1. Mathematics of finance
L3-Banque et Finance
Christian Haddad

2. Contents Simple interest and simple discount
Compound interest and compound discount
Simple annuities
General and other annuities
Capital budgeting and depreciation

3. Simple interest and simple discount
Key Terms
Interest: an amount paid or earned for the use of money.
Simple interest: interest earned when a loan or investment is repaid in a lump sum.
Principal: the amount of money borrowed or invested.
Rate: the percent of the principal paid as interest per time period.
Time: the number of days, months or years that the money is borrowed or invested.

4. Simple interest and simple discount
The Simple Interest Formula
The interest formula shows how interest, rate, and time are related and gives us a way of finding one of these values if the other three values are known.
I = P x R x T
And the amount S is given by
S = P + I = P + Prt = P(1+ rt)
P = S/(1+rt) = present or discounted value at rate r of S due in t years. The factor (1+rt)^-1 is called a discount factor at simple interest, and the process of calculating P from S is called discounting at simple interest.

5. Simple interest and simple discount
Find the simple interest using the simple interest formula

6. Simple interest and simple discount
Identify the principal, rate and time
The interest is a percentage.
Principal is the amount borrowed or invested.
Rate of interest is a percent for a given time period, usually one year.
Time must be expressed in the same unit of time as the rate. (i.e. one year)

7. Simple interest and simple discount
Find the interest paid on a loan
Principal = (P) $1,200
Interest rate = 8% (or 0.08)
Time = 1 year
Interest = P x R x T
Interest = 1,200 x 0.08 x 1
Interest = $96
The interest on the loan is $96.

8. Simple interest and simple discount
Find the Maturity Value of a Loan
Maturity value: the total amount of money due by the end of a loan period; the amount of the loan and interest.
If the principal and the interest are known, add them.
MV = principal + PRT
MV = P(1+RT)

9. Simple interest and simple discount
Marcus Logan can purchase furniture on a 2-year simple interest loan at 9% interest per year.
What is the maturity value for a $2,500 loan?
Solution
MV = P (1 + RT) Substitute known values.
MV = $2,500 ( x 2)

10. Simple interest and simple discount
What is the maturity value?
MV = $2,500 ( x 2)
MV = $2,500 ( )
MV = $2,500 (1.18)
MV = $2,950
Marcus will pay $2,950 at the end of two years.

11. Simple interest and simple discount
Try these examples Terry Williams is going to borrow $4,000 at 7.5% interest. What is the maturity value of the loan after three years? $4,900 Jim Sherman will invest $3,000 at 8% for 5 years. What is the maturity value of the investment? $4,200

12. Simple interest and simple discount
Convert Months to a Fractional or Decimal Part of a Year
Convert 9 months and 15 months, respectively, to years, expressing both as fractions and decimals.
9/12 = ¾ = 0.75
9 months = ¾ or 0.75 of a year
15/12 = 1 3/12 = 1 ¼ = 1.25
15 months = 1 ¼ or 1.25 of a year.

13. Simple interest and simple discount
Look at this example To save money, Stan Wright invested $2,500 for 42 months at 4 ½ % simple interest. How much interest did he earn? 42 months = 42/12 = 3.5 I = P x R x T I = $2,500 x x 3.5 I = $ Stan will earn $393.75

14. Simple interest and simple discount
Find the Principal, Rate or Time Using the Simple Interest Formula

15. Simple interest and simple discount
Find the principal using the simple interest formula P = I / RT Judy paid $108 in interest on a loan that she had for 6 months. The interest rate was 12%. How much was the principal? Substitute the known values and solve. P = 108/ 0.12 x 0.5 P = $1,800

16. Simple interest and simple discount
Find the rate using the simple interest formula R = I / PT Sam wants to borrow $1,500 for 15 months and will have to pay $225 in interest. What is the rate he is being charged? Substitute the known values and solve. R = 225/ $1,500 x 1.25 R = .12 or 12% The rate Sam will pay is 12%.

17. Simple interest and simple discount
Find the time using the simple interest formula T = I / RP Shelby borrowed $10,000 at 8% and paid $1,600 in interest. What was the length of the loan? Substitute the known values and solve. T = $1,600/0.08 x $10,000 T = 2 The length of the loan was two years.

18. Simple interest and simple discount
Find Exact Time Ordinary time: time that is based on counting 30 days in each month. Exact time: time that is based on counting the exact number of days in a time period.

19. Simple interest and simple discount
Examples The ordinary time from July 12 to September 12 is 60 days. To find the exact time from July 12 to September 12, add the following: Days in July ( =) 19 Days in August 31 Days in September days

20. Simple interest and simple discount
Progressions Arithmetic Progressions A sequence of numbers, called terms, such that any two consecutive numbers in the sequence are separated by a fixed common difference, is an arithmetic progression. 5, 8, 11, 14, … is an arithmetic progression with common difference 3 24, 20, 16, 12, … is an arithmetic progression with common difference -4 Consider the arithmetic progression with first term t1 and common difference d

21. Simple interest and simple discount
The nth term of the progression is
The sum of the first n terms of the progression is

22. Simple interest and simple discount
Do this example
Find the 15th term and the sum of the first 15 terms of the arithmetic progression -1, 2, 5, 8, …
Solution
T1 = -1, d = 3, and n = 15;
T15 = t1 + 14d = -1+14(3) = 41
S15 = 15/2(t1+t15) = 15/2(-1+41) = 300

23. Simple interest and simple discount
Do this example An individual borrows $5000, agreeing to reduce the principal by $200 at the end of each month and to pay 15% interest per annum, that is, 1 ¼ % per month, on all unpaid balances. Find the sum of all interest payments. Solution In all there are 5000/200 = 25 payments. Interest payments are calculated as 1 ¼ % of the balances 5000, 4800, 4600, …, 200; they form the arithmetic progression 62.50, 60, 57.50, …, The sum of all interest payments is the sum of 25 terms of an arithmetic progression with t1 = and t2 = 2.50 S25 = 25/2( ) = $812.50

24. Simple interest and simple discount
Geometric Progressions A sequence of numbers, called terms, such that any two consecutive numbers in the sequence are in a fixed common ratio, is a geometric progression. The nth term of the progression is
The sum of the first n terms of the progression is

25. Simple interest and simple discount
Do this example The value of a certain machine at the end of each year is 80% as much as its value at the beginning of the year. If the machine originally costs $10 000, find its value at the end of 10 years. Solution Given t1= , r = 0.80, n = 11, we calculate T11 = * (0.80)^10 = $

26. Simple interest and simple discount
Do this example
Find the 10th term and the sum of the first 10 terms of the geometric progressions
1, 3, 9, 27
Solution
19683
29524

27. Simple interest and simple discount
Problems to solve At what rate of simple interest will $1200 accumulate interest of $72 in 6 months ? Solution First transform time to months (6/12). Second search for the formula to calculate the rate. I = P * R * T R is unknown = > 12%
Page 20 chapter 3 : mathematics of finance

28. Simple interest and simple discount
Problems to solve How long will it take for 500$ to accumulate to at least $560 at 13 ¼ % ordinary simple interest. Solution First , ordinary simple interest means we need to use days as 360. Then, apply the formula we get 327 days.
Ordinary simple interest : T = numbers of days/360 this is what banks apply.

29. Simple interest and simple discount
Problems to solve Eighty days after borrowing money, a person pays back exactly $850. How much was borrowed if the $850 payment includes principal and simple interest at 9 ¾ %? Solution Formula to use : P = S(1+ rt)^-1 = $831.97

30. Simple interest and simple discount
Problems to solve Find the discounted value of $1000 due in 3 months if the rate is 11%. Solution P = S(1+ rt)^-1 = $973.24

31. Simple interest and simple discount
The time between dates Problem to solve Using the Banker’s Rule, find the simple on $ 3260 at 12 ¼ % from April 21 to December 24 of the same year. Solution The exact time is = 247 days, and the simple interest is I = 3260 * ( ) (247/360) = $274

32. Simple interest and simple discount
Equations of value At a simple interest rate 12%, $100 due in 1 year is considered to be equivalent to $112 due in 2 years, since $100 would accumulate to $112 in the second year. In the same way, 100( )^-1 = $89.29 would be considered an equivalent sum at present. In general, we compare dated values by the following definition of equivalence : $ X due on a given date is equivalent at a given simple interest rate r to $Y due t years later if Y = X (1 + rt) or X = Y(1 +rt)^-1 The sum of a set of dates values, due on different dates, has no meaning. We have to replace all the dated values by equivalent dated values due on the same date. The sum of the equivalent values is called the dated value of the set.
When we move money forward we accumulate, ex : multiply the sum by an accumulation factor 1+ rt; when we move money backward, we discount, ex : multiply the sum by a discount factor (1+ rt) ^-1

33. Simple interest and simple discount
Solve the problem
An obligation of $1500 is due in 6 months with interest at 11%. At 15% simple interest, find the value of the obligation a) at the end of 3 months, b) at the end of 12 months
Solution
The value of the obligation in 6 months is 1500 [1 + (0.11) * (6/12) ] = $
X = [ 1 + (0.15) (3/12)]^-1 = $
Y = [1+ (0.15) (6/12)] = $

34. Simple interest and simple discount
Solve the problem Ms. Hill owes $500 due in 4 months and $700 due in 9 months. What single payment (a) now, (b) in 6 months, (c) in 1 year, will liquidate these obligations if money is worth 11%? Solution a) X1 = 500[1 + (0.11) * (4/12)]^ [1 + (0.11) * (9/12)]^-1 = $ b) X2 = 500[1 + (0.11) * (2/12)] + 700[1 + (0.11) * (3/12)]^-1 = $ c) X3 = 500[1 + (0.11) * (8/12)] + 700[1 + (0.11) * (3/12)] = $

35. Simple interest and simple discount
Solve the problem Blake borrowed $5000 on January 1, He paid $2000 on April 30, 1995, and $2000 on August 31, The final payment was made on December 15, Find the size of the final payment if the rate of interest was 7% and the focal date was (a) December 15, 1995; b) January 1, a) 2000*[1 + (0.07) * (229/360)] * [1 + (0.07) * (106/360) + X = 5000*[1+ (0.07) * (348/360)] = $ b) 2000*[1 + (0.07) * (119/360)]^ *[1 + (0.07)*(242/360)^-1 + X[1 + (0.07) * (348/360)^-1 = 5000 X = $

36. Simple interest and simple discount
Partial payments Financial obligations are sometimes liquidated by a series of partial payments during the term of obligation. Then it is necessary to determine the balance due on the final due date. There are two ways to allow interest credit on short-term transactions. We will employ the United States Rule. The interest on the unpaid balance of the debt is computed each time a partial payment is made. If the payment is greater than the interest due, the difference is used to reduce debt. If the payment is less than the interest due, it is held without interest until other partial payments are made whose sum exceeds the interest due at the time of the last of these partial payments.

37. Simple interest and simple discount
Solve the problem Gordon borrowed $1000 on January 15, 1995, at 16%. He paid $350 on April 12, 1995; $20 on August 10, 1995; and $400 on October 3, What is the balance due on December 1, 1995, under the United States Rule?
Solution page 29

38. Simple interest and simple discount
Simple Discount at a Discount Rate The calculation of present value over durations of less than one year is sometimes based on a simple discount rate. The annual simple discount rate d is the ratio of the discount D for the year to the amount S on which the discount is given. The simple discount D on an amount S for t years at the discount rate d is calculated by means of the formula : D = S*d*t And the discounted value P of S, or the proceeds P, is given by P = S – D By Substituting for D = S*d*t we obtain P in terms of S, d, and t ; P = S(1 – d*t)
(1- dt) is called a discount factor at a simple discount rate d.

39. Simple interest and simple discount
Some financial institutions offer what are referred to as discounted loans. For these types of loans, the interest charge is based on the final amount, S, rather than on the present value. The lender calculates the interest D, using D = S*d*t and deducts this amount from S. The difference, P, is the amount the borrower actually receives, even though the actual loan amount is considered to be S. The borrower pays back S on the due date. The interest charge on discounted loans is referred to as interest in advance. S= P(1 – d*t)^-1 accumulated sum using simple discount.
(1 – dt) accumulation factor at a simple discount rate d.

40. Simple interest and simple discount
Example
A person takes out a discount loan with a face value of $500 for 6 months from a lender who charges a 9 ½ % discount rate.
What is the discount, and how much money does the borrower receive?
What size loan should the borrower ask for if he wants to receive $500 cash?
Solution
a) We have S = 500, d = 9 ½ %, t=1/2 and calculate
Discount D = Sdt = $500*0.095*1/2 = $23.75
Proceeds P = S – D = $( ) = $476.25

41. Simple interest and simple discount
Alternatively, we could calculate the proceeds by P = S(1 – dt) = $500[1 – (0.095) * (1/2)] = $ What is happening in this situation is that the borrower receives a loan of $500, but he must pay $23.75 in interest charges today. He ends up with a net of $ in his pocket. Six months later, he repays the original loan amount, $500, but does not pay any interest at that time, as he has already paid the interest up front. b) We have P = 500, d = 9 ½ %, t=1/2 and calculate the maturity value of the loan S = P/(1-dt) = $500/1-(0.095)*(1/2) = $ The borrower should ask for a loan of $ to receive proceeds of $500

42. Simple interest and simple discount
Example
Calculate the discounted value of $1000 due in 1 year: a) at a simple interest rate of 7%; b) at a simple discount rate of 7%.
We have S = 1000, r = 7%, t=1 and calculate the discounted value P
P = S/1+ rt = $1000/1+ (0.07)*(1) = $934.58
b) We have S = 1000, d = 7%, t =1 and calculate the discounted value P by
P = S(1 –dt) = $1000[1- (0.07)*(1)] = $1000*(0.93) = $930

43. Simple interest and simple discount
Note the difference of $4.58 between the discounted value at a simple interest rate and the discounted value at a simple discount rate. We can conclude that a given simple discount rate results in a larger money return to a lender than the same simple interest rate. Note also that in a), the borrower is taking out a loan for $ At the end of 1 year, he/she pays back the original principal of $ plus interest of $65.42 for a total of $1000. In b), the borrower is taking out a loan for $1000. He/she pays $70 in interest up front, receives $930 and repays the original principal of $1000 at the end of 1 year.

44. Simple interest and simple discount
Example 3 If $1200 is the present value of $1260 due at the end of 9 months, determine a) the annual simple interest rate, and b) the annual simple discount rate. a) We have P = 1200, I = 60, t = 9/12 and calculate R = I/pt = 60/1200*(9/12) = 6.67% b) We have S = 1260, D = 60, t = 9/12, and calculate d = D/St = 60/1260*(9/12) = 6.35%

45. Simple interest and simple discount
Promissory note A promissory note is a written promise by a debtor, called the maker of the note, to pay to; or to the order of, the creditor, called the payee of the note, a sum of money, with or without interest, on a specific date. $ 1500 New York City, May 11, 1995 Ninety Days after date, I promise to pay to the order of J.D. Green Fifteen hundred and 00/100 dollars for value received with interest at 8% per annum. J.B. Smith

46. Simple interest and simple discount
Promissory note Example 1- Find the maturity value of the note 2- On July 2, 1995 Mr. Green sold the note to a bank at 9% bank discount rate. a) How much money did Mr. Green receive? b) What rate of interest did Mr. Green realize on his investment, c) What rate of interest did the bank realize on its investment if it holds the note till maturity?

47. Simple interest and simple discount
Solution
1- The maturity value S of the note is the accumulated value of $1500 for 90 days at 8% simple interest
S = P(1+rt) = 1500 [1+(0.08) (90/360)] = $ 1530 2-
P = S(1-dt) = 1530 [1-(0.09) (38/360)] = $
Mr Green realized a profit of $ on his investment of $1500 for the 52 days he held the note. The rate of interest Mr. Green realized was
r = I/Pt = 15.46/1500(52/360) = 7.14%
The bank received $1530 from Mr. Smith on August 9, 1995, thereby realizing a profit of = $ on its investment of $ for 38 days. The rate of interest the bank realized was
r = I/Pt = 14.54/ (38/360) = 9.09%

48. Simple interest and simple discount
Treasury Bills (or T-Bills) Treasury bills are popular short-term and low-risk securities issued by the Federal government to meet their short-term financing needs. Treasury bills are issued at face values, of $1000, $5000, $25000, $100000, and $ The face value of a T-bill is the amount the government guarantees it will pay on the maturity date. There is no interest stated on a T-Bill. Instead, to determine the purchase price of a T-bill, you need to discount the face value to the date of sale at an interest rate that is determined by market conditions.
Maturities = 13,26,52 weeks all less than one year.

49. Simple interest and simple discount
Treasury Bills (or T-Bills) Examples 1- A 182-day T-bill with a face value of $25000 is purchased by an investor who wishes to yield 3.8%. What price is paid ? 2- A 91-day T-bill with a face value of $ is purchased for $ What rate of interest is assumed?

50. Simple interest and simple discount
Treasury Bills (or T-Bills) Solutions 1- We have S= 25000, r=0.038, t =182/365 P =S[1+rt]^-1 = 25000[ (182/365)^-1 = The investor will pay for the T-bill and receive $25000 in 182 days time. 2- We have S = , P=97250, t=91/365 R=I/Pt = 2750/97250(91/365) = = 11.34%

51. Simple interest and simple discount
Applications 1-What principal will accumulate to a) $5100 in 6 months at 9% simple interest? b) $580 in 120 days at 18% exact simple interest ? 2-A couple borrows $ The annual interest rate is 10 1/2%, payable monthly, and the monthly payment is $200. How much of the first payment goes to interest and how much to principal? 3-A bank pays 10% per annum on savings accounts. Interest is credited quarterly on March 31, June30, September 30, and December 31, based on the minimum quarterly balance. If a person opens an account with a deposit of $200 on January 1 and withdraws $100 on August 8, how much interest is earned in the first year ?

52. Simple interest and simple discount
4- Paula owes $100 due in 6 months and $150 due in 1 year. She and the lender agree that she can pay off both debts today using a simple interest rate of 16% and putting the focal date now. How much will be paid in cash today ? 5-Carl owes $300 due in 3 months and $500 due in 8 months. What single payment a)now, b) in 6 months, c) in 1 year will liquidate these obligations, if money is worth 8% and the focal date is the time of the single payment? 6-Jackie borrows $1000 at 11%. She is to repay the loan in equal payments at 3 months, 6 months, and 9 months. Find the size of the payments, putting the focal date a) at the present, b) at the end of 9 months.

53. Simple interest and simple discount
7-Last night Marion won $5000 in a lottery. She was given two options. She can take $5000 today or $X every six months ( beginning six months from now) for 2 years. If the options are equivalent and the simple interest rate is 10%, find X using a focal date of 2 years. 8-A loan of 1000 is due in one year with interest at 14 ¼%. The debtor pays $200 in 3 months and $400 in 7 months. Find the balance due in one year . 9-A loan of $1400 is due in one year with simple interest at 12%. Partial payments of $400 in 2 months, $30 in 6 months, and $600 in 8 months are made. Find the balance due in one year.