Chapter-2 Interest-and-Money-Time-Relationships PDF

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BES 004: Engin Engine eerin eringg EEcono cono conomy my

Topic 2: Intere nterest-M st-M st-Mone one oney-Tim y-Tim y-Time eR Relati elati elationship onship onshipss

IN INTEREST TEREST A AND ND M MONEY ONEY – TIM TIME E RELA RELATION TION TIONSSHIP HIPSS Inter Interest est – is the amount of money paid for the use of borrowed capital. For the lender, interest is the income produced by the money which he lent. It is therefore clear that interest must be added to the amount borrowed when full payment of the debt is made.

Sim Simple ple int interest erest – the interest to be paid which is proportional to the length of time the principal sum has been borrowed. The interest earned maybe found in the following manner. Let I represent the interest earned, P the principal amount, n the interest period, and i the interest rate. Then,

I = Pin The total amount F to be repaid is equal to the sum of the principal and the total interest and is given by the formula.

F=P+I = P + Pin = P ( 1 + in ) (a) Ordinary simple interest is computed on the basis of one banker’s year which is

1 banker’s year = 12 months, each consisting of 30 days = 360 days (b) Exact simple interest is based on the exact number of days, 365 days for an ordinary year. The leap years are those which are exactly divisible by 4, but excluding the century years such as the years 1900, 2100, etc.

If d is the number of days in the interest period, then d Ordinary simple interest = Pi360 d  ( for ordinary year ) Pi 365 Exact simple interest =  d Pi ( for leap year )  366 Sa Sam mple PPrroblems oblems:: 1. Determine the ordinary simple interest on P10,000 for 9 months and 10 days if the rate of interest is 12%. 2. Determine the ordinary and exact simple interest on P 5,000 for the period from January 15 to June 20, 1978, if the rate of simple interest is 14%. 3. Annie buys a television set from a merchant who ask P 1250 at the end of 60 days ( cash in 60 days). Annie wishes to pay immediately and the merchant offers to compute the cash price on the assumption that money is worth 8% simple interest. What is the cash price today? 4. A man borrows P 10,000 from a loan firm. The rate of simple interest is 15%, but the interest is to be deducted from the loan at the time the money is borrowed. At the end of one year he has to pay back P 10,000. What is the actual rate of interest? Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

Cas Cash-Fl h-Fl h-Flow ow Di Diagra agra agrams ms A cash-flow diagram is simply a graphical representation of cash flows drawn on a time scale. Cash-flow diagram for economic analysis problems is analogous to that of free body diagram for mechanics problems. receipt ( positive cash flow or cash inflow ) disbursement ( negative cash flow or cash outflow ) A loan of P200 at simple interest of 10% will become P320 after 6 years. P 320 0 1

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P 200 Cash flow ddiagra iagra iagram m on th thee vi viewpoi ewpoi ewpoint nt of tth he len lender der P 200 6 0

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5 P 320

Cash flow ddiag iag iagram ram on th thee vi viewpoi ewpoi ewpoint nt of tth he bo borrow rrow rrower er

Com Compoun poun pound d Inte Interest rest In compound interest, the interest earned by the principal is not paid at the end of each interest period but is considered as added to the principal, and therefore will also earn interest for the succeeding periods. The interest earned by the principal when invested at compound interest is much more than that earned by the same principal when invested at simple interest for the same number of periods. Using the same nomenclature as that for simple interest, the total amount due after n interest periods for compound interest is given by formula

F  P1  in , which is derived in accordance with the following manner: Period No.1 Principal at beginning = P Interest earned

= Pi

Principal at end

= P + Pi F1 = P ( 1 + i )

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

Period No.2 Principal at beginning = P + Pi Interest earned

= P ( 1 + i )i

Principal at end

= P + Pi + P ( 1 + i )i = P ( 1 + i )( 1 + i ) F2 = P ( 1 + i )2

Period No.3 Principal at beginning = P ( 1 + i )2 = P ( 1 + i )2i

Interest earned

= P ( 1 + i )2 + P ( 1 + i )2i

Principal at end

= P ( 1 + i )2 ( 1 + i ) F3 = P ( 1 + i )3

In general, therefore,

F = P ( 1 + i )n

The factor ( 1 + i )n is called the “ single-payment compound-amount factor “ and is designated by ( F/P, i% n). Thus F = P ( F/P, i%, n)

Pres Valu Present ent Valu lue e The principal P in the formula F = P ( 1 + i )n may be considered as the value of the compound amount F at present or it is the amount which, when invested at present, will become F after n interest periods. P is called the present value of the amount F, and is given by the formula,

F P = F ( 1 + i )-n = (1+i)n The factor ( 1 + i )-n =

1 (1+i)n is called the “ Single-Payment Present-Worth Factor “ and is

designated (P/F, i%, n). Thus

( 1 + i )-n =

1 (1+i)n = ( P/F, i%, n )

and formula becomes F P = F ( 1 + i )-n = (1+i)n = F ( P/F, i%, n )

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

Rat Rate eo off in interest terest (a) Nominal rate of interest is expressed on an annual basis and is determined by multiplying the actual or effective interest rate per interest period by the number of compounding periods per year. r i= m where: i = rate of interest per interest period r = nominal interest rate m = number of compounding periods per year (b) Effective rate of interest is the actual rate of interest on the principal for one year. It is equal to nominal rate of interest if the interest is compounded annually, but it is bigger than the nominal rate of interest if the interest is compounded semi-annually, quarterly or monthly. Example: P 1.00 is the principal which is invested at a nominal rate of 8% semi-annually. After one year this will become F = P1 ( 1 + i )m 8 i = = 4% 2 m = 1(2) = 2 F = P 1 ( 1 + 0.04 )2 = P 1.0816 The actual interest earned is P 0.0816, this corresponds to an effective rate of interest of 8.16% which is bigger than the nominal interest of 8%. Hence, Effective rate of interest = F – 1 = ( 1 + i )m - 1 m

r   = 1   - 1  m Sam Sample ple PProblem roblem roblemss 1. Suppose that you borrow P 8,000 now, promising to repay the loan principal plus accumulated interest in four years at i= 10% per year. How much would you repay at the end of four year? 2. An investor (owner) has an option to purchase a tract of land that will be worth P 10,000 in six years. If the value of the land increases at 8% each year, how much should the investor be willing to pay now for this property? 3. If P 1,000 becomes P 1,811.36 after 5 years when invested at an unknown rate of interest compounded bimonthly (every two months), determine the unknown nominal rate and the corresponding effective rate. 4. Find the nominal rate compounded monthly which is equivalent to 12% compounded quarterly. What is the corresponding effective rate? 5. The present worth of several future cash payments maybe defined as the sum of the values of the future payments discounted at a given rate for the corresponding periods to the present. Find the present value of installment payments P 1,000 now; P 2,000 at the end of first year; P 3,000 at the end of second year; P 4,000 at the end of the third year and P 5,000 at the end of the fourth year; if money is worth 10% compounded annually.

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

Equ Equaati tion on o off Val Value ue An 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. Sample Problems: 1. A company expects to retire an existing machine at the end of 1982 and will replace it with a new machine for the same task at an estimated cost of P 60,000. The old machine is expected to be sold for P 5,000 when it is replaced. To provide for replacement, the company intends to deposit the following amounts in an account earning interest at 8% compounded quarterly: P 20,000 at the end of 1979, P 15,000 at the end of 1980, P 10,000 at the end of 1981. What additional amount is needed at the end of 1982 to purchase the new machine? 2. A owes B the following amounts: P 4,000 due 3 years hence, P 5,600 due 4 years hence, P 3,800 due 5 years hence. The agreed rate of interest is 12% compounded monthly. A has just won a major prize in a lottery and decides to liquidate these debts now. How much should B be willing to accept in full payment? Conti Continuo nuo nuous us Com Compou pou poundin ndin ndingg an and d Di Discret scret screte e PPayme ayme ayments nts In discrete compounding, the interest is compounded at the end of each finite-length period, such as month, a quarter or a year. In continuous compounding, it is assumed that cash payments occur once per year, but the compounding is continuous throughout the year. F 0 1

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3 n years

mn

P Continu Continuous ous Co Compou mpou mpounding nding ( Lend Lender’s er’s Vie iewpo wpo wpoint int ) r = nominal rate of interest per year r m = rate of interest per period m = number of interest periods per year mn = number of interest periods in n years

r   F = P 1    m

mn

m Let r = k, then m = rk, as m increases so must k

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

r   1    m  

The limit of 1 

mn

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

 1 = 1    k

rkn  

1 =  1    k 

k

rn

 

k

r   as k approaches infinite is e m

 1  k   1     k  

rn

= ern

Thus, F = Pern P = Fe-rn Disc Discount ount Discount on a negotiable paper is the difference between what it is worth in the future and its present worth. Thus Discount = Future value – Present value For example, if a negotiable paper such a bond can be sold for P100, six months from now, but it sold for P95 at present, then the discount is P5. The rate of discount is the discount on one unit of principal per unit of time. If d is the rate of discount, then 1 d = 1 - 1 + i = 1 – (P/F, i%, 1 ) i d = 1 + i = ( P/F, i%, 1 )i For the equivalent rate of interest corresponding to a rate of interest I, w have d d i = 1 - d = (P/F, i%, 1)

Sam Sample ple PProblem roblem roblemss

1. Compare the accumulated values at the of 10 years if P 100 is invested at the rate of 12% per year compounded annually, semi-annually, quarterly, monthly, daily and continuously. 2. Mr. J. de la Cruz borrowed money from a bank. He received from the bank P 1,340 and promised to pay P 1,500 at the end of 9 months. Determine the following: a. Simple interest rate b. The corresponding discount rate or often referred to as the “ Bankers discount”. 3. At 12% discount, find the present worth of a. P 1,000 due in 1 year from today b. P 1,200 due in 6 months

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

ANNU ANNUITIES ITIES Ann Annuity uity – a series of equal payments made at equal intervals of time. TYPE TYPESS OF AN ANNUIT NUIT NUITYY 1. Ordin Ordinary ary aannu nnu nnuity ity – one where equal payments are made at the end of each payment period starting from the first period. 0

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A

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P

 1 i n  1 P = A  n  1  i i  The quantity brackets is called the “uniform series present worth factor” and is designated by the functional symbol P/A, i%, n , read as “ P given A at i percent in n interest periods.”The equation can be expressed as P = A(P/A, i%, n) 0

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A F

 1  i n  1 F=A  i   The quantity brackets is called the “uniform series compound amount factor” and is designated by the functional symbol F/A, i%, n, read as “ F given A at i percent in n interest periods.”The equation can be expressed as F = A(F/A, i%, n) Sa Sam mple PPrroblems 1.1 For having been loyal, trusthworthy and efficient, the company has offered a supervisor a yearly gratuity pay of P20,000.00 for 10 years with the first payment to be made one year after his retirement. The supervisor, instead, requested that he be paid a lump sum on the date of his retirement less interest that the company would have earned if the gratuity is to be paid on yearly basis. If interest is 15%, what is the equivalent lump sum that he could get?

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

1.2 Rianer Wandrew borrowed P 50,000 from a Social Security System, in the form of calamity loan, with interest at 8%, compounded quarterly, payable in equal quarterly installments for 10 years. Find the quarterly payments. 1.3 In anticipation of a much bigger volume of business after 10 years, a fabrication company purchased an adjacent lot for its expansion program where it hopes to put up a building projected to cost P4,000,000.00 when it will be constructed 10 years after. To provided for the required capital expenses, it plans to put up a sinking fund for the purpose.How much must the company deposit each year if interest to be earned is computed at 15%. 2. Def Deferr err erred ed annuit annuityy – it is also an ordinary annuity but the payment of the first amount is deferred a certain number periods after the first. m periods

0

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0 m

A(P/A,i%,n)(P/F,ib%,m )

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A(P/A,i%,n)

P = A(P/A, i%, n)(P/F,i%,m)

 1 i n  1  P = A  1  i  m n  1  i i  Sa Sam mple PPrroblems 2.1 A new generator has just been installed. It is expected that there will be no maintenance charges until the end of the sixth year, when P300 will be spent on the generator and P300 will be spent at the end of each successive years until the generator is scrapped at the end of its fourteenth year of service. What sum of money set aside at the time of installation of the generator at 6% interest will take care of all maintenance expenses? 2.2 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? 2.3 The purchaser of a tractor paid P10,000 cash and agreed to pay P3000 at the end of 6 months for 10 years. He failed to make the first 5 payments of P3000 each. At the end of 3 years he desires to pay the tractor by a single payment which will cancel both his accumulated liabilities and his future liabilities. What must he pay if money is worth 6% per annum compounded semi-annually? 2.4 A man wishes to provide a fund for his retirement such that from his 60th to 70th birthdays he will be able to withdraw equal sums of P18,000 each for his yearly expenses. He invests equal amounts from his 41st to 59th birthdays in a fund earning 10% compounded annually. How much should each of these amounts be?

Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

3. Ann Annuity uity du due e – is one where payments are made at the start of each period, beginning from the first period.

P

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n A

 1  i n  1  P = A + A  n  1  i  i   1 i n  1 F = A 1  i n + A   i   Sa Sam mple PPrroblems 3.1 A farmer bought a tractor costing P25,000 payable in 10 semi-annual payments, each installment at the beginning of each period. If the rate of interest is 26% compounded semiannually, determine the amount of each installment. 3.2 A farmer bought a tractor costing P12,000 if paid in cash. The tractor may also be purchased by installment to be paid within 5 years. Money is worth 8% compounded annually. Determine the amount of each annual payment if all payments are made a. at the end of each of the 5 years b. at the beginning of each of the 5 years. 3.3 a. Juan bought a car with a cash value of P14,000 on the installment plan under the following terms: P4,000 cash upon delivery and the balance payable in 12 equal monthly payments, each payment combining an amortization installment and 6% on the previously unpaid balance.Solve for his monthly payment. b. At the end of 8 months, he was forced to sell the car in order to pay a P7,000 debt. At what price must he sell his car so that he can completely pay the car and also pay his P7,000 debt? 4. Perp Perpe etuity – is one where the payments periods extend forever or in which periodic payments continue indefinitely.

P

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1  1  i    P = A(P/A, i%, n) = A   i   P= P A

Where: P is called the capitalized value or cost of A. Prepared by: Diann Dianne e Mae M. Mag Magomn omn omnang ang

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BES 004: Engin Engineerin eerin eeringg EEcono cono conomy my

Topic 2: Intere nterest-Mo st-Mo st-Money-Ti ney-Ti ney-Time me R Relati elati elationship onship onshipss

Sa Sam mple PPrroblems 1. An endowment fund is to provide an annual scholarship of P4,000 for the first 5 years; P6,...