Engineering Economics Formula Sheet

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  power - sizing exponent

x

Learning Curve: TN  Tinitial  N b b

log(learning curve rate) log 2

TN  time to make Nth unit Tinitial  time to make first unit N  number of finished units b  learning curve exponent

Simple Interest:

Interest earned on amount P : I  Pin Maturity value : F  P (1  in)

ieq  interest rate for payment period p  number of payment periods per year ic  interest rate for compounding period c  number of compounding periods per year

Ordinary Simple Annuity (Uniform Series):  (1  i ) n  1 Compound Amount F  A  i     i A  F  n  (1  i )  1

Sinking Fund

 i (1  i ) n  A  P  n  (1  i )  1

Capital Recovery

 (1  i ) n  1 P  A  n   i (1  i ) 

1  (1  i )  n  A  i  

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

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

r (1  ieff )  1  m

r  nominal interest rate per year

Geometric Gradient Annuity: 1  (1  g ) n (1  i )  n  P  A1  ; i  g ig  

m  number of compounding periods per year

P

m

i  periodic interest rate

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 ig

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

Present Worth

Relationships: NPV/W  net present va lue/worth NPW  PWc  PWb A  Pi for n   Capitalized cost  P 

A

i EACF  equivalent annual cash 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/PW 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