Title | 261611929 Engineering Economics Formula Sheet |
---|---|
Author | Jonathan Zhang |
Course | Engineering Management Principles and Economics |
Institution | Concord University |
Pages | 2 |
File Size | 65.2 KB |
File Type | |
Total Downloads | 101 |
Total Views | 158 |
Download 261611929 Engineering Economics Formula Sheet PDF
Cost Indexes:
General Annuity:
Cost at time A Index value at time A Cost at time B Index value at time B
(1 ieq) p (1 ic )c
Power sizing:
ieq (1 i ) c / p 1
Cost of asset A Size (capacity) of asset A Cost of asset B Size (capacity) of asset B
x
ieq interest rate for payment period p number of payment periods per year
x power - sizing exponent
ic interest rate for compounding period
Learning Curve: TN Tinitial N b
c number of compounding periods per year
b
log(learning curve rate) log 2
TN time to make Nth unit
Ordinary Simple Annuity (Uniform Series): (1 i) n 1 Compound Amoun F A i i A F n (1 i) 1
Sinking Fund
i(1 i) n A P n (1 i ) 1
Capital Recovery
Maturity value : F P (1 in )
1 (1 i ) n (1 i ) n 1 P A A n i i(1 i)
Present Worth
i interest rate per time period
A periodic payment (end of period)
n number of time periods
P, F, i, n as above for compound interest
Compound Interest (Single Payment):
Arithmetic Gradient Annuity: 1 n Aeq G n i (1 i) 1
Tinitial time to make first unit N number of finished units b learning curve exponent
Simple Interest:
Interest earned on amount P : I Pin
F P (1 i )n F future value P present value i periodic interest rate n number of periods
Effective Interest Rates: r i m
(1 i)n in 1 P G 2 n i (1 i) Aeq equivalent periodic payment G gradient amount (periodic increment) P , i , n as above for compound interest
(1 ieff ) 1 mr
m
i periodic interest rate r nominal interest rate per year m number of compounding periods per year ieff effective interest rate (compounded annually)
Annuity Due: [Appropriate formula for question](1+ i)
Perpetual Annuities: A P i A Geometric Growth : P ;i g ig
Geometric Gradient Annuity: 1 (1 g) n (1 i) n P A1 ; i g i g P
nA1 ;i g (1 i )
(1 i ) n (1 g ) n F A1 ; i g i g n 1 F nA1 (1 i) ; i g A1 payment in first period (end) g periodic rate of growth P, F, i, n as above for compound interest
Relationships: NPV/W net present va lue/worth NPW PWc PW b A Pi for n Capitalized cost P
A
i EACF equivalent annualcash flow EACF EUAC EUAB NPW EACF(Capital recovery factor) EACF NPW(Presen t worth factor) IRR internal rate of return MARR minimum acceptable rate of return
To find rate of return : PW of benefits – PW of costs 0 PW of benefits/P W of costs 1 Present worth Net present worth 0 EUCF EUAB – EUAC 0 PW of costs PW of benefits
Salvage Value (using capital recovery formula): EUAC = P(A/P, i, n) – S(A/F, i, n) EUAC = (P – S)(A/F, i, n) + Pi EUAC = (P – S)(A/P, i, n) + Si
S Salvage value P, F , i, n as above for compound interest...