CHAPTER IIINTEREST AND MONEY-TIME RELATIONSHIPS INTEREST - the amount of money paid for the use of borrowed capital or the income produced by the money which has been loaned SIMPLE INTEREST - calculated using the principal only, ignoring any interest that have been accrued in preceding periods - paid on a short-term loans in which the time of the loan is measured in day I Pni where: I= interest P=principal or present worth n=number of interest period F P (1 ni ) i= rate of interest per interest period F= accumulated amount or future worth Determine the exact simple interest on P500 for the period from Jan. 10 to Oct. 28,1996 at 16% interest. Solution: Jan – 21 I = Pni Feb – 29 Mar – 31 Apr – 30 = P500(292/366)(0.16) May – 31 = P63.83 Jun – 30 Ans Jul – 31 Aug – 31 Sep – 30 Oct – 28 TOTAL: 292 days SAMPLE PROBLEM a. Ordinary Simple Interest –computed on the basis of 12 months of 30 days a year 1 interest period= 360 days b. Exact Simple Interest –based on the exact number of days in a year, 365 days for an ordinary year and 366 days for a leap year 1 interest period= 365 or 366 days SIMPLE INTEREST Determine the ordinary simple interest on P700 for 8 months and 15 days if the rate of interest is 15%. Solution: n= (8)(30)+15= 255 days I = Pni =P700(255/360)(0.15) = P74.38 Ans SAMPLE PROBLEM Cash Flow Diagram - simply a graphical representation of cash flows drawn on a time scale Receipt (positive cash flow or cash inflow) Disbursement (negative cash flow or cash outflow) A loan of P100 at simple interest of 10% will become P150 after 5 years. 150 0 1 2 3 4 5 100 Cash flow diagram on the viewpoint of the lender 100 0 1 2 3 4 5 150 Cash flow diagram on the viewpoint of the borrower Cash Flow Diagram COMPOUND INTEREST - In calculations of compound interest, the interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous periods. - means “interest on top of interest” COMPOUNDING FREQUENCIES AND PERIODS FREQUENCY NO. PER YEAR PERIOD Annually 1 1year Semi-annually 2 6months Quarterly 4 3months Monthly 12 1month Daily 365 1day Interest Principal at Interest Amount at End of Period the Earned Period Beginning of During Period Period 1 P Pi P+Pi=P(1+ni) 2 P(1+i) Pi(1+i) P(1+i)+Pi(1+i) = P(1+i)^2 3 P(1+i)^2 Pi(1+i)^2 P(1+i)^2+Pi(1+i)^2 =P(1+i)^3 ... ... ... ... n P(1+i)^(n-1) Pi(1+i)^(n-1) P(1+i)^n COMPOUND INTEREST - The quantity is commonly called the “single payment compound amount factor” and is designated by functional symbol F/P, i%, n. - this symbol is read as “F given P at i per cent in n interest periods” - The quantity is commonly called the “single payment present worth factor” and is designated by functional symbol P/F, i%, n. - this symbol is read as “P given F at i per cent in n interest periods” COMPOUND INTEREST Rate of Interest a. Nominal rate of interest - specifies the rate of interest and a number of interest periods in one year where: i=rate of interest per interest period r=nominal interest rate m=number of compounding periods per year b. Effective rate of interest – the actual or exact rate of interest on the principal during one year. It is equal to the nominal rate if the interest is compounded annually, but greater than the nominal rate if the number of interest periods per year exceeds one (i.e. Quarterly, monthly, etc.) COMPOUND INTEREST Find the nominal rate if converted quarterly could be used instead of 12% compounded monthly. What is the corresponding effective rate? ANSWER: 12.12% compounded quarterly SAMPLE PROBLEM A P2000 loan was originally made at 8% simple interest for 4 years. At the end of this period the loan was extended for 3 years, without the interest being paid, but the new interest rate was made 10% compounded semi—annually. How much should the borrower pay at the end of 7 years? ANSWER: P3,537.86 SAMPLE PROBLEM Find the amount at the end of two years and seven months if P1000 is invested at 8% compounded quarterly using simple interest for anytime less than a year interest period. ANSWER: P1,226.834 SAMPLE PROBLEM EQUATION OF VALUE - is obtained by setting the sum of the values on a certain comparison or focal date of one set of obligations equal to the sum of the values on the same date of another set of obligations A man bought a lot worth P1,000,000 if paid in cash. On the instalment basis, he paid a down payment of P200,000; P300,000 at the end of one year; P400,000 at the end of three years and a final payment at the end of five years. What was the final payment if the interest was 20%? ANSWER: P792,576 SAMPLE PROBLEM Over the past 10 years, Gentrack has placed varying sums of money into a special capital accumulation fund. The company sells compost produced by garbage-to-compost plants in the United States & Vietnam. At the start, $1000 was placed, $3000 at the 4th year and $1500 on its 6th year. Find the amount in the account now (after 10 years) at an interest rate of 12% per year, compounded semiannually using [a] semiannual CP and [b] effective annual rate. ANSWER: $11,634 SAMPLE PROBLEM CONTINUOUS COMPOUNDING AND DISCRETE PAYMENTS - In discrete compounding, the interest is compounded at the end of each finite- length period, such as a month, a quarter or a year - In continuous compounding, it is assumed that cash payments occur once a year, but the compounding is continuous throughout the year Compare the accumulated amounts after 5 years of PhP1,000 invested at the rate of 10% per year compounded a. Annually b. Semi-annually c. Quarterly d. Monthly e. Continuously ANSWER: a. P1,610.51, b. P1,628.89, c. P1,638.62, d. P1,645.31, e. P1, 648.72 Sample Problem DISCOUNT -the difference between the present worth and the worth at some time in the future -interest paid in advance • Rate of discount – the discount on one unit of principal for one unit of time where: d= rate of discount for the period involved i= rate of interest for the same period A man borrowed P5000 from a bank and agreed to pay the loan at the end of 9 months. The ban discounted the loan and gave him P4000 in cash. a)What was the rate of discount? b)What was the rate of interest? c)What was the rate of interest for one year? ANSWER: a. 20%, b. 25%, c. 33.33% SAMPLE PROBLEM INFLATION - the increase in the prices for goods and services from one year to another , thus decreasing the purchasing power of money where: PC= present cost of the commodity FC= future cost of the same commodity f= annual inflation rate n= number of years -In an inflationary economy, the buying power of money decreases as cost increases. where: P= present worth F= future worth f= annual inflation rate n= number of years -If interest is being compounded at the same time that inflation is occurring, the future worth will be INFLATION An economy is experiencing inflation at an annual rate of 8%. If this continues, what will P1000 be worth two years from now in terms of today’s pesos? ANSWER: P857.34 SAMPLE PROBLEM A man invested P10000 at an interest rate of 10% compounded annually. What will be the final amount of his investment, in terms of today’s pesos, after five years, if inflation remains the same at the rate of 8% per year? ANSWER: P10,960.86 SAMPLE PROBLEM ANNUITIES - a series of equal payments occurring at equal periods of time. - annuities occur in the following instances: - payment of a 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. - ANNUITIES - a series of equal payments occurring at equal periods of time - • Ordinary Annuity - one where the payments is made at the end of each period where: P= value or sum of money at present F= value or sum of money at some future time A= a series of periodic, equal amounts of money n= number of interest period i= interest rate per interest period • Finding P when A is given P 0 1 2 3 n-1 n A A A A A A(P/F,i%,1) A(P/F,i%,2) A(P/F,i%,3) A(P/F,i%,n-1) A(P/F,i%,n) Cash flow diagram to find P given A ANNUITIES • Finding F when A is given F 0 1 2 3 n-1 n A A A A A A(F/P,i%,1) A(F/P,i%,n-3) A(F/P,i%,n-2) A(F/P,i%,n-1) Cash flow diagram to find F given A ANNUITIES - The quantity is called the “uniform series present worth factor” and designated by the functional symbol P/A, i %, n. - The quantity is commonly called the “uniform series compound amount factor” and designated by the functional symbol F/A, i%, n. - The quantity is commonly called the “capital recovery factor” and denoted by the functional symbol A/P, i%, n. - The quantity is commonly called the “sinking fund factor” and denoted by the functional symbol A/F, i%, n. ANNUITIES What are the present worth and the accumulated amount of a 10-year annuity paying P10,000 at the end of each year, with interest at 15% compounded annually? ANSWER: P = P50,188, F = P203,037 SAMPLE PROBLEM What is the present worth of P500 deposited at the end at every 3 months for 6 years if the interest rate is 12% compounded semi-annually? ANSWER: P8,504 SAMPLE PROBLEM DEFERRED ANNUITY • Deferred Annuity – one where the payment of the first amount is deferred a certain number of periods after the first. If P10,000 is deposited each year for nine years, how much annuity can a person get annually from the bank every year for 8 years starting one year after the 9th deposit is made. Cost of money is 14%. ANSWER: A = 34,675 SAMPLE PROBLEM A debt of P40000 , whose interest rate is 15% compounded semi-annually, is to be discharge by a series of 10 semi-annual payments, the first payment to be made six months after consummation of the loan. The first 6 payments will be P6000 each, while the other four 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? ANSWER: P5,454 SAMPLE PROBLEM ANNUITY DUE - one where the payments are made at the beginning of each period P • Finding P when A is given 0 1 2 3 n-1 n A A A A A • Finding F when A is given F 0 1 2 3 n-1 n A A A A A ANNUITY DUE A certain property is being sold and the owner received two bids. The first bidder offered to pay P400000each year for five years, each payment is to be made at the beginning of each year. The second bidder offered to pay P240000 first year, P360000 the second year and P540000 each year for the next three years, all payments will be made at the beginning of each year. If money is worth 20% compounded annually, which bid should the owner of the property accept? SAMPLE PROBLEM PERPETUITY - an annuity in which the payments continue indefinitely P • Finding P when A is given 0 1 2 3 n A A A A What amount of money invested today at 115% interest can provide the following scholarship: P30000 at the end of each year for six years; P40000 for the next six years ad P50000 thereafter. SAMPLE PROBLEM CAPITAL COST - one of the most important application of perpetuity - the sum of the first cost and the present worth of all costs of replacement, operation and maintenance for a long time or forever PERPETUITY AMORTIZATION - Any method of repaying a debt, the principal and interest included, usually by a series of equal payments at equal interval of time A debt of P5000 with interest AT 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. SAMPLE PROBLEM Uniform Arithmetic Gradient - In a certain cases, economic analysis problems involve receipts or disbursements that increase or decrease by a uniform amount each period. This is known as the uniform arithmetic gradient. (n-1)(G) 3G 2G A A A A A 1G 0 1 2 3 4 5 0 1 2 3 4 5 P P A loan was amortized by a group of four end-of- year payments forming an ascending arithmetic progression. The initial payment was to be P5000 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 the of the loan was 15%, what was the annual payment? SAMPLE PROBLEM END OF CHAPTER 2 THANK YOU AND GOD BLESS YOU