capitalized cost and arithmetic gradient.

1. CAPITALIZED COST, AMORTIZATION AND
UNIFORM ARITHMETIC GRADIENT
ESECON230

2. Capitalized Cost
One of the most important applications of perpetuity is in capitalized
cost. The capitalized cost of any property is the sum of the first cost and
the present worth of all costs of replacement, operation and maintenance for a
long time or forever.
Capitalized cost is an application for perpetuity. It is one method
used in comparing alternatives. It is defined as the sum of the first cost
(FC) and the present worth of all perpetual maintenance and
replacement cost.

3. Capitalized Cost

5. Capitalized Cost
k 2k 3k 4k
X
S S S S
0

6. Difference between P and X in a perpetuity:
k 2k 3k 4k
X
S S S S
0
1 2 3 4
P
A A A A
0
P is the amount invested now at i% per period whose interest at the end of
every period forever is A, while X is the amount invested now at i% per period
whose interest at the end of every k periods is S. if k = 1, then X = P.

7. 15 30 45 60
0
FC=P300,000
S=300,000-30,000=P270,000
S S S
S
FC=P300,000
Capitalized Cost

8. Capitalized Cost

9. Capitalized Cost

10. OPERATION:
Capitalized Cost

11. Capitalized Cost

12. AMORTIZATION
Amortization is any method of repaying a debt, the
principal and interest included, usually by a series of
equal payments at equal interval of time.
Amortization Schedule
-Is a table of periodic loan payments, showing the
amount of principal and interest that comprise each level
payment until the loan is paid off.

13. AMORTIZATION PROBLEM
A debt of P5,000 with interest of 12%
compounded semiannually is to be amortized by
equal semiannual payments over the next 3 years,
the first due in 6 months. Find the semiannual
payment and construct an amortization schedule.

14. AMORTIZATION PROBLEM
A debt of P5,000 with interest of 12%
compounded semiannually is to be amortized by
equal semiannual payments over the next 3 years,
the first due in 6 months. Find the semiannual
payment and construct an amortization schedule.

15. Period (A) Outstanding
principal at
beginning of
period (B)
Interest due
at end of
period
(C)
Payment (D) Principal
repaid at end
of period (E)
1 P5,000
2
3
4
5
6
Totals
AMORTIZATION SCHEDULE

16. Period Outstanding
principal at
beginning of
period (B)
Interest due
at end of
period
(C)
Payment (D) Principal
repaid at end
of period (E)
1 P5,000 P300.00 P1,016.82 P716.82
2 P4,283.18
COLUMN B = OUTSTANDING PRINCIPAL AT BEGINNING OF THE PERIOD. AT THE FIRST
PERIOD, IT IS THE LOAN AMOUNT RECEIVED.
B2= B1 - E1
COLUMN C = INTEREST DUE AT THE END OF THE PERIOD.
C1 = B1i
i = interest rate per period
COLUMN D = PAYMENT = AMOUNT OF ANNUITY
COLUMN E = PRINCIPAL REPAID AT END OF THE PERIOD
E1 = D-C1

17. AMORTIZATION PROBLEM
Amortization Schedule
Period Outstanding
principal at
beginning of
period
Interest due at
end of period
Payment Principal repaid
at end of period
1 P5,000 P300.00 P1,016.82 P716.82
2 P4,283.18 P256.99 P1,016.82 P759.83
3 P3,523.35 P211.40 P1,016.82 P805.42
4 P2,717.93 P163.08 P1,016.82 P853.74
5 P1,864.19 P111.85 P1,016.82 P904.97
6 P 959.22 P 57.55 P1,016.82 P959.27
Totals P1,100.87 P6,100.92 P5,000.05

18. GRADIENT SERIES
- series of cash flow where the amounts
chnages every period

19. Uniform Arithmetic Gradient
In certain cases, economic analysis problems involved
receipts or disbursement that increase or decrease by a
uniform amount each period. For example, maintenance and
repair expenses on specific equipment or property may
increase by a relatively constant amount each period. This is
known as a uniform arithmetic gradient.
Some problems involve cash flows that are projected to
increase or decrease by a uniform amount each period,
thus constituting a arithmetic sequence of cash flows.

20. Uniform Arithmetic Gradient
Suppose that the maintenance expense on a certain
machine is P1000 at the end of the first year and increasing
at constant rate of P500 each year for the next four years.

21. Uniform Arithmetic Gradient
This cash flow may be resolved into two
components
= +
G is known as the uniform gradient amount

22. Uniform Arithmetic Gradient

23. P = PA+PG
where:
A = initial payment
G = uniform gradient amount
Uniform Arithmetic Gradient

24. Example Problem
A loan was to be amortized by a group of four end-of-
the-year payments forming as ascending arithmetic
progression. The initial payment was to be P5,000
and the difference between successive payments
was to be P400. But the loan was renegotiated to
provide for the payment of equal rather than
uniformly varying sums. If the interest rate of the loan
was 15%, what was the annual payment?

25. Example Problem

26. Example Problem
Given:
A = P 5000
G = P 400
n = 4
i = 15%
Req’d:
A’ = Equivalent annual
payment= ?
Sol’n:
P = PA+PG= PA’
P = A(P/A,i%,n) + G(P/G,i%,n)
= A’

27. Excercise Problem 1
Find the equivalent annual payment of the following
obligations at 20% interest.
End of Year Payment
1 P8,000
2 P7,000
3 P6,000
4 P5,000
Hint: uniform gradient is negative