# midterm - formula b P/A =(1 i N-1(i(1 i N = 6.637 A=...

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Question 2 A F= 15,000(F/P, 10.5%, 3) = 15,000*1.3492=20,238 Or by using the formula B P=2400(P/F,8%,5)+3000(P/F,8%,4)+3200(P/F,8%,3)+1540(P/F,8%,2) =2400*0.6806 +3000*0.7350+3200*0.7938+1540*0.8573 =7698.842 C F=3P=P(1+i) N 3P=P(1+0.11) N log 3 = N log 1.11 N= 10.53 years Qustion 3 A G(P/G,12%,5)*(P/F,12%,1) = 800(F/A,12%,4)+(500-200 (P/G,12%,4))(F/P,12%,4) G*6.397*0.8929=800*4.7793+ ((500-(200*4.1273))*1.5735 G=579.63 B 175(F/P, 8.5%, 1)+175 = 175(P/F, 8.5%, 1)+ 75(P/F, 8.5%, 2)+ 75(P/F, 8.5%, 3)+ 275(P/F, 8.5%, 4) =283.1279 0 10 500 800 600 400 200 G 3G 4G 2G 5 4 3 2 1 6 9 8 7

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Question 4 8% compounded quarterl b i=0.08/4 =0.02 25 years B 100 quarter i a = (1+0.08/4) 4 -1 = 0.08243 b 8.243% A(F/A,2%,100)=45,000(P/A,8.243%,10) (P/A,8.243,10) could be found by using the tables and interpolation or by using the
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Unformatted text preview: formula b P/A =( (1+i) N-1 )/(i(1+i) N = 6.637 A= 45,000*6.637/312.2323 A=956.547 Question 5 First compute the present equivalent of the energy cost during the first operating cycle i= 9%/12 = 0.75% P= 25(P/A,0.75%,3)(P/F,0.75,1)+40(P/A,0.75,3)(P/F,0.75%,7) =185.54 Then compute the total present worth of the energy cost over 3 operating cycles P= 185.54+185.54(P/F,0.75%,12) + 185,54(P/F,0.75,24) = 510.25 Question 6 A P= 300(P/F,0.5%,12)+300(P/F,0.75,12)(P/F,0.5,12)+500(P/f,0.75%,24)(P/F,0.5%,12)+500(P /f,0.5%,12)(P/f,0.75%,24)(P/F,o.5%,12) =1,305.26 B 1,305.26 = 300(P/A, i, 2) +500(P/A, i, 2)(P/f, i,2) This could be solved by using trail and error Using the formula instead of the factors required a trail and error solution at the end...
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## This note was uploaded on 11/25/2010 for the course SYSC 44509 taught by Professor Mattdamomn during the Fall '10 term at Carleton CA.

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midterm - formula b P/A =(1 i N-1(i(1 i N = 6.637 A=...

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