Cost analysis and estimating for engineering and management ch10

Average Annual Rate of Return Yearly After Tax Profit from Activity  Total Investment in the Activity  Alternately Eq 10.1 Eq 10.2 % , 100 value investment year per life economic in

Cost Analysis and Estimating

for Engineering and Management

Chapter 10 Engineering Economy

 Methods to Determine Project Returns

 Impact of Time Value of Money (Interest)

 Various Rates of Return

 Evaluation of Replacement Alternatives

 Tax Effects on Decisions

 All Projects Require Money

 All Should Return Money (Payback)

 When Large Sums of Money or

Longer Times Are Involved –

Interest Becomes Important

 Decisions Are Based on Money

 Monetary Assets

 Should Be Employed to Earn a Return

 Invested

 Need to Evaluate Investments

Average Annual Rate of Return

 Yearly After Tax Profit from Activity

 Total Investment in the Activity

 Alternately

Eq 10.1

Eq 10.2

% ,

100 value

investment

year per

life economic investment

total -

earnings

average

=

, earnings after tax

annual

investment

net

= payback

Payback Caveats

 Assumes Equal Annual Earnings

 Does Not Account for

 Choice of Investments

 Time Value of Money

 Interest Calculations

Compound Interest

 Interest Adds to Principal

Various Interest Calculations

 Net Present Worth

 Net Future Worth

 Net Equivalent Annual Worth

 Rate of Return

 All Generally Yield Consistent

Recommendations

Set Up for Discussions

 Example Situation

 Factors Can Be Calculated or Found In Tables (10% and 20% in Appendices)

Year Cost Revenue

0 $1025 $0

Net Present Worth

 Convert All Amounts to Present Value

Period n

PW Factor

at 10%

Cash Flow Amount Year 1 to zero 1 0.9091 $450 $409 Year 2 to zero 2 0.8264 425 351 Year 3 to zero 3 0.7513 400 301

Net Future Worth

 Convert All Amounts to Future Value

Period n

Compound Amount Factor 10%

Cash Flow Amount Year 1 to 3 2 1.2100 $450 $545 Year 2 to 3 1 1.1000 425 468 Year 3 0 1.0000 400 400

Net Equivalent Annual Worth

 Find Present Value of Receipts

 Calculate Annual Equivalent

 Calculate Net Annual Equivalent Worth

1061 0.7513

+ 0.8264 +

2

1025 427

Net Equivalent Annual Worth

Example

Amount

Anticipated annual receipts $427 Less equal annual equivalent worth 14 Equal annual receipts $413

Rate of Return

 Other Methods Assume an Interest Rate

 Find the Interest Rate to Yield Same

Return as Investment

 Found By Trial-and-Error or Interpolation

Example at 5%

Example at 15%

Interpolate

Compare the Methods

 Compare Present and Future Worths

 Rate of Return Finds Actual Interest

 Other Methods Assume an Interest

Method Amount Net present worth at 10% $36

Net future worth at 10% $49

Net annual equivalent worth at 10% $14

Rate of return 12.3%

Single Payment Methods

Single Payment Diagrams

Uniform Payment Series

A P i i

i

n n

P A i

i i

n n

Uniform Series Diagrams

Uniform Series Diagrams

 Cash Flow at Year End (Lump Sum)

 Ignore Inflation

 Special Handling of Taxes

 Constant Interest Rate

 Predetermined MARR

 Save Non-Quantifiable Effects for End

 Cost of Using Capital

 Internal Rate of Return (IRR)

 Minimum Acceptable Rate of Return

(MARR)

 External Rate of Return (ERR)

 Percentage Showing Yield on Different Uses of Capital

 Based Solely on Project’s Cash Flow

 Find IRR for Present Worth of $0

 Required IRR < the Projected IRR

 Compare Alternatives

 Rate Set By Management

 Sets Lower Limit on Acceptable Return

 Rations Capital to Avoid Unproductive Investments

Comparison of Alternatives

 Feasible Alternatives

 Mutually Exclusive

 “Do Nothing” Is an Alternative

 May Omit Costs/Revenues If the Same

 Find Alternative with Highest Return or Lowest Cost

 Equal/Unequal Lives Handled Differently

Equal Life Example (i=10%)

Cash Flows

Comparison for Example

Unequal Life

 Comparison Must Be for Equal Output

 Different Useful Lives for Alternatives

Useful Lives

 Example I1 – 3 Years, I2 – 4 Years

Unequal Life Example

12 Year Cash Flow

 Present Worth for 12 Years

PW (I1) = – $800,520

PW (I2) = – $629,116

 Select Investment 2, Lower Cost

 Assuming Revenues Are Equal

 Study Period May Be Excessive

 Situation May Vary from Assumptions

Restricted Life

 Project Limited to 2 Years

 Assume Salvage Value = $0

PW(I1) = – $273,388 PW(I2) = – $320,000

 Compare to Include Salvage Value

(320,000-273,338)/(P/F,10%,2)=$56,400

 Decision Depends on Salvage Cost of I2

 Existing Alternative In Operation

 Economic Facts Are Different for

Challenger

 Equipment Life – Period of Lower Cost

 Need Value for Existing Equipment

Equivalent Annual Cost for

)

%, ,

/ )(

)

%, /

( )

%, ,

/

P

 i = 20%, Life = 4 Years

Year

Defender (D) Challenger (C) Operating

cost

Salvage value

Operating cost

Salvage value

After Tax MARR  (Before Tax MARR)(1-t)

Before Tax MARR = 20%, t = 38%

After Tax MARR = About 12.4%

 Timing of Cash Flow Has Impacts

Considerations for After Tax

Analysis

 Costs, Savings, Revenues

 Depreciation

 Taxable Income

 Cash Flow Effects

 Engineering Economy Analysis

 Decision Process

 Consider Non-Economic Factors

After Tax Example

Yr

End

Before Tax Dep.

Deductible Charges

35% Tax Savings

After Tax Cash Flow

 Costs Represent Disbursements

 Select Challenger

 How to Evaluate Replacements

 What Is the Impact of Taxes for

Economic Decisions