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Bank Interest, Loan, Bank Interest, Loan, and Investment and Investment Computation Computation

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Page 1: Bank Interest, Loan, and Investment Computation

Bank Interest, Loan, Bank Interest, Loan, and Investment and Investment

ComputationComputation

Page 2: Bank Interest, Loan, and Investment Computation

Interest Interest – is a fee charged by a lender

to a borrower for the use of borrowed money usually expressed as an annual percentage of the principal; the rate is dependent upon the time value of money, the credit risk of the borrower, and the inflation rate. Interest is also defined as the return earned on an investment.

Page 3: Bank Interest, Loan, and Investment Computation

Interest Rate The Interest Rate – is the cost of borrowing

money or the price paid for the rental funds; the ratio of interest to the amount lent.

Interest rates have an impact on the overall health of the economy because they affect consumers willingness to spend, save, or make business investment decisions.

Ex. Suppose that a $100 is lent and, at the end of the year, and $110 must be paid back. The interest paid is $10 and the interest rate is 10% (10÷100=0.10)

Page 4: Bank Interest, Loan, and Investment Computation

Bank Interest Bank interest – is the amount of

money that banks receive when they extend credit.

Note: The rate of interest is the price of credit.

Page 5: Bank Interest, Loan, and Investment Computation

Bank Interest Rates When banks quote an interest rate, they typically use a

“benchmark” to calculate that interest rate. Most of the time, that benchmark is the prime interest rate.

The prime interest rate – is the interest rate a bank charges its most creditworthy customers. It is the base rate on corporate loans posted by majority of banks.

The prime interest rate is relevant to small businesses because bank's use it as the starting interest rate from which to calculate the interest rate on bank loans.

Another important interest rate is the Londan Interbank Offered Rate (LIBOR) for import/export business or another business with an international presence; LIBOR generally moves right along with the prime rate.

Page 6: Bank Interest, Loan, and Investment Computation

Calculating Interest Rate Before making a decision to borrow

money, the prospective borrower considers the rate of interest and should be convinced that it is not too harsh.

The lender on the other hand must be convinced that the reward for his sacrifice (postponing expenditure) is worth it.

Interest rates also serve as a basis for comparing the returns offered by various financial instruments.

Page 7: Bank Interest, Loan, and Investment Computation

The Simple Interest RateSimple Interest – is a form of computing interest on an annual basis.Formula:

iPFA 2)

Ex. If Mr. X borrows $10,000 from Ms. Y at a simple interest of 10% per year, and is due for repayment in 3 years, how much does Mr. X owe Ms. Y on repayment date?

Where: FA = Final amount (Total Amount Due) r = Rate of interest

P = Principal t = Time

i = Interest

Prti 1)

$3,000

3 x 0.10 x 10,000i

$13,000

3,00010,000FA

Solving for Interest

Solving for Total Amount Due

Page 8: Bank Interest, Loan, and Investment Computation

Loan Price

Loan Price (LP):

– Corresponds to the total amount a borrower will pay for what he borrowed; the total amount due.

Formula: LP = P+i

Where: P = Principal

i = interest

Page 9: Bank Interest, Loan, and Investment Computation

Simple interest where time (t) is expressed in months:

Ex.: Compute the simple interest of a 6-months P100,000 loan with 7%.

Solution: i = Prt

= 100,000 . 0.07 . 6/12

= P3,500

Calculation of the Loan Price (LP) and Simple Interest on Loans

Simple interest where time (t) is expressed in weeks:

Ex.: Compute the simple interest and LP of a 6-months P100,000 loan with 1% per week.

Solution: i = Prt

= 100,000 . 0.01 . 26

= P26,000

Note: There are 52 weeks in 12 months

Page 10: Bank Interest, Loan, and Investment Computation

Simple interest where time is expressed between two (2) dates:

Ex.: Compute the simple interest of a P100,000 LOAN at 7% granted from 10/15/09 until 1/20/10.

i = Prt

=100,000 . 0.07 . 97/360

= P1,886.11

LP = P + i

= 100,000+1,886.11

= 101,886.11

Month No. of days

Oct. (31days) 16

Nov. (30days) 30

Dec.(31days) 31

Jan. 20

Total 97

(31–15)

Calculation of the Loan Price (LP) and Simple Interest on Loans

Note: Banks use 360 days for loans

Page 11: Bank Interest, Loan, and Investment Computation

Simple interest where time is expressed between two (2) dates:

Ex.: Compute the simple interest of a P100,000 DEPOSIT at 7% granted from 10/15/09 until 1/20/10.

i = Prt

=100,000 . 0.07 . 97/365

= P1,860.27

F = P+i

= 100,000+1,886.11

= 101,860.27

Month No. of days

Oct.(31days) 16

Nov.(30days) 30

Dec.(31days) 31

Jan. 20

Total 97

(31–15)

Calculation of the Final Amount and Simple Interest on Deposits

Note: Banks use 365 days for deposits

Page 12: Bank Interest, Loan, and Investment Computation

Different Time Factors

Note: No. of days in a year used by banks: For loans = 360 days in 1 year (as much as possible exact

time) For deposits = 365 days in 1 year (as much as possible

ordinary time)

Different time factors:

1. Exact time ÷ 3602. Ordinary time ÷ 3603. Exact time ÷ 3654. Ordinary time ÷ 365

Page 13: Bank Interest, Loan, and Investment Computation

Different time factors: Ex. July 5, 2009 – Sep. 17, 20091. Exact time ÷ 360

July(31days)26 (31-5) Aug(31days) 31 Sep 17

742. Ordinary time ÷ 360

July(30days) 25 (30-5) Aug(30days) 30 Sep 17

72

Calculation of the Loan Price (LP) and Simple Interest on Loans

ET÷360: 74 ÷ 360 = 0.2056

OT÷360: 72 ÷ 360 = 0.2000

Only 30 days every month

Page 14: Bank Interest, Loan, and Investment Computation

Different time factors: Ex. July 5, 2009 – Sep. 17, 2009Cont’d…3. Exact time ÷ 365

July(31days) 26 (31-5) Aug(31days) 31 Sep 17

744. Ordinary time ÷ 365

July(30days) 25 (30-5) Aug(30days) 30 Sep 17

72

Calculation of the Loan Price (LP) and Simple Interest on Loans

ET÷365: 74 ÷ 365 = 0.2027

OT÷365: 72 ÷ 365 = 0.1973

Only 30 days every month

Page 15: Bank Interest, Loan, and Investment Computation

Summary

ET÷360 OT÷360

ET÷365 OT÷365

0.2056 0.2000 0.2027 0.1973

No. of days in a year used in loans and deposits: For loans: ET÷360 – this will give them the greatest

interest income For deposits: OT÷365 – this will give them the least

interest expense

Page 16: Bank Interest, Loan, and Investment Computation

Derivation of rate (r), time (t), and principal (P)

Based on the above formula, the following can be derived:

i = Prt

To find rate r r = i/ Pt To find time t t = i/ Pr To find Principal P P = i/ r t

Simple interest formula:

Page 17: Bank Interest, Loan, and Investment Computation

Time (t):

Ex. Ben opened a time deposit in a bank amounting to P100,000 with a 4% simple interest rate. If he received P8,000 at the maturity date, how many years had his money been deposited in the bank?

Solution:

t = i/ Pr

= 8,000 ÷ (100,000 . 0.04 )

= 2 years

Consider i = P8,000

To find time (t)

Page 18: Bank Interest, Loan, and Investment Computation

Rate (r):

Ex. Flor invested P70,000 in a cooperative bank where the interest was P14,000 after two and a half years. What is the rate of her investment?

Solution: r = i/ Pt

= 14,000 ÷ (70,000 . 2.5 )

= 8%

Principal (P):

Ex. Lito received an interest of P12,500 in his savings account for 1 year at 5%. What was the original deposit?

Solution: P = i/ r t

= 12,500 ÷ (0.05 . 1 )

= P250,000

To find rate (r) and principal (P)

Page 19: Bank Interest, Loan, and Investment Computation

The two (2) formulas we have to review are i=Prt and F=P+i

F = P + i

= P + Prt since i=Prt

F = P . (1+rt)

Derivation of final amount (F):

P= F/(1+rt)

Derivation of principal (P):

Derivation of F and P

Page 20: Bank Interest, Loan, and Investment Computation

Final amount (F):

Ex. Luis deposited P200,000 to a bank which gives 5% simple interest. How much would he received at the end of 48 months?

Solution: Consider t = 4 (48 ÷12) F = P . (1+rt)

= 200,000 . (1+0.05 . 4 )

= P240,000 Principal (P):

Ex. How much must be invested in order to have P240,000 at the end of 4 years if money’s worth is 5% simple interest?

Solution: P = F/(1+r t)

= 240,000 ÷ (1+0.05 . 4 )

= P200,000

Derivation of F and P, Examples

Page 21: Bank Interest, Loan, and Investment Computation

Calculation of Basic Elements of Loan Pricing

2 Types of interest computation:

1. Ordinary interest – interest is paid at maturity date.

2. Discounted interest – interest is paid in advance.

Note: In both instances, one may use the simple interest rate formula.

Page 22: Bank Interest, Loan, and Investment Computation

Simple Discount Interest discount (i):

Formula:

i = Fdt

Where: F = Final amount

d = discount rate

t = time

Proceeds (Pr):

Formula:

Pr = F–i

Where: Pr = Proceeds

F = Final amount

i = Interest

Page 23: Bank Interest, Loan, and Investment Computation

i and Pr, Examples Interest discount (i) :

Formula: i = Fdt

Ex: How much interest will be collected in advance of a P114,000 loan for a term of 5 years if the discount rate is 12%.

Sol. : i = 114,000 . 0.12 . 5

= P68,400 Proceeds (Pr):

Formula: Pr = F–i

Ex: Compute the proceeds of a 1-year discounted loan amounting to P100,000 loan with 7%.

Sol. : Consider i = Prt = 100,000.0.07.1= 7,000

Proceeds = 100,000 – 7,000

= P93,000

Page 24: Bank Interest, Loan, and Investment Computation

Simple Discount Interest discount (i):

Formula: i = Fdt

To find maturity value (F) F = i/ dt or F=P-i To find time (t) t = i/ Fd To find rate (r) d = i/ Ft

Based on above formula, we can derive the following formulas:

Page 25: Bank Interest, Loan, and Investment Computation

The two (2) formulas we have to review are i=Fdt and Pr=F– i

Pr = F – i

= F – Fdt since i=Fdt

Pr = F . (1– dt)

Derivation of proceeds (Pr):

F= P/(1–dt)

Derivation of the Final amount (F):

Derivation of Pr and F

Page 26: Bank Interest, Loan, and Investment Computation

Derivation of Pr, Example Proceeds (Pr):

Formula: Pr= F. (1– dt)

Ex: What are the proceeds and the discount on P400,000 for 2 years at 10% simple discount.

Sol. : Pr = F. (1– dt)

= 400,000 (1– 0.10 . 2)

= P320,000

i=F–Pr or i=Fdt

= 400,000–320,000 = 400,000 . 0.10 . 2

= P80,000 = P80,000

Page 27: Bank Interest, Loan, and Investment Computation

Comparison Simple interest Simple discount

Interest i=Prt i=Fdt

Principal P=F/(1+rt) Pr=F(1–dt)

Amount F=P(1+rt) F=Pr/(1–dt)

Summarizing the Comparison

Page 28: Bank Interest, Loan, and Investment Computation

The Compound Interest Rate

The Final Amount (FA or simply F) in a compound interest is said to be the loan price (in a loan transaction) or future value (in an investment).

Formula: FA = P (1 + r)n

Ex. In a $10,000 loan, if the 10% is compounded annually for 3 years, what is the total amount due on maturity date?

FA = 10,000 (1 + 0.10)3

= $13,310

Where: P = Principalr = Rate of interestn = Compounding period

Page 29: Bank Interest, Loan, and Investment Computation

Ex.: Compute the compounded amount of a 2-year loan amounting to P100,000 with 7% interest rate compounded quarterly.

Sol. F = P . (1+j/m) t.m

= 100,000 . (1+0.07/4) 2.4

= 100,000 . (1.0175)8

= P114,888.18

Computing for r and n

Formula: r = j/m

Where j = nominal rate

m = frequency of conversion

n = tm

Where t = time

m = no. of times

compounded

Compound Interest – Computing for r and n

Page 30: Bank Interest, Loan, and Investment Computation

Compound Interest – is the method of calculating interest on the original capital invested on interest earned on previous periods.

Formula:

Compounded once a year

i = P. [(1+r) n – 1]

Where:

i = Compound interest

P = Principal

r = rate

t = No. of years

Compounded n times a year

i = P. [(1+ j/m) tm – 1]

Where:

A =Amount

P = Principal

j = nominal rate

m = Frequency of conversion

t = No. of years

Compound Interest

Page 31: Bank Interest, Loan, and Investment Computation

Compound interest

Ex.: Compute the compounded interest of a 2-year loan amounting to P100,000 with 7% interest rate.

Sol. i = P . [(1+ r) n – 1]

= 100,000 . [(1.07)2 – 1]

= P14,490

Compound interest (w/ no. of times compounded)

Ex.: Compute the compounded interest of a 2-year loan amounting to P100,000 with 7% interest rate compounded quarterly.

Sol. i = P . [(1+ j/m) tm – 1]

= 100,000 . [(1+0.07/4)2 . 4 – 1]

= P14,888

Compound Interest

Page 32: Bank Interest, Loan, and Investment Computation

Total deposits after n payments have been made:

Formula: Bn = A (1+i)n + {P÷i . [(1+i)n – 1]}

Ex. At the end of every month, you put P10,000 into a mutual fund that pays 6%, compounded monthly. How much will you have at the end of five years?

Solution: Consider P = 10,000, i = 6%/12 = .005, A = 0 (because you start with nothing in the account), n = 60 (5x12)

B60 = 0 + {10,000÷0.005 . [(1+0.005)60 – 1]}

= P697,700

Where:B = Balance P = PrincipalA = Amount n = No. of monthsi = interest rate/month

Compound Interest

Page 33: Bank Interest, Loan, and Investment Computation

Loan balance after n payments have been made:

Formula: Bn = A (1+i)n – {P÷i . [(1+i)n – 1]}

Ex. You have a $18,000 car loan at 14.25% for 36 months. You have just made your 24th payment of $617.39 and would like to know the payoff amount

Solution: Consider P = 18,000, i = 14.25%/12 = .011875, n = 24

B24 = 18,000 . 1.01187524 – {617.39÷0.011875 . (1.01187524 – 1)} = $6,866.97

Compound Interest

Page 34: Bank Interest, Loan, and Investment Computation

Computing monthly payment on a loan:

Formula:

P = iA

1-(1+i)-n

Ex. What is the monthly payment if you bough a P2.5M house, with 10% down, on a 30-year mortgage at a fixed rate of 7.8%?

Sol. n = 360 (30x12); i = 0.0065 (0.078/12); A = 2.25M (2.5M x 0.90)

P = 0.0065 x 2,250,000

1-(1+0.0065)-360

= P16,197.09

Compound Interest

Page 35: Bank Interest, Loan, and Investment Computation

Original loan amount:

Formula:

A = P÷i . [1 – (1+i)–n]

Ex1. You want to purchase a 20-year annuity that will pay $500 a month. If the guaranteed interest rate is 4%, how much will the annuity cost?

Solution: Consider P = 500, i = 4%/12 = .0033, n = 240 (20x12)

A = 500÷0.0033 . [1 – (1.0033)–240]

= $82,798.67

Compound Interest

Page 36: Bank Interest, Loan, and Investment Computation

Ex2. You’re looking to buy furniture for your living room. You can afford to pay about $60 a month over the next three years, and your credit card charges 4% interest. How much furniture can you buy?

Solution: Consider P = 60, i = 4%/12 = .0033, n = 36 (3x12)

A = P ÷i . [1 – (1+i)–n]

= 60/0.0033 . [1 – (1.0033)–36]

= $2,033

Compound Interest

Page 37: Bank Interest, Loan, and Investment Computation

Ex3. (Continued from Ex2). But you have your eye on a set that’s on sale for $1850. The saleswoman offers you a store credit card with a special promotional rate of 12% for three years. Now can you afford the furniture?

Solution: Consider P = 60, i = 12%/12 = .01, n = 36 (3x12)

A = P ÷ i . [1 – (1+i)–n]

= 60/0.01 . [1 – (1.01)–36]

= $1,806.45

You cannot afford the furniture. Either you raise your monthly payment or lower the interest rate.

Compound Interest

Page 38: Bank Interest, Loan, and Investment Computation

Number of payments on a loan:

Formula:

N = -log(1-iA÷P)

log (1+i)

Ex1. Sally offers to lend you P35,000 at 6% for that new home theater system you want. If you pay her back P1,000 a month, how long will it take?

Sol. Consider i = 0.005 (0.06/12)

N = -log(1-0.005 . 35,000÷1,000)

log (1.005)

= 38.57 months

Compound Interest

Page 39: Bank Interest, Loan, and Investment Computation

Ex2. (Continued from Ex1). How much is the final balance you owe Sally?

Formula:

Bn = A (1+i)n – {P÷i . [(1+i)n – 1]}

B38 = 35,000 . 1.00538 – {1,000÷0.005 . (1.00538 – 1]}

= P568.25

Interpretation:

You will pay Sally P1,000/month for 38 months and finish-off the loan on the 39th month paying P568.25.

Compound Interest

Page 40: Bank Interest, Loan, and Investment Computation

Number of payments on a loan:

Formula:

N = -log(1-iA÷P)

log (1+i)

Ex. You have $15,000 in a 5% savings account, which is compounded monthly. How long will it take to run down the account if you withdraw $100 a month?

Sol. Consider i = 0.004167 (0.05/12)

N = -log(1-0.004167 . 15,000÷100)

log (1.004167)

= 235.89 months

Compound Interest

Page 41: Bank Interest, Loan, and Investment Computation

Nominal Versus Real Interest Rate

• Nominal Interest Rate – refers to the interest rate that takes inflation into account.

Ex. If the market, or nominal, rate of interest is 10% per annum, then a dollar today can be exchanged for $1.10 a year from now.

EIRRIRNIR

Where: RIR =Real int. rateNIR = Nominal int. rateEIR = Expected inflation

rate

Formula:

Page 42: Bank Interest, Loan, and Investment Computation

Nominal Versus Real Interest Rate

• Real Interest Rate – is the rate adjusted for expected changes in the price level (inflation) so that it more accurately reflect the true cost of borrowing.

The real rate of interest is adjusted for expected future price-level changes; it is the rate of exchange between goods and services today, and goods and services in the future.Formula:

Note: An inflation rate of 10% will off-set the 10% nominal interest rate or a real interest rate of zero.

EIRNIRRIR

Where: RIR =Real int. rateNIR = Nominal int. rateEIR = Expected inflation

rate

Page 43: Bank Interest, Loan, and Investment Computation

Nominal Versus Real Interest Rate

Cont’d.Ex.1: Mr. X is considering lending his $100,000 for 1 year to Ms. Y with a promised interest payment of $8,000. As the inflation rate for next 12 months is forecasted at 10%, will it be wise for Mr. X to lend his money to Ms. Y?

%2

%10)000,100000,8(

EIRNIRRIR

Answer: The intended loan is not a good proposition

because he will stand to lose 2% of the purchasing power of his money instead of adding 8% to it.

Page 44: Bank Interest, Loan, and Investment Computation

Cont’d.Ex.2: If the nominal interest rate on a bond is 4.50%, and a reliable estimate suggests that inflation over the term will average 2.75%, the real interest rate could be estimated as follows:

Thus, the real rate of return in this case is just 1.75%.

Nominal Versus Real Interest Rate

%75.1

%75.2%50.4

EIR-NIRRIR

Page 45: Bank Interest, Loan, and Investment Computation

Nominal Versus Real Interest Rate

The Equation

This equation can be arranged to show that:

EIRRIRNIR

EIRNIRRIR

Page 46: Bank Interest, Loan, and Investment Computation

Effective Interest Rate (EIR) on an Ordinary Interest

Effective interest rate on an ORDINARY INTEREST loan:

Formula:

EIR = i/ P

Where: eir = Effective int. rate

i = Interest

P = Principal

Ex. Compute for the effective interest rate of a P100,000 1-year with a P7,000 interest.

Sol.EIR = i/ P

= 7,000 ÷ 100,000

= 7%

Page 47: Bank Interest, Loan, and Investment Computation

Effective interest rate on a DISCOUNTED INTEREST loan:

Formula:

Ex. Compute for the effective interest rate of a 1-year discounted loan amounting to P100,000 at 7%.

Sol. Consider i = 7,000, Proceeds = 93,000 (100,000 –7,000)

EIR = i/ Pr

= 7,000 ÷ 93,000

= 7.53%

EIR = i/ Pr

Where: EIR = Effective int. rate

i = Interest

EIR on a Discounted Interest

Page 48: Bank Interest, Loan, and Investment Computation

Computing the effective interest rate on a loan with a term of less than 1 year.

Formula: EIR = (i/ P) . 360/days loan outstanding

Where: EIR = Effective int. rate

i = Simple interest rate

P = Principal

Ex. Compute for the effective interest of a P100,000 120- days at 7% loan with a P7,000 interest.

Sol. Consider i = 7,000

EIR = (i/P) . 360/days loan outstanding = (7,000 ÷ 100,000) . 360/120

= 21%

EIR on a Discounted Interest (Less than 1 yr)

Page 49: Bank Interest, Loan, and Investment Computation

Comparative Analysis Which has the lower effective interest rate between an

ORDINARY and DISCOUNTED INTERES RATE?

Ex. Bank A (Ordinary) 20% 1 year

Bank B (Discounted) 19% 1 year

Sol. Bank A = 20%

Bank B 0.19 = 24%

0.81

Bank A has the lower interest rate.

Page 50: Bank Interest, Loan, and Investment Computation

Banks, under a line of credit arrangement, may require borrowers to deposit money that will not earn interest known as the compensating balance.

Compensating balance (C/B):

Formula: C/B = P . r

Where r = rate

Ex. A company borrows P200,000 and is required to keep a 12% compensating balance. It also has an unused line of credit in the amount of P100,000 for which a 10% compensating balance is required. Compute the minimum balance the company must maintain.

Sol. Consider C/B1 = 24,000 (200,000 x 0.12), C/B2 = 10,000 (100,000 X 0.10)

C/Btotal = 24,000 + 10,000

= P34,000

Compensating Balance

Page 51: Bank Interest, Loan, and Investment Computation

Effective interest rate of a loan with a C/B:

Formula:

EIR (w/ C/B) = r . P

Pr, % . P

Ex. Compute the effective interest rate of a P100,000 1-year ordinary interest loan at 7% requiring a 15% C/B.

Sol.

EIR w/ C/B

EIR (w/ C/B) = 0.07 . 100,000

(1 – 0.15) . 100,000

= 8.24%

Page 52: Bank Interest, Loan, and Investment Computation

Effective interest rate with a line of credit.

Formula: EIR (w/ line of credit) = r (on loan) . P

P – C/B

Ex. A co. has a line of credit amounting P400,000 and a C/B of 13% on the outstanding loan and 10% C/B on the unused credit. The interest rate on the loan is 18%. The company borrows P275,000. What is the effective interest rate on the loan?

Sol. Consider C/B1 = 35,750 (275,000 x 0.13), C/B2= 12,500 (125,000 x 0.10)

EIR w/ Line of Credit

EIR (w/ line of credit) = 0.18 . 275,000

275,000 – 48,250

= 21.8%

Page 53: Bank Interest, Loan, and Investment Computation

Prime Rate and Compensating Balance

The Prime Rate• A Prime Rate – is the banks charge on short-

term loans made to large corporations with impeccable financial credentials – their most creditworthy customers (i.e., with little risk of non-repayment on loans).

This rate is typically the lowest that creditworthy customers pay for short-term loans (since there are fewer expenses incurred by the lending bank to investigate the creditworthiness of the borrower).

Through the years, bank require borrowers to leave “compensating balance” to obtain a loan.

Page 54: Bank Interest, Loan, and Investment Computation

Prime Rate and Compensating Balance

Compensating Balance• Compensating Balance – is the amount of

money a bank requires a customer to maintain in a non-interest bearing account, in exchange for which the bank provides otherwise free services.

Note: The compensating balance should be publicly known, thus the actual interest rate paid cannot be determined by anyone but the parties involved only.

Page 55: Bank Interest, Loan, and Investment Computation

Prime Rate and Compensating Balance

Cont’d.Ex. A US steel desires to borrow $10M from ABC

Bank that charges 10% but requires the borrower to leave the $2.5M balance in its (non-interest bearing) current account.

a. Compensating balance = $2,500,000b. Withdrawable amount = $7,500,000 (10,000,000–

2,500,000= 7,500,000)c. Interest = $1,000,000 (10,000,000 x 0.10)d. Real interest rate = 13.5% (1,000,000÷7,500,000)

Page 56: Bank Interest, Loan, and Investment Computation

Different Types of Interest Rates

The Corporate Bond Rate• A Corporate Bond – is a bond issued by corporations to

raise money. The term is usually applied to longer-term debt instruments, generally with a maturity date falling at least a year after their issue date.

• A Corporate Bond Rate – is the interest rate paid on high-grade (low risk) corporate bonds. Formula: Corporate Bond Rate = Annual Interest Payment ÷ Price of the Bond.

Ex. DEF Corp. wants to expand its production facilities and must borrow money to do this. It sells the DEF Corporate Bonds for $1,000 apiece (10,000 units) and agrees to pay back the principal to the lender after 3 years. During those years, the corporation also promises to pay 5% interest rates annually.

Page 57: Bank Interest, Loan, and Investment Computation

To answer the question, it is helpful to understand the concept of discounting – the process of finding the value today of dollars in the future.Formula:

Where:

P = present value, value today, market price of the assetR1 = the amount of money to be received 1 yr. henceR2 = the amount of money to be received 2 yr. henceRn = the amount of money to be received in n year.

i = the market rate of interest

nn

33

221

)1(

R...

)1(

R

)1(

R

)1(

RP

iiii

The Present Value of Bonds

Page 58: Bank Interest, Loan, and Investment Computation

• Explanation: There is an inverse relationship that exists between price of the bond and the market interest rate. For a given R-stream, the higher the market rate of interest, the less the bond will sell for today. The lower the interest rate, the higher the value today of a given future R-stream.

Suppose that the DEF Corporate Bond with a face value of $1,000 pays $50 per annum (or nominal yield of 5%) and matures in 3 years. If the market rate is 10%, what will be the PV of the bond today?65.875$

)10.01(

1,050

)10.01(

50

)10.01(

50P

32

• Interpretation: A bond that gives $50/annum for 3 years has a present value at the end of 3 years of $875.65 when market int. rate is 10%.

The Present Value of Bonds

Page 59: Bank Interest, Loan, and Investment Computation

1. Nominal Yield – is the income received from a security, expressed as a percentage of its par value. “Yield” is synonymous with interest rate).

Formula:

Ex.If a bond is issued for $1,000 with an agreement to pay, say $100 in interest every year, then it has an annual coupon rate of interest of 10%.

F

Ci n

Bonds may be subjected to calculation of interest based on several conditions, namely: (1) Nominal Yield (2) Yield to Maturity (3) Current Yield, and (4) Zero Coupon.

Where: in = Nominal yieldC = Annual coupon int.

paymentF = Face amount of the bond

Finding the Interest Rate on a Bond

Page 60: Bank Interest, Loan, and Investment Computation

Finding the Interest Rate on a Bond

2. Yield to Maturity – refers to the average return on a debt security if kept until maturity, taking into account the income provided by interest payments as well as capital gains or losses.

Formula:

tPFn

P-F i

YTM

Where: P = Present ValueF = Maturity valuei = Nominal interestn = no. of years to

maturityt = no. of times int. is

paid

Page 61: Bank Interest, Loan, and Investment Computation

Cont’d. (Yield to Maturity)Ex. Mr. X paid $9,000 for a $10,000 face value, 10%

coupon bond that will mature in 10 years and which he expects to hold until maturity. What annual interest rate will Mr. X be getting on the security considering that interest payments are made twice a year?

%57.11

2000,9000,10

10000,9000,10

1,000YTM

Finding the Interest Rate on a Bond

Page 62: Bank Interest, Loan, and Investment Computation

3. Current Yield – refers to the promised annual interest payment from a bond divided by the market value of the bond. Formula:

Ex. A $10,000 face value bond with maturity of 10 years, promises to pay $1,000 interest per year. If it were to sell at a market price of $9,000, what is the current yield of the bond?%11.11

9,000

1,000i

C

Finding the Interest Rate on a Bond

F

Ci

C

Where: ic = Current yieldC = Annual coupon int.

paymentF = Face amount of the bond

Page 63: Bank Interest, Loan, and Investment Computation

Bonds are often issued (and resold) at a price different from their face value. Thus, a 6% bond currently selling at $900 would have a nominal yield of 6% ($60÷$1,000) but a current yield of 6.67% ($60÷$900).

Nominal vs. Current Yield

1000

60

%6

Nominal Yield

Current Yield

F

Ci

C

900

60

%67.6

F

Ci

n

Page 64: Bank Interest, Loan, and Investment Computation

4. Zero Coupon Bonds – are bonds which carry no interest but are issued at a deep discount which provides capital gains when they are redeemed at face value.

The face value of the bond is the price at which it will be redeemed and which is written on the bond certificate. The discount provided corresponds to the interest paid to the bond.

Formula: 1)P(Fr /1 n

Where: r = Rate of interestF = Face value of the

bondP = current price of the

bondn = no. of payment

periods, i.e., 5 years x 2

Finding the Interest Rate on a Bond

Page 65: Bank Interest, Loan, and Investment Computation

Cont’d.Ex. A bond with a face value of $10,000 maturing in 5 years, with no interest is sold today at $4,000. What rate of interest (compounded semi-annually) is applicable to the bond?

%6.91)4,000(10,000r 2x5)/(1

Finding the Interest Rate on a Bond