Example 1

Ms. Smith loans Mr. Brown $10,000 with interest compounded at a rate of 8% per year.

How much will Mr. Brown owe Ms. Smith if he repays the loan at the end of 5 years?

Example 1 - Solution

Since the problém is of the form "find F when given P" the formula to use is

F = P(F/P, 8%, 5)

= $10,000(1.4693)

= $14,693.

Example 2

Mr. Lee wishes to accumulate $10,000 in a savings account in 10 years.

If the bank pays 5% compounded annually on deposits of this size, how much should Mr. Lee deposit ir the account?

Example 2 - Solution

This problem is of the form "find P when given F," and the formula to use is

P = F(P/F,5%,10)

= $10,000(0.6139)

= $6,139.

Example 3

An individual has been making equal annual payments of $2,000 to repay a loan. The individual wishes to pay off the loan immediately after having made an annual payment. Four payments remain to be paid.

With an interest rate of 8%, how much should be paid in the final payment?

Example 3 - Solution

P = A(P/A,8%,4)

= $2,000(3.3121) = $6,624

Example 4

A person borrows $10,000 at 6% compounded annually.

If the loan is repaid in ten equal annual payments, what will be the size of the payments if the first payment is made l year after borrowing the money?

Example 4 - Solution

A = P(A/P,6%,10)

= $10,000(0.1359) = $1,359

Example 5

If $800 is deposited annually for 10 years in an account that pays 6% compounded annually, how much money will be in the fund immediately after the tenth deposit?

Example 5 - Solution

F = A(F/A,6%,10)

= $800(13.1808) = $10,545

Example 6

An individual wishes to accumulate $1,000,000 in 30 years. If 30 end-of-year deposits are made into an account that pays interest at a rate of 10% compounded annually, what size deposit is required to meet the stated objective?

Example 6 - Solution

A = F(A/F, 10%, 30)

= $1,000,000(0.0061) = $6,100

Example 7

It is expected that a machine will incur operatíng costs of $4,000 the first year and that these costs will increase by $500 each year thereafter for the 10-year life of the machine.

If money is worth 15% per year to the firm, what is the equivalent annual worth of the operating costs?

Example 7 - Solution

The problem reduces to a constant $4,000 per year plus the $500 gradient.

A = $4,000 + G(A/G, i%, N) = $4,000 + $500(A/G,15%,10) = $4,000 + $500(3.3832) = $5,692

Example 8

Work Problem 7 if the timing of the costs is reversed as shown in the following diagram:

Example 8 - Solution

The problem reduces to a constant $8,500 per year minus a $500 gradient of payments.

A = $8,500 - 500(A/(7,15%, 10)

= $8,500 - 500(3.3832) = $6,808

Example 9

In Problem 7 suppose the operating cost the first year is $4,000 and that each year thereafter for the 10-year life of the machine operating costs increase by 6% per year. If money is worth 15% per year to the finn, what is the equivalent annual worth of the operatíng costs?

Example 9 - Solution

Since j = 6% and i = 15%,

F = Aj [ (F/P,15%,10) - (F/P, 6%, 10)] / 0.15 - 0.06 = $4,000[4.0456 - 1.7908] / 0.09 = $100,213

and

A = F(A/F, 15%, 10) = $100,213(0.0493) = $4,941

Example 10

A person makes six end-of-year deposits of $1,000 in an account paying 5% compounded annually. If the accumulated rund is withdrawn 4 years after the last deposit, how much money will be withdrawn?

Example 10 - Solution

Since F does not occur at the ume of the last A, it is necessary that the solution proceed in two steps. The amount of money in the account at the ume of the last deposit may be computed as

F6 = A(F/A,5%,6) = $1,000(6.8019) = $6,802.

The problem now is to find F10 given F6, = P6 = $6,802.

F10 = F6 (F/P,5%,4) = $6,802(1.2155) = $8,268

Example 11 - Solution

What is the effective interest rate for 4.75% compounded annually and 4.60% compounded quarterly?

Example 11 - Solution

Effectíve rate = (1+ rIM)M - 1

= (1+ 0.0475)1 - 1

= 4.75%

Effective rate = (1+ rIM)M - 1

= (1+ 0.0476/4)4 - 1

= 4.68%

Example 12

A loan company advertises that it will loan $1,000 to be repaid in 30 monthly installments of $44.60.

What is the cffective interest rate?

Example 12 - Solution

A = P(A/P,i%,30)

$44.60 / $1,000 =(A/P,i%,30) =0.0446

By inspection (with interpolation in tables), i = 2%, and

Effective rate = (F/P,2%,12) - 1

= 0.2682 = 26.82%.

Example 13

Annual deposite of $1,000 are made in an account that pays 4% compounded quarterly.

How much money should be in the account immediately after the fifth deposit?

Example 13 - Solution

Effective rate = (F/P, 1%, 4) – 1 = 4.06% F = A(F/A,4.06%,5) = $1,000 [(1+ 0.0406)5 - 1] / 0.0406 = $5 423

or A = $1,000(A/F,1%,4) = $1,000(0.2463) = $246.30 F20 = $246.30(F/A,1%,20) = 246.30(22.019) = $5,423

An alternate solution method is to treat the five annual deposits as single sums of money.

Example 13 - Solution

Therefore,

F = $1,000(A/F,1%,16)+(F/P,1%,12)+(F/P,1%,8)+(F/P,1%,4)+1]

= $1,000[1.1726 + 1.1268 + 1.0829 + 1.0406 + 1.000] = $5,423

Example 14

Given the payments shown in the following cash flow diagram, what is the equivalent worth in 2005 with interest at 6%?

Example 14 - Solution

X = [$500(F/A, 6%, 5) + $100(F/A, 6%, 3) + $200](F/P, 6%, 1)

= [$500(5.6371) + $100(3.1836) + $200](1.060)

= $3,537

Example 15

With interest at 8% compounded annually, how long does it take for a certain amount to double in magnitude?

(F/P, 8%, N) = 2.00

By inspection of 8% interest tables, N= 9 years.

Example 16

An individua! approaches the Loan Shark Agencv for $ 1,000 to be repaid in 24 monthly installments. The agency advertises interest at 1.5% per month. They proceed to calcu-ate the slze of his payment in the following manner:

Amount requested: $1,000Credit investigation: 25Credit risk insurance: 5

Total: $1,030 Interest: (1030) (24) (0.015) = $371 Total owed: $1,030 + $371 = $1,401Payment: $1,401/ 24 = $58.50

What effective interest rate is the individua! paying?

Example 16 - Solution

A =$58.50

A = P(AIP, 1%, 24)

$58.50 = $1,000(A/P, í%, 24)

By interpolation in tables, i = 2.9% per month, and

Effective rate = (F/P,2.9%,12) - 1

= 1.41 - 1= 41%

Example 17

Money is to be invested for a chilďs college expenses. Annual deposits of $2,000 are made in a fund that pays 5% compounded annually. If the first deposit is made on the chilďs 5th birthday and the last on the chilďs 15th biithday, what is the size of4 equal withdrawals on the chilďs 18th, 19th,20th, and 21st birthdays that will just deplete the account?

Example 17 - Solution

Amount in fund at t = 15:

F15 = A(F/A,5%,11) = $2,000(14.2068) = $28,414.

Amount in fund at t = 17:

F17 = P15(F/P, 5%, 2) = $28,414(1.1025) = $31,326

Amount of withdrawals:

A = P15(A/P,5%,4) = $31,326(0.2820) = $8,834

Example 18

An university student borrows money in her senior year to buy a car. She defers payments for 6 months and makes 36 beginning-of-month payments thereafter.

If the original note is for $12,000 and interest is 0.5% per month on the unpaid balance, how much will her payments be?

Example 18 - Solution

Amount owed at t = 5:

F5 = Po(F/P,0.5%,5)

= $12,000(1.0253) = $12,304

Amount of monthly payments:

A = P,(A/P, 0.5%,36) = $12,304 [i(1 + i)N / (1 + i)N - 1] = $12,304 [(1,005)(1,005)36 / (1,005)36 - 1]

= $372.45

Example 19

What is the present wonh of $100,000 ten years hence if interest is

a) 15% compounded annually?

b) 15% compounded continuously?

Example 19 - Solution

a. P = F(P/F,15%10) = $100,000(0.2472) = $24,720

b. P= F(P/F,15%,10) = $100,000(0.2231) = $22,310