Eeco Chapter 2

ANNUITIES An annuity is a series of equal payments occurring at equal periods of time 

Ordinary annuity o Payments are made at the end of each period [

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Ex. a.What are the present worth and the accumulated amount of a 10 year annuity paying 10,000 at the end of each year with interest as 15% compounded annually? Ans. F = 203, 037 b. A chemical engineer wishes to set –up a special fund by making a uniform semiannual end- of period deposits for 20 years. The fund is to provide a 100,000 at the end of each of the last five years of a 20 year period. If interest is 8% compounded semiannually, what is the required semi-annual deposit to be made? Ans. A = 6,193.99 

Deferred annuity o The first payment is made several periods after the beginning of the annuity [

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Ex. a. If 10,000 is deposited each year for 9 years, how much annuity can a person get annually from the bank for 8 years starting 1 year after the 9th deposit is made. Cost of money is 14% Ans. A = 34, 675 b. A debt of 40,000 whose interest rate is 15% compounded semi-annually, is to be discharged by a series of 10 semi-annual payment, the first payment is to be made 6 months after consummation of the loan. The first 6 payemnts will be 6,000 each while the remaining 4 payments will be equal and of such amount that the final payment will liquidate the debt. What is the amount of the last 4 payments? Ans. A = 5454

PERPETUITY An annuity in which the payments continue indefinitely

Ex. What amount of money invested today at 15% interest can provide the following scholarships; 30,000 at the end of each year for 6 years; 40,000 for the next 6 years and 50,000 thereafter? Ans. 241,277. CAPITALIZED COST  One of the important applications of perpetuity  The sum of the first cost and the present worth of all costs of replacement, operation and maintenance for a long time of forever. Case 1: no replacement, only maintenance and or operation every period Capitalized cost = first cost + present worth of perpetual operation and or maintenance Ex. Determine the capitalized cost of a structure that requires an initial investment of 1,5000,000 and an annual maintenance of 150,000. Interest is 15% Ans. 2,500,000

Case 2: replacement only, no maintenance and or operation Capitalized cost = first cost + present worth of perpetual replacement Ex. Determine the capitalized cost of each research laboratory which requires 5M for original construction; 100,000 at the end of each year for the first 6 years and then 120,000 each year thereafter for operating expenses, and 500,000 every 5 years for replacement of equipment with interest at 12% per annum? Ans; 6,753,650

AMORTIZATION  Any method of repaying debt, the principal and the interest included, usually by a series of equal payments at equal interval of time. Ex. A debt of 5,000 with interest of 12% compounded semi-annually is to be amortized by equal semi-annual payments over the next 3 years, the first due in 6 months. Find the semi-annual payment and construct an amortization schedule. Ans. A = 1,016.82

UNIFORM ARITHMETIC GRADIENT  Maintenance and repair expenses on specific equipment of property may increase by a relatively constant amount each period. P = Pa + Pg Ex. A loan was to be amortized by a group of four end of year payments forming an ascending arithmetic progression. The initial payment was to be 5,000 and the difference between successive payments was to be 400. But the loan was

renegotiated to provide for the payment of equal rather tha uniformly varying sums. If the interest rate of the loan was 15%, what was the annual payment? Ans. 5530.51