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Derivations of Interest Formulas: A Comprehensive Guide with Examples - Prof. Rao, Cheat Sheet of Engineering Economy

A detailed explanation of the derivations of various interest formulas used in finance. It covers single-payment compound amount, single-payment present worth amount, equal-payment series compound amount, equal-payment series sinking fund, equal-payment series present worth amount, equal-payment series capital recovery amount, uniform gradient series annual equivalent amount, and effective interest rate. Each formula is explained step-by-step with clear derivations and practical examples to illustrate their application in real-world financial scenarios.

Typology: Cheat Sheet

2023/2024

Uploaded on 11/07/2024

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Derivations of Interest Formulas
Deriving financial formulas involves understanding the accumulation and discounting processes
based on the time value of money. Below are the derivations for each of the formulas:
3.3.1 Single-Payment Compound Amount
Formula:
[ FV = PV(1 + i)^n ]
Derivation:
1. Start with a present value ((PV)).
2. After one period, the value grows by the interest rate ((i)), so the value is (PV \times (1 + i)).
3. After two periods, the value grows again: ((PV \times (1 + i)) \times (1 + i) = PV \times (1 + i)^2).
4. Continuing this process for (n) periods gives (FV = PV \times (1 + i)^n).
3.3.2 Single-Payment Present Worth Amount
Formula:
[ PV = \frac{FV}{(1 + i)^n} ]
Derivation:
1. Start with a future value ((FV)) and discount it to the present using the interest rate ((i)) over (n)
periods.
2. The present value is obtained by dividing the future value by the accumulated factor: (PV =
\frac{FV}{(1 + i)^n}).
3.3.3 Equal-Payment Series Compound Amount
Formula:
[ FV = A \left[\frac{(1 + i)^n - 1}{i}\right] ]
Derivation:
1. Consider a series of equal payments ((A)) made at the end of each period.
2. The first payment accumulates interest for (n-1) periods, the second for (n-2), and so on until the
last payment which accumulates no interest.
3. Sum the future values of all payments:
[ FV = A \left[(1 + i)^{n-1} + (1 + i)^{n-2} + \ldots + (1 + i) + 1 \right] ]
4. This series is a geometric progression with a common ratio of ((1 + i)).
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Derivations of Interest Formulas

Deriving financial formulas involves understanding the accumulation and discounting processes based on the time value of money. Below are the derivations for each of the formulas: 3.3.1 Single-Payment Compound Amount Formula: [ FV = PV(1 + i)^n ] Derivation:

  1. Start with a present value ((PV)).
  2. After one period, the value grows by the interest rate ((i)), so the value is (PV \times (1 + i)).
  3. After two periods, the value grows again: ((PV \times (1 + i)) \times (1 + i) = PV \times (1 + i)^2).
  4. Continuing this process for (n) periods gives (FV = PV \times (1 + i)^n). 3.3.2 Single-Payment Present Worth Amount Formula: [ PV = \frac{FV}{(1 + i)^n} ] Derivation:
  5. Start with a future value ((FV)) and discount it to the present using the interest rate ((i)) over (n) periods.
  6. The present value is obtained by dividing the future value by the accumulated factor: (PV = \frac{FV}{(1 + i)^n}). 3.3.3 Equal-Payment Series Compound Amount Formula: [ FV = A \left[\frac{(1 + i)^n - 1}{i}\right] ] Derivation:
  7. Consider a series of equal payments ((A)) made at the end of each period.
  8. The first payment accumulates interest for (n-1) periods, the second for (n-2), and so on until the last payment which accumulates no interest.
  9. Sum the future values of all payments: [ FV = A \left[(1 + i)^{n-1} + (1 + i)^{n-2} + \ldots + (1 + i) + 1 \right] ]
  10. This series is a geometric progression with a common ratio of ((1 + i)).
  1. Using the sum formula for a geometric series, the expression simplifies to: [ FV = A \left[\frac{(1 + i)^n - 1}{i}\right] ] 3.3.4 Equal-Payment Series Sinking Fund Formula: [ A = FV \left[\frac{i}{(1 + i)^n - 1}\right] ] Derivation:
  2. Rearrange the formula for the compound amount of equal payments to solve for (A): [ A = \frac{FV}{\frac{(1 + i)^n - 1}{i}} = FV \left[\frac{i}{(1 + i)^n - 1}\right] ]
  3. This equation expresses the equal payment required to accumulate a future value ((FV)) over (n) periods at interest rate ((i)). 3.3.5 Equal-Payment Series Present Worth Amount Formula: [ PV = A \left[\frac{1 - (1 + i)^{-n}}{i}\right] ] Derivation:
  4. Calculate the present value of each equal payment ((A)) at the end of each period: [ PV = A \left[\frac{1}{(1 + i)^1} + \frac{1}{(1 + i)^2} + \ldots + \frac{1}{(1 + i)^n}\right] ]
  5. This series is a geometric progression with a ratio of (\frac{1}{(1 + i)}).
  6. Using the sum formula for a geometric series, the expression simplifies to: [ PV = A \left[\frac{1 - (1 + i)^{-n}}{i}\right] ] 3.3.6 Equal-Payment Series Capital Recovery Amount Formula: [ A = PV \left[\frac{i(1 + i)^n}{(1 + i)^n - 1}\right] ] Derivation:
  7. Rearrange the formula for the present worth amount of equal payments to solve for (A): [ A = \frac{PV}{\frac{1 - (1 + i)^{-n}}{i}} = PV \left[\frac{i(1 + i)^n}{(1 + i)^n - 1}\right] ]
  8. This equation expresses the payment required to recover a present value ((PV)) over (n) periods at interest rate ((i)). 3.3.7 Uniform Gradient Series Annual Equivalent Amount

Using the formula: [ FV = PV(1 + i)^n = 5000 \times (1 + 0.06)^5 ] [ FV = 5000 \times (1.06)^5 \approx 5000 \times 1.3382 = 6691 ] Future Value = $6,691.

3.3.2 Single-Payment Present Worth Amount

Problem: You want to have $10,000 in your savings account 10 years from now. If the bank offers an annual interest rate of 5%, what is the present value you need to deposit today? Solution: Given: Future Value ((FV)) = $10, Interest Rate ((i)) = 5% = 0. Number of Periods ((n)) = 10 years Using the formula: [ PV = \frac{FV}{(1 + i)^n} = \frac{10000}{(1 + 0.05)^{10}} ] [ PV = \frac{10000}{(1.05)^{10}} \approx \frac{10000}{1.6289} = 6139 ] Present Value = $6,139.

3.3.3 Equal-Payment Series Compound Amount

Problem: You make an annual deposit of $1,000 into an account that pays 4% interest compounded annually. How much will you have in the account after 8 years? Solution: Given: Equal Payment ((A)) = $1, Interest Rate ((i)) = 4% = 0. Number of Periods ((n)) = 8 years Using the formula: [ FV = A \left[\frac{(1 + i)^n - 1}{i}\right] = 1000 \left[\frac{(1 + 0.04)^8 - 1} {0.04}\right] ] [ FV = 1000 \left[\frac{(1.04)^8 - 1}{0.04}\right] \approx 1000 \left[\frac{1.3686 - 1}{0.04}\right] = 1000 \times 9.215 ]

Future Value = $9,215.

3.3.4 Equal-Payment Series Sinking Fund

Problem: You want to accumulate $15,000 in 6 years by making annual payments into an account that pays 3% interest compounded annually. What should be your annual payment? Solution: Given: Future Value ((FV)) = $15, Interest Rate ((i)) = 3% = 0. Number of Periods ((n)) = 6 years Using the formula: [ A = FV \left[\frac{i}{(1 + i)^n - 1}\right] = 15000 \left[\frac{0.03}{(1 + 0.03)^6 - 1}\right] ] [ A = 15000 \left[\frac{0.03}{(1.03)^6 - 1}\right] \approx 15000 \left[\frac{0.03}{1.194 - 1}\right] = 15000 \times 0.142 ] Annual Payment = $710.

3.3.5 Equal-Payment Series Present Worth Amount

Problem: You plan to receive $2,000 annually for 5 years from an investment. If the interest rate is 7% per year, what is the present value of this series of payments? Solution: Given: Equal Payment ((A)) = $2, Interest Rate ((i)) = 7% = 0. Number of Periods ((n)) = 5 years Using the formula: [ PV = A \left[\frac{1 - (1 + i)^{-n}}{i}\right] = 2000 \left[\frac{1 - (1 + 0.07)^{-5}} {0.07}\right] ] [ PV = 2000 \left[\frac{1 - (1.07)^{-5}}{0.07}\right] \approx 2000 \left[\frac{1 - 0.713}{0.07}\right] = 2000 \times 4.1 ] Present Value = $8,200.

3.3.8 Effective Interest Rate

Problem: A savings account offers a nominal interest rate of 8% compounded quarterly. What is the effective annual interest rate? Solution: Given: Nominal Interest Rate ((i_{\text{nom}})) = 8% = 0. Compounding Periods ((m)) = 4 (quarterly) Using the formula: [ i_{\text{eff}} = \left(1 + \frac{i_{\text{nom}}}{m}\right)^m - 1 = \left(1 + \frac{0.08} {4}\right)^4 - 1 ] [ i_{\text{eff}} = \left(1 + 0.02\right)^4 - 1 \approx (1.02)^4 - 1 = 1.0824 - 1 = 0.0824 ] Effective Annual Interest Rate = 8.24%.