Money-Time Relationships and Equivalence

Chapter 4. Money-Time

Relationships and Equivalence

Engineering Economy (Ed.13), Sullivan et. al., 2006, P. H.

Outline z z z z z z

Return of capital Origins of interest – Simple interest and compound interest Equivalence Cash-flow diagrams / tables Interest formulas Cash flows z

z z z

arithmetic sequences and geometric sequences

Interest rates that vary with time Nominal versus effective interest rates Continuous compounding

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4.1 Introduction z

Capital z

z

refers to wealth in the form of money or property that can used to produce more wealth

Money has time value because of the interest

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4.2 Why Consider Return to Capital? z

Capital in the form of money for the people, machines, materials, energy, and other things needed in the operation of an organization. z z

z

Equity capital – owned by individuals Debt capital (borrowed capital) – obtained form lenders

Reasons z

z

Interest and profit pay the providers of capital for forgoing its use during the time the capital is being used Interest and profit are payments for the risk the investor takes in permitting another person, or an organization, to use his or her capital

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4.3 The Origins of Interest z z

z

z

In Babylon in 2000 B.C. The idea of interest: International bankers in 575 B.C. for financing international trade Typical annual rates: 6-25%, or 40% (usury found in the Bible – Exodus 22:21-27) John Calvin, 1536, Protestant theory of usury was established

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4.4 - 4.5 Simple and Compound Interest z z

Simple Interest Compound Interest z

$1,000 loaned for three periods at an interest rate of 10% compound each period Period

Amount Owed at Beginning of Period

Interest Amount for Period

Amount Owed an End of Period

1

$1,000

$100

$1,100

2

$1,100

$110

$1,210

3

$1,210

$121

$1,331

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4.6 The Concept of Equivalence z

z

Comparison of alternatives should consider the equivalence Equivalent basis dependent on z z z z

(1) the interest rate, (2) the amount of money involved, (3) the timing of monetary receipts or expenses, and (4) the interest (profit) on invested capital and the initial capital recovered

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Example z

Borrow $17,000 and repay in 4 months with 1% interest per month on the unpaid balance of principal z

z

z

z

Plan 1 z Pay interest due at end of each month and principal at end of forth month. Plan 2 z Pay off the debt in for equal end-of-month installments (principal and interest) Plan 3 z Pay principal and interest in one payment at end of forth month. The three plans are equivalent

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4.7 Notation and Cash-Flow Diagrams and Tables z

Notations z z z z z

z

i = effective interest rate per interest period; N = number of compounding periods; P = present sum of money; F = future sum of money; A = end-of-year period cash flows

Cash flow diagram: an example z z

↑ : Cash inflows (positive cash flow), e.g. receipts ↓ : Cash outflows (negative cash flow), e.g. expense

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F = $17,690.27

0

1

2

3

4=N

Months i = 1% per month P = $17,000

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Example 4-1 z

Cash-flow diagramming z

An investment of $10,000 will produce uniform annual revenue of $5,310 for 5 years. The market value will be 2,000 at the end of year 5. Annual expenses will be $3,000 at the end of each year for operating and maintaining the project. Use the cooperation’s viewpoint to draw a cash-flow diagram.

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Example 4-2 Developing a net cash flow table z

Two feasible alternative for upgrading the heating, ventilation, and air conditioning (HAVC) system z Alternative A - Rebuild the existing HVAC system z z z

z

$18,000 32,000 2,400

Alternative B - Install a new HAVAC system that utilizes existing ductwork z z z z

z

Equipment, labor, and materials to rebuild Annual cost of electricity Annual maintenance expenses

Equipment, labor, and materials to install Annual cost of electricity Annual maintenance Replacement of a major component 4 years hence

$60,000 9,000 16,000 9,400

At the end of 8 years, the estimated market value for Alternative A is $2,000, and for Alternative B is $8,000

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4.8 Interest Formulas Relating Present and Future Equivalent Values of Single Cash Flows z

General Cash-Flow Diagram Relating Present Equivalent and Future Equivalent of Single Payments F = Future Equivalent (Find)

i = Interest Rate per Period 0 1

2

3 Period

N-2

N-1

N

P = Present Equivalent (Given) 9530IEEM281: Engineering Economics, NTHU. C.-Y. Kuo, Lab. Soft Computing and Fuzzy Optimization.

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4.8.1 Finding F when Given P z

F = P (1 + i ) z

z

z

N

The (1 + i ) is called single payment compound amount factor with interest rate i%. The functional symbol (F/P, i%, N) can be used for (1 + i )N . Hence, N

F = P(F/P, i%, N).

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4.8.2 Finding P when Given F z

P = F (1 + i )

−N

z

z

z

−N The (1 + i ) is called single payment present worth factor and the functional symbol is (P/F, i%, N). Hence,

P = F(P/F, i%, N).

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4.9 Interest Formulas Relating a Uniform Series (Annuity) to Its Present and Future Equivalent Values A = Uniform Amount (Given) A A A A A

A

0 1

2

3 N-2 N-1 Period i = Interest Rate per Period

P = Present Equivalent (Find)

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N

F = Future Equivalent (Find)

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4.9.1 Finding F when Given A z

⎡ (1 + i )N − 1⎤ F = A⎢ ⎥ i ⎣ ⎦ z

z

The quantity [(1 + i ) − 1 i ] is called the uniform series compound amount factor and the functional symbol is (F/A, i%, N), hence, N

F = A(F/A, i%, N)

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z

Example 4-5 z

z

21

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

Example 4-6 z

„If you are 20 years of age and save $1.00 each day, you can become a millionaire.“ Assumption: live to 80 years old, and the annual interest rate is 10%.

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4.9.2 Finding P when Given A ⎡ (1 + i )N − 1⎤ P = A⎢ N ⎥ ( ) i i 1 + ⎦ ⎣

z

z

z

The quantity in brackets is called the uniform series present worth factor and the functional symbol is (P/A, i%, N), hence,

P = A(P/A, i%, N)

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z

Example 4-7 z

z

23

i = 15% per year. Maintain a machine can add $20,000 revenue at the end of each year of 5 years. How much is the investment? z P = A(P/A, i%, N) = 20,000(P/A, 15%, 5) = 20,000(3.3522) = 67,044

Example 4-8 z

Suppose that your 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, z How many years will it take to completely deplete the account? z How long will it take if the account earns 8% interest per year instead of 6%? ƒ

1,000,000 = 100,000(P/A, i%, N) = 10,000(P/A, 6%, N) (P/A, 6%, N) = 10, (P/A, 6%, 15) = 9.7122 , (P/A, 6%, 16) = 10.1059

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4.9.3 Finding A when Given F ⎡ ⎤ i A = F⎢ ⎥ N ( ) i 1 + − 1 ⎣ ⎦

z

z

z

The quantity in brackets is called the sinking fund factor and the functional symbol is (A/F, i%, N), hence,

A = F(A/F, i%, N).

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z

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Example 4-9 z

One who is 20 years old now hopes to have 1,000,000 savings when she at age 65. what equal end-of-year amount must she save over the next 45 years with annual 7% interest rate?

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4.9.4 Finding A when Given P z

⎡ i (1 + i )N ⎤ A = P⎢ ⎥ N ( ) i + − 1 1 ⎣ ⎦ z

z

The quantity in brackets is called the capital recovery factor and the functional symbol is (A/P, i%, N), hence,

A = P(A/P, i%, N) z

Ex. Plan 2 in Table 4-1 z

A = $17,000(A/P, 1%, 4) = $17,000(0.2563) = $4,357.10

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4.10 Interest Formulas for Discrete Compound and Discrete Cash Flows

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4.11 Deferred Annuities (Uniform Series Time Present 0

1

J-1 Period

J

J+1

J+2

N-1

N

i% z

z

All annuities discussed to this point involve the first cash flow being made at the end of the first period, and they are called ordinary annuities. If the cash flow does begin until some later date, the annuity is known as a deferred annuity.

P0 = A(P/A, i%, N - J)(P/F, i%, J)

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z

Example 4-10 z

z

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12% per year interest. One day a new baby born. His father wishes to determine what lump amount could provide $2,000 on his 18th, 19th, 20th, and 21st birthdays.

Example 4-11 z

Determine the equivalent worth of the four $2,000 as of the son´s 24th birthday.

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4.12 Equivalence Calculations Involving Multiple Interest Formulas

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Example 4-12 z

Cash flows over 8 years. The amounts are $100 for the first year, $200 for the second year, $500 for the third year, and $400 for each year from the 4th through the 8th.

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Example 4-13 z

Transform the cash flows on the left-hand side to their equivalent cash flows on the right-hand side. 2H

Q H

0

1

2

3

4

5

6

7

8

0

1

End of Year (EOY)

2

3

4

5

7

6

End of Year (EOY) Q

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4.13 Interest Formulas for A Uniform (Arithmetic) Gradient of Cash Flows z

Receipts or expenses are projected to increase or decrease by a uniform amount each period, thus constituting an arithmetic sequence of cash flows. This situation can be modeled as a uniform gradient of cash flows z Notice: the first cash flow occurs at the end of period 2 (N-1)G (N-2)G (N-3)G 3G 2G G 0

1

2

3

4

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N-2

N-1

N 34

z

4.13.1 Finding F when Given G z

z

z

F = G(F/A, i%, N - 1) + G(F/A, i%, N - 2) + … + G(F/A, i%, 2) + G(F/A, i%, 1) F=

G NG ( F/A, i%, N ) − i i

4.13.2 Finding A when Given G z

z

The quantity in brackets is called the gradient to uniform series conversion factor and the functional symbol is (A/G, i%, N), hence A = G(A/G, i%, N)

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4.13.3 Finding P when Given G ⎧⎪1 ⎡ (1 + i )N − 1 N − P = G⎨ ⎢ N (1 + i )N ⎪⎩ i ⎣ i (1 + i )

z

z

The quantity in braces is called the gradient to present equivalent conversion factor and the functional symbol is z

z z

⎤ ⎫⎪ ⎥ ⎬. ⎦ ⎪⎭

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

also can be expressed as (P/G, i%, N), hence, P = G(P/G, i%, N)

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Example 4-15 z

End-of-year cash flows are expected to be $1,000 for the 2nd year, $2,000 for the 3rd year, and $3,000 for the 4th year. If the interest is 15% per year, find that z z

(a) present equivalent value at the beginning of the first year, (b) uniform annual equivalent value at the end of the four years.

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Example 4-16 z

Calculate the present equivalent at i = 15% per year, using arithmetic gradient interest factors End of Year

Cash Flows ($)

1 2 3 4

-8,000 -7,000 -6,000 -5,000

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4.14 Interest Formulas Relating A Geometric Sequence of Cash Flows to Its Present and Annual Equivalents z

The cash flow patterns change at an average rate f . And the resultant end-of-period cash flow pattern is referred to a geometric gradient series.

⎧ A1 [1 − ( P / F , i %, N )( F / P, f %, N )] ⎪ P=⎨ i− f ⎪ A N ( P / F , i %,1) ⎩ 1

f ≠i f = i.

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Example 4-20 z

Consider the end-of-year geometric sequence of cash flow and determine the P, A, and F equivalent values. The rate of increase is 20% per year after the first year, and the $1,000(1.2) interest rate is 25% per year.

3

$1,000(1.2)2 $1,000(1.2)1 $1,000

0

1

2

3

4

End of Year

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Example 4-19 z

As the same in Example 4-18, suppose that the geometric gradient decreases by 20% per year after the first year. Determine P, A, and F under this condition.

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4.15 Interest Rates That Vary with Time z

Example 4-21 z

A person has made an arrangement to borrow $1,000 now and another $1,000 two years hence. If the projected interest rates in years one, two, three, and four are 10%, 12%, 12%, and 14%, respectively, how much will be repaid as a lump-sum amount at the end of four years.

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4.15 Interest Rates That Vary with Time z

Obtaining the present equivalent of a series of future cash flows subjected to vary interest rates P=

z

FN . N ∏k =1 (1 + ik )

For instance, if F4 = $1,000 and i1 = 10%, i2 = 12%, i3 = 13%, i4 = 10%, then z

P = $1,000[(P/F, 10%, 1)(P/F, 12%, 1) (P/F, 13%, 1)(P/F, 10%, 1)]

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4.16 Nominal and Effective Interest Rate z z

The interest period is less than one year E.g. 6% interest rate per half-year v.s. 12% per year z

12% is the nominal, but effective rate is larger than 12% z 1st half-year $1,000 + $1,000 × (0.12/2) = $1,000 + $60 = $1,060 z 2nd half-year $1,060 + $1,060 × (0.12/2) = $1,060 + $63.6 = $1,123.6 z The effective annual interest rate for the entire year is

$123.6 × 100% = 12.36%. $1,000 9530IEEM281: Engineering Economics, NTHU. C.-Y. Kuo, Lab. Soft Computing and Fuzzy Optimization.

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