tm 661 engineering economics for managers unit 2 multiple/continuous compounding
TRANSCRIPT
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TM 661TM 661Engineering Economics Engineering Economics
for for ManagersManagers
Unit 2Unit 2
Multiple/Continuous Multiple/Continuous CompoundingCompounding
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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
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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
Bank A F10 = 1000 (F/P, 12, 10)
= $3106
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Multiple Multiple CompoundingCompounding
Bank B F10 = 1000(F/P,6,20)
= $3207
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Multiple Multiple CompoundingCompounding
Bank B F10 = 1000(F/P,6,20)
= $3207
Bank C F10 = 1000(F/P,1,120)
= $3300
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Multiple Multiple CompoundingCompounding
Consider investing $100 @ 12% per annum
Bank A compounds AnnuallyBank B compounds QuarterlyBank C compounds Monthly
Find F1yr for Banks A, B, C
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Effective Interest Effective Interest RateRate
Bank A F1yr = 100 (1+.12)1
= 112.00 ieff = 12.00%
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Effective Interest Effective Interest RateRate
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%
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Effective Interest Effective Interest RateRate
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%
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Effective InterestEffective Interest
Let r = annual interest rate m = # compounding periods / year
P = amount invested
F1yr = P(1+r/m)m
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Effective InterestEffective Interest
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]
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Effective InterestEffective Interest
Now, Interest rate = Interest Earned in Period
Principal Started
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Effective InterestEffective Interest
Now, Interest rate = Interest Earned in PeriodPrincipal Started
ieff = Interest / P
= P[(1+r/m)m - 1] / P
= ( 1 + r/m)m - 1
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Class ProblemClass Problem
Find effective interest rate of 12% compounded monthly.
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Class ProblemClass Problem
Find effective interest rate of 12% compounded monthly.
ieff = ( 1+.12/12)12 - 1
= (1 +.01)12 -1 = .1268 = 12.68%
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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.
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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
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Solution; Solution; Continuous Continuous Comp.Comp.
100,000
30,000
1 2 3 4 5
P = -100,000 + 30,000(P/A, ieff, 5)
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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
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Continuous Continuous CompoundingCompounding
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?
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Continuous Continuous CompoundingCompounding
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)
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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
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Continuous Continuous CompoundingCompounding
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
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Continuous Continuous CompoundingCompounding
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
But,
limm
limm
emr rm
m
/)1(lim
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Continuous Continuous CompoundingCompounding
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
F = Penr
P = Fe-nr
limm
limm
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Continuous Effective Continuous Effective InterestInterest
Recall, F = Penr
Interest Earned1yr = F - P
= Per - P
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Continuous Effective Continuous Effective InterestInterest
Recall, F = Penr
Interest Earned1yr = F - P
= Per - PNow, Interest rate = Interest Earned
Principal
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Continuous Effective Continuous Effective InterestInterest
Now, Interest rate = Interest Earned
Principal
ieff = Per - P
P
ieff = er - 1
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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%
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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
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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
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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
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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
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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.
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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.
In today’s dollars$1.00 $1.10
$1.05 $1.10
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Combined Interest Combined Interest RateRate
That is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
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Combined Interest Combined Interest RateRate
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
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Combined Interest Combined Interest RateRate
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)
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Solving for d, the real interest earned after inflation,
wherei = interest rate (combined)j = inflation rated = real interest rate (after inflation
rate)
Combined Interest Combined Interest RateRate
j
jid
1
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ExampleExample
Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?
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Example (cont.)Example (cont.)
How much is $67,275 20 years from now worth if the inflation rate is 3%?
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Alternate: Recall
= (.1 - .03)/(1+.03) = .068
Example (cont.)Example (cont.)
jjid
1
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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.
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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)
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Break TimeBreak Time