Lecture Note 2 Time Value of Money

Lecture Note 2Lecture Note 2

Time Value of MoneyTime Value of Money

Department of Systems Engineering and Engineering ManagementThe Chinese University of Hong Kong

Seg2510 – Management Principles for Engineering Managers

Interest: The Cost of Money

• In financial world, money itself is a commodity.– Money costs money

• Cost of Money– measured by an interest rate

• Interest– The cost of having money available

The Time Value of Money

• Example: Buy today or buy later?

The Time Value of Money (Cont’d)

• The principle of the Time value of money– The economic value of a sum depends on when the sum is

received.

• Money has both earning power and purchasing powerover time.– A dollar received today has a greater value than a dollar

received at some future time.

• The operation of interest and the time value of money have to be considered for deciding among alternative proposals.

Elements of Transactions Involving Interest

• Types of transactions involve interest:– E.g. borrowing money, investing money, or purchasing machinery on

credit

• Common elements in transactions:– Principle (P)

• The initial amount of money invested or borrowed

– Interest rate (i)• Measures the cost or price of money• Expressed as a percentage per period of time.

– Interest period (n)• Determines how frequently interest is calculated.

– Number of interest periods (N)– A plan for receipts or disbursements (An)

• Yields a particular cash flow pattern over a specified length of time.• E.g. a series of equal monthly payments that repay a loan.

– Future amount of money (F)• Results from the cumulative effects of the interest rate over a number of

interest periods.

Example of an Interest Transaction

• An electronics manufacturing company borrows $20,000 from a bank at 9% annual interest rate to buy a machine. The company also pays a $200 loan origination fee when the loan commences.

• Two repayment plans are offered by the bank:• Principle amount, P, is $20,000.• Interest rate, i, is 9%

– Plan 1:• Interest period, n, is one year.• Duration of the transaction is five years.• Number of interest periods, N = 5

Repayment Plans Offered by the Lender

End of Year Receipts Payments

Plan 1 Plan 2

Year 0 $20,000.00 $200.00 $200.00

Year 1 5,141.85

Year 2 5,141.85

Year 3 5,141.85

Year 4 5,141.85

Year 5 5,141.85 30,772.48

Both payment plans are based on a rate of 9% interest.

– Plan 2:• A single future repayment F.

• What is it?– The flow of cash!

• End-of-period convention– The assumption of placing all cash flow transactions at the end of an

interest period

• When to use it?– Cash flow diagrams is strongly recommended for situations in which the

analyst needs to clarify or visualize what is involved when flows of money occur at various time.

Cash Flow Diagrams

1 2 3 4 50

$19,800

$5,141.85 $5,141.85 $5,141.85 $5,141.85 $5,141.85

Years

$20,000

$200

Cash flow at n=0Is a net cash flowAfter summing$20,000 and takingaway $200

i = 9%

Simple Interest

• The total interest earned is linearly proportional to the principal (loan).

• The interest do not accumulate– Mathematically,

where P = PrincipleN = number of periods (e.g. year)i = interest rate per interest period

– Question:• How much interest we need to pay if we have borrowed $1000 for 3

years with an annual interest rate of 10%?

))()(( iNPI =

Compound Interest

• The interest earned for any period is based on the remaining principal amount plus any accumulated interest charges up to the beginning of that period.– Mathematically,

– Question:• What is the interest we need to pay if we have borrowed $1000 for 3

years with an annual interest rate of 10%?

( ) ]1)1[(1)(1

1 −+=+= ∑=

− NN

n

n iPiPiI

$121$110$100Interest

$1,210$1,100$1,000Borrowed

Year 3Year 2Year 1

Compound Interest (cont’d)

• Suppose that the interest is accumulated in six months (twice a year)…– Question:

• What is the interest we need to pay if we have borrowed $1000 for 3 years with a 5% interest rate for every six months?

• Note that the interest that we have to pay is more than 10% a year!

…

…

Year 2.5

$55.1

$1152.5

Year 1.5

$50

$1,000

Year 0.5

…$60.4$52.5Interest

…$1207.6$1,050Borrowed

Year 3Year 2Year 1

Economic Equivalence

• Consider the following two option:– (a) Receive $20,000 today, and $50,000 ten years from now– (b) Receive $8,000 each year for the next 10 yearsWhich one to choose?

Economic Equivalence (Cont’d)

• Economic equivalence exists between cash flows that have the same economic effect, and could therefore be traded for one another.

• Economic equivalence refers to the fact that a cash flow – whether a single payment or a series of payments – can be converted to an equivalent cash flow at any point in time.

• Even though the amounts and timing of the cash flows may differ,the appropriate interest rate makes them equal.


• Comparison of alternatives– Compare the value of alternatives by finding the equivalent value

of alternatives at any common point in time

• Using compound interest to establish economic equivalence:


• A simple example– We invest $1,000 at 12% annual interest for five years– At the end of the investment period, our sums grow to:

– We could trade $1,000 now for the promise of receiving $1,762.34 in five years.

34.762,1$)12.01(000,1$)1( 5 =+=+= NiPF


• Which plan you would like to choose?– Plan 1: Pay interest at the end of each month and principal at the

end of fourth month

– Plan 2: Pay principal and interest in one payment

$17,000

$170 $170 $170

$170 + $17,000

0

1 2 3 4

End of the month


• Which plan you would like to choose?– Plan 3: Pay the debt in for equal end of month installments

Types of Cash Flow

• Single Cash Flow

• Equal (Uniform) Series

• Linear Gradient Series

• Geometric Gradient Series

• Irregular (Mixed) Series

$100

0 1 2 3 4 5

$100

1 2 3 4 5

$100 $100$100$100

0

$50+2G

1 2 3 4 5

$50+3G $50+4G

$50+G

$50

0

$50(1+g)2

1 2 3 4 5

$50(1+g)3

$50(1+g)4

$50(1+g)$50

0

$60

1 2 3 4 5

$50 $100$100$70

0

Interest Formulas

• List of factors for major formulas and their names

– Example:• The single-payment compound-amount factor

– Find the Future Worth (F), given the Present Worth (P), interest rate (i%) and the number of periods (N).

),,/( NiPF

)1 P(F/P,i,N i) P(F N =+=

recovery-Capital1)1(

)1(),,/(

fund-Sinking1)1(

),,/(

worth-present series -paymentEqual)1(

1)1(),,/(

amount-compound series -paymentEqual1)1(

),,/(

worth-presentpayment -Single)1(

1),,/(

amount-compoundpayment -Single)1(),,/(

−++=

−+=

−+

−+=

−−+=

+=

+=

N

N

N

N

N

N

N

N

i

iiNiPA

i

iNiFA

ii

iNiAP

i

iNiAF

iNiFP

iNiPFSingle Payment

Series

Equal-Payment (Uniform) Series

Formulas (cont’d)

• The previous six equations is enough for us to solve most of the problem.

• Remember the following steps1. Identify the problem carefully2. Draw a cash flow diagram3. Decompose the cash flow diagram such that it can be solved by

some of the previous equations

• Note:– The interest rates for all equations are “effective interest rates”

Single Cash Flows

A Loan Plan

• Suppose you borrowed $8,000 now, promising to repay the loan andaccumulated interest after 9 years at 10% interest rate, how much you need to pay at that time?

• Cash flow diagram:

0 1 2 …

9

$P = $100,000

$FWhat is this amount?

In the question, we are given:- P ($8,000), i (10%) and N (9 years)

And we need to find:- F (the future worth)

So we use:-

Answer-

NiNiPF )1(),,/( +=

58.18863$

%)101(000,8$

)1(000,8$

)9%,10,/(000,8$

9

=+=

+=

×=Ni

PFF

Single Cash Flows

An Investment Decision

• How much money you need to put into the bank now, with annual interest rate 7%, in order to save $1,000,000 in 45 years?


0 1 2 …

45

$P

$F = $1,000,000

Answer: P = $47,613

Purchasing a Tract of Land

• An investor has an option 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?


Answer: $6,302

Equal-Payment Series

• Compound-amount factor– Find F, given A, i, and N

• Present-worth factor– Find P, given A, i, and N

An Investment Plan

• If you deposit $1,000 in HSBC every year starting from the next year. How much would this amount be after 15 years if the interest rate is 5% p.a.?


0

1 2…

15

$F

14

A = $10,000

Answer: F = $21,578.60

NoteWhen using the component (F / A, i%, N), the last deposit and F are coincident at the same time. Also, A must begin from thefirst year (not now).

Borrowing Money

• Your friend wants to borrow money from you. He agrees to pay you$20,000 each year with annual interest rate of 15% for 5 years (very generous!). How much money at most should you lend to him?


Answer: P = $67,044

Lottery Problem

• A couple won a lottery game. The couple could choose between a single lump sum of $104 million jackpot or a total of $198 million paid out over 25 years (or $7.92 million per year).

• Suppose the couple could invest the money at 8% annual interest,which option should the couple choose?


Equal-Payment Series

• Sinking-Fund factor– Find A, given F, i, and N

• Capital-Recovery factor– Find A, given P, i, and N

Savings Plan

• To set up a saving plan of $100,000 in 8 years, how much do one need to save each year if the current rate of interest is 7%?


Answer: A = $9,746.78

Become A Millionaire in 30 Years

• How much do you need to invest in a bank every year if you want to yield $1,000,000 in 45 years? Assume that the interest rate is 7%


Answer: A = $3,500

Become a Millionaire By Saving $1 a Day!

• If you invest $1 each day, how many years would the money become $1,000,000? Assume that the interest rate is 10%.


Answer: N = 58.93

Mortgage Plan

• You want to buy an apartment at the price of $4,000,000. You will do this with a mortgage from Heng Seng Bank at (P-2%) for 30 years. What is your monthly payment?


Answer: A = $X (Monthly payment = $X/12)

When N is large…

• Present Value of Perpetuities• For (A / P, i, N):

– When N is large, the annual payment will be less and less, and eventually:

i

AP =

More About the Formulas

• For most of the problems, they are very complex, and we have to “combine the formulas” and “reformulate the cash flow diagram” to solve these problems.– That’s why cash flow diagram is very useful.

Investment Plan

• A father, on the day his son is born, wishes to determine what amount would have to be put into an account with 12% interest rate, so that he can withdraw $2,000 during his son’s 18th, 19th, 20th and 21st birthdays.


01 2 … 21

$P

$A = $2,000

20191817

Answer: P = $884.46

21

$A = $2,000

20191817

$F’

0 1 2 …

$P

17

$F’

Step 1 Step 2

Mortgage Plans

– Bank 1: • (P – 0.5%) for the whole duration +

4% cash rebate

– Bank 2: • (P – 2.125%) for the fist 3 years +

(P – 1.75%) for the rest of the time + $5,000 administrative fee

– Bank 3:• (P – 2%) for the whole duration

– Developer:• (P – 2%) for the whole duration for the first 70% of the price +

(P + 2%) for the rest of the purchasing price + first ten years are interest fee

Which plan should you choose (Borrow $3,000,000)?

Two Saving Plans

• Plan 1:– You have to deposit $1,000 each year for 15 years. The annual

interest rate is 10%. You can take your money out when you are 65. You can join this plan when you are 22.

• Plan 2:– You have to deposit $2,000 each year when you are 32 until 65.

The annual interest rate is 10%.

Answer: $504,010 versus $444,500!

Equivalent Cash Flow Diagram

• Determine Q in terms of H, so as to make both cash flow diagram equivalent

Linear Gradient Series

• Sometimes, engineers encounter situations involving periodic payments that increase or decrease by a constant amount, G.

0 1 2 … N3 N-1N-2

.........

4

G

2G

3G

(N - 3)G

(N - 2)G

(N - 1)G

NoteBoth Time 0 and Time 1 DO NOT contain any cash flows

Linear Gradient Series (Cont’d)

• The formulas for the uniform gradient series:

�⌋

⌉�⌊

⌈

−+−=

=

�⌋

⌉�⌊

⌈

+�

��

+−=

=

−−+=

−+=

−+++−++−+=

++−+−=

∑−

=

−−

1)1(

1

),,/)(,,/(),,/(

)1(

1111

),,/)(,,/(),,/(

]1)1[(1

)1(1

1)1(1)1(1)1(

)1,,/()2,,/()1,,/(),,/(

2

1

0

121

N

N

N

N

n

n

NN

i

N

i

NiFANiGFNiGA

iiN

ii

NiFPNiGFNiGPi

Ni

i

i

Ni

i

i

i

i

i

i

i

iAFNiAFNiAFNiGF

L

L

A Simple Example

• Suppose that the expenses are: $1,000 for the second year, $2,000 for the third year and $3,000 for the forth year. What is the present worth of the expenses? Assume that the interest rate is 15%.


Answer: P = $3,790 / A = $1,326.3

A Complicated Example

• Suppose we have a project which will expense $8,000, $7,000, $6,000 and $5,000 in Year 1, Year 2, Year 3 and Year 4, respectively. Assume that the interest rate is 15%. What is the total present expense of the project?

• Cash flow:

$8,000

$7,000

$6,000

$5,000

0 1 2 3 4

$P

Answer: P = -$19,050

Geometric Gradient Series

• Similar to the idea of linear gradient series, the geometric gradient series is used to model the situation that a fixed percentage is increased or decreased in every period

Geometric Gradient Series (cont’d)

• Find P when you are given A:

��

��

�

=

≠−

−=

��

��

�

=+

≠−

++−

=

�

��

++

+=

++=

++++++++=

++++++=

+++=

−

=

−

=

−−

−−−−

−−−

∑

∑

giiFPNA

gigi

NiPFNiFPA

gii

NA

gigi

giA

i

g

i

A

igA

igAigAiA

iAiAiA

NiFPAiFPAiFPAP

NN

N

n

n

N

n

nn

NN

NN

N

)1,,/(

)],,/)(,,/(1[

)1(

])1()1(1[

1

1

1

)1()1(

)1()1()1)(1()1(

)1()1()1(

),,/()2,,/()1,,/(

1

1

1

1

1

1

1

1

11

11

21

11

22

11

21

L

L

L

Geometric Gradient Series – An Example

• Suppose the retirement pension is $50,000 each year and • the cost of living increases at an annual rate of 5% due to inflation,

• what is the additional savings required to meet the future increase in cost of living over 35 years? (interest rate at 7%).

Geometric Gradient Series – An Example (cont’d)

• References– Lecture 4: Money-Time Relationships & Equivalence, by Gabriel Fung– Chan S. Park, Fundamentals of Engineering Economics. Prentice Hall

Lecture Note 2 Time Value of Money

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