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Chapter 5

- Mathematics of Finance

Compound Interest

- In the last class session, we covered the basics

regarding compound interest - Tonight, we build on these results to get some

help for working with financial investments. - Let P principal invested, let S the compound

amount (i.e., the amount of money we have after n

time periods) and let r interest rate.

Compound Interest (cont.)

- Then S P(1 r)n
- Example Suppose that we purchase a certificate

of deposit for 1000 to mature in 4 years with an

interest rate of 4.5 - S 1000(1 0.045)n 1000(1.1925)
- S 1,192.50
- Note that in this problem, the idea is to find S,

given P, r, and n

Quarterly Compounding

- Suppose now compounding is quarterly. How long

does it take for 1500 to become 2100 - 2100 1500(1 0.0125)n
- ln(2100/1500)/ln(1.0125) n
- So, n 27.088 quarters or 6.77 years
- Shows that quarterly compounding decreases the

length of time it takes for 1500 to become 2100

at 5 interest.

Power of Compound Interest

- Suppose a person at age 23 puts 1000 aside at 6

interest for retirement. How much money will be

available at age 65 - S 1000(1 0.06)42 11,560
- Suppose a person at age 60 puts 1000 aside at 6

interest for retirement. How much money will be

available at age 65 - S 1000(1 0.06)5 1,340

Power of Compound Interest

- Suppose that you have outstanding credit card

debt of 4700. The interest rate on the card is

23 per year. How much would be owed at the end

of one year if there was no need to make the

minimum payments - S 4700(1 0.24) 5828.00
- After 5 years it would be
- S 4700(1 0.24)5 13,778.63
- After 5 years with monthly compounding
- S 4700(1 0.02)60 15,420.84

Two Variations on Compound Interest

- Suppose we know P, S, and n and want to find r.
- Example Assume that we buy a CD for 600 and

after 5 years it is worth 900. What interest

rate was earned - Then, 900 600(1 r)5
- So, (900/600)1/5 (1 r)
- Or, 1.0845 1 r
- This means that r 0.0845

Two Variations (cont.)

- Another use for the compound interest formula

comes up if we know S, P, and r and want to find

n. - Suppose we know that a CD purchased for 1500

will eventually be valued at 2100. At 5

interest, how long will it take for this to

happen. - 2100 1500(1.05)n
- ln(2100/1500) n(ln(1.05))
- n ln(2100/1500)/(ln(1.05)) 6.897 years

Doubling Time of Money

- Given that S P(1 r)n then the principal

invested, P, doubles when S 2P - Example How long does it take for 1500 to

double at 5 - 3000 1500(1 0.05)n
- ln(3000/1500) ln(2) n(ln(1.05))
- n ln(2)/ln(1.05) 14.21
- n 14.21 years

Doubling Time of Money

- With quarterly compounding, how long does it take

for the 1500 to double - 3000 1500(1 0.0125)n
- n ln(2)/ln(1.0125)
- Daily compounding
- n ln(2)/ln(1.0001)
- Continuous compounding Use rule of 70, so at 5

interest it takes 14 years(i.e., 5x14 70

Effective Rate of Interest

- Definition The effective rate re that is

equivalent to a nominal (annual) rate of interest

r compounded n times per year is - re (1 (r/n))n 1
- Example suppose that compounding occurs 4 times

per year - re (1 (r/4))4 - 1

Effective Interest Rates

- Suppose compounding is quarterly and we want to

know the equivalent annual rate of interest - Again, let P 1500, let the given (nominal)

interest rate equal 5 and suppose money is

invested for one year - Annual compounding S 1500(1.05) 1575
- Quarterly compounding S 1500(1.0125)4

1500(1.0509) 1576.41

Effective Rate (cont.)

- So, the effective interest rate calculation asks

What annual rate would bring 1500 to 1576.41

This would be re 0.0509 or 5.09 - Another way to say this is that 5 interest

compounded quarterly is the same as 5.09

compounded annually.

Another example

- What effective rate is equivalent to a nominal

rate of 6 compounded monthly - re (1 (0.06/12))12 1.0617
- What if compounding occurs daily
- re (1 (0.06/365))365 1.0618
- What if compounding occurs continuously
- In this case, as we saw in Chapter 4, the

effective rate of interest is e0.06 1.0618

27 p. 222

- A zero-coupon bond is a bond that is sold for

less than its face value and has no periodic

interest payments instead the bond is redeemed

at face value at maturity. Suppose that such a

bond sells today for 420 and can be redeemed in

14 years for 1000. The bond earns what nominal

rate compounded semiannually

Solution

- 1000 420(1 (r/2))28
- Let x r/2
- 2.381 (1 x)28
- ln(2.381) 28ln(1 x)
- Ln(1 x ) ln(2.381)/28 0.031
- So, (1 x) e0.031 1.0315
- This means that x 0.0315 r 0.0630

28, p. 222

- Suppose that 1000 is hidden under a mattress for

safekeeping. Each year the purchasing power of

money is 96.5 of what it was the previous year.

After 6 years, what is the purchasing power of

the 1000 - S 1000(1 (-0.035))6
- S 1000(1 0.035)6 807.54

Homework

- Pp. 221-222 7,9,11,15,21,25

Present Value

- We know that the future sum, S, that is obtained

by investing the principal, P, at 100r for n

years is given by - S P(1 r)n
- So, if we want to find the value today of a

future sum to be paid in n years when the

interest rate is 100r, we would calculate - P S/(1 r)n
- Or, P S(1 r)-n

Example 1

- Find the present value of 1000 due after 3 years

if the interest rate is 9 - P 1000(1.09)-3 1000/1.295772.18
- Same problem but with monthly compounding
- P 1000(1.0075)-36 1000/1.309764.15

Example 2

- Suppose that a trust fund for a childs education

is set up by a single payment today so that at

the end of 15 years 50,000 will be available for

college tuition payments. If the interest rate

is 7 compounded semiannually, how much should be

paid into the fund today - P 50,000(1.035)-30 17,813.92

Equations of Value

- Suppose that Mr. Smith owes Mr. Jones two sums of

money 600 due in 5 years and 1000 due in 2

years. The interest rate is 8 compounded

quarterly. How much should Mr. Smith pay today

to retire both of the debts - x 1000(1.02)-8 600(1.02)-20
- x 1000/1.17 600/1.49 854.70 403.78

1,258.48

Equations of Value

- A debt of 3000 due in 6 years is to be paid off

by three payments, 500 now, 1500 in 3 years and

a final payment after 5 years. How large should

the final payment be The interest rate is 6

compounded annually. - 3000 500(1.06)6 1500(1.06)3 x(1.06)
- Divide both sides by (1.06)
- x 3000(1.06)-1 500(1.06)5 1500(1.060)2
- x 2830.19 669.11 1685.40 475.68

Net Present Value

- Suppose that you can invest 20,000 in a business

that will guarantee you a cash flow of 10,000 in

year 2, 8,000 in year 3, and 6,000 in year 4.

the interest rate 7 compounded annually. What

is the net present value of the business venture

This is the value today of the cash flow less

investment cost. - NPV 10,000(1.07)-2 8,000(1.07)-3

6,000(1.07)-4 20,000 -471.31 - With a negative NPV, the investment should not be

undertaken.

Homework

- P. 226-227 1, 13, 15, 19, 21

Geometric Sequence

- Definition The sequence of n numbers
- a, ar, ar2, ar3, ,arn-1 where a 0 is called

a geometric sequence with first term a and common

ratio r - Example Suppose 100 is invested at 6 for 4

years. Then the compound amounts at the end of

each year is a geometric sequence and can be

written as 100(1.06), 100(1.06)2, 100(1.06)3,

100(1.06)4

Geometric Series

- Definition A geometric series is defined as the

sum of terms in a geometric sequence. - Example 100(1.06) 100(1.06)2 100(1.06)3

100(1.06)4 - In general, a geometric series can be written as

a ar ar2 ar3 . arn-1

Geometric Series

- Suppose we wish to compute the sum of terms in a

geometric series - s a ar ar2 ar3 . arn-1
- rs ar ar2 ar3 ar4 . arn
- s rs a arn
- s(1 r) a(1 rn)
- s a(1 rn)/(1 r)

Examples

- Suppose a 1, r ½, and n 6. Find s.
- s a(1 rn)/(1 r)
- s (1 0.57)/(1 0.5) .9922/0.5 1.9844

127/64 (see p. 230). - Suppose we wish to evaluate 35 36 37 38

39 310 311 - a 35 243, r 3, n 7
- s 243(1 37)/(1 3) 243(-2,186)/(-2)

265,599

Annuities

- An annuity pays a fixed amount of money per

period (year, quarter, month) for a fixed number

of years. - Suppose that the payment per period is R, a

period is one year long, the interest rate is r,

and there are n payments. In general, how much

would we pay today for this income stream - A R(1 r)-1 R(1 r)-2 R(1 r)-n
- A R(1 r)-1 1 - (1 r)-n/1 - (1 r)-1

Annuities (cont.)

- With a little algebra, The annuity value can be

written as - A R1 (1 r)-n/r
- This formula gives the present value A of an

annuity of R dollars per payment for n periods at

the interest rate of r per period. - Example Find the present value of an annuity of

100 per month for 3.5 years at interest rate 6

compounded monthly - Using Appendix B , p. 954, this is 100(37.7983)

3779.83

Annuities (cont.)

- Or, this value can be found approximately using a

calculator. Let anr 1 (1 r)-n/r - For r 0.005 and n 42, anR 37.8, so A

3780 - Value of a consol (promise to pay in perpetuity).

In this case, anr 1/r. So, a promise to pay

100 each year in perpetuity would be valued at

2000 if the interest rate is 5. If the

interest rate is 10, A 1000

Example 6, p. 231

- Given an interest rate of 5 compounded annually,

find the present value of an annuity of 2000 due

at the end of each year for three years and 5000

due at thereafter at the end of each year for

four years. - A 5000(5.786) - 3000(2.723) 2885.91
- where a7,0.05 5.786 and a3,0.05 2.723

Example 7, p. 232

- If 10,000 is used to purchase an annuity

consisting of equal annual payments for four

years and the interest rate is 6 compounded

annually, find the amount of each payment. - Know that A Ranr so, R A/anr
- R 10,000/a4,0.06 10,000/3.465 2885.91

Another Example

- Suppose that a home mortgage of 150,000 is to be

retired by making 48 monthly payments. The

interest rate is 6. Determine the monthly

payment. - The way to look at this is that the bank is

purchasing an annuity valued today at 150,000

with monthly payment for 4 years at 6 compounded

monthly. - R A/anr 150,000/a48,0.005 150,000/42.580

3522.78

Bond Values

- Suppose that you buy a 20-year corporate bond for

10,000. The coupon rate is 5 paid

semiannually. How much would you pay for the

bond if the interest rate is now 6 - S Ranr 10,000/(1 r)40
- R 250, n 40, r 0.03, anr 23.114772
- R 5778.69 S 5778.69 3065.57 8844.26

Bond Prices (cont.)

- Suppose now that interest rates are 4. What

would you pay for the same bond - S Ranr 10,000/(1 r)40
- R 250, n 40, r 0.02, anr27.355479
- S 6838.87 4528.90 11,367.77
- Two conclusions
- When interest rates are above (below) the coupon

rate, bonds trade at a discount (premium) - Bond prices vary inversely with the interest rate

Amount of an annuity

- Thus far, we have considered the value of an

annuity. This is what should be paid today for a

stream of future periodic payments - Now we wish to consider the amount of an annuity

The amount of money available at the end of the

period of a stream of payments compounded forward - The formula for the amount of an annuity is
- S R(1 r)n 1/r Rsnr

Amount of an Annuity

- Find the amount of an annuity consisting of

payments of 50 at the end of every 3 months for

three years at 6 compounded quarterly. Also,

find the compound interest that would accrue. - S Rsnr
- S 50s12,0.015 50(13.041) 652.06
- Compound interest accruals are 52.06.
- 52.06 652.06 - 600

Sinking Fund

- A sinking fund is a fund into which periodic

payments are made to satisfy a future obligation.

Suppose that a machine costing 7000 is to be

replaced at the end of 8 years, when it will have

a salvage value of 700. How much must be set

aside each quarter to replace the machine if the

interest rate is 8 compounded quarterly

Solution

- The amount needed after 8 years is 7000 - 700

6300 - The required quarterly payment is
- R S/snr
- R 6300/s32,0.02 6300/44.227 142.45
- So, a quarterly payment of 142.45 will

accumulate to 6300 after 8 years at 8 interest

compounded quarterly

Retirement Planning

- A woman at age 25 has 17,000 in her retirement

plan and wishes to contribute 2000 each year

until she retires at age 65. Assume an interest

rate of 6 compounded annually. How much will

she have available for retirement - S Rsnr 17000(1 r)40
- S 2000(154.762) 17000(10.29)
- S 309,254 174,857.21 484,381.21

Homework

- P. 236-238 1, 5, 7, 13, 19, 23, 25, 27, 31,33,

41, 43

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