tm 661 engineering economics for managers unit 2 multiple/continuous compounding

TM 661TM 661Engineering Economics Engineering Economics

for for ManagersManagers

Unit 2Unit 2

Multiple/Continuous Multiple/Continuous CompoundingCompounding

Multiple Multiple CompoundingCompounding

Example: Consider investment of $1000 at 12% per annum

Bank A compounds annuallyBank B compounds semi-annuallyBank C compounds monthly

Find F10 for bank A, B, and C


Example: Consider investment of $1000 at 12% per annum

Bank A compounds annuallyBank B compounds semi-annuallyBank C compounds monthly

Find F10 for bank A, B, and C

Bank A F10 = 1000 (F/P, 12, 10)

= $3106


Bank B F10 = 1000(F/P,6,20)

= $3207


Bank B F10 = 1000(F/P,6,20)

= $3207

Bank C F10 = 1000(F/P,1,120)

= $3300


Consider investing $100 @ 12% per annum

Bank A compounds AnnuallyBank B compounds QuarterlyBank C compounds Monthly

Find F1yr for Banks A, B, C

Effective Interest Effective Interest RateRate

Bank A F1yr = 100 (1+.12)1

= 112.00 ieff = 12.00%


Bank A F1yr = 100 (1+.12)1

= 112.00 ieff = 12.00%

Bank B F1yr = F4 = 100 (1 + .03)4

= 112.55 ieff = 12.55%


Bank A F1yr = 100 (1+.12)1

= 112.00 ieff = 12.00%

Bank B F1yr = F4 = 100 (1 + .03)4

= 112.55 ieff = 12.55%

Bank C F1yr = F12 = 100 ( 1 + .01)12

= 112.68 ieff = 12.68%

Effective InterestEffective Interest

Let r = annual interest rate m = # compounding periods / year

P = amount invested

F1yr = P(1+r/m)m


Let r = annual interest rate m = # compounding periods / year

P = amount invested

F1yr = P(1+r/m)m

Int Earned = F1 - P

= P(1+r/m)m - P = P[(1+r/m)m - 1]


Now, Interest rate = Interest Earned in Period

Principal Started


Now, Interest rate = Interest Earned in PeriodPrincipal Started

ieff = Interest / P

= P[(1+r/m)m - 1] / P

= ( 1 + r/m)m - 1

Class ProblemClass Problem

Find effective interest rate of 12% compounded monthly.

Class ProblemClass Problem

Find effective interest rate of 12% compounded monthly.

ieff = ( 1+.12/12)12 - 1

= (1 +.01)12 -1 = .1268 = 12.68%

Nominal vs. Effective Nominal vs. Effective Int.Int.

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

Period Cash Flow Period Cash Flow0 - $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.

Solution; Solution; Compound Compound MonthlyMonthly

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; Solution; Continuous Continuous Comp.Comp.

100,000

30,000

1 2 3 4 5

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

Continuous Continuous CompoundingCompounding

Consider r = 12% /yr m= 12 compounding periods

n = 10 yrsThen

F = P (1+r/m)mn

= P (1+.12/12)12*10

= P (1+.01)120


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?


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? Choose n = B.A.N.

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

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

Formal DerivationFormal Derivation

In General

F = P (1 + r/m)mn

= P (1 + r/m)m/r *nr

= P [(1 + r/m)m/r]nr


Now suppose we use an infinite # of compounding periods (continuous)

F = P[( 1 + r/m)m/r]nr

= P [ ( 1 + r/m)m/r]nr

limm

limm



F = P[( 1 + r/m)m/r]nr

= P [ ( 1 + r/m)m/r]nr

But,

limm

limm

emr rm

m

/)1(lim



F = P[( 1 + r/m)m/r]nr

= P[ ( 1 + r/m)m/r]nr

F = Penr

P = Fe-nr

limm

limm

Continuous Effective Continuous Effective InterestInterest

Recall, F = Penr

Interest Earned1yr = F - P

= Per - P


Recall, F = Penr

Interest Earned1yr = F - P

= Per - PNow, Interest rate = Interest Earned

Principal


Now, Interest rate = Interest Earned

Principal

ieff = Per - P

P

ieff = er - 1

Continuous iContinuous ieffeff

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

Soln: ieff = e.06 - 1

= .0618= 6.18%

Solution; Solution; RevistedRevisted

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; Solution; Continuous Continuous Comp.Comp.

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; Solution; Continuous Continuous Altern.Altern.

100,000

30,000

1 2 3 4 5

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

= $6,164

100 000 30 0001

1, ,

( )

e

e e

rn

rn r

InflationInflation

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

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

10% inflation

Combined Interest Combined Interest RateRate

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.


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

ExampleExample

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

Example (cont.)Example (cont.)

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

Alternate: Recall

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

Example (cont.)Example (cont.)

jjid

1

Retirement a.Retirement a.Stu 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, he wishes to have enough money saved so that he 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 he will continue to earn 6% throughout the life of the account, how much does Stu 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 b.Retirement b.

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)

Break TimeBreak Time

tm 661 engineering economics for managers unit 2 multiple/continuous compounding

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