multiple/continuous compounding. understand effective interest rates figure out how to use...

Multiple/Continuous Compounding

ENGM 661Engineering Economics for

Managers

Understand Effective Interest RatesFigure out how to use Inflation/Deflation in

your decisions

Learning Objectives for tonight:

Summary of discrete compounding interest factors.

P F (1 + i)-n (P| F i%,n ) Single sum, present worth factor

F P (1 + i)n (F| P i%,n ) Single sum, compound amount factor

(1 + i)n - 1

i(1 + i)n

i(1 + i)n

(1 + i)n - 1

(1 + i)n - 1

i

i

(1 + i )n - 1

[1 - (1 + ni )(1 + i )-n ]

i 2

(1 + i)n - (1 + ni )

i [(1 + i )n - 1]

1 - (1 + j )n (1 + i )-n

i - j

(1 + i )n - (1 + j) n

i - jF A 1,j (F| A 1 i%,j%,n ) Geometric series, future worth factor

A G (A| G i%,n ) Gradient series, uniform series factor

P A 1,j (P| A 1 i%,j%,n ) Geometric series, present worth factor

A F (A| F i%,n ) Uniform series, sinking fund factor

P G (P| G i%,n ) Gradient series, present worth factor

A P (A| P i%,n ) Uniform series, capital recovery factor

F A (F| A i%,n ) Uniform series, compound amount factor

Name

P A (P| A i%,n ) Uniform series, present worth factor

To Find Given Factor Symbol

for i ≠ j

for i ≠ j

To Find Given Worksheet Function Cell EntryP i, n, F PV =PV(i%,n,,-F )

P i, n, A PV =PV(i%,n,-A )

P i, n, F, A PV =PV(i%,n,-A,-F )

P i, A 1 , A 2 , …, A n NPV =NPV(i%,A 1 ,A 2 ,…,A n )

F i, n, P FV =FV(i%,n,,-P )

F i, n, A FV =FV(i%,n,-A )

F i, n, P, A FV =FV(i%,n,-A,-P )

A i, n, P PMT =PMT(i%,n,-P )

A i, n, F PMT =PMT(i%,n,,-F )

A i, n, P, F PMT =PMT(i%,n,-P,-F )

i n, P,A RATE =RATE(n,A,-P )

i n, P,F RATE =RATE(n,,-P,F )

i n, A, F RATE =RATE(n,A,-F )

i n, P, A, F RATE =RATE(n,A,-P,F )

n i, P,A NPER =NPER(i%,A,-P )

n i, P,F NPER =NPER(i%,,-P,F )

n i, A, F NPER =NPER(i%,-A,,F )

n i, P, A, F NPER =NPER(i%,A,-P,F )

i eff r, m EFFECT =EFFECT(r%,m )

Summary of Selected Excel® Worksheet Financial Functions

Interest Rate Terms…Interest Rate Terms…● Compounding Period (cp) – the time between points

when interest is computed and added to the initial amount.

● Payment Period (pp) – the shortest time between payments. Interest is earned on payment money once per period (cost of money)

● Nominal Rate ( r ) – is a simplified expression of the annual cost of money. It means nothing, unless the compounding period is stated along with it.

● Annual Percentage Rate (APR) – is the nominal interest rate on a yearly basis (credit cards, bank loans, …). It, too, should have a compounding period stated.

● Effective Rate ( i ) – is the rate that is used with the table factors or the closed form equations, and it converts the nominal rate taking into account both the compounding period and the payment period so that the blocks match.

Consider the discrete End-of-Year cash flow tables below:

Period Cash Flow PeriodCash Flow

0 - $100,000 3 $30,0001 30,000 4 30,0002 30,000 5 30,000

Determine the Present Worth equivalent ifa. the value of money is 12% compounded annually.b. the value of money is 12% compounded monthly.c. the value of money is 12% compounded continuously.

Nominal vs. Effective Int.

Solution; Annual Rate

100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .12, 5)

Solution; Nominal/Effective

100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .12, 5)

= -100,000 + 30,000(3.6048)

= $8,144

Solution; Compound Monthly

100,000

30,000

1 2 3 4 5

ir

meff

m

= + -1 1


100,000

30,000

1 2 3 4 5

ir

meff

m

= + -1 1

= + -

= + -= =

112

121

1 01 1

1268 12 68%

12

12

.

( . )

. .


100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .1268, 5)


100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .1268, 5)

= -100,000 + 30,000(3.5449)

= $6,346

Solution; Continuous Comp.

100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

ieff = ????

Now suppose we use an infinite # of compounding periods (continuous). How might we find an answerto our problem of r=12% per year compounded on a continuous basis?

Continuous Compounding

Now suppose we use an infinite # of compounding periods (continuous). How might we find an answerto our problem of r=12% per year compounded on a continuous basis

F = P(1+.12/9999)9999 (one year period)

= P(1.1275)= P(1+.1275)

Continuous Compounding

Continuous CompoundingContinuous

Compoundingi = e( r )(# of years) – 1

Examples:r = 12% per year compounded continuously

ia = e( .12 )(1) – 1 = 12.75%

What would be an effective six month interest rate for r = 12% per year compounded continuously?

i6 month = e( .12 )(.5) – 1 = 6.184%


100,000

30,000

1 2 3 4 5

i e

e

effr= -

= -= - =

1

1

11275 1 12 75%

1

. .


100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .1275, 5)


100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .1275, 5)

= -100,000 + 30,000(3.5388)

= $6,164

Solution; Revisted

100,000

30,000

1 2 3 4 5

P = -100,000 + 30,000(P/A, ieff, 5)

= -100,000 + 30,000(P/A, .1275, 5)

= -100,000 + 30,000(3.5388)

= $6,164

Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate?

Continuous ieff


Soln: ieff = e.06 - 1

= .0618= 6.18%

Continuous ieff


Soln: ieff = e.06 - 1

= .0618= 6.18%

Continuous ieff

Check: Let r=6%, m=999

ieff = ( 1 + r/m)m - 1

= (1+.06/999)999 - 1

= .0618= 6.18%

EFFECTIVE INTEREST RATE

ie = effective interest rate per payment period

= ( 1 + interest rate per cp)(# of cp per pay

period) – 1

= 1 + r me – 1 m

Example:r = 12% APR, compounded monthly, payments quarterly

imonth = 12% yearly = 1 % compounded monthly

12 months

ie = (1 + .01)3 – 1 = .0303 – or – 3.03% per payment

Compounding Period is More Frequent than the Payment Perod

Compounding Period is More Frequent than the Payment Perod

An “APR” or “% per year” statement is a Nominal interest rate – denoted r – unless there is no compounding period stated

The Effective Interest rate per period is used with tables & formulas

Formulas for Effective Interest Rate:

If continuous compounding, usey is length of pp, expressed in decimal years

If cp < year, and pp = 1 year, usem is # compounding periods per year

If cp < year, and pp = cp, usem is # compounding periods per year

If cp < year, and pp > cp, useme is # cp per payment period

11

m

a m

ri

m

ri

1ei )y(r

11

em

e m

ri

CRITICAL POINTCRITICAL POINT

When using the factors,

n and i must always match!

Use the effective interest rate formulas to make sure that i

matches the period of interest(sum any payments in-between compounding

periods so that n matches i before using formulas or tables)

Note:Note:Interest doesn’t start

accumulating until the money has been invested for the full period!

0 1 2 periods

Shows up here on CFD…

(End of Period Convention)

X

Deposit made here …

i

Returns interest here!

Problem 1Problem 1The local bank branch pays interest on savings accounts at the rate of 6% per year, compounded monthly. What is the effective annual rate of interest paid on accounts?

GIVEN: r = 6%/yrm =

12mo/yrFIND ia:

DIAGRAM: NONE

NEEDED!

%17.6112

06.01

11

12

m

a m

ri

Problem 2Problem 2What amount must be deposited today in an account paying 6% per year, compounded monthly in order to have $2,000 in the account at the end of 5 years? GIVEN:

F5 = $2 000

r = 6%/yr

m = 12 mo/yrFIND P:

0 1 25 yrs

P?

$2 000

DIAGRAM: 12

5

5

0 061 1 1 1 6 17

12

2000 6 17 5

2000 1 0 0617 2000 74129

1482 59

.. % /

$ ( | , . %, )

$ ( . ) $ (. )

$ .

m

a

n yrs

ri yr

m

P P F

Problem 2 – Alternate Soln

Problem 2 – Alternate SolnWhat amount must be deposited today in an

account paying 6% per year, compounded monthly in order to have $2,000 in the account at the end of 5 years? GIVEN:

F5 = $2 000

r = 6%/yr

m = 12 mo/yrFIND P:

0 1 260mos

P?

$2 000

DIAGRAM:

80.1482$

)7414.0(2000$)60%,5.0,F|P(2000$P

mo/%5.0mo12

yr1

yr

06.0

m

ri

mos60)yrs5(yr

mos12)yrs)(#m(n

Problem 3Problem 3A loan of $5,000 is to be repaid in equal monthly payments over the next 2 years. The first payment is to be made 1 month from now. Determine the payment amount if interest is charged at a nominal interest rate of 12% per year, compounded monthly.

0

1 2 yrs$5 000

A ?

DIAGRAM:

GIVEN:P = $5

000 r =

12%/yrm = 12

mo/yrFIND A:

Problem 4Problem 4You have decided to begin a savings plan in order to make a down payment on a new house. You will deposit $1000 every 3 months for 4 years into an account that pays interest at the rate of 8% per year, compounded monthly. The first deposit will be made in 3 months. How much will be in the account in 4 years?

0 1 2 3 4 yrs

$1 000

DIAGRAM: F ?

Problem 5Problem 5Determine the total amount accumulated in an account paying interest at the rate of 10% per year, compounded continuously if deposits of $1,000 are made at the end of each of the next 5 years.

0 1 2 3 5 yrs

$1 000

DIAGRAM: F ?

Problem 6Problem 6A firm pays back a $10 000 loan with quarterly payments over the next 5 years. The $10 000 returns 4% APR compounded monthly. What is the quarterly payment amount?

01 2 3 5 yrs = 20

qtrs

$A

DIAGRAM:

$10 000

Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years,

Price = 1000(1.1)5 = $1,610.51

But we still only have 1 ton of copper

Inflation


Price = 1000(1.1)5 = $1,610.51


$1,610 5 years from now buys the same as $1,000 now

Inflation


Price = 1000(1.1)5 = $1,610.51



10% inflation

Inflation


Price = 1000(1.1)5 = $1,610.51



10% inflation (deflation = neg. inflation)

Inflation

Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year

Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.

Combined Interest Rate

Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year

Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.

In today’s dollars$1.00 $1.10

$1.05 $1.10


That is

$1.10 = 1.05 (1+d)1

1(1+.10) = 1(1+.05)(1+d)


That is

$1.10 = 1.05 (1+d)1

1(1+.10) = 1(1+.05)(1+d)

1+i = (1+j)(1+d)

i = d + j + dj


That is

$1.10 = 1.05 (1+d)1

1(1+.10) = 1(1+.05)(1+d)

1+i = (1+j)(1+d)

i = d + j + dj

i = interest rate (combined)j = inflation rated = real interest rate (after inflation

rate)


Solving for d, the real interest earned after inflation,

wherei = interest rate (combined)j = inflation rated = real interest rate (after inflation

rate)


j

jid

+-=1

Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?

Example

Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?

Solution: F = 10,000(1+.1)20

= $67,275

Example

How much is $67,275 20 years from now worth if the inflation rate is 3%?

Example (cont.)

How much is $67,275 20 years from now worth if the inflation rate is 3%?

Solution: FT = 67,275(P/F,3,20)

= 67,275(1.03)-20

= $37,248

Example (cont.)

Alternate: Recall

= (.1 - .03)/(1+.03) = .068

Example (cont.)

jjid +

-=1

Alternate: Recall

= (.1 - .03)/(1+.03) = .068

FT = 10,000(1+d)20

= 10,000(1.068)20

= $37,248

Example (cont.)

jjid +

-=1

Alternate: Recall

= (.1 - .03)/(1+.03) = .068

FT = 10,000(1+d)20

= 10,000(1.068)20

= $37,248

Note: This formula will not work with annuities.

Example (cont.)

jjid +

-=1

Superwoman wishes to deposit a certain amount of money at the end of each month into a retirement account that earns 6% per annum (1/2% per month). At the end of 30 years, she wishes to have enough money saved so that she can retire and withdraw a monthly stipend of $3,000 per month for 20 years before depleting the retirement account. Assuming there is no inflation and that she will continue to earn 6% throughout the life of the account, how much does Superwoman have to deposit each month? You need only set up the problem with appropriate present worth or annuity factors. You need not solve but all work must be shown.

Retirement a.

Solution; Retirement a.

Take everything to time period 360

FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)

0 1 2 3 4 360

1 2 3 4 240A

3,000FP



FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)

A(1,004.52) = 3,000(139.58)

0 1 2 3 4 360

1 2 3 4 240A

3,000FP



FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240)

A(1,004.52) = 3,000(139.58)

A = $416.82

0 1 2 3 4 360

1 2 3 4 240A

3,000FP

Suppose that the solution to the above problem results in monthly deposits of $200 with an amassed savings of $350,000 by the end of the 30th year. For this problem assume that inflation is 3% per annum. Compute the value of the retirement account in year 30 before funds are withdrawn (in today’s dollars)

Retirement b.


0 1 2 3 4 360

1 2 3 4 240200

3,000FP = 350,000

FP = 350,000

FPT = 350,000(1+j)-n


0 1 2 3 4 360

1 2 3 4 240200

3,000FP = 350,000

FP = 350,000

FPT = 350,000(1+j)-n

= 350,000(1+0.03)-30

= $144,195


0 1 2 3 4 360

1 2 3 4 240200

3,000FP = 418,195

FP = 418,195

FPT = 418,195(1+j)-n

= 418,195(1+0.03)-30

= $172,290

multiple/continuous compounding. understand effective interest rates figure out how to use...

Documents

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solution revisted100

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