lecture 6 exam in one week, will cover chapters 1 and 2. do chapter 2 self test

Lecture 6

Exam in One week, will cover Chapters 1 and 2. Do Chapter 2 Self test.

Review

Review Problems 2.28, 2.30(b) (reviewed other problems, took the entire class)

2.28 Compute P for:

Problem 2.30

Lecture 7

Due TuesdayRead Chapter 3 115-136Problems 3.1, 3.2, 3.5, 3.6, 3.7

Chapter 3

Nominal interest rate or annual percentage rate (APR)

r = the nominal interest rate per year M = the compounding frequency or the

number of interest periods per year r/M = interest rate per compounding

period Effective interest rate = the rate that truly

represents the amount of interest earned in a year or some other time period

ia = (1 + r/M)M – 1

ia = effective annual interest rate

Example If a savings bank pays 1 ½% interest

every three months, what are the nominal and effective interest rates per year,

Nominal %/year, r = 1 1/2% x 4 = 6% Effective interest rate per year, ia = ( 1 + 0.06/4)4 –1 = 0.061 = 6.1% 

Notice that when M=1, ia = r

Example

A loan shark lends money on the following conditions,

Gives you $50 on Monday, you owe $60 the following Monday

Calculate nominal interest rate , r, ? Calculate effective interest rate, ia? If the loan shark started with $50, and

stayed in business for one year, how much money would he have in one year?

Example

F=P(F/P,i,n) 60=50(F/P,i,1) (F/P,i,1)= 1.2, Therefore, i = 20% per

week Nominal interest rate per year = 52

weeks x 0.20 = 10.40, 1040% = r Effective interest rate per year ia = ( 1+

10.40/52)52 –1 = 13,104 = 1,310,400% F = P(1+i)n = 50(1+0.2)52 = $655,200

Effective interest rate

Who said crime doesn’t pay? To calculate the effective interest rate

for any time duration we have the equation,

  ia = (1 + r/M)C – 1

ia = (1 + r/CK)C – 1

where M = number of interest periods per year (ie quarterly compounding, M = 4; monthly

compounding, M = 12) C = number of interest periods per

payment period K = number of payment periods per year

(ie weekly payments, K = 52, monthly payments K = 12)

Effective Interest

Notice that M = CK or M/K = C  Simple case – compounding and

payment are the same

Example Borrow $10,000 at yearly nominal rate of 9%.

Compounding monthly, payment monthly. You pay on the loan for 6 years. What is your monthly payment?

M = 12 (monthly payments),r/M = 0.09/12 = 0.0075 per month, n = 12 months * 6 years = 72 A = P(A/P, i, N) = 10,000 (A/P,

0.0075, 72) = $180/ month

Example Just using equivalence here.

Note that you are really paying.

(1.0075)12 - 1 = 9.38% and not really 9% as stated.

Harder - cases when compounding and payment occur at different time periods.

  Must convert one to the same time

period.

Example Invest at yearly nominal of 9%. Compounding monthly, payment

quarterly. You will invest for 8 years. If you want to have a fund of $100,000

at the end of the 8 years, how much do you have to invest in each quarter?

Solution

M = 12 (monthly compound), K = 4 (quarterly payments). Since we compound more frequently

than we pay, we use the CK method. C = number of compound periods per

payment period = 3. iper = [1 + r / (CK)]C - 1 = [1 + .09/12]3 - 1 = .022

Solution N = 4 * 8 years = 32 payments.

A = F (A/F, i, N) = 100,000 (A/F, .0227, 32) = 2160

Example

Invest at yearly nominal 12%. Compounding semi annually,

payment quarterly and you will invest for 10 years.

If you invest $12,000 per quarter, how much will you have at the end of the 10th year?

Solution M = 2 (semi-annual), K = 4 (quarterly

payments). Two alternate approaches for

compounding less frequently that payment.

(1) Bank gives us interest on the dollars invested from the point of investment, we use the CK method.

This transforms the compound period to the payment period!

Solution Here C = number of compound periods

per payment period = ½ iper = [1 + r / (CK)]C - 1 = [1 + 0.12/2]1/2 -

1 = .0296 compute N = 10 years * 4 payments per

year = 40 payments. F = A (F/A, i, N) = 12,000 (F/A, 0.0296,

40) = 896,654

Solution (2)

(2) In the case where the bank does not give interest on middle of period deposits we use the lumping method.

Lump all payments in an interest period at the end of the interest period.

2 payments in each semi-annual interest period.

Payment is now $24,000 semi-annually. This transforms the payment period to

the compound period!

Solution (2) Now, use the r/M formula. r/M = 0.12/2

= .06. N = 10 years * 2 = 20 payments. F = A (F | A, i, N) = 24,000 (F/A, .06, 20) =

882,854 Note that the bank's strategy in the

second case has cost you about $14,000!!

Continuous Compounding As an incentive in investment, some

institutions offer frequent compounding.Continuous Compounding – as M approaches infinity and r/M approaches zero

Continuous Compounding

1

71828.2)1(lim

limit theCalculus from recall

11lim

11lim

/

/1

x

Kr

x

CK

CK

C

CK

ei

ex

CKri

CKri

Continuous Compounding

When K = 1, to find the effective annual interest of continuous compounding

ia = er – 1

Example $2000 deposited in a bank that pays

5% nominal interest, compounded continuously, how much in two years?

ia = e0.05 – 1 = 5.127%  F = 2000(1 +0.05127)2 = 2210

Now when compounding and payment periods coincide

1. Identify number of compounding periods (M) per year

2. Compute effective interest rate per payment period, i = r/M

3. Determine number of compounding periods, N = M x (number of years)

When compounding and payment periods don’t coincide, they must be made uniform before equivalent analysis can continue.

1. Identify M, K, and C.2. Compute effective interest rate

per payment periodFor discrete compounding,

i = (1 + r/M)C – 1

For continuous compounding, i = er/K - 1

Equivalence

3. Find total number of payment periods, N = K x (number of years)

4. Use i and N with the appropriate interest formula

Example Equal quarterly deposits of $1000, with

r = 12% compounded weekly, find the balance after five years

M = 52 compounding periods/year K = 4 payment periods per year C = 13 interest periods/payment period

Example

i = (1 + .12/52)13 – 1 =3.042% per quarter

N = K x (5) = 4 x 5 = 20

F = A(F/A, 3.042%,20) = $26,985

Example You are deciding whether to invest

$20,000 into your home at 6.5% continuously compounding, or the same amount into a CD compounded semi-annually at 7%, which is the wiser investment, assume 10 years?

Home Investment

r = 6.5% K = 1 ia = er/K – 1 = e0.065 –1 = 6.7%

F = 20,000(1+0.067)10 = $38,254

CD Investment r = 7% M = 2 ia = (1 + r/M)M – 1 = (1 + 7%/2)2 – 1 = 7.12% F = 20,000(1+0.0712)10 = $39,787