Download INTEREST

INTEREST
SIMPLE INTEREST:
I=P * i * n
I : Total interest paid
P : Principal or Capital Borrowed
i : Interest rate for one period of time
n : Number of periods.
Repayment is
S = P + I = P * (1 + i * n)
Usually : 1 period = 1 year . For less than 1
year we have:
Ordinary Simple Interest = P* i * d/360
Exact Simple Interest
= P* i * d/365
COMPOUNDED INTEREST
At the end of each interest period the
interest is added to the principal
Period
Principal at
start of period
Interest earned
Compound
amount S at the
end of period
1
2
3
……..
n
P
P(1+i)
P(1+i)2
Pi
P(1+i)i
P(1+i)2i
P(1+i)
P(1+i)2
P(1+i)3
P(1+i)n-1
P(1+i)n-1i
P(1+i)n
S=P*(1 + i )n
(1 + i )n : Discrete single payment compound-amount factor
NOMINAL INTEREST
Interest rate for 1-year period but
compounded for periods different than
one year.
Example : P=1000, at 6% compounded every 6 months. At the
end of six months the interest is:
Iafter 1/2 year= P 0.06 / 2 = 30
Safter 1/2 year= P (1 + 0.06/2) = 1030
Then
Iafter 1 year= P (1 + 0.06/2) 0.06 / 2 = 30.90
Safter 1 year= P(1 + 0.06/2)(1 + 0.06/2) = 1060.9
General Formula:
Safter 1 year= P*(1 + r/m)m
r : Nominal annual Interest
m: Number of periods of compounding
Safter n years= P*(1 + r/m)m*n
EFFECTIVE ANNUAL INTEREST
Simple interest that will produce the
same total interest at the end of one
year.
Safter 1 year = P*(1 + r/m)m
Nominal
= P*(1 + ieff)
Then
ieff=(1+r/m)m-1
Effective
EXAMPLE :
P
= $1000
Interest
= 2% monthly
Total Time = 2 years.
Simple :
S=1000 (1+0.02*24) = $1480
Compounded :
S= 1000(1+0.02)24 =$1608
Nominal Interest Rate:
2 x 12 = 24 % (annual)
Effective Rate
ieff=(1+0.02)12-1=0.268
(26.8%)
PRESENT WORTH
Amount of money needed NOW to meet a
financial obligation in the future.
P =
1
(1  i )
n
S
(1  i ) n
: Discrete single payment presentworth value factor.
ANNUITIES
Series of equal payments occurring at
equal intervals of time.
Common type : Payment at the end of each interest
interval
ANNUITY TERM = Time from beginning
of the first to the end of the last payment
periods.
AMOUNT OF ANNUITY = Sum of all the
payments and the interest accumulated
at the end of the period.
PAYMENT CALCULATION
Let
R : Payment
S : Amount of Annuity
i : Interest
Total
Interest
accumulated
0
n
1
2
3
4
5
(n-1)
Payment
R
R
R
R
R
R
R
R (1+I)n-1
R (1+in)-2
R (1+i)
n
S= R
 (1  i)
nk
k 1
Then
n
S(1+i)=R
n  k 1
(
1

i
)

k 1
Subtract. (All except first and last term cancel)
Obtain:
S i  R(1  i) n  R
Then
RS
i
(1  i) n  1
If S corresponds to a loan, then S=P(1+i)n.
RP
i(1  i) n
(1  i) n  1
This is the same formula obtained assuming a
revolving loan with zero principal at the end of n
years.
0
n
1
2
3
4
5
(n-1)
Principal left
P(1+i)-R
(P(1+i)-R)(1+i)-R
P(1+i)n-R(1+i)n-1-....- R(1+i)-R
PRESENT WORTH OF AN
ANNUITY
Principal needed to be invested NOW at
compounded interest i to yield the amount
of the annuity, S, at the end of the annuity
term.
P R
(1  i) n  1
i(1  i) n
PERPETUITY
Annuity where periodic payments are
indefinite.
The
annuity
repeats
indefinitely.
EXAMPLE :
Asssume a piece of equipment costs
$12,000. It lasts for 10 years, after
which it is going to be replaced. The
scrap value (VS) is $2,000.
Therefore, there is a one time cost of
CV=$12,000 now and an amount P
needed to obtain 10,000 every 10 years.
The interest is 6%
Example:
Need S=10,000 +P at the end of 10 years. Then:
S=10,000+P = P (1+i)10
Then, P=$12,650 (present value of the renewable
perpetuity)
In general:
S=P(1+i)n
P=S-CR
CR :Replacement cost =CV-VS
Then
C
P  (1  i)Rn  1
The total cost now is : CV+ P= $24,650.
This is called:
CAPITALIZED COST (K)
K= CV+P= CV +
CR
(1  i)n  1
EQUIPMENT COST
Comparison based on Capitalized cost.
Reactor costs:
Life
Mild Steel : 5,000
SS
: 15,000
3 years
12 years
Scrap value=0 (Both reactors)
Which one should be installed?
Capitalized Costs:
Mild Steel:K=5,000 + 5,000
=$ 31,180
(106
. )3  1
SS
:K=15,000+ 15,000
=$ 31,180
n
(1.06)  1
SS reactor is preferred. If SS reactor < 11.3
years, Mild Steel reactor is preferred.
MUST BE
(Again)
CONCEPT:
DONE
PROJECTS
You borrow the FCI.
payments are the fixed cost annual cost.
The
(1  i ) n i
AFC  FCI
(1  i ) n  1
Thus
(1  i ) n i
TAC  FCI
 OC
n
(1  i )  1
annual