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TIME VALUE OF MONEY

TIME VALUE OF MONEY. CHAPTER 2. Time Value of Money. The Cost of Money is established and measured by an interest rate, a percentage that is periodically applied and added to an amount of money over a specified length of time. Economic Equivalence Interest Formulas – Single Cash Flows

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TIME VALUE OF MONEY

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  1. TIME VALUE OF MONEY CHAPTER 2

  2. Time Value of Money • The Cost of Money is established and measured by an interest rate, a percentage that is periodically applied and added to an amount of money over a specified length of time. • Economic Equivalence • Interest Formulas – Single Cash Flows • Equal-Payment Series • Dealing with Gradient Series • Composite Cash Flows.

  3. Time Value of Money • Money has a time value because it can earn more money over time (earning power). • Money has a time value because its purchasing power changes over time (inflation). • Time value of money is measured in terms of interest rate. • Interest is the cost of having money available for use - a cost to the borrower and an earning to the lender

  4. Elements of Transactions involve Interest • Initial amount of money in transactions involving debt or investments is called the principal (P). • The interest rate ( i ) measures the cost or price of money and is expressed as a percentage per period of time. • A period of time, called the interest period (n), determines how frequently interest is calculated. • A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods (N). • A plan for receipts or disbursements (An) that yields a particular cash flow pattern over a specified length of time. [monthly equal payment] • A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.

  5. EXAMPLE OF INTEREST TRANSACTION • Suppose that you apply for an education loan of $30,000 from a bank at a 9% annual interest rate. In addition you pay a $300 loan origination fee when the loan begins. • The bank offers two repayment plans, one with equal payments made at the end of every year for the next five years (installment plan) and the other with a single payment made after the loan period of five years (deferment plan).

  6. Which Repayment Plan? Table 2.1 Repayment plan offered by the lender

  7. Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example

  8. = $7,712.77 = $7,712.77 Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example

  9. Methods of Calculating Interest • Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). • Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. • Engineering economic analysis uses the compound interest scheme exclusively, as it is most frequently practiced in real world.

  10. Practice Problem Problem Statement If you deposit $100 now (n= 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?

  11. Solution F 0 1 2 3 4 5 6 7 8 9 10 $100 $200

  12. Economic Equivalence • Whatdo we mean by “economic equivalence?” • Whydo we need to establish an economic equivalence? • Howdo we establish an economic equivalence?

  13. Economic Equivalence How do we know, whether we should prefer to have $20,000 today and $50,000 ten years from now, or $8,000 each year for the next ten years? Figure 2-3 Which option would you prefer?

  14. Equivalence Calculation: A Simple example Figure 2-4 Using compound interest to establish economic equivalence

  15. Equivalence relation between P and F.

  16. EXAMPLE 2.3 Equivalence Calculation (Consider cash flow in Fig 2.6. Compute the equivalence at n=3) FIND:V3 (equivalent worth at n = 3) and i= 10%. Step 1: $100(1+0.1)3+ $80(1+0.1)2+$120(1+0.1)1+$150 = $511.90 Step 2: $200(1+0.1)-1+ $100(1+0.1)-2 = $264.46 Step 3: V3= $511.90 + $264.46 = $776.36 Figure 2-6 Equivalent worth calculation at n = 3

  17. Interest Formulas for Single Cash Flows Compound Amount Factor Figure 2-7 Compounding process: Find F, given P, i, and N

  18. Interest Rate Factors (10 %)

  19. Example 2.4Ifyou had $1,000 now and invested it at 7% interest compounded annually, how much would it be worth in 8 years? Given: P = $1,000, i = 7 %, and N = 8 years; Find: F F = $1,000 (1+0.07)8 = $1,718.19orusing this F = P (F/P, i, N) factor notation together with table value F = $1,000 (1.7182) = $1,718.19 Figure 2-8 Cash flow diagram

  20. Present -Worth Factor

  21. Example 2.5 Given: F = $1000, i= 6%, and N = 8 years Find:P P = $1,000 (1+0.06)-8 = $1,000 (0.6274) = $627.40 Figure 2-10 Cash flow diagram

  22. Example 2.6Given: P = $10, F = $20 and N = 5 years, Find: i F = $20 = $10 (1+ i )5 $2 = (1+ i)5= (F / P, i, 5) 1+ i== 1.14869 i= 1.14869 – 1 = 0.14869 or 0.1487 = 14.87%

  23. Example 2.7 Given: P = $3,000, F = $6,000 and i= 12%, Find: N F = P (1+ i)N = P (F/P, i, N) …… 6,000 = $3,000 (1+ 0.12)N 2 = (1.12)N ………… Log 2 = N log 1.12 solve for N gives N = log 2 / log 1.12 = 0.301 / 0.049 = 6.116 ≈ 6.12 years Figure 2-12 Cash flow diagram

  24. $1,000 $500 A 0 1 2 3 C C B 0 1 2 3 Practice Problem Given: i= 10%, Find: C that makes the two cash flow streams to be indifferent

  25. $1,000 $500 A 0 1 2 3 C C B 0 1 2 3 Approach Step 1: Select the base period to use, say n= 2 Step 2: Find the equivalent lump sum value at n= 2 for both A and B. Step 3: Equate both equivalent values and solve for unknown C.

  26. $1,000 $500 A 0 1 2 3 C C B 0 1 2 3 Solution For A: For B: To Find C:

  27. $1,000 $500 A 0 1 2 3 $502 $502 $502 B 0 1 2 3 Practice Problem At what interest rate would you be indifferent between the two cash flows?

  28. Approach Step 1: Select the base period to compute the equivalent value (say, n= 3) Step 2: Find the net worth of each at n= 3. $1,000 $500 A 0 1 2 3 $502 $502 $502 B 0 1 2 3

  29. Establish Equivalence at n= 3 Find the solution by trial and error, say i= 8%

  30. Practice Problem • You want to set aside a lump sum amount today in a savings account that earns 7% annual interest to meet a future expense in the amount of $10,000 to be incurred in 6 years. • How much do you need to deposit today?

  31. SolutionF = $10,000; N = 6 years; i = 7 %; Find P $10,000 0 6 P

  32. Figure 2.13 Decomposition of uneven cash flow series

  33. Equal - Payment Series Figure 2-14 Equal payment series: Find equivalent P or F

  34. Equal Payment Series Compound Amount Factor:Find F, Given A, i, and N Figure 2-15 Cash flow diagram of the relationship between A and F

  35. Equal Payment Series Compound Amount Factor (Future Value of an annuity) F 0 1 2 3 N A Example 2.9:Suppose you make an annual contribution of $5,000 to your savings account at the end of each year for five years. If your savings account earns 6% interest annually, how much can be withdrawn at the end of five years? • Given: A = $5,000, N = 5 years, and i= 6% Find: F • Solution: F = $5,000(F/A, 6%, 5) = $28,185.46 or use interest factor table and get the value, F = $5,000 (5.6371) = $28,185.46

  36. Validation

  37. Example 2.10 Handling Time Shifts in a Uniform SeriesFirst deposit of the five deposit series was made at the end of period one and the remaining four deposits were made at the end of each following period. Suppose that all deposits were made at the beginning of each period instead. Compute the balance at the end of period five. F = ? i= 6% First deposit occurs at n= 0 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000

  38. Finding an Annuity Value F 0 1 2 3 N A =? Example: • Given: F = $5,000, N= 5 years, and i= 7% Find: A (0.1739) • Solution: A = $5,000 (A/F, 7%, 5) = $869.50

  39. Sinking fund • A fund created by making periodic deposits (usually equal) at compound interest in order to accumulate a given sum at a given future time for some specific purpose.

  40. Sinking Fund Factor is an interest-bearing account into which a fixed sum is deposited each interest period; The term within the colored area is called sinking-fund factor. F 0 1 2 3 N A Example 2.11 – College Savings Plan: • Given: F = $100,000, N = 8 years, and i= 7 % Find: A • Solution: A = $100,000 (A/F, 7%, 8) = $9,746.78

  41. Given: F = $100,000 i = 7% N = 8 years Find: A from table (0.0975) Solution:A = $100,000 (A/F, 7%, 8) = $9,746.78

  42. Capital Recovery Factor (Annuity Factor) Annuity: • An amount of money payable to a recipient at regular intervals for a prescribed period of time out of a fund reserved for that purpose. • A series of equal payments occurring at equal periods of time. Annuity factor: • The function of interest rate and time that determines the amount of periodic annuity that may be paid out of a given fund.

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