Chapter 4

Capital equipment

Payback period

• 1. Calculating Payback What is the payback period for the following set of cash flows.

• To calculate the payback period, we need to find the time that the project has recovered its initial investment. After two years, the project has created:

• $1,200 + 2,500 = $3,700 in cash flows. • The project still needs to create another: • $4,800 – 3,700 = $1,100 in cash flows.• During the third year, the cash flows from the project will be

$3,400. So, the payback period will be 2 years, plus what we still need to make divided by what we will make during the third year.

• The payback period is: Payback = 2 + ($1,100 / $3,400) = 2.32 years

Return On Investment

• The benefit (return) of an investment is divided by the cost of the investments; the result is expressed as a percentage or a ratio. This is referred to as “simple ROI”.

ROI= Gains from investment – Cost of investment Cost of Investment

$700,000 - $500,000 = 40% $500,000

Return On Investemnt

• ROI is used to compare returns on investment where the money gained or lost—or the money invested—are not easily compared using monetary values. For example, a $1,000 investment that earns $50 in interest obviously generates more cash than a $100 investment that earns $20 interest, but the $100 investment earns a higher return.

$50/$1,000 $1050 - $1000 = 50 = 5% ROI $1000 $1000$20/$100 $120-100 = 20 = 20% ROI $100 $100

Cash flow method

–Cash Defined -- refers to cash and cash equivalents.

–Cash equivalents are short-term, highly liquid investments that are (1) readily convertible to known amounts of cash, and (2) near maturity (typically within 3 months) with limited risk of price changes due to interest rate shifts.

–Cash is the beginning and the end of a company’s operating cycle.

–Net cash flow is the end measure of –profitability.

–Cash repays loans, replaces equipment, expands facilities, and pays dividends.

–Analyzing cash inflows and outflows helps assess liquidity, solvency, and financial flexibility.

• Liquidity is the nearness to cash of assets and liabilities.

• Solvency is the ability to pay liabilities when they mature.

• Financial flexibility is the ability to react to opportunities and adversities.

Developing Project Cash Flows

• Estimating Cost/Benefit for Engineering Projects

• Incremental Cash Flows• Developing Cash Flow Statements• Generalized Cash Flow Approach

10

Classification of Investment Projects

Project

Profit-addingproject

Profit-maintainingproject

Expansion project

Product Improvement project

Necessity project

Replacement Project

Cost Improvementproject

Cash Flow Diagram

Cash Flow ElementOperating activities:

Other Terms Used in Business

Gross income Gross revenue, Sales revenue, Gross Profit, Operating revenue

Cost savings Cost reduction

Manufacturing expenses Cost of goods sold, Cost of revenue

O&M cost Operating expenses

Operating income Operating profit, Gross margin

Interest expenses Interest payments, Debt cost

Income taxes Income taxes owed

Investing activities

Capital investment Purchase of new equipment, Capital expenditure

Salvage value Net selling price, Disposal value, Resale value

Investment in working capital Working capital requirement

Working capital release Working capital recovery

Gains taxes Capital gains taxes, Ordinary gains taxes

Financing activities:

Borrowed funds Borrowed amounts, Loan amount

Principal repayments Loan repayment

(c) 2001 Contemporary Engineering Economics

13

A Typical Format used for Presenting Cash Flow Statement

Income statement Revenues Expenses Cost of goods sold Depreciation Debt interest Operating expensesTaxable incomeIncome taxesNet income

Cash flow statement

+ Net income+Depreciation

-Capital investment+ Proceeds from sales of depreciable assets- Gains tax- Investments in working capital+ Working capital recovery

+ Borrowed funds-Repayment of principal Net cash flow

Operatingactivities

Investing activities

Financingactivities

+

+

When Projects Require only Operating and Investing Activities

• Project Nature: Installation of a new computer control system • Financial Data:

– Investment: $125,000– Project life: 5 years– Salvage value: $50,000– Annual labor savings: $100,000– Annual additional expenses:

• Labor: $20,000• Material: $12,000• Overhead: $8,000

– Depreciation Method: 7-year MACRS– Income tax rate: 40%

(c) 2001 Contemporary Engineering Economics

15

Example 12.1 - Net Cash Flow Table Generated by Traditional Method Using Approach 2

A B C D E F G H I JYear End

Investment & Salvage Value

Revenue Labor Expenses Materials

Overhead

Depreciation

Taxable Income

Income Taxes

Net Cash Flow

0 -$125,000 -$125,000

1 $100,000 20,000

12,000 8,000 $17,863 42,137 16,855 $43,145

2 100,000 20,000

12,000 8,000 30,613 29,387 11,755 $48,245

3 100,000 20,000

12,000 8,000 21,863 38,137 15,255 $44,745

4 100,000 20,000

12,000 8,000 15,613 44,387 17,755 $42,245

5 100,000 20,000

12,000 8,000 5,581 54,419 21,678 $38,232

50,000* 16,525 6,613 $43,387

Information required tocalculate the income taxes*Salvage value

Note thatH = C-D-E-F-GI = 0.4 * HJ= B+C-D-E-F-I

Time value money• The Interest Rate

Obviously, $10,000 today.

You already recognize that there is TIME VALUE TO MONEY!!

Which would you prefer -- $10,000 today or $10,000 in 5 years?

Why time

• TIME allows you the opportunity to postpone consumption and earn INTEREST.– Types of Interest Simple Interest

• Interest paid (earned) on only the original amount, or principal, borrowed (lent).

– Compound Interest• Interest paid (earned) on any previous interest earned,

as well as on the principal borrowed (lent).

Simple Interest Formula

Formula SI = P0(i)(n)SI: Simple InterestP0: Deposit today (t=0)

i: Interest Rate per Periodn: Number of Time Periods

Example

• Assume that you deposit $1,000 in an account earning 7% simple interest for 2 years. What is the accumulated interest at the end of the 2nd year?

• SI = P0(i)(n)= $1,000(.07)(2)= $140

Simple Interest (FV)

• What is the Future Value (FV) of the deposit?FV = P0 + SI

= $1,000 + $140= $1,140

• Future Value is the value at some future time of a present amount of money, or a series of payments, evaluated at a given interest rate.

The Present Value is simply the $1,000 you originally deposited. That is the value today!

• Present Value is the current value of a future amount of money, or a series of payments, evaluated at a given interest rate.

Simple Interest (PV)Simple Interest (PV)

• What is the Present Value (PV) of the previous problem?

0

5000

10000

15000

20000

1st Year 10thYear

20thYear

30thYear

Future Value of a Single $1,000 Deposit

10% SimpleInterest

7% CompoundInterest

10% CompoundInterest

Why Compound Interest?Why Compound Interest?Fu

ture

Val

ue (U

.S. D

olla

rs)

Assume that you deposit $1,000 at a compound interest rate of 7% for 2 years.

Future ValueSingle Deposit (Graphic)

Future ValueSingle Deposit (Graphic)

0 1 2

$1,000FV2

7%

FV1 = P0 (1+i)1 = $1,000 (1.07)= $1,070

Compound InterestYou earned $70 interest on your $1,000

deposit over the first year.This is the same amount of interest you would

earn under simple interest.

Future ValueSingle Deposit (Formula)

Future ValueSingle Deposit (Formula)

Future ValueSingle Deposit (Formula)Future ValueSingle Deposit (Formula)

FV1 = P0 (1+i)1 = $1,000 (1.07)

= $1,070

FV2 = FV1 (1+i)1

= P0 (1+i)(1+i) = $1,000(1.07)(1.07) = P0 (1+i)2

= $1,000(1.07)2

= $1,144.90

You earned an EXTRA $4.90 in Year 2 with compound over simple interest.

FV1 = P0(1+i)1

FV2 = P0(1+i)2

General Future Value Formula:FVn = P0 (1+i)n

or FVn = P0 (FVIFi,n) -- See Table I

General Future Value Formula

General Future Value Formula

etc.

FVIFi,n is found on Table I

at the end of the book.

Valuation Using Table IValuation Using Table I

Period 6% 7% 8%1 1.060 1.070 1.0802 1.124 1.145 1.1663 1.191 1.225 1.2604 1.262 1.311 1.3605 1.338 1.403 1.469

FV2 = $1,000 (FVIF7%,2)= $1,000 (1.145)

= $1,145 [Due to Rounding]

Using Future Value TablesUsing Future Value Tables

Period 6% 7% 8%1 1.060 1.070 1.0802 1.124 1.145 1.1663 1.191 1.225 1.2604 1.262 1.311 1.3605 1.338 1.403 1.469

Julie Miller wants to know how large her deposit of $10,000 today will become at a compound annual interest rate of 10% for 5 years.

Story Problem ExampleStory Problem Example

0 1 2 3 4 5

$10,000FV5

10%

• Calculation based on Table I:FV5 = $10,000 (FVIF10%, 5)

= $10,000 (1.611)= $16,110 [Due to Rounding]

Story Problem SolutionStory Problem Solution

Calculation based on general formula: FVn = P0 (1+i)n

FV5 = $10,000 (1+ 0.10)5

= $16,105.10

3-31

Types of AnnuitiesTypes of Annuities

Ordinary Annuity: Payments or receipts occur at the end of each period.

Annuity Due: Payments or receipts occur at the beginning of each period.

An Annuity represents a series of equal payments (or receipts) occurring over a specified number of equidistant periods.

3-32

Examples of Annuities

Student Loan Payments Car Loan Payments Insurance Premiums Mortgage Payments Retirement Savings

3-33

Parts of an AnnuityParts of an Annuity

0 1 2 3

$100 $100 $100

(Ordinary Annuity)End of

Period 1End of

Period 2

Today Equal Cash Flows Each 1 Period Apart

End ofPeriod 3

3-34

Parts of an AnnuityParts of an Annuity

0 1 2 3

$100 $100 $100

(Annuity Due)Beginning of

Period 1Beginning of

Period 2

Today Equal Cash Flows Each 1 Period Apart

Beginning ofPeriod 3

3-35

FVAn = R(1+i)n-1 + R(1+i)n-2 + ... + R(1+i)1 +

R(1+i)0

Overview of an Ordinary Annuity -- FVAOverview of an Ordinary Annuity -- FVA

R R R

0 1 2 n n+1

FVAn

R = Periodic Cash Flow

Cash flows occur at the end of the period

i% . . .

3-36

FVA3 = $1,000(1.07)2 + $1,000(1.07)1 + $1,000(1.07)0

= $1,145 + $1,070 + $1,000 = $3,215

Example of anOrdinary Annuity -- FVAExample of anOrdinary Annuity -- FVA

$1,000 $1,000 $1,000

0 1 2 3 4

$3,215 = FVA3

7%

$1,070

$1,145

Cash flows occur at the end of the period

3-37

Hint on Annuity Valuation

The future value of an ordinary annuity can be viewed as

occurring at the end of the last cash flow period, whereas the future value of an annuity due can be viewed as occurring at the beginning of the last cash

flow period.

3-38

FVAn = R (FVIFAi%,n) FVA3 = $1,000 (FVIFA7%,3)

= $1,000 (3.215) = $3,215

Valuation Using Table IIIValuation Using Table III

Period 6% 7% 8%1 1.000 1.000 1.0002 2.060 2.070 2.0803 3.184 3.215 3.2464 4.375 4.440 4.5065 5.637 5.751 5.867

3-39

FVADn = R(1+i)n + R(1+i)n-1 + ... + R(1+i)2 +

R(1+i)1 = FVAn (1+i)

Overview View of anAnnuity Due -- FVADOverview View of anAnnuity Due -- FVAD

R R R R R

0 1 2 3 n-1 n

FVADn

i% . . .

Cash flows occur at the beginning of the period

3-40

FVAD3 = $1,000(1.07)3 + $1,000(1.07)2 + $1,000(1.07)1

= $1,225 + $1,145 + $1,070 = $3,440

Example of anAnnuity Due -- FVADExample of anAnnuity Due -- FVAD

$1,000 $1,000 $1,000 $1,070

0 1 2 3 4

$3,440 = FVAD3

7%

$1,225

$1,145

Cash flows occur at the beginning of the period

3-41

FVADn = R (FVIFAi%,n)(1+i)FVAD3 = $1,000 (FVIFA7%,3)(1.07)

= $1,000 (3.215)(1.07) = $3,440

Valuation Using Table IIIValuation Using Table III

Period 6% 7% 8%1 1.000 1.000 1.0002 2.060 2.070 2.0803 3.184 3.215 3.2464 4.375 4.440 4.5065 5.637 5.751 5.867

3-42