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259 Lecture 3 Spring 2013

259 Lecture 3 Spring 2013. Finance Applications with Excel – Annuities an Amortization. Example 1. Suppose we invest $500 at the end of each year for 5 years, at an interest rate of 4%, compounded annually. How much money will we have at the end of year 5?. Example 1 (cont.).

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259 Lecture 3 Spring 2013

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  1. 259 Lecture 3 Spring 2013 Finance Applications with Excel – Annuities an Amortization

  2. Example 1 • Suppose we invest $500 at the end of each year for 5 years, at an interest rate of 4%, compounded annually. • How much money will we have at the end of year 5?

  3. Example 1 (cont.)

  4. Example 1 (cont.) • At the end of year 5, we will have a total investment S as follows: S = 500 + 500(1.04)1 + 500(1.04)2 + 500(1.04)3 + 500(1.04)4 dollars • To add up this sum, we can use the following idea!

  5. Example 2 • Find a formula for the following sum for non-zero real numbers a and r: Sn = a + ar + ar2 + … + arn-1 • Solution: Write down the equation for Sn, then multiply both sides of the equation by r to get: Sn = a + ar + ar2 + … + arn-1 rSn = ar + ar2 + … + arn-1 + arn • Subtract the first equation from the second and solve for Sn: rSn – Sn = arn – a (r-1)Sn = arn - a Sn = a(rn – 1)/(r-1)

  6. Sequences and Series • We’ll see ideas like those used in Example 2 appear in many mathematical settings! • Here is some useful terminology! • A sequence is a function whose domain is the positive integers. • We can think of a sequence as a list of numbers written in a definite order: a1, a2, a3, … • For example, the Fibonacci sequence is: 1, 1, 2, 3, 5, 8, 13, 21, … • Another type of sequence is a geometric sequence, such as: 1/2, 1/4, 1/8, 1/16, …

  7. Sequences and Series (cont.) • An idea related to sequences is series. • A series is an infinite sum constructed from a sequence. • Given the sequence a1, a2, a3, … a corresponding series is a1 + a2 + a3 + …

  8. Sequences and Series (cont.) • We study series by looking at sequences of partial sums S1 = a1 S2 = a1 + a2 … Sn = a1 + a2 + … + an • In Example 2, we were finding a closed-form expression for the nth partial sum of a geometric series.

  9. Example 1 (cont.) • Using the result from Example 2, with a = 500, r = 1.04, and n = 5, we get • S = 500 + 500(1.04) + 500(1.04)2 + 500(1.04)3 + 500(1.04)4 = 500(1.045 – 1)/(1.04 – 1) = 2708.16 dollars • An investment of this type is called an annuity!

  10. Annuities • An annuity is a sequence of equal payments made at equal periods of time. • An ordinary annuity is an annuity in which payments are made at the end of each time period, with interest compounded each time period. • We call the time between payments the payment period and the time from the beginning of the first payment period to the end of the last period the term of the annuity. • The future value of the annuity is the final sum after all of the payments have been made – it is the sum of the compound amounts of all payments, compounded to the end of the term.

  11. Annuities (cont.) • Example 1 is an example of an ordinary annuity, with payments of $500, payment period of 1 year, and term of 5 years. • In general, the future value of an annuity of n payments of R dollars each at the end of each consecutive interest period, with interest compounded at a rate i per period, is

  12. Annuities Due • Annuities due are annuities in which payment is made at the beginning of a payment period. • One can show that the future value of an annuity due is

  13. Sinking Funds • One application of annuities is to make periodic payments to guarantee a certain amount of money is available at some future date. • Such an investment is known as a sinking fund.

  14. Example 3 • A 45-year-old teacher puts $1000 in a retirement account at the end of each quarter until the teacher reaches the age of 60 and makes no further deposits. • If the account pays 8% interest compounded quarterly, how much will be in the account when the teacher retires at age 65?

  15. Example 3 (cont.) • Solution: This investment is a combination of an ordinary annuity with a fixed principal earning compound interest! • The ordinary annuity’s future value S after 15 years is found with R = 1000 i = 0.08/4 n = 15*4 S = R[((1+i)n-1)/i] • This is followed by investing principal S for 5 more years, earning compound interest with P = S i = 0.08/4 n = 5*4 A = P(1+i)n • We can use the built-in Excel function FV to help with the calculations (see next page)! • We find that the investment yields $169,474.59.

  16. Example 3 (cont.)

  17. Example 4 • How much money would I have to invest at the end of each month, compounded monthly, at annual interest rate of 9% to have $1,000,000 in twenty years? • Repeat with investment terms of 30 years and 40 years.

  18. Example 4 (cont.) • Solution: This is an example of a sinking fund with S = 1,000,000 i = 0.09/12 n = 20*12 • Using Excel’s PMT function, we find that monthly payments of R = $1497.26 will yield $1,000,000 in 20 years. • For 30 years, R = $546.23 and for 40 years, R = $213.61. • Moral of Example 4: Invest as soon as possible!

  19. Present Value of an Annuity • Suppose instead of investing money in an ordinary annuity, by making n payments of R dollars each at an interest rate i per period, we want to invest P dollars once at the beginning of term, at the same compound interest rate, to get the same amount as the annuity would yield. • To find P, all we need to do is to set the formulas for the future values of compound interest and an ordinary annuity equal to each other!

  20. Present Value of an Annuity (cont.) • Solving for P and simplifying (check!) we find that • We call P the present value of the annuity. • One of the biggest uses of the present value of an annuity formula is figure out periodic payments on a loan!

  21. Example 5 • Our offer to purchase a house for $130,000 has been accepted! • We can afford a down payment of 20% of the sale price and want to take out a mortgage to cover the balance. • A local bank is offering a fixed 30-year loan at an annual interest rate of 6.618%. • What would our monthly payment be for this loan? • How much interest would we pay over the lifetime of the loan?

  22. Example 5 (cont.) • Solution: Our loan amount is 130,000 - 0.20*130,000 = 104,000 dollars. • Paying a lump sum of $104,000 at the beginning of the loan would pay off the loan, so if we take the loan amount to be the present value of an ordinary annuity, we can figure out our monthly payment!

  23. Example 5 (cont.) • With P = 104,000 i = 0.06618/12 n = 30*12 = 360 we can find the monthly payment R from the present value of an annuity formula by solving for R: • This calculation can be done in Excel with the PMT function! • Our payment is $665.44 per month. • For this loan, we’ll end up paying 665.44*360 - 104,000 = 135,558.40 dollars in interest!

  24. Amortization • A loan is amortized if both the principal and interest are paid by a sequence of equal payments. • Thus, the loan in Example 5 is amortized by making 360 monthly payments of $665.44. • A useful way to work with a loan is via an amortization schedule, which is a table that keeps track of how much of each payment is used to pay for interest and how much is used to pay off the principal loan amount. • Remark: When loans for property such as a house are made, usually the home is used as a security against the borrower being unable to make loan payments. Such an arrangement is called a mortgage.

  25. Example 6 • Make an Amortization Schedule for the house loan in Example 5. • Use this schedule to determine how long it will take to pay off half of the loan (i.e. to owe only $52,000).

  26. Example 6 (cont.)

  27. Example 6 (cont.) • The 258th payment (21.5 years) is the one at we will have paid off half of the house loan! • Technically, we still owe $51,805.91 at this point. • Notice that for each payment, the amount going towards interest decreases and the amount going towards the principal increases!

  28. References • Finite Mathematics and Calculus with Applications (4th edition), by Margaret Lial, Charles Miller, and Raymond Greenwell.

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