Chapter 3 Lecture 2 Notes: Economic Equivalence, Time Value of Money, and Cash Flow Comparisons

Economic Equivalence and Time Value of Money (Engineering Economics)

• Context: Chapter 3, Lecture 2 in EBGN 321 (Colorado School of Mines).
• Central ideas: economic equivalence; valuing cash flows at different times using interest rates; converting between present and future values; handling single and multiple payments; comparing different cash-flow patterns; and using standard factors and formulas to compute values.

Berkshire Hathaway example (growth assumption and future value)

• Berkshire Hathaway publicly traded since 1965 at $18 per share (historical reference).
• Current context (May 29, 2015): market value per share ~$214,800; annual compound growth ~ 20.65%.
• Projection: assume stock continues to appreciate at 20.65% for the next 15 years (to age 100 for Warren Buffett).
• Calculation (future price at Buffett’s 100th birthday):
  F = 214{,}800igl(1 + 0.2065igr)^{15} \
  \,\approx \$3{,}588{,}758
• Demonstrates how a current value can be escalated to a future value using compounding at rate i.

1626 Manhattan sale and long-term value (Example 3.2)

• Historical fact: Manhattan Island sold by American Indians to Peter Minuit for $24 in 1626.

• Question: If they saved $24 in a bank account paying 8% interest, how much would it be in 2010?

• Perspective: In 2015, total U.S. population ~ 308 million; if the 2010 amount were distributed equally in 2015, how much per person?

• Solution highlights:

  • Future value: $F = P(1+i)^N = 24(1+0.08)^{384} \approx 1.64033801073200\times 10^{14}$
  • Shared among 308,000,000 people:
    $\text{per person} = \frac{F}{308{,}000{,}000} \approx \$532{,}577$

• Significance: illustrates exponential growth and the scale of long time horizons; also underscores the concept of economic equivalence across time.

What is “Economic Equivalence?”

• Economic equivalence exists between cash flows that have the same economic effect and could be traded for one another.
• Cash flows may differ in amount/timing, but an appropriate interest rate makes them equal in an economic sense.
• Why it's important: enables comparison and substitution among different cash-flow patterns; underpins investment decisions, project comparisons, and financing choices.

Equivalence concepts: compounding and discounting (Concepts and Formulas)

• Compounding (finding a future value from a current cash payment): $F = P(1+i)^N$
  • Here, P is the present amount, i is the interest rate per period, N is the number of periods, and F is the future value.
  • Equivalent factor form: $\frac{F}{P} = (1+i)^N, \quad F = P(1+i)^N.$
• Discounting (finding a present value from a future cash payment): $P = \frac{F}{(1+i)^N}$
  • Equivalent factor form: $\frac{P}{F} = \frac{1}{(1+i)^N}$

Example 3.3: Future value and equivalent worth

• Given: deposit $2{,}042 today at 8% for 5 years.
• Find: future value at end of 5 years.
• Calculation:
  $F = 2{,}042(1+0.08)^5 \approx 3{,}000$
• Also consider the equivalent value: how much would $3{,}000 in five years be worth today at 8%?
  $P = \frac{3{,}000}{(1+0.08)^5} \approx 2{,}042$
• Takeaway: a given amount today is equivalent to a larger amount in the future when invested at a positive interest rate; conversely, a future amount has a present value today.

Example 3.4: Cash-flow equivalence across time

• Given: $2{,}042 today is equivalent to receiving $3{,}000 in five years at 8%.
• Question: Are these two cash flows equivalent at the end of year 3?
• Answer: Yes. Equivalent cash flows remain equivalent at any common point in time if evaluated with the same interest rate (here, 8%).
• Practical takeaway: the equivalence relation is time-consistent when the same rate is used.

Finding an equivalent value for a cash-flow series (Example 3.5)

• Objective: find the equivalent value of a cash-flow series at a common time (n = 3) using i = 10%.
• Given cash flows at times 0–5: $100, $80, $120, $150, $200, $100.
• Compounding to time 3:
  $V_3 = 100(1+0.10)^{3} + 80(1+0.10)^{2} + 120(1+0.10)^{1} + 150(1+0.10)^{0} + 200(1+0.10)^{-1} + 100(1+0.10)^{-2} \ \approx 776.36$
• Discounting back to time 0 (alternative check):
  $V_0 = \frac{100}{(1+0.10)^{0}} + \frac{80}{(1+0.10)^{1}} + \frac{120}{(1+0.10)^{2}} + \frac{150}{(1+0.10)^{3}} + \frac{200}{(1+0.10)^{4}} + \frac{100}{(1+0.10)^{5}} \approx 264.46$
• Note: The transcript shows a compounding result 776.36 and a discounting result 264.46; both are valid depending on the chosen common time (3 vs 0) for equivalence.

Comparing two different cash flows (Example 3.6)

• Objective: find a third cash flow C that makes two given payment streams equivalent at a chosen baseline period.
• Approach:
  1) Choose a base period n (e.g., n = 2).
  2) Compute the equivalent lump-sum value at that base period for both cash-flow patterns A and B.
  3) Solve for the unknown C so that the two equivalent values are equal at i = 10%.
• Given cash-flow patterns A and B across periods 0–3, the derived C makes the streams equivalent at i = 10%.
• Result illustrated in the slides: the method yields C such that the two streams are economically equivalent (the example shows the algebra culminating in equalized future values at the base period). The key point is setting the present/ future-value equivalents equal and solving for C.

Finding an interest rate that establishes economic equivalence (Example 3.7–3.8 style)

• Objective: determine the rate i for which two cash-flow patterns are indifferent (i.e., yield the same value).
• Approach:
  1) Select a base period (e.g., n = 3).
  2) Compute the equivalent worth of each cash-flow series at that base period as a function of i.
  3) Solve for i such that FA = FB.
• Illustration (from the slides): with cash flows A and B, i = 8% makes the two streams equivalent, giving FA = FB = $1,630.
  • Example form (indicative): Option A: 500(1+i)^3 + 1000, Option B: 502(1+i)^3 + 502(1+i)^2 + 502; setting equal and solving yields i = 8% and F = 1,630.
• Takeaway: there exists an internal rate of return-like threshold where two different cash-flow structures are economically interchangeable.

Interest Formulas for Single Cash Flows

• Types of common cash flows:
  • Single cash flow
  • Equal (uniform) payment series at regular intervals
  • Linear gradient series
  • Geometric gradient series
  • Irregular (random) payment series

Important formulas for single cash flows

• Compound Amount (Future Value) Factor: $F = P(1+i)^N$
• Future Value Factor (F/P,i,N): $\dfrac{F}{P} = (1+i)^N$
• Present Worth (Present Value) Factor: $P = \dfrac{F}{(1+i)^N}$
• Present Value Factor (P/F,i,N): $\dfrac{P}{F} = \dfrac{1}{(1+i)^N}$

A note on the compound-interest table (handy tool)

• A typical compound-interest table helps compute factors like (F/A,i,N) or (P/F,i,N) for common i and N without performing repeated multiplications.
• Example reference: a 12% table can show (F/A,12%,N) for N up to 10.

Example 3.7: Find F given P, i, N (compound amount) (Excel reference)

• Given: P = $2{,}000, i = 10%, N = 8 years.
• Compute: $F = P(1+i)^N = 2000(1.10)^8 \approx 4287.18$
• Excel/lookup approach yields same result via FV function or (F/P,i,N) factor.

Example 3.8: Find P given i, N, and F

• Given: F = $1{,}000, i = 12%, N = 5 years.
• Compute: $P = \frac{F}{(1+i)^N} = \frac{1000}{(1.12)^5} \approx 567.43$

Example 3.9: Find i given P, F, and N

• Given: F = $40, P = $20, N = 5 years.
• Task: solve for i in $F = P(1+i)^N$.
• Method: use algebra or a calculator/Excel (Goal Seek) to solve for i. Result: i ≈ 14.87% per year.

Example 3.10: Find N given P, F, and i

• Given: P = $6,000, F = $12,000, i = 20%.
• Solve for N from $F = P(1+i)^N$.
• Result: $N = \frac{\ln(F/P)}{\ln(1+i)} = \frac{\ln(2)}{\ln(1.2)} \approx 3.80 \text{ years}$

Rule of 72 (approximation for doubling time)

• Rule of 72: estimate the number of years N required to double an investment at annual interest rate i (in percent).
• Formula: $N \approx \frac{72}{i}$
• Example: at 6% annual rate, doubling time ≈ 12 years; at 12%, doubling time ≈ 6 years.

Equal-Payment Series (Annuities): F in terms of A

• For an equal-payment (uniform series) A over N periods at rate i, the future value is:
  $F = A \cdot (F/A,i,N)$
• The corresponding present value (P) is:
  $P = A \cdot (P/A,i,N)$
• The factors (F/A,i,N) and (P/A,i,N) are the annuity factors derived from the same series.

Example 3.11: Uniform Series: Find F given A, i, N

• Given: A = $3{,}000, N = 10 years, i = 7%.
• Compute: $F = A \cdot (F/A,7$
• From tables or calculator: (F/A,7%,10) ≈ 13.8164.
• Result: $F \approx 3{,}000 \times 13.8164 = 41{,}449.20$
• (Excel reference shows similar table-row progression and final FV ~ $41,449.34 depending on rounding.)

Example 3.12: Handling time shifts (payments shifted in time)

• Scenario: Three cash flows of $3,000 each? with timing shifts; base rate i = 7%.
• Key idea: if the first payment is made at time 0, but the base evaluation is at time 9, you shift each payment to the base time and compound accordingly; or equivalently, use a combination of (F/A,i,n) and (F/P,i,m) factors.
• Given approach (as shown): shift payments to year 9 and compute the future value: $3{,}000(F/A,7%,9) + 3{,}000(F/P,7%,1) ≈ $44{,}350.98$ (using tables).
• Lesson: time-shifting requires adjusting each cash flow to the chosen base time before summing.

Finding A given i, N, and F (Example 3.13 style)

• Given: F = $5,000, N = 5 years, i = 7%.
• Find: A (the equivalently regular payment).
• Formula: A = \frac{F}{(F/A,i,N)} = \frac{5000}{(F/A,7%,5)}
• Excel/financial calculator reference: using PMT or related annuity function to solve for A.

Example 3.14: Comparison of three different retirement plans

• Scenario: Three retirement plans (A, B, C) evaluated at i = 8% with a common endpoint at age 65.
• Method: use (F/A,i,N) and (F/P,i,N) factors to project balances to the target time, allowing apples-to-apples comparison.
• Reported results (from the slides):
  • Plan A balance at 65: approximately $314{,}870.34 (derived from $2{,}000$ contributions with (F/A,8%,10) and then accumulated to age 65 with (F/P,8%,31)).
  • Plan B balance at 65: approximately $246{,}691.74.
  • Plan C balance at 65: approximately $561{,}562.08.
• Takeaway: different retirement strategies yield very different future balances; longer accumulation or larger periodic contributions with compounding dramatically affect final wealth.

Practical notes on practice and tools

• Excel tools commonly used: FV (future value), PV (present value), PMT (payment), and Goal Seek for solving for i, N, or A where the formula is not easily isolated.
• The material emphasizes choosing a common time, calculating equivalent values (present or future), and solving for unknowns by equating the values of alternative cash-flow streams.
• Units and scale: big numbers are common (e.g., trillions over centuries), so rounding and significant figures matter for comparisons.
• Ethical/practical relevance: understanding when different payment structures (lump sums vs. annuities vs. time-shifted payments) are economically interchangeable informs investment choices, financing decisions, and retirement planning.

Summary of key concepts and formulas

• Economic equivalence: same economic effect despite different timing/amounts, when evaluated with the same interest rate.
• Core formulas:
  • Future value: $F = P(1+i)^N$
  • Present value: $P = \frac{F}{(1+i)^N}$
  • Equivalent factors: $\frac{F}{P} = (1+i)^N, \quad \frac{P}{F} = \frac{1}{(1+i)^N}$
  • Annuity future value: $F = A \cdot (F/A,i,N)$
  • Annuity present value: $P = A \cdot (P/A,i,N)$
• Time-consistency: equivalent cash flows at one time remain equivalent at any other common time when the same rate is used.
• Time-shift technique: shift all payments to a chosen base time and then apply compounding/discounting to compute the equivalent value.
• Industry practice: use standard factors (F/P, P/F, F/A, P/A) and financial calculators or software to compute values efficiently.

Key numerical references (for quick review)

• Berkshire Hathaway projection: $F = 214{,}800(1+0.2065)^{15} \approx \$3{,}588{,}758$
• Manhattan 1626: $F = 24(1+0.08)^{384} \approx 1.640338010732 \times 10^{14}$; per-person share in 2015: ≈ $\frac{F}{308{,}000{,}000} \approx \$532{,}577$
• Example 3.3: $F = 2042(1+0.08)^5 \approx 3000$
• Example 3.7: $F = 2000(1.10)^8 \approx 4287.18$
• Example 3.8: $P = \dfrac{1000}{(1.12)^5} \approx 567.43$
• Example 3.9: i = 2^{1/5} - 1 \approx 0.1487 \text{ or } 14.87\%
• Example 3.10: N = \dfrac{\ln(F/P)}{\ln(1+i)} = \dfrac{\ln 2}{\ln 1.2} \approx 3.80\text{ years}
• Rule of 72: N \approx \dfrac{72}{i}
• Example 3.11: F = 3000\cdot (F/A,7\%,10) \approx 3{,}000\cdot 13.8164 \approx 41{,}449.20
• Example 3.12: time-shifted value to base year 9: F = 3000\,(F/A,7\%,9) + 3000\,(F/P,7\%,1) \approx 44{,}350.98
• Example 3.13: solving for A: A = \dfrac{F}{(F/A,i,N)} = \dfrac{5000}{(F/A,7\%,5)}$$
• Example 3.14: retirement plan balances (illustrative values)
  • Plan A: F65 ≈ 314{,}870.34
  • Plan B: F65 ≈ 246{,}691.74
  • Plan C: F65 ≈ 561{,}562.08