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# Mathematics of Finance

Module by: Rupinder Sekhon. E-mail the author

Summary: This chapter covers principles of finance. After completing this chapter students should be able to: solve financial problems that involve simple interest; solve problems involving compound interest; find the future value of an annuity; find the amount of payments to a sinking fund; find the present value of an annuity; and find an installment payment on a loan.

## Chapter Overview

In this chapter, you will learn to:

1. Solve financial problems that involve simple interest.
2. Solve problems involving compound interest.
3. Find the future value of an annuity, and the amount of payments to a sinking fund.
4. Find the present value of an annuity, and an installment payment on a loan.

## Simple Interest and Discount

### Section Overview

In this section, you will learn to:

1. Find simple interest.
2. Find present value.
3. Find discounts and proceeds.

### SIMPLE INTEREST

It costs to borrow money. The rent one pays for the use of money is called the interest. The amount of money that is being borrowed or loaned is called the principal or present value. Simple interest is paid only on the original amount borrowed. When the money is loaned out, the person who borrows the money generally pays a fixed rate of interest on the principal for the time period he keeps the money. Although the interest rate is often specified for a year, it may be specified for a week, a month, or a quarter, etc. The credit card companies often list their charges as monthly rates, sometimes it is as high as 1.5% a month.

### SIMPLE INTEREST

If an amount PP size 12{P} {} is borrowed for a time tt size 12{t} {} at an interest rate of rr size 12{r} {} per time period, then the simple interest is given by

I = P r t I = P r t size 12{I=P cdot r cdot t} {}

The total amount AA size 12{A} {} also called the accumulated value or the future value is given by

{} A = P + I = P + Pr t A = P + I = P + Pr t size 12{A=P+I=P+"Pr"t} {}

or A=P1+rtA=P1+rt size 12{A=P left (1+ ital "rt" right )} {}

Where interest rate rr size 12{r} {} is expressed in decimals.

### Example 1

#### Problem 1

Ursula borrows $600 for 5 months at a simple interest rate of 15% per year. Find the interest, and the total amount she is obligated to pay? ##### Solution The interest is computed by multiplying the principal with the interest rate and the time. I=PrtI=Prt size 12{I="Pr"t} {} (1) I=$6000.15512 = $37.50I=$6000.15512 = $37.50 (2) The total amount is A =$600+$37.50 =$637.50 A = $600+$37.50 = $637.50 size 12{ matrix { A=$"600"+$"37" "." "50" {} ## =$"637" "." "50" } } {}
(3)

Incidentally, the total amount can be computed directly as

A = P1+rt = $600 1 + .15 5/12 =$600 1 + .0625 = $637.50 A = P1+rt =$600 1 + .15 5/12 = $600 1 + .0625 =$637.50 size 12{ matrix { A=P left (1+ ital "rt" right ) {} ## =$"600" left [1+ left ( "." "15" right ) left (5/"12" right ) right ] {} ## =$"600" left (1+ "." "0625" right ) {} ## =$"637" "." "50" } } {} (4) ### Example 2 #### Problem 1 Jose deposited$2500 in an account that pays 6% simple interest. How much money will he have at the end of 3 years?

##### Solution

The total amount or the future value is given by A=P1+rtA=P1+rt size 12{A=P left (1+ ital "rt" right )} {}.

A = $2500 1 + .06 3 =$ 2950 A = $2500 1 + .06 3 =$ 2950 size 12{ matrix { A=$"2500" left [1+ left ( "." "06" right ) left (3 right ) right ] {} ## =$"2950" } } {}
(5)

### Example 4

#### Problem 1

A Visa credit card company charges a 1.5% finance charge each month on the unpaid balance. If Martha owes $2350 and has not paid her bill for three months, how much does she owe? ##### Solution Before we attempt the problem, the reader should note that in this problem the rate of finance charge is given per month and not per year. The total amount Martha owes is the previous unpaid balance plus the finance charge. A =$2350 + $2350 .015 3 =$ 2350 + $105.75 =$ 2455.75 A = $2350 +$2350 .015 3 = $2350 +$105.75 = $2455.75 size 12{ matrix { A=$"2350"+$"2350" left ( "." "015" right ) left (3 right ) {} ## =$"2350"+$"105" "." "75" {} ## =$"2455" "." "75" } } {}
(7)

Once again, we can compute the amount directly by using the formula A=P1+rtA=P1+rt size 12{A=P left (1+ ital "rt" right )} {}

A = $2350 1 + .015 3 =$ 2350 1.045 = $2455.75 A =$2350 1 + .015 3 = $2350 1.045 =$ 2455.75 size 12{ matrix { A=$"2350" left [1+ left ( "." "015" right ) left (3 right ) right ] {} ## =$"2350" left (1 "." "045" right ) {} ## =$"2455" "." "75" } } {} (8) ### DISCOUNTS AND PROCEEDS Banks often deduct the simple interest from the loan amount at the time that the loan is made. When this happens, we say the loan has been discounted. The interest that is deducted is called the discount, and the actual amount that is given to the borrower is called the proceeds. The amount the borrower is obligated to repay is called the maturity value. ### DISCOUNT AND PROCEEDS If an amount MM size 12{M} {} is borrowed for a time tt size 12{t} {} at a discount rate of rr size 12{r} {} per year, then the discount DD size 12{D} {} is D = M r t D = M r t size 12{D=M cdot r cdot t} {} The proceeds PP size 12{P} {}, the actual amount the borrower gets, is given by P = M D P = M D size 12{P=M - D} {} P = M Mrt P = M Mrt size 12{P=M - ital "Mrt"} {} or P=M1rtP=M1rt size 12{P=M left (1 - ital "rt" right )} {} Where interest rate rr size 12{r} {} is expressed in decimals. ### Example 5 #### Problem 1 Francisco borrows$1200 for 10 months at a simple interest rate of 15% per year. Determine the discount and the proceeds.

##### Solution

The discount DD size 12{D} {} is the interest on the loan that the bank deducts from the loan amount.

D=MrtD=Mrt size 12{D= ital "Mrt"} {}
(9)
D = $1200 .15 10 12 =$150 D = $1200 .15 10 12 =$150 size 12{ matrix { D=$"1200" left ( "." "15" right ) left ( { {"10"} over {"12"} } right ) {} ## =$"150" } } {}
(10)

Therefore, the bank deducts $150 from the maturity value of$1200, and gives Francisco $1050. Francisco is obligated to repay the bank$1200.

In this case, the discount D=$150D=$150 size 12{D=$"150"} {}, and the proceeds P=$1200$150=$1050P=$1200$150=$1050 size 12{P=$"1200" - $"150"=$"1050"} {}.

### Example 6

#### Problem 1

If Francisco wants to receive $1200 for 10 months at a simple interest rate of 15% per year, what amount of loan should he apply for? ##### Solution In this problem, we are given the proceeds P and are being asked to find the maturity value MM size 12{M} {}. We have P=$1200P=$1200 size 12{P=$"1200"} {}, r=.15r=.15 size 12{r= "." "15"} {}, t=10/12t=10/12 size 12{t="10"/"12"} {} . We need to find MM size 12{M} {}.

We know

P=MDP=MD size 12{P=M - D} {}
(11)

but

D=MrtD=Mrt size 12{D= ital "Mrt"} {}
(12)

therefore

P=MMrtP=MMrt size 12{P=M - ital "Mrt"} {}
(13)
P=M1rtP=M1rt size 12{P=M left (1 - ital "rt" right )} {}
(14)
$1200=M1.151012$1200=M1.125$1200=M.875$1200.875=M$1371.43=M$1200=M1.151012$1200=M1.125$1200=M.875$1200.875=M$1371.43=M size 12{ matrix { $"1200"=M left [1 - left ( "." "15" right ) left ( { {"10"} over {"12"} } right ) right ] {} ##$"1200"=M left (1 - "." "125" right ) {} ## $"1200"=M left ( "." "875" right ) {} ## { {$"1200"} over { "." "875"} } =M {} ## $"1371" "." "43"=M } } {} (15) Therefore, Francisco should ask for a loan for$1371.43.

The bank will discount $171.43 and Francisco will receive$1200.

## Compound Interest

### Section Overview

In this section you will learn to:

1. Find the future value of a lump-sum.
2. Find the present value of a lump-sum.
3. Find the effective interest rate.

In the Section 2, we did problems involving simple interest. Simple interest is charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, the interest is paid (or charged) not only on the principal, but also on the past interest, and we say the interest is compounded.

Suppose we deposit $200 in an account that pays 8% interest. At the end of one year, we will have$200+$200.08=$2001+.08=$216$200+$200.08=$2001+.08=$216 size 12{$"200"+$"200" left ( "." "08" right )=$"200" left (1+ "." "08" right )=$"216"} {}. Now suppose we put this amount,$216, in the same account. After another year, we will have $216+$216.08=$2161+.08=$233.28$216+$216.08=$2161+.08=$233.28 size 12{$"216"+$"216" left ( "." "08" right )=$"216" left (1+ "." "08" right )=$"233" "." "28"} {}.

So an initial deposit of $200 has accumulated to$233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of$16.08=$1.28$16.08=$1.28 size 12{$"16" left ( "." "08" right )=$1 "." "28"} {}, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest. Now suppose we leave this amount,$233.28, in the bank for another year, the final amount will be $233.28+$233.28.08=$233.281+.08=$251.94$233.28+$233.28.08=$233.281+.08=$251.94 size 12{$"233" "." "28"+$"233" "." "28" left ( "." "08" right )=$"233" "." "28" left (1+ "." "08" right )=$"251" "." "94"} {}.

Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.

$200 1 + . 08 =$ 216 $200 1 + . 08 =$ 216 size 12{$"200" left (1+ "." "08" right )=$"216"} {}
(16)

$216 1 + . 08$ 216 1 + . 08 size 12{$"216" left (1+ "." "08" right )} {} (17) But$216=$2001+.08$216=$2001+.08 size 12{$"216"=$"200" left (1+ "." "08" right )} {}, therefore, the above expression becomes$ 200 1 + . 08 1 + . 08 = $233 . 28$ 200 1 + . 08 1 + . 08 = $233 . 28 size 12{$"200" left (1+ "." "08" right ) left (1+ "." "08" right )=$"233" "." "28"} {} After three years, we get$ 200 1 + . 08 1 + . 08 1 + . 08 $200 1 + . 08 1 + . 08 1 + . 08 size 12{$"200" left (1+ "." "08" right ) left (1+ "." "08" right ) left (1+ "." "08" right )} {}
(18)

Which can be written as

$200 1 + . 08 3 =$ 251 . 94 $200 1 + . 08 3 =$ 251 . 94 size 12{$"200" left (1+ "." "08" right ) rSup { size 8{3} } =$"251" "." "94"} {}
(19)

Suppose we are asked to find the total amount at the end of 5 years, we will get

$200 1 + . 08 5 =$ 293 . 87 $200 1 + . 08 5 =$ 293 . 87 size 12{$"200" left (1+ "." "08" right ) rSup { size 8{5} } =$"293" "." "87"} {}
(20)

We summarize as follows:

The original amount $200 =$200 The amount after one year $200 1 + . 08 =$ 216 The amount after two years $200 1 + . 08 2 =$ 233 . 28 The amount after three years $200 1 + . 08 3 =$ 251 . 94 The amount after five years $200 1 + . 08 5 =$ 293 . 87 The amount after t years $200 1 + . 08 t The original amount$200 = $200 The amount after one year$200 1 + . 08 = $216 The amount after two years$200 1 + . 08 2 = $233 . 28 The amount after three years$200 1 + . 08 3 = $251 . 94 The amount after five years$200 1 + . 08 5 = $293 . 87 The amount after t years$200 1 + . 08 t size 12{ matrix { "The original amount" {} # "$200"="$200" {} ## "The amount after one year" {} # "$200" left (1+ "." "08" right )=$"216" {} ## "The amount after thwo years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{2} } =$"233" "." "28" {} ## "The amount after three years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{3} } =$"251" "." "94" {} ## "The amount after five years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{5} } =$"293" "." "87" {} ## "The amount after "t" years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{t} } {} } } {} Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to one-fourth of 8%, or 2%. Now if we deposit$200 in the bank, after one quarter we will have $2001+.084$2001+.084 size 12{$"200" left (1+ { { "." "08"} over {4} } right )} {} or$204.

After two quarters, we will have $2001+.0842$2001+.0842 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{2} } } {} or$208.08.

After one year, we will have $2001+.0844$2001+.0844 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4} } } {} or$216.49.

After three years, we will have $2001+.08412$2001+.08412 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"12"} } } {} or$253.65, etc.

The original amount $200 =$200 The amount after one quarter $200 1 + . 08 4 =$ 204 The amount after two quarters $200 1 + . 08 4 2 =$ 208 . 08 The amount after one year $200 1 + . 08 4 4 = 216 . 49 The amount after two years$200 1 + . 08 4 8 = $234 . 31 The amount after three years$200 1 + . 08 4 12 = $253 . 65 The amount after five years$200 1 + . 08 4 20 = $297 . 19 The amount after t years$200 1 + . 08 4 4t The original amount $200 =$200 The amount after one quarter $200 1 + . 08 4 =$ 204 The amount after two quarters $200 1 + . 08 4 2 =$ 208 . 08 The amount after one year $200 1 + . 08 4 4 = 216 . 49 The amount after two years$200 1 + . 08 4 8 = $234 . 31 The amount after three years$200 1 + . 08 4 12 = $253 . 65 The amount after five years$200 1 + . 08 4 20 = $297 . 19 The amount after t years$200 1 + . 08 4 4t size 12{ matrix { "The original amount" {} # "$200"="$200" {} ## "The amount after one quarter" {} # "$200" left (1+ { { "." "08"} over {4} } right )=$"204" {} ## "The amount after two quarters" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{2} } =$"208" "." "08" {} ## "The amount after one year" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4} } ="216" "." "49" {} ## "The amount after two years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{8} } =$"234" "." "31" {} ## "The amount after three years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"12"} } =$"253" "." "65" {} ## "The amount after five years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"20"} } =$"297" "." "19" {} ## "The amount after "t" years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4t} } {} } } {}
(21)

Therefore, if we invest a lump-sum amount of PP size 12{P} {} dollars at an interest rate rr size 12{r} {}, compounded nn size 12{n} {} times a year, then after tt size 12{t} {} years the final amount is given by

A = P 1 + r n nt A = P 1 + r n nt size 12{A=P left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } } {}

### Example 8

#### Problem 1

##### Solution

This time we know the future value, but we need to find the principal. Applying the formula A=P1+rnntA=P1+rnnt size 12{A=P left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } } {}, we get

$5000=P1+.09365365×5$5000=P1.568225$3188.32=P$5000=P1+.09365365×5$5000=P1.568225$3188.32=P size 12{ matrix { $"5000"=P left (1+ { { "." "09"} over {"365"} } right )"365" times 5 {} ##$"5000"=P left (1 "." "568225" right ) {} ## $"3188" "." "32"=P } } {} (23) For comparison purposes, the government requires the bank to state their interest rate in terms of effective yield or effective interest rate. For example, if one bank advertises its rate as 7.2% compounded monthly, and another bank advertises its rate as 7.5%, how are we to find out which is better? Let us look at the next example. ### Example 10 #### Problem 1 If a bank pays 7.2% interest compounded monthly, what is the effective interest rate? ##### Solution Suppose we deposit$100 in this bank and leave it for a year, we will get

$1001+.0721212=$107.44$1001+.0721212=$107.44 size 12{$"100" left (1+ { { "." "072"} over {"12"} } right ) rSup { size 8{"12"} } =$"107" "." "44"} {}
(24)

Which means we earned an interest of $107.44$100=$7.44$107.44$100=$7.44 size 12{$"107" "." "44" -$"100"=$7 "." "44"} {} But this interest is for$100, therefore, the effective interest rate is 7.44%.

Interest can be compounded yearly, semiannually, quarterly, monthly, daily, hourly, minutely, and even every second. But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts. When interest is compounded "infinitely many times", we say that the interest is compounded continuously. Our next objective is to derive a formula to solve such problems, and at the same time put things in proper perspective.

Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be$11+1=$2$11+1=$2 size 12{$1 left (1+1 right )=$2} {}. If the interest is compounded semiannually, in one year we will have$11+1/22=$2.25$11+1/22=$2.25 size 12{$1 left (1+1/2 right ) rSup { size 8{2} } =$2 "." "25"} {} If the interest is compounded quarterly, in one year we will have$11+1/44=$2.44$11+1/44=$2.44 size 12{$1 left (1+1/4 right ) rSup { size 8{4} } =$2 "." "44"} {}, etc. We show the results as follows:  Frequency of compounding Formula Total amount Annually$ 1 1 + 1 $1 1 + 1 size 12{$1 left (1+1 right )} {} $2 Semiannually$ 1 1 + 1 / 2 2 $1 1 + 1 / 2 2 size 12{$1 left (1+1/2 right ) rSup { size 8{2} } } {} $2.25 Quarterly$ 1 1 + 1 / 4 4 $1 1 + 1 / 4 4 size 12{$1 left (1+1/4 right ) rSup { size 8{4} } } {} $2.44140625 Monthly$ 1 1 + 1 / 12 12 $1 1 + 1 / 12 12 size 12{$1 left (1+1/"12" right ) rSup { size 8{"12"} } } {} $2.61303529 Daily$ 1 1 + 1 / 365 365 $1 1 + 1 / 365 365 size 12{$1 left (1+1/"365" right ) rSup { size 8{"365"} } } {} $2.71456748 Hourly$ 1 1 + 1 / 8760 8760 $1 1 + 1 / 8760 8760 size 12{$1 left (1+1/"8760" right ) rSup { size 8{"8760"} } } {} $2.71812699 Every second$ 1 1 + 1 / 525600 525600 $1 1 + 1 / 525600 525600 size 12{$1 left (1+1/"525600" right ) rSup { size 8{"525600"} } } {} $2.71827922 Continuously$ 1 2 . 718281828 . . . $1 2 . 718281828 . . . size 12{$1 left (2 "." "718281828" "." "." "." right )} {} $2.718281828... We have noticed that the$1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name "ee size 12{e} {}" after the great mathematician Euler.

In mathematics, we say that as nn size 12{n} {} becomes infinitely large the expression 1+1nn1+1nn size 12{ left (1+ { {1} over {n} } right ) rSup { size 8{n} } } {} equals ee size 12{e} {}.

Therefore, it is natural that the number ee size 12{e} {} play a part in continuous compounding. It can be shown that as nn size 12{n} {} becomes infinitely large the expression 1+rnnt=ert1+rnnt=ert size 12{ left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } =e rSup { size 8{ ital "rt"} } } {}.

Therefore, it follows that if we invest $P$P size 12{$P} {} at an interest rate rr size 12{r} {} per year, compounded continuously, after tt size 12{t} {} years the final amount will be given by A=PertA=Pert size 12{A=P cdot e rSup { size 8{ ital "rt"} } } {}. ### Example 11 #### Problem 1 If$3500 is invested at 9% compounded continuously, what will the future value be in four years?

##### Solution

Using the formula for the continuous compounding, we get A=PertA=Pert size 12{A= ital "Pe" rSup { size 8{ ital "rt"} } } {}.

##### Solution

There are 60 deposits made in this account. The first payment stays in the account for 59 months, the second payment for 58 months, the third for 57 months, and so on.

The first payment of $500 will accumulate to an amount of$5001+.08/1259$5001+.08/1259 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } } {}.

The second payment of $500 will accumulate to an amount of$5001+.08/1258$5001+.08/1258 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } } {}.

The third payment will accumulate to $5001+.08/1257$5001+.08/1257 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"57"} } } {}. And so on. The last payment is taken out the same time it is made, and will not earn any interest. To find the total amount in five years, we need to add the accumulated value of these sixty payments. In other words, we need to find the sum of the following series.$5001+.08/1259+$5001+.08/1258+$5001+.08/1257+...+$500$5001+.08/1259+$5001+.08/1258+$5001+.08/1257+...+$500 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"57"} } + "." "." "." +$"500"} {} (37) Written backwards, we have$500+$5001+.08/12+$5001+.08/122+...+$5001+.08/1259$500+$5001+.08/12+$5001+.08/122+...+$5001+.08/1259 size 12{$"500"+$"500" left (1+ "." "08"/"12" right )+$"500" left (1+ "." "08"/"12" right ) rSup { size 8{2} } + "." "." "." +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } } {} (38) This is a geometric series with a=500a=500 size 12{a="500"} {}, r=1+.08/12r=1+.08/12 size 12{r= left (1+ "." "08"/"12" right )} {}, and n=59n=59 size 12{n = "59"} {}. Therefore the sum is$5001+.08/12601.08/12=$50073.47686=$36738.43$5001+.08/12601.08/12=$50073.47686=$36738.43 size 12{ matrix { { {$"500" left [ left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "08"/"12"} } {} ## =$"500" left ("73" "." "47686" right ) {} ## =$"36738" "." "43" } } {}
(39)

When the payments are made at the end of each period rather than at the beginning, we call it an ordinary annuity.

### Future Value of an Ordinary Annuity

If a payment of mm size 12{m} {} dollars is made in an account nn size 12{n} {} times a year at an interest rr size 12{r} {}, then the final amount AA size 12{A} {} after tt size 12{t} {} years is

A = m 1 + r / n nt 1 r / n A = m 1 + r / n nt 1 r / n size 12{A= { {m left [ left (1+r/n right ) rSup { size 8{ ital "nt"} } - 1 right ]} over {r/n} } } {}
(40)

(41)

### Example 17

x1+.08/12361.08/12=$5000x40.5356=$5000x=500040.5356x=$123.35x1+.08/12361.08/12=$5000x40.5356=$5000x=500040.5356x=$123.35 size 12{ matrix { { {x left [ left (1+ "." "08"/"12" right ) rSup { size 8{"36"} } - 1 right ]} over { "." "08"/"12"} } =$"5000" {} ## x left ("40" "." "5356" right )=$"5000" {} ## x= { {"5000"} over {"40" "." "5356"} } {} ## x=$"123" "." "35" } } {} (43) ### Example 18 #### Problem 1 A business needs$450,000 in five years. How much should be deposited each quarter in a sinking fund that earns 9% to have this amount in five years?

##### Solution

Again, suppose that xx size 12{x} {} dollars are deposited each quarter in the sinking fund. After five years, the future value of the fund should be $450,000. Which suggests the following relationship: x1+.09/4201.09/4=$450,000x24.9115=450,000x=45000024.9115x=18,063.93x1+.09/4201.09/4=$450,000x24.9115=450,000x=45000024.9115x=18,063.93 size 12{ matrix { { {x left [ left (1+ "." "09"/4 right ) rSup { size 8{"20"} } - 1 right ]} over { "." "09"/4} } =$"450","000" {} ## x left ("24" "." "9115" right )="450","000" {} ## x= { {"450000"} over {"24" "." "9115"} } {} ## x="18","063" "." "93" } } {}
(44)

If the payment is made at the beginning of each period, rather than at the end, we call it an annuity due. The formula for the annuity due can be derived in a similar manner. Reconsider Example 15, with the change that the deposits be made at the beginning of each month.

### Example 19

#### Problem 1

If at the beginning of each month a deposit of $500 is made in an account that pays 8% compounded monthly, what will the final amount be after five years? ##### Solution There are 60 deposits made in this account. The first payment stays in the account for 60 months, the second payment for 59 months, the third for 58 months, and so on. The first payment of$500 will accumulate to an amount of $5001+.08/1260$5001+.08/1260 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } } {}. The second payment of$500 will accumulate to an amount of $5001+.08/1259$5001+.08/1259 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } } {}. The third payment will accumulate to$5001+.08/1258$5001+.08/1258 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } } {}.

And so on.

The last payment is in the account for a month and accumulates to $5001+.08/12$5001+.08/12 size 12{$"500" left (1+ "." "08"/"12" right )} {} To find the total amount in five years, we need to find the sum of the following series.$5001+.08/1260+$5001+.08/1259+$5001+.08/1258+...+$5001+.08/12$5001+.08/1260+$5001+.08/1259+$5001+.08/1258+...+$5001+.08/12 size 12{$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"59"} } +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"58"} } + "." "." "." +$"500" left (1+ "." "08"/"12" right )} {} (45) Written backwards, we have$5001+.08/12+$5001+.08/122+...+$5001+.08/1260$5001+.08/12+$5001+.08/122+...+$5001+.08/1260 size 12{$"500" left (1+ "." "08"/"12" right )+$"500" left (1+ "." "08"/"12" right ) rSup { size 8{2} } + "." "." "." +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } } {}
(46)

If we add $500 to this series, and later subtract that$500, the value will not change. We get

$500+$5001+.08/12+$5001+.08/122+...+$5001+.08/1260$500$500+$5001+.08/12+$5001+.08/122+...+$5001+.08/1260$500 size 12{$"500"+$"500" left (1+ "." "08"/"12" right )+$"500" left (1+ "." "08"/"12" right ) rSup { size 8{2} } + "." "." "." +$"500" left (1+ "." "08"/"12" right ) rSup { size 8{"60"} } - $"500"} {} (47) Not considering the last term, we have a geometric series with a=$500a=$500 size 12{a=$"500"} {}, r=1+.08/12r=1+.08/12 size 12{r= left (1+ "." "08"/"12" right )} {}, and n=60n=60 size 12{n="60"} {}. Therefore the sum is

$5001+.08/12611.08/12$500=$50074.9667$500=$37483.35$500=$36983.35$5001+.08/12611.08/12$500=$50074.9667$500=$37483.35$500=$36983.35 size 12{ matrix { { {$"500" left [ left (1+ "." "08"/"12" right ) rSup { size 8{"61"} } - 1 right ]} over { "." "08"/"12"} } -$"500" {} ## =$"500" left ("74" "." "9667" right ) -$"500" {} ## =$"37483" "." "35" -$"500" {} ## =$"36983" "." "35" } } {} (48) So, in the case of an annuity due, to find the future value, we increase the number of periods nn size 12{n} {} by 1, and subtract one payment. The Future Value of an Annuity due = m 1 + r / n nt + 1 1 r / n m The Future Value of an Annuity due = m 1 + r / n nt + 1 1 r / n m size 12{"The Future Value of an Annuity due"= { {m left [ left (1+r/n right ) rSup { size 8{ ital "nt"+1} } - 1 right ]} over {r/n} } - m} {} Most of the problems we are going to do in this chapter involve ordinary annuity, therefore, we will down play the significance of the last formula. We mentioned the last formula only for completeness. Finally, it is the author's wish that the student learn the concepts in a way that he or she will not have to memorize every formula. It is for this reason formulas are kept at a minimum. But before we conclude this section we will once again mention one single equation that will help us find the future value, as well as the sinking fund payment. ### The Equation to Find the Future Value of an Ordinary Annuity, Or the Amount of Periodic Payment to a Sinking Fund If a payment of mm size 12{m} {} dollars is made in an account nn size 12{n} {} times a year at an interest rr size 12{r} {}, then the future value AA size 12{A} {} after tt size 12{t} {} years is A = m 1 + r / n nt 1 r / n A = m 1 + r / n nt 1 r / n size 12{A= { {m left [ left (1+r/n right ) rSup { size 8{ ital "nt"} } - 1 right ]} over {r/n} } } {} ## Present Value of an Annuity and Installment Payment ### Section Overview In this section, you will learn to: 1. Find the present value of an annuity. 2. Find the amount of installment payment on a loan. In Section 4, we learned to find the future value of a lump sum, and in Section 6, we learned to find the future value of an annuity. With these two concepts in hand, we will now learn to amortize a loan, and to find the present value of an annuity. Let us consider the following problem. ### Example 20 #### Problem 1 Suppose you have won a lottery that pays$1,000 per month for the next 20 years. But, you prefer to have the entire amount now. If the interest rate is 8%, how much will you accept?

##### Solution

This classic present value problem needs our complete attention because the rationalization we use to solve this problem will be used again in the problems to follow.

Consider for argument purposes that two people Mr. Cash, and Mr. Credit have won the same lottery of $1,000 per month for the next 20 years. Now, Mr. Credit is happy with his$1,000 monthly payment, but Mr. Cash wants to have the entire amount now. Our job is to determine how much Mr. Cash should get. We reason as follows: If Mr. Cash accepts xx size 12{x} {} dollars, then the xx size 12{x} {} dollars deposited at 8% for 20 years should yield the same amount as the $1,000 monthly payments for 20 years. In other words, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like the future values to equal. Since Mr. Cash is receiving a lump sum of xx size 12{x} {} dollars, its future value is given by the lump sum formula we studied in Section 4, and it is x1+.08/12240x1+.08/12240 size 12{x left (1+ "." "08"/"12" right ) rSup { size 8{"240"} } } {} (49) Since Mr. Credit is receiving a sequence of payments, or an annuity, of$1,000 per month, its future value is given by the annuity formula we learned in Section 6. This value is

$10001+.08/122401.08/12$10001+.08/122401.08/12 size 12{ { {$"1000" left [ left (1+ "." "08"/"12" right ) rSup { size 8{"240"} } - 1 right ]} over { "." "08"/"12"} } } {} (50) The only way Mr. Cash will agree to the amount he receives is if these two future values are equal. So we set them equal and solve for the unknown. x1+.08/12240=$10001+.08/122401.08/12x4.9268=$1000589.02041x4.9268=$589020.41x=$119,554.36x1+.08/12240=$10001+.08/122401.08/12x4.9268=$1000589.02041x4.9268=$589020.41x=$119,554.36 size 12{ matrix { x left (1+ "." "08"/"12" right ) rSup { size 8{"240"} } = { {$"1000" left [ left (1+ "." "08"/"12" right ) rSup { size 8{"240"} } - 1 right ]} over { "." "08"/"12"} } {} ## x left (4 "." "9268" right )=$"1000" left ("589" "." "02041" right ) {} ## x left (4 "." "9268" right )=$"589020" "." "41" {} ## x=$"119","554" "." "36" } } {} (51) The reader should also note that if Mr. Cash takes his lump sum of$119,554.36 and invests it at 8% compounded monthly, he will have $589,020.41 in 20 years. We have just found the present value of an annuity of$1,000 each month for 20 years at 8%.

We now consider another problem that involves the same logic.

### Example 21

#### Problem 1

Find the monthly payment for a car costing $15,000 if the loan is amortized over five years at an interest rate of 9%. ##### Solution Again, consider the following scenario: Two people, Mr. Cash and Mr. Credit, go to buy the same car that costs$15,000. Mr. Cash pays cash and drives away, but Mr. Credit wants to make monthly payments for five years. Our job is to determine the amount of the monthly payment. We reason as follows: If Mr. Credit pays xx size 12{x} {} dollars per month, then the xx size 12{x} {} dollar payment deposited each month at 9% for 5 years should yield the same amount as the $15,000 lump sum deposited for 5 years. Again, we are comparing the future values for both Mr. Cash and Mr. Credit, and we would like them to be the same. Since Mr. Cash is paying a lump sum of$15,000, its future value is given by the lump sum formula, and it is

$15,0001+.09/1260$15,0001+.09/1260 size 12{$"15","000" left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } } {} (52) Mr. Credit wishes to make a sequence of payments, or an annuity, of xx size 12{x} {} dollars per month, and its future value is given by the annuity formula, and this value is x1+.09/12601.09/12x1+.09/12601.09/12 size 12{ { {x left [ left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "09"/"12"} } } {} (53) We set the two future amounts equal and solve for the unknown.$15,0001+.09/1260=x1+.09/12601.09/12$15,0001.5657=x75.4241$311.38=x$15,0001+.09/1260=x1+.09/12601.09/12$15,0001.5657=x75.4241$311.38=x size 12{ matrix {$"15","000" left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } = { {x left [ left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "09"/"12"} } {} ## $"15","000" left (1 "." "5657" right )=x left ("75" "." "4241" right ) {} ##$"311" "." "38"=x } } {}
(54)

Therefore, the monthly payment on the loan is $311.38 for five years. ### The Equation to Find the Present Value of an Annuity, Or the Installment Payment for a Loan If a payment of mm size 12{m} {} dollars is made in an account nn size 12{n} {} times a year at an interest rr size 12{r} {}, then the present value PP size 12{P} {} of the annuity after tt size 12{t} {} years is P 1 + r / n nt = m 1 + r / n nt 1 r / n P 1 + r / n nt = m 1 + r / n nt 1 r / n size 12{P left (1+r/n right ) rSup { size 8{ ital "nt"} } = { {m left [ left (1+r/n right ) rSup { size 8{ ital "nt"} } - 1 right ]} over {r/n} } } {} where the amount PP size 12{P} {} is also the loan amount, and mm size 12{m} {} the periodic payment. ## Miscellaneous Application Problems We have already developed the tools to solve most finance problems. Now we use these tools to solve some application problems. One of the most common problems deals with finding the balance owed at a given time during the life of a loan. Suppose a person buys a house and amortizes the loan over 30 years, but decides to sell the house a few years later. At the time of the sale, he is obligated to pay off his lender, therefore, he needs to know the balance he owes. Since most long term loans are paid off prematurely, we are often confronted with this problem. Let us consider an example. ### Example 22 #### Problem 1 Mr. Jackson bought his house in 1975, and financed the loan for 30 years at an interest rate of 9.8%. His monthly payment was$1260. In 1995, Mr. Jackson decided to pay off the loan. Find the balance of the loan he still owes.

##### Solution

The reader should note that the original amount of the loan is not mentioned in the problem. That is because we don't need to know that to find the balance.

As for the bank or lender is concerned, Mr. Jackson is obligated to pay $1260 each month for 10 more years. But since Mr. Jackson wants to pay it all off now, we need to find the present value of the$1260 payments. Using the formula we get,

x1+.098/12120=$12601+.098/121201.098/12x2.6539=$255168.94x=$96,149.65x1+.098/12120=$12601+.098/121201.098/12x2.6539=$255168.94x=$96,149.65 size 12{ matrix { x left (1+ "." "098"/"12" right ) rSup { size 8{"120"} } = { {$"1260" left [ left (1+ "." "098"/"12" right ) rSup { size 8{"120"} } - 1 right ]} over { "." "098"/"12"} } {} ## x left (2 "." "6539" right )=$"255168" "." "94" {} ## x=$"96","149" "." "65" } } {} (55) The next application we discuss deals with bond problems. Whenever a business, and for that matter the U. S. government, needs to raise money it does it by selling bonds. A bond is a certificate of promise that states the terms of the agreement. Usually the businesses sells bonds for the face amount of$1,000 each for a period of 10 years. The person who buys the bond, the bondholder, pays $1,000 to buy the bond. The bondholder is promised two things: First that he will get his$1,000 back in ten years, and second that he will receive a fixed amount of interest every six months. As the market interest rates change, the price of the bond starts to fluctuate. The bonds are bought and sold in the market at their fair market value. The interest rate a bond pays is fixed, but if the market interest rate goes up, the value of the bond drops since the money invested in the bond can earn more elsewhere. When the value of the bond drops, we say it is trading at a discount. On the other hand, if the market interest rate drops, the value of the bond goes up, and it is trading at a premium.

### Example 23

#### Problem 1

The Orange computer company needs to raise money to expand. It issues a 10-year $1,000 bond that pays$30 every six months. If the current market interest rate is 7%, what is the fair market value of the bond?

##### Solution

A bond certificate promises us two things – An amount of $1,000 to be paid in 10 years, and a semi-annual payment of$30 for ten years. Therefore, to find the fair market value of the bond, we need to find the present value of the lump sum of $1,000 we are to receive in 10 years, as well as, the present value of the$30 semi-annual payments for the 10 years.

The present value of the lump-sum $1,000 is x1+.07/220=$1,000x1+.07/220=$1,000 size 12{x left (1+ "." "07"/2 right ) rSup { size 8{"20"} } =$1,"000"} {}
(56)

Note that since the interest is paid twice a year, the interest is compounded twice a year.

x1.9898=$1,000x=$502.57x1.9898=$1,000x=$502.57 size 12{ matrix { x left (1 "." "9898" right )=$1,"000" {} ## x=$"502" "." "57" } } {}
(57)

The present value of the $30 semi-annual payments is x1+.07/220=$301+.07/2201.07/2x=$426.37x1+.07/220=$301+.07/2201.07/2x=$426.37 size 12{ matrix { x left (1+ "." "07"/2 rSup { size 8{"20"} } = { {$"30" left [ left (1+ "." "07"/2 right ) rSup { size 8{"20"} } - 1 right ]} over { "." "07"/2} } right ) {} ## x=$"426" "." "37" } } {} (58) Once again, The present value of the lump-sum$1,000=$502.57The present value of the lump-sum$1,000=$502.57 size 12{"The present value of the lump-sum "$1,"000"=$"502" "." "57"} {} The present value of the$30 semi-annual payments=$426.37The present value of the$30 semi-annual payments=$426.37 size 12{"The present value of the$30 semi-annual payments"=$"426" "." "37"} {} Therefore, the fair market value of the bond is$502.57+$426.37=$928.94the fair market value of the bond is $502.57+$426.37=$928.94 size 12{"the fair market value of the bond is$502" "." "57"+"$426" "." "37"="$928" "." "94"} {}

### Example 24

#### Problem 1

An amount of $500 is borrowed for 6 months at a rate of 12%. Make an amortization schedule showing the monthly payment, the monthly interest on the outstanding balance, the portion of the payment contributing toward reducing the debt, and the outstanding balance. ##### Solution The reader can verify that the monthly payment is$86.27.

The first month, the outstanding balance is $500, and therefore, the monthly interest on the outstanding balance is outstanding balancethe monthly interest rate=$500.12/12=$5outstanding balancethe monthly interest rate=$500.12/12=$5 size 12{ left ("outstanding balance" right ) left ("the monthly interest rate" right )= left ($"500" right ) left ( "." "12"/"12" right )=$5} {} (59) This means, the first month, out of the$86.27 payment, $5 goes toward the interest and the remaining$81.27 toward the balance leaving a new balance of $500$81.27=$418.73$500$81.27=$418.73 size 12{$"500" -$"81" "." "27"=$"418" "." "73"} {}. Similarly, the second month, the outstanding balance is$418.73, and the monthly interest on the outstanding balance is $418.73.12/12=$4.19$418.73.12/12=$4.19 size 12{ left ($"418" "." "73" right ) left ( "." "12"/"12" right )=$4 "." "19"} {}. Again, out of the $86.27 payment,$4.19 goes toward the interest and the remaining $82.08 toward the balance leaving a new balance of$418.73$82.08=$336.65$418.73$82.08=$336.65 size 12{$"418" "." "73" - $"82" "." "08"=$"336" "." "65"} {}. The process continues in the table below.

 Payment # Payment Interest Debt Payment Balance 1 $86.27$5 $81.27$418.73 2 $86.27$4.19 $82.08$336.65 3 $86.27$3.37 $82.90$253.75 4 $86.27$2.54 $83.73$170.02 5 $86.27$1.70 $84.57$85.45 6 $86.27$0.85 $85.42$0.03

Note that the last balance of 3 cents is due to error in rounding off.

Most of the other applications in this section's problem set are reasonably straight forward, and can be solved by taking a little extra care in interpreting them. And remember, there is often more than one way to solve a problem.

## Classification of Finance Problems

We'd like to remind the reader that the hardest part of solving a finance problem is determining the category it falls into. So in this section, we will emphasize the classification of problems rather than finding the actual solution.

We suggest that the student read each problem carefully and look for the word or words that may give clues to the kind of problem that is presented. For instance, students often fail to distinguish a lump-sum problem from an annuity. Since the payments are made each period, an annuity problem contains words such as each, every, per etc.. One should also be aware that in the case of a lump-sum, only a single deposit is made, while in an annuity numerous deposits are made at equal spaced time intervals.

Students often confuse the present value with the future value. For example, if a car costs $15,000, then this is its present value. Surely, you cannot convince the dealer to accept$15,000 in some future time, say, in five years. Recall how we found the installment payment for that car. We assumed that two people, Mr. Cash and Mr. Credit, were buying two identical cars both costing $15, 000 each. To settle the argument that both people should pay exactly the same amount, we put Mr. Cash's cash of$15,000 in the bank as a lump-sum and Mr. Credit's monthly payments of xx size 12{x} {} dollars each as an annuity. Then we make sure that the future values of these two accounts are equal. As you remember, at an interest rate of 9%

the future value of Mr. Cash's lump-sum was $15,0001+.09/1260$15,0001+.09/1260 size 12{$"15","000" left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } } {}, and the future value of Mr. Credit's annuity was x1+.09/12601.09/12x1+.09/12601.09/12 size 12{ { {x left [ left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "09"/"12"} } } {}. To solve the problem, we set the two expressions equal and solve for xx size 12{x} {}. The present value of an annuity is found in exactly the same way. For example, suppose Mr. Credit is told that he can buy a particular car for$311.38 a month for five years, and Mr. Cash wants to know how much he needs to pay. We are finding the present value of the annuity of $311.38 per month, which is the same as finding the price of the car. This time our unknown quantity is the price of the car. Now suppose the price of the car is yy size 12{y} {}, then the future value of Mr. Cash's lump-sum is y1+.09/1260y1+.09/1260 size 12{y left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } } {}, and the future value of Mr. Credit's annuity is$311.381+.09/12601.09/12$311.381+.09/12601.09/12 size 12{ { {$"311" "." "38" left [ left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "09"/"12"} } } {}.

Setting them equal we get,

y 1 + . 09 / 12 60 = $311 . 38 1 + . 09 / 12 60 1 . 09 / 12 y 1 . 5657 =$ 311 . 38 75 . 4241 y 1 . 5657 = $23 , 485 . 57 y =$ 15 , 000 . 04 y 1 + . 09 / 12 60 = $311 . 38 1 + . 09 / 12 60 1 . 09 / 12 y 1 . 5657 =$ 311 . 38 75 . 4241 y 1 . 5657 = $23 , 485 . 57 y =$ 15 , 000 . 04 size 12{ matrix { y left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } = { {$"311" "." "38" left [ left (1+ "." "09"/"12" right ) rSup { size 8{"60"} } - 1 right ]} over { "." "09"/"12"} } {} ## y left (1 "." "5657" right )= left ($"311" "." "38" right ) left ("75" "." "4241" right ) {} ## y left (1 "." "5657" right )=$"23","485" "." "57" {} ## y=$"15","000" "." "04" } } {}
(60)

We now list six problems that form a basis for all finance problems. Further, we classify these problems and give an equation for the solution.

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