# Financial Math FM/Time Value of Money

## Interest

A basic definition of interest is the price paid for obtaining, or price received for providing, money or capital in a credit transaction. For example, today Brian borrows $10,000 from Ivan and agrees to pay him$10,500 in one year. In this example the $10,000 is the principal that Ivan invested over a period of 1 year. After 1 year, the accumulated value is$10,500. $500 is the amount of interest Brian paid Ivan. ### Interest Rate (Rate of Interest) An interest rate or the rate of interest is how much is paid by a borrower as expressed as a percent of the borrowed capital. In the previous example$500 dollars of interest accrued at the end of 1 year over a $10,000 loan. ${\displaystyle {\operatorname {\} 500 \over \operatorname {\} 10,000}=5\%}$ Thus the interest rate for this loan was 5% and unless otherwise stated in the examination question, rates are expressed as annual rates.[1] Loans can and often are over multiple periods and the exact terms of the loan can depend on the interest type. #### Simple Interest Brian and Ivan can agree to use simple interest. The final amount of interest paid is simply the original loan amount times the interest rate times the number of periods. ${\displaystyle I_{simp}=k\cdot i\cdot t}$ Example Ivan invests$10000 into a bank account which pays simple interest with an annual rate of 5%. Find the balance in Ivan's account after 2 years.

Solution
${\displaystyle I_{simp}=k\cdot i\cdot t}$
${\displaystyle \ I=10,000(.05\cdot 2)}$
${\displaystyle \ I=\1,000}$
${\displaystyle \ \10,000+\1,000=\11,000}$


The final balance in Ivan's account will be his original investment plus the interest accrued over two years. This comes out to $11,000. #### Compound Interest Compound interest arises when interest is added to the principal, so that from that moment on, the interest that has been added also itself earns interest. This addition of interest to the principal is called compounding. A bank account, for example, may have its interest compounded every year: in this case, an account with$1000 initial principal and 20% interest per year would have a balance of $1200 at the end of the first year,$1440 at the end of the second year, and so on. In the case of compound interest, the accumulated value at the end of t periods is

${\displaystyle \ A(t)=k(1+i)^{t}}$
Example
Ivan invests $10000 into a bank account which pays compound interest with an annual rate of 5%. Find the balance in Ivan's account after 2 years. Solution ${\displaystyle \ A(t)=k(1+i)^{t}}$ ${\displaystyle \ A(t)=10,000(1+.05)^{2}}$ ${\displaystyle \ A(t)=11025}$  The final balance in Ivan's account comes to$11,025.

## Amount and Accumulation function

### Amount function

A(t) is defined as the amount function where A(t) denotes the value of an investment at time t. For this function there are two general assumptions.

• For each ${\displaystyle t\geq 0,A(t)>0}$.
• A is nondecreasing.

It should also be noted that the amount of interest earned over the period ${\displaystyle \ [s,t]}$ is ${\displaystyle \ A(t)-A(s)}$.
The effective rate of interest earned in the period ${\displaystyle \ [s,t]}$ is :${\displaystyle {\frac {A(t)-A(s)}{A(s)}}}$

Example
Ivan invests $2000 into a bank account with an interest rate of 5% and has an amount function of ${\displaystyle \ A(t)=ke^{0.5\cdot t\cdot i}}$ What is the effective rate of interest in the third year? Solution ${\displaystyle i_{[s,t]}={\frac {A(t)-A(s)}{A(s)}}}$ ${\displaystyle i_{[s,t]}={\frac {ke^{0.5\cdot t\cdot i}-ke^{0.5\cdot s\cdot i}}{ke^{0.5\cdot s\cdot i}}}}$ ${\displaystyle i_{[3,4]}={\frac {e^{0.5\cdot 4\cdot 0.05}-e^{0.5\cdot 3\cdot 0.05}}{e^{0.5\cdot 3\cdot 0.05}}}}$ ${\displaystyle i_{[3,4]}=0.0253}$  The effective interest rate in the third year is 2.53% ### Accumulation function a(t) is defined as the accumulation function where a(t) denotes the value of an investment at time t where the initial investment is 1. This is simply a special case of the amount function. The amount function and accumulation function share a simple relationship. ${\displaystyle A(t)=k\cdot a(t)}$ Example An investment's accumulation function is ${\displaystyle \ a(t)={t^{2}}}$ . The value of the investment at time t=4 is 80. What was the value of the initial investment? Solution ${\displaystyle A(t)=k\cdot a(t)}$ ${\displaystyle A(4)=k\cdot a(4)}$ ${\displaystyle 80=k\cdot 4^{2}}$ ${\displaystyle k=5}$  The initial investment was 5. ## Present and Future value As we've seen money has a time value so that$1 today is worth more than $1 a year from now. Given an effective annual interest rate of ${\displaystyle i}$ compounded annually over ${\displaystyle t}$ years: ${\displaystyle PV(1+i)^{t}=FV}$ where ${\displaystyle PV}$ is the present value and ${\displaystyle FV}$ is the future value. ${\displaystyle (1+i)}$ is called the interest factor. Note that ${\displaystyle PV=FV(1+i)^{-t}}$ ${\displaystyle v={\frac {1}{1+i}}}$ is called the discount factor and is denoted as ${\displaystyle v}$. Example In exchange for your car John has promised to pay you$5,000 in 1 year and $10,000 in 3 years. Using an annual effective interest rate of 5% find the present value of these payments. Solution ${\displaystyle PV=5,000v+10,000v^{3}}$ ${\displaystyle v=1.05^{-1}}$ ${\displaystyle PV=13400.28}$  Using accumulation function ${\displaystyle a(t)}$ notation • ${\displaystyle {\frac {a(n)}{a(n-1)}}=}$The n-th year interest factor. • ${\displaystyle i_{n}={\frac {a(n)-a(n-1)}{a(n-1)}}=}$the effective rate of interest in the n-th year. • ${\displaystyle v_{n}={\frac {a(n-1)}{a(n)}}=}$ the n-th year discount factor. • ${\displaystyle d_{n}={\frac {a(n)-a(n-1)}{a(n)}}=}$ the effective rate of discount in the n-th year. Note that • ${\displaystyle v_{n}=1-d_{n}}$ • ${\displaystyle d_{n}={\frac {i_{n}}{1+i_{n}}}}$ • ${\displaystyle i_{n}={\frac {d_{n}}{1-d_{n}}}}$ ### Discount Interest has been so far defined as the payment required at the end of the period, however there is also the rate of discount, denoted by ${\displaystyle d}$, which is a measure of the interest paid at the beginning of the period. For example, if Brian borrows$100 at a rate of interest of 10% over a period, then he will receive $100 now and he will have an obligation of$100 dollars at the end of the period plus $10 in interest. However, if Brian borrows$100 at a 10% rate of discount, he will only receive $90 now and he'll have an obligation of$100 at the end of the period.

${\displaystyle d_{n}={\frac {A(n)-A(n-1)}{A(n)}}={\frac {I_{n}}{A(n)}}}$

Example
Brian borrows $45 at a 5% annual rate of interest. Find the annual rate of discount. Solution ${\displaystyle d={\frac {I_{n}}{A(n)}}={\frac {45*0.05}{45*1.05}}=0.0476}$ ### Examples Example 1 If i = 6.5%, what are ${\displaystyle d}$ and ${\displaystyle v}$? Solution ${\displaystyle d={\frac {i}{1+i}}={\frac {0.065}{1.065}}=6.103}$ ${\displaystyle v=(1+i)^{-1}=1.065^{-1}=.93896}$ OR ${\displaystyle v=1-d=1-0.06103=.93896}$  Example 2 What is the present value of$9,000 to be received in 10 years at an annual effective rate of discount of 8%.
Solution
${\displaystyle PV=v^{n}FV}$
${\displaystyle PV=(1-d)^{n}\cdot 9,000}$
${\displaystyle PV=(1-.08)^{1}0\cdot 9,000}$
${\displaystyle PV=3,909.5}$

Example 3
At time ${\displaystyle t=0}$, Ivan deposits $260 into a fund with an annual discount factor of 0.95. Find the fund's value at time ${\displaystyle t=3.75}$. Solution ${\displaystyle PV=v^{n}FV}$ ${\displaystyle 260=0.95^{3.75}FV}$ ${\displaystyle FV=315.15}$  Example 4 Thor invests$250 into a bank account. One year later, his account is $260. 1. Find the effective annual interest rate Thor earned that year. 2. Find the effective annual discount rate Thor earned that year. Solution ${\displaystyle i={\frac {260-250}{250}}}$ ${\displaystyle i=.04}$ ${\displaystyle d={\frac {260-250}{260}}}$ ${\displaystyle d=.03846}$  ## Nominal vs Effective rates So far when dealing with compound interest we've only seen the interest compounded annually. That is to say, the interest is only added to the principal once a year. To illustrate this, Ivan invests$1000 into a bank account that offers 10% interest compounded annually. At the end of the first year, the account is $1000 +$100 in interest. At this point the principal earning interest is $1100 so that at the end of the second year the interest earned is$111 and the principal is $1100. However, the interest could be compounded more frequently. Suppose that instead of being compounded annually, the interest is compounded semi-annually. At the end of 6 months, the interest is$50 and added to the $1000 principal. The new principal is$1,050. At the end of the first year, the interest earned over the last 6 months is $52.5. After one year, the bank account contains$1,102.5. Notice that the interest rate is still 10% annually, however the effective annual rate of interest is 10.25%.

${\displaystyle i_{n}={\frac {1,102.5-1000}{1000}}=0.01025}$

10% is the nominal interest rate and 10.25% is the effective rate. A nominal rate of interest compounded m times a year is denoted as ${\displaystyle i^{(m)}}$. The relationship between the effective rate and nominal rate of interest is:

${\displaystyle i_{effective}={(1+{\frac {i^{(m)}}{m}})}^{m}-1}$
Example 1
Loki borrows $200 from Thor and repays him$205 in one month.
1. Find the monthly effective interest rate, which Loki is charged for the loan.
2. Find the annual nominal interest rate compounded monthly.
3. Find the annual effective interest rate.
Solution
${\displaystyle i={\frac {205-200}{200}}=0.025}$
${\displaystyle i^{(m)}=12\cdot 0.025=0.3}$
${\displaystyle i=(1.025)^{12}-1=0.3448}$


The monthly effective rate was 2.5%. The nominal rate will simply be 12 times the monthly effective rate, which is 30%. Finally, the effective annual rate is the monthly effective rate compounded 12 times, which is about 34.5%

Example2
Xiao Yu invests $7,500 into a bank account with a nominal interest rate of 8% convertible quarterly. How much does she have in the account after 6 years? Solution ${\displaystyle 7,500(1+{\frac {0.08}{4}})^{6\cdot 4}=12,063.28}$  ## Inflation and real rate of interest ## Force of interest So far we've said that the effective rate of interest earned in a period ${\displaystyle n}$ is ${\displaystyle i_{n}={\frac {a(n)-a(n-1)}{a(n-1)}}}$. This gives the interest rate earned over a period however. We can also find the rate of interest over an interval smaller than the period. ${\displaystyle i={\frac {a(t)-a(t-h)}{h\cdot a(t-h)}}}$. If we have a continuous accumulation function we can also measure the interest acting at a given moment called the force of interest which is denoted as ${\displaystyle \delta }$. ${\displaystyle \delta _{t}=\lim _{h\to 0}{\frac {a(t)-a(t-h)}{h\cdot a(t-h)}}={\frac {a'(t)}{a(t)}}={\frac {d}{dt}}\ln a(t)}$ ${\displaystyle e^{\int _{0}^{t}\delta _{r}dr}=a(t)}$ ### Examples Example1 Ivan invests$10,000 at a force of interest of 3%. Find the accumulated value at the end of 7.25 years.
Solution

${\displaystyle e^{\int _{0}^{t}\delta _{r}dr}=a(t)}$ ${\displaystyle 10,000\cdot e^{.03*7.25}=12,429.65}$