial principal. Each period just adds the same slice:

$$F = P(1 + iN)$$

Plotted against time this is a straight line, $y = mx + c$ with intercept $P$ and slope set by $iN$.

Compound interest is applied on the principal and the accrued interest. After year one, $F_1 = P(1+i)$. That becomes the new principal, so $F_2 = F_1(1+i) = P(1+i)^2$, and continuing to $N$ periods:

$$F = P(1 + i)^N$$

The $N$ has moved from a factor to an exponent, and that drastically changes the dynamics. Exponentials grow, and decay, rapidly. For the first few periods the two models barely differ; let time pass and the gap widens enormously. This long-term growth is the idea often called the eighth wonder of the world, and many people build their financial decisions on it.

One equation, four parameters ($F$, $P$, $i$, $N$): know any three and you can find the fourth. That is the entire problem catalog of this topic.

• Find F: \$2,500 invested 4 years at 6%: $F_4 = 2500(1.06)^4 \approx \$3{,}156$.
• Find i: \$1,200 grows to \$2,000 in 6 years: $(2000/1200)^{1/6} = 1+i \Rightarrow i \approx 8.9\%$.
• Find N: \$50 doubles to \$100 at 4% monthly: $N = \ln 2 / \ln 1.04 \approx 17.7$, roughly 18 months. (Watch the time unit of the rate.)

A cash flow diagram is a graphic representation of the magnitude, direction, and timing of cash flows. Cash received is an up arrow (positive side); cash paid out is a down arrow. Arrow length scales with the amount, like drawing vectors in statics. Time 0 is the start of year one; tick $n$ marks the end of period $n$.

Why bother? Because good decisions face decision dilemmas. You win a lottery: take \$1 million as a lump sum at time 0, or \$100k per year for ten years? Same nominal total, very different value, because the money arrives at different times. The diagram makes the dynamics visible at a glance: one tall arrow at zero versus a row of equal-height arrows marching across the years. The same goes for comparing loan repayment plans: five equal payments of \$2,400, or \$500 up front and one large payment at the end.

This concept is similar to equivalent resistance from circuits: two 100 Ω resistors in series can be replaced by one 200 Ω resistor — different components, same effect. Likewise, two cash flows with the same economic effect are economically equivalent. Borrow \$2,000 at 8%: the bank does not care whether you pay \$2,160 after one year or \$2,332.80 after two. The numbers look different, but each is the same money updated for the privilege of holding it longer.

Four cash flow rules govern all of it:

1. Money has a time value: you cannot add or subtract amounts unless they occur at the same point in time.
2. Cash flows are equivalent if their computed worths at a common time period are the same.
3. To move money forward one time unit, multiply by $(1+i)$ — forward marching.
4. To move money backward one time unit, divide by $(1+i)$ — backward marching. (Quite a bit of time travel of money here.)

Forward is the compounding problem; backward is the inverse, or discounting, problem (same $i$, but called the discount rate when used in reverse):

$$F_N = P(1+i)^N \qquad\Longleftrightarrow\qquad P = \frac{F_N}{(1+i)^N}$$

Worked example. Receipts of \$4,000 at time 0, \$6,000 at end of year 2, and \$9,000 at end of year 5, with $i = 6\%$. What single amount $F_3$ at end of year 3 is equivalent? March each flow to year 3: the \$4,000 forward three periods, the \$6,000 forward one, the \$9,000 backward two:

$$F_3 = 4000(1.06)^3 + 6000(1.06) + \frac{9000}{(1.06)^2} \approx \$19{,}134$$

To avoid repeated calculation, the conversion functions are tabulated (appendix of Park; one table per interest rate). The standard notation reads "find this, given that":

• $(F/P, i, N)$ — find $F$ given $P$: the single-payment compound amount factor, $(1+i)^N$.
• $(P/F, i, N)$ — find $P$ given $F$: the single-payment present worth factor, $(1+i)^{-N}$.

Example: \$3,000 invested 12 years at 5%. Look up $(F/P, 5\%, 12) = 1.7959$, so $F_{12} = 3000 \times 1.7959 \approx \$5{,}388$. Inverse: \$2,000 maturing in 9 years at 5% is worth $2000 \times (P/F, 5\%, 9) = 2000 \times 0.6446 \approx \$1{,}289$ today. Writing solutions in factor notation is the standardized habit that pays off in later topics (and on the FE exam).

Investing. Everything in security analysis rests on this topic. A stock's intrinsic value is the present worth of its expected future cash flows discounted at a required rate of return. It is the same $(P/F, i, N)$ machinery, applied to dividends or free cash flow instead of machine costs. And the refrigerator lesson generalizes: an investment only builds purchasing power when its earning rate beats inflation, so track real (inflation-adjusted) returns, not nominal ones. When rates rise, the present worth of distant cash flows falls. Long-duration growth stocks feel it more than mature dividend payers.

Engineering. Any design decision with costs and benefits spread over time (buy the efficient pump now, or pay higher energy bills for a decade?) is an equivalence problem. Discount every alternative's cash flows to the same point in time before comparing; comparing raw dollar totals across different years violates rule 1 and quietly biases you toward whichever option defers its costs.

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