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Introduction to Engineering Economy

Engineering Economics

Topic 6: The Time Value of Money and Economic Equivalence

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

The Time Value of Money

The Economic Equivalence

Some Useful Formulas for Establishing Economic Equivalence

Miscellaneous Topics of Interest Rates and Compounding

Contents of Topic 6

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Cost-benefit Analysis and Time Value of Money

To make rational decisions as an engineer, you should apply the cost-benefit analysis (CBA) to study the cash flows of an investment project.

Usually benefits (B) are cash inflows (e.g. receipts) and costs (C) are cash outflows (e.g. expenses).

As these cash flows occur at different points in time, to compare B and C properly, you need to consider the time value of money by asking:

“How much will $100 today be worth one year from now?” or “What is today’s worth of the $100 I will receive or have to pay one year later?”

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Causes and Effects of Time Value of Money

Determinants of Time Value of Money:

People’s preference: The more they prefer current consumption over future consumption the higher the time value of money

The productivity of financial capital (money): The higher its return in the future the higher the time value of money

The time value of money is reflected by the interest rate at which the value of such financial capital grows over time (denoted by i in Topics 6-9).

Effects of Time Value of Money on CBA:

To properly conduct CBA for an investment project, its cash flows at different time points must be converted to their corresponding values (economic equivalence) at the same point in time based on the time value of money.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

An investment project requires the purchase of physical capital.

It must be financed by money (financial capital).

Such an amount cannot be available for alternative use throughout the project life.

Supplier(s) of this financial capital (entrepreneur or creditors) must be sufficiently compensated for the time value of money.

Recall the factors of production and their respective rewards in Topic 3.

The Return to Capital

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

The Return to Capital

An investment project requires the purchase of physical capital.

The amount paid back to the supplier(s) must also include the return to financial capital.

At equilibrium, return to physical capital = cost of financial capital to the entrepreneur = return to financial capital for supplier(s) = actual interest payment (explicit cost) or forgone interest incomes (implicit cost)

Recall the factors of production and their respective rewards in Topic 3.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest Rate and Economic Equivalence

Interest rate is the rate at which an amount of financial capital (money) grows in value over a specific period of time. Usually the period is a year and the interest rate quoted is annual percentage rate (APR).

If the APR = i, then “$100 today” (a present value, P) is worth or economically equivalent to “$100(1+i) one year from now” (a future value, F);

If i = 10% and P = $100, then its economic equivalence one year from now is F = P (1+i) = $100(1 + 0.10) = $110.

F = P (1+i)

and “$100/(1+i) today” (P) is the economic equivalence of “$100 one year from now” (F).

P = F /(1+i)

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Suppose you are given the following two options of student aid for the new academic year.

Option 1: You receive $10,000 on September 1 when the academic year begins.

Option 2: You receive $5,000 on September 1 and $5,000 on March 1 the following year.

APR is 10% and the interest is compounded every 6 months

the corresponding interest rate for a six-month period (i) is 5%.

Which option will you prefer?

Interest Rate and Economic Equivalence

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Two ways to compare the two options:

Comparing their economic equivalence at March 1 the following year (future value):

Option 1: Derive the future value (F1) at March 1 the following year (6 months later) of $10,000 received on September 1 (present value, P1) according to the equation

F1 = P1(1 + i)= $10,000(1.05)= $10,500

Option 2: Derive the future value at March 1 the following year of the two amounts received as follows.

F2 = $5,000(1.05) + $5,000 = $10,250

F1 > F2 Option 1 is preferred to Option 2.

Interest Rate and Economic Equivalence

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Two ways to compare the two options:

Comparing their economic equivalence at September 1 (present value):

Option 1: P1= $10,000

Option 2: Derive the present value of the two amounts received as follows.

P2 = $5,000 + $5,000/1.05 = $9,761.90

P1 > P2 Option 1 is preferred to Option 2.

Interest Rate and Economic Equivalence

The two methods should give consistent results. Which method is better?

The second method is better because it is easier.

Always choose the easier method.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Notations and Cash-Flow Diagrams

Economic equivalence of the cash flows that characterizes a given investment project depends on:

1. the interest rate;

2. the amounts of cash flows involved; and

3. the timing of the cash inflows and outflows.

Cash flow diagrams with the time dimension (e.g. Figure 6.1) help us visualise the flows of money occurring at various time points.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Usual conventions for cash flow diagrams:

The horizontal line is a time scale with the number denoting the end of a specific period (e.g. 0 is now and 1 is the end of period 1)

Cash flows are depicted by arrows placed at the end of their corresponding periods on the time scale.

Larger (smaller) amount of cash flow is depicted by longer (shorter) arrow.

Downward arrows stand for cash outflows (expenses, deposits in bank accounts, etc.)

Upward arrows stand for cash inflows (receipts, withdrawals from bank accounts, etc.)

3. The arrows for known amounts are in solid lines. The arrows for unknown amounts are in dotted lines.

Notations and Cash-Flow Diagrams

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Suppose the following cash flow diagram describes the deposits you will put into your bank account at the end of each of the coming six years. If your bank offers you an annual interest rate of 10%, how much money will you have in your bank account at the end of the 6th year?

Notations and Cash-Flow Diagrams

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Notations used in formulas to derive economic equivalence:

i: effective interest rate per interest (or compounding) period

N: number of interest (or compounding) periods

P: present value, which is the equivalent value of one or more cash flows at a reference time point called the “present”

F: future value, which is the equivalent value of one or more cash flows at a reference time point called the “future”

A: uniform end-of-period cash flows that begin at the end of period 1 and continue for a specific duration (N periods)

Notations and Cash-Flow Diagrams

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

The variables i, N, P, and F are related to one another in the establishment of economic equivalence.

The unknown value of one of these four variables can be derived when the values of the other three are given.

Three simple rules for cash flow calculations:

Cash flows cannot be added or subtracted unless they occur at the same point in time.

To move a cash flow forward in time, we perform compounding.

To move a cash flow backward in time, we perform discounting.

Deriving Present and Future Equivalent Values

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

We can apply compound interest formulas to “move” cash flows along the cash flow diagram.

In a similar way we can find P given F by

Deriving Present and Future Equivalent Values

Figure 4.5 in textbook

Future Equivalent of Single Payment

A present amount, P, can grow into a future amount, F, in N time periods at interest rate i according to this formula:

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Deriving Present and Future Equivalent Values

Suppose the following cash flow diagram describes the deposits you will put into your bank account at the end of each of the coming six years. If your bank offers you an annual interest rate of 10%, how much money will you have in your bank account at the end of the 6th year?

F1 = $200(1+10%)5 = $322.1

F2 = $150(1+10%)4 = $219.62

F3 = $100(1+10%)3 = $133.1

F4 = $100(1+10%)2 = $121

F5 = $150(1+10%) = $165

F = F1 + F2 + F3 + F4 + F5 + F6 = $1,160.82

Redo this question later using the interest factor table. Which method is better?

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 10%

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Finding the Unknown Values of F, P, i or N

Finding F when given P, N, and i:

Finding P when given F, N, and i:

This is called the single payment compound amount factor. The term on the right is read “F given P at i% interest per period for N interest periods.”

This is called the single payment present worth factor. The term on the right is read “P given F at i% interest per period for N interest periods.”

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Finding the Unknown Values of F, P, i or N

Finding i when given P, F, and N:

Given that , we can obtain the following expression for i:

Using logarithms,

Finding N when given P, F, and i:

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

$2,500 at time zero is equivalent to how much after six years if the interest rate is 8% per year?

$3,000 at the end of year seven is equivalent to how much today (time zero) if the interest rate is 6% per year?

Finding the Unknown Values of F, P, i or N

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 8%

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

$2,500 at time zero is equivalent to how much after six years if the interest rate is 8% per year?

F = $2,500 (F/P, 8%, 6) = $2,500 (1.587) = $3,968

Finding the Unknown Values of F, P, i or N

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 6%

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

$3,000 at the end of year seven is equivalent to how much today (time zero) if the interest rate is 6% per year?

Finding the Unknown Values of F, P, i or N

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Self-assessment (Self Exercise 1, Unit 6)

F1, F2 and F4 in the following cash flow diagram are the payments you have to make one year, two years and four years from now respectively. To prepare for such foreseeable payments, you have decided to put aside enough money and keep it in your bank deposit. If your bank pays you 10% interest per year for your deposit, how much money do you have to put aside now?

P = F1(P/F, 10%, 1) + F2(P/F, 10%, 2) + F4(P/F, 10%, 4)

= $25,000(1+0.1)-1 + $3,000(1+0.1)-2 + $5,000(1+0.1)-4

= $25,000(0.9091) + $3,000(0.8264) + $5,000(0.6830)

= $28,621.7

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 10%

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Some useful formulas for Establishing Economic Equivalence

Figure 4-6 in textbook

General Cash-Flow Diagram Relating Uniform Series (Ordinary Annuity) to Its Present Equivalent and Future Equivalent Values

Note that when the number of cash flows increases, it will be very time-consuming to calculate the present or future equivalence of the individual cash flows.

If they follow some regular pattern (e.g. a uniform series as shown in the following diagram), then we can simplify the calculations tremendously by using some formulas.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Future Equivalence of a Uniform Series (Annuity)

How much will you have in 40 years if you save $3,000 each year and your bank deposits earn 8% interest each year?

F = $3,000 (F/A, 8%, 40) = $3,000 (259.057) = $777,171

Some useful formulas for Establishing Economic Equivalence

Uniform series compound amount factor:

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 8%

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Present Equivalence of a Uniform Series (Annuity)

Some useful formulas for Establishing Economic Equivalence

How much would you need today to provide an annual amount of $50,000 each year for 20 years, at 9% interest each year?

Uniform series present worth factor:

P = $50,000(P/A, 9%, 20) = $50,000(9.129) = $456,450

Upper Saddle River, 07458

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 9%

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Finding A when given F, i and N:

How much would you need to set aside each year for 25 years, at 10% interest, to have accumulated $1,000,000 at the end of the 25 years?

Some useful formulas for Establishing Economic Equivalence

Sinking fund factor:

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

If you had $500,000 today in a bank account earning 10% each year, how much could you withdraw each year for 25 years?

Some useful formulas for Establishing Economic Equivalence

Finding A when given P, i and N:

Capital recovery factor:

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 10%

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

More examples from the textbook

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Please do Self Exercises 2, 3 and 4 in Self-study Notes Unit 6 for more practice.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Summary table in textbook

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Deferred Annuities

Deferred annuities are uniform series that do not begin until some time in the future.

If the annuity is deferred J periods, then the first payment (cash flow) begins at the end of period J+1.

Figure 4-9 (in textbook)

General Cash-Flow Representation of a Deferred Annuity (Uniform Series)

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Finding the value at time 0 of a deferred annuity

Use (P/A, i%, N-J) find the value of the deferred annuity at the end of period J (where there are N-J cash flows in the annuity).

Use (P/F, i%, J) to find the value of the deferred annuity at time zero.

It involves a two-step process.

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Engineering Economy, Fifteenth Edition

By William G. Sullivan, Elin M. Wicks, and C.

Irene just purchased a new sports car and wants to also set aside cash for future maintenance expenses. The car has a bumper-to-bumper warranty for the first five years. Irene estimates that she will need approximately $2,000 per year in maintenance expenses for years 6-10, at which time she will sell the vehicle. How much money should Irene deposit into a bank account today, at 8% per year, so that she will have sufficient funds in that account to cover her projected maintenance expenses?

Exercise on deferred annuities:

P0 = $2,000 (P/A, 8%, 10-5) (P/F, 8%, 5)

= $2,000(3.993)(0.6806)

= $5,435.3

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Engineering Economy, Fifteenth Edition

By William G. Sullivan,

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