Money-Time Money-Time RelationshipsRelationships

PrinciplesPrinciples

Sep. 9 1997

Simple InterestSimple Interest

Total interest earned or paid I = P*N*i

where P = principle lent or borrowedN = Number of interest periods i = interest rate per interest period

Example: If $1,000 was deposited in the bank with simple interest rate 10%. What’s total amount after 3 years?

I = P*N*i = 1,000*3*0.1 = 300

P3 = P + I = 1,000 + 300 = 1,300

Sep. 9 1997

Compound InterestCompound Interest

Year Start Balance Interest for Yr. End BalanceYear Start Balance Interest for Yr. End Balance

(at 10%) (Start Bal.+Interest)(at 10%) (Start Bal.+Interest)

1 1,000.00 100.00 1,100.001 1,000.00 100.00 1,100.00

2 1,100.00 110.00 1,210.002 1,100.00 110.00 1,210.00

3 1,210.00 121.00 1,331.003 1,210.00 121.00 1,331.00

4 1,331.00 133.10 1,464.104 1,331.00 133.10 1,464.10

Sep. 9 1997

Compound InterestCompound Interest

Year Start Balance Interest for Yr. End Balance

(at 10%) (Start Bal.+Interest)

1 1,000.00 1000*0.1 1000*(1+.1)

2 1000*(1+.1) 1000*(1+.1)*(.1) 1000*(1+.1)2

3 1000*(1+.1)2 (.1) 1000*(1+.1)2 1000*(1+.1)3

4 1000 *(1+.1)3(.1) 1000 *(1+.1)3 1000*(1+.1)4

Sep. 9 1997

Compound InterestCompound Interest

Year Start Balance Interest for Yr. End Balance (at 10%) (Start Bal.+Interest)

1 P P*i P(1+i) 2 P(1+i) P(1+i)i P(1+i)2

3 P*(1+i)2 P*(1+i)2(i) P(1+i)3

4 P*(1+i)3 P*(1+i)3(i) P(1+i)4

......................................…

n P*(1+i)n-1 P*(1+i)n-1(i) P(1+i)n

Sep. 9 1997

NotationsNotations• i = effective interest rate per interest period • N = number of compounding periods• P = present sum of money; the equivalent value of one or more cash flows at a reference point in time called the present• F = future sum of money; the equivalent value of one or more cash flows at a reference point in time called the future• A = end-of-period cash flows in a uniform series continuing for a specified number of periods, starting at the end of the first period and continuing through the last period

Sep. 9 1997

F given P F given P (F/P, i%, N)(F/P, i%, N)P given F P given F (P/F, i%, N)(P/F, i%, N)Balance at the end of year n = P(1+i)n

F = P(1+i)F = P(1+i)nn

F = P(F/P, i%, N) => (F/P, i%, N) = (1+i)(1+i)nn

=> P = F(1+i)=> P = F(1+i)-n-n

P = F(P/F, i%, N) => (P/F, i%, N) = (1+i )(1+i )-n-n

Sep. 9 1997

Cash Flow DiagramCash Flow Diagram

|---------+---------+---------+---------+---------|

0 1 2 3 4 5 Years.0 1 2 3 4 5 Years. Start End End EndStart End End End of of of …... ofof of of …... of Study Yr.1 Yr.2 StudyStudy Yr.1 Yr.2 Study

Negative Cash Flow

Positive Cash Flow

Sep. 9 1997

Example Example

The XYZ Corporation insists that its engineers develop a cash flow diagram of the proposal. An investment of $10,000 can be made that will produce uniform annual revenue of $5,310 for five years and then have a market value of $2,000 at the end of year five. Annual expenses will be $3,000 at the end of each year for operating and maintaining the project. Draw a cash flow diagram for the five-year life of the project. Use the corporation's viewpoint

Sep. 9 1997

DiagramDiagram

10,000

3,000 3,000 3,000 3,000 3,000

5,310 5,310 5,310 5,310 5,310

2,000

1 2 3 4 5

Sep. 9 1997

Example - Finding F Given PExample - Finding F Given P

Suppose that you borrow $8,000 now, promising to repay the loan principal plus accumulated interest in four years at i = 10% per year. How much would you repay at the end of four years?

F4 = P(1 + i)N = 8,000(1 + 0.1)4 = 11,713 or (by Table) F4 = P(F/P,i%,N) = 8,000(1.4641) = 11,713

Sep. 9 1997

Graphically ...Graphically ...

Sep. 9 1997

Example - Finding P Given FExample - Finding P Given F

An investor wishes to purchase a tract of land that will be worth $10,000 in six years. If the value of the land increases at 8% each year, how much should the investor be willing to pay now for this property?

The purchase price (Present Value) PP = $10,000(P/F, 8%, 6)

P = $10,000(0.6302) = $6,302

Sep. 9 1997

Graphically ...Graphically ...

Sep. 9 1997

Uniform Series (Annuity) to Its Present & Future Equivalent Values1. P (present equivalent value) occurs one interest

period before the first A (uniform amount).

2. F (future equivalent value) occurs at the same time as the last A, and N periods after P.

3. A (annual equivalent value) occurs at the end of periods 1 through N, inclusive.

Sep. 9 1997

Uniform Series (Annuity) to Its Present & Future Equivalent Values

0 1 2 3 N-1 N

P0 = Present Worth FN = Future Worth

i = Interest Rate per PeriodN = Number of Interest Periods

Sep. 9 1997

Uniform Series Compound Amount Uniform Series Compound Amount FactorFactorF = A(F/P,i%,N -1) + A(F/P,i%,N - 2) +A(F/P,i%,N - 3) +

… + A(F/P, i%, 1) + A(F/P, i%, 0)

= A[(1 + i)N—1 + (1 + i)N—2 + (1 + i)N -3 + ... + (1 + i)+ 1]

= A [ (1 + i)N - 1] / i

F = A(F/A, i%, N) => (F/A, i%, N) = [(1 + i)N - 1]/i is called the uniform series compound amount factor

0 1 2 3 N-1 N

P0 = Present Worth FN = Future Worth

Sep. 9 1997

ExampleExample

Suppose you make 15 equal annual deposits of $1,000 each into a bank account paying 5% interest per year. The first deposit will be made one year from today. How much money can be withdrawn from this bank account immediately after the 15th deposit?

F= $1,000(F/A,5%,15) = $1,000(21.5786) = $21,578.60

Sep. 9 1997

ExampleExample

If you are 20 years of age and save $1.00 each day for the rest of your life, you can become a millionaire." Let's assume you live to age 80, and the annual interest rate is 10% (i = 10%). Under these specific conditions, we compute the future compound amount (F) to be

F = $365/yr. (F/A, 10%, 60 years) = $365 (3,034.81)= $1,107,706

Sep. 9 1997

Uniform Series Present Worth Factor Finding P Given AFinding P Given A

We Know that F = A [ (1 + i)N - 1] / iAlso,

F = P(1 + i)N => P(1 + i)N = A[(1 + i)N - 1 ] / i

Dividing both sides by (1 + i)N,

P = A [ (1 + i)N - 1 ] / [ i (1 + i)N ] = A(P/A, i%, N)

=> (P/A, i%, N) = [ (1 + i)N - 1 ] / [ i (1 + i)N ]

Sep. 9 1997

ExampleExample If a certain machine undergoes a major overhaul

now, its output can be increased by 20% - which translates into additional cash flow of $20,000 at the end of each year for five years. If i = 15% per year, how much can we afford to invest to overhaul this machine?

Sol: The increase in cash flow is $20,000 per year, and it continues for five years at 15% annual interest. The upper limit on what we can afford to spend is P = $20,000(P/A,15%,5) = $20,000(3.3522) = $67,044

Sep. 9 1997

ExampleExample

Suppose that your rich uncle has $1,000,000 that he wishes to distribute to his heirs at the rate of $100,000 per year. If the $1,000,000 is deposited in a bank account that earns 6% interest per year, how many years will it take to completely deplete the account? How long will it take if the account earns 8% interest per year instead of 6%?

Sol: P = A(P/A, i%, N) => $1,000,000 = $100,000(P/A, 6%,N) N = 15.7 years. $1,000,000 = $100,000(P/A, 8%,N); N = 20.9 years.

Sep. 9 1997

Finding A Given F Finding A Given F Sinking Fund FactorSinking Fund Factor P = A [ (1 + i)N - 1 ] / [ i (1 + i)N ] P = F(1 + i)-N A [ (1 + i)N - 1 ] / [ i (1 + i)N ] = F(1 + i)-N A = F { i / [ (1 + i)N - 1 ] } = F (A/F, i%, N) (A/F, i%, N) = { i / [ (1 + i)N - 1 ] }

Sep. 9 1997

ExampleExample

A student is planning to have personal savings $1,000,000 when she retires at age 65. She is now 20 years old. If the annual interest rate will average 7% over the next 45 years on her saving account, what equal end-of-year amount must she save to accomplish her goal?

Sol: A = F(A/F, i%, N) = 1,000,000(A/F, 7%,45) = 1,000,000 (0.0035) = 3,500

Sep. 9 1997

Finding A Given P Finding A Given P Capital Recovery FactorCapital Recovery Factor (P/A, i%, N) = [ (1 + i)N - 1 ] / [ i (1 + i)N ] (A/P, i%, N) = 1 / (P/A, i%, N) (A/P, i%, N) = [ i (1 + i)N ] / [ (1 + i)N - 1 ] A student borrowed $800,000 to pay for tuition. He is

planning to pay back in next four years. If the annual interest rate will average 10% over the next 4 years. what equal end-of-year amount must he pay each year?

Sol: A = P (A/P, i%, N) = $800,000 (A/P, 10%, 4) = 800,000 (0.3155) = 25,240

Sep. 9 1997

Graphically ...Graphically ...

Sep. 9 1997

Discrete Compounding Interest Discrete Compounding Interest FactorsFactors

Sep. 9 1997

More Examples: Example 1More Examples: Example 1

How much interest is payable each year on a loan of $2,000 if the interest rate is 10% per year when half of the loan principal will be repaid as a

lump sum at the end of four years and the other half will be repaid in one lump-sum amount at the end of eight years? How much interest will be paid over the eight-year period?

Sep. 9 1997

Sol:Sol:

Year

1 2 3 4 5 6 7 8

Amount Owed atBeginning ofYear

$2,0002,0002,0002,0001,0001,000

1,0001,000

InterestAccruedfor Year

$ 200200200200100100

100100

Total AmountOwed atEnd of Year

$2,2002,2002,2002,2001,1001,1001,1001,100

PrincipalPayment

$ 0 0 0

1,000 0

0 01,000

TotalEnd of Year

Payment

$ 200 200 2001,200 100 100 1001,100

Sep. 9 1997

Example 2Example 2

You can buy a machine for $100,000 that will produce a net income, after operating expenses, of $10,000 per year. If you plan to keep the machine for four years, what must the market (resale) value be at the end of four years to justify the investment? You must make a 12% annual return on your investment.

Sep. 9 1997

Sol:Sol:Equivalent receipts = Equivalent expenditures

F4 + $10,000 (F/A, 12%, 4) = $100,000 (F/P, 12%, 4)

so, F4 = $100,000 (F/P, 12%, 4) - $10,000 (F/A,12%, 4) = $100,000 (1.5735) - $10,000 (4.7793) = 109,557

100,000

F4 = ?

1 2 3 4

A = 10,000

Sep. 9 1997

Homework # 1 - Chapter 3Homework # 1 - Chapter 3

Page 122Problem Number:5, 7, 8, 9, 13, 18, 22Due Date: Sep. 17 (Thursday)