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Problems of

Engineering

EconomyPrepared by:Engr. Glenn Ortiz

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LECTURE 2: Interest and the Time Value of Money

Elements of Transactions Involving Interest

1. Many types of transactions involve interest-e.g.,borrowing money, investing money, or purchasingmachinery on credit-but certain elements are commonto all of these types of transactions:

2. The initial amount of money invested or borrowed intransactions is called the principal (P).

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3.The interest rate (i)measures the cost or price of moneyand is expressed as a percentage per period of time. It is theamount paid for the use of borrowed capital or it is the

money earned by an investment.4. A period of time called the interest period(n)determines how frequently interest is calculated. (Notethat, even though the length of time of an interest period can

vary, interest rates are frequently quoted in terms of an

annual percentage rate.5. A specified length of time marks the duration of thetransaction and thereby establishes a certain number ofinterest periods (N).

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6. A plan for receipts or disbursements (A) that yields aparticular cash flow pattern over a specified length of time. (Forexample, we might have a series of equal monthly payments thatrepay a loan.)

7. A future amount of money (F)results from the cumulativeeffects of the interest rate over a number of interest periods.

8. Problems involving the time value of money can be convenientlyrepresented in graphic form with a cash flow diagram. Cash flow

diagrams represent time by a horizontal line marked off with thenumber of interest periods specified. Arrows represent the cashflows over time at relevant periods:

9. Upward arrows represent positive flows (receipts), anddownward arrows represent negative flows (disbursements).

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Types of Interest

Simple interest-the type of interest that follows theprinciple only the principal earned interest.

Simple InterestI = Pin

Future worth or accumulated amount

F= P + I

F = P (1 + in)where:

i= interest rate in decimal

n= number of years

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Types of Simple Interest

Ordinary simple interest is computed based on 1banker year = 360 days

Exact simple interest- is computed based on exactnumber of days in one year as follows:

1 ordinary year (not divisible by 4) = 365 days

1 leap year (divisible by 4) = 366 days

1 century years = 100 years

If the century year is divisible by 400 it is a leap year,otherwise it is an ordinary year

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If the kind of simple interest is not mentioned in theproblem, assume it is an ordinary simple interest.

If exact simple interest, 30 days has September, April,June and November. February has 28 or 29 days, the rest31 days.

Compound Interestis type of interest that follows the

principle interest on top of an interest i.e. bothprincipal and interest earn interest.

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Continuous Compounding assumes that compounding

continuously,

Nominal rate of interest (r)is the interest rate whereconversion is allowed.

Number of Compounding (m)is the number ofpayments in one year.

Effective rate of interest (e)is the actual rate of interestin one year. It is the interest rate for any compounding that

will give the same accumulation as computed compoundedannually.

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Problem #1:

If P1000 accumulates to P1500when invested at a simple interestfor three years, what is the rate of

interest?

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Problem #2:A loan of P5000 is made for a

period of 15 months, at a simpleinterest rate of 15%, what futureamount is due at the end of the

loan period?

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Problem #3:If you borrowed money from your

friend with simple interest of 12%,find the present worth of P50000,which is due at the end of 7

months.

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Problem #4:Mr. J. Reyes borrowed money from

the bank. He received from thebank P1842 and promised to repayP2000 at the end of 10 months.

Determine the rate of simpleinterest.

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Problem #5:A man borrowed money from a

loan shark. He receives from theloan sharkand amount of P1342.00and promised to repay P1500.00 at

the end of 3 quarters. What is thesimple interest rate?

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Problem #6:Determine the exact simple interest

on P5000 invested for the periodfrom January 15,1996 to October12,1996, if the rate of interest is

18%.

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Problem #7:

The exact simple interest of P5000invested from June 21,1995 todecember 25,1995 is P100. What is

the rate of interest?

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Problem #8:

Nicole has P20,400 in cash. She

invested it at 7% from March 1,2006to November 1,2006 at 7% interest.

How much is the interest using theBankers rule(360 days)?

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Problem #9:The amount of P20000 was

deposited in a bank earning aninterest of 6.5% per annum.Determine the total amount at the

end of 7 years if the principal andinterest were not withdrawn duringthis period.

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Problem #10:

A loan for P50,000 is to be paid in 3years at the amount of P65,000.What is the effective rate of

money?

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Problem #11:

Find the present worth of a futurepayment of P80,000 to be made insix years with an interest of 12%

compounded annually.

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Problem #12:

What is the effective ratecorresponding to 18% compoundeddaily? Take 1 year is equal to 360

days.

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Problem #13:

What nominal rate, compoundedsemi annually, yields the sameamount as 16% compounded

quarterly?

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Problem #14:

What rate of interest compounded

annully is the same as the rate ofinterest of 8% compounded

quarterly?

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Problem #15:Find the compound amount if

P2500 is invested at 8%compounded quarterly for 5yearsand 6 months.

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Problem #16:About how many years will

P100,000 earn a compound interestof P50,000 if the interest rate is 9%compounded quarterly?

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Problem #17:

Compute the equivalent rate of 6%compounded semi-annually to arate compounded quarterly.

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Problem #18:P20,0000 was deposited on January

1,1988 at an interest rate of 24%compounded semi annually. Howmuch would the sum be on January

1,1993?

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Problem #19:

What is the present worth P100payments at the end of the thirdyear and fourth year? The annual

interest rate is 8%.

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Problem #20:A firm borrows P2000 for 6 years at

8% . At the end of 6 years, it renewsthe loan for the amount due plusP2000 more for 2 years at 8%.What

is the total future worth? What isthe lump sum due?

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Problem #21:Find the present value of installment

payments of P1,000 now, P2,000 at the

end of the first year, P3,000 at the endof the second year, P4,000 at the end ofthe third year and P5,000 at the end of

the fourth year, if money is worth 10%compounded annually.

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Problem #22:If money is worth 5%

compounded quarterly, find theequated time for paying a loanof P 150,000 due in 1 year and

P280,000 due in 2 years.

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Problem #23:BPI advertises 9.5% account thatyields 9.84% annually. Find howoften the interest is compounded.