Q.4 (1 Mark)
Fairmont Textile has a plant in which employees have been having trouble with carpal
tunnel syndrome (CTS. an inflammation of the nerves that pass through the carpal tunnel,
a tight space at the base of palm), resulting from long-term repetit
Q.7 (1 Mark)
Suppose that an oil well is expected to produce 100,000 barrels of oil during its first year
in production. However, its subsequent production (yield) is expected to decrease by 10%
over the previous years production. The oil well has a prove
Equivalence Calculation with Changing
Interest Rates
F=?
Find the balance at the end of year 5.
6%
6%
5%
4%
4%
0
1
2
3
4
5
$400
$300
$500
Solution
n = 1:
$300( F / P , 5% ,1) = $315
n = 2:
$315( F / P , 6% ,1) + $500 = $833.90
n = 3:
$833.90( F / P , 6% ,
Personal Savings Point of View
!
Situation 1: If you make
four annual deposits of
$100 in your savings
account which earns 10%
annual interest, what equal
annual amount (A) can be
withdrawn over 4
subsequent years?
Economic Equivalence Point of View
!
Sit
Equal Cash Flow Series
Compound amount factor (Equal Cash Flow
Series)- Find F, Given A, i , N.
F=?
0
1
A
A
N-1
3
2
A
N
A
A
A
F = A+A(1+i) +A(1+i)2+A(1+i)3+.+A(1+i)N-1
F = A[(1+i)N-1)/i] = A(F/A, i, N)
1
Example 1
You deposit $ 500 in a credit union at th
Approach: Modify the Interest Rate
! Idea:
Since the cash flows occur every other year,
let's find out the equivalent compound interest rate
that covers the two-year period.
! How: If interest is compounded 10% annually, the
equivalent interest rate for t
o
o
Annual interest rate computed as the product of
interest rate per period and the number of periods
per year
Interest rates are normally quoted on an annual
basis
Or Compounding several times within a year:
!
!
!
!
!
Semiannually
Quarterly
Monthly
Week
Arithmetic - Gradient
Gradient series: A pattern of receipts or disbursements
(expenditure) for a series that has a uniform linear
(arithmetic or constant) or geometric increase in each
time period
( N 1)G
A
2G 3G
G
01
N
234
1
N
A = G
N
i (1 + i ) 1
Per
Summary
PF
F P i , N Compound Amount Factor
FP
P Fi , N Present Worth Factor
A F
F Ai , N Compound Amount Factor (U. Series)
FA
A F i , N Sinking Fund Factor (U. Series)
PA
A P i , N Capital Recovery Factor (U. Series)
A P
P Ai , N Series Present Worth Fa
Example
Calculate P. Given A=$1000, i=10%, g=8%, N=5
for i g
1 (1 + g ) N (1 + i ) N
P = A
ig
g%
P
A
1 (1.08) 5 (1 + 0.1) 5 0 1 2
P = $1000
0.1 0.08
P = $4,383
If i = 8%, g = 10%. P = $4,804
NA
If i = g = 10%
= 5($1000) / 1.1 = $4,545
(1 + i )
N
15
Exam
Arithmetic Gradient Conversion
to Uniform Series
Another Form
A G A 2G
A
0123
N 1
N
The pattern :
A, A G , A 2 G , A 3G ,., A ( N 1)G
N , is the duration of the series
3
Example
A man has purchased a new car. He
wishes to set a side enough money in a
bank
23/09/2013
Example
Example 6 Multiple Payments Tuition
Tuition
Prepayment
Prepayment Option
Check to See if $107,690 is Enough to
Meet the Future Tuition Needs
0
1
2
3
$107,690
$80,482
$56,599
$29,873
0
$4,829
$3,396
$1,792
Payment
-$27,208
-$28,712
-$30,
Example
Find Future Worth F. Given: i = 10%, N = 5 yrs.
G = $200
$1200
01
Solution
F
F
F
F
2
3
4
5
= [A(F / A,10%,5)] [$G ( A / G,10%,5)]* (F / A,10%,5)
= 1200(6.105) $200(1.8100)(6.1050)
= $7,326.00 $2,210.01
= $5,115.99
5
Example: Cash Flow Analysis
Fin
Example 5
Suppose we decided to pay the 6,800 for the time
payment purchase contract in the previous
example (4) what monthly rate of return would
we obtain on our investment?
9
Example
Example 6 Handling Time Shifts in a
Uniform
Uniform Series
10
5
Example
!
On a certain piece of machinery, it is Year
estimated that the maintenance
expense will be as shown. What is
1
the equivalent uniform annual
maintenance cost for the machinery 2
if 6% interest is used?
Maintenance
$ 100
200
3
300
4
400
Geometric
Geometric Series
End of Yr.
Amount
1
A(1 + g )0
2
A(1 + g )1
:
:
N
A(1 + g ) N 1
Present Worth
A(1 + g )0
(1 + i)1
A(1 + g )1
(1 + i) 2
A(1 + g ) N 1
(1 + i) N
13
Note:
P / F , i, N =
1
(1 + i ) N
N
1
P = A (1 + g ) k 1
(1 + i) k
k =1
:
:
From the above
Nominal vs. Effective Interest Rates:
Note Statements:
8% Compounded Quarterly
cfw_It is a nominal rate, r
8% nominal
= 2% effective interest rate per quarter
4 quarters
ieff quarter = 2%
Effective Interest Rate ieff :
Is the one rate that truly represen
Continuous Compounding
Continuous Compounding is the ultimate limit for the
number of compounding periods in 1 year.(m infinity)
Starting from:
r
i = 1 +
m
m
1
r
i = lim 1 +
m
m
m
m
m
r r
1 = lim 1 +
m
m
r
1
r
m
r r
r
lim 1 +
=e
m
m
r
i = e 1 ( where
Tabular Method
B0 is the starting balance
I1 = B0*i =P *i
PP1 = A - P *i
B1 = B0 - (A-P *i) =P-PP1
At the end of second period
I2 = B1 i = (P-PP1)i
P2 = A (P-PP1)i = (A-Pi) + PP1 i = PP1 (1+i)
B2 = B1 PP2 = P-(PP1+PP2)
Bn = P- (PP1+PP2+.+PPN)
= P- (PP1+PP
Example
Example 7
Suppose you make equal quarterly deposit of
$1,000 into a fund that pays interest at a rate of
12% compounded monthly. Find the balance at
the end of year three.
Solution
Solution
!
!
Step 1: M = 12 compounding periods/year
K = 4 payment
Solution
!
!
!
In = (Bn -1) i
I1 = 5000 (12%/12) = 50.0
PP1 = A In
= 235.37 50
PP1 = 185.37
B1 = 5000 185.37 = 4814.63
I2 =4814.63 *0.1 = 48.15
PP2 = 235.37-48.15
PP2 = 187.22
B2 = 4814.63 187.22= 4627.41
5
Add on loans
In this type of loan , the total in
Example
!
!
Suppose that you borrow $5000 with add-on rate
of 12% for 2 years. You will make a 24 equal
monthly payment.
Determine the amount of the monthly installment
Compute the nominal and effective annual
interest rate on the loan
Add- on rate = 12%
Mortgage
!
!
1.
2.
3.
4.
"
A mortgage is generally a long-term amortized loan used primarily to the purpose of
purchasing a piece of property such as home
Types of Mortgages
National Housing Act (NHA) mortgage
Conventional mortgage
High- Ratio mortgage
Co
Example Closed mortgage with prepayment
privileges
Home price = 125,000
Down payment 25,000
Conventional mortgage = 100,000
3 years term 8% per annum
Amortization period of 25 years
Closed mortgage with payment privileges once each 1- year extra payment o