Ce40 - Co1 (3q1718).pdf

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INTEREST - is the amount of money paid for the use of borrowed capital or the income produced by money which has been loaned

INTEREST

Engr. Charity Hope Gayatin

1. SIMPLE INTEREST- calculated using the principal only, ignoring any interest that had been accrued in preceding periods I = Pin I – interest P – present worth n – number of interest period i – rate of interest per interest period

PROBLEMS 1. Find the missing value P = P20,500 r = 4 1/2% n = 3yrs P = P19,350.75 I = P4,500 n = 2yrs P = P5,250 I = P775 r = 9 3/4% I = P5,630 r = 8 1/4% n = 4yrs & 6mos

2. Compute for the simple interest and the amount on a P15,000 salary loan at 12% simple interest for 3years.

4. Mr. ABCD obtained a loan of P45,000 from Pagibig at 15% interest rate. When was the loan due if he paid P68,000 on the maturity date?

3. If Mrs. ABCD paid an interest of P4,259 for a bank loan payable in 2 and a half years at 15% per annum, how much was the original loan?

5. Ms. ABCD borrowed P80,000 in a cooperative. She paid an interest of P2,500 for 11months. At what rate is the interest charged?

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TIME BETWEEN 2 DATES EXACT TIME- is determined by counting each day of each month of the term except the origin date.

PROBLEMS 1.

Find the approximate number of days between

a)

February 14 to May 11, 2009

BANKER’S RULE- uses the exact time / 360

b)

October 1 to Dec 25, 2008

2.

Find the actual number of days between

a) ORDINARY SIMPLE INTEREST- computed on the basis of 12 months of 30 days each or 360 days a year

a)

February 14 to May 11, 2009

b)

October 1 to Dec 25, 2008

APPROXIMATE TIME- is determined by assuming each month to have 30days and then counting each day of each month except the origin date.

b) EXACT SIMPLE INTEREST- based on the exact number of days in a year, 365 days for ordinary year and 366 days for leap year

ACCUMULATION AT SIMPLE INTEREST PROBLEMS 1. Determine the ordinary simple interest on P7,000 for 8 months and 15 days if the rate of interest 15%. 2. Determine the exact simple interest of P15,000 for the period of January 10 to October 28, 2012 at 16% interest.

The Principal, P accumulates to the value F, Future Amount at the end of n years. P refers to the discounted value of F on the day which is n years before the due date of F. The current value of a future amount is the present value of a loan or investment.

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F  P  I  P (1  in)

P 

F (1  in )

PROBLEMS 1. If the money is worth 14% simple interest which is due in 6years, what is the amount of a P250,000 loan? 2. Discount P45,310 at 11.5% simple interest for 175days.

P– present worth i – interest F– future worth

3. Accumulate P70,000 at 13.5% simple interest for 2years and 10months.

5. Ms. ABCD borrowed P20,500 from Mr. EFGH at 20% simple interest. She promised to pay the principal and the corresponding interest on October 13, 2009. If the loan was made last January 2, 2008, how much is a)

4. What sum will accumulate to P72,000 in 8years at 12.45% simple interest?

the interest on the loan?

b) the amount October 13, 2009?

Mr.

EFGH

would

receive

on

Introduction  Re-investing

Compound Interest Engr. Charity Hope Gayatin

your interest income from an investment makes your money grow faster over time! This is what compound interest does.  Compound interest uses the same information as simple interest, but what is new is the frequency of compounding n.  n=1 annual, n=2 semi-annual, n=4 quarterly, n=12 monthly, n=52 weekly, n=365 daily.

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COMPOUND INTEREST - is the interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous periods

NOMINAL RATE OF INTEREST - specifies the rate of interest and a number of interest periods in one year i=r/m

Single payment at i present interest compound amount in n interest period factor n F  P(1  i ) , F / P, i %, n

i r m per year

Single payment present worth n P  F (1  i ) , P / F, i %, n

EFFECTIVE RATE OF INTEREST - exact rate of interest on the principal during one year

rate of interest per interest period nominal interest rate number of compounding periods

Compound Interest Formula  If

( 1 + r / m) m – 1

P represents the present value, r the annual interest rate, n the time in years, and m the frequency of compounding, then the future value is given by the formula: F = P( 1 + r/m)mn

Example  Suppose

you invest $32,000 into a certificate of deposit that has an annual interest rate of 5.2% compounded annually for 3 years. Determine its accrued value.

Example  Suppose

you invest $32,000 into a certificate of deposit that has an annual interest rate of 5.2% compounded annually for 3 years. Determine its accrued value.  ANSWER: Use the compound interest formula.  F = 32000(1+.052/4)(4)(3) = 32000(1.013)12 = $37,364.86

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Annual Yield

Derivation of yield

 To

•

compare different savings plans, you need to have a common basis for making the comparisons.  The annual yield of a compound interest investment is the simple interest rate that has the same future value the compound rate would have in one year.

• • • • • •

Future Value Compound = Future value simple P(1 + r/m)mn = P(1 + in) Since this computation is done for 1 year, we set n = 1. P(1+ r/m)m = P(1 + i) Since P appears on both side, we divide by P and P disappears. (1 + r/m)m = 1 + i, now solve for i by subtracting 1 from both sides. The formula for yield is i = (1 + r/m)m – 1.

Example yield calculation

Example yield calculation

 Find

 Find

the annual yield for an investment that has an annual interest rate of 8.4% compounded monthly.

yield will usually be greater than the interest rate.  Note the interest rate is sometimes called the nominal interest rate.

the annual yield for an investment that has an annual interest rate of 8.4% compounded monthly.  ANSWER: y = (1 + .084/12)12 – 1  y = (1.007)12 – 1 = 0.087310661 = 8.73%  The yield will usually be greater than the interest rate.  Note the interest rate is sometimes called the nominal interest rate.

Continuous Compounded Interest

Example of Continuous Compound interest.

 The

 What

would happen if we let the frequency of compounding get very large. That is we would compound not just every hour, or every minute or every second but for every millisecond!  What happens is that the expression (1 +r/m)mn goes to ern. This e is the famous Euler number. It’s value is the irrational number 2.7182818 …  The future value formula is F = Pern.  The annual yield for continuously compounded interest is i = er – 1.

 Consider

the $32,000 from the earlier example. Now we will invest the money in an account that has 5.2% annual interest compounded continuously for 3 years. What is the future value? What is the yield for this investment?

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Example of Continuous Compound interest.  Consider

the $32,000 from the earlier example. Now we will invest the money in an account that has 5.2% annual interest compounded continuously for 3 years. What is the future value?  ANSWER: F = 32000e(.052)(3) = $37,402.44  Note this investment option is only greater by $37.58.  What is the yield for this investment?  ANSWER: y = e.052 – 1 = 0.05337 = 5.34%

Learning Unit Objectives

DISCOUNT

Engr. Charity Hope Gayatin

Structure of a Promissory Note

Structure of Promissory Notes; the Simple Discount Note 1. Calculate bank discount and proceeds for simple discount notes. 2. Calculate and compare the interest, maturity value, proceeds, and effective rate of a simple interest note with a simple discount note. Finding the Face Value, Rate, and Time for a Simple Interest Note and a Simple Discount Note 1. Find the face value (principal), rate, and time of a simple interest note. 2. Find the face value (maturity value), bank discount rate, and time of a simple discount note.

Simple Discount Note Terminology

Simple Discount Note

Simple Discount Note -- A note in which the loan interest is deducted in advance. Bank Discount -- The interest that banks deduct in advance. Maturity Value – The total amount due at the end of the loan (the sum of the face value, or principal, and interest). Proceeds -- The amount the borrower receives after the bank deducts its discount from the loan’s maturity value. Bank Discount Rate -- The percent of interest.

Pete Runnels has a choice of two different notes that both have a face value (principal) of $14,000 for 60 days. One note has a simple interest rate of 8%, while the other note has a simple discount rate of 8%. For each type of note, calculate a) interest owed, b) maturity value, c) proceeds, and d) effective rate:

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Simple Interest & Simple Discount Simple Interest

Simple Discount

Interest Owed

Interest Owed

I

= Pin D = Fdt = Principal * Interest Rate * Time = Maturity Value * Discount Rate * Time

F

=P+I = Principal + Interest

Maturity Value

Maturity Value F

= FV = Face Value

Proceeds P = FV = Face Value Effective Rate ER = =

I P∗n Interest Principal∗Time

EQUATION of VALUE

Proceeds

Engr. Charity Hope Gayatin

P =F −D = Maturity Value − Discount Effective Rate ER = =

D P∗n Interest Proceeds∗Time

Cash Flow

Cash Flow Example

Cash Flow is the sum of money recorded as receipts or disbursements in a project’s financial records.

A mechanical device will cost $20,000 when purchased. Maintenance will cost $1,000 per year. The device will generate revenues of $5,000 per year for 5 years. The salvage value is $7,000.

A cash flow diagram presents the flow of cash as arrows on a time line scaled to the magnitude of the cash flow, where expenses are down arrows and receipts are up arrows.

Economic Equivalence and Equation of Value Economic Equivalence is the process of comparing two different cash amounts at different point in time. Economic Equivalence exists between cash flows that have the same economic effect and could be traded for one another in the financial marketplace, which we assume to exist. Economic equivalence refers to the fact that a cash flow – whether a single payment or a series of payments – can be converted to an equivalent cash flow at any point in time. For example, we could find the equivalent future value F of a present amount P at interest I at period n; or we could determine the equivalent present value P of N equal payments A.

Discount Factors and Equivalence Present Worth (P) – Present Amount at t = 0 Future Worth (F) – Equivalent Future Amount at t = n of any present amount at t = 0 Annual Amount (A) – Uniform Amount that repeats at the end of each years for n years Uniform Gradient Amount (G) – Uniform Gradient Amount that repeats at the end of each year, starting at the end of the second year and stopping at the end of year n

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

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Discount Factors and Equivalence

Equivalence Calculations General Principles 1. Equivalence calculations made to compare alternatives require a common time basis – equivalent cash flows are equivalent at any common point in time 2. Equivalence depends on interest rate – changing the interest rate destroys equivalence 3. Equivalence calculations may require the conversion of multiple payment cash flows to a single cash flow – Equivalence calculations with multiple payments 4. Equivalence is maintained regardless of point of view

Equivalence Example A man bought a lot worth P1million if paid in cash. On installment basis, he paid a down payment of P200,000, P300,000 at the end of 1year, P400,000 at the end of 3years and a final payment at the end of 5years. What was the final payment if interest was 20%?

ANNUITIES

Engr. Charity Hope Gayatin

Learning Outcome

Annuities ANNUITY are series of equal payments occurring at equal periods of time

Derive and use factors for uniform series - present worth (P/A) and capital recovery factors (A/P)

ORDINARY ANNUITY are series of equal payments made at the end of each period

Derive and use factors for uniform series – compound amount (F/A) and sinking fund (A/F)

DEFERRED ANNUITY are series of equal payments where the first payment is made several periods after the beginning of the annuity PERPETUITY are series of equal payments in which the payments continue indefinitely P=A/i

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Uniform Series Present Worth (P/A) and Capital Recovery Factor (A/P)

Uniform Series Present Worth (P/A) and Capital Recovery Factor (A/P)

The Uniform Series Present Worth Factor (P/A) is used to calculate the equivalent P value in year 0 for a uniform end-of-period series of A values beginning at the end period 1 and extending for n periods. The Capital Recovery Factor (A/P) is used to calculate the equivalent uniform annual worth A over n years for a given P in year 0, when the interest rate is i.

Uniform Series Present Worth (P/A) and Capital Recovery Factor (A/P)

Sinking Fund Factor (A/F) and Uniform Series Compound Amount Factor (F/A) The Sinking Fund Factor (A/F) determines the uniform annual series A that is equivalent to a given future amount F. The uniform series A begins at the end of year (period) 1 and continues through the year of the given F. The last A value and F occur at the same time. The Uniform Series Compound Amount Factor (F/A) is used to yield the future worth of the uniform series. The future amount F occurs in the same period as the last A.

Sinking Fund Factor (A/F) and Uniform Series Compound Amount Factor (F/A)

Sinking Fund Factor (A/F) and Uniform Series Compound Amount Factor (F/A)

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Problem 1.

What are the present worth and the accumulated amount of a 10 year annuity paying P10000 at the end of each year, with interest at 15% compounded annually?

2.

On the day his grand son was born, a man deposited to a trust company a sufficient amount of money so that the boy could receive 5 annual payments of P10000 each for his college tuition fees, starting with his 18th birthday. Interest at the rate of 12% per annum was to be paid on all amounts on deposit. There was also a provision that the grandson could elect to withdraw no annual payments and receive a single lump amount on his 25th birthday. The grandson chose this option. a) How much did the boy receive as single payment? b) How much did the grandfather deposit?

3.

GRADIENT

Engr. Charity Hope Gayatin

What amount of money invested today at 15% interest can provide the following scholarship: P30000 at the end of each year for 6 years, P40000 for the next 6 years and P50000 thereafter?

Learning Outcome

Use the present worth (P/G) and uniform annual series (A/G) factors for arithmetic gradient

Arithmetic Gradient Factors (P/G and A/G) An Arithmetic Gradient Series is a cash flow series that either increases or decreases by a constant amount each period. The amount of change is called the Gradient.

Use the geometric gradient series factor (P/A, g) to find the present worth

It is important to realize that the base amount defines a uniform cash flow series of the size A that occurs each time period.

Arithmetic Gradient Factors (P/G and A/G)

Arithmetic Gradient Factors (P/G and A/G)

If the base amount is ignored, a generalized arithmetic (increasing) gradient cash flow diagram is as shown in the figure. Note that the gradient begins between years 1 and 2. This is called Conventional Gradient.

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Arithmetic Gradient Factors (P/G and A/G)

Arithmetic Gradient Factors (P/G and A/G)

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Problem 1.

Suppose that the maintenance expense on a certain machine is P10T at the end of the first year and increasing at a constant rate of P500 each year for the next four years. Determine the present worth of the total maintenance expense if interest rate is 10%.

2.

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 P5T 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 rate of the loan was 15%, what was the annual payment?

3.

Find the equivalent annual payment of the following obligations at 20% interest. End of Year Payment 1 P8000 2 P7000 3 P6000 4 P5000

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