Engineering Economy HW 1

Engineering Economy HW 1 - 4-1 I= P*N*i I= $10,000*4.25*0.1...

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Unformatted text preview: 4-1 I= P*N*i I= $10,000*4.25*0.1 I= $4,250 The amount of interest paid on November 1, 2012 would be $4,250 4-2 I= P*N*i I= 200*6*0.01 I= 12 I would owe $212 after 6 months and $12 would be the interest owed 4-3 F= P(1 + i)^N F= $1000(1+ 0.08) ^ 2.5 F= $1,212.16 The future equivalence of the $1,000 after 2.5 years is $1,212.16. 4-4 I= P*N*i I= P* 4-7 EOYK 1 2 3 Interest Paid $800 $553.60 $287.49 Principal Repayment $3,080 $3,326.40 $3,592.51 b. The amount of principle owed at the beginning of year 3 would be $3,593.60 4-8 F= P(1 + i)^N F= $10,000(1 + 0.06) ^ 5 F= $13,382.3 Juanita will owe Jim $13,382.3 if she repays the entire loan at the end of five years 4-15 72/10 = 7.2 Ashlea's money would double every 7.2 years 4-28 F= A * [(((1+i)^N) -1)/ i ] F= $730 * [(((1 + 0.06) ^ 35) -1) / 0.06 ] F= $117,732 I will accumulate $117,732 by saving $2 a day for 35 years if the annual interest rate is 7% ...
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This note was uploaded on 09/09/2009 for the course ISE 13715 taught by Professor Koelling during the Fall '09 term at Virginia Tech.

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