LECTURE-5-CAPITALIZED-COST-AMORTIZATION-GRADIENT.pptx

capitalized cost

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

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

5. Difference between P and X in a perpetuity: k 2k 3k 4k X S S S S 1 2 3 4 P A A A A 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.

6. 15 30 45 60 FC=P300,000 S =300,000-30,000=P270,000 S S S S FC =P300,000 Capitalized Cost

7. Capitalized Cost

8. Capitalized Cost

9. OPERATION: Capitalized Cost

10. Capitalized Cost

11. 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.

12. 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.

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. 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 Total s AMORTIZATION SCHEDULE

15. 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. B 2 = B 1 - E 1 COLUMN C = INTEREST DUE AT THE END OF THE PERIOD. C 1 = B 1 i i = interest rate per period COLUMN D = PAYMENT = AMOUNT OF ANNUITY COLUMN E = PRINCIPAL REPAID AT END OF THE PERIOD E 1 = D-C 1

16. 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

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

18. 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.

19. 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.

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

21. Uniform Arithmetic Gradient

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

23. 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?

24. Example Problem

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

26. 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

27. Excercise Problem 2 Compute the values of x and F