Fe Review
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FE Review for Engineering Economics
FE Review for Engineering Economics http://ristroph.louisiana.edu/EngrEcon/Overhead/FEReview.pdf Present Economics $ Revenue Revenue = ($/un ($/ unit) it) X Profit = Revenue − Total ot al Cost Total Cost = Fixed Fixed Cost Cost + Variable Variable Cost Cos t Variable = ($/un ($ /unit) it) X Cost
Total Cost
Fixed Cost
Breakeven
X (units)
Breakeven Diagram
f(x) $
df(x) =0 dx
x*
x (units)
Nonlinear Revenue or Profit
© 8/17/08 J H Ristroph
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FE Review for Engineering Economics
$ TC1 TC2 Method 1
Method 2 x (units)
Breakeven
Alternative Production Methods
Expected Profit or Cost P(100) = 30%
P(200) = 20%
P(300) = 50%
Expected value = 220 = 100(0.30) + 200(0.20) + 300(0.50)
Equivalence Single Payments E 4 55
65
3
6
2 4
5
100 Single Cash Flows E4 = -100(F |P,i,4-2) + 55(P|F ,i,5-4) + 65(P|F ,i,6-4)
© 8/17/08 J H Ristroph
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FE Review for Engineering Economics
Uniform Series E 3
E 20
100 •
3
4
•
•
5
20
Simple Series E3 = 100(P |A,i,20-3) E20 = 100(F |A,i,20-3)
E -2 E 3
E 20
100 •
3
-2
4
•
E 30
•
5
20
30
Two Step Equivalents to 100 Series E-2 = E3(P|F ,i,3-(-2)) E30 = E20(F |P,i,30-20)
E -2 E 3
E 20
100 •
-6 -5 -4
-3
-2
3
4
5
•
E 30
•
20
30 31 32 33 34
Three Step Equivalents to 100 Series E -5 to -2 = E-2( A|F ,i, -2-(-6)) E 31 to 33 = E30( A|P,i,34-30)
© 8/17/08 J H Ristroph
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FE Review for Engineering Economics
Arithmetic Gradients 70
54 50 52 •
•
9 10 11 12
•
B = 50 =
20
•
•
G = +2
9 10 11 12
Linear Trend
+
•
20
2 4 •
•
•
9 10 11 12
20
Base Series
20
Arithmetic Gradient
Components of a Linear Trend
9 10 11 12
20
9 10 11 12 =
•
•
9 10 11 12
20 •
•
+
•
•
B = -50
70
•
•
2 4 20 G = -2
•
50 52 54
20
a) Negative Base Series and Negative Gradient B = 70
70 68 66
50 •
•
•
•
•
•
9 10 11 12
20
=
9 10 11 12
20
+
9 10 11 12
2
b) Positive Base Series and Negative Gradient
20 •
•
•
4 20 G = -2 G = +2
9 10 11 12
20 •
66 70 68
•
=
9 10 11 12
20
2 4 +
9 10 11 12
•
50
•
•
•
B = -70
c) Negative Base Series and Positive Gradient Types of Linear Trends
© 8/17/08 J H Ristroph
•
•
20
•
20
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FE Review for Engineering Economics
E r = G (P | G, i, s-r )
(s-r -1) G G 2G r r +1 r +2 r+3
•
•
•
s
Arithmetic Gradient Present Worth Factor
(s-r -1) G G 2G r r +1 r +2 r+3
•
•
E U = G ( A | G, i, s-r ) =
•
s
•
•
•
r r +1 r +2 r+3
s
Arithmetic Gradient Uniform Series Factor
Effective Interest r = nominal rate, such as 12% per year compounded monthly
P = number of short periods in the long period, such as 12 months in a year iP = r / P = rate per short period, such as a month ie = (1 + iP)P – 1 = effective (or true) interest rate for the long period,
such as a year Example: 18% per year compounded monthly P = 12 im = 1.5% per month = 18% / 12 = true monthly rate ie = (1 + 0.015)12 – 1 = 19.56% effective (or true) interest rate per year
© 8/17/08 J H Ristroph
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FE Review for Engineering Economics
Example: 12% per year compounded monthly with quarterly cash flows => The monthly interest rate is 1% Q M •
•
Q3
Q6
M
•
M •
0
Q33
•
M
Q36
•
0 1
2
12
1
2
3
4
5
i = 3.0301% / qr
1,000
6
31 32
33 34 35 36
i = 1% / mo
1,000 Monthly Rate, Quarterly Payments
If draw cash flow diagram on a quarterly basis, must use quarterly rate 3.0301% = (1 + 0.01) 3 – 1
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