# Engineering Economics

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1. Single Payment Compound Amount
- Future Amount from Past Amount
• (F/P, i%, n )
• (1+I)n
2. Single Payment Present Worth
- Past Amount from Future Amount
• ( P/F, i%.n )
• (1+ i )- n
3. Uniform Series Sinking Fund
- Annuity from Future
• ( A/F, i%, n )
• i / ( 1+i )n - 1
4. Capital Recovery
Annuity from Present Value
• ( A / P, i%, n )
• i( 1+i)n / (1 + i )n -1
5. Uniform Series Compound Amount
- Future Value from Annuity
• ( F / A, i%, n )
• (1 + i )n - 1 / i
6. Uniform Series Present Worth
- Present Value from Annuity
• ( P / A, i, n )
• ( 1+ i )n - 1 / i( 1 + i )n
• ( P / G, i%, n )
• [( 1 + i )n - 1 / i2 ( 1 + I )n ] - [ n / i(1+i)n ]
• ( F / G, i%, n )
• [ (1+ i )n -1 / i2 ] -[ n / i ]
• ( A / G, i%, n )
• [ 1 / i ] - [ n / (1 + i )n -1 ]
10. Continuous Compounding
• ( F/P,r%,n) = ern
• ( P/F,r%,n) = e-rn
• ( A/F,r%,n) = ( er - 1) / ( ern -1 )
• ( F/A,r%,n) = ( ern - 1) / ( er -1 )
• ( A/P,r%,n) = ( er - 1) / (1 - e-rn )
• ( P/A,r%,n) = (1 - e-rn ) / ( er -1 )
11. Straight - Line Depreciation
Dj = ( C - Sn ) / n
12. ACRS and MACRS
• The initial cost used is not reduced by the asset's salvage value. jth year:
• Dj = C x factor
• Recovery Period Years:
• -----3 ---------5 ----------7 ---------10
• 1-- 33.3--- 20.0------- 14.3 ------10.0
• 2-- 44.5--- 32.0 -------24.5-------18.0
• 3-- 14.8--- 19.2 -------17.5------ 14.4
• 4--- 7.4 ---11.5 -------12.5------- 11.5
• 5----------- 11.5------ - 8.9 ---------9.2
• 6------------ 5.8-------- 8.9 ---------7.4
• 7------------------------- 8.9--------- 6.6
• 8------------------------- 4.5----------6.6
• 9--------------------------------------- 6.5
• 10------------------------------------- 6.5
• 11------------------------------------- 3.3
13. Book Value
• BVj = initial cost - accumulated depreciation
• = C - Sigma tj=1 Dj
14. Capitalized Cost
• The Present worth of a project with an infinite life is known as Capitalized Cost
• Capitalized Cost = P = A / i [ infinite series ]
15. Bonds
• ..a method of obtaining long-term financing commonly used by governments, states, and municipalities
• Because of "discounting" the bonds Face Value doesn't generally equal the Purchase Price.
• - they call for quarterly or semiannual interest payments and the face value of the bond at the maturity date.
• By convention, Bond Yield is specified as a Nominal Rate ( rate per annum ), not as an effective rate per year.
• The Bond Yield should be determined by determined by finding the effective rate of return per payment period ( eg per semiannual interest payment )
16. Inflation
• One alternative is to replace the effective annual interest rate, i, with an interest rate adjusted for inflation:
• d = i + f + if
17. Probablistic Problems
- Where multiple alternatives with different probabilities that they will occur
• Typically these use annual costs and expected values
• E(x) = p1(Cost1) + p2(Cost2) + .....
18. MARR
Minimum Attractive Rate of Return
• ROR - Rate of Return - the effective annual interest rate at which an investment accrues income
• Where the company does not know what interest rate, i, to use in engineering economic analysis, the company can establish a MARR
• MARR - Minimum Attractive Rate of Return
• Finding the rate of return can be a long, iterative process using either interpolation or trial and error.
19. Benefit Cost Ratio Analysis
• The present worth of all benefits is devided by the present value of all costs. The ratio should be at least 1.0.
• Disbursements by the initiators or sponsors are costs. Disbursements by the users are known as Disbenefits. It is often dificult to tell whether a cash flow is a disbenefit ( whether to place it in the numerator or denominator of the ratio )
• For ranking project 1 and 2, use:
• (Benefit2 - Benefit1) / (Cost2 - Cost1) >= 1
20. Break Even Analysis
• If the manufactured quantity is less than the break even quantity, a loss is incurred.
• An alternative form of the Break Even analysis is to find the quantity at which two alternatives are the same total cost.
• Fixed costs are spread over a period longer than one year using the EUAC concept.
• The payback period, PBP is defined as the length of time, n, usually in years, for the cumulative net annual profit to equal the initial investment. It is tempting to introduce equivalence into the payback period but the convention is not to.
• C - (PBP)(net annual profit) = 0

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 Author: timothyowens ID: 114818 Filename: Engineering Economics Updated: 2011-11-06 23:32:02 Tags: FE Engineering Economics Folders: Description: FE exam preparation - Engineering Economics Section Show Answers:

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