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1 Warm-Up Review Homepage Rule of 72 Single Sum Compounding Annuities.

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Presentation on theme: "1 Warm-Up Review Homepage Rule of 72 Single Sum Compounding Annuities."— Presentation transcript:

1 1 Warm-Up Review Homepage Rule of 72 Single Sum Compounding Annuities

2 2 Homepage is Important Homepage has everything –All Slides ……… (important) –Lecture outline –Quiz and project..time. location –Class/Quiz/Tutorial information –Teaching Group (1+4) contact www.ece.uvic.ca/~wli

3 3 The Rule of 72 Estimates how many years an investment will take to double in value Number of years to double = 72 / annual compound interest rate Example -- 72 / 8 = 9 therefore, it will take 9 years for an investment to double in value if it earns 8% annually

4 4 72 Approx. Years to Double = 72 / i% 726 Years 72 / 12% = 6 Years [Actual Time is 6.12 Years] Quick! How long does it take to double $5,000 at a compound rate of 12% per year? Example: Double Your Money!!!

5 5 Given: Amount of deposit today (PV):$50,000 Interest rate: 11% Frequency of compounding: Annual Number of periods (5 years): 5 periods What is the future value of this single sum? FV n = PV(1 + i) n $50,000 x (1.68506) = $84,253 Single Sum Problems: Future Value

6 6 Given: Amount of deposit end of 5 years: $84,253 Interest rate (discount) rate: 11% Frequency of compounding: Annual Number of periods (5 years): 5 periods What is the present value of this single sum? FV n = PV(1 + i) n $84,253 x (0.59345) = $50,000 Single Sum Problems: Present Value

7 7 An annuity requires that: the periodic payments or receipts (rents) always be of the same amount, the interval between such payments or receipts be the same, and the interest be compounded once each interval. Annuity Computations

8 8 If one saves $1,000 a year at the end of every year for three years in an account earning 7% interest, compounded annually, how much will one have at the end of the third year? Example of Annuity $1,000 $1,000 $1,000 3 0 1 2 3 4 $3,215 = FVA 3 End of Year 7% $1,070 $1,145 FVA3 = $1,000(1.07) 2 + $1,000(1.07) 1 + $1,000(1.07) 0 = $3,215

9 9 Given: Deposit made at the end of each period: $5,000 Compounding:Annual Number of periods:Five Interest rate:12% What is future value of these deposits? F = A[(1+i) n - 1] / i $5,000 x (6.35285) = $ 31,764.25 Annuities: Future Value

10 10 Given: Rental receipts at the end of each period: $6,000 Compounding:Annual Number of periods (years):5 Interest rate:12% What is the present value of these receipts? F = A[(1+i) n - 1] / i $6,000 x (3.60478) = $ 21,628.68 Annuities: Present Value

11 11 Key of Annuity Calculation Fv = Pv[(1+i) n - 1] / i

12 12 Single Payment, Present/Future Value Factor Sinking Factor, Capital Recovery Factor Conversion for Arithmetic Gradient Series Conversion for Geometric Gradient Series Topics Today

13 13 Compound Amount Factor (Single Payment) This factor finds the equivalent future worth, F, of a present investment, P, held for n periods at i rate of interest. Example: What is the value in 9 years of $1,200 invested now at 10% interest

14 14 Compound Amount Factor (Single Payment) P = $1,200 12349 F

15 15 Present Worth Factor (Single Payment) This factor finds the equivalent present value, P, of a single future cash flow, F, occurring at n periods in the future when the interest rate is i per period. Example: What amount would you have to invest now to yield $2,829 in 9 years if the interest rate per year is 10%?

16 16 Present Worth Factor (Single Payment) P 12349 F = $2,829

17 17 Compound Amount Factor (Uniform Series) This factor finds the equivalent future value, F, of the accumulation of a uniform series of equal annual payments, A, occurring over n periods at i rate of interest per period. Example: What would be the future worth of an annual year-end cash flow of $800 for 6 years at 12% interest per year?

18 18 Compound Amount Factor (Uniform Series) 12346 F 5 $800

19 19 Sinking Fund Factor This factor determines how much must be deposited each period in a uniform series, A, for n periods at i interest per period to yield a specified future sum. Example: If a $1.2 million bond issue is to be retired at the end of 20 years, how much must be deposited annually into a sinking fund at 7% interest per year?

20 20 Sinking Fund Factor 123420 F = $1,200,000 AAAAAA

21 21 Capital Recovery Factor This factor finds an annuity, or uniform series of payments, over n periods at i interest per period that is equivalent to a present value, P. Example: What savings in annual manufacturing costs over an 8 year period would justify the purchase of a $120,000 machine if the firm’s minimum attractive rate of return (MARR) were 25%?

22 22 Capital Recovery Factor $120,000 1238 AAAAA

23 23 Present Worth Factor (Uniform Series) This factor finds the equivalent present value, P, of a series of end- of-period payments, A, for n periods at i interest per period. Example: What lump sum payment would be required to provide $50,000 per year for 30 years at an annual interest rate of 9%?

24 24 Present Worth Factor (Uniform Series) P 12330 $50,000

25 25 Series and Arithmetic Series A series is the sum of the terms of a sequence. The sum of an arithmetic progression (an arithmetic series, difference between one and the previous term is a constant) Can we find a formula so we don’t have to add up every arithmetic series we come across?

26 26 Sum of terms of a finite AP

27 27 Arithmetic Gradient Series A series of N receipts or disbursements that increase by a constant amount from period to period. Cash flows: 0G, 1G, 2G,..., (N–1)G at the end of periods 1, 2,..., N Cash flows for arithmetic gradient with base annuity: A', A’+G, A'+2G,..., A'+(N–1)G at the end of periods 1, 2,..., N where A’ is the amount of the base annuity

28 28 Arithmetic Gradient to Uniform Series Finds A, given G, i and N The future amount can be “converted” to an equivalent annuity. The factor is: The annuity equivalent (not future value!) to an arithmetic gradient series is A = G(A/G, i, N)

29 29 Arithmetic Gradient to Uniform Series The annuity equivalent to an arithmetic gradient series is A = G(A/G, i, N) If there is a base cash flow A', the base annuity A' must be included to give the overall annuity: A total = A' + G(A/G, i, N) Note that A' is the amount in the first year and G is the uniform increment starting in year 2.

30 30 Arithmetic Gradient Series with Base Annuity

31 31 Example 3-8 A lottery prize pays $1000 at the end of the first year, $2000 the second, $3000 the third, etc., for 20 years. If there is only one prize in the lottery, 10 000 tickets are sold, and you can invest your money elsewhere at 15% interest, how much is each ticket worth, on average?

32 32 Example 3-8: Answer Method 1: First find annuity value of prize and then find present value of annuity. A' = 1000, G = 1000, i = 0.15, N = 20 A = A' + G(A/G, i, N) = 1000 + 1000(A/G, 15%, 20) = 1000 + 1000(5.3651) = 6365.10 Now find present value of annuity: P = A (P/A, i, N) where A = 6365.10, i = 15%, N = 20 P = 6365.10(P/A, 15, 20) = 6365.10(6.2593) = 39 841.07 Since 10 000 tickets are to be sold, on average each ticket is worth (39 841.07)/10,000 = $3.98.

33 33 Arithmetic Gradient Conversion Factor (to Uniform Series) The arithmetic gradient conversion factor (to uniform series) is used when it is necessary to convert a gradient series into a uniform series of equal payments. Example: What would be the equal annual series, A, that would have the same net present value at 20% interest per year to a five year gradient series that started at $1000 and increased $150 every year thereafter?

34 34 Arithmetic Gradient Conversion Factor (to Uniform Series) 1234512345 AAAAA $1000 $1150 $1300 $1450 $1600

35 35 Arithmetic Gradient Conversion Factor (to Present Value) This factor converts a series of cash amounts increasing by a gradient value, G, each period to an equivalent present value at i interest per period. Example: A machine will require $1000 in maintenance the first year of its 5 year operating life, and the cost will increase by $150 each year. What is the present worth of this series of maintenance costs if the firm’s minimum attractive rate of return is 20%?

36 36 Arithmetic Gradient Conversion Factor (to Present Value) $1000 $1150 $1300 $1450 $1600 12345 P

37 37 Geometric Gradient Series A series of cash flows that increase or decrease by a constant proportion each period Cash flows: A, A(1+g), A(1+g) 2, …, A(1+g) N–1 at the end of periods 1, 2, 3,..., N g is the growth rate, positive or negative percentage change Can model inflation and deflation using geometric series

38 38 Geometric Series The sum of the consecutive terms of a geometric sequence or progression is called a geometric series. For example: Is a finite geometric series with quotient k. What is the sum of the n terms of a finite geometric series

39 39 Sum of terms of a finite GP Where a is the first term of the geometric progression, k is the geometric ratio, and n is the number of terms in the progression.

40 40 Geometric Gradient to Present Worth The present worth of a geometric series is: Where A is the base amount and g is the growth rate. Before we may get the factor, we need what is called a growth adjusted interest rate:

41 41 Geometric Gradient to Present Worth Factor: (P/A, g, i, N) Four cases: (1) i > g > 0: i° is positive use tables or formula (2) g < 0:i° is positive use tables or formula (3) g > i > 0:i° is negative Must use formula (4) g = i > 0:i° = 0

42 42 Compound Interest Factors Discrete Cash Flow, Discrete Compounding

43 43 Compound Interest Factors Discrete Cash Flow, Discrete Compounding

44 44 Compound Interest Factors Discrete Cash Flow, Continuous Compounding

45 45 Compound Interest Factors Discrete Cash Flow, Continuous Compounding

46 46 Compound Interest Factors Continuous Uniform Cash Flow, Continuous Compounding

47 47 Calculator Talk No programmable No economic Function Simple the best Trust your ability Trust your teaching group

48 48 Summary Single Sum Compounding Annuities Conversion for Arithmetic Gradient Series Conversion for Geometric Gradient Series Key: Compound Interests Calculation


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