![FORMULAS FOR UNIFORM ARITHMETIC
GRADIENT
(A/G,i%,n) = 1/i – n/[(1+i)^n - 1)]
(P/G,i%,n) = 1/i [{1-(1+i)^-n /i} – {n/(1+i)^n}]
(F/G,i%,n) = 1/i [{(1+i)^n – 1}/i –{n}]
A = G(A/G,i%,n)
P = G(P/G,i%,n)
F = G(F/G,i%,n)](https://image.slidesharecdn.com/annuitiesandgradient-150716165635-lva1-app6892/85/Annuities-and-gradient-24-320.jpg)
Annuities and gradient
- 1. ANNUITY
- 2. ANNUITY consists of a series of equal payments made at equal intervals of time
- 3. OCCURRENCES OF ANNUITY Payment of debt by a series of equal payments at equal intervals of time. Accumulation of a certain amount by setting equal amounts periodically Substitution of a series of equal amounts periodically in lieu of a lump sum at retirement of an individual.
- 4. TYPES OF ANNUITIES Ordinary annuity Deferred annuity Annuity due Perpetuity
- 5. ORDINARY ANNUITY One where equal payments are made at the end of each payment period starting form the first period
- 6. 4 ESSENTIAL ELEMENTS OF ORDINARY ANNUITY The amount of payments are equal The payments made at equal intervals of time The first payment is made at the end of the first period and all payments thereafter are made at the end of the corresponding period Compound interest is paid on all amounts in the annuity
- 7. 1 2 n-2 n-1 n $1 $1 $1 $1 $1 (P/A, i%,n) (F/A, i%,n)
- 8. EXAMPLE FOR ORDINARY ANNUITY A steam boiler is purchased on the basis of guaranteed performance. However, initial tests indicate that the only operating cost will be P 400 more per year than guaranteed. If the expected life is 25 years and money is worth 10%, what deduction from the purchase price would compensate the buyer for the additional operating cost?
- 9. EXAMPLE ON ORDINARY ANNUITY How much money would you have to deposit for five consecutive years starting one year from now if you want to be able to withdraw P50,000 ten years from now? Assume interest is 14% compounded annually.
- 10. DEFERRED ANNUITY One where the payment of the first amount is deferred a certain number of periods after the first. It will be noted that the first payment is made at a period later than the first. After the first payment is made, all succeeding payments are paid at the end of the periods extending to the end of the annuity.
- 11. 1 2 k k+1 k+2 k+n A 0 1 2 n-1 n k/(P/A,i%,n ) (P/A,i%,n) Deferment, k periods Ordinary Annuity, n periods Deferred Annuity, (k+n) periods
- 12. EXAMPLE ON DEFERRED PAYMENT A lathe for a machine shop costs P60,000 if paid in cash. On the instalment plan, a purchaser should pay P20,000 down payment and 10 quarterly instalments, the first due at the end of the of the first year purchase. If the money is worth 15% compounded quarterly, determine the quarterly instalment.
- 13. EXAMPLE ON DEFERRED ANNUITY A man invests P10,000 now for the college education of his 2 year old son. If the fund earns 14% effective, how much will the son get each year starting from his 18th to the 22nd birthday?
- 14. ANNUITY DUE One where the payments are made at the start of each period, beginning from first period.
- 15. EXAMPLE ON ANNUITY DUE A farmer bought a tractor costing P25,000 payable in 10 semi-annual payments, each instalment payable at the beginning of each period. If the rate of interest is 26% compounded semi-annually, determine the amount of each instalment.
- 16. PERPETUITY An annuity where the payment periods extend FOREVER or in which the periodic payments continue indefinitely
- 17. CAPITALIZED COST The capitalized cost of any structure or property is the sum of its first cost and the present worth of all cost for replacement, operation, and maintenance for a long period of time or forever.
- 18. EXAMPLE ON PERPETUITY If money is worth 8% compounded quarterly, compare the present values of the following: Annuity of P1000 payable quarterly for 50 years Annuity of P1000 payable quarterly for 100 years A perpetuity of P1000 payable quarterly
- 19. EXAMPLES ON CAPITALIZED COST AND PERPETUITY The capitalized cost of a piece of equipment was found to be P142,000. the rate of interest used in the computations was 12%, with a salvage value of P10,000 at the end of a service life of 8 years. Assuming that the cost of perpetual replacement remains constant, determine the original cost of the equipment.
- 20. EXAMPLES ON CAPITALIZED COST AND PERPETUITY Compare the capitalized costs of the following road pavements: An asphalt pavement costing P 100,000 which would last for 5 years with negligible repairs. At the end of 5 years, P 5,000 would spent to remove the old surface before P 100,000 is spent again for a new surface. A thick concrete pavement costing P 250,000 which would last indefinitely, with a cost of P 20,000 for minor repairs at the end of every 3 years. Money is worth 8% compounded annually.
- 21. EXAMPLE ON CAPITALIZED COST AND PERPETUITY It costs P 50,000 at the end of each year to maintain a section of Kennon Road in Baguio City. If money is worth 10% how much would it pay to spend immediately to reduce the annual cost to P10,000?
- 22. UNIFORM GRADIENT
- 23. GRADIENT A series of disbursement or receipts that increase or decrease in each succeeding periods.
- 24. FORMULAS FOR UNIFORM ARITHMETIC GRADIENT (A/G,i%,n) = 1/i – n/[(1+i)^n - 1)] (P/G,i%,n) = 1/i [{1-(1+i)^-n /i} – {n/(1+i)^n}] (F/G,i%,n) = 1/i [{(1+i)^n – 1}/i –{n}] A = G(A/G,i%,n) P = G(P/G,i%,n) F = G(F/G,i%,n)
- 25. EXAMPLES ON ARITHMETIC GRADIENT Find the value of each of the following: (A/G,14.53%,23) (A/G,31%,50) (P/G,12%,10) (P/G,15.6%,35) (F/G,7.8%,21) F/G,12.5%,18)
- 26. EXAMPLES ON ARITHMETIC GRADIENT Compute the value of the amount of C. C 0 50 100 150 i =10%
- 27. SOLUTION C = G(P/G, i%, n) G = 50, i = 10%, n =4 , (P/G, i%, n) =4.378 C = 50(4.378) = 218.91 0 50 C 100 150 i =10%
- 28. EXAMPLES ON ARITHMETIC GRADIENT Compute the value of the amount of F 50 F 200 100 150 i = 10%
- 29. EXAMPLES ON ARITHMETIC GRADIENT Mr. Marcelo Santos, author of Para Sa Hopeless Romantic was offered he following alternatives by Star Cinema for the rights to make his novel into a movie. A single lump sum payment of P 500,000 or An initial payment of P 250,000 plus 2% of the movie’s gross receipts for the next 5 years estimated to be as follows: After the 5th year, the author will not receive further royalties. If money is worth 14%, which alternative should he select. Disregard income tax considerations. End of year Gross Receipts In Millions 2% of Gross Receipts in Thousands 1 P 10 P200 2 8 160 3 6 120 4 4 80 5 2 40
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