Money Time Low

Prepared by:

MS. KAREN S. TAFALLA

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

PRINCIPAL – is the amount of money used on which interest is charge.

SIMPLE INTEREST - is the interest to be paid which is directly proportional to the principal involved, the interest period and the interest rate.

I = PniWhere: I = total interest earned or charged

P = principal amount lent or borrowed n = number of interest periods i = interest rate per interest period

ORDINARY SIMPLE INTEREST = an interest based on the banker’s year1 banker’s year = 12 months, each consisting of 30 days = 360 days

EXACT SIMPLE INTEREST = an interest based on the exact number of days, 365 days for ordinary year and 366 days for leap year

NOTE: Unless otherwise stated on the problem, the ordinary simple interest is used.

COMPOUND INTEREST – the interest earned by the principal is considered as added to the principal, and therefore will also earn interest for the succeeding periods

Future Value Formula:for simple interest

for compound interest

INTEREST RATE:The cost of borrowing money. It refers to the amount earned by a unit principal per unit time.

Nominal Rate of interest is the basic annual rate of interest.Effective rate of interest is defined as the actual or the exact rate of

interest earned on the principal during a one-year period

where m is the number of interest period per year

niPF

niPPniPIPF

)1(

)1(

11

m

m

rER

CASH FLOW DIAGRAMA cash flow diagram is a graphical representation of cash flows on a time scale.

Receipts ( cash inflow )

Disbursement ( cash outflow)  

EXERCISES 1. Determine the ordinary and exact simple interest

on P1,200 for the period from January 12 to November 26,1998. If the rate is simple interest is 12.5% per year.

2. Determine the ordinary and exact simple interest on P1,500 for 10 months and 15 days if the rate of simple interest is 15% per year.

3. Determine the simple interest rate if an investment of P37,500 accumulates to P45,973.50 in 18 months.

4. The repayment on a loan was P12,100. If the loan was for 15 months at 16.8% interest a year, how much was the principal?

5. Annie buys a television set from a merchant who asks P1,250 at the end of 60 days (2 months). 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.

6. A deposit of P110,000 was made for 21 days. The net interest after deducting 20% withholding tax is P890.36. Find the simple rate of return.

7. A savings deposit of P500 will amount in 5 years to what, if interest is at 15% compounded annually?

8. After 12 years a certain investment accumulated to P10,120. If interest was at 10% compounded annually how much was the original investment?

9. How many years will be required for an investment of P30,000 to increase to P40,000 at an interest rate of 5% per year compounded annually?

 10. What rate of interest compounded annually is involved if an

investment of P15,000 made now will result 10 years hence in a receipt of P28,600?

11. P200,000 was deposited on January 1, 1988 at an interest rate of 24% compounded semiannually. How much would be the sum be on January 1, 1993?

12. What is the effective rate corresponding to 18% compounded daily? Take 1 year equal to 360 days.

13. How long will it take money to double itself if invested at 5% compounded bimonthly?

14. A national credit card carries an interest rate of 2% per month on the unpaid balance (a) calculate the effective rate per semiannual period. (b) if the interest rate is stated 5% per quarter, fine the effective rates per semiannual and annual time periods.

15. If 500,000 is deposited at a rate of 11.25% compounded monthly, determine the compounded interest after 7 years an 9 months.

A N N U I T I E SAn annuity consists of a series of equal payments made at

equal intervals of time. There are four types of annuity- ordinary annuity, deferred annuity, annuity due and perpetuity.

1. ORDINARY ANNUITYAn ordinary annuity is the one where the equal payments are made at the end of each payment period starting from the first period.

2. DEFERRED ANNUITYDeferred annuity is also an ordinary annuity but the payment of the first amount is deferred a certain number of periods after the first.

3. ANNUITY DUE An annuity due is one where the payments are made at the start of each period, beginning from the first period.

4. PERPETUITYA type of annuity where the payment period extend forever or in which the periodic payment continues indefinitely

FORMULAS

UNIFORM SERIES PRESENT WORTH FACTOR (P/A)P = A (P/A, i%, n)

UNIFORM SERIES CAPITAL RECOVERY FACTOR (A/P)A = P ( A/ P, i%, n)

UNIFORM SERIES COMPOUND AMOUNT FACTOR (F/A)F = A ( F/ A, i%, n)

UNIFORM SERIES SINKING FACTOR (P/A)A = F (A/F, i%, n)

i

iAP

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ni

iPA

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i

iAF

n 1)1(

1)1( ni

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Examples:1. Ms. Mercado bought a house thru the SSS housing loan.

She is required to pay P80,000 at the end of each year for 25 years. What is the original cost of the house if money is worth 16% per year compounded annually.

 2.A one bagger concrete mixer can be purchased with a down

payment of P200,000 and equal installments of P60,000 each paid at the end of year for the next 12 years. If money is worth 12 % compounded monthly, determine the equivalent cash price of the mixer.

3.How much would you have to deposit for five consecutive years starting one year from now if you want to be able to withdraw P50,000 ten years from now? Assume the interest rate to be 14% compounded semi-annually?

4.A man borrows P50,000, which agrees to repay in installments of P10,000 at the end of each year. How long will it take to pay the principal and interest at 7% compounded annually?

 5.A grove of mango trees is expected to yield P15,000 a

year for 25 years after the tree come into full bearing. Find the present value of the entire expected yield at an interest rate of 7% per year compounded semi-annually, assuming the first income of P15,000 to be due at the end of 5 years?

 6.A highway contractor purchased a grader for P350,000.

The annual operation cost was P18,000 a year. Five years after purchase, it was overhauled at a cost of P35,000. Three years after the overhaul, it was sold for P180,000. If interest rate is 15%, what was the equivalent uniform annual cost?

7. A farmer bought a tractor costing P250,000 payable 10 annual payments, 10 each installments payable at the beginning of each period. If the rate of interest is 10% compounded annually, determine the amount of each installment.

8. If a woman deposits P500 every 6 months for 7 years, how much will she have in her account after she makes her last deposit if the interest rate is 20% per year compounded quarterly?

10. A man paid 10% down payment of P200,000 for a house and lot and agreed to pay the 90% balance on monthly installment for 60 months at an interest rate of 15% compounded monthly. Compute the amount of the monthly payment

12. A loan association lends money at the rate of 15% compounded semiannually. A man borrows P50,000 payable in 16 semiannual installment, the first payment due at the end of two years reckoned from the date of the loan. How much is each of the semiannual installments?

13. Christine deposits P150 a month into an account paying 8% per year compounded quarterly. Twelve months deposits were made. Determine how much will be accumulated in the account one year after the last deposit.

14. Calculate the equivalent lump sum receipt now for the following cashflows: You invest $2,000 today, another $200 one year from now, and still another $800 two years from now. You then receive $1,000 each year for 10 years starting 4 years from now. The interest rate is 8% per year.

14. P0 = -$2,000 - $200 (P/F, 8%, 1) - $800 (P/F,

8%, 2) + $1,000 (P/A, 8%, 10)(P/F, 8%, 3)

= - $2,000 - $200 (0.9259) - $800 (0.8573) + $1,000 (6.7101) (0.7938)

= $2,455.46 (lump-sum receipt now)

15. What lump sum of money must be deposited into a bank account at present time so that $500 per month can be withdrawn for five years, with the first withdrawal scheduled for six years from today? The interest rate is ¾% per month

15. This is a deferred annuity, the time periods are months, and i = 3/4 % per month:

P71 = $500 (P/A, 3/4%, 60) = $500 (48.1733) = $24,086.65

P0 = $24,086.65 (P/F, 3/4% ,71) = 24,086.65 (0.58836) = $14,171.62

16. An individual is borrowing $100,000 at 8% interest compounded annually. The loan is to be repaid in equal annual payments over 30 years. However, just after the eighth payment is made, the lender allows the borrower to triple the annual payment. The borrower agrees to this increased payment. If the lender is still charging 8% per year, compounded annually, on the unpaid balance of the loan, what is the balance still owed just after the twelfth payment.

16 Original Payments = A = $100,000 (A/P, 8%, 30) = $100,000 (0.0888) = $8,880

Balance at EOY 8= $100,000 (F/P, 8%, 8) - $8,880 (F/A, 8%, 8) = $100,000 (1.8509) - $8,880 (10.6366)

= $90,636.99 or P8 = $8,880 (P/F, 8%, 22) = 8,880 (10.2007) = $90,636.99

New Payment = $8,880(3) = $26,640   Balance at EOY 12 = $90,636.99 (F/P, 8% , 4) - $26,640 (F/A, 8% , 4) = $90,636.99 (1.3605) - $26,640 (4.5061) = $3,269.12

17. An individual makes six annual deposits of $2,000 in a savings account that pays interest at a rate of 4% compounded annually. Two years after making the last deposit, the interest rate changes to 7% compounded annually. Twelve years after the last deposit, the accumulated money is withdrawn from the account. How much is withdrawn,

4-40 A woman arranges to repay $1,000 bank loan in 10 equal payments at a 10% effective annual interest rate. Immediately after her third payment, she borrows another $500, also at 10% per year. When she borrows the $500, she talks the banker into letting her repay the remaining debt of the first loan and the entire amount of the second loan in 12 equal annual payments. The first of these payments would be one year after she receives the $500. Compute the amount of each of payments.

4-38 Using time = 0 as the reference point, set P0(LHS) = P0(RHS):

 $2,000(P/F,8%,2) + $5,000(P/F,8%,6) =

Z(P/F,8%,4) - 2Z(P/F,8%,5)+ 3Z(P/F,8%,6)$2,000(0.8573) + $5,000(0.6302) =

Z(0.7350) - 2Z(0.6806) + 3Z(0.6302)$4,865.6 = 1.2644 ZZ = $3,848.15

UNIFORM GRADIENT SERIES OF CASH FLOWS

Arithmetic gradient series is a series of payments in which each payment is greater than or less than previous one by a constant amount G

n0 1 2 3 4

G2G

3G

(n-1)G

1. Determine the equivalent present worth, future worth and uniform annual worth of the following cash flow diagram. The interest rate is 15% per year compounded semiannually.

0 1 2 3 4

2500

5000

2000

3000

3500

7 EOY

2.3. Land is purchased for $25,000. It is agreed that

land will be paid over a five year period with annual payments and using a 12% annual compound interest rate. Each payment is to be $2000 greater than the previous payment. Determine the size of the last payment.

4. A company borrows P25,000 at an interest rate of 15% compounded annually with the agreement that the loan will be repaid in 8 installments. The repayment scheme will be such that each payment will be P600 larger than the preceding one, with the first payment to be made 3 years after the loan is negotiated. Determine the amount of the third payment?

5. A company must make a license payment for a process that they have adopted for a new plant. The payment will begin at P10,000, and the first payment is expected to be made 3 years from the present when the plant is completed and in production. Payment will be made every three months hereafter, and the license payments are expected to increase by P500 each quarter. What single present amount is equivalent to the series of license payments made over an 8-year period if the interest rate is 8% compounded quarterly?

6. Mr. Smith borrows P45,000 at 10% compounded bi-monthly, he pays a 7 year period with semiannual payments. Each successive payment P200 greater than each of the previous payments. If the first payment starts the 2nd year how much was the first payment? The last payment?

7. An increasing annual uniform gradient series begins at the end of the second year and ends after the fifteenth year. What is the value of the gradient G that makes the gradient series equivalent to a uniform flow of payments of P900 per year for 7 years at 12% per year compounded semiannually?

8. Suppose that the parents of young child decide to make annua deposits into a savings accoun, with the first deposit being made on the child’s 5th birthday and the last deposit being made on the 15th birthday. Then starting on the child’s 18th , 19th, 20th , 21st birthday, the withdrawals of $2000, $2400, $2800 and $3200 are to be made respectively. If the effective annual interest rate is 8% during this period of time, what are the annual deposits in years 5 though 15? Use uniform gradient amount in your solution

8. A = [$2,000 (P/A,8%,4) + $400 (P/G,8%,4)] (P/F,8%,2) (A/F,8%,11)

= [$2,000 (3.3121) + $400 (4.650)] (0.8573) (0.0601) = $437.14

4-62 Equivalent cash outflows = Equivalent cash inflows  Using time 9 as the reference point, set

F9(outflows) = F9(inflows)

  A(F/A,8%,5) = [$400(P/A,8%,4)-$100(P/G,8%,4)](F/P,8%,10) + $500(F/A,8%,3)

A(5.8666) = [$400(3.3121) - $100(4.65)](2.1589) + $500(3.2464)

A(5.8666) = $3,479.51 A = $593.10

GEOMETRIC SERIES OF CASH FLOWS◦ Geometric gradient series is series of payments

where annual payments increase or decrease over time, by a constant percentage.

0 1 32 n-1 n

A1

A2

A3

An-1

An

1. Consider the end-of-year geometric sequence of cash flows in the figure and determine the P, A, A0, and F equivalent values. The rate of increase is 15% per year after the first year, and annual interest rate is 20%.

2. Suppose that the salary for a recent graduate is expected to increase by 12% per year from a base of P32,000 over the next five years. If the interest rate is expected to be 10% per year compounded annually find the present worth of the said earnings.

3. Suppose that a shallow oil well is expected to produce 12,000 barrels of oil during its first year at P21 per barrel. If its yield is expected to decrease by 10% per year over the next seven years, what is the present worth of the anticipated gross revenue? The interest rate is 17% per year compounded annually.

4. In a geometric sequence of annual cash flows starting at end of year zero, the value of A0 is P1,304.35. The value of the last term in the series, A10, is P5,276.82. What is the equivalent value of A1-10 ? Let i= 20% per year compounded annually?

5. An End of Year (EOY) geometric gradient lasts for 10 years, whose initial value at EOY three is $5,000 and f = 6.04% per year thereafter. Find the equivalent uniform gradient amount (G) over the same time period if the initial value of the uniform gradient at EOY one is $4,000. The interest rate is 8% nominal, compounded semiannually.

PERPETUITIES AND CAPITALIZED COSTS

A perpetuity is an annuity where the periodic payments (withdrawals) continue indefinitely.

P is the capitalized value of A A capitalized costs of any property is the sum

of the first cost and present worth of all costs of replacement, operation and/or maintenance for a long time or forever.

0 1 2 ∞

3

A ( perpetuity)

1. I f the fund pays 12% compounded annually, what deposit is required today such that P1000 be withdrawn every year forever?

2. Find the present value of a perpetuity of P780 payable at the end of each year if money is worth (a) 6% effective (b) 6% compounded semiannually, (c) 6% compounded quarterly.

3. The XYZ Company is expected to pay P50 every 6 months indefinitely on a share of its preferred stock. If money is worth 65 per year compounded continuously to Co. B, what should be willing to pay for a share of the stock?

4. What is the capitalized cost of a structure that will require construction cost of P1,000,000 immediately and P80,000 each year for the next 4 years and annual year – end maintenance of P36,000 plus the expenditure of P200,000 at the end of each 10 year period for replacement? Assume a 12% interest rate per year compounded bimonthly?

5. A manufacturing plant installed a new boiler at a total cost of P150,000 and is estimated to have a useful life of 10 years. It is estimated to have a scrap value at the end of its useful life of P5,000. If the interest rate is 10% compounded semiannually, determine its capitalized cost.

6. A woman is considering giving an endowment to a university in order to provide payments of P5,000, P4,000, P3,000, and P2,000, respectively, at the end of the first, second, third and fourth quarters during a year. If the interest rate is 12% compounded quarterly, what is the capitalized equivalent that must be deposited now so that the quarterly payment can be repeated forever?

7. A firm can invest in a venture which cost P200,000 and returns P120,000 per year at the end of each year for 4 years. This investment can be renewed perpetually every 4 years. If the firms rate of return is 20% per year compounded annually, determine the capitalized worth of an infinite series of this investment.