ECON
slides session 2 TVOM class Pt1 chrt 3.ppt

P 100000 5941845 principles of engineering

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P =NPV(10%,50000,40000,30000,40000,50000)-100000 = \$59,418.45

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Principles of Engineering Economic Analysis , 5th edition Examples 2.13 & 2.16 Determine the present worth equivalent of the following series of cash flows. Use an interest rate of 6% per interest period. P = \$300(P|F 6%,1)- \$300(P|F 6%,3)+\$200(P|F 6%,4)+\$400(P|F 6%,6) +\$200(P|F 6%,8) = \$597.02 P =NPV(6%,300,0,-300,200,0,400,0,200) P =\$597.02 End of Period Cash Flow 0 \$0 1 \$300 2 \$0 3 -\$300 4 \$200 5 \$0 6 \$400 7 \$0 8 \$200
Principles of Engineering Economic Analysis , 5th edition Computing the Net Future Worth of Computing the Net Future Worth of Multiple Cash flows Multiple Cash flows

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Principles of Engineering Economic Analysis , 5th edition Example 2.15 Determine the future worth equivalent of the CFD shown below, using an interest rate of 10% compounded annually. F =10000*FV(10%,5,,-NPV(10%,5,4,3,4,5)+10) = \$95,694.00
Principles of Engineering Economic Analysis , 5th edition Examples 2.14 & 2.16 Determine the future worth equivalent of the following series of cash flows. Use an interest rate of 6% per interest period. F = \$300(F|P 6%,7)-\$300(F|P 6%,5)+\$200(F|P 6%,4)+\$400(F|P 6%,2)+\$200 F = \$951.59 F =FV(6%,8,,-NPV(6%,300,0,-300,200,0,400,0,200)) F =\$951.56 End of Period Cash Flow 0 \$0 1 \$300 2 \$0 3 -\$300 4 \$200 5 \$0 6 \$400 7 \$0 8 \$200 (The 3¢ difference in the answers is due to round-off error in the tables in Appendix A.)

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition Common Cash Flow Series Common Cash Flow Series

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Principles of Engineering Economic Analysis , 5th edition Uniform Series Gradient Series Geometric Series
Principles of Engineering Economic Analysis , 5th edition DCF Uniform Series Formulas DCF Uniform Series Formulas

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Principles of Engineering Economic Analysis , 5th edition Uniform Series A t = A t = 1,…,n
Principles of Engineering Economic Analysis , 5th edition P occurs 1 period before first A P A A A A A A P = A[(1+i) n -1]/[i(1+i) n ] P = A(P|A i%,n) P = [ =PV(i%,n,-A) ] A = Pi(1+i) n /[(1+i) n -1] A = P(A|P i%,n) A = [ =PMT(i%,n,-P) ]

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Principles of Engineering Economic Analysis , 5th edition P occurs 1 period before first A P A A A A A A P = A[(1+i) n -1]/[i(1+i) n ] P = A(P|A i%,n) P =PV(i%,n,-A) A = Pi(1+i) n /[(1+i) n -1] A = P(A|P i%,n) A =PMT(i%,n,-P)
Principles of Engineering Economic Analysis , 5th edition F occurs at the same time as last A F A A A A A A F = A[(1+i) n -1]/i F = A(F|A i%,n) F = [ =FV(i%,n,-A) ] A = Fi/[(1+i) n -1] A = F(A|F i%,n) A = [ =PMT(i%,n,,-F) ]

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Principles of Engineering Economic Analysis , 5th edition F occurs at the same time as last A F A A A A A A F = A[(1+i) n -1]/i F = A(F|A i%,n) F =FV(i%,n,-A) A = Fi/[(1+i) n -1] A = F(A|F i%,n) A =PMT(i%,n,,-F)
Principles of Engineering Economic Analysis , 5th edition P = A(P|A i%,n) = A (2.22) A = P(A|P i%,n) = P (2.25) P occurs one period before the first A F = A(F|A i%,n) = A (2.28) A = F(A|F i%,n) = F (2.30) F occurs at the same time as the last A n n i i i ) 1 ( 1 ) 1 ( 1 ) 1 ( ) 1 ( n n i i i i i n 1 ) 1 ( 1 ) 1 ( n i i Summary Summary

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Principles of Engineering Economic Analysis , 5th edition Example 2. 17 Troy Long deposits a single sum of money in a savings account that pays 5% compounded annually. How much must he deposit in order to withdraw \$2,000/yr for 5 years, with the first withdrawal occurring 1 year after the deposit?
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• Fall '17
• Mike Heny

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