Engineering Economics. November 3, 2004. Engineering Economy. It deals with the concepts and techniques of analysis useful in evaluating the worth of systems, products, and services in relation to their costs. Engineering Economy. It is used to answer many different questions
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November 3, 2004
End of year Cash flow
1 -$580 (-$500 - $80)
2 -$540 (-$500 - $40)
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9
I2 = $100 x (1+.09)2 - $100 = $18.81
F = $100 (1 + .09)2 = $118.81
F=P(1+i)n is the
single payment compound amount factor.
single payment present worth factor.
Interpretation of (P/F, i, n): a present sum P, given a future sum, F, n interest periods hence at an interest rate i per interest period
P = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means end-of-period cash payments, Type = 1 means beginning-of-period payments, guess is a guess value of the interest rate
Both at the same interest rate and at the same time point.
To make a choice the cash flows must be altered so a comparison may be made.