chapter 2

TIME VALUE OF MONEY

CHAPTER 2

1

Time Value of Money

The Cost of Money is established and measured by an interest rate, a percentage that is periodically applied and added to an amount of money over a specified length of time.

Economic Equivalence

Interest Formulas – Single Cash Flows

Equal-Payment Series

Dealing with Gradient Series

Composite Cash Flows.

2

Time Value of Money

Money has a time value because it can earn more money over time (earning power).

Money has a time value because its purchasing power changes over time (inflation).

Time value of money is measured in terms of interest rate.

Interest is the cost of having money available for use - a cost to the borrower and an earning to the lender

3

Elements of Transactions involve Interest

1. Initial amount of money in transactions involving debt or investments is called the principal (P).

2. The interest rate ( i ) measures the cost or price of money and is expressed as a percentage per period of time.

3. A period of time, called the interest period (n), determines how frequently interest is calculated.

4. A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods (N).

5. A plan for receipts or disbursements (An) that yields a particular cash flow pattern over a specified length of time. [monthly equal payment]

6. A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.

5

EXAMPLE OF INTEREST TRANSACTION

Suppose that you apply for an education loan of $30,000 from a bank at a 9% annual interest rate. In addition you pay a $300 loan origination fee when the loan begins.

The bank offers two repayment plans, one with equal payments made at the end of every year for the next five years (installment plan) and the other with a single payment made after the loan period of five years (deferment plan). 6

Which Repayment Plan?Table 2.1 Repayment plan offered by the lender

End of Year Receipts Payments

Plan 1 Plan 2

Year 0 $30,000.00 $300.00 $300.00

Year 1 $7,712.77 0

Year 2 $7,712.77 0

Year 3 $7,712.77 0

Year 4 $7,712.77 0

Year 5 $7,712.77 46,158.72

The amount of loan = $30,000 & origination fee = $300 & interest rate = 9% APR F = P (1 + i )N = P (1+0.09)5 = $30,000 x (1.09)5 = $46,158.72

7

Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example

8

Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example

1)1(

)1(),,/(

N

N

i

iiPANiPA

1)09.01(

)09.01(09.0)000,30(

5

5

PA = $7,712.77

1)1(

),,/(Ni

iFANiFA

1)09.01(

09.072.158,46$

5A =

$7,712.77

9

Methods of Calculating Interest

10

1

22 1

0 :

1: (1 )

2 : (1 ) (1 )

: (1 )N

n P

n F P i

n F F i P i

n N F P i

Practice Problem

Problem Statement

If you deposit $100 now (n = 0) and $200 two years from

now (n = 2) in a savings account that pays 10% interest,

how much would you have at the end of year 10?

14

Solution

0 1 2 3 4 5 6 7 8 9 10

$100$200

F

10

8

$100(1 0.10) $100(2.59) $259

$200(1 0.10) $200(2.14) $429

$259 $429 $688F

15


What do we mean by “economic equivalence?”

Why do we need to establish an economic equivalence?

How do we establish an economic equivalence?

16

Which option would you prefer?


How do we know, whether we should prefer to have $20,000 today and

$50,000 ten years from now, or $8,000 each year for the next ten years?

17

Figure 2-4 Using compound interest to establish economic equivalence

Equivalence Calculation: A Simple example

18

Equivalence relation between P and F.

19

Figure 2-6 Equivalent worth calculation at n = 3

EXAMPLE 2.3 Equivalence Calculation

FIND: V3 (equivalent worth at n = 3) and i = 10%.

22

Step 1: $100(1+0.1)3+ $80(1+0.1)2+$120(1+0.1)1+$150

= $511.90

Step 2: $200(1+0.1)-1+ $100(1+0.1)-2 = $264.46

Step 3: V3= $511.90 + $264.46 = $776.36

Figure 2-7 Compounding process: Find F, given P, i, and N

Interest Formulas for Single Cash FlowsCompound Amount Factor

23

Interest Rate Factors (10 %)

24

Figure 2-8 Cash flow diagram

Example 2.4 If you had $1,000 now and invested it at 7% interest compounded annually, how much would it be worth in 8 years?

25

Given: P = $1,000, i = 7 %, and N = 8 years; Find: F

F = $1,000 (1+0.07)8 = $1,718.19 or using this

F = P (F/P, i, N) factor notation together with table value

F = $1,000 (1.7182) = $1,718.19

Present -Worth Factor

26


Example 2.5A zero coupon (installment) bond is a popular variation on the bond theme for some investors. What should be the price of an eight year (maturity) zero-coupon with a face value (future value) of $1,000 if similar, nonzero coupon bonds are yielding 6% annual interest?

27

Given: F = $1000, i = 6%, and N = 8 years Find: P P = $1,000 (1+0.06)-8 = $1,000 (0.6274) = $627.40

Example 2.6 Suppose you buy a share of stock for $10 and sell it for $20; your profit is thus $10. it takes five years to gain this profit, what would be the rate of return on your investment?

1+ i = = 1.14869 i = 1.14869 – 1 = 0.14869 or 0.1487 = 14.87%

28

Given: P = $10, F = $20 and N = 5 years, Find: i

F = $20 = $10 (1+ i )5 $2 = (1+ i )5 = (F / P, i, 5)

Example 2.7You have just purchased 200 shares of GE stock at $15 per share. You will sell the stock when its market price doubles. If you expect the stock price to increase 12% per year, how long do you expect to wait before selling the stock?


29

Given: P = $3,000, F = $6,000 and i = 12%, Find: N F = P (1+ i )N = P (F/P, i, N) …… 6,000 = $3,000 (1+ 0.12)N

2 = (1.12) N ………… Log 2 = N log 1.12 solve for N

gives

N = log 2 / log 1.12 = 0.301 / 0.049 = 6.116 ≈ 6.12 years

Given: i = 10%,

Find: C that makes the two cash flow streams to be indifferent

Practice Problem

$500

$1,000

0 1 2 3

0 1 2 3

A

B

C C

30

Approach

Step 1:

Select the base period to use, say n = 2

Step 2:

Find the equivalent lump sum value at n = 2 for both A and B.

Step 3:

Equate both equivalent values and solve for unknown C.

$500

$1,000

0 1 2 3

0 1 2 3

A

B

C C

31

Solution

For A:

For B:

To Find C:

2 12 $500(1 0.10) $1,000(1 0.10)

$1,514.09

V $500

$1,000

0 1 2 3

0 1 2 3

A

B

C C

2 (1 0.10)

2.1

V C C

C

2.1 $1,514.09

$721

C

C

32

At what interest rate

would you be

indifferent between

the two cash flows?

$500

$1,000

0 1 2 3

0 1 2 3

$502 $502 $502

A

B

Practice Problem

33

Approach

Step 1:

Select the base period to compute

the equivalent value (say, n = 3)

Step 2:

Find the net worth of each at n = 3.

$500

$1,000

0 1 2 3

0 1 2 3

$502 $502 $502

A

B

34

Establish Equivalence at n = 3

33

23

Option A : $500(1 ) $1,000

Option B : $502(1 ) $502(1 ) $502

F i

F i i

33

23

Option A : $500(1.08) $1,000

$1,630

Option B : $502(1.08) $502(1.08) $502

$1,630

F

F

Find the solution by trial and error, say i = 8%

35

Practice Problem

• You want to set aside a lump sum amount

today in a savings account that earns 7%

annual interest to meet a future

investment in the amount of $10,000 to

be incurred in 6 years.

• How much do you need to deposit today?

36

Solution

F = $10,000; N = 6 years; i = 7 %; Find P

0

6

$10,000

P

6$10,000(1 0.07)

$10,000( / ,7%,6)

$6,663

P

P F

37

chapter 2

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