Matakuliah
Tahun
: D0762 – Ekonomi Teknik
: 2009
Factors - Extra Problems
Course Outline 3
Outline
•
•
•
•
Problems - Single Payment (P and F)
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Problems - Uniform Series
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Problems – Gradient
Problems – Nominal and Effective Interest
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References :
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Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.44-99
Engineering Economic Analysis, Donald G. Newman, p. 41-86
Engineering Economy, William G. Sulivan, p.137-194, p. 135-140
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Mathematical Factor
• Single Payment Compound Formula
If you put P in the bank now at an interest rate of i% for n years,
the future amount you will have after n years is given by
F = P (1+i)n
The term (1+i)n is called the single payment compound factor.
The factor is used to compute F, given P, and given i and n.
Handy Notation.
(F/P,i,n) = (1+i)n
F = P (1+i)n = P (F/P,i,n).
Single Payment Present Worth Formula
P = F/(1+i)n = F(1+i)-n
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Problems – Single Payment
Define the symbol for the problems :
1. What amount would have to be invested for five years, earning an annual
interest rate of 10 %, to have $805.26? In other words, what is the present
value of 805.26 discounted back five years at an annual rate of 10 %.
2. If you put the money $45,000 at the Bank right now, earning 10% of annual
interest, what would you have in 10 years?
4
Answer no.2
answer
no. 1 -F = $805.26, n = 5, I 10%, P? ; P = $45,000, n =10years, i = 10%, F?
F = P (1+i)n
P = F/(1+i)n = F(1+i)-n
Problems – Single Payment
3. Suppose at year 0 (now) you offered a piece of paper
that guaranteed you would be paid $400 at the end of
three years and $600 at the end of five years. How much
would you be willing to pay for this piece of paper if you
wanted your money to produce a 12% interest?
Answer
Fn=3 = $400; Fn=5=$600, i =12%, P?
P = (1+i)-3 + (1+i)-5
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Problem-Single Payment
• If you were to deposit $2000 in a bank that pays 5%
nominal interest, compounded continuously, how much
would be in account a the end of two years?
F = Pern
R = nominal interest rate = 0,05
N = number of years = 2
F = 2.000e(0,05x2) = 2000(1,1052)
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Problem-Single Payment
• A Bank offers to sell saving certificates that will pay the
purchaser $5000 at the end of ten years but will pay
nothing to the purchaser in the meantime. If interest is
computed at 6%, compounded continuously, at what
price is the bank selling the certificate?
P = Fe-rn
P = 5000e-(0,06x10)=5000(0,5488) =$2744
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Uniform Series
Given F, Find A
Uniform series compound amount factor :
(F/A,i,n) = [(1+i)n – 1]/i, i > 0
Uniform series sinking fund:
(A/F,i,n) = i/[(1+i)n – 1]
Uniform series capital recovery :
(A/P,i,n) = [i (1 + i)n]/[(1+i)n – 1]
Uniform series present worth:
(P/A,i,n) = [(1+i)n – 1]/[i (1 + i)n]
Given A, Find F
Given A, Find P
Given P, Find A
8
9
Answer
Alternative 1
P = Fn=2 (P/F,i,2) + Fn=3(P/F,Ii,3) + Fn=4(P/F,i,4)
Alternative 2
First - Find total F for the cash flow and then Find P
P = Total F (P/F,i,4)
= [Pn=2 (F/P,I,2) + Pn=3(F/P,I,3), Pn=4](P/F,i,4)
Alternative 3
Find A, and then Find P
P= [An=2,3,4(P/A,i,3) + Fn=2(P/F,i,2)](P/F,i,1)
P = [20(P/A,15%,3) + 10(P/F,15%,2)](P/F,15%,1)
• If you will receive $20 in the end year two, $30 in the end
of year three, and $20 in the end of year 4, compute the
cash flow value in year 0 (annual interest = 15%)
Uniform Series Problem
Uniform Series Problem
• If company agrees to pay a machine for $12,000 in five
equal annual payment, what would be the payment each
year? (i=4%)
• If immediately after the second payment, the terms of the
agreement are changed to allow the balance due to be
paid in a single payment the next year, what is the final
single payment?
Answer
A = P(A/P,5,4%)
Answer
Final payment = A + A(P/A,4%,2)
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Problem – Uniform Series
• A man deposited $500 per year into a credit union that
paid 5% interest, compounded annually. At the end of
five years, he had $2763 in the credit union. How much
would he have if they paid 5% interest compounded
continuously?
F=A(F/A,r,n)
 e rn  1 
 e0,05(5)  1 
  500 0,05
  $2769.84
F  A r
e

1
e

1




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Problem - Gradient
• Find the present worth value of the cash flow on the tables below :
• Table 1
year
1
2
3
4
5
6
Value
500
480
460
440
420
400
5
6
7
8
9
1000
900
800
750
700
• Table 2
year
• Table
3 1-4
Value
500
year
1
2
3
4
5
6
Value
500
470
440
390
360
330
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Problem – Nominal & Effective Interest
• A department store charges 1 ¾ % interest per month,
compounded continuously, on its customer’s charge
accounts. What the nominal interest rate? What is the
effective interest rate?
Answer :
Nominal = 1 ¾ % x 12 month = 21%
Effective = (1+i)m – 1 = (1+0.0175)12-1 = 0.2314 = 23,14%,
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Problem – Nominal & Effective Interest
• If someone borrow $2000 and repay $51 for the next fifty
months, beginning thirty days after receiving the moneycompute the nominal annual interest for this loan. What
is the effective interest rate?
Annual nominal = (51:2000) x 12 month = 0,306 ≈ 30,6%
Effective : [1+(51:2000)]12-1 = 0,356 ≈ 35,6%
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Problem – Nominal & Effective Interest
• If bank A offers 5% interest, compounded annually for
deposit, and bank B pays 5% interest compounded
quarterly, and you have $3000 and want to put the in a
savings account and leave the money for two year,
compare how much additional interest would you obtain
from bank A and bank B?
Bank A : F = P (1+i)n = 3000(1+5%)2
Bank B : F = P (1+i)n = 3000(1+(5%/3))6
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• What are the nominal and the effective interest rate for
6% compounded continuously?
Nominal : 6%
Effective : er-1 = e0.06-1 = 6.18%
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Continuous Compounding
We will pay little attention to continuous compounding in this course. You are
supposed to read the material on continuous compounding in the book, but it
will not be included in the homework or tests.
r = nominal interest rate per year
m = number of compounding sub-periods per year
i = r/m = effective interest rate per compounding sub-period.


r
r

lim 1    1  lim 1  
m 
 m  m 
 m 
m
r
m
r
r

r

 1  lim 1  
m / r  
 m

ia = (1 + i)m – 1 =
= (1 + r/m)m – 1
m
r
r

 1 


r
1
1
r




 lim 1  x  x   1  lim 1  x  x   1   e  1
1/ x 


 x 0

Continuous compounding can sometimes be used to simplify computations, and for
theoretical purposes. The table above illustrates that er - 1 is a good approximation
of (1 + r/m)m for large m. This means there are continuous compounding versions of
the formulas we have seen earlier.
For example,
F = P ern is analogous to F = P (F/P,r,n):
(F/P,r,n)inf= ern
P = F e-rn is analogous to P = F (P/F,r,n):
(P/F,r,n)inf= e-rn
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Summary Notation
i:
n:
P:
F:
A:
G:
g:
r:
ia:
m:
effective interest rate per interest period (stated as a decimal)
number of interest periods
present sum of money
future sum of money: an amount, n interest periods from the present, that is equivalent to P with interest
rate i
end-of-period cash receipt or disbursement amount in a uniform series, continuing for n periods, the
entire series equivalent to P or F at interest rate i.
arithmetic gradient: uniform period-by-period increase or decrease in cash receipts or disbursements
geometric gradient: uniform rate of cash flow increase or decrease from period to period
nominal interest rate per interest period (usually one year)
effective interest rate per year (annum)
number of compounding sub-periods per period
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Summary: Formulas
•
Single Payment formulas:
Compound amount: F = P (1+i)n = P (F/P,i,n)
Present worth:
P = F (1+i)-n = F (P/F,i,n)
•
Uniform Series Formulas
Compound Amount: F
Sinking Fund:
Capital Recovery:
A
Present Worth:
= A{[(1+i)n –1]/i}
= A (F/A,i,n)
n
A
= F {i/[(1+i) –1]}
= F (A/F,i,n)
= P {[i(1+i)n]/[(1+i)n – 1]
= P (A/P,i,n)
P
= A{[(1+i)n – 1]/[i(1+i)n]}
= A (P/A,i,n)
Arithmetic Gradient Formulas
Present Worth
Uniform Series
P
A
Geometric Gradient Formulas
If i  g,
P
If i = g,
P
= A {[1 – (1+g)n(1+i)-n]/(i-g)}
= A [n (1+i)-1]
•
•
= G {[(1+i)n – i n – 1]/[i2 (1+i)n]} = G (P/G,i,n)
= G {[(1+i)n – i n –1]/[i (1+i)n – i]} = G (A/G,i,n)
= A (P/A,g,i,n)
= A (P/A,g,i,n)
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Summary: Formulas
•
Nominal interest rate per year, r: the annual interest rate without considering the
effect of any compounding
•
Effective interest rate per year, ia:
ia = (1 + r/m)m – 1 = (1+i)m – 1 with i = r/m
•
Continuous compounding, :
r – one-period interest rate, n – number of periods
(P/F,r,n)inf= e-rn
(F/P,r,n)inf= ern
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